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
       Digital-Logic-Design       GATE 2019
Question 1 Explanation: 
+28 = 0000 0000 0001 1100

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
       Digital-Logic-Design       GATE 2019
Question 2 Explanation: 
Correct answer is 0.502
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
       Digital-Logic-Design       GATE 2019
Question 3 Explanation: 
In case of addition of two numbers with the same sign, there is a chance of overflow.
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')'
       Digital-Logic-Design       GATE 2019
Question 4 Explanation: 
Let x=1, y=1, z=0.
(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)
       Digital-Logic-Design       GATE 2019
Question 5 Explanation: 
f = Σ(0, 2, 5, 7, 8, 10, 13, 15)

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)
       Digital-Logic-Design       GATE 2019
Question 6 Explanation: 
f1*f2 = ∑(2,8,14)
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
       Digital-Logic-Design       Logic-Gates       Gate 2018
Question 7 Explanation: 
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
       Digital-Logic-Design       Sequential-Circuits       Gate 2018
Question 8 Explanation: 
Let,

Now lets draw characteristic table,
D1 = Q0
D0 = in
Question 9

Consider the unsigned 8-bit fixed point binary number representation below,

    b7b6b5b4b3 ⋅ b2b1b0

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
       Digital-Logic-Design       Number-Systems       Gate 2018
Question 9 Explanation: 
(i) (31.5)10 = (11111.100)2 = 24 + 23 + 22 + 21 + 20 + 2-1
= 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
D
6
       Digital-Logic-Design       K-Map       Gate 2018
Question 10 Explanation: 
f = Σ(0, 2, 5, 7, 9, 11) + d(3, 8, 10, 12, 14)

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 )
       Digital-Logic-Design       Number-Systems       Gate 2017 set-01
Question 11 Explanation: 
Size of the fixed point number → n-bits
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
       Digital-Logic-Design       Ripple-Carry-Adder       Gate 2017 set-01
Question 12 Explanation: 
⇾ Overflow may occur when numbers of same sign are added
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
       Digital-Logic-Design       K-Map       Gate 2017 set-01
Question 13 Explanation: 
Given K-Map represents the function f(a, b, c, d) = a' c = a'(c' )' = (a + c')'
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
B
C
D
       Digital-Logic-Design       Sequential-Circuits       Gate 2017 set-01
Question 14 Explanation: 

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
       Digital-Logic-Design       Number-Systems       GATE 2017(set-02)
Question 15 Explanation: 
X = (BCA9)16
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:

00111110011011010000000000000000

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
       Digital-Logic-Design       Number-Systems       GATE 2017(set-02)
Question 16 Explanation: 

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
       Digital-Logic-Design       Number-Systems       GATE 2017(set-02)
Question 17 Explanation: 
x2 - 13x + 36 = 0 ⇾(1)
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
       Digital-Logic-Design       Boolean-Expressions       GATE 2017(set-02)
Question 18 Explanation: 
Option-A:
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
       Digital-Logic-Design       K-Map       GATE 2017(set-02)
Question 19 Explanation: 
f(w,x,y,z) = Σm(0,1,2,3,7,8,10) + Σd(5,6,11,15)
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:

00000, 01011, 10101, 11110

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
       Digital-Logic-Design       Number-Systems       GATE 2017(set-02)
Question 20 Explanation: 
Hamming distance of a code is minimum distance between any two code words.
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
       Digital-Logic-Design       Sequential-Circuits       GATE 2017(set-02)
Question 21 Explanation: 

By using above excitation table,
Question 22

Consider the Boolean operator with the following properties:

Then x#y is equiavlent to

   
A
B
C
D
       Digital-Logic-Design       Boolean-Algebra       2016 set-01
Question 22 Explanation: 


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
       Digital-Logic-Design       Number-Systems       2016 set-01
Question 23 Explanation: 
Given number is 1111 1111 1111 0101.
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
       Digital-Logic-Design       Sequential-Circuits       2016 set-01
Question 24 Explanation: 
Given sequence is 0-1-0-2-0-3
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
       Digital-Logic-Design       Multiplexer       2016 set-01
Question 25 Explanation: 
Output of 1st MUX is

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)
       Digital-Logic-Design       Carry-Look-Ahead-Buffer       2016 set-01
Question 26 Explanation: 
Formula: θ(logk (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
       Digital-Logic-Design       Adder       GATE 2016 set-2
Question 27 Explanation: 
In the question, longest LATENCY means longest DELAY for the sum to get settle.
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
       Digital-Logic-Design       Boolean-Algebra       GATE 2016 set-2
Question 28 Explanation: 
Given expression is,
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
       Digital-Logic-Design       Number-Systems       GATE 2016 set-2
Question 29 Explanation: 
X = 216
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
       Digital-Logic-Design       Sequential-Circuits       GATE 2015 (Set-01)
Question 30 Explanation: 
In a Johnson’s counter LSB is complemented and a circular right shift operation has to be done to get the next state.

The state sequence is 0,8,12,14,15,7,3,1,0.
Question 31
A
Both commutative and associative
B
Commutative but not associative
C
Not commutative but associative
D
Neither commutative nor associative
       Digital-Logic-Design       Truth Table and Boolean Expressions       GATE 2015 (Set-01)
Question 31 Explanation: 
It is clear that from the truth table, the binary operation # is equivalent to XOR i.e., ⊕, which satisfies both commutative and associative i.e., p#q q#p and p#(q#r) (p#q)#r
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...
       Digital-Logic-Design       Sequential-Circuits       GATE 2015 (Set-01)
Question 32 Explanation: 
The circuit for the given data is

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
       Digital-Logic-Design       Sequential-Circuits       GATE 2015 -(Set-2)
Question 33 Explanation: 
Count sequence mentioned is
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.

Now and every count is unique, occuring only once.
So finally 3-flip flops is required.
Question 34

A
1
B
2
C
3
D
4
       Digital-Logic-Design       Boolean-Algebra       GATE 2015 -(Set-2)
Question 34 Explanation: 
Lets simplify it
[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
       Digital-Logic-Design       Adder       GATE 2015 -(Set-2)
Question 35 Explanation: 

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
       Digital-Logic-Design       K-Map       GATE 2015(Set-03)
Question 36 Explanation: 

Total 3 prime implicants are there.
Question 37
 
A
B
C
D
       Digital-Logic-Design       Boolean-Algebra       GATE 2014(Set-01)
Question 37 Explanation: 
PQ + P’QR + P’QR’S
= 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
 
A
5
B
6
C
7
D
8
       Digital-Logic-Design       Number-Systems       GATE 2014(Set-01)
Question 38 Explanation: 
Let base of the number system is r.
(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
A
B
C
D
       Digital-Logic-Design       Multiplexer       GATE 2014(Set-01)
Question 39 Explanation: 
F(P,Q,R)= P’Q’ (0) + P’Q (1) + PQ’ (R) + PQ(R’)
= 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) )
       Digital-Logic-Design       Self-Dual-Function       Gate 2014 Set -02
Question 40 Explanation: 
Number of possible minterms = 2n.
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.
       Digital-Logic-Design       Sequential-Circuits       Gate 2014 Set -02
Question 41 Explanation: 
A ring counter is a circular shift register with only one flip-flop being set at any particular time and all others are cleared.
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
       Digital-Logic-Design       Number-Systems       Gate 2014 Set -02
Question 42 Explanation: 
First we have to fullfill all the conditios,
(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
A
B
C
D
       Digital-Logic-Design       Number-Systems       Gate 2014 Set -03
Question 43 Explanation: 
Question 44
A
Full adder
B
Priority encoder
C
Multiplexor
D
Flip-flop
       Digital-Logic-Design       Boolean-Variables       Gate 2014 Set -03
Question 44 Explanation: 
A 2x1 Multiplexer is most suitable for implementing the function.
x is the select line, I0 is 'b' and I1 is a.
The output line, y = xa + x’b
Question 45
 
A
001, 010, 011
B
111, 110, 101
C
100, 110, 111
D
100, 011, 001
       Digital-Logic-Design       Flip-Flops       Gate 2014 Set -03
Question 45 Explanation: 
Question 46
A
P+Q
B
C
P⨁Q
D
       Digital-Logic-Design       Number-Systems       Gate 2014 Set -03
Question 46 Explanation: 
((1 ⊕ P ) ⊕ (P ⊕ Q)) ⊕ ((P ⊕ Q) ⊕ (Q ⊕ 0))
⊕ 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
       Digital-Logic-Design       Number-Systems       Gate 2013
Question 47 Explanation: 
The range of 8-bit signed numbers representable is – 2n-1 to 2n-1 -1.
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
 
A
Priority encoder
B
Decoder
C
Multiplexer
D
Demultiplexer
       Digital-Logic-Design       Combinational-Circuits       Gate 2013
Question 48 Explanation: 
It is a 22 × 2 encoder. The inputs have priorities. So, it is a priority encoder.
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'
       Digital-Logic-Design       Number-Systems       Gate 2013
Question 49 Explanation: 
x ⊕ y = x’y + xy’
x’ ⊕ y’ = xy’ + x’y = x⊕y. Hence option D is correct.
Question 50
A
X
B
X + Y
C
X ⊕ Y
D
Y
       Digital-Logic-Design       Boolean-Algebra       Gate 2012
Question 50 Explanation: 
f(X,Y)=XY’ + XY = X(Y’ + Y) = X
Question 51
The decimal value 0.5 in IEEE single precision floating point representation has
A
fraction bits of 000…000 and exponent value of 0
B
fraction bits of 000…000 and exponent value of −1
C
fraction bits of 100…000 and exponent value of 0
D
no exact representation
       Digital-Logic-Design       Number-Systems       Gate 2012
Question 51 Explanation: 
(0.5)10 = (1.0)2 × 2–1
So, value of the exponent = -1
and
fraction is 000…000 (Implicit representation)
Question 52
The amount of ROM needed to implement a 4 bit multiplier is
A
64 bits
B
128 bits
C
1 Kbits
D
2 Kbits
       Digital-Logic-Design       Combinational-Circuits       Gate 2012
Question 52 Explanation: 
To implement a 4-bit multiplier we need to store all the possible combinations of 24 x 24 inputs and their corresponding 8 output bits. The total ROM size needed = 28 x 8 bits = 211 bits = 2 Kbits.
Hence option D is the answer.
Question 53
A
B
C
D
       Digital-Logic-Design       K-Map       Gate 2012
Question 53 Explanation: 
Question 54
       
