Digital-Logic-Design
Question 1 |
In 16-bit 2’s complement representation, the decimal number -28 is:
A | 1111 1111 1110 0100 |
B | 1111 1111 0001 1100 |
C | 0000 0000 1110 0100 |
D | 1000 0000 1110 0100 |
1’s complement = 1111 1111 1110 0011
2’s complement = 1’s complement + 1
2’s complement = 1111 1111 1110 0100 = (-28)
Question 2 |
Two numbers are chosen independently and uniformly at random from the set {1, 2, …, 13}. The probability (rounded off to 3 decimal places) that their 4-bit (unsigned) binary representations have the same most significant bit is ______.
A | 0.502 |
B | 0.461 |
C | 0.402 |
D | 0.561 |
1 – 0001
2 – 0010
3 – 0011
4 – 0100
5 – 0101
6 – 0110
7 – 0111
8 – 1000
9 – 1001
10 – 1010
11 – 1011
12 – 1100
13 – 1101
The probability that their 4-bit binary representations have the same most significant bit is
= P(MSB is 0) + P(MSB is 1)
= (7×7)/(13×13) + (6×6)/(13×13)
= (49+36)/169
= 85/169
= 0.502
Question 3 |
Consider Z = X – Y, where X, Y and Z are all in sign-magnitude form. X and Y are each represented in n bits. To avoid overflow, the representation of Z would require a minimum of:
A | n bits |
B | n + 2 bits |
C | n – 1 bits |
D | n + 1 bits |
To store overflow/carry bit there should be extra space to accommodate it.
Hence, Z should be n+1 bits.
Question 4 |
Which one of the following is NOT a valid identity?
A | (x + y) ⊕ z = x ⊕ (y + z) |
B | (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) |
C | x ⊕ y = x + y, if xy = 0 |
D | x ⊕ y = (xy + x’y’)’ |
(x+y) ⊕ z = (1+1)⊕ 0 = 1 ⊕ 0 = 1
x ⊕ (y+z) = 1⊕(1+0) = 1 ⊕ 1 = 0
So,
(x+y) ⊕ z ≠ x ⊕ (y+z)
Question 5 |
What is the minimum number of 2-input NOR gates required to implement a 4-variable function function expressed in sum-of-minterms form as f = Σ(0, 2, 5, 7, 8, 10, 13, 15)? Assume that all the inputs and their complements are available.
A | 2 |
B | 4 |
C | 7 |
D | 1 |
E | 3(Option not given) |


Question 6 |
Consider three 4-variable functions f1, f2 and f3, which are expressed in sum-of-minterms as
f1 = Σ(0, 2, 5, 8, 14), f2 = Σ(2, 3, 6, 8, 14, 15), f3 = Σ(2, 7, 11, 14)
For the following circuit with one AND gate and one XOR gate, the output function f can be expressed as:
A | Σ (2, 14) |
B | Σ (7, 8, 11) |
C | Σ (2, 7, 8, 11, 14) |
D | Σ (0, 2, 3, 5, 6, 7, 8, 11, 14, 15) |
f3 = ∑(2,7,11,14)
f1*f2 ⊕ f3 = ∑(2,8,14) ⊕ ∑(2,7,11,14)
= ∑(8,7,11)
(Note: Choose the terms which are not common)
Question 7 |
Let ⊕ and ⊙ denote the Exclusive OR and Exclusive NOR operations, respectively. Which one of the following is NOT CORRECT?
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |

Question 8 |
Consider the sequential circuit shown in the figure, where both flip-flops used are positive edge-triggered D flip-flops.

The number of states in the state transition diagram of this circuit that have a transition back to the same state on some value of “in” is ______
A | 2 |
B | 3 |
C | 4 |
D | 5 |

