Digital Logic and Design CS302 Solution of Assignment 2 (Graded) Semester Fall 2013

[chakwal]

ye lo sarkar solution cs 302 assignment 2

Question 1

Using a karnaugh map, minimize the following POS expression:

(B+C+D) (A+B’+C+D’) (A’+B’+D’) (A+B’+C’+D’)(A+B+C’+D)(A’+B+C’+D)
Ans:
Firstly convert this expression into standard POS
(A+B+C+D) (A’+B+C+D) (A+B’+C+D’) (A’+B’+C+D’) (A’+B’+C’+D’) (A+B’+C’+D’)
(A+B+C’+D’) (A’+B+C’+D)
THEN write binary values of these expression
(0000)(1000)(0101)(1101)(1111)(0111)(0010)(1010)
Now create K.map
AB/CD 00 01 11 10
00 0 1 1 0
01 1 0 0 1
11 1 0 0 1
10 0 1 1 0

Question 2

Convert the expression A’B + AC into standard SOP form and then complete the X column in the truth table mentioned below.

ANS:
A’B+AC
A’B(C+C’)+A(B+B’)C
A’BC+A’BC’+ABC+AB’C
Now write binary values of these expressions

(011)(010)(111)(101)






A B C X
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1

Question 3

Convert Gray code 01101111 to Binary code by showing all steps of conversion and mention which method is used for conversion?
Ans;

Step 1
Gray code 0 1 1 0 1 1 1 1
Binary code 0
Step 2
Gray code 0 1 1 0 1 1 1 1
Binary code 0 1
Step 3
Gray code 0 1 1 0 1 1 1 1
Binary code 0 1 0
Step 4
Gray code 0 1 1 0 1 1 1 1
Binary code 0 1 0 0
Step 5
Gray code 0 1 1 0 1 1 1 1
Binary code 0 1 0 0 1
Step 6
Gray code 0 1 1 0 1 1 1 1
Binary code 0 1 0 0 1 0
Step 7
Gray code 0 1 1 0 1 1 1 1
Binary code 0 1 0 0 1 0 1
Step8
Gray code 0 1 1 0 1 1 1 1
Binary code 0 1 0 0 1 0 1 0


END