A
B
C
D
       Digital-Logic-Design       Boolean-Algebra       Gate 2011
Question 54 Explanation: 
Question 55
The minimum number of D flip-flops needed to design a mod-258 counter is
A
9
B
8
C
512
D
258
       Digital-Logic-Design       Sequential-Circuits       Gate 2011
Question 55 Explanation: 
Let n is the number of flip-flops.
The max Mod values is 2n.
So 2n ≥ 258 ⇒ n = 9
Question 56
Which one of the following circuits is NOT equivalent to a 2-input XNOR (exclusive NOR) gate?
A
B
C
D
       Digital-Logic-Design       Logic-Gates       Gate 2011
Question 56 Explanation: 
All options except option ‘D’ gives EX-NOR gates
Question 57
 
A
3
B
4
C
5
D
6
       Digital-Logic-Design       Sequential-Circuits       Gate 2011
Question 57 Explanation: 

So total no. of distinct output (states) are 4.
Question 58
A
000
B
001
C
010
D
011
       Digital-Logic-Design       Sequential-Circuits       Gate 2011
Question 58 Explanation: 

So, after 010 it moves to 011.
Question 59
     
A
m2+m4+m6+m7
B
m0+m1+m3+m5
C
m0+m1+m6+m7
D
m2+m3+m4+m5
       Digital-Logic-Design       Boolean-Algebra       2010
Question 59 Explanation: 
Convert PQ + QR' + PR' into canonical form
= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'
= PQR + PQR' + P'QR' + PQ'R'
=m7 + m6 + m2 + m4
Question 60
A main memory unit with a capacity of 4 megabytes is built using 1M 1-bit DRAM chips. Each DRAM chip has 1K rows of cells with 1K cells in each row. The time taken for a single refresh operation is 100 nanoseconds. The time required to perform one refresh operation on all the cells in the memory unit is
A
100 nanoseconds
B
100*210 nanoseconds
C
100*220 nanoseconds
D
3200*220 nanoseconds
       Digital-Logic-Design       Memory-Interfacing       2010
Question 60 Explanation: 
Each chip capacity = 1M x 1-bit
Required capacity = 4MB
Number of chips needed = 4M*8 bits / 1M x 1-bit = 32 (1M x 1-bit)/(1M x 1-bit) = 32
Irrespective of the number of chips, all chips can be refreshed in parallel.
And all the cells in a row are refreshed in parallel too. So, the total time for refresh will be number of rows times the refresh time of one row.
Here we have 1K rows in a chip and refresh time of single row is 100ns.
So total time required =1K×100
=100 ×210 nanoseconds
Question 61
P is a 16-bit signed integer. The 2's complement representation of P is (F87B)16. The 2's complement representation of 8*P is
A
(C3D8)16
B
(187B)16
C
(F878)16
D
(987B)16
       Digital-Logic-Design       Number-Systems       2010
Question 61 Explanation: 
(F87B)16 is 2's complement representation of P.
(F87B)16=(1111 1000 0111 1011)2. (It is a negative number which is in 2's complement form)
P=1111 1000 0111 1011 (2's complement form)
8 * P = 23* P= 1100 0011 1101 1000. ( NOTE: Left shift k times is equivalent to Multiplication by 2k)
Hence, 1100 0011 1101 1000 is 2's complement representation of 8P.
1100 0011 1101 1000 = (C3D8)16.
Question 62
     
A
B
P⊕Q⊕R
C
P+Q+R
D
       Digital-Logic-Design       Multiplexer       2010
Question 62 Explanation: 
f= P’Q’ R + P’Q R’ + PQ’ R’ + PQR
= (P’Q’ + PQ)R + (P’Q+PQ’)R’
= (P⊕Q)’ R + (P⊕Q)R’
= (P⊕Q⊕R)
Question 63
 
A
B
C
D
       Digital-Logic-Design       Logic-Gates       2010
Question 63 Explanation: 
f = ((P’Q’ + Q’R’)’ + ( P’R’ + Q’R’)’ )’
= (P’Q’ + Q’R’)( P’R’ + Q’R’)
= (P’Q’P’R’ + P’Q’Q’R’ + Q’R’P’R’ + Q’R’Q’R’)
= (P’Q’R’ + P’Q’R’ + P’Q’R’ + Q’R’)
= (P’Q’R’ + Q’R’)
= (Q’R’)
= (Q+R)’
Question 64
   
A
11, 10, 01, 00
B
10, 11, 01, 00
C
10, 00, 01, 11
D
11, 10, 00, 01
       Digital-Logic-Design       Sequential-Circuits       2010
Question 64 Explanation: 

The next four values of Q1Q0 are 11, 10, 01, 00.
Question 65
(1217)8 is equivalent to
A
(1217)16
B
(028F)16
C
(2297)10
D
(0B17)16
       Digital-Logic-Design       Number-Systems       2009
Question 65 Explanation: 
(1217)8= (001 010 001 111)2
Divide the bits into groups, each containing 4 bits.
=(0010 1000 1111)2
=(28F)16
Question 66
What is the minimum number of gates required to implement the Boolean function (AB+C) if we have to use only 2-input NOR gates?
A
2
B
3
C
4
D
5
       Digital-Logic-Design       Logic-Gates       2009
Question 66 Explanation: 
NOR is Complement of OR
AB+C
= (A+C)(B+C) ← Distribution of + over
= ((A+C)’+(B+C)’)’
1st NOR- (A+C)’. Let X = (A+C)’
2nd NOR- (B+C)’. Let Y = (B+C)’
3rd NOR- (X+Y)’
Question 67
How many 32K × 1 RAM chips are needed to provide a memory capacity of 256K-bytes?
A
8
B
32
C
64
D
128
       Digital-Logic-Design       Memory-Interfacing       2009
Question 67 Explanation: 
Each chip capacity = 32K×1- bit
Needed memory capacity = 256K-bytes = 256K*8 bits
Number of chips needed = 256K*8 / 32K×1= 64
Question 68
In the IEEE floating point representation, the hexadecimal value 0×00000000 corresponds to
A
the normalized value 2-127
B
the normalized value 2-126
C
the normalized value +0
D
the special value +0
       Digital-Logic-Design       Number-Systems       Gate-2008
Question 68 Explanation: 
Value is ±0 if M=0 and E=0.
Question 69
       
A
B
C
D
       Digital-Logic-Design       K-Map       Gate-2008
Question 69 Explanation: 

Question 70
A
decimal 10
B
decimal 11
C
decimal 10 and 11
D
any value >2
       Digital-Logic-Design       Number-Systems       Gate-2008
Question 70 Explanation: 

Any value of r will satisfy the above equation. But the radix should be greater than 2 because the 121 has 2. So r >2 is correct.
Question 71
 
A
B
C
D
       Digital-Logic-Design       Boolean-Algebra       Gate-2008
Question 71 Explanation: 
Question 72
A set of Boolean connectives is functionally complete if all Boolean functions can be synthesized using those. Which of the following sets of connectives is NOT functionally complete?
A
EX-NOR
B
implication, negation
C
OR, negation
D
NAND
       Digital-Logic-Design       Boolean-Functions       Gate 2008-IT
Question 72 Explanation: 
→ EX-NOR is not functionally complete.
→ NOR and NAND are the functionally complete logic gates, OR, AND, NOT only logic gate can be implemented by using them.
→ And (Implication, Negation) is also functionally complete.
Question 73
 
A
-10
B
-13
C
-26
D
None of these
       Digital-Logic-Design       Number-Systems       Gate 2008-IT
Question 73 Explanation: 
Sign bit is 1 then given number is negative.
Exponent bits - 10000011
Exponent can be added with 127 bias in IEEE single precision format then outval exponent
= 10000011 - 127
= 131 - 127
= 4
→ In IEEE format, an implied 1 is before mantissa, and hence the outval number is
→ 1.101 × 24 = -(11010)2 = -26
Question 74
 
A
independent of one variable
B
independent of two variables
C
independent of three variable
D
dependent on all the variables
       Digital-Logic-Design       Boolean-Functions       Gate 2008-IT
Question 74 Explanation: 
f(A, B, C, D) = Σ(2, 3, 6, 7, 8, 9, 10, 11, 12, 13)

Independent of one variable '0'.
Question 75
 
A
xz+x’z’
B
xz’+x’z
C
x’y’+yz
D
xy+y’z’
       Digital-Logic-Design       Decoder       Gate 2008-IT
Question 75 Explanation: 
f = (x’y’z’ + x’yz’+xy’z+xyz)’
= (x’z’ + xz)’
= x’z + xz’
Question 76
A processor that has carry, overflow and sign flag bits as part of its program status word (PSW) performs addition of the following two 2's complement numbers 01001101 and 11101001. After the execution of this addition operation, the status of the carry, overflow and sign flags, respectively will be:
A
1, 1, 0
B
1, 0, 0
C
0, 1, 0
D
1, 0, 1
       Digital-Logic-Design       Number-Systems       Gate 2008-IT
Question 76 Explanation: 

Carry flag = 1
Overflow flag = 0
Sign bit = 0 (MSB bit is 0)
Overflow flag:
In computer processors, the overflow flag is usually a single bit in a system status register used to indicate when an arithmetic overflow has occurred in an operation.
Question 77
   
A
(x⊕y)’ and x’⊕y’
B
(x⊕y)’ and x⊕y
C
x⊕y and (x⊕y)’
D
x⊕y and x⊕y
       Digital-Logic-Design       Sequential-Circuits       Gate 2008-IT
Question 77 Explanation: 
From the given statement:

Excitation table of JK:
Question 78
 
A
6
B
8
C
10
D
12
       Digital-Logic-Design       Number-Systems       Gate 2008-IT
Question 78 Explanation: 
Take the multiples and add 0 to the LSB.
Now we have some values defined for pair of bits in Booth’s Algorithm,
00 → 0
11 → 0
01 → -1
10 → 1
Now after adding 0 to the LSB of the multiplier, start traversing from left to right and accordingly put the values defined above.

