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Quad 2: ( m +m +m +m ) = B'D'
10
8
2
0
Quad 3: (m +m +m +m ) = AC'
9
8
13
12
Hence, F (A, B, C, D) = C'D' + B'D' + AC'
40. Given the Boolean function: F (A, B, C, D) = π(3,4,5,6,7,10,11,14,15).
Reduce the above expression by using 4-variable Karnaugh map, showing the various groups (i.e. octal, quads and pairs). [ISC 2018]
Ans. F (A, B, C, D) = π(3,4,5,6,7,10,11,14,15)
C+D C+D' C'+D' C'+D
A+B 0 1 3 2
1 1 0 1
A+B' 4 0 5 0 7 0 6 0
A'+B' 12 13 15 14
1 1 0 0
A'+B 8 1 9 1 11 0 10 1
= There are three quads:
Quad 1: (M +M +M +M ) = A+B'
7
6
4
5
Quad 2: (M +M +M +M ) = C'+D'
7
11
15
3
Quad 3: (M +M +M +M ) = A'+C'
11
15
10
14
Hence, F (A, B, C, D) = (A + B').(C' + D').(A' + C')
41. (a) A training institute intends to give scholarships to its students as per the criteria given below: [ISC 2018]
The student has excellent academic record but is financially weak.
OR
The student does not have an excellent academic record and belongs to a backward class.
OR
The student does not have an excellent academic record and is physically impaired.
The inputs are:
INPUT
A Has excellent academic record
F Financially sound
C Belongs to a backward class
I Is physically impaired
(In all the above cases 1 indicates Yes and 0 indicates No).
Output: X [1 indicates Yes, 0 indicates No for all cases]
Draw the truth table for the inputs and outputs given above and write the SOP expression for X(A, F, C, I).
Ans. A F C I X
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 1
0 1 0 0 0
0 1 0 1 1
0 1 1 0 1
0 1 1 1 1
1 0 0 0 1
1 0 0 1 1
1 0 1 0 1
1 0 1 1 1
1 1 0 0 0
1 1 0 1 0
1 1 1 0 0
1 1 1 1 0
6060 Touchpad Computer Science-XII