Now lets draw characteristic table,
D1 = Q0
D0 = in

Question 9 |
b7 b6 b5 b4 b3 b2 b1 b0
where the position of the binary point is between b3 and b2 . Assume b7 is the most significant bit.
Some of the decimal numbers listed below cannot be represented exactly in the above representation:
(i) 31.500
(ii) 0.875
(iii) 12.100
(iv) 3.001
Which one of the following statements is true?
A | None of (i), (ii), (iii), (iv) can be exactly represented
|
B | Only (ii) cannot be exactly represented |
C | Only (iii) and (iv) cannot be exactly represented |
D | Only (i) and (ii) cannot be exactly represented |
= 16 + 8 + 4 + 2 + 1 + 0.5
= (31.5)10
(ii) (0.875)10 = (00000.111)2
= 2-1 + 2-2 + 2-3
= 0.5 + 0.25 + 0.125
= (0.875)10
(iii) (12.100)10
It is not possible to represent (12.100)10
(iv) (3.001)10 It is not possible to represent (3.001)10
Question 10 |
Consider the minterm list form of a Boolean function F given below.
- F(P, Q, R, S) = Σm(0, 2, 5, 7, 9, 11) + d(3, 8, 10, 12, 14)
Here, m denotes a minterm and d denotes a don’t care term. The number of essential prime implicants of the function F is _______.
A | 3 |
B | 4 |
C | 5 |

There are 3 prime implicant i.e., P’QS, Q’S’ and PQ’ and all are essential.
Because 0 and 2 are correct by only Q’S’, 5 and 7 are covered by only P’QS and 8 and 9 are covered by only PQ’.
Question 11 |
The n-bit fixed-point representation of an unsigned real number X uses f bits for the fraction part. Let i = n-f. The range of decimal values for X in this representation is
A | 2-f to 2i |
B | 2-f to (2i – 2-f) |
C | 0 to 2i |
D | 0 to (2i – 2-f ) |
Number of bits in fraction part → f-bits
Number of bits in integer part → (n – f) bits

Minimum value:
000…0.000…0 = 0
Maximum value:

= (2 n-f – 1) + (1 – 2 -f
= (2n-f – 2 -f)
= (2i – 2 -f )
Question 12 |
When two 8-bit numbers A7…A0 and B7…B0 in 2’s complement representation (with A0 and B0 as the least significant bits) are added using a ripple-carry adder, the sum bits obtained are S7…S0 and the carry bits are C7…C0. An overflow is said to have occurred if
A | the carry bit C7 is 1 |
B | all the carry bits (C7,…,C0) are 1 |
C | ![]() |
D | ![]() |
i.e., A7 = B7
⇾ Overflow can be detected by checking carry into the sign bits (Cin) and carry out of the sign bits (Cout).
⇾ Overflow occurs iff A7 = B7 and Cin ≠ Cout
These conditions are equivalent to

Consider

Here A7 = B7 = 1 and S7 = 0
This happens only if Cin = 0

Carry out Cout=1 when

Similarly, in case of
Cin=1 and Cout will be 0.

Question 13 |
Consider the Karnaugh map given below, where X represents “don’t care” and blank represents 0.
Assume for all inputs , the respective complements
are also available. The above logic is implemented using 2-input NOR gates only. The minimum number of gates required is _________.
A | 1 |
B | 2 |
C | 3 |
D | 4 |
As all variables and their complements are available we can implement the function with only one NOR Gate.
Question 14 |
Consider a combination of T and D flip-flops connected as shown below. The output of the D flip-flop is connected to the input of the T flip-flop and the output of the T flip-flop is connected to the input of the D flip-flop.
Initially, both Q0 and Q1 are set to 1 (before the 1st clock cycle). The outputs
A | Q1Q0 after the 3rd cycle are 11 and after the 4th cycle are 00 respectively |
B | Q1Q0 after the 3rd cycle are 11 and after the 4th cycle are 01 respectively |
C | Q1Q0 after the 3rd cycle are 00 and after the 4th cycle are 11 respectively |
D | Q1Q0 after the 3rd cycle are 01 and after the 4th cycle are 01 respectively |


Question 15 |
The representation of the value of a 16-bit unsigned integer X in hexadecimal number system is BCA9. The representation of the value of X in octal number system is
A | 136251 |
B | 736251 |
C | 571247 |
D | 136252 |
Each hexadecimal digit is equal to a 4-bit binary number. So convert
X = (BCA9)16 to binary