Hence, total 8 additions / subtractions required.
Question 79
What is the maximum number of different Boolean functions involving n Boolean variables?
A
n2
B
2n
C
22n
D
2n2
       Digital-Logic-Design       Boolean-Algebra       Gate-2007
Question 79 Explanation: 
Each “boolean” variable has two possible values i.e 0 and 1.
Number of variables= n
Number of input combinations is 2n.
Each “boolean” function has two possible outputs i.e 0 and 1.
Number of boolean functions possible is 22n.
Formula: The number of m-ary functions possible with n k-ary variables is mkn.
Question 80
How many 3-to-8 line decoders with an enable input are needed to construct a 6-to-64 line decoder without using any other logic gates?
A
7
B
8
C
9
D
10
       Digital-Logic-Design       Decoder       Gate-2007
Question 80 Explanation: 
Each 3-to-8 lines decoder has 8 output lines.
So, we can say that
8 lines covered by ----- 1 decoder
1 line covered by ----- 1/8 decoder
64 lines covered by ----- 64/8 = 8 decoders
8 lines covered by ----- 8/8 = 1 decoder
Hence total no. of decoder needed is,
8 + 1 = 9 decoders.
Question 81
 
A
independent of one variable.
B
independent of two variables.
C
independent of three variables.
D
dependent on all the variables.
       Digital-Logic-Design       Boolean-Variables       Gate-2007
Question 81 Explanation: 

w and y are not needed to represent the function f. So f is independent of two variables.
Question 82
 
A
P only
B
Q and S
C
R and S
D
S only
       Digital-Logic-Design       Boolean-Expressions       Gate-2007
Question 82 Explanation: 

(P), (Q), (R) cover all the minterms and are equivalent to f(w,x,y,z) = Σ(0,4,5,7,8,9,13,15).
(S) covers the minterms m0, m8, m9, m2, m3, m6, m7.
(S) is not covering the minterms m4, m5, m13, m15.
Question 83
 
A
Only P and Q are valid.
B
Only Q and R are valid.
C
Only P and R are valid.
D
All P, Q, R are valid.
       Digital-Logic-Design       Boolean-Expressions       Gate-2007
Question 83 Explanation: 
P: Y * Z =YZ + Y’Z’
=Y(XY + X’Y’) + Y’(XY+X’Y’)’
=XY+Y’(X ⊕ Y)
=XY+Y’(XY’+X’Y)
=XY+XY’
=X(Y+Y’) =X
Q: X*Z = (XZ + X’Z’)
= X(XY + X’Y’) + X’(XY + X’Y’)’
=XY+X’(X’Y+XY’)
=XY+X’Y
=(X+X’)Y = Y
R: X* Y*Z
= X*X Since P: Y*Z= X
=XX + X’X’
= 1
Question 84
Suppose only one multiplexer and one inverter are allowed to be used to implement any Boolean function of n variables. What is the minimum size of the multiplexer needed?
A
2n line to 1 line
B
2n+1 line to 1 line
C
2n-1 line to 1 line
D
2n-2 line to 1 line
       Digital-Logic-Design       Multiplexer       Gate-2007
Question 84 Explanation: 
Both true and complement forms of all variables are necessary to implement any function of n variables.
A 2n X 1 multiplexer can implement any function of n variables. As n variables are given to select lines, so that true and complement forms of all variables get generated inside the MUX.
As one inverter is available, we can generate complement of one variable outside of the Multiplexer. And remaining (n-1) variables are given to select lines. With this we have true and complement form of all n variables.
So, the answer is 2n-1 X 1 MUX.
Question 85
 
A
6, 3
B
10, 4
C
6, 4
D
10, 5
       Digital-Logic-Design       Carry-Look-Ahead-Adder       Gate-2007
Question 85 Explanation: 
Formula: n(n+1)/2 AND gates and n OR gates are needed for an n-bit carry look ahead circuit for addition of two binary numbers.
Question 86
 
A
0, 3, 4
B
0, 3, 4, 5
C
0, 1, 2, 3, 4
D
0, 1, 2, 3, 4, 5
       Digital-Logic-Design       Circuits-Output       Gate-2007
Question 86 Explanation: 
A counter goes through a sequence of states. Given counter resets when A1=A3=1.
Here, initial state is 0000. It goes through 0001,0010,0011,0100 and 0101. When the state is 5(0101) it immediately resets to initial state 0. Here, state 5 is not considered as valid state.
So valid states are 0,1,2,3, and 4 and hence it is a Mod5 counter.
Question 87
Which of the following is TRUE about formulae in Conjunctive Normal Form?
A
For any formula, there is a truth assignment for which at least half the clauses evaluate to true.
B
For any formula, there is a truth assignment for which all the clauses evaluate to true.
C
There is a formula such that for each truth assignment, at most one-fourth of the clauses evaluate to true.
D
None of the above.
       Digital-Logic-Design       Conjunctive-Normal-Form       Gate-2007
Question 87 Explanation: 
For n=2, (means two variables a and b)
Formula: a ∧ b
Truth table:

Conjunctive normal form : (a ∨ b) ∧ (a ∨ ~b) ∧ (~a ∨ b)

Similarly,
For n=1-----TRUE=1, FALSE=1 (1/2 ARE TRUE)
For n=2-----TRUE=3, FALSE=1 (3/4 ARE TRUE)
For n=3-----TRUE=7, FALSE=1 (7/8 ARE TRUE)
(1-2-n) are TRUE.
Looking at options,
Question 88
Which of the following input sequences for a cross-coupled R-S flip-flop realized with two NAND gates may lead to an oscillation?
A
11, 00
B
01, 10
C
10, 01
D
00, 11
       Digital-Logic-Design       Sequential-Circuits       Gate 2007-IT
Question 88 Explanation: 
RS slip-flop using NAND gates.
So, 00 input cause indeterminate state which may lead to oscillation.
Question 89
 
A
X1 = b, X2 = 0, X3 = a
B
X1 = b, X2 = 1, X3 = b
C
X1 = a, X2 = b, X3 = 1
D
X1 = a, X2 = 0, X3 = b
       Digital-Logic-Design       Multiplexer       Gate 2007-IT
Question 89 Explanation: 
F = (bX1' + aX1)X3 + X2X3'
If we put
X1 = b
X2 = 0
X3 = a
Then we get,
F = ab
Question 90
 
A
1
B
a’ + b’ + c’ + d’
C
a’ + b + c’ + d’
D
a’ + b’ + c + d’
       Digital-Logic-Design       Boolean-Functions       Gate 2007-IT
Question 90 Explanation: 
(a⋅b)⋅c + (a'⋅c)⋅d + (b⋅c)⋅d + a⋅d
= ((ab)'c)' + ((a'c)'d)' + ((bc)'d)' + (ad)'
= ab + c' + a'c + d' + bc + d' + a' + d'
= ab + c' + a'c + bc + a' + d'
= ab + c' + bc + a' + d'
= b + c' + bc + a' + d'
= a' + b + c' + d'
Question 91
A
0110
B
1011
C
1101
D
1111
       Digital-Logic-Design       Shift-register       Gate 2007-IT
Question 91 Explanation: 
Question 92
(C012.25)H – (10111001110.101)B =
A
(135103.412)O
B
(564411.412)O
C
(564411.205)O
D
(135103.205)O
       Digital-Logic-Design       Number-Systems       Gate 2007-IT
Question 92 Explanation: 
(C012.25)H – (10111001110.101)B
= 1100000000010010.00100101 - 0000010111001110.10100000
= 1011101001000011.10000101
= 1011101000011.100001010
= (135103.412)O
Question 93
 
A
0000
B
0111
C
1111
D
None of these
       Digital-Logic-Design       Circuits-Output       Gate 2007-IT
Question 93 Explanation: 

Since the problem is in the link T which is connected as input to NOR gate. So to check link T we have to make the output dependent on T by deactivating link M. So to deactivate link M, the output at M should be 0, as link M is input to NOR gate. So, to output at M as 0,
X1 = 1
X2 = 1
X3 = 1
X4 = 0
∴ None of the given option is correct.
Question 94
 
A
B
C
D

Hence, option (A) matches.
Question 95
 
A
c’d’+ ad’ + abc’ + (ac)’d
B
(ac)’ + c’d’ + ad’ + abc’d
C
(ac)’ + ad’ + abc’ + c’d
D
b’c’d’ + acd’ + (ac)’ + abc’
       Digital-Logic-Design       K-Map       Gate 2007-IT
Question 95 Explanation: 
Let's check for option (C):
a'c' + ad' + abc' + c'd

Not equivalent to the K-map, we get in previous question.
Question 96
You are given a free running clock with a duty cycle of 50% and a digital waveform f which changes only at the negative edge of the clock. Which one of the following circuits (using clocked D flip-flops) will delay the phase of f by 180°?
A
B
C
D
       Digital-Logic-Design       Sequential-Circuits       Gate-2006
Question 96 Explanation: 
Duty cycle is the period of time where the signal high, i.e. 1.
50% of duty cycle means, the wave is 1 for half of the time and 0 for the other half of the time. It is a usual digital signal with 1 and 0.
The waveform f changes for every negative edge, that means f value alters from 1 to 0 or 0 to 1 for every negative edge of the clock.
Now the problem is that we need to find the circuit which produces a phase shift of 180, which means the output is 0 when f is 1 and output is 1 when f is 0.
Like the below image.

