https://www.trccompsci.online/mediawiki/api.php?action=feedcontributions&user=DannyDaEpic&feedformat=atomTRCCompSci - AQA Computer Science - User contributions [en-gb]2024-03-29T09:16:37ZUser contributionsMediaWiki 1.31.6https://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4870Boolean Algebra2018-05-09T12:54:45Z<p>DannyDaEpic: /* Example 16 */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A.1 = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.<br />
<br />
==OR Identities==<br />
<math> 0+A = A </math><br />
<br />
0 or A can be simplified as just A.<br />
<br />
<math> 1+A = 1 </math><br />
<br />
1 or A can be simplified as just 1.<br />
<br />
<math> A+A=A</math><br />
<br />
A or A can be simplified as just A.<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
NOT A or A can be simplified as just 1.<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<br />
<math> A+B = B+A </math><br />
<br />
or<br />
<br />
<math> A.B = B.A </math><br />
<br />
==Associate Law==<br />
If all of the symbols are the same it doesn't matter which order the equation is evaluated.<br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = C + (A+B) </math><br />
<br />
So:<br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = C . (A.B) </math><br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
This is essentially factorising or expanding the brackets, but you can also:<br />
<br />
<math> A.B + A.C = A.(B+C)</math><br />
<br />
<math> A+B.A+C = A+(B.C) </math><br />
<br />
==Redundancy Law==<br />
Law 1 :<br />
<math> A + \overline{A} B = A + B </math><br />
<br />
Proof : <br />
<br />
<math>= A + \overline{A} B \\<br />
= (A + \overline{A})(A + B) \\<br />
= 1 . (A + B) \\<br />
= A + B </math><br />
<br />
<br />
Law 2:<br />
<math> A.(\overline{A} + B) = A.B</math><br />
<br />
Proof : <br />
<br />
<math>= A.(\overline{A} + B) \\<br />
= A.\overline{A} + A.B \\<br />
= 0 + A.B \\<br />
= A.B </math><br />
<br />
==Identity Law==<br />
This is also in the identities section:<br />
<br />
<math> A.A = A </math><br />
<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
Just like in any other logic negating a negative is a positive so:<br />
<br />
<math> \overline{ \overline{A} } = A </math><br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<math> \overline{\overline{A}} . \overline{(B+C)} = A. \overline{B} . \overline{C} </math><br />
<br /><br /><br />
<math> \overline{\overline{A}} = A </math> ''Use Negation Law''<br />
<br /><br /><br />
<math> \overline{(B+C)} = (\overline{B} . \overline{C}) </math> ''Use De Morgan's Law''<br />
<br /><br /><br />
<math> A . (\overline{B} . \overline{C}) = A . \overline{B} . \overline{C} </math> ''Use Associate Law''<br />
<br />
==Example 2==<br />
<math> (\overline{A} + A) . (\overline{A} + C) </math><br />
<br />
OR Identity<br />
<br />
<math>(\overline{A} + A) = 1</math><br />
<br />
AND Identity<br />
<br />
<math>= 1.(\overline{A} + C)</math><br />
<br />
Simplify<br />
<br />
<math>= (\overline{A} + C)</math><br />
<br />
==Example 3==<br />
<math>(X + Y) . (X + \overline{Y})</math><br />
<br>Distributive: <br />
<br><math>X . (Y + \overline{Y})</math><br />
<br>Identity laws: <br />
<br><math>Y + \overline{Y} = 1</math><br />
<p><math> X.1 = X</math><br />
<br />
==Example 4==<br />
<br />
Expression Rule(s) Used<br />
C + BC Original Expression<br />
C + (B + C) DeMorgan's Law.<br />
(C + C) + B Commutative, Associative Laws.<br />
T + B Complement Law.<br />
T Identity Law.<br />
<br />
==Example 5==<br />
{| class="wikitable"<br />
|-<br />
! Simplify: !! AB(A + B)(B + B):<br />
|-<br />
| Expression || Rule(s) Used<br />
|-<br />
| AB(A + B)(B + B) || Original Expression<br />
|-<br />
| AB(A + B) || Complement law, Identity law.<br />
|-<br />
| (A + B)(A + B) || DeMorgan's Law<br />
|-<br />
| A + BB || Distributive law. This step uses the fact that or distributes over and. It can look a bit strange since addition does not distribute over multiplication<br />
|-<br />
| A || Complement, Identity<br />
|}<br />
<br />
==Example 6==<br />
<math><br />
\overline{A.B}(\overline{A} + B)(\overline{B} + B) </math> Original Expression<br />
<br />
<math><br />
\overline{AB}(\overline{A} + B) </math> Complement law, Identity law.<br />
<br />
<math><br />
(\overline{A} + \overline{B})(\overline{A} + B) </math> DeMorgan's Law<br />
<br />
<math><br />
\overline{A} + \overline{B}.B </math> Distributive law. This step uses the fact that or distributes over and. <br />
<br />
<math><br />
\overline{A} </math> Complement, Identity.<br />
<br />
==Example 7==<br />
<br />
<br />
SIMPLIFY <math>(A + C)A + AC + C</math><br />
<br />
<br />
<math>(A + C)A + AC + C</math> Complement, Identity.<br />
<br />
<br />
<math>A((A + C) + C) + C</math> Commutative, Distributive.<br />
<br />
<br />
<math>A(A + C) + C</math> Associative, Idempotent.<br />
<br />
<br />
<math>AA + AC + C</math> Distributive.<br />
<br />
<br />
<math>A + (A + T)C</math> Idempotent, Identity, Distributive.<br />
<br />
<br />
<math>A + C</math> Identity, twice.<br />
<br />
==Example 8==<br />
<br />
<br />
===Simplify=== <br />
<math> (X+Y).(X+\overline{Y}) </math> <br />
<br />
===solution===<br />
<math> X.X + X.\overline{Y} + Y.X + Y.\overline{Y} </math> Expanding the brackets<br />
<br />
<math> X + X.\overline{Y} + Y.X + 0 </math> Use of <math> X.X = X </math> and <math> Y.\overline{Y} = 0 </math><br />
<br />
<math> X + X(\overline{Y}+Y) </math> Taking X out of the brackets<br />
<br />
<math> X + X(1) </math> Use of <math> Y + \overline{Y} = 1 </math><br />
<br />
<math> X(1) </math><br />
<br />
<math> X </math><br />
<br />
==Example 9==<br />
<math> (X + Y) . (X + \overline{Y}) </math><br />
<br />
<math> X + (Y.\overline{Y}) </math> after distributive law applied<br />
<br />
<math> (Y.\overline{Y}) = 0 </math><br />
<br />
<math> X + 0 </math><br />
<br />
<math> X </math><br />
<br />
<math> (X + Y) . (X + \overline{Y}) = X </math><br />
<br />
==Example 10==<br />
Simplify:<br />
<br />
<math> \overline {AB} (\overline {A}+B)(\overline {B}+B) </math> <br />
<br />
Complement law, Identity law <br />
<br />
<math> \overline {AB} (\overline {A}+B).1 </math> <br />
<br />
DeMorgan's law <br />
<br />
<math> (\overline {A}+\overline {B})(\overline {A}+B) </math><br />
<br />
Distributive law<br />
<br />
<math> \overline {A}+\overline {B}B </math> <br />
<br />
Complement, Identity<br />
<br />
<math> \overline {A} </math><br />
<br />
==Example 11==<br />
<br />
==Example 12==<br />
<br />
==Example 13==<br />
Simplify<br />
<br />
<math>\overline{\overline{A} + \overline{(B.A)}}</math><br />
<br />
First, using De Morgan’s Law simplify it to<br />
<br />
<math> \overline{\overline{A} + \overline{B} + \overline{A}} </math><br />
<br />
This can then be simplified to<br />
<br />
<math> \overline{\overline{A} + \overline{B}} </math><br />
<br />
Then using De Morgan’s Law again, you get<br />
<br />
<math> \overline{\overline{A}} + \overline{\overline{B}} </math><br />
<br />
Then there are double negatives, so you end up with<br />
<br />
<math> A + B </math><br />
<br />
==Example 14==<br />
'''Simplify:'''<br />
<math> X.(\overline{X}+Y) </math><br />
<br />
<math> (X.\overline{X})+(X.Y) </math> using the Distributive Law<br />
<br />
<math> 0 + (X.Y) </math> using the identity <math>A.\overline{A}=0</math><br />
<br />
<math> X.Y </math> using the identity <math>A+0=A</math><br />
<br />
Fully simplified the equation<br />
<br />
==Example 15==<br />
Simplify the following:<br />
<math> \overline{ \overline{(A.A)}. \overline{(B.B)}} </math><br />
<br />
Using De Morgan's law we can put the equation into another form:<br />
<br />
1. Change All operations,<br />
<br />
<math> \overline{ \overline{(A+A)}+ \overline{(B+B)}} </math><br />
<br />
2. Negate letters:<br />
<br />
<math> \overline{ (A+A)+(B+B) } </math><br />
<br />
3. Negate the whole expression:<br />
<br />
<math> (A+A)+(B+B) </math><br />
<br />
Using the identity law, we can then simplify this expression to:<br />
<br />
<math> A+B </math><br />
<br />
==Example 16==<br />
For this example the Boolean Algebra itself is shown above while the description/the rules that have been used are listed below it:<br />
<br />
<Math>\overline{A}(A + B) + (B + AA)(A + \overline{B})</Math><br />
<br />
Original expression<br />
<br />
<Math>\overline{A}A + \overline{A}B + (B + A)A + (B + A)\overline{B}</Math><br />
<br />
Idempotent (AA to A), then Distributive, used twice.<br />
<br />
<Math>\overline{A}B + (B + A)A + (B + A)\overline{B}</Math><br />
<br />
Complement, then Identity. (Strictly speaking, we also used the Commutative Law for each of these applications.)<br />
<br />
<Math>\overline{A}B + BA + AA + B\overline{B} + A\overline{B}</Math><br />
<br />
Distributive, two places.<br />
<br />
<Math>\overline{A}B + BA + A + A\overline{B}</Math><br />
<br />
Idempotent (for the A's), then Complement and Identity to remove BB.<br />
<br />
<Math>\overline{A}B + AB + AT + A\overline{B}</Math><br />
<br />
Commutative, Identity; setting up for the next step.<br />
<br />
<Math>\overline{A}B + A(B + T + \overline{B})</Math><br />
<br />
Distributive.<br />
<br />
<Math>\overline{A}B + A</Math><br />
<br />
Identity, twice (depending how you count it).<br />
<br />
<Math>A + \overline{A}B</Math><br />
<br />
Commutative.<br />
<br />
<Math>(A + \overline{A})(A + B)</Math><br />
<br />
Distributive.<br />
<br />
<Math>A + B</Math><br />
<br />
Complement, Identity.<br />
<br />
==Example 16==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4858Boolean Algebra2018-05-09T12:46:15Z<p>DannyDaEpic: /* Example 16 */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A.1 = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.<br />
<br />
==OR Identities==<br />
<math> 0+A = A </math><br />
<br />
0 or A can be simplified as just A.<br />
<br />
<math> 1+A = 1 </math><br />
<br />
1 or A can be simplified as just 1.<br />
<br />
<math> A+A=A</math><br />
<br />
A or A can be simplified as just A.<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
NOT A or A can be simplified as just 1.<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<br />
<math> A+B = B+A </math><br />
<br />
or<br />
<br />
<math> A.B = B.A </math><br />
<br />
==Associate Law==<br />
If all of the symbols are the same it doesn't matter which order the equation is evaluated.<br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = C + (A+B) </math><br />
