Description

Boolean Algebra
Task 1.1

Z1(x1, x2, x3, x4) = 𝑿𝟏̅̅̅̅ 𝑿𝟐̅̅̅̅ 𝑿𝟑̅̅̅̅ 𝑿𝟒̅̅̅̅ + 𝑿𝟏̅̅̅̅ 𝑿𝟐̅̅̅̅ 𝑿𝟑̅̅̅̅ X4 +𝑿𝟏̅̅̅̅ 𝑿𝟐̅̅̅̅ X3 X4 + 𝑿𝟏̅̅̅̅ X2 𝑿𝟑̅̅̅̅ ̅𝑿𝟒̅̅̅ +𝑿𝟏̅̅̅̅ X2 𝑿𝟑 𝑿𝟒̅̅̅̅ + 𝑿𝟏̅̅̅̅ X2 X3 X4 + X1 𝑿𝟐̅̅̅̅ 𝑿𝟑̅̅̅̅ 𝑿𝟒̅̅̅̅+ X1 𝑿𝟐̅̅̅̅ X3 X4+ 𝑿𝟏 𝑿𝟐 ̅𝑿𝟑̅̅̅ 𝑿𝟒̅̅̅̅ + X1 X2 X3 𝑿𝟒̅̅̅̅

Z2(X1, X2, X3, X4) = 𝑿𝟏̅̅̅̅ 𝑿𝟐̅̅̅̅ 𝑿𝟑̅̅̅̅ X4 + 𝑿𝟏̅̅̅̅ 𝑿𝟐̅̅̅̅ X3 X4 + 𝑿𝟏̅̅̅̅ 𝑿𝟐 𝑿𝟑̅̅̅̅ 𝑿𝟒 + 𝑿𝟏̅̅̅̅ X2 X3 X4 +
X1 𝑿𝟐̅̅̅̅ 𝑿𝟑̅̅̅̅ 𝑿𝟒̅̅̅̅ + 𝑿𝟏 𝑿𝟐̅̅̅̅ 𝑿𝟑̅̅̅̅ X4 + 𝑿𝟏 𝑿𝟐̅̅̅̅ X3 ̅𝑿𝟒̅̅̅ + 𝑿𝟏 𝑿𝟐 𝑿𝟑̅̅̅̅ 𝑿𝟒̅̅̅̅ + X1 X2 𝑿𝟑̅̅̅̅ 𝑿𝟒+ X1 X2 X3 𝑿𝟒̅̅̅̅

Task 1.2

In this part, we have X1-4, therefore we create 4 input inside the Logical circuit. Then drawing 2 lines in every input, one is original, the other one has to give it the not gate.
Because we have 10 expressions for Z1, so create 10 and gates, and all and gate should have 4 inputs. The line connection should follow the table. For example, if 𝑿𝟏̅̅̅̅ 𝑿𝟐̅̅̅̅ 𝑿𝟑̅̅̅̅ X4, then connecting not gate’s line of X1X2X3, and connecting normal X4 line to the or gate. For Z2, because some expressions are repeated, according to my table, we just need to create more 4 gates. In the end, creating an or gate for z2 and connect with other and gate. Here is the testing case.

Case 1

Case 2

Case 3

Task 1.3

1， K-Map

K map solves the equation is according to the equation, put 1 in blocks in K map and the other place is 0. If number change, ignore the match, if number not change, 1 means stay normal (x), 0 means change to ‘not’ (!x).

Z1 = 𝑿𝟑̅̅̅̅ 𝑿𝟒̅̅̅̅ + 𝑿𝟏̅̅̅̅ X2 X3 +𝑿𝟏̅̅̅̅ 𝑿𝟐̅̅̅̅ 𝑿𝟑̅̅̅̅+ X2 𝑿𝟒̅̅̅̅ + 𝑿𝟐̅̅̅̅ X3 X4

Z2 = X1̅𝑿𝟑̅̅̅+𝑿𝟏̅̅̅̅X4+ X1𝑿𝟒̅̅̅̅

Boolean Identity

Testing

Case 1

Case 2

Case 3

https://docs.google.com/document/d/1DZ5T9WFKkITCO-o-f5OeWeZqC9_0lf-JbVEr0voWDI/edit#

MARIE