Now to find the answer we can choose elimination method.
F changes for negative edge, so that output too should change at negative edge. i.e if f becomes 0, then at the same time output should become 1, vice versa.
So, whenever input changes, at the same point of time output too should change. As input changes on negative edge, the output should be changed at negative edge only.
To have the above behaviour, the second D flip-flop which produces the final output should be negative edge triggered. because whatever the 2nd flip-flop produces, that is the output of the complete circuit.
So, we can eliminate option a, d.
Now either b or c can be answer.
How the flip-flop chain works in option b and c is as below.
—> F changes at negative edge.
—> But flip-flop1 responds at next positive edge.
—> After this flip-flop2 responds at next negative edge.
That means flip-flop2 produces the same input which is given to flip-flop now after a positive edge and a negative edge, that means a delay of one clock cycle, which is 180 degrees phase shift for the waveform of f.
Option b) we are giving f’, so that the output is f’ with 180 degrees phase shift.
Option c) we are giving f, so that the output is f with 180 degrees phase shift.
Hence option C is the answer.
Question 97
We consider the addition of two 2’s complement numbers bn-1bn-2...b0 and an-1an-2a0. A binary adder for adding unsigned binary numbers is used to add the two numbers. The sum is denoted by cn-1cn-2c0 and the carry-out by cout. Which one of the following options correctly identifies the overflow condition?  
A
B
C
D
       Digital-Logic-Design       Carry-Generator       Gate-2006
Question 97 Explanation: 
There is an overflow if
1. The sign bits are same i.e MSB bits are same.
2. Carry_in ≠ Carry_out.
In option B, the MSB are equal.
Question 98
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) (modulo 16) value of the number. Which one of the following functions is correct?
A
g0(h3h2h1h0) = Σ(1,2,3,6,10,13,14,15)
B
g1(h3h2h1h0) = Σ(4,9,10,11,12,13,14,15)
C
g2(h3h2h1h0) = Σ(2,4,5,6,7,12,13,15)
D
g3(h3h2h1h0) = Σ(0,1,6,7,10,11,12,13)
       Digital-Logic-Design       Number-Systems       Gate-2006
Question 98 Explanation: 

g2(h3h2h1h0) = Σ(2,4,5,6,7,12,13,15)
Question 99
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 = w1x1y1z1 and i2 w2x2y2z2〉, we would like the function to remain true as the input changes from vectors 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?
A
B
C
D
       Digital-Logic-Design       K-Map       Gate-2006
Question 99 Explanation: 
Static hazard is the situation where, when one input variable changes, the output changes momentarily before stabilizing to the correct value. The most commonly used method to eliminate static hazards is to add redundant logic (consensus terms in the logic expression).
f = X1 * X2 + X1' * X3
If (X1,X2,X3) = (1,1,1) then f=1 because X1 * X2 =1 X1' * X3 = 0.
Let the input is changed from 111 to 011 , then f = 1 because X1 * X2 = 0 X1' * X3 =1.
The output f will be momentarily 0 if AND gate X1 * X2 is faster than the AND gate X1' * X3.
This Hazard can be avoided by adding the term X2 * X3 (because X1 is in true form in first term and in complement form in the second term . So pick the fixed terms X2 and X3 from both terms) to f i.e f = X1 * X2 + X1' * X3 + X2 * X3
Option D is equivalent to f(w, x, y, z) = ∑(5,7,11,12,13,15)
Question 100
   
A
B
C
D
       Digital-Logic-Design       Multiplexer       Gate-2006
Question 100 Explanation: 
f = yx + y’ (zy’+z’x)
= xy + zy’ + y’z’x
= x(y+y’z’) + zy’
= x(y+z’) + y’z
= xy + xz’ + y’z
Question 101
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:
A
B
C
D
       Digital-Logic-Design       Carry-Generator       Gate-2006
Question 101 Explanation: 
Initial Carry c is not included in any option. Hence c=0.
Carry c1 = a0b0
Carry c2 = a2b2 + c1(a2 ⊕ b2 )
= a1b1 +c1 (a1 b’1+ a’1 b1 )
= a1b1 +c1 a1 b’1+ c1 a’1 b1
= (a1b1 + c1a1 b’1)+ (c1 a’1 b1 + a1b1 )
= a1(b1+c1) +b1 (c1 + a1)
= a1b1+b1c1+a1c1
Carry c3 = a2b2 + c2(a2 ⊕ b2)
= a2b2 + c2(a’2b2 + a2b’2 )
= a2b2 + b2c2 + a2c2
= a2b2+a2a1b1+a2a1a0b0+a2a0b1b0+a1b2b1+a1a0b2b0+a0b2b1b0
Question 102
A
000
B
001
C
010
D
101
       Digital-Logic-Design       Sequential-Circuits       Gate-2006
Question 102 Explanation: 
q0N = Data, q1N = q0q22N = q1
Question 103
The addition of 4-bit, two’s complement, binary numbers 1101 and 0100 results in
A
0001 and an overflow
B
1001 and no overflow
C
0001 and no overflow
D
1001 and an overflow
       Digital-Logic-Design       Number-Systems       Gate 2006-IT
Question 103 Explanation: 
2's complement of 1101 = 0011
2's complement of 1100 = 1100
Add = 1111
Now convert 1111 to normal form.
⇒ 0000 (1's complement)
⇒ 0001 (2's complement) No carry bit.
Question 104
 
A
QRS
B
PQS
C
PQ'S'
D
Q'S'
       Digital-Logic-Design       K-Map       Gate 2006-IT
Question 104 Explanation: 
Essential prime implicants which are grouped only by only one method or way. So, in the given question cornor's ones are grouped by only one method.
Question 105
A
XOR, AND
B
XOR, XOR
C
OR, OR
D
OR, AND
       Digital-Logic-Design       Boolean-Functions       Gate 2006-IT
Question 105 Explanation: 

Thus we have OR and AND which gives different outputs on (0,0) and (1,1).
The encodes can be hence select from the two and decide output of the function according to x.
Question 106
 
A
2Y and Y
B
-2Y and 2Y
C
-2Y and 0
D
0 and Y
       Digital-Logic-Design       Number-Systems       Gate 2006-IT
Question 106 Explanation: 

⇒ -2Y and 0
Question 107
 
A
f is independent of X
B
f is independent of Y
C
f is independent of Z
D
None of X, Y, Z is redundant
       Digital-Logic-Design       Circuits-Output       Gate-2005
Question 107 Explanation: 
f(X,Y,Z) = ((XY’)’ (YZ))’
= ((X’+Y) YZ)’
= (X’YZ + YZ)’
= ((X’+1) YZ)’
= (YZ)’
Question 108
The range of integers that can be represented by an n bit 2's complement number system is:
A
- 2n-1 to (2n-1 - 1)
B
- (2n-1 - 1) to (2n-1 - 1)
C
- 2n-1 to 2n-1
D
- (2n-1 + 1) to (2n-1 - 1)
       Digital-Logic-Design       Number-Systems       Gate-2005
Question 108 Explanation: 
The maximum (positive) n bit number is 011….1 (i.e., 0 followed by n-1 ones) which is equal to 2n-1 - 1.
The smallest (negative) n bit number is 100..0 (i.e., 1 followed by n-1 zeros) which is equal to - 2n-1.
1000...00
0111...11 <- 1’s complement
1000..00 <- 2’s complement
= - 2n-1
Question 109
 
A
0D 24
B
0D 4D
C
4D 0D
D
4D 3D
       Digital-Logic-Design       Number-Systems       Gate-2005
Question 109 Explanation: 
Sign Bit = 0
Convert 0.239 to binary
0.239 * 2 = 0.478
0.478 * 2 = 0.956
0.956 * 2 = 1.912
0.912 * 2 = 1.824
0.824 * 2 = 1.648
0.648 * 2 = 1.296
0.296 * 2 = 0.512
0.512 * 2 = 1.024
Mantissa = (0. 00111101)2
Bias= 64. So biased exponent is 13+64 = 77= (1001101)2
0.239 × 213 = 0 1001101 00111101
= 0100 1101 0011 1101
= 4 D 3 D
Question 110
 
A
0A 20
B
11 34
C
4D D0
D
4A E8
       Digital-Logic-Design       Number-Systems       Gate-2005
Question 110 Explanation: 
Sign Bit = 0
Convert 0.239 to binary
0.239 * 2 = 0.478
0.478 * 2 = 0.956
0.956 * 2 = 1.912
0.912 * 2 = 1.824
0.824 * 2 = 1.648
0.648 * 2 = 1.296
0.296 * 2 = 0.512
0.512 * 2 = 1.024
Mantissa = (0. 00111101)2
0.239 × 213 = 1.11101000 x 210 <- Normalized Mantissa
Bias = 64. So biased exponent is 10+64 = 74 = (1001010)2
0.239 × 213 = 0 1001010 11101000
= 0100 1010 1110 1000
= (4 A E 8)16
Question 111
       
A
B
C
D
       Digital-Logic-Design       Sequential-Circuits       Gate-2005
Question 111 Explanation: 
Question 112
 
A
A0 A1 A1' A3 A4
B
A0 A1 A2' A3 A4
C
A1 A2 A2' A3 A4
D
A1 A2' A3 A4 A5'
       Digital-Logic-Design       Circuits-Output       Gate-2005
Question 112 Explanation: 

Input will be accepted by the flipflop after the cycle gets finished, because +ve edge is occuring at the end of the clock cycle only.
Question 113
The hexadecimal representation of 6578 is
A
1AF
B
D78
C
D71
D
32F
       Digital-Logic-Design       Number-Systems       Gate-2005
Question 113 Explanation: 
(657)8 = (110 101 111)2
Make 3 zeros on the left side so that the number of bits is multiple of 4.
= (0001 1010 1111)2
= (1 A F)16
Question 114
The switching expression corresponding to f(A, B, C, D) = Σ (1, 4, 5, 9, 11, 12) is:
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'
       Digital-Logic-Design       Minimum-Sum-Of-Product       Gate-2005
Question 114 Explanation: 

f(A,B,C,D) = A'C'D + BC'D' + AB'D
Question 115

A
B
C
D
       Digital-Logic-Design       Circuits-Output       Gate 2005-IT
Question 115 Explanation: 
The bulb will be on when both the switch S1 and S2 are in same state, either OFF (or) ON:

