DigitalLogicDesign
Question 1 
In 16bit 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 4bit (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 4bit 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 signmagnitude 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 2input NOR gates required to implement a 4variable function function expressed in sumofminterms 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 4variable functions f_{1}, f_{2} and f_{3}, which are expressed in sumofminterms as
f_{1} = Σ(0, 2, 5, 8, 14), f_{2} = Σ(2, 3, 6, 8, 14, 15), f_{3} = 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 flipflops used are positive edgetriggered D flipflops.
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,
D_{1} = Q_{0}
D_{0} = in
Question 9 
b_{7} b_{6} b_{5} b_{4} b_{3} b_{2} b_{1} b_{0}
where the position of the binary point is between b_{3} and b_{2} . Assume b_{7} 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 nbit fixedpoint representation of an unsigned real number X uses f bits for the fraction part. Let i = nf. The range of decimal values for X in this representation is
A  2^{f} to 2^{i} 
B  2^{f} to (2^{i} – 2^{f}) 
C  0 to 2^{i} 
D  0 to (2^{i} – 2^{f }) 
Number of bits in fraction part → fbits
Number of bits in integer part → (n – f) bits
Minimum value:
000…0.000…0 = 0
Maximum value:
= (2^{ nf } – 1) + (1 – 2 ^{f}
= (2^{nf} – 2 ^{f})
= (2^{i} – 2 ^{ f })
Question 12 
When two 8bit numbers A_{7}…A_{0} and B_{7}…B_{0} in 2’s complement representation (with A_{0} and B_{0} as the least significant bits) are added using a ripplecarry adder, the sum bits obtained are S_{7}…S_{0} and the carry bits are C_{7}…C_{0}. An overflow is said to have occurred if
A  the carry bit C_{7} is 1 
B  all the carry bits (C_{7},…,C_{0}) are 1 
C  
D 
i.e., A_{7} = B_{7}
⇾ Overflow can be detected by checking carry into the sign bits (C_{in}) and carry out of the sign bits (C_{out}).
⇾ Overflow occurs iff A_{7} = B_{7} and C_{in} ≠ C_{out}
These conditions are equivalent to
Consider
Here A_{7} = B_{7} = 1 and S_{7} = 0
This happens only if C_{in} = 0
Carry out C_{out}=1 when
Similarly, in case of
C_{in}=1 and C_{out} 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 2input 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 flipflops connected as shown below. The output of the D flipflop is connected to the input of the T flipflop and the output of the T flipflop is connected to the input of the D flipflop.
Initially, both Q_{0} and Q_{1} are set to 1 (before the 1^{st} clock cycle). The outputs
A  Q_{1}Q_{0} after the 3^{rd} cycle are 11 and after the 4^{th} cycle are 00 respectively 
B  Q_{1}Q_{0} after the 3^{rd} cycle are 11 and after the 4^{th} cycle are 01 respectively 
C  Q_{1}Q_{0} after the 3^{rd} cycle are 00 and after the 4^{th} cycle are 11 respectively 
D  Q_{1}Q_{0} after the 3^{rd} cycle are 01 and after the 4^{th} cycle are 01 respectively 
Question 15 
The representation of the value of a 16bit 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 4bit 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 3bit 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 32bit (single precision) IEEE754 format:
The decimal value closest to this floatingpoint number is
A  1.45 × 10^{1} 
B  1.45 × 10^{1} 
C  2.27 × 10^{1} 
D  2.27 × 10^{1} 
For singleprecision floatingpoint representation decimal value is equal to (1)^{5} × 1.M × 2^{(E127)}
S = 0
E = (01111100)_{2} = (124).
So E – 127 = – 3
1.M = 1.11011010…0
= 2^{0} + 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^{(E127)}
= 1^{0} × 1.847 × 2^{3}
≈ 0.231
≈ 2.3 × 10^{1}
Question 17 
Consider a quadratic equation x^{2} – 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
x^{2} – (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
5_{b} = 5_{10}
6_{10} = 6_{10}
13_{b} = b+3
11 = b+3
b = 8
Now check with ab = 36
5_{b} × 6_{b} = 36_{b}
Convert them into decimals
5_{b} × 6_{b} = (b×3) + 6_{10}
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
OptionB:
OptionC:
OptionD:
(w + y)(wxy + wyz) = wxy + wyz + wxy + wyz = wxy + wyz
Question 19 
Given f(w,x,y,z) = Σ_{m}(0,1,2,3,7,8,10) + Σ_{d}(5,6,11,15), where d represents the don’tcare condition in Karnaugh maps. Which of the following is a minimum productofsums (POS) form of f(w,x,y,z)?
A  
B  
C  
D 
KMap for the function f is
Consider maxterms in Kmap to represent function in productofsums (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 = (p1)/2 = (31)/2 = 1
∴ p=3 and q=1
Question 21 
The next state table of a 2bit saturating upcounter is given below.
The counter is built as a synchronous sequential circuit using T flipflops. The expressions for T_{1} and T_{0} 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 
ExOR satisfies all the properties. Hence,
Question 23 
The 16bit 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 010203 and then repeats. The minimum number of JK ﬂipﬂops required to implement this counter is __________.
A  4 
B  5 
C  6 
D  7 
There are 3 transitions from 0.
Hence ⌈log_{2}^{3}⌉ = 2 bits have to be added to the existing 2 bits to represent 4 unique states.