Divide the binary data into groups 3 bits each because each octal digit is represented by 3-bit binary number.
X = (001 011 110 010 101 001)2
Note: Two zeroes added at host significant position to make number bits of a multiple of 3 (16 + 2 = 18)
X = (136251)8
Question 16 |
Given the following binary number in 32-bit (single precision) IEEE-754 format:
The decimal value closest to this floating-point number is
A | 1.45 × 101 |
B | 1.45 × 10-1 |
C | 2.27 × 10-1 |
D | 2.27 × 101 |

For single-precision floating-point representation decimal value is equal to (-1)5 × 1.M × 2(E-127)
S = 0
E = (01111100)2 = (124).
So E – 127 = – 3
1.M = 1.11011010…0
= 20 + 2(-1) + 2(-1) + 2(-4) + 2(-5) + 2(-7)
= 1+0.5+0.25+0.06+0.03+0.007
≈ 1.847
(-1)5 × 1.M × 2(E-127)
= -10 × 1.847 × 2-3
≈ 0.231
≈ 2.3 × 10-1
Question 17 |
Consider a quadratic equation x2 – 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b=________.
A | 8 |
B | 9 |
C | 10 |
D | 11 |
Generally if a, b are roots.
(x – a)(x – b) = 0
x2 – (a + b)x + ab = 0
Given that x=5, x=6 are roots of (1)
So, a + b = 13
ab=36 (with same base ‘b’)
i.e., (5)b + (6)b = (13)b
Convert them into decimal value
5b = 510
610 = 610
13b = b+3
11 = b+3
b = 8
Now check with ab = 36
5b × 6b = 36b
Convert them into decimals
5b × 6b = (b×3) + 610
30 = b × 3 + 6
24 = b × 3
b = 8
∴ The required base = 8
Question 18 |
If w, x, y, z are Boolean variables, then which one of the following is INCORRECT?
A | wx + w(x + y) + x(x + y) = x + wy |
B | ![]() |
C | ![]() |
D | (w + y)(wxy + wyz) = wxy + wyz |
wx + w(x + y) + x(x + y)
= (wx + wx) + wy + (x + xy)
= wx + wy + x(1 + y)
= wx + wy + x
= (w + 1)x + wy
= x + wy
Option-B:

Option-C:

Option-D:
(w + y)(wxy + wyz) = wxy + wyz + wxy + wyz = wxy + wyz
Question 19 |
Given f(w,x,y,z) = Σm(0,1,2,3,7,8,10) + Σd(5,6,11,15), where d represents the don’t-care condition in Karnaugh maps. Which of the following is a minimum product-of-sums (POS) form of f(w,x,y,z)?
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |
K-Map for the function f is

Consider maxterms in K-map to represent function in product-of-sums (POS) form
f(w,x,y,z) = (w’ + z’)(x’ + z)
Question 20 |
Consider a binary code that consists of only four valid code words as given below:
Let the minimum Hamming distance of the code be p and the maximum number of erroneous bits that can be corrected by the code be q. Then the values of p and q are
A | p=3 and q=1 |
B | p=3 and q=2 |
C | p=4 and q=1 |
D | p=4 and q=2 |
Minimum Distance = p = 3

Error bits that can be corrected = (p-1)/2 = (3-1)/2 = 1
∴ p=3 and q=1
Question 21 |
The next state table of a 2-bit saturating up-counter is given below.

The counter is built as a synchronous sequential circuit using T flip-flops. The expressions for T1 and T0 are
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |

By using above excitation table,

Question 22 |
Consider the Boolean operator with the following properties:

Then x#y is equivalent to
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |


Ex-OR satisfies all the properties. Hence,

Question 23 |
The 16-bit 2’s complement representation of an integer is 1111 1111 1111 0101; its decimal representation is __________.
A | -11 |
B | -12 |
C | -13 |
D | -14 |
It is a negative number because MSB is 1.
Magnitude of 1111 1111 1111 0101 is 2’s complement of 1111 1111 1111 0101.
1111 1111 1111 0101
0000 0000 0000 1010 : 1’s Complement
0000 0000 0000 1011 : 2’s complement
= (11)10
Hence, 1111 1111 1111 0101 = -11
Question 24 |
We want to design a synchronous counter that counts the sequence 0-1-0-2-0-3 and then repeats. The minimum number of J-K flip-flops required to implement this counter is __________.
A | 4 |
B | 5 |
C | 6 |
D | 7 |
There are 3 transitions from 0.
Hence ⌈log23⌉ = 2 bits have to be added to the existing 2 bits to represent 4 unique states.