From this we can clearly know that given is EX-NOR operation i.e.,
(S1⊙S2) = (S1⊕S2)'
Question 116
How many pulses are needed to change the contents of a 8-bit up counter from 10101100 to 00100111 (rightmost bit is the LSB)?
A
134
B
133
C
124
D
123
       Digital-Logic-Design       Sequential-Circuits       Gate 2005-IT
Question 116 Explanation: 
The 8 bit counter will be 0-255 to move from 10101100 (172) to 1000111 (39).
→ First counter is move from 172 to 255 = 83 pulses
→ 255 to 0 = 1 pulse
→ 0 to 39 = 39 pulses
Total = 83 + 1 + 39 = 123 pulses
Question 117

A line L in a circuit is said to have a stuck-at-0 fault if the line permanently has a logic value 0. Similarly a line L in a circuit is said to have a stuck-at-1 fault if the line permanently has a logic value 1. A circuit is said to have a multiple stuck-at fault if one or more lines have stuck at faults. The total number of distinct multiple stuck-at faults possible in a circuit with N lines is

A
3N
B
3N - 1
C
2N - 1
D
2
       Digital-Logic-Design       Sequential-Circuits       Gate 2005-IT
Question 117 Explanation: 
Answer should be 3N-1.
This is because the total possible combinations (i.e., a line may either be at fault (in 2 ways i.e., stuck at 0 or 1) or it may not be, so there are only 3 possibilities for a line) is 3N. In only one combination the circuit will have all lines to be correct (i.e., not a fault). Hence, total combinations in which distinct multiple stuck-at-faults possible in a circuit with N lines is 3N - 1.
Question 118
(34.4)8 × (23.4)8 evaluates to
A
(1053.6)8
B
(1053.2)8
C
(1024.2)8
D
None of these
       Digital-Logic-Design       Number-Systems       Gate 2005-IT
Question 118 Explanation: 
First convert (34.4)8 and (23.4)8 to decimal.
(34.4)8 = 3×81 + 4×80 + 4×8-1
= 24 + 4 + 0.5
= (28.5)10
(23.4)8 = 2×81 + 3×80 + 4×8-1
= 16 + 3 + 0.5
= (19.5)10
Now,
(28.5)10 × (19.5)01 = (555.75)10
Now,
(555.75)10 = ( ? )8
To convert the integer part,

We get, 1053.
To convert the fractional part, keep multiplying by 8 till decimal part becomes 0,

∴ (555.75)10 = (1053.6)8
Question 119
   
A
1, 0, B
B
1, 0, A
C
0, 1, B
D
0, 1, A
       Digital-Logic-Design       Sequential-Circuits       Gate 2005-IT
Question 119 Explanation: 
In MUX1, the equation is
g = Ax + Bz'
In MUX2, the equation is
f = xg + yg'
= x(Az+Bz') + y(Az+Bz')'
Function f should be equal to (A+B)'.
Just try to put the values of option (D), i.e., x=0, y=1, z=A,
f = 0(AA+BA') +1(AA+BA')'
= (A+B)'
∴ Option (D) is correct.
Question 120
A
0
B
103
C
22
D
55
       Digital-Logic-Design       CPU-Control-Design-and-Interfaces       Gate 2005-IT
Question 120 Explanation: 
In horizontal microprogramming we need 1 bit for every control word, therefore total bits in horizontal microprogramming
= 20 + 70 + 2 + 10 + 23
= 125
Now lets consider vertical microprogramming. In vertical microprogramming no. of bits required to activate 1 signal in group of N signals, is ⌈log2 N⌉. And in the question 5 groups contains mutually exclusive signals,
group 1 = ⌈log2 20⌉ = 5
group 2 = ⌈log2 70⌉ = 7
group 3 = ⌈log2 2⌉ = 1
group 4 = ⌈log2 10⌉ = 4
group 5 = ⌈log2 23⌉ = 5
Total bits required in vertical microprogramming
= 5 + 7 + 1 + 4 + 5
= 22
So, number of bits saved is
= 125 - 22
= 103
Question 121
The Boolean function x'y' + xy + x'y is equivalent to
A
x' + y'
B
x + y
C
x + y'
D
x' + y
       Digital-Logic-Design       Boolean-Algebra       Gate-2004
Question 121 Explanation: 
x'y' + xy + x'y
= x'y' + x'y + xy
= x'(y'+y)+xy
= x'⋅1+xy
= x'+xy
= (x'+x)(x'+y)
= 1⋅(x'+y)
= x'+y
Question 122
In an SR latch made by cross-coupling two NAND gates, if both S and R inputs are set to 0, then it will result in
A
Q = 0, Q' = 1
B
Q = 1, Q' = 0
C
Q = 1, Q' = 1
D
Indeterminate states
       Digital-Logic-Design       Sequential-Circuits       Gate-2004
Question 122 Explanation: 

Truth table for the SR latch by cross coupling two NAND gates is

So, Answer is Option (D).
Question 123
If 73x (in base-x number system) is equal to 54y (in base-y number system), the possible values of x and y are
A
8, 16
B
10, 12
C
9, 13
D
8, 11
       Digital-Logic-Design       Number-Systems       Gate-2004
Question 123 Explanation: 
(73)x = (54)y
7x+3 = 5y+4
7x-5y = 1
Only option (D) satisfies above equation.
Question 124
 
A
9.51 and 10.0 respectively
B
10.0 and 9.51 respectively
C
9.51 and 9.51 respectively
D
10.0 and 10.0 respectively
       Digital-Logic-Design       Number-Systems       Gate-2004
Question 124 Explanation: 
(113. + -111.) + 7.51
= (2) + 7.51
= 9.51 (✔️)
113. + (-111. + 7.51)
= 113. + (-103.51)
= 113. + -103
= 10 (✔️)
Question 125
A circuit outputs a digit in the form of 4 bits. is represented by 0000, 1 by 0001, ..., 9 by 1001. A combinational circuit is to be designed which takes these 4 bits as input and outputs 1 if the digit ≥ 5, and 0 otherwise. If only AND, OR and NOT gates may be used, what is the minimum number of gates required?
A
2
B
3
C
4
D
5
       Digital-Logic-Design       Combinational-Circuits       Gate-2004
Question 125 Explanation: 

= A + BD + BC
= A + B (D + C)
So minimum two OR gates and 1 AND gate is required. Hence, in total minimum 3 gates is required.
Question 126
 
A
aȼc and acȼ
B
aȼc and bȼc
C
aȼc only
D
acȼ and bcȼ
       Digital-Logic-Design       K-Map       Gate-2004
Question 126 Explanation: 
From given function 'f' we can draw,

There are two EPI,
A'C and AC'.
Question 127
Consider a multiplexer with X and Y as data inputs and Z as control input. Z = 0 selects input X, and Z = 1 selects input Y. What are the connections required to realize the 2-variable Boolean function f = T + R, without using any additional hardware?  
A
R to X, 1 to Y, T to Z
B
T to X, R to Y, T to Z
C
T to X, R to Y, 0 to Z
D
R to X, 0 to Y, T to Z
       Digital-Logic-Design       Multiplexer       Gate-2004
Question 127 Explanation: 
Given,

f = z'x + zy
Put z=T, x=R, y=1 in f
f = T'R + T = (T+T') (R+T) = T+R
Hence, correct option is (A).
Question 128
 
A
Q2c
B
Q2 + Q1
C
(Q1 + Q2)c
D
Q1 ⊕ Q2
       Digital-Logic-Design       Sequential-Circuits       Gate-2004
Question 128 Explanation: 
Sequence given is
0 - 2 - 3 - 1 - 0
or
00 - 10 - 11 - 01 - 00
From the given sequence, we have state table as,

Now we have present state and next state, so we should use excitation table of T flip-flop,

From state table,
Question 129

A 4-bit carry lookahead adder, which adds two 4-bit numbers, is designed using AND, OR, NOT, NAND, NOR gates only. Assuming that all the inputs are available in both complemented and uncomplemented forms and the delay of each gate is one time unit, what is the overall propagation delay of the adder? Assume that the carry network has been implemented using two-level AND-OR logic.

A
4 time units
B
6 time units
C
10 time units
D
12 time units
       Digital-Logic-Design       Carry-Look-Ahead-Adder       Gate-2004
Question 129 Explanation: 
The 4-bit addition will be calculated in 3 stages:
1) (2 time units) In 2 time units we can compute Gi and Pi in parallel, 2 time units for Pi since its an XOR operation and 1 time unit for Gi sinceits an AND operation.
2) (2 time units) Once Gi and Pi are available, we can calculate the carries, Ci, in 2 time units.
Level-1 we can compute all the conjunctions (AND). Example P3G2, P3P2G1, P3P2P1G0 and P3P2P1P0C0 which are required for C4.
Level-2 we get the carries by computing the disjunction (OR).
3) (2 time units) Finally, we compute the sum in 2 time units, as its a XOR operation.
Hence, the total is 2+2+2=6 time units.
Question 130
Let A = 1111 1010 and B = 0000 1010 be two 8-bit 2's complement numbers. Their product in 2's complement is
A
1100 0100
B
1001 1100
C
1010 0101
D
1101 0101
       Digital-Logic-Design       Number-Systems       Gate-2004
Question 130 Explanation: 
A = 1111 1010 = -610 [2's complement number]
B = 0000 1010 = 1010 [2's complement number]
A×B = -6×10 = - 6010
⇒ -6010 = 101111002
= 110000112 (1's complement)
= 110001002 (2's complement)
Question 131
What is the minimum number of NAND gates required to implement a 2-input EXCLUSIVE-OR function without using any other logic gate?
A
3
B
4
C
5
D
6
       Digital-Logic-Design       Logic-Gates       Gate 2004-IT
Question 131 Explanation: 

To create 2-input Exclusive-OR function we require 4 NAND gates.
Question 132
What is the minimum size of ROM required to store the complete truth table of an 8-bit × 8-bit multiplier?
A
32 K × 16 bits
B
64 K × 16 bits
C
16 K × 32 bits
D
64 K × 32 bits
       Digital-Logic-Design       Multiplexer       Gate 2004-IT
Question 132 Explanation: 
Input: 2 lines, 8 bits each
Possible combination in ROM = (28 × (28) [size of truth table]
= 216
= 64 KB
= 64 K ×16 bits
Question 133
 
