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Digital-Logic-Design

Question 1

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
Question 1 Explanation: 

To create 2-input Exclusive-OR function we require 4 NAND gates.
Question 2

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
Question 2 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 3

Using a 4-bit 2’s complement arithmetic, which of the following additions will result in an overflow?

1. 1100 
  +1100 
2. 0011 
  +0111 
3. 1111 
  +0111 
A
1 only
B
2 only
C
3 only
D
1 and 3 only
Question 3 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 4

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
Question 4 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 5

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’
Question 5 Explanation: 
For given min term the K-map is,

⇒ A'C + AC' + AB'
Question 6

Consider a parity check code with three data bits and four parity check bits. Three of the code words are 0101011, 1001101 and 1110001. Which of the following are also code words?

1. 0010111             
2. 0110110         
3. 1011010             
4. 0111010 
A
1 and 3
B
1, 2 and 3
C
2 and 4
D
1, 2, 3 and 4
Question 6 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 7

A multiplexer is placed between a group of 32 registers and an accumulator to regulate data movement such that at any given point in time the content of only one register will move to the accumulator. The minimum number of select lines needed for the multiplexer is _____.

A
5
Question 7 Explanation: 
Number of registers is 32. Only one register has to be selected at any instant of time.
A 25x1 Multiplexer with 5 select lines selects one of the 32(= 25) registers at a time depending on the selection input.
The content from the selected register will be transferred through the output line to the Accumulator.
Question 8

If there are m input lines and n output lines for a decoder that is used to uniquely address a byte addressable 1 KB RAM, then the minimum value of m + n is ____.

A
1034
Question 8 Explanation: 
The size of the decoder required is 10 x 210 i.e., 10 x 1024.
Each output line of the decoder is connected to one of the 1K(= 1024) rows of RAM.
Each row stores 1 Byte.
m=10 and n=1024
Question 9

Consider the Boolean function z(a,b,c).

Which one of the following minterm lists represents the circuit given above?

A
Z = ∑(0,1,3,7)
B
Z = ∑(2,4,5,6,7)
C
Z = ∑(1,4,5,6,7)
D
Z = ∑(2,3,5)
Question 9 Explanation: 
The output of the given circuit is a + b’c.
Convert a+b’c into canonical form which is sum of minterms.
a + b’c = a(b + b’)(c + c’) + (a + a’)b’c
= abc + abc’ + ab’c + ab’c’ + ab’c + a’b’c
= Σ(7,6,5,4,1)
Question 10

Consider three registers R1, R2 and R3 that store numbers in IEEE-754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively.

If R3 = R1/R2, what is the value stored in R3?

A
0x40800000
B
0x83400000
C
0xC8500000
D
0xC0800000
Question 10 Explanation: 
Given numbers are 0x42200000 and 0xC1200000 which are stored in the registers R1 and R2, respectively.

R1 = 1.0100..0 X 2132-127
= 1.0100..0 X 25
= 101.0 X 23
= 5 X 8
= 40

R2 = (-1) x 1.0100..0 X 2130-127
= (-1) x 1.0100..0 X 23
= (-1) x 101.0 X 21
= (-1) x5 X 2
= -10
R3 = R1/R2
= -4
= (-1)x 1.0 x 22
Sign = 1
Mantissa = 000..0
Exponent = 2+127 = 129

R3 = 1100 0000 1000 000..0
= 0x C 0 8 0 0 0 0 0
Question 11
Let R1 and R2 be two 4-bit registers that store numbers in 2’s complement form. For the operation R1+R2, which one of the following values of R1 and R2 gives an arithmetic overflow?
A
R1 = 1011 and R2 = 1110
B
R1 = 1100 and R2 = 1010
C
R1 = 0011 and R2 = 0100
D
R1 = 1001 and R2 = 1111
Question 11 Explanation: 
Question 12
Consider a digital display system (DDS) shown in the figure that displays the contents of register X. A 16-bit code word is used to load a word in X, either from S or from R. S is a 1024-word memory segment and R is a 32-word register file. Based on the value of mode bit M, T selects an input word to load in X. P and Q interface with the corresponding bits in the code word to choose the addressed word. Which one of the following represents the functionality of P, Q, and T?
A
P is 10:1 multiplexer; Q is 5:1 multiplexer; T is 2:1 multiplexer
B
P is 10:2 ^10 decoder; Q is 5:2^ 5 decoder; T is 2:1 encoder
C
P is 10:2^ 10 decoder; Q is 5:2^ 5 decoder; T is 2:1 multiplexer
D
P is 1:10 de-multiplexer; Q is 1:5 de-multiplexer; T is 2:1 multiplexer
Question 12 Explanation: 
P is a 10:2^10 decoder that takes a 10-bit address from S-address as input and enable one of the 1024 words of the S memory.
Q is a 5:2^5 decoder that takes a 5-bit address from R-address as input and enable one of the 32 words of the R memory.
T is a 2x1 Multiplexer that select one of the 2 inputs and transmit it as output.
Question 13
Consider three floating point numbers A , B and C stored in registers R A , R B and R C , respectively as per IEEE-754 single precision floating point format. The 32-bit content stored in these registers (in hexadecimal form) are as follows.