Question 25 |
Consider the two cascaded 2-to-1 multiplexers as shown in the figure.

The minimal sum of products form of the output X is
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |

Now

Question 26 |
Consider a carry lookahead adder for adding two n-bit integers, built using gates of fan-in at most two. The time to perform addition using this adder is __________.
A | Θ(1) |
B | Θ(log(n)) |
C | Θ(√n) |
D | Θ(n) |
Where n is number of bits added
and k is fan-in of the gates.
As we are adding n-bit numbers and fan-in is at most 2,
the solution is θ(log2 (n)).
Question 27 |
Consider an eight-bit ripple-carry adder for computing the sum of A and B, where A and B are integers represented in 2’s complement form. If the decimal value of A is one, the decimal value of B that leads to the longest latency for the sum to stabilize is _________.
A | -1 |
B | -2 |
C | -3 |
D | -4 |
If we do 2’s complement of 1 = 0000 0001, we get -1 = “1111 1111”

So, if B = -1, every carry bit is 1.
Question 28 |
Let, x1⊕x2⊕x3⊕x4 = 0 where x1, x2, x3, x4 are Boolean variables, and ⊕ is the XOR operator. Which one of the following must always be TRUE?
A | x1x2x3x4 = 0 |
B | x1x3+x2 = 0 |
C | ![]() |
D | x1 + x2 + x3 + x4 = 0 |
x1 ⊕ x2 ⊕ x3 ⊕ x4 = 0 —–(1)
A) x1x2x3 x4 = 0
Put x1 = 1, x2 = 1, x3 = 1, x4 = 1
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
But,
x1x2x3 x4 ≠ 0
So, false.
B) x1x3 + x2 = 0
Put x1 = 1, x2 = 1, x3 = 0 , x4 = 0
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
But,
x1x3 + x2 ≠ 0
So, false.
D) x1 + x2 + x3 + x4 = 0
Let x1=1, x2=1, x3=0, x4=0
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
But,
x1 + x2 + x3 + x4 ≠ 0
So, false.
(i) True.
Question 29 |
Let X be the number of distinct 16-bit integers in 2’s complement representation. Let Y be the number of distinct 16-bit integers in sign magnitude representation.
Then X-Y is _________.
A | 1 |
B | 2 |
C | 3 |
D | 4 |
Since range is – 215 to 215 – 1
Y = 216 – 1
Here, +0 and -0 are represented separately.
X – Y = 216 – (216 – 1)
= 1
Question 30 |
Consider a 4-bit Johnson counter with an initial value of 0000. The counting sequence of this counter is
A | 0, 1, 3, 7, 15, 14, 12, 8, 0 |
B | 0, 1, 3, 5, 7, 9, 11, 13, 15, 0 |
C | 0, 2, 4, 6, 8, 10, 12, 14, 0 |
D | 0, 8, 12, 14, 15, 7, 3, 1, 0 |

The state sequence is 0,8,12,14,15,7,3,1,0.
Question 31 |
The binary operator ≠ is defined by the following truth table

Which one of the following is true about the binary operator ≠?
A | Both commutative and associative |
B | Commutative but not associative |
C | Not commutative but associative |
D | Neither commutative nor associative |
Question 32 |
A positive edge-triggered D flip-flop is connected to a positive edge-triggered JK flipflop as follows. The Q output of the D flip-flop is connected to both the J and K inputs of the JK flip-flop, while the Q output of the JK flip-flop is connected to the input of the D flip-flop. Initially, the output of the D flip-flop is set to logic one and the output of the JK flip-flop is cleared. Which one of the following is the bit sequence (including the initial state) generated at the Q output of the JK flip-flop when the flip-flops are connected to a free-running common clock? Assume that J = K = 1 is the toggle mode and J = K = 0 is the state-holding mode of the JK flip-flop. Both the flip-flops have non-zero propagation delays.
A | 0110110… |
B | 0100100… |
C | 011101110… |
D | 011001100… |

The characteristic equations are
QDN=D=QJK

The state table and state transition diagram are as follows:

Consider QDQJK=10 as initial state because in the options QJK=0 is the initial state of JK flip-flop.
The state sequence is