A
1 only
B
2 only
C
3 only
D
1 and 3 only
       Digital-Logic-Design       Number-Representation       Gate 2004-IT
Question 133 Explanation: 
In 2's complement arithmetic, overflow happens only when
1) Sign bit of two input numbers is 0, and the result has sign bit 1.
2) Sign bit of two input numbers is 1, and the result has sign bit 0.
So, only (2) causes overflow.
Question 134
The number (123456)8 is equivalent to
A
(A72E)16 and (22130232)4
B
(A72E)16 and (22131122)4
C
(A73E)16 and (22130232)4
D
(A62E)16 and (22120232)4
       Digital-Logic-Design       Number-Systems       Gate 2004-IT
Question 134 Explanation: 
(123456)8 = (001 010 011 100 101 110)2
= (00 1010 0111 0010 1110)2
= (A72E)16
Also,
(001 010 011 100 101 110)2
= (00 10 10 01 11 00 10 11 10)2
= (22130232)4
Question 135
The function AB’C + A’BC + ABC’ + A’B’C + AB’C’ is equivalent to
A
AC’+ AB + A’C
B
AB’+ AC’+ A’C
C
A’B+ AC’+ AB’
D
A’B + AC + AB’
       Digital-Logic-Design       Minimization       Gate 2004-IT
Question 135 Explanation: 
For given min term the K-map is,

⇒ A'C + AC' + AB'
Question 136
A
I and III
B
I, II and III
C
II and IV
D
I, II, III and IV
       Digital-Logic-Design       Number-Systems       Gate 2004-IT
Question 136 Explanation: 
Let x1, x2, x3 are data bits, and c1, c2, c3 and c4 are parity check bits.
Given transmitted codewords are

By inspection we can find the rule for generating each of the parity bits,

Now from above we can see that (I) and (III) are only codewords.
Question 137
Assuming all numbers are in 2’s complement representation, which of the following numbers is divisible by 11111011?  
A
11100111
B
11100100
C
11010111
D
11011011
       Digital-Logic-Design       Number-Systems       Gate-2003
Question 137 Explanation: 
Given: Binary numbers = 11111011
MSB bit is '1' then all numbers are negative
1's complement = 00000100
2's complement = 00000100 + 00000001 = 00000101 = -5
(A) 11100111 - (-25)10
(B) 11100100 - (-28)10
(C) 11010111 - (-41)10
(D) 11011011 - (-37)10
Answer: Option A (-25 is divisible by -5)
Question 138
Consider an array multiplier for multiplying two n bit numbers. If each gate in the circuit has a unit delay, the total delay of the multiplier is  
A
Θ (1)
B
Θ (log n)
C
Θ (n)
D
Θ (n2)
       Digital-Logic-Design       Multiplexer       Gate-2003
Question 138 Explanation: 
Each bit in Multiplier is ANDed with a bit in Multiplicand which produce n n-bit numbers. The multiplication takes n units of time. The n n-bit numbers are added by using (n-1) n-bit adders. The time taken by (n-1) n-bit adders is k*(n-1) units.
The total time is n+kn-k = Θ(n)
Question 139
 
A
2-40
B
2-9
C
222
D
231
       Digital-Logic-Design       Number-Systems       Gate-2003
Question 139 Explanation: 
Largest gap will be in between two most largest numbers.
The largest number is 1.111111111× 262-31 = (2−2−9)×231
Second largest number is 1.111111110×262-31 = (2−2-8)×231
Difference = (2−2−9)×231 - (2−2-8)×231
= (2-8−2−9) ×231
= 2−9×231
= 222
Question 140

A
5
B
6
C
7
D
8
       Digital-Logic-Design       Sequential-Circuits       Gate-2003
Question 140 Explanation: 
Let q is the initial state.

q0 ← Number of zeros is one more than number of ones.
q1 ← Number of ones is one more than number of zeros.
q00 ← Number of zeros is two more than number of ones.
q11 ← Number of ones is two more than number of zeros.
Question 141
   
A
(11, 9)
B
(9, 13)
C
(9, 10)
D
(11, 11)
       Digital-Logic-Design       K-Map       Gate-2003
Question 141 Explanation: 
For SOP,

⇒ w'y' + z'wx' + xyz'
Total 8 literals are there.
For POS,

⇒ (z' + w')(z' + y')(w' + x')(x + z + w)
Total 9 literals are there.
Question 142

A
A + B, and A – B, but not A + 1
B
A + B, and A + 1, but not A – B
C
A + B, but not A – B or A + 1
D
A + B, and A – B, and A + 1
       Digital-Logic-Design       Adder       Gate-2003
Question 142 Explanation: 
The circuits performs
1) A+B when K=0 and C0 = 0. It is binary adder which performs addition of two binary numbers.
2) A - B = A+ B' + 1 when K=1 and C0 = 1 ;
Here XOR gates produce B' if K=1. Since 1⊕b= b'.
"1" in (A+B+1) is coming from C0.
Note: 2's complement of B is (B'+1). 3) A+1 when B=0, K=0, C0= 1.
Increments A.
Question 143
 
A
1
B
2
C
3
D
4
       Digital-Logic-Design       Logic-Gates       Gate-2003
Question 143 Explanation: 

⇒ a will always be equal to A.
Question 144
 
A
xz+y'z
B
xz'+zx'
C
x'y+zx'
D
None of the above
       Digital-Logic-Design       K-Map       Gate-2002
Question 144 Explanation: 

⇒ xz' + zx'
Question 145
The decimal value 0.25
A
is equivalent to the binary value 0.1
B
is equivalent to the binary value 0.01
C
is equivalent to the binary value 0.00111…
D
cannot be represented precisely in binary
       Digital-Logic-Design       Number-Systems       Gate-2002
Question 145 Explanation: 
1st Multiplication iteration:
Multiply 0.25 by 2.
0.25×2 = 0.50 (product)
Fractional part = 0.50
Carry = 0
2nd Multiplication iteration:
Multiply 0.50 by 2.
0.50×2 = 1.00 (product)
Fractional part = 0.00
Carry = 1
The fractional part in the 2nd iteration becomes zero and so we stop the multiplication iteration.
Carry from 1st multiplication iteration becomes MSB and carry from 2nd iteration becomes LSB. So the result is 0.01.
Question 146
The 2’s complement representation of the decimal value -15 is
A
1111
B
11111
C
111111
D
10001
       Digital-Logic-Design       Number-Systems       Gate-2002
Question 146 Explanation: 
15 = 1111
-15 = 11111
1's complement = 10000
2's complement = 10001
Question 147
Sign extension is a step in
A
floating point multiplication
B
signed 16 bit integer addition
C
arithmetic left shift
D
converting a signed integer from one size to another
       Digital-Logic-Design       Number-Systems       Gate-2002
Question 147 Explanation: 
Sign extension is a step in converting a signed integer from on size to another.
Question 148
In 2’s complement addition, overflow
A
is flagged whenever there is carry from sign bit addition
B
cannot occur when a positive value is added to a negative value
C
is flagged when the carries from sign bit and previous bit match
D
None of the above
       Digital-Logic-Design       Number-Systems       Gate-2002
Question 148 Explanation: 
The left most bit of positive value is zero. And left most bit for negative value is one. The value of 0+1 becomes 1. Then overflow never occurs.
Question 149
 
A
Σ(1,4,5)
B
Σ(6,7)
C
Σ(0,1,3,5)
D
None of the above
       Digital-Logic-Design       Logic-Circuit       Gate-2002
Question 149 Explanation: 
f(x,y,z) = (f1', (x,y,z) ⋅ f2'(x,y,z) + f3'(x,y,z))
= (Σ(0,1,3,5) ⋅ Σ(6,7) + Σ(1,4,5))
[Σ(0,1,3,5) and Σ(6,7) ⇒ No common terms]
= (Σ(1,4,5))
Question 150
   
A
xyz'
B
xy+z
C
x+y
D
None of the above
       Digital-Logic-Design       Multiplexer       Gate-2002
Question 150 Explanation: 
F = (A'A0'10 + A'A0'11 + A'A0'12 + A1A013) EN
F = (xyz' + xyz + y'zy + zy')z'
= (xyz' + xyz + y'z(y+1))z'
= (xyz' + xyz + y'z)z'
= (xy(z+z') + y'z)z'
= (xy + y'z)z'
= (xyz' + y'zz')
= (xyz')
Question 151
Let f(A,B) = A'+B. Simplified expression for function f (f (x + y, y), z) is
A
x'+z
B
xyz
C
xy'+z
D
None of the above
       Digital-Logic-Design       Boolean-Expressions       Gate-2002
Question 151 Explanation: 
f(A,B) = A' +B
⇒ f(f((x+y), y), z)
⇒ f(((x+y)' + y), z)
⇒ f(((x'⋅y') + y), z)
⇒ f((x'⋅y') + y), z)
⇒ ((x'⋅y') + y)' + z
⇒ (x'⋅y')⋅y' + z
⇒ (x+y)⋅y' + z
⇒ (xy'+yy') + z
⇒ xy' + z
Question 152
   
A
Theory Explanation is given below.
       Digital-Logic-Design       Boolean-Expression-and-Logic-Gates       Gate-2002
Question 152 Explanation: 
A
Theory Explanation is given below.
       Digital-Logic-Design       Clock-Pulse       Gate-2002
Question 154
 
A
Theory of Explantion is given below.
       Digital-Logic-Design       Number-Systems       Gate-2002
Question 155
The number 43 in 2’s complement representation is
A
01010101
B
11010101
C
00101011
D
10101011
       Digital-Logic-Design       Number-Systems       Gate-2000
Question 155 Explanation: 
Positive integers are represented in its normal binary form while negative numbers are represented in its 2′s complement form. Binary representation of 43 is 00101011.
Question 156
       
A
0 1 0 0
B
1 1 0 1
C
1 0 1 1
D
1 0 0 0
       Digital-Logic-Design       Boolean-Algebra       Gate-2000
Question 156 Explanation: 
Just put the values of each options in the equation and check it.
Question 157
   