<br />
So:<br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = C . (A.B) </math><br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
This is essentially factorising or expanding the brackets, but you can also:<br />
<br />
<math> A.B + A.C = A.(B+C)</math><br />
<br />
<math> A+B.A+C = A+(B.C) </math><br />
<br />
==Redundancy Law==<br />
Law 1 :<br />
<math> A + \overline{A} B = A + B </math><br />
<br />
Proof : <br />
<br />
<math>= A + \overline{A} B \\<br />
= (A + \overline{A})(A + B) \\<br />
= 1 . (A + B) \\<br />
= A + B </math><br />
<br />
<br />
Law 2:<br />
<math> A.(\overline{A} + B) = A.B</math><br />
<br />
Proof : <br />
<br />
<math>= A.(\overline{A} + B) \\<br />
= A.\overline{A} + A.B \\<br />
= 0 + A.B \\<br />
= A.B </math><br />
<br />
==Identity Law==<br />
This is also in the identities section:<br />
<br />
<math> A.A = A </math><br />
<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
Just like in any other logic negating a negative is a positive so:<br />
<br />
<math> \overline{ \overline{A} } = A </math><br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<math> \overline{\overline{A}} . \overline{(B+C)} = A. \overline{B} . \overline{C} </math><br />
<br /><br /><br />
<math> \overline{\overline{A}} = A </math> ''Use Negation Law''<br />
<br /><br /><br />
<math> \overline{(B+C)} = (\overline{B} . \overline{C}) </math> ''Use De Morgan's Law''<br />
<br /><br /><br />
<math> A . (\overline{B} . \overline{C}) = A . \overline{B} . \overline{C} </math> ''Use Associate Law''<br />
<br />
==Example 2==<br />
<math> (\overline{A} + A) . (\overline{A} + C) </math><br />
<br />
OR Identity<br />
<br />
<math>(\overline{A} + A) = 1</math><br />
<br />
AND Identity<br />
<br />
<math>= 1.(\overline{A} + C)</math><br />
<br />
Simplify<br />
<br />
<math>= (\overline{A} + C)</math><br />
<br />
==Example 3==<br />
<math>(X + Y) . (X + \overline{Y})</math><br />
<br>Distributive: <br />
<br><math>X . (Y + \overline{Y})</math><br />
<br>Identity laws: <br />
<br><math>Y + \overline{Y} = 1</math><br />
<p><math> X.1 = X</math><br />
<br />
==Example 4==<br />
<br />
Expression Rule(s) Used<br />
C + BC Original Expression<br />
C + (B + C) DeMorgan's Law.<br />
(C + C) + B Commutative, Associative Laws.<br />
T + B Complement Law.<br />
T Identity Law.<br />
<br />
==Example 5==<br />
{| class="wikitable"<br />
|-<br />
! Simplify: !! AB(A + B)(B + B):<br />
|-<br />
| Expression || Rule(s) Used<br />
|-<br />
| AB(A + B)(B + B) || Original Expression<br />
|-<br />
| AB(A + B) || Complement law, Identity law.<br />
|-<br />
| (A + B)(A + B) || DeMorgan's Law<br />
|-<br />
| A + BB || Distributive law. This step uses the fact that or distributes over and. It can look a bit strange since addition does not distribute over multiplication<br />
|-<br />
| A || Complement, Identity<br />
|}<br />
<br />
==Example 6==<br />
<math><br />
\overline{A.B}(\overline{A} + B)(\overline{B} + B) </math> Original Expression<br />
<br />
<math><br />
\overline{AB}(\overline{A} + B) </math> Complement law, Identity law.<br />
<br />
<math><br />
(\overline{A} + \overline{B})(\overline{A} + B) </math> DeMorgan's Law<br />
<br />
<math><br />
\overline{A} + \overline{B}.B </math> Distributive law. This step uses the fact that or distributes over and. <br />
<br />
<math><br />
\overline{A} </math> Complement, Identity.<br />
<br />
==Example 7==<br />
<br />
<br />
SIMPLIFY <math>(A + C)A + AC + C</math><br />
<br />
<br />
<math>(A + C)A + AC + C</math> Complement, Identity.<br />
<br />
<br />
<math>A((A + C) + C) + C</math> Commutative, Distributive.<br />
<br />
<br />
<math>A(A + C) + C</math> Associative, Idempotent.<br />
<br />
<br />
<math>AA + AC + C</math> Distributive.<br />
<br />
<br />
<math>A + (A + T)C</math> Idempotent, Identity, Distributive.<br />
<br />
<br />
<math>A + C</math> Identity, twice.<br />
<br />
==Example 8==<br />
<br />
<br />
===Simplify=== <br />
<math> (X+Y).(X+\overline{Y}) </math> <br />
<br />
===solution===<br />
<math> X.X + X.\overline{Y} + Y.X + Y.\overline{Y} </math> Expanding the brackets<br />
<br />
<math> X + X.\overline{Y} + Y.X + 0 </math> Use of <math> X.X = X </math> and <math> Y.\overline{Y} = 0 </math><br />
<br />
<math> X + X(\overline{Y}+Y) </math> Taking X out of the brackets<br />
<br />
<math> X + X(1) </math> Use of <math> Y + \overline{Y} = 1 </math><br />
<br />
<math> X(1) </math><br />
<br />
<math> X </math><br />
<br />
==Example 9==<br />
<math> (X + Y) . (X + \overline{Y}) </math><br />
<br />
<math> X + (Y.\overline{Y}) </math> after distributive law applied<br />
<br />
<math> (Y.\overline{Y}) = 0 </math><br />
<br />
<math> X + 0 </math><br />
<br />
<math> X </math><br />
<br />
<math> (X + Y) . (X + \overline{Y}) = X </math><br />
<br />
==Example 10==<br />
Simplify:<br />
<math> \overline {AB} (\overline {A}+B)(\overline {B}+B) </math> <br />
<math> \overline {AB} (\overline {A}+B) </math> <br />
Complement law, Identity law <br />
<math> (\overline {A}+\overline {B})(\overline {A}+B) </math><br />
DeMorgan's law <br />
<math> \overline {A}+/overline {B}B </math> <br />
Distributive law. <br />
<math> \overline {A} </math><br />
<br />
==Example 11==<br />
<br />
==Example 12==<br />
<br />
==Example 13==<br />
<br />
<math>\overline{\overline{A} + \overline{(B.A)}}</math><br />
<br />
First simplify using De Morgam's Law<br />
<br />
<math> \overline{\overline{A} + \overline{B} + \overline{A}} </math><br />
<br />
This can be simplified to<br />
<br />
<math> \overline{\overline{A} + \overline{B}}<br />
<br />
Then use De Morgans Law Again<br />
<br />
<math> \overline{\overline{A}} + \overline{\overline{B}} </math><br />
<br />
Finaly there are double negatives so you end up with<br />
<br />
<math> A + B </math><br />
<br />
==Example 14==<br />
'''Simplify:'''<br />
<math> X.(\overline{X}+Y) </math><br />
<br />
<math> (X.\overline{X})+(X.Y) </math> using the Distributive Law<br />
<br />
<math> 0 + (X.Y) </math> using the identity <math>A.\overline{A}=0</math><br />
<br />
<math> X.Y </math> using the identity <math>A+0=A</math><br />
<br />
Fully simplified the equation<br />
<br />
==Example 15==<br />
Simplify the following:<br />
<math> \overline{ \overline{(A.A)}. \overline{(B.B)}} </math><br />
<br />
Using De Morgan's law we can put the equation into another form:<br />
<br />
1. Change All operations,<br />
<br />
<math> \overline{ \overline{(A+A)}+ \overline{(B+B)}} </math><br />
<br />
2. Negate letters:<br />
<br />
<math> \overline{ (A+A)+(B+B) } </math><br />
<br />
3. Negate the whole expression:<br />
<br />
<math> (A+A)+(B+B) </math><br />
<br />
Using the identity law, we can then simplify this expression to:<br />
<br />
<math> A+B </math><br />
<br />
==Example 16==<br />
For this example the Boolean Algebra itself is shown above while the description/the rules used are listed below it:<br />
<br />
<Math>\overline{A}(A + B) + (B + AA)(A + \overline{B})</Math><br />
<br />
Original expression<br />
<br />
<Math>\overline{A}A + \overline{A}B + (B + A)A + (B + A)\overline{B}</Math><br />
<br />
Idempotent (AA to A), then Distributive, used twice.<br />
<br />
<Math>\overline{A}B + (B + A)A + (B + A)\overline{B}</Math><br />
<br />
Complement, then Identity. (Strictly speaking, we also used the Commutative Law for each of these applications.)<br />
<br />
<Math>\overline{A}B + BA + AA + B\overline{B} + A\overline{B}</Math><br />
<br />
Distributive, two places.<br />
<br />
<Math>\overline{A}B + BA + A + A\overline{B}</Math><br />
<br />
Idempotent (for the A's), then Complement and Identity to remove BB.<br />
<br />
<Math>\overline{A}B + AB + AT + A\overline{B}</Math><br />
<br />
Commutative, Identity; setting up for the next step.<br />
<br />
<Math>\overline{A}B + A(B + T + \overline{B})</Math><br />
<br />
Distributive.<br />
<br />
<Math>\overline{A}B + A</Math><br />
<br />
Identity, twice (depending how you count it).<br />
<br />
<Math>A + \overline{A}B</Math><br />
<br />
Commutative.<br />
<br />
<Math>(A + \overline{A})(A + B)</Math><br />
<br />
Distributive.<br />
<br />
<Math>A + B</Math><br />
<br />
Complement, Identity.<br />
<br />
==Example 16==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4856Boolean Algebra2018-05-09T12:45:21Z<p>DannyDaEpic: /* Example 16 */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A.1 = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.<br />
<br />
==OR Identities==<br />
<math> 0+A = A </math><br />
<br />
0 or A can be simplified as just A.<br />
<br />
<math> 1+A = 1 </math><br />
<br />
1 or A can be simplified as just 1.<br />
<br />
<math> A+A=A</math><br />
<br />
A or A can be simplified as just A.<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
NOT A or A can be simplified as just 1.<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<br />
<math> A+B = B+A </math><br />
<br />
or<br />
<br />
<math> A.B = B.A </math><br />
<br />
==Associate Law==<br />
If all of the symbols are the same it doesn't matter which order the equation is evaluated.<br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = C + (A+B) </math><br />
<br />
So:<br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = C . (A.B) </math><br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
This is essentially factorising or expanding the brackets, but you can also:<br />
<br />
<math> A.B + A.C = A.(B+C)</math><br />
<br />
<math> A+B.A+C = A+(B.C) </math><br />
<br />
==Redundancy Law==<br />
Law 1 :<br />
<math> A + \overline{A} B = A + B </math><br />
<br />
Proof : <br />
<br />
<math>= A + \overline{A} B \\<br />
= (A + \overline{A})(A + B) \\<br />
= 1 . (A + B) \\<br />
= A + B </math><br />
<br />
<br />
Law 2:<br />
<math> A.(\overline{A} + B) = A.B</math><br />
<br />
Proof : <br />
<br />
<math>= A.(\overline{A} + B) \\<br />
= A.\overline{A} + A.B \\<br />
= 0 + A.B \\<br />
= A.B </math><br />
<br />
==Identity Law==<br />
This is also in the identities section:<br />
<br />
<math> A.A = A </math><br />
<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
Just like in any other logic negating a negative is a positive so:<br />
<br />
<math> \overline{ \overline{A} } = A </math><br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<math> \overline{\overline{A}} . \overline{(B+C)} = A. \overline{B} . \overline{C} </math><br />