Which one of the following is FALSE?
A
A + C = 0
B
C = A + B
C
B = 3 C
D
( B - C ) > 0
Question 13 Explanation: 

A= -12, B= +36 and C= +12
A+C= 0
B=3C
(B-C)>0
C≠A+B
Question 14

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
Question 14 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 possible = 28 × 28 = 216 = 64K
Then total size of ROM = 64K × 16
Question 15

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
Question 15 Explanation: 
When the pairs 01 (or) 10 occur frequently in the multiplier. In that case Booth multiplication gives worst performance.
Question 16

Consider the following floating point number representation

The exponent is in 2's complement representation and mantissa is in the sign magnitude representation. The range of the magnitude of the normalized numbers in this representation is

A
0 to 1
B
0.5 to 1
C
2-23 to 0.5
D
0.5 to (1-2-23)
Question 16 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 17

Consider the circuit given below which has a four bit binary number b3b2b1b0 as input and a five bit binary number d4d3d2d1d0 as output. The circuit implements:

A
Binary of Hex conversion
B
Binary to BCD conversion
C
Binary to grey code conversion
D
Binary to radix-12 conversion
Question 17 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 18

Consider the circuit in below figure. f implements

A
B
A + B + C
C
A ⊕ B ⊕ C
D
AB + BC + CA
Question 18 Explanation: 
Question 19

What is the equivalent Boolean expression in product-of-sums form for the Karnaugh map given below.

A
B
C
D
E
None of the above
Question 19 Explanation: 
Correct option is

Question 20

A logic network has two data inputs A and B, and two control inputs C0 and C1. It implements the function F according to the following table.

Implement the circuit using one 4 to 1 Multiplexer, one 2-input Exclusive OR gate, one 2-input AND gate, one 2-input OR gate and one Inverter.

A
Theory Explanation.
Question 21

Consider the synchronous sequential circuit in the below figure.

(a) Draw a state diagram, which is implemented by the circuit. Use the following names for the states corresponding to the values of flip-flops as given below.

(b) Given that the initial state of the circuit is S4, identify the set of states, which are not reachable.

A
Theory Explanation.
Question 22

Let * be defined as x * y = x' + y. Let z = x * y. Value of z * x is

A
x'+y
B
x
C
0
D
1
Question 22 Explanation: 
Question 23

An N-bit carry look ahead adder, where N is a multiple of 4, employs ICs 74181 (4 bit ALU) and 74182 (4 bit carry look ahead generator).

The minimum addition time using the best architecture for this adder is

A
proportional to N
B
proportional to log N
C
a constant
D
None of the above
Question 24

Let f(x, y, z) = x' + y'x + xz be a switching function. Which one of the following is valid?

A
B
xz is a minterm of f
C
xz is an implicant of f
D
y is a prime implicant of f
Question 24 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 25

Given √224)r = 13)r.
The value of the radix r is:

A
10
B
8
C
5
D
6
Question 25 Explanation: 
(√224)r = (13)r
Convert r base to decimal.
√2r2 + 25 + 4 = r + 3
Take square both sides,
2r2 + 2r + 4 = r2 + 6r + 9
r2 - 4r - 5 = 0
r2 - 5r + r - 5 = 0
r(r - 5) + (r - 5) = 0
r = -1, 5
r cannot be -1,
So r = 5 is correct answer.
Question 26

Consider a logic circuit shown in figure below. The functions f1, f2 and f (in canonical sum of products form in decimal notation) are:

f1(w,x,y,z) = ∑8,9,10
f2(w,x,y,z) = ∑7,8,12,13,14,15
f(w,x,y,z) = ∑8,9 

The function f3 is

A
Σ9,10
B
Σ9
C
Σ1,8,9
D
Σ8,10,15
Question 26 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 27

