Digital Logic & Number representation
Question 1 |

Priority encoder | |
Decoder | |
Multiplexer | |
Demultiplexer |
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Question 2 |
xy+x'y' | |
x⊕y' | |
x'⊕y | |
x'⊕y' |
Discuss it




Question 3 |
X | |
X+Y | |
X xor Y | |
Y |
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= XY’+XY = X(Y’+Y) = X *1 = XWe see from truth table –
Column x = f(x,y) So , f(x,y)=xOption (A) is correct.
Question 4 |
What is the minimal form of the Karnaugh map shown below? Assume that X denotes a don’t care term.
b'd' | |
b'd' + b'c' | |
b'd' + a'b'c'd' | |
b'd' + b'c' + c'd' |
Discuss it
There are two prime implicants in the following K-Map- Prime Implicant highlighted in Green =
Prime Implicant highlighted in Orange =
So the Boolean expression is-
Therefore option (B) is correct. This explanation is provided by Chirag Manwani.
Question 5 |
What is the minimal form of the Karnaugh map shown below? Assume that X denotes a don’t care term.
b'd' | |
b'd' + b'c' | |
b'd' + a'b'c'd' | |
b'd' + b'c' + c'd' |
Discuss it
There are two prime implicants in the following K-Map- Prime Implicant highlighted in Green =
Prime Implicant highlighted in Orange =
So the Boolean expression is-
Therefore option (B) is correct. This explanation is provided by Chirag Manwani.
Question 6 |
A | |
B | |
C | |
D |
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Question 7 |
(P'.Q + R') | |
(P + Q'.R') | |
(P'.Q + R) | |
(P.Q + R) |
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Question 8 |

000 | |
001 | |
010 | |
011 |
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D | Q(t+1) |
0 | 0 |
1 | 1 |





Question 9 |
3 | |
4 | |
5 | |
6 |
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Question 10 |
m2 + m4 + m6 + m7 | |
m0 + m1 + m3 + m5 | |
m0 + m1 + m6 + m7 | |
m2 + m3 + m4 + m5 |
Discuss it

= PQ + QR’ + PR’ = PQ(R+R’) + (P+P’)QR’ + P(Q+Q’)R’ = PQR + PQR’ +PQR’ +P’QR’ + PQR’ + PQ’R’ = PQR(m7) + PQR'(m6)+P’QR'(m2) +PQ’R'(m4) = m2 + m4 + m6 + m7Option (A) is correct.