<br /><br /><br />
<math> \overline{\overline{A}} = A </math> ''Use Negation Law''<br />
<br /><br /><br />
<math> \overline{(B+C)} = (\overline{B} . \overline{C}) </math> ''Use De Morgan's Law''<br />
<br /><br /><br />
<math> A . (\overline{B} . \overline{C}) = A . \overline{B} . \overline{C} </math> ''Use Associate Law''<br />
<br />
==Example 2==<br />
<math> (\overline{A} + A) . (\overline{A} + C) </math><br />
<br />
OR Identity<br />
<br />
<math>(\overline{A} + A) = 1</math><br />
<br />
AND Identity<br />
<br />
<math>= 1.(\overline{A} + C)</math><br />
<br />
Simplify<br />
<br />
<math>= (\overline{A} + C)</math><br />
<br />
==Example 3==<br />
<math>(X + Y) . (X + \overline{Y})</math><br />
<br>Distributive: <br />
<br><math>X . (Y + \overline{Y})</math><br />
<br>Identity laws: <br />
<br><math>Y + \overline{Y} = 1</math><br />
<p><math> X.1 = X</math><br />
<br />
==Example 4==<br />
<br />
Expression Rule(s) Used<br />
C + BC Original Expression<br />
C + (B + C) DeMorgan's Law.<br />
(C + C) + B Commutative, Associative Laws.<br />
T + B Complement Law.<br />
T Identity Law.<br />
<br />
==Example 5==<br />
{| class="wikitable"<br />
|-<br />
! Simplify: !! AB(A + B)(B + B):<br />
|-<br />
| Expression || Rule(s) Used<br />
|-<br />
| AB(A + B)(B + B) || Original Expression<br />
|-<br />
| AB(A + B) || Complement law, Identity law.<br />
|-<br />
| (A + B)(A + B) || DeMorgan's Law<br />
|-<br />
| A + BB || Distributive law. This step uses the fact that or distributes over and. It can look a bit strange since addition does not distribute over multiplication<br />
|-<br />
| A || Complement, Identity<br />
|}<br />
<br />
==Example 6==<br />
<math><br />
\overline{A.B}(\overline{A} + B)(\overline{B} + B) </math> Original Expression<br />
<br />
<math><br />
\overline{AB}(\overline{A} + B) </math> Complement law, Identity law.<br />
<br />
<math><br />
(\overline{A} + \overline{B})(\overline{A} + B) </math> DeMorgan's Law<br />
<br />
<math><br />
\overline{A} + \overline{B}.B </math> Distributive law. This step uses the fact that or distributes over and. <br />
<br />
<math><br />
\overline{A} </math> Complement, Identity.<br />
<br />
==Example 7==<br />
<br />
<br />
SIMPLIFY <math>(A + C)A + AC + C</math><br />
<br />
<br />
<math>(A + C)A + AC + C</math> Complement, Identity.<br />
<br />
<br />
<math>A((A + C) + C) + C</math> Commutative, Distributive.<br />
<br />
<br />
<math>A(A + C) + C</math> Associative, Idempotent.<br />
<br />
<br />
<math>AA + AC + C</math> Distributive.<br />
<br />
<br />
<math>A + (A + T)C</math> Idempotent, Identity, Distributive.<br />
<br />
<br />
<math>A + C</math> Identity, twice.<br />
<br />
==Example 8==<br />
<br />
<br />
===Simplify=== <br />
<math> (X+Y).(X+\overline{Y}) </math> <br />
<br />
===solution===<br />
<math> X.X + X.\overline{Y} + Y.X + Y.\overline{Y} </math> Expanding the brackets<br />
<br />
<math> X + X.\overline{Y} + Y.X + 0 </math> Use of <math> X.X = X </math> and <math> Y.\overline{Y} = 0 </math><br />
<br />
<math> X + X(\overline{Y}+Y) </math> Taking X out of the brackets<br />
<br />
<math> X + X(1) </math> Use of <math> Y + \overline{Y} = 1 </math><br />
<br />
<math> X(1) </math><br />
<br />
<math> X </math><br />
<br />
==Example 9==<br />
<math> (X + Y) . (X + \overline{Y}) </math><br />
<br />
<math> X + (Y.\overline{Y}) </math> after distributive law applied<br />
<br />
<math> (Y.\overline{Y}) = 0 </math><br />
<br />
<math> X + 0 </math><br />
<br />
<math> X </math><br />
<br />
<math> (X + Y) . (X + \overline{Y}) = X </math><br />
<br />
==Example 10==<br />
Simplify:<br />
<math> \overline {AB} (\overline {A}+B)(\overline {B}+B) </math> <br />
<math> \overline {AB} (\overline {A}+B) <\math> <br />
Complement law, Identity law <br />
<math> (\overline {A}+\overline {B})(\overline {A}+B) <\math><br />
DeMorgan's law <br />
<math> \overline {A}+/overline {B}B <\math> D<br />
istributive law. <br />
<math> \overline {A} <\math><br />
<br />
==Example 11==<br />
<br />
==Example 12==<br />
<br />
==Example 13==<br />
<br />
<math>\overline{\overline{A} + \overline{(B.A)}}</math><br />
<br />
First simplify using De Morgam's Law<br />
<br />
<math> \overline{\overline{A} + \overline{B} + \overline{A}} </math><br />
<br />
This can be simplified to<br />
<br />
<math> \overline{\overline{A} + \overline{B}}<br />
<br />
Then use De Morgans Law Again<br />
<br />
<math> \overline{\overline{A}} + \overline{\overline{B}} </math><br />
<br />
Finaly there are double negatives so you end up with<br />
<br />
<math> A + B </math><br />
<br />
==Example 14==<br />
'''Simplify:'''<br />
<math> X.(\overline{X}+Y) </math><br />
<br />
<math> (X.\overline{X})+(X.Y) </math> using the Distributive Law<br />
<br />
<math> 0 + (X.Y) </math> using the identity <math>A.\overline{A}=0</math><br />
<br />
<math> X.Y </math> using the identity <math>A+0=A</math><br />
<br />
Fully simplified the equation<br />
<br />
==Example 15==<br />
Simplify the following:<br />
<math> \overline{ \overline{(A.A)}. \overline{(B.B)}} </math><br />
<br />
Using De Morgan's law we can put the equation into another form:<br />
<br />
1. Change All operations,<br />
<br />
<math> \overline{ \overline{(A+A)}+ \overline{(B+B)}} </math><br />
<br />
2. Negate letters:<br />
<br />
<math> \overline{ (A+A)+(B+B) } </math><br />
<br />
3. Negate the whole expression:<br />
<br />
<math> (A+A)+(B+B) </math><br />
<br />
Using the identity law, we can then simplify this expression to:<br />
<br />
<math> A+B </math><br />
<br />
==Example 16==<br />
For this example the Boolean Algebra itself is shown above while the description/the rules used are listed below it:<br />
<br />
<Math>\overline{A}(A + B) + (B + AA)(A + \overline{B})</Math><br />
Original expression<br />
<br />
<Math>\overline{A}A + \overline{A}B + (B + A)A + (B + A)\overline{B}</Math><br />
Idempotent (AA to A), then Distributive, used twice.<br />
<br />
<Math>\overline{A}B + (B + A)A + (B + A)\overline{B}</Math><br />
Complement, then Identity. (Strictly speaking, we also used the Commutative Law for each of these applications.)<br />
<br />
<Math>\overline{A}B + BA + AA + B\overline{B} + A\overline{B}</Math><br />
Distributive, two places.<br />
<br />
<Math>\overline{A}B + BA + A + A\overline{B}</Math><br />
Idempotent (for the A's), then Complement and Identity to remove BB.<br />
<br />
<Math>\overline{A}B + AB + AT + A\overline{B}</Math><br />
Commutative, Identity; setting up for the next step.<br />
<br />
<Math>\overline{A}B + A(B + T + \overline{B})</Math><br />
Distributive.<br />
<br />
<Math>\overline{A}B + A</Math><br />
Identity, twice (depending how you count it).<br />
<br />
<Math>A + \overline{A}B</Math><br />
Commutative.<br />
<br />
<Math>(A + \overline{A})(A + B)</Math><br />
Distributive.<br />
<br />
<Math>A + B</Math><br />
Complement, Identity.<br />
<br />
==Example 16==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4847Boolean Algebra2018-05-09T12:41:24Z<p>DannyDaEpic: /* Example 16 */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A.1 = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.<br />
<br />
==OR Identities==<br />
<math> 0+A = A </math><br />
<br />
0 or A can be simplified as just A.<br />
<br />
<math> 1+A = 1 </math><br />
<br />
1 or A can be simplified as just 1.<br />
<br />
<math> A+A=A</math><br />
<br />
A or A can be simplified as just A.<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
NOT A or A can be simplified as just 1.<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<br />
<math> A+B = B+A </math><br />
<br />
or<br />
<br />
<math> A.B = B.A </math><br />
<br />
==Associate Law==<br />
If all of the symbols are the same it doesn't matter which order the equation is evaluated.<br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = C + (A+B) </math><br />
<br />
So:<br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = C . (A.B) </math><br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
This is essentially factorising or expanding the brackets, but you can also:<br />
<br />
<math> A.B + A.C = A.(B+C)</math><br />
<br />
<math> A+B.A+C = A+(B.C) </math><br />
<br />
==Redundancy Law==<br />
Law 1 :<br />
<math> A + \overline{A} B = A + B </math><br />
<br />
Proof : <br />
<br />
<math>= A + \overline{A} B \\<br />
= (A + \overline{A})(A + B) \\<br />
= 1 . (A + B) \\<br />
= A + B </math><br />
<br />
<br />
Law 2:<br />
<math> A.(\overline{A} + B) = A.B</math><br />
<br />
Proof : <br />
<br />
<math>= A.(\overline{A} + B) \\<br />
= A.\overline{A} + A.B \\<br />
= 0 + A.B \\<br />
= A.B </math><br />
<br />
==Identity Law==<br />
This is also in the identities section:<br />
<br />
<math> A.A = A </math><br />
<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
Just like in any other logic negating a negative is a positive so:<br />
<br />
<math> \overline{ \overline{A} } = A </math><br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<math> \overline{\overline{A}} . \overline{(B+C)} = A. \overline{B} . \overline{C} </math><br />
<br /><br /><br />
<math> \overline{\overline{A}} = A </math> ''Use Negation Law''<br />
<br /><br /><br />
<math> \overline{(B+C)} = (\overline{B} . \overline{C}) </math> ''Use De Morgan's Law''<br />
<br /><br /><br />
<math> A . (\overline{B} . \overline{C}) = A . \overline{B} . \overline{C} </math> ''Use Associate Law''<br />
<br />
==Example 2==<br />
<math> (\overline{A} + A) . (\overline{A} + C) </math><br />
<br />
OR Identity<br />
<br />
<math>(\overline{A} + A) = 1</math><br />
<br />
AND Identity<br />
<br />
<math>= 1.(\overline{A} + C)</math><br />
<br />
Simplify<br />
<br />
<math>= (\overline{A} + C)</math><br />
<br />
==Example 3==<br />
<math>(X + Y) . (X + \overline{Y})</math><br />
<br>Distributive: <br />
<br><math>X . (Y + \overline{Y})</math><br />
<br>Identity laws: <br />
<br><math>Y + \overline{Y} = 1</math><br />
<p><math> X.1 = X</math><br />
<br />
==Example 4==<br />
<br />
Expression Rule(s) Used<br />
C + BC Original Expression<br />
C + (B + C) DeMorgan's Law.<br />
(C + C) + B Commutative, Associative Laws.<br />
T + B Complement Law.<br />
T Identity Law.<br />
<br />
==Example 5==<br />
{| class="wikitable"<br />
|-<br />
! Simplify: !! AB(A + B)(B + B):<br />
|-<br />
| Expression || Rule(s) Used<br />
|-<br />