0 → 1 → 1 → 0 → 1 → 1
∴ Option (a) is the answer.
Question 33 |
The minimum number of JK flip-flops required to construct a synchronous counter with the count sequence (0, 0, 1, 1, 2, 2, 3, 3, 0, 0,……) is ___________.
A | 2 |
B | 3 |
C | 4 |
D | 5 |
00
00
01
01
10
10
11
11
In the above sequence two flip-flop’s will not be sufficient. Since we are confronted with repeated sequence, we may add another bit to the above sequence.
000
100
001
101
010
110
011
111
Now and every count is unique, occurring only once.
So finally 3-flip flops is required.
Question 34 |
The number of min-terms after minimizing the following Boolean expression is ______.
[D′ + AB′ + A′C + AC′D + A′C′D]′
A | 1 |
B | 2 |
C | 3 |
D | 4 |
[D’ + AB’ + A’C + AC’D + A’C’D]’
[D’ + AB’ + A’C + C’D (A + A’)’]’ (since A+A’ = 1)
[AB’ + A’C + (D’ + C’) (D’ + D)]’ (since D’ + D =1)
[AB’ + A’C + D’ + C’]’
[AB’ + (A’ + C’) (C + C’) + D’]’
[AB’ + A’ + C’ + D’]’
[(A + A’) (A’ + B’) + C’ + D’]’
[A’ + B’ + C’ + D’]’
Apply de-morgan’s law,
ABCD
Question 35 |
A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate. The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.2 microseconds. A 4-bit ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 4-bit binary adder in microseconds is ____________.
A | 19.1 |
B | 19.2 |
C | 18.1 |
D | 18.2 |

Here, each Full Adder is taking 4.8 microseconds. Given adder is a 4 Bit Ripple Carry Adder. So it takes 4*4.8 = 19.2 microseconds.
Question 36 |
The total number of prime implicants of the function f(w,x,y,z) = Σ(0, 2, 4, 5, 6, 10) is ______.
A | 3 |
B | 4 |
C | 2 |
D | 1 |

Total 3 prime implicants are there.
Question 37 |
Consider the following Boolean expression for F:
F(P, Q, R, S) = PQ + P'QR + P'QR'S
The minimal sum-of-products form of F is
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |
= Q(P+P’R) + P’QR’S
= Q(P+R) + P’QR’S
= QP + QR + P’QR’S
= QP + Q(R + P’R’S)
= QP + Q( R + P’S)
= QP + QR + QP’S
= Q(P+P’S) + QR
= Q(P+S)+ QR
= QP + QS + QR
Question 38 |
The base (or radix) of the number system such that the following equation holds is_________.
312/20 = 13.1
A | 5 |
B | 6 |
C | 7 |
D | 8 |
(3r2 + r + 2) / 2r= (r+3+1/r)
(3r2 + r + 2) / 2r= (r2+3r+1) / r
(3r2 + r + 2) = (2r2+6r+2)
r2 -5r = 0
Therefor r = 5
Question 39 |
Consider a 4-to-1 multiplexer with two select lines S1 and S0, given below