A
B
C
D
None of the above
       Digital-Logic-Design       K-Map       Gate-2000
Question 157 Explanation: 
Given k-map gives xy + xy + wz

⇒ wy + wz + xy
Question 158
 
A
1, 0
B
1, 1
C
0, 0
D
0, 1
       Digital-Logic-Design       Sequential-Circuits       Gate-2000
Question 158 Explanation: 
Here clocks are applied to both flip flops simultaneously.
When 11 is applied to Jk flip flop it toggles the value of P so op at P will be 1.
Input to D flip flop will be 0(initial value of P) so op at Q will be 0
Question 159
 
A
X = 1.0, Y = 1.0
B
X = 1.0, Y = 0.0
C
X = 0.0, Y = 1.0
D
X = 0.0, Y = 0.0
       Digital-Logic-Design       Number-Systems       Gate-2000
Question 159 Explanation: 
Given: 32 bits representation. So, the maximum precision can be 32 bits (In 32-bit IEEE representation, maximum precision is 24 bits but we take best case here). This means approximately 10 digits.
A = 2.0 * 1030, C = 1.0
So, A + C should make the 31st digit to 1, which is surely outside the precision level of A (it is 31st digit and not 31st bit). So, this addition will just return the value of A which will be assigned to Y.
So, Y + B will return 0.0 while X + C will return 1.0.
Question 160
 
A
Theory Explanation is given below.
       Digital-Logic-Design       Descriptive       Gate-2000
Question 161
   
A
x NAND X
B
x NOR x
C
x NAND 1
D
x NOR 1
       Digital-Logic-Design       Logic-Gates       Gate-1999
Question 161 Explanation: 
Question 162
 
A
B
C
D
       Digital-Logic-Design       K-Map       Gate-1999
Question 162 Explanation: 

⇒ CD+AD = D(A+C)
Question 163
Booth’s coding in 8 bits for the decimal number –57 is
A
0 – 100 + 1000
B
0 – 100 + 100 -1
C
0 – 1 + 100 – 10 + 1
D
00 – 10 + 100 - 1
       Digital-Logic-Design       Number-Systems       Gate-1999
Question 163 Explanation: 
Option-B:
Question 164
The maximum gate delay for any output to appear in an array multiplier for multiplying two n bit number is
A
On2
B
O(n)
C
O(log n)
D
O(1)
       Digital-Logic-Design       Multiplexer       Gate-1999
Question 164 Explanation: 
Total no. of gates being used for 'n' bit multiplication in an array multiplier (n*n) = (2n-1)
Total delay = 1 * 2n - 1 = O(2n - 1) = n
Question 165
The number of full and half-adders required to add 16-bit numbers is
A
8 half-adders, 8 full-adders
B
1 half-adder, 15 full-adders
C
16 half-adders, 0 full-adders
D
4 half-adders, 12 full-adders
       Digital-Logic-Design       Adder       Gate-1999
Question 165 Explanation: 
For Least Significant Bit we do not need a full adder since initially carry is not present.
But for rest of bits we need full address since carry from previous addition has to be included into the addition operation.
So, in total 1 half adder and 15 full adders are required.
Question 166
Zero has two representations in
A
Sign magnitude
B
1’s complement
C
2’s complement
D
None of the above
E
Both A and B
       Digital-Logic-Design       Number-Systems       Gate-1999
Question 166 Explanation: 
Sign magnitude:
+0 = 0000
-0 = 1000
1's complement:
+0 = 0000
-0 = 1111
Question 167
 
A
complements when n is even
B
complements when n is odd
C
divides by 2n always
D
remains unchanged when n is even
       Digital-Logic-Design       EX-OR       Gate-1998
Question 167 Explanation: 
B⊕(B⊕(B⊕...) n times
Consider:
B⊕(B⊕B)
= B⊕0
= 0 (if consider n times it remains unchanged)
Question 168
A multiplexor with a 4 bit data select input is a
A
4:1 multiplexor
B
2:1 multiplexor
C
16:1 multiplexor
D
8:1 multiplexor
       Digital-Logic-Design       Multiplexer       Gate-1998
Question 168 Explanation: 
For 'n' bit data it selects 2n : 1 input
For 4 bit data it selects 24 : 1 = 16: 1 input
Question 169
The threshold level for logic 1 in the TTL family is
A
any voltage above 2.5 V
B
any voltage between 0.8 V and 5.0 V
C
any voltage below 5.0 V
D
any voltage below Vcc but above 2.8 V
       Digital-Logic-Design       Threshold-Voltage       Gate-1998
Question 169 Explanation: 
Voltage is to be below Vcc = 5V but above 2.8V
Question 170
The octal representation of an integer is (342)8. If this were to be treated as an eight-bit integer is an 8085 based computer, its decimal equivalent is
A
226
B
-98
C
76
D
-30
       Digital-Logic-Design       Number-Systems       Gate-1998
Question 170 Explanation: 
(342)8 = (011 100 010)2 = (1110 0010)2
If this can be treated as 8 bit integer, then the first becomes sign bit i.e., '1' then the number is negative.
8085 uses 2's complement then

⇒ -30
Question 171
   
A
A⋅B
B
AB+BC+CA
C
D
None of the above
       Digital-Logic-Design       K-Map       Gate-1998
Question 171 Explanation: 
Question 172
Which of the following operations is commutative but not associative?
A
AND
B
OR
C
NAND
D
EXOR
       Digital-Logic-Design       Logic-Gates       Gate-1998
Question 172 Explanation: 
NAND operation is commutative but not associative.
Question 173

Suppose the domain set of an attribute consists of signed four digit numbers. What is the percentage of reduction in storage space of this attribute if  it is stored as an integer rather than in character form?

A
80%
B
20%
C
60%
D
40%
       Digital-Logic-Design       Number-Systems       Gate-1998
Question 173 Explanation: 
We assume byte addressable memory - nothing smaller than a byte can be used.
We have four digits. So to represent signed 4 digit numbers we need 5 bytes, 4 bytes for four digits and 1 for the sign.
So required memory = 5 bytes.
Now, if we use integer, the largest no. needed to represent is 9999 and this requires 2 bytes of memory for signed representation.
9999 in binary requires 14 bits. So, 2 bits remaining and 1 we can use for sign bit.
So, memory savings,
= 5 - 2/5 × 100
= 60%
Question 174
   
A
B
x
C
0
D
1
       Digital-Logic-Design       Boolean-Expressions       Gate-1997
Question 174 Explanation: 
Question 175
 
A
proportional to N
B
proportional to log N
C
a constant
D
None of the above
       Digital-Logic-Design       Adder       Gate-1997
Question 176
 
A
B
xz is a minterm of f
C
xz is an implicant of f
D
y is a prime implicant of f
       Digital-Logic-Design       K-Map       Gate-1997
Question 176 Explanation: 
In sum of terms,any term is an implicant because it implies the function. So xz is an implicant and hence 'C' is the answer.
Question 177
   
A
10
B
8
C
5
D
6
       Digital-Logic-Design       Number-Systems       Gate-1997
Question 177 Explanation: 
Question 178
 
A
Σ9,10
B
Σ9
C
Σ1,8,9
D
Σ8,10,15
       Digital-Logic-Design       Logic-Circuit       Gate-1997
Question 178 Explanation: 
f = f1⋅f2 + f3
Since, f1 and f2 are in canonical sum of products form, f1⋅f2 will only contain their common terms that is f1⋅f2 = Σ8.
Now,
Σ8 + f3 = Σ8,9
So, f3= Σ9
Question 179
A ROM is sued to store the table for multiplication of two 8-bit unsigned integers. The size of ROM required is
A
256 × 16
B
64 K × 8
C
4 K × 16
D
64 K × 16
       Digital-Logic-Design       ROM       Gate-1996
Question 179 Explanation: 
When we multiply the two 8 bit numbers result will reach upto 16 bits. So we require 16 bits for each multiplication output.
No. of results possibe = 28 × 28 = 216 = 64K
Then total size of ROM = 64K × 16
Question 180
Both’s algorithm for integer multiplication gives worst performance when the multiplier pattern is
A
101010 …..1010
B
100000 …..0001
C
111111 …..1111
D
011111 …..1110
       Digital-Logic-Design       Booth's-Algorithm       Gate-1996
Question 180 Explanation: 
When the pairs 01 (or) 10 occur frequently in the multiplier. In that case Booth multiplication gives worst performance.
Question 181
 
A
0 to 1
B
0.5 to 1
C
2-23 to 0.5
D
0.5 to (1-2-23)
       Digital-Logic-Design       Number-Systems       Gate-1996
Question 181 Explanation: 
Maximum value of mantissa will be 23, is where a decimal point is assumed before first 1. So the value is 1 - 2-23.
Question 182
 
A
Binary of Hex conversion
B
Binary to BCD conversion
C
Binary to grey code conversion
D
Binary to radix-12 conversion
       Digital-Logic-Design       Adder       Gate-1996
Question 182 Explanation: 
Here ф means 0.
Whenever, b2 = b3 = 1, then only 0100, i.e., 4 is added to the given binary number. Lets write all possibilities for b.