| AB(A + B)(B + B) || Original Expression<br />
|-<br />
| AB(A + B) || Complement law, Identity law.<br />
|-<br />
| (A + B)(A + B) || DeMorgan's Law<br />
|-<br />
| A + BB || Distributive law. This step uses the fact that or distributes over and. It can look a bit strange since addition does not distribute over multiplication<br />
|-<br />
| A || Complement, Identity<br />
|}<br />
<br />
==Example 6==<br />
<math><br />
\overline{A.B}(\overline{A} + B)(\overline{B} + B) </math> Original Expression<br />
<br />
<math><br />
\overline{AB}(\overline{A} + B) </math> Complement law, Identity law.<br />
<br />
<math><br />
(\overline{A} + \overline{B})(\overline{A} + B) </math> DeMorgan's Law<br />
<br />
<math><br />
\overline{A} + \overline{B}.B </math> Distributive law. This step uses the fact that or distributes over and. <br />
<br />
<math><br />
\overline{A} </math> Complement, Identity.<br />
<br />
==Example 7==<br />
<br />
<br />
SIMPLIFY <math>(A + C)A + AC + C</math><br />
<br />
<br />
<math>(A + C)A + AC + C</math> Complement, Identity.<br />
<br />
<br />
<math>A((A + C) + C) + C</math> Commutative, Distributive.<br />
<br />
<br />
<math>A(A + C) + C</math> Associative, Idempotent.<br />
<br />
<br />
<math>AA + AC + C</math> Distributive.<br />
<br />
<br />
<math>A + (A + T)C</math> Idempotent, Identity, Distributive.<br />
<br />
<br />
<math>A + C</math> Identity, twice.<br />
<br />
==Example 8==<br />
<br />
<br />
===Simplify=== <br />
<math> (X+Y).(X+\overline{Y}) </math> <br />
<br />
===solution===<br />
<math> X.X + X.\overline{Y} + Y.X + Y.\overline{Y} </math> Expanding the brackets<br />
<br />
<math> X + X.\overline{Y} + Y.X + 0 </math> Use of <math> X.X = X </math> and <math> Y.\overline{Y} = 0 </math><br />
<br />
<math> X + X(\overline{Y}+Y) </math> Taking X out of the brackets<br />
<br />
<math> X + X(1) </math> Use of <math> Y + \overline{Y} = 1 </math><br />
<br />
<math> X(1) </math><br />
<br />
<math> X </math><br />
<br />
==Example 9==<br />
<math> (X + Y) . (X + \overline{Y}) </math><br />
<br />
<math> X + (Y.\overline{Y}) </math> after distributive law applied<br />
<br />
<math> (Y.\overline{Y}) = 0 </math><br />
<br />
<math> X + 0 </math><br />
<br />
<math> X </math><br />
<br />
<math> (X + Y) . (X + \overline{Y}) = X </math><br />
<br />
==Example 10==<br />
Simplify:<br />
<math> \overline {AB} (\overline {A}+B)(\overline {B}+B) </math> <br />
<math> \overline {AB} (\overline {A}+B) <\math> Complement law, Identity law <br />
<math> (\overline {A}+\overline {B})(\overline {A}+B) <\math> DeMorgan's law <br />
<math> \overline {A}+/overline {B}B <\math> Distributive law. <br />
<math> \overline {A} <\math><br />
<br />
==Example 11==<br />
<br />
==Example 12==<br />
<br />
==Example 13==<br />
Using De Morgam's Law<br />
<br />
<math>\overline{\overline{A} + \overline{(B.A)}}</math><br />
<br />
<br />
<br />
<math> \overline{\overline{A} + \overline{B} + \overline{A}} </math><br />
<br />
Using De Morgans Law Again<br />
<br />
<math> \overline{\overline{A}} + \overline{\overline{B}} </math><br />
<br />
Then there are double negatives so you end up with<br />
<br />
<math> A + B </math><br />
<br />
==Example 14==<br />
'''Simplify:'''<br />
<math> X.(\overline{X}+Y) </math><br />
<br />
<math> (X.\overline{X})+(X.Y) </math> use the Distributive Law<br />
<br />
<math> 0 + (X.Y) </math> using the identity <math>A.\overline{A}=0</math><br />
<br />
<math> X.Y </math> using the identity <math>A+0=A</math><br />
<br />
Fully simplified the equation<br />
<br />
==Example 15==<br />
Simplify the following:<br />
<math> \overline{ \overline{(A.A)}. \overline{(B.B)}} </math><br />
<br />
==Example 16==<br />
<br />
<Math>\overline{A}(A + B) + (B + AA)(A + \overline{B})</Math><br />
<br />
<Math>\overline{A}A + \overline{A}B + (B + A)A + (B + A)\overline{B}</Math><br />
<br />
<Math>\overline{A}B + (B + A)A + (B + A)\overline{B}</Math><br />
<br />
<Math>\overline{A}B + BA + AA + B\overline{B} + A\overline{B}</Math><br />
<br />
<Math>\overline{A}B + BA + A + A\overline{B}</Math><br />
<br />
<Math>\overline{A}B + AB + AT + A\overline{B}</Math><br />
<br />
<Math>\overline{A}B + A(B + T + \overline{B})</Math><br />
<br />
<Math>\overline{A}B + A</Math><br />
<br />
<Math>A + \overline{A}B</Math><br />
<br />
<Math>(A + \overline{A})(A + B)</Math><br />
<br />
<Math>A + B</Math><br />
<br />
==Example 16==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4826Boolean Algebra2018-05-09T12:18:09Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A.1 = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.<br />
<br />
==OR Identities==<br />
<math> 0+A = A </math><br />
<br />
0 or A can be simplified as just A.<br />
<br />
<math> 1+A = 1 </math><br />
<br />
1 or A can be simplified as just 1.<br />
<br />
<math> A+A=A</math><br />
<br />
A or A can be simplified as just A.<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
NOT A or A can be simplified as just 1.<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<br />
<math> A+B = B+A </math><br />
<br />
or<br />
<br />
<math> A.B = B.A </math><br />
<br />
==Associate Law==<br />
If all of the symbols are the same it doesn't matter which order the equation is evaluated.<br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = B + (A+C) </math><br />
<br />
<math> A+(B+C) = C + (A+B) </math><br />
<br />
So:<br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = B . (A.C) </math><br />
<br />
<math> A.(B.C) = C . (A.B) </math><br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
This is essentially factorising or expanding the brackets, but you can also:<br />
<br />
<math> A.B + A.C = A.(B+C)</math><br />
<br />
<math> A+B.A+C = A+(B.C) </math><br />
<br />
==Redundancy Law==<br />
Law 1 :<br />
<math> A + B \overline{A} = A + B </math><br />
<br />
Proof : <br />
<br />
<math>= A + B \overline{A} \\<br />
= (A + \overline{A})(A + B) \\<br />
= 1 . (A + B) \\<br />
= A + B </math><br />
<br />
<br />
Law 2:<br />
<math> A.(\overline{A} + B) = A.B</math><br />
<br />
Proof : <br />
<br />
<math>= A.(\overline{A} + B) \\<br />
= A.\overline{A} + A.B \\<br />
= 0 + A.B \\<br />
= A.B </math><br />
<br />
==Identity Law==<br />
This is also in the identities section:<br />
<br />
<math> A.A = A </math><br />
<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
Just like in any other logic negating a negative is a positive so:<br />
<br />
<math> \overline{ \overline{A} } = A </math><br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<math> \overline{\overline{A}} . \overline{(B+C)} = A. \overline{B} . \overline{C} </math><br />
<br /><br /><br />
<math> \overline{\overline{A}} = A </math> ''Use Negation Law''<br />
<br /><br /><br />
<math> \overline{(B+C)} = (\overline{B} . \overline{C}) </math> ''Use De Morgan's Law''<br />
<br /><br /><br />
<math> A . (\overline{B} . \overline{C}) = A . \overline{B} . \overline{C} </math> ''Use Associate Law''<br />
<br />
==Example 2==<br />
<math> (\overline{A} + A) . (\overline{A} + C) </math><br />
<br />
OR Identity<br />
<br />
<math>(\overline{A} + A) = 1</math><br />
<br />
AND Identity<br />
<br />
<math>= 1.(\overline{A} + C)</math><br />
<br />
Simplify<br />
<br />
<math>= (\overline{A} + C)</math><br />
<br />
==Example 3==<br />
<math>(X + Y) . (X + \overline{Y})</math><br />
<br>Distributive: <br />
<br><math>X . (Y + \overline{Y})</math><br />
<br>Identity laws: <br />
<br><math>Y + \overline{Y} = 1</math><br />
<p><math> X.1 = X</math><br />
<br />
==Example 4==<br />
<br />
Expression Rule(s) Used<br />
C + BC Original Expression<br />
C + (B + C) DeMorgan's Law.<br />
(C + C) + B Commutative, Associative Laws.<br />
T + B Complement Law.<br />
T Identity Law.<br />
<br />
==Example 5==<br />
{| class="wikitable"<br />
|-<br />
! Simplify: !! AB(A + B)(B + B):<br />
|-<br />
| Expression || Rule(s) Used<br />
|-<br />
| AB(A + B)(B + B) || Original Expression<br />
|-<br />
| AB(A + B) || Complement law, Identity law.<br />
|-<br />
| (A + B)(A + B) || DeMorgan's Law<br />
|-<br />
| A + BB || Distributive law. This step uses the fact that or distributes over and. It can look a bit strange since addition does not distribute over multiplication<br />
|-<br />
| A || Complement, Identity<br />
|}<br />
<br />
==Example 6==<br />
<math><br />
\overline{A.B}(\overline{A} + B)(\overline{B} + B) </math> Original Expression<br />
<br />
<math><br />
\overline{AB}(\overline{A} + B) </math> Complement law, Identity law.<br />
<br />
<math><br />
(\overline{A} + \overline{B})(\overline{A} + B) </math> DeMorgan's Law<br />
<br />
<math><br />
\overline{A} + \overline{B}.B </math> Distributive law. This step uses the fact that or distributes over and. <br />
<br />
<math><br />
\overline{A} </math> Complement, Identity.<br />
<br />
==Example 7==<br />
<br />
<br />
SIMPLIFY <math>(A + C)A + AC + C</math><br />
<br />
<br />
<math>(A + C)A + AC + C</math> Complement, Identity.<br />
<br />
<br />
<math>A((A + C) + C) + C</math> Commutative, Distributive.<br />
<br />
<br />
<math>A(A + C) + C</math> Associative, Idempotent.<br />
<br />
<br />
<math>AA + AC + C</math> Distributive.<br />
<br />
<br />
<math>A + (A + T)C</math> Idempotent, Identity, Distributive.<br />
<br />
<br />
<math>A + C</math> Identity, twice.<br />
<br />
==Example 8==<br />
<br />
<br />
===Simplify=== <br />
<math> (X+Y).(X+\overline{Y}) </math> <br />
<br />
===solution===<br />
<math> X.X + X.\overline{Y} + Y.X + Y.\overline{Y} </math> Expanding the brackets<br />
<br />
<math> X + X.\overline{Y} + Y.X + 0 </math> Use of <math> X.X = X </math> and <math> Y.\overline{Y} = 0 </math><br />
<br />
<math> X + X(\overline{Y}+Y) </math> Taking X out of the brackets<br />
<br />
<math> X + X(1) </math> Use of <math> Y + \overline{Y} = 1 </math><br />
<br />
<math> X(1) </math><br />
<br />
<math> X </math><br />
<br />
==Example 9==<br />
<math> (X + Y) . (X + \overline{Y}) </math><br />
<br />
<math> X + (Y.\overline{Y}) </math><br />
<br />
<math> (Y.\overline{Y}) = 0 </math><br />
<br />
<math> X + 0 </math><br />
<br />
<math> X </math><br />
<br />
<math> (X + Y) . (X + \overline{Y}) = X </math><br />
<br />
==Example 10==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4767Boolean Algebra2018-05-08T08:41:55Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