The minimal sum-of-products form of the Boolean expression for the output F of the multiplexer is
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |
= P’Q + PQ’R + PQR’
= Q(P’ + P R’) + PQ’R
= Q(P’ + R’) + PQ’R
= P’Q + QR’ + PQ’R
Question 40 |
The dual of a Boolean function F(x1, x2, …, xn, +, ⋅, ‘), written as FD, is the same expression as that of F with + and ⋅ swapped. F is said to be self-dual if F = FD. The number of self-dual functions with n Boolean variables is
A | 2n |
B | 2(n-1) |
C | 2(2n ) |
D | 2(2(n-1) ) |
Number of mutually exclusive pairs of minterms = 2n-1.
There are 2 choices for each pair i.e., we can choose one of the two minterms from each pair of minterms for the function.
Therefore number of functions = 2 x 2 x …. 2n-1 times.
= 2(2(n-1))
Question 41 |
Let k = 2n. A circuit is built by giving the output of an n-bit binary counter as input to an n-to-2n bit decoder. This circuit is equivalent to a
A | k-bit binary up counter. |
B | k-bit binary down counter. |
C | k-bit ring counter. |
D | k-bit Johnson counter. |
A n x 2n decoder is a combinational circuit with only one output line has one and all others (2n-1) have zeros.
A n-bit binary Counter produces outputs from 0 to 2n i.e 000…00 to 111…11 and repeats.
The n x 2n Decoder gets the input (000..00 to 111…11 ) from the binary counter and only one output line has one and rest have zeros.
This circuit is equivalent to a 2n – bit ring counter.
Question 42 |
Consider the equation (123)5 = (x8)y with x and y as unknown. The number of possible solutions is __________.
A | 3 |
B | 5 |
C | 6 |
D | 7 |
(123)5 = (x8)y
In R.H.S. since y is base so y should be greater than x and 8, i.e.,
y > x
y > 8
Now, to solve let’s change all the above bases number into base 10 number,
52 × 1 +2 × 5 + 3 = y × x + 8
38 = xy + 8
xy = 30
⇒ yx = 30
So the possible combinations are
(1,30), (2,15), (3,10), (5,6)
But we will reject (5,6) because it violates the condition (y > 8).
So, total solutions possible is 3.
Question 43 |
Consider the following minterm expression for F:
F(P,Q,R,S) = Σ0,2,5,7,8,10,13,15
The minterms 2, 7, 8 and 13 are ‘do not care’ terms. The minimal sum-of-products form for F is:
A | ![]() |
B | ![]() |
C | ![]() |
D | ![]() |

Question 44 |
Consider the following combinational function block involving four Boolean variables x, y, a, b where x, a, b are inputs and y is the output.
f (x, y, a, b) { if (x is 1) y = a; else y = b; }
Which one of the following digital logic blocks is the most suitable for implementing this function?
A | Full adder |
B | Priority encoder |
C | Multiplexor |
D | Flip-flop |
x is the select line, I0 is ‘b’ and I1 is a.
The output line, y = xa + x’b
Question 45 |

The above synchronous sequential circuit built using JK flip-flops is initialized with Q2Q1Q0 = 000. The state sequence for this circuit for the next 3 clock cycles is
A | 001, 010, 011 |
B | 111, 110, 101 |
C | 100, 110, 111 |
D | 100, 011, 001 |

Question 46 |
Let ⊕ denote the Exclusive OR (XOR) operation. Let ‘1’ and ‘0’ denote the binary constants. Consider the following Boolean expression for F over two variables P and Q:
F(P,Q) = ((1⊕P)⊕(P⊕Q))⊕((P⊕Q)⊕(Q⊕0))
The equivalent expression for F is
A | P+Q |
B | ![]() |
C | P⨁Q |
D | ![]() |
⊕ is associative i.e P ⊕ (Q ⊕ R) = (P⊕Q) ⊕ R.
P ⊕ P = 0, 1 ⊕ P = P’ and 0 ⊕ Q = Q
(1 ⊕ P) ⊕ ((P ⊕ Q) ⊕ (P ⊕ Q)) ⊕ (Q ⊕ 0)
= P’⊕ (0) ⊕ Q
= P’ ⊕ Q
= (P ⊕ Q)’
Question 47 |
The smallest integer that can be represented by an 8-bit number in 2’s complement form is
A | -256 |
B | -128 |
C | -127 |
D | 0 |
The smallest 8-bit 2’s complement number is 1000 0000.
MSB is 1. So it is a negative number.
To know the magnitude again take 2’s complement of 1000 0000.
1000 0000
0111 1111 ← 1’s complement
1000 0000 ← 2’s complement (1’s complement +1)
= 128
-128 is 1000 0000 in 2’s complement representation.
Question 48 |
In the following truth table, V = 1 if and only if the input is valid.

What function does the truth table represent?
A | Priority encoder |
B | Decoder |
C | Multiplexer |
D | Demultiplexer |
Question 49 |
Which one of the following expressions does NOT represent exclusive NOR of x and y?
A | xy+x’y’ |
B | x⊕y’ |
C | x’⊕y |
D | x’⊕y’ |
x’ ⊕ y’ = xy’ + x’y = x⊕y. Hence option D is correct.
Question 50 |
The truth table

represents the Boolean function
A | X |
B | X + Y |
C | X ⊕ Y |
D | Y |