Note that the last 4 combinations leads to b3 and b2 as 1. So, in these combinations only 0010 will be added.
1100 is 12
1101 is 13
1110 is 14
1111 is 15
in binary unsigned number system.
1100 + 0100 = 10000
1101 + 0100 = 10001, and so on.
This is conversion to radix 12.
Question 183
 
A
B
C
D
       Digital-Logic-Design       Multiplexer       Gate-1996
Question 183 Explanation: 
Question 184
   
A
B
C
D
E
None of the above
       Digital-Logic-Design       K-Map       Gate-1996
Question 184 Explanation: 
Correct option is

Question 185
   
A
A = 1, B = 0, C = 0, D = 1
B
A = 1, B = 1, C = 0, D = 0
C
A = 1, B = 0, C = 1, D = 1
D
A = 1, B = 0, C = 0, D = 0
       Digital-Logic-Design       Boolean-Algebra       Gate-1995
Question 185 Explanation: 
For verification, just put up the values and check for AND, OR operations and their outputs.
Question 186
 
A
8
B
9
C
10
D
12
       Digital-Logic-Design       Number-Systems       Gate-1995
Question 186 Explanation: 
Question 187
 
A
B
C
D
E
None of the above.
       Digital-Logic-Design       Logic-Gates       Gate-1994
Question 187 Explanation: 
Question 188
The number of flip-flops required to construct a binary modulo N counter is __________
A
⌈log2 N⌉
       Digital-Logic-Design       Flip-Flops       Gate-1994
Question 188 Explanation: 
For mod-N counter we need ⌈log2 N⌉ flip flops.
Question 189

Consider n-bit (including sign bit) 2’s complement representation of integer number. The range of integer values, N, that can be represented is _________ ≤ N ≤ _________

A
-2n-1 to 2n-1 - 1
       Digital-Logic-Design       Number-Systems       Gate-1994
Question 190
 
A
exclusive OR
B
exclusive NOR
C
NAND
D
NOR
E
None of the above
       Digital-Logic-Design       Logic-Gates       Gate-1993
Question 190 Explanation: 

So finally, we can write
Question 191
 
A
Shift Register
B
Mod-3 Counter
C
Mod-6 Counter
D
Mod-2 Counter
E
Both A and C
       Digital-Logic-Design       J-K-Flip-Flop       Gate-1993
Question 191 Explanation: 

Circuit behaves as shift register and mod-6 counter. Note that this is the Johnson counter which is the application of shift register. And Johnson counter is mod-2N counter.
Question 192
 
A
(a) 6.625, (b) (45E)H
       Digital-Logic-Design       Number-Systems       Gate-1993
Question 192 Explanation: 
(a) 1*22 + 1*21 + 0*20 + 1*2-1 + 0*2-2 + 1*2-3
= 4 + 2 + 0 + 0.5 + 0 + 0.125
= 6.625
(b) 1118 mod 16 = E, quotient = 69
69 mod 16 = 5, quotient = 4
4 mod 16 = 4
Writing the mods result in reverse order gives (45E)H.
Question 193

A ROM is used to store the Truth table for a binary multiple unit that will multiply two 4-bit numbers. The size of the ROM (number of words × number of  bits) that is required to accommodate the Truth table is M words × N bits. Write the values of M and N.

A
M=256, N = 8
       Digital-Logic-Design       Truth Table       Gate-1993
Question 193 Explanation: 
Input will consist of 8 bit (two 4-bit numbers) = 28 address.
Output will be of 8 bits.
So memory will be of 28 × 8.
So, M = 256, N = 8.
Question 194
 
A
ABC + B'C' + A'C'
       Digital-Logic-Design       K-Map       Gate-1992
Question 194 Explanation: 
We can write this as

⇒ ABC + B'C' + A'C'
Question 195
Consider a 3-bit error detection and 1-bit error correction hamming code for 4-bit date. The extra parity bits required would be ________ and the 3-bit error detection is possible because the code has a minimum distance of ________
A
Fill in the blanks
       Digital-Logic-Design       Parity-Bits       Gate-1992
Question 196
The operation which is commutative but not associative is:  
A
AND
B
OR
C
EX-OR
D
NAND
       Digital-Logic-Design       Operators       Gate-1992
Question 196 Explanation: 
NAND and NOR operation follow commutativity but do not follow associativity.
Question 197
All digital circuits can be realized using only
A
Ex-OR gates
B
Multiplexers
C
Half adders
D
OR gates
E
Both B and C
       Digital-Logic-Design       Operators       Gate-1992
Question 197 Explanation: 
NOR gate, NAND gate, Multiplexers and Half adders can also be used to realize all digital circuits.
Question 198
 
A
xy+y'z
B
wx'y'+xy+xz
C
w'x+y'z+xy
D
xz+y
       Digital-Logic-Design       K-Map       Gate-2001
Question 198 Explanation: 

⇒ y'z + xy
Question 199

A
B
C
D
       Digital-Logic-Design       Sequential-Circuits       Gate-2001
Question 199 Explanation: 

Given clock is +edge triggered.
See the first positive edge. X is 0, and hence the output is 0, because
Y = Q1N = D1×Q0' = 0⋅Q0' = 0
At second +edge, X is 1 and Q0' is also 1. So output is 1 (when second +ve edge of the clock arrives, Q0' would surely be 1 because the setup time of flip flop is given as 20ns and clock period is ≥ 40ns).
At third +ve edge, X is 1 and Q0' is 0, so output is 0.
Now output never changes back to 1 as Q0' is always 0 and when Q0' finally becomes 1, X is 0.
Hence option (A) is the correct answer.
Question 200
The 2’s complement representation of (-539)10 in hexadecimal is
A
ABE
B
DBC
C
DE5
D
9E7
       Digital-Logic-Design       Number-Systems       Gate-2001
Question 200 Explanation: 
(539)10 = (0010 0001 1011)2
For (-539)10 = (1101 1110 0100)2
1's complement = (1101 1110 0100)2
2's complement = (1101 1110 0101)2
= (DE5)16
Question 201
 
A
f = x1' + x2
B
f = x1'x2 + x1x2'
C
f = x1x2 + x1'x2'
D
f = x1 + x2'
       Digital-Logic-Design       Number-Systems       Gate-2001
Question 201 Explanation: 
g = (a and x1′) or (b and x1)
g = (1 and x1’) or (0 and x1)
g = x1’
f = ac’ + bc
f = (a and x2′) or (b and x2)
f = (g and x2′) or (x1 and x2)
f = x1’x2’ + x1x2
Question 202
 
A
1,3,4,6,7,5,2
B
1,2,5,3,7,6,4
C
1,2,7,3,5,6,4
D
1,6,5,7,2,3,4
       Digital-Logic-Design       Sequential-Circuits       Gate-2001
Question 202 Explanation: 
Question 203
 
A
Theory Explanation is given below.
       Digital-Logic-Design       Descriptive       Gate-2001
Question 203 Explanation: 
(a)
The function is self dual because
→ There is no mutually exclusive pair.
→ No. of minterms = No. of maxterms
(b)
Write Minimal POS.
Question 204
 
A
Theory Explanation is given below.
       Digital-Logic-Design       Sequential-Circuits       Gate-2001
Question 204 Explanation: 

Question 205
 
A
Out of syllabus.
       Digital-Logic-Design       Circuits       Gate-1991
Question 206
   
A
9
       Digital-Logic-Design       Number-Systems       Gate-1991
Question 206 Explanation: 
Hexadecimal representation of a given no. is,
(9753)16
It's binary representation is,
1001011101010011
∴ The no. of 1's is 9.
Question 207
When two 4-bit binary number A = a3a2a1a0 and B = b3b2b1b0 are multiplied, the digit c1 of the product C is given by _________
A
c1 = b1a0 ⊕ a1b0
       Digital-Logic-Design       Number-Systems       Gate-1991
Question 207 Explanation: 

⇒ c1 = b1a0 ⊕ a1b0
Question 208
A
faster operation
B
ease of avoiding problems due to hazards
C
lower hardware requirement
D
better noise immunity
E
none of the above
       Digital-Logic-Design       Sequential-Circuits       Gate-1991
Question 208 Explanation: 
In synchronization, there is a less chance of hazards but it can increase the delay. Then the advantage is ease of avoiding problems due to hazards.
Question 209
 
A
Y = ABC + DE
B
Y = ABC + DE
C
Y = ABC . DE
D
Y = ABC . DE
E
Both (C) and (D)
       Digital-Logic-Design       Circuits-Output       Gate-1990
Question 209 Explanation: 
There should be bubbled connection between two gates
Y = ((ABC)' + (DE)')'
Y = ABC . DE
Note: Open gate works as NOR gate.
Question 210
 
A
True
B
False
       Digital-Logic-Design       Combinational-and-Sequential-Circuits       Gate-1990
Question 210 Explanation: 
1) RAM is not a combinational circuit. For RAM, the input is the memory location selector and operation (read or write) and another byte (which can be input for write operation or output for read operation), and the output is either a success indicator (for write operation) or the byte at the selected location (for read operation). It does depend on past inputs, or rather, on the past write operations at the selected byte. This is a sequential logic circuit.
2) PLA is a combination circuit as ROM. PLA is a programmable AND array and a programmable OR array. A PLA with n inputs has fewer than 2n AND gates (otherwise there would be no advantage over a ROM implementation of the same size). A PLA only needs to have enough AND gates to decode as many unique terms as there are in the functions it will implement it.
Question 211
The total number of Boolean functions which can be realised with four variables is:
A
4
B
17
C
256
D
65,536
       Digital-Logic-Design       Boolean-Functions       GATE-1987
Question 211 Explanation: 
Total no. of Boolean functions which can be realized with four variables is:
Question 212
 
A
1111 1111 0000 0000
B
1111 0000 1111 000
C
1111 0001 0011 010
D
1010 1010 1010 1010
       Digital-Logic-Design       Sequential-Circuits       GATE-1987
Question 212 Explanation: 
Let us suppose initially output of all JK flip flop is 1.
So we can draw below table to get the output Q3.

From the above table Q3 that is output is 1111 0001 0011 010.
So, answer is (C).
Question 213
       
A
B
A⊕B⊕C
C
A⊕B
D
       Digital-Logic-Design       Multiplexer       GATE-1987
Question 213 Explanation: 
Question 214
The exponent of a floating-point number is represented in excess-N code so that:
A
The dynamic range is large.
B
The precision is high.
C
The smallest number is represented by all zeros.
D
Overflow is avoided.
       Digital-Logic-Design       Number-Systems       GATE-1987
Question 214 Explanation: 
To avoid extra work, excess-N code is used so that all exponent can be represented in positive numbers, starting with 0.
Question 215
The refreshing rate of dynamic RAMs is in the range of
A
2 microseconds
B
2 milliseconds
C
50 milliseconds
D
500 milliseconds
       Digital-Logic-Design       DRAM       GATE-1987
Question 215 Explanation: 
During a 2 millisecond interval all dynamic RAM memory is refreshed.
There are 215 questions to complete.