Here the output will be 0, regardless of A's value. A would have to be 1 and 0 for the output to be 1. This means we can simplify this to just 0.<br />
<br />
==OR Identities==<br />
<math> 0+A = A </math><br />
<br />
0 or A can be simplified as just A.<br />
<br />
<math> 1+A = 1 </math><br />
<br />
1 or A can be simplified as just A.<br />
<br />
<math> \overline{A}+1=1</math><br />
<br />
NOT A or 1 can be simplified as just 1.<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
NOT A or A can be simplified as just 1.<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example.<br />
<math> A+B = B+A </math><br />
or.<br />
<math> A.B = B.A </math><br />
<br />
=Laws=<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
==Redundancy Law==<br />
Law 1 :<br />
<math> A + A \overline{B} = A + B </math><br />
<br />
Proof : <br />
<br />
<math>= A + A \overline{B} \\<br />
= (A + \overline{A})(A + B) \\<br />
= 1 . (A + B) \\<br />
= A + B </math><br />
<br />
<br />
Law 2:<br />
<math> A.(\overline{A} + B) = A.B</math><br />
<br />
Proof : <br />
<br />
<math>= A.(\overline{A} + B) \\<br />
= A.\overline{A} + A.B \\<br />
= 0 + A.B \\<br />
= A.B </math><br />
<br />
<br />
==Identity Law==<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4760Boolean Algebra2018-05-08T08:36:17Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
==OR Identities==<br />
<math> 0+A = A </math><br />
<br />
If 0 or A goes in, A is the output<br />
<br />
<math> 1+A = 1 </math><br />
<br />
If 1 or A goes in, 1 is the output<br />
<br />
<math> \overline{A}+1=1</math><br />
<br />
If NOT A or 1 goes in, the output is 1<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
If NOT A or A goes in, the output is 1<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
==Redundancy Law==<br />
<math> \overline{A} = \overline{A} </math><br />
or<br />
<math> \overline{\overline{A}} = A </math><br />
<br />
==Identity Law==<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4752Boolean Algebra2018-05-08T08:32:03Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
==OR Identities==<br />
<math> 0+A = A </math><br />
<br />
"If 0 or A goes in, A is the output"<br />
<br />
<math> 1+A = 1 </math><br />
<br />
"If 1 or A goes in, 1 is the output"<br />
<br />
<math> \overline{A}+1=1</math><br />
<br />
"If NOT A or 1 goes in, the output is 1"<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
"If NOT A or A goes in, the output is 1"<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
==Redundancy Law==<br />
<math> \overline{A} = \overline{A} </math><br />
or<br />
<math> \overline{\overline{A}} = A </math><br />
<br />
==Identity Law==<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4750Boolean Algebra2018-05-08T08:31:20Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
==OR Identities==<br />
<br />
The logic gate 'OR' in Boolean algebra is represented by a '+' symbol. The identities for the 'OR' gate in Boolean algebra is as follows:<br />
<br />
<math> 0+A = A </math><br />
"If 0 or A goes in, A is the output"<br />
<br />
<math> 1+A = 1 </math><br />
"If 1 or A goes in, 1 is the output"<br />
<br />
<math> \overline{A}+1=1</math><br />
"If NOT A or 1 goes in, the output is 1"<br />
<br />
<math> \overline{A}+A=1</math><br />
"If NOT A or A goes in, the output is 1"<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
==Redundancy Law==<br />
<math> \overline{A} = \overline{A} </math><br />
<br />
==Identity Law==<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4749Boolean Algebra2018-05-08T08:30:52Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
==OR Identities==<br />
<br />
The logic gate 'OR' in Boolean algebra is represented by a '+' symbol. The identities for the 'OR' gate in Boolean algebra is as follows:<br />
<br />
<math> 0+A = A </math><br />
<br />
"If 0 or A goes in, A is the output"<br />
<br />
<math> 1+A = 1 </math><br />
<br />
"If 1 or A goes in, 1 is the output"<br />
<br />
<math> \overline{A}+1=1</math><br />
<br />
"If NOT A or 1 goes in, the output is 1"<br />
<br />
<math> \overline{A}+A=1</math><br />
<br />
"If NOT A or A goes in, the output is 1"<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
==Redundancy Law==<br />
<math> \overline{A} = \overline{A} </math><br />
<br />
==Identity Law==<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4747Boolean Algebra2018-05-08T08:30:25Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
==OR Identities==<br />
<br />
The logic gate 'OR' in Boolean algebra is represented by a '+' symbol. The identities for the 'OR' gate in Boolean algebra is as follows:<br />
<br />
<math> 0+A = A </math><br />
"If 0 or A goes in, A is the output"<br />
<br />
<math> 1+A = 1 </math><br />
"If 1 or A goes in, 1 is the output"<br />
<br />
<math> \overline{A}+1=1</math><br />
"If NOT A or 1 goes in, the output is 1"<br />
<br />
<math> \overline{A}+A=1</math><br />
"If NOT A or A goes in, the output is 1"<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
==Redundancy Law==<br />
<br />
==Identity Law==<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4746Boolean Algebra2018-05-08T08:29:34Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
==OR Identities==<br />
<br />
The logic gate 'OR' in Boolean algebra is represented by a '+' symbol. The identities for the 'OR' gate in Boolean algebra is as follows:<br />
<br />
<math> 0+A = A </math><br />
"If 0 or A goes in, A is the output"<br />
<br />
<math> 1+A = 1 </math><br />
"If 1 or A goes in, 1 is the output"<br />
<br />
<math> /overline{A}+1=1</math><br />
"If NOT A or 1 goes in, the output is 1"<br />
<br />
<math> /overline{A}+A=1</math><br />
"If NOT A or A goes in, the output is 1"<br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
The distributive law is these two equations.<br />
<br />
<math> A.(B+C) = A.B + A.C </math><br />
<br />
<math> A+(B.C) = (A+B).(A+C) </math><br />
<br />
==Redundancy Law==<br />
<br />
==Identity Law==<br />
<math> A+A = A </math><br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4743Boolean Algebra2018-05-08T08:25:55Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
<math> A = A </math><br />
<br />
This equation means that the output is determined by the value of A. So if A =0, The output is 0, and vice versa.<br />
<br />
<math> 0.A = 0 </math><br />
<br />
Because there is a 0 in this equation, the output of this will always be 0 regardless of the value of A.<br />
<br />
<math> A.A = A</math><br />
<br />
The output is determined by A alone in this equation. This can be simplified to just "A".<br />
<br />
<math> A.\overline{A}=0 </math><br />
<br />
==OR Identities==<br />
<br />
The logic gate 'OR' in Boolean algebra is represented by a '+' symbol. The identities for the 'OR' gate in Boolean algebra is as follows:<br />
<br />
<math> 0+A = A </math><br />
"If 0 or A go in, A is the output"<br />
<br />
<math> 1+A = 1 </math><br />
"If 1 or A is the output, 1 is the output"<br />
<br />
<math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
The Commutative Law is where equations are the same no matter what way around the letters are written. For example<br />
<math> A+B=B+A </math><br />
or<br />
<math> A.B=B.A </math><br />
<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
<br />
==Redundancy Law==<br />
<br />
==Identity Law==<br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Boolean_Algebra&diff=4735Boolean Algebra2018-05-08T08:18:16Z<p>DannyDaEpic: /* OR Identities */</p>
<hr />
<div><br />
Any equation must be within the <nowiki><math> </math></nowiki> tags. For Boolean alegbra the main issue is how to negate a term like:<br />
<br />
<math> \overline{a}</math> or <math> \overline{\overline{a}+b}</math><br />
<br />
this can be done by adding the following around any term you wish to negate.:<br />
<br />
<nowiki><math> \overline{} </math></nowiki> <br />
<br />
<math> \overline{a}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{a} </math></nowiki><br />
<br />
<math> \overline{\overline{a}+b}</math><br />
<br />
is<br />
<br />
<nowiki> <math> \overline{\overline{a}+b} </math></nowiki>.<br />
<br />
=Identities=<br />
==AND Identities==<br />
<br />
The logic gate AND is represented by the "." symbol. Some examples of an equation containing this operation is:<br />
<br />
<math> A.B </math><br />
<br />
This expression means "A AND B = 1".<br />
<br />
<math> \overline{A.B} </math><br />
<br />
The line above the equation means "NOT", therefore this expression means " NOT A AND B = 1".<br />
<br />
==OR Identities==<br />
<br />
The logic gate 'OR' in Boolean algebra is represented by a '+' symbol. For example, if I was to represent "A or B" in Boolean algebra, it would look like this:<br />
<math> a+b </math><br />
<br />
=Laws=<br />
==Commutative Law==<br />
<br />
==Associate Law==<br />
<br />
==Distributive Law==<br />
<br />
==Redundancy Law==<br />
<br />
==Identity Law==<br />
<br />
==Negation Law==<br />
<br />
=Equations=<br />
Solving equations is a matter of applying the laws of boolean algrebra, followed by any of the identities you can find:<br />
<br />
==Example 1==<br />
<br />
==Example 2==<br />
<br />
==Example 3==<br />
<br />
==Example 4==<br />
<br />
==Example 5==<br />
<br />
==Example 6==<br />
<br />
==Example 7==</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=4116Error Correction2017-11-15T14:31:23Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
|type="()"}<br />
+Even parity<br />
||<br />
-Odd parity<br />
||<br />
<br />
{Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
|type="()"}<br />
-Even parity<br />
||<br />
+Odd parity<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
<br />
{Perform an even parity check on 1101110111011110<br />
|type="()"}<br />
+First and second bytes are valid<br />
||This is the correct answer as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First byte is invalid, second byte is valid<br />
||This answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First and second bytes are invalid<br />
||This answer is incorrect as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-first byte is valid, second byte is invalid<br />
||this answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 20 digit number, "12657894562545896716"<br />
{ 2 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 11 digit number, "9997586738824"<br />
