Ques 21: The switching expression corresponding to f(A,B,C,D) =Σ(1,4,5,9,11,12) is:
GATE 2005
(A) BC′D′+A′C′D+AB′D
(B) ABC′+ACD+B′C′D
(C) ACD′+A′BC′+AC′D′
(D) A′BD+ACD′+BCD′
Ans: (A) BC′D′+A′C′D+AB′D
Solution:
![](https://gargsacademy.com/wp-content/uploads/2022/03/img26.jpg)
Ques 22: Given two three bit numbers a2a1a0 and b2b1b0 and c the carry in, the function that represents the carry generate function when these two numbers are added is:
GATE 2006
(A) a2b2+a2a1b1+a2a1a0b0+a2a0b1b0+a1b2b1+a1a0b2b0+a0b2b1b0
(B) a2b2+a2b1b0+a2a1b1b0+a1a0b2b1+a1a0b2+a1a0b2b0+a2a0b1b0
(C) a2+b2+(a2⊕b2)(a1+b1+(a1⊕b1)+(a0+b0))
(D) a2b2+a2‘a1b1+(a2a1)’a0b0+a2‘a0b1‘b0+a1b2‘b1+a1‘a0b2‘b0+a0(b2b1)’b0
Ans: A) a2b2+a2a1b1+a2a1a0b0+a2a0b1b0+a1b2b1+a1a0b2b0+a0b2b1b0
Solution: c1=a0b0
c2=a1b1+a1c1+b1c1
c3=a2b2+a2c2+b2c2
=a2b2+a2a1b1+a2a1c1+a2b1c1+b2a1b1+b2a1c1+b2b1c1
=a2b2+a2a1b1+a2a1a0b0+a2b1a0b0+b2a1b1+b2a1a0b0+b2b1a0b0
Option is A.
Ques 23: Consider a Boolean function f(w,x,y,z). Suppose that exactly one of its inputs is allowed to change at a time. If the function happens to be true for two input vectors i1=⟨w1,x1,y1,z1⟩ and i2=⟨w2,x2,y2,z2⟩ , we would like the function to remain true as the input changes from i1 to i2 (i1 and i2 differ in exactly one bit position) without becoming false momentarily. Let f(w,x,y,z)= ∑(5,7,11,12,13,15). Which of the following cube covers of f will ensure that the required property is satisfied?
GATE 2006
(A) w’xz, wxy’, xy’z, xyz, wyz
(B) wxy, w’xz, wyz
(C) wx(yz)’, xz, wx’yz
(D) wxy’, wyz, wxz, w’xz, xy’z, xyz
Ans: (A) w’xz, wxy’, xy’z, xyz, wyz
Solution:
![](https://gargsacademy.com/wp-content/uploads/2022/03/img28.png)
Ques 24: We consider the addition of two 2′s complement numbers bn−1bn−2…b0 and an−1an−2…a0. A binary adder for adding unsigned binary numbers is used to add the two numbers. The sum is denoted by cn−1cn−2…c0 and the carry-out by cout. Which one of the following options correctly identifies the overflow condition?
GATE 2006
(A) cout(an−1⊕bn−1)’
(B) an−1bn−1c’n−1+(an−1bn−1)’cn−1
(C) cout⊕cn−1
(D) an−1⊕bn−1⊕cn−1
Ans: (B) an−1bn−1c’n−1+(an−1bn−1)’cn−1
Solution: Number representation in 2’s complement representation:
- Positive numbers as it is
- Negative numbers in 2′s complement form.
So, the overflow conditions are
- When we add two positive numbers (sign bit 0) and we get a sign bit 1
- When we add two negative numbers (sign bit 1) and we get sign bit 0
- Overflow is relevant only for signed numbers and carry is used for unsigned numbers
- When the carryout bit and the carryin to the most significant bit differs
When we add one positive and one negative number we won’t get a carry. Also,points 1 and 2 are leading to point 4. So, if we see the options, B is the correct one here as the first part takes care of case 2 (negative numbers) and the second part takes care of case 1 (positive numbers) – point 4. We can see counter examples for other options:
A – Let n=4 and we do 0111+0111=1110. This overflows as in 2′s complement representation we can store only up to 7. But the overflow condition in A returns false as cout=0
C – This works for the above example. But fails for 1001+0001=1010 where there is no actual overflow (−7+1=−6), but the given condition gives an overflow as cout=0 and cn−1=1
D – This works for both the above examples, but fails for 1111+1111=1110 (−1+−1=−2) where there is no actual overflow but the given condition says so.
Ques 25: Consider numbers represented in 4-bit Gray code. Let h3h2h1h0 be the Gray code representation of a number n and let g3g2g1g0 be the Gray code of (n+1)(modulo16) value of the number. Which one of the following functions is correct?
GATE 2006
(A) g0(h3h2h1h0)=∑(1,2,3,6,10,13,14,15)
(B) g1(h3h2h1h0)=∑(4,9,10,11,12,13,14,15)
(C) g2(h3h2h1h0)=∑(2,4,5,6,7,12,13,15)
(D) g3(h3h2h1h0)=∑(0,1,6,7,10,11,12,13)
Ans: (C) g2(h3h2h1h0)=∑(2,4,5,6,7,12,13,15)
Solution: We need to map min terms of g3g2g1g0 with respect to h3h2h1h0.
Hence g2 matches with option C.