Question 25 
Consider the two cascaded 2to1 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 nbit integers, built using gates of fanin 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 fanin of the gates.
As we are adding nbit numbers and fanin is at most 2,
the solution is θ(log_{2} (n)).
Question 27 
Consider an eightbit ripplecarry 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, x_{1}⊕x_{2}⊕x_{3}⊕x_{4} = 0 where x_{1}, x_{2}, x_{3}, x_{4} are Boolean variables, and ⊕ is the XOR operator. Which one of the following must always be TRUE?
A  x_{1}x_{2}x_{3}x_{4} = 0 
B  x_{1}x_{3}+x_{2} = 0 
C  
D  x_{1} + x_{2} + x_{3} + x_{4} = 0 
x_{1} ⊕ x_{2} ⊕ x_{3} ⊕ x_{4} = 0 —–(1)
A) x_{1}x_{2}x_{3} x_{4} = 0
Put x_{1} = 1, x_{2} = 1, x_{3} = 1, x_{4} = 1
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 1 ⊕ 1 = 0
But,
x_{1}x_{2}x_{3} x_{4} ≠ 0
So, false.
B) x_{1}x_{3} + x_{2} = 0
Put x_{1} = 1, x_{2} = 1, x_{3} = 0 , x_{4} = 0
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
But,
x_{1}x_{3} + x_{2} ≠ 0
So, false.
D) x_{1} + x_{2} + x_{3} + x_{4} = 0
Let x_{1}=1, x_{2}=1, x_{3}=0, x_{4}=0
The given equation will be zero, i.e.,
1 ⊕ 1 ⊕ 0 ⊕ 0 = 0
But,
x_{1} + x_{2} + x_{3} + x_{4} ≠ 0
So, false.
(i) True.
Question 29 
Let X be the number of distinct 16bit integers in 2’s complement representation. Let Y be the number of distinct 16bit integers in sign magnitude representation.
Then XY is _________.
A  1 
B  2 
C  3 
D  4 
Since range is – 2^{15} to 2^{15} – 1
Y = 2^{16} – 1
Here, +0 and 0 are represented separately.
X – Y = 2^{16} – (2^{16} – 1)
= 1
Question 30 
Consider a 4bit 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 edgetriggered D flipflop is connected to a positive edgetriggered JK flipflop as follows. The Q output of the D flipflop is connected to both the J and K inputs of the JK flipflop, while the Q output of the JK flipflop is connected to the input of the D flipflop. Initially, the output of the D flipflop is set to logic one and the output of the JK flipflop is cleared. Which one of the following is the bit sequence (including the initial state) generated at the Q output of the JK flipflop when the flipflops are connected to a freerunning common clock? Assume that J = K = 1 is the toggle mode and J = K = 0 is the stateholding mode of the JK flipflop. Both the flipflops have nonzero propagation delays.
A  0110110… 
B  0100100… 
C  011101110… 
D  011001100… 
The characteristic equations are
Q_{DN}=D=Q_{JK}
The state table and state transition diagram are as follows:
Consider Q_{D}Q_{JK}=10 as initial state because in the options Q_{JK}=0 is the initial state of JK flipflop.
The state sequence is
0 → 1 → 1 → 0 → 1 → 1
∴ Option (a) is the answer.
Question 33 
The minimum number of JK flipflops 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 flipflop’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 3flip flops is required.
Question 34 
The number of minterms 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 demorgan’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 4bit ripplecarry binary adder is implemented by using four full adders. The total propagation time of this 4bit 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 sumofproducts 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 
(3r^{2} + r + 2) / 2r= (r+3+1/r)
(3r^{2} + r + 2) / 2r= (r^{2}+3r+1) / r
(3r^{2} + r + 2) = (2r^{2}+6r+2)
r^{2} 5r = 0
Therefor r = 5
Question 39 
Consider a 4to1 multiplexer with two select lines S1 and S0, given below
The minimal sumofproducts 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(x_{1}, x_{2}, …, x_{n}, +, ⋅, ‘), written as F^{D}, is the same expression as that of F with + and ⋅ swapped. F is said to be selfdual if F = F^{D}. The number of selfdual functions with n Boolean variables is
A  2^{n} 
B  2^{(n1)} 
C  2^{(2n )} 
D  2^{(2(n1) )} 
Number of mutually exclusive pairs of minterms = 2^{n1}.
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 …. 2^{n1} times.
= 2^{(2(n1)) }
Question 41 
Let k = 2^{n}. A circuit is built by giving the output of an nbit binary counter as input to an nto2^{n} bit decoder. This circuit is equivalent to a
A  kbit binary up counter. 
B  kbit binary down counter. 
C  kbit ring counter. 
D  kbit Johnson counter. 
A n x 2^{n} decoder is a combinational circuit with only one output line has one and all others (2^{n}1) have zeros.
A nbit binary Counter produces outputs from 0 to 2^{n} i.e 000…00 to 111…11 and repeats.
The n x 2^{n} 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 2^{n} – 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,
5^{2} × 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 sumofproducts 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  Flipflop 
x is the select line, I_{0} is ‘b’ and I_{1} is a.
The output line, y = xa + x’b
Question 45 
The above synchronous sequential circuit built using JK flipflops 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 8bit number in 2’s complement form is
A  256 
B  128 
C  127 
D  0 
The smallest 8bit 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 