{ 5 }<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=4115Error Correction2017-11-15T14:30:27Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
|type="()"}<br />
+Even parity<br />
||<br />
-Odd parity<br />
||<br />
<br />
{Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
|type="()"}<br />
-Even parity<br />
||<br />
+Odd parity<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
<br />
{Perform an even parity check on 1101110111011110<br />
|type="()"}<br />
+First and second bytes are valid<br />
||This is the correct answer as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First byte is invalid, second byte is valid<br />
||This answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First and second bytes are invalid<br />
||This answer is incorrect as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-first byte is valid, second byte is invalid<br />
||this answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 20 digit number, "12657894562545896716"<br />
{ 2 }<br />
<br />
|type="{}"}<br />
Find the check digit for the 11 digit number, "9997586738824"<br />
{ 5 }<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=4110Error Correction2017-11-15T14:17:10Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
|type="()"}<br />
+Even parity<br />
||<br />
-Odd parity<br />
||<br />
<br />
{Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
|type="()"}<br />
-Even parity<br />
||<br />
+Odd parity<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
<br />
{Perform an even parity check on 1101110111011110<br />
|type="()"}<br />
+First and second bytes are valid<br />
||This is the correct answer as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First byte is invalid, second byte is valid<br />
||This answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First and second bytes are invalid<br />
||This answer is incorrect as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-first byte is valid, second byte is invalid<br />
||this answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 20 digit number, "12657894562545896716"<br />
{ 2 }<br />
<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=4109Error Correction2017-11-15T14:16:41Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
|type="()"}<br />
+Even parity<br />
||<br />
-Odd parity<br />
||<br />
<br />
{Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
|type="()"}<br />
-Even parity<br />
||<br />
+Odd parity<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
<br />
{Perform an even parity check on 1101110111011110<br />
|type="()"}<br />
+First and second bytes are valid<br />
||This is the correct answer as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First byte is invalid, second byte is valid<br />
||This answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First and second bytes are invalid<br />
||This answer is incorrect as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-first byte is valid, second byte is invalid<br />
||this answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
<br />
{<br />
|type="()"}<br />
Find the check digit for the 20 digit number, "12657894562545896716"<br />
{ 2 }<br />
<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=4108Error Correction2017-11-15T14:16:04Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
|type="()"}<br />
+Even parity<br />
||<br />
-Odd parity<br />
||<br />
<br />
{Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
|type="()"}<br />
-Even parity<br />
||<br />
+Odd parity<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
<br />
{Perform an even parity check on 1101110111011110<br />
|type="()"}<br />
+First and second bytes are valid<br />
||This is the correct answer as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First byte is invalid, second byte is valid<br />
||This answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First and second bytes are invalid<br />
||This answer is incorrect as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-first byte is valid, second byte is invalid<br />
||this answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
<br />
Find the check digit for the 20 digit number, "12657894562545896716"<br />
|type="()"}<br />
Find the check digit for the 20 digit number, "12657894562545896716"<br />
{ 2 }<br />
<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=4107Error Correction2017-11-15T14:14:01Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
|type="()"}<br />
+Even parity<br />
||<br />
-Odd parity<br />
||<br />
<br />
{Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
|type="()"}<br />
-Even parity<br />
||<br />
+Odd parity<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
<br />
{Perform an even parity check on 1101110111011110<br />
|type="()"}<br />
+First and second bytes are valid<br />
||This is the correct answer as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First byte is invalid, second byte is valid<br />
||This answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First and second bytes are invalid<br />
||This answer is incorrect as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-first byte is valid, second byte is invalid<br />
||this answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
<br />
{Find the check digit for the 20 digit number, "12657894562545896716"<br />
|type="()"}<br />
{ 2 }<br />
}<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=4106Error Correction2017-11-15T14:13:27Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
|type="()"}<br />
+Even parity<br />
||<br />
-Odd parity<br />
||<br />
<br />
{Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
|type="()"}<br />
-Even parity<br />
||<br />
+Odd parity<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
<br />
{Perform an even parity check on 1101110111011110<br />
|type="()"}<br />
+First and second bytes are valid<br />
||This is the correct answer as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First byte is invalid, second byte is valid<br />
||This answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First and second bytes are invalid<br />
||This answer is incorrect as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-first byte is valid, second byte is invalid<br />
||this answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
<br />
{Find the check digit for the 20 digit number, "12657894562545896716"<br />
|type="()"}<br />
{ 2 }<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=4104Error Correction2017-11-15T14:12:53Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
|type="()"}<br />
+Even parity<br />
||<br />
-Odd parity<br />
||<br />
<br />
{Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
|type="()"}<br />
-Even parity<br />
||<br />
+Odd parity<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
<br />
{Perform an even parity check on 1101110111011110<br />
|type="()"}<br />
+First and second bytes are valid<br />
||This is the correct answer as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First byte is invalid, second byte is valid<br />
||This answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-First and second bytes are invalid<br />
||This answer is incorrect as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
-first byte is valid, second byte is invalid<br />
||this answer is half correct as the first byte has 5 ones and a parity bit of one making it valid by even parity, the second byte has 6 ones and a parity bit of zero meaning it is also valid by even parity.<br />
<br />
{Find the check digit for the 20 digit number, "12657894562545896716"<br />
|type="()"}<br />
{ 6 }<br />
</quiz><br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4097Bitmap Graphics2017-11-15T14:01:46Z<p>DannyDaEpic: /* Daniel */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
}<br />
<br />
{ In a .PNG file, each pixel is usually represented with 1 to 64 bits including an extra bit. What could that extra bit be for? <br />
| type ="()" }<br />
- additional resolution<br />
- anti-aliasing (FXAA, MSAA etc.)<br />
+ Transparency<br />
- The NSA<br />
}<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4096Bitmap Graphics2017-11-15T14:01:12Z<p>DannyDaEpic: /* Daniel */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
}<br />
<br />
{ In a .PNG file, each pixel is usually represented with 1 to 64 bits including an extra bit. What could that extra bit be for? }<br />
| type ="()" }<br />
- additional resolution<br />
- anti-aliasing (FXAA, MSAA etc.)<br />
+ Transparency<br />
- The NSA<br />
}<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4095Bitmap Graphics2017-11-15T14:00:13Z<p>DannyDaEpic: /* Daniel */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
}<br />
<br />
{ In a .PNG file, each pixel is usually represented with 1 to 64 bits including an extra bit. What could that extra bit be for? }<br />
|type="()"}<br />
- additional resolution<br />
- anti-aliasing (FXAA, MSAA etc.)<br />
+ Transparency<br />
- The NSA<br />
}<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4094Bitmap Graphics2017-11-15T13:59:37Z<p>DannyDaEpic: /* Daniel */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
<br />
{ In a .PNG file, each pixel is usually represented with 1 to 64 bits including an extra bit. What could that extra bit be for? }<br />
|type="()"}<br />
- additional resolution<br />
- anti-aliasing (FXAA, MSAA etc.)<br />
+ Transparency<br />
- The NSA<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4093Bitmap Graphics2017-11-15T13:59:13Z<p>DannyDaEpic: /* Daniel */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
<br />
{ In a .PNG file, each pixel is usually represented with 1 to 64 bits including an extra bit. What could that extra bit be for? }<br />
|type="()"}<br />
-additional resolution<br />
-anti-aliasing (FXAA, MSAA etc.)<br />
+Transparency<br />
-The NSA<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4092Bitmap Graphics2017-11-15T13:57:13Z<p>DannyDaEpic: /* Daniel */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
<br />
{ In a .PNG file, each pixel is usually represented with 1 to 64 bits including an extra bit. What could that extra bit be for? }<br />
|type="()"}<br />
-additional resolution<br />
-anti-aliasing (FXAA, MSAA etc.)<br />
-Transparency<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4091Bitmap Graphics2017-11-15T13:56:38Z<p>DannyDaEpic: /* Daniel */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