What happens when a bit-string is XORed with itself n-times as shown:

  [B⊕(B⊕(B⊕(B........ n times)]  
A
complements when n is even
B
complements when n is odd
C
divides by 2n always
D
remains unchanged when n is even
Question 27 Explanation: 
B⊕(B⊕(B⊕...) n times
Consider:
B⊕(B⊕B)
= B⊕0
= 0 (if consider n times it remains unchanged)
Question 28

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
Question 28 Explanation: 
For 'n' bit data it selects 2n : 1 input
For 4 bit data it selects 24 : 1 = 16: 1 input
Question 29

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
Question 29 Explanation: 
Voltage is to be below Vcc = 5V but above 2.8V
Question 30

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
Question 30 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 31

The function represented by the Karnaugh map given below is:

A
A⋅B
B
AB+BC+CA
C
D
None of the above
Question 31 Explanation: 
Question 32

Which of the following operations is commutative but not associative?

A
AND
B
OR
C
NAND
D
EXOR
Question 32 Explanation: 
NAND operation is commutative but not associative.
Question 33

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%
Question 33 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 34

(a) The implication gate shown below, has two inputs (x and y), the output is 1 except when x=1 and y=0. Realize f = x'y + xy' using only four implication gates.

(b) Show that the implication gate is functionally complete.

A
Theory Explanation.
Question 35

Design a synchronous counter to go through the following states:

  1, 4, 2, 3, 1, 4, 2, 3, 1, 4,........... 

A
Theory Explanation.
Question 36
Which of the following expressions is not equivalent to
?
A
x NAND X
B
x NOR x
C
x NAND 1
D
x NOR 1
Question 36 Explanation: 
Question 37
Which of the following functions implements the Karnaugh map shown below?
A
B
C
D
Question 37 Explanation: 

⇒ CD+AD = D(A+C)
Question 38

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
Question 38 Explanation: 
Question 39

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)
Question 39 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 40

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
Question 40 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 41

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
Question 41 Explanation: 
Sign magnitude:
+0 = 0000
-0 = 1000
1's complement:
+0 = 0000
-0 = 1111
Question 42

The number 43 in 2’s complement representation is

A
01010101
B
11010101
C
00101011
D
10101011
Question 42 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 43

The simultaneous equations on the Boolean variables x, y, z and w,

      x + y + z = 1
             xy = 0
         xz + w = 1
      xy +  = 0

have the following solution for x, y, z and w, respectively.

A
0 1 0 0
B
1 1 0 1
C
1 0 1 1
D
1 0 0 0
Question 43 Explanation: 
Just put the values of each options in the equation and check it.
Question 44

Which function does NOT implement the Karnaugh map given below?

A
(w + x)y
B
xy + yw
C
D
None of the above
Question 44 Explanation: 
Given k-map gives xy + xy + wz

⇒ wy + wz + xy
Question 45

The following arrangement of master-slave flip flops

has the initial state of P, Q as 0, 1 (respectively). After three clock cycles the output state P, Q is (respectively),

A
1, 0
B
1, 1
C
0, 0
D
0, 1
Question 45 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 46

Consider the values A = 2.0 x 1030, B = -2.0 x 1030, C = 1.0, and the sequence

             X: = A + B    Y: = A + C
             X: = X + C    Y: = Y + B  

executed on a computer where floating-point numbers are represented with 32 bits. The values for X and Y will be

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
Question 46 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 47

Design a logic circuit to convert a single digit BCD number to the number modulo six as follows (Do not detect illegal input):
(a) Write the truth table for all bits. Label the input bits I1, I2, …. With I1 as the least significant bit. Label the output bits R1, R2, …. With R1 as the least significant bit. Use 1 to signify truth.
(b) Draw one circuit for each output bit using, altogether, two two-input AND gates, one two-input gate and two NOT gates.

A
Theory Explanation is given below.
Question 48

Minimum sum of product expression for f(w,x,y,z) shown in Karnaugh-map below is

A
xz+y'z
B
xz'+zx'
C
x'y+zx'
D
None of the above
Question 48 Explanation: 

⇒ xz' + zx'
Question 49

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
Question 49 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 50

The 2’s complement representation of the decimal value -15 is

A
1111
B
11111
C
111111
D
10001
Question 50 Explanation: 
15 = 1111
-15 = 11111
1's complement = 10000
2's complement = 10001
There are 50 questions to complete.

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