<br />
<quiz display=simple><br />
{ In a .PNG file, each pixel is usually represented with 1 to 64 bits including an extra bit. What could that extra bit be for? }<br />
|type="()"}<br />
-additional resolution<br />
-anti-aliasing (FXAA, MSAA etc.)<br />
-Transparency<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4083Bitmap Graphics2017-11-15T13:47:27Z<p>DannyDaEpic: /* Daniel */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
<br />
<quiz display=simple><br />
{ <br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4075Bitmap Graphics2017-11-15T13:35:00Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Bitmap_Graphics&diff=4074Bitmap Graphics2017-11-15T13:34:14Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Definitions ==<br />
{| class="wikitable"<br />
|-<br />
! Name !! Definition<br />
|-<br />
| <tt>Bitmap image</tt> || Bitmap images are stored as a series of pixels. <br />
|-<br />
| <tt>Pixel</tt> || A pixel is a single point in a graphic image with an assigned colour. Many of them on a screen are assigned a colour in a specific place to recreate a bitmap image.<br />
|}<br />
<br />
== Colour Depth ==<br />
Colour depth is either the number of bits used to indicate the colour of a single pixel, in a bitmap image, or the number of bits used for each colour component of a single pixel. This means the number of bits needed to code an image. For example an image will only need 1 bit colour depth for black and white, i.e. a 0 for white or 1 for black, such as the following image... <br />
<br />
[[File:Black-White.jpg|500px]] <br />
<br />
although if an image has 24 bit colour depth, it can code up to 16777216 colours. This uses a 24bit binary number to represent each pixel, and the highest value you can represent using 24bits is 16777215 + 1 for 0. such as the following image...<br />
<br />
[[File:Colour.jpg|500px]]<br />
<br />
== Resolution ==<br />
Resolution is determined by the height and width of an image file, for example:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
This square has a total of 10000 pixels because it is 100 pixels high and 100 wide. You multiply the height by the width. This image was downsized from a larger image to this resolution.<br />
<br />
[[File:200px-leopardpic.jpg]] <br />
<br />
The image above is the same image but downsized to the larger resolution of 200 x 200. This resolution is double the previous image so it looks of equal quality even though it is double the resolution.<br />
<br />
== File Size ==<br />
File Size of a bitmap file links nicely to this all as it is essentially the [[Bitmap_Graphics#Colour%20Depth|Colour Depth]] multiplied by the [[Bitmap_Graphics#Resolution|Resolution]]. This is because for every pixel there needs to be a colour assigned to it. Hence, an image, 100 pixels high, 500 pixels wide, with a colour depth of 24, bits would have the size of (100*500*24=) 1,200,000 bits or (100*500*25/8=) 150,000 bytes if it would be a bitmap image.<br />
<br />
== Problems with Bitmap Graphics ==<br />
One of the major problems with bitmap graphics is that when an image is zoomed into, you can see all of the pixels used for that image, hence some of the initial quality is lost, for example, here is a bitmap image when not zoomed in. This image is 100 x 100:<br />
<br />
[[File:100px-leopardpic.jpg]]<br />
<br />
However, when enlarged, what's known as 'pixelation' begins to occur, for example, in this enlarged version of the previous image. The image can also become blurry due to pixelation, this image is 100 x 100 but is been displayed at 500 x 500:<br />
<br />
[[File:100px-leopardpic.jpg|500px]]<br />
<br />
==Compared With Vector Graphics==<br />
Bitmaps are stored pixel by pixel However, in a vector graphic the image is stored be calculating the points, lines and shapes used to create the image. A drawing list is created to recreate the image. Vector graphics can be resized using mathematics so increasing the size has no effect on quality of the image.<br />
<br />
The file size of a vector graphic can be significantly smaller than a bitmap. the drawing list is essentially the instructions to rebuild the image. However if you attempted to capture a real world photograph, you would essentially have so many shapes to represent the changes in colour that the file size might not be any smaller.<br />
<br />
=Revision Questions=<br />
==Ardi==<br />
<quiz display=simple><br />
{ What is a jpeg?<br />
|type="()"}<br />
+ Joint Photographic Experts Group A format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
- Joint Photographic Expo Group B format that uses lossy compression to store bitmap images. JPEG (pronounced "JAY-peg") files have a .jpg extension.<br />
<br />
{ similiar to the gif what bitmap file format is used for moving images?<br />
|type="()"}<br />
- jpeg<br />
+ bmp<br />
|| The bmp file format stands for bitmap image file <br />
- tiff<br />
|| TIFF or Tagged image file format stores still images<br />
<br />
{from the following What program are vector images created in?<br />
|type="()"}<br />
- Paint<br />
|| Paint creates bitmap images <br />
- FL studios<br />
+ Adobe illustrator<br />
|| Correct adobe illustrator is used to create vector images<br />
</quiz><br />
<br />
==Ahsan==<br />
<quiz display=simple> <br />
When are bitmap graphics used?<br />
| type="()"}<br />
+ They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can be changed and edited<br />
- They are produced when a camera or a scanner is used and most clip art is saved as a bitmap- meaning that they can not be changed and edited<br />
<br />
<br />
{What is a pixel?<br />
| type="()"}<br />
- The largest element of a picture <br />
+ the smallest element of a picture <br />
- Resolution<br />
|| Resolution is determined by the number of pixels in an image <br />
<br />
{What do you call the number of pixels in a screen ?<br />
| type="()"}<br />
- Pixelation<br />
|| Pixelation occurs when a bitmap image is enlarged <br />
- Colour <br />
+ Resolution <br />
- Pelanyo <br />
<br />
{What is meant by color depth ?<br />
| type="()"}<br />
- The Number of pixels on screen<br />
|| This is known as resolution<br />
+ The Number of colors that can be represented by an image<br />
- The size of the file<br />
- How good the Image looks on a scale of 1 to 10<br />
|| What where you thinking?<br />
<br />
{ What happens to the file size when you increase the resolution ?<br />
<br />
| type="()"}<br />
+ Increases<br />
- Decreases<br />
|| Increasing resolution increases number of pixels and therefore increases the amount of data stored<br />
</quiz><br />
<br />
==Daniel==<br />
<quiz display=simple><br />
{ How would the color white be represented in one pixel of an image with a 2 bit color depth? <br />
|type="()"}<br />
+ 00<br />
- 01<br />
- 10<br />
- 11</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=3842Error Correction2017-11-14T10:08:32Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
{ Even parity }<br />
<br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
{ Odd parity }<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
<br />
{<br />
|type="{}"}<br />
Find the check digit for the 13 digit number, "1338564796583"<br />
{ 6 }<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=3834Error Correction2017-11-14T09:57:26Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
{ Even parity }<br />
<br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
{ Odd parity }<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{ 01101110 }<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{ 259 }<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=3833Error Correction2017-11-14T09:56:50Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
<br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
{Even parity}<br />
<br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
{Odd parity}<br />
<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{01101110}<br />
<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{259}<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=3830Error Correction2017-11-14T09:55:36Z<p>DannyDaEpic: /* Revision Questions */</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
{Even parity}<br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
{Odd parity}<br />
{<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{01101110}<br />
{<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{259}</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Error_Correction&diff=3825Error Correction2017-11-14T09:54:18Z<p>DannyDaEpic: 1</p>
<hr />
<div>== Parity Checks ==<br />
One method of error checking would be parity bits which appends a 1 or 0 to the end of 7 bit packet. This additional bit is for even parity and odd parity error checking. Parity checks are a method of error detection only.<br />
<br />
{| class="wikitable"<br />
|-<br />
| <b>Even parity</b> || Even parity will check to see if there is an even number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 1 appended to the packet, otherwise 0 would be appended to the end of the packet. If there would have been an error in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "0" parity bit at the end of the packet or an odd number of 1's in the packet but a "1" party bit.<br />
|-<br />
| <b>Odd parity</b> || Odd parity will check to see if there is an odd number of 1's in the first 7 bit of the packet. If there is an even amount, there would be a number 0 appended to the end of the packet, otherwise 1 would be appended. If an error occurred in the transmission, the receiving party would receive the packet with either an even number of 1's in the packet but "1" parity bit at the end of the packet or an odd number of 1's in the packet but a "0" party bit.<br />
|}<br />
<br />
However, if there were 2 bits that were transmitted with errors, then the check would not find the errors and it will be sent regardless, causing the parity check to not work.<br />
<br />
== Majority Voting ==<br />
Majority voting sends the same bit 3 times and goes with the most common bit for example instead of sending a 1 it would send 1 1 1 and if there was an error it would be 1 1 0 but as there are 2 ones That is what will be used.<br />
<br />
Sending 0110 would be sent 000 111 111 000. This means if an error occurs it wont change the data as an error in 0110 might be 1110, Which would be completely different however if an error occurs in 000 111 111 000 It might be 010 111 110 000 But the data wouldn't be different as it would still be read as 0110.<br />
<br />
If 2 errors occur in the same 3 bits, the result of the majority voting will result in the wrong bit, changing the output completely but failing to detect the error. For example, the binary number 010 would ideally be transmitted as 000 111 000. However, if two errors occurred in close proximity, such as 011 111 000, it would be read as 110. This is a downside to majority voting.<br />
<br />
Another disadvantage of majority voting is that it can triple the amount of data sent, as each bit needs to be transmitted three times.<br />
<br />
== Check Sums ==<br />
<br />
A check sum is another form of error correction. A check sums is a method or algorithm to calculate a check value, this value is transmitted separately from the data. <br />
<br />
For example, websites which allow you to download large files often provide a checksum value on the page, once you have downloaded a file you can run the checksum method or algorithm and compare your check value with the original value on the site.<br />
<br />
The method or algorithm can be simple or complex, a simple method could be to add together every byte of data and then divide by the number of bytes. Remember check sums are trying to verify the data and could be applied to any type of file.<br />
<br />
==Check Digits==<br />
<br />
A check digit is an additional digit included at the end of the transmitted data. Check digits are used for error detection, and can be used to ensure that the data received is valid and correct.<br />
<br />
A common use of check digits is in International Standard Book Numbers (ISBN) found under bar codes on books. An ISBN is a 10 or 13 digit number; the last digit in the sequence is a check digit generated from the other 12 and is used to check they are correct.<br />
<br />
=Revision Questions=<br />
<quiz display=simple><br />
{<br />
|type="{}"}<br />
Is the 8-bit binary number "01101001" valid using Even parity or Odd parity?<br />
{Even parity}<br />
|type="{}"}<br />
Is the 8-bit binary number "11111110" valid using Even parity or Odd parity?<br />
{Odd parity}<br />
|type="{}"}<br />
Using Majority Voting, what would the transmission "010110111000110101111001" be interpreted as?<br />
{01101110}<br />
|type="{}"}<br />
Calculate the checksum for "11000101 00001111 00101110".<br />
{259}</div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Multiplication&diff=3392Multiplication2017-09-20T12:46:02Z<p>DannyDaEpic: /* Revision */</p>
<hr />
<div>=Binary Multiplication=<br />
Binary Multiplication uses a combination of multiplying by one, shifting and addition. When multiplying a binary number by 10 it is simply shifted to the left into the next column, this multiplies the original number by 2. Multiplying by 100 causes a shift of two places to the left which multiplies the original number by four.<br />
<br />
The rules for multiplying a binary number by another binary number is:<br />
<br />
For every 1 in the multiplier repeat the number being multiplied with as many zero's to the right of it as there are digits before the 1 in the multiplier.<br />
For every 0 in the multiplier nothing is written.<br />
<br />
For example multiply 22 by 5, which in binary is 10110 by 101<br />
<br />
starting from right to left there is a 1 in the multiplier so we write the number being multiplied normally 10110.<br />
<br />
The second number in the multiplier is a 0 so we write nothing.<br />
<br />
The last number in the multiplier is a 1 and there are two digits before it a 0 and a 1, 2 digits, so we write out the number being multiplied with 2 zeros to the right of it, 10110'''00'''<br />
<br />
After the number has been multiplied by all the digits in the multiplier we simply add them up using binary addition, be careful when writing the numbers out so they are lined up correctly for addition.<br />
<br />
..10110 + <br />
1011000<br />
=1101110 converting 1101110 to denary gives us 110, which is 22x5.<br />
<br />
=Revision=<br />
<quiz display=simple><br />
{What is 1101 x 11?<br />
| type="()" }<br />
<br />
+100111<br />
||Correct.<br />
-10000<br />
||You added them together.<br />
-10101<br />
||<br />
<br />
{<br />
|type="{}"}<br />
Multiply 10101 by 11. Leave your answer as an 8-bit binary number.<br />
{ 00111111 }<br />
|| The first unit is a 1, so we add the first number once.<br />
|| 10101<br />
|| The second unit is also a 1, so we add the first number but moved to the left by one unit.<br />
|| 101010<br />
|| Then, we add those two numbers together.<br />
|| 111111<br />
|| Finally, add the extra bits to make it 8-bit<br />
|| 00111111<br />
<br />
{<br />
|type="{}"}<br />
Multiply 10111 by 101. Leave your answer as an 8-bit binary number.<br />
{ 1110011 }<br />
<br />
{<br />
|type="{}"}<br />
Multiply 11111 by 101. Leave your answer as an 8-bit binary number.<br />
{ 10011011 }<br />
|| The first unit is a 1, so we add the first number once.<br />
|| 11111<br />
|| The second unit is a 0, so we do nothing.<br />
|| The third unit is a 1, so we add the first number once but moved to the left by two units.<br />
|| 1111100<br />
|| Then we add the two binary numbers together.<br />
|| 00011111<br />
|| +<br />
|| 01111100<br />
|| 10011011<br />
<br />
{Multiply 11001 by 1011<br />
<br />
| type="()"}<br />
-100100<br />
|| You added them together<br />
+100010011<br />
|| Correct<br />
-1100100<br />
<br />
{<br />
|type="{}"}<br />
Multiply 110111 by 011. Leave your answer as an 8-bit binary number.<br />
{ 10100101 }<br />
|| The first unit is a 1, so we add the first number once.<br />
|| 110111<br />
|| The second unit is a 1, so we add the first number once but move it to the left once.<br />
|| 1101110<br />
|| The third unit is a 0, so we do nothing.<br />
|| Add the two numbers together.<br />
|| 110111 + 1101110<br />
|| 10100101<br />
<br />
<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Multiplication&diff=3254Multiplication2017-09-19T09:29:58Z<p>DannyDaEpic: /* Revision 2 */</p>
<hr />
<div>=Binary Multiplication=<br />
Binary Multiplication uses a combination of multiplying by one, shifting and addition. When multiplying a binary number by 10 it is simply shifted to the left into the next column, this multiplies the original number by 2. Multiplying by 100 causes a shift of two places to the left which multiplies the original number by four.<br />
<br />
The rules for multiplying a binary number by another binary number is:<br />
<br />
For every 1 in the multiplier repeat the number being multiplied with as many zero's to the right of it as there are digits before the 1 in the multiplier.<br />
For every 0 in the multiplier nothing is written.<br />
<br />
For example multiply 22 by 5, which in binary is 10110 by 101<br />
<br />
starting from right to left there is a 1 in the multiplier so we write the number being multiplied normally 10110.<br />
<br />
The second number in the multiplier is a 0 so we write nothing.<br />
<br />
The last number in the multiplier is a 1 and there are two digits before it a 0 and a 1, 2 digits, so we write out the number being multiplied with 2 zeros to the right of it, 10110'''00'''<br />
<br />
After the number has been multiplied by all the digits in the multiplier we simply add them up using binary addition, be careful when writing the numbers out so they are lined up correctly for addition.<br />
<br />
..10110 + <br />
1011000<br />
=1101110 converting 1101110 to denary gives us 110, which is 22x5.<br />
<br />
=Revision 2=<br />
<quiz display=simple><br />
{What is 1101 x 11?<br />
| type="()" }<br />
<br />
+100111<br />
||Correct.<br />
-10000<br />
||You added them together.<br />
-10101<br />
||<br />
</quiz><br />
<br />
=Revision=<br />
<br />
<quiz display=simple><br />
{<br />
|type="{}"}<br />
Multiply 10101 by 11. Leave your answer as an 8-bit binary number.<br />
{ 00111111 }<br />
<br />
<br />
Multiply 110101 by 101. Leave your answer as an 8-bit binary number.<br />
{ 100001001 }<br />
</quiz></div>DannyDaEpichttps://www.trccompsci.online/mediawiki/index.php?title=Multiplication&diff=3252Multiplication2017-09-19T09:29:22Z<p>DannyDaEpic: /* Binary Multiplication */</p>
<hr />
<div>=Binary Multiplication=<br />
Binary Multiplication uses a combination of multiplying by one, shifting and addition. When multiplying a binary number by 10 it is simply shifted to the left into the next column, this multiplies the original number by 2. Multiplying by 100 causes a shift of two places to the left which multiplies the original number by four.<br />
<br />
The rules for multiplying a binary number by another binary number is:<br />
<br />
For every 1 in the multiplier repeat the number being multiplied with as many zero's to the right of it as there are digits before the 1 in the multiplier.<br />
For every 0 in the multiplier nothing is written.<br />
<br />
For example multiply 22 by 5, which in binary is 10110 by 101<br />
<br />
starting from right to left there is a 1 in the multiplier so we write the number being multiplied normally 10110.<br />
<br />
The second number in the multiplier is a 0 so we write nothing.<br />
<br />
The last number in the multiplier is a 1 and there are two digits before it a 0 and a 1, 2 digits, so we write out the number being multiplied with 2 zeros to the right of it, 10110'''00'''<br />
<br />
After the number has been multiplied by all the digits in the multiplier we simply add them up using binary addition, be careful when writing the numbers out so they are lined up correctly for addition.<br />
<br />
..10110 + <br />
1011000<br />
=1101110 converting 1101110 to denary gives us 110, which is 22x5.<br />
<br />
=Revision 2=<br />
<quiz display=simple><br />
{What is 1101 x 11?<br />
| type="()" }<br />
<br />
+the correct answer<br />
-100111<br />
||Correct.<br />
-10000<br />
||You added them together.<br />
-10101<br />
||<br />
</quiz><br />
<br />
=Revision=<br />
<br />
<quiz display=simple><br />
{<br />
|type="{}"}<br />
Multiply 10101 by 11. Leave your answer as an 8-bit binary number.<br />
{ 00111111 }<br />
<br />
<br />
Multiply 110101 by 101. Leave your answer as an 8-bit binary number.<br />
{ 100001001 }<br />
</quiz></div>DannyDaEpic