Digital Logic Test 2

Result

Digital Logic Test 2

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
2 / -0.33

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
7

B
8

C
9

D
10

Solution

Formula:
In quadratic equation, ax²+ bx + c = 0
Sum of root = \(- \frac{b}{a}\)
Product of roots = \(\frac{c}{a}\)
Calculation:
Quadratic equation: x² – 13x + 36 = 0
Given base is b.
Roots of the equation with base b is 5 and 6
\({5_{b\;}} + \;{6_b} = \;{\left( {13} \right)_b}\)
\(11 = b + 3\)
b = 8
Also, by considering the product of roots,
\({5_{b\;}} \times {6_b} = {\left( {36} \right)_b}\)
30 = 3b + 6,
24 = 3b, b = 8
In both cases, base b = 8
Question 2
2 / -0.33

A function (A, B, C) defined by three boolean variables A, B, and C when expressed as the sum of products is given by:
F = A̅.B̅.C̅ + A̅.B.C̅ + A.B̅.C̅
where, A̅, B̅, and C̅ are the complements of the respective variables. The product of sums (POS) form of the function F is

A
F = (A + B + C) . (A + B̅ + C) . (A̅ + B + C)

B
F = (A̅ + B̅ + C̅) . (A̅ + B + C̅) + (A + B̅ + C̅)

C
F = (A + B + C̅) . (A + B̅ + C̅) . (A̅ + B + C̅) . (A̅ + B̅ + C) . (A̅ + B̅ + C̅)

D
F = (A̅ + B̅ + C) . (A̅ + B + C) . (A + B̅ + C) . (A + B + C̅) . (A + B + C)

Solution

F = A̅.B̅.C̅ + A̅.B.C̅ + A.B̅.C̅
In terms of minterms, this can be represented as:
F = ∑m (0, 2, 4)
The equivalent maxterm will contain the terms not present in the minterm representation, i.e.
F = ∑m (0, 2, 4) = π(1, 3, 5, 6, 7) = M₁. M₃. M₅. M₆. M₇
⇒ (A + B + C̅) (A + B̅ + C̅) (A̅ + B + C̅) (A̅ + B̅ + C) (A̅ + B̅ + C̅)
Question 3
2 / -0.33

What are the first eight bits of mantissa part and the decimal value stored in exponent if (-28.375)10 is represented in single-precision IEEE 754 format?

A
Mantissa = 11000110

B
Exponent = 132

C
Mantissa = 11100011

D
Exponent = 131

Solution

Single precision IEEE 754 format
Signed (S)
Mantissa (M)
Exponent (E)
1 bit
23 bits
8 bits
1 (negative)
11000110
131 (decimal)
Since (-28.375)10 is negative therefore S = 1.
(28.375)10 = (11100.0110)2
(11101.011)2 = 1.11000110 × 24
Mantissa = 11000110
Since it is excess 127
Exponent = 4 + 127 + 131
Question 4
2 / -0.33

In the following truth table, V = 1 if and only if the input is valid.
Inputs
Outputs
D₀
D₁
D₂
D₃
X₀
X₁
V
0
0
0
0
x
x
0
1
0
0
0
0
0
1
x
1
0
0
0
1
1
x
x
1
0
1
0
1
x
x
x
1
1
1
1

What function does the truth table represent?

A
Priority encoder

B
Decoder

C
Multiplexer

D
Demultiplexer

Solution

Concept:
A multiplexer or MUX is a combination circuit that contains more than one input line, one output line and more than one selection line.
Simple encoder takes 2ⁿ input bits and produces n output bits.
Explanation:
V = 1
When input D₃ = 1, then it does not care about the input D₀, D₁, and D₂
Output put will be 1, 1
When input D₃ = 0 and D₂ = 1, then it does not care about the input D₀, and D₁
Output put will be 1, 0
When input D₃ = 0, D₂ = 0 and D₁ = 1, then it does not care about the input D₀,
Output put will be 0, 1
When input D₃ = 0, D₂ = 0, D₁ = 0 and D₀ = 1,
Output put will be 0, 0
Here priority has been assigned to input and hence it is a Priority encoder.
Question 5
2 / -0.33

The clock frequency of 12 MHz is applied to a cascaded counter of modulus-4 counter, modulus-5 counter and modulus-6 counter.

A
The overall modulus of the cascaded counter is 60

B
The lowest output frequency of the cascaded counter is 100 kHz

C
The overall modulus of the cascaded counter is 120

D
The lowest output frequency of the cascaded counter is 200 kHz

Solution

The overall modulus of the cascaded counter is the multiplication of the modulus of the individual counters.

Overall modulus (N) = 4 × 5 × 6 = 120

The clock frequency (f_clk) = 12 MHz

The lowest output frequency \(= \frac{{{f_{clk}}}}{N}\)

\(= \frac{{12 \times {{10}^6}}}{{120}} = 100\;kHz\)
Question 6
2 / -0.33

The simplified Sum of Product form of the following Boolean expression:
\((P + \bar Q + \bar R )(P + Q+R)(P+Q+\bar R )\) is

A
\(\bar P Q+R\)

B
\(P+Q\bar R \)

C
\(P\bar Q +R\)

D
\(PQ+ R\)

Solution

\((P+\bar Q+ \bar R).(P+Q+R).(P+Q+\bar R)\)
\((P+\bar Q+\bar R).(P+(Q+R)(Q+\bar R))\)
\((P+\bar Q+\bar R).(P+Q)\)
\((P+(\bar Q+\bar R)Q)\)
\(P+Q\bar R\)
Question 7
2 / -0.33

Consider the unsigned 8-bit point binary number representation below:
X₇X₆X₅X₄ .X₃X₂X₁X₀
where the position of the binary point is between X₄ and X₃. X₇ is the MSB. Which of the decimal numbers cannot be represented exactly in the above representation:
(a) 14.1875
(b) 16.125
(c) 3.03125
(d) 1.6875

A
Only a and c

B
Only b and c

C
Only b and d

D
All (a), (b), (c) and (d)

Solution

(a) 14.1875 = 1110. 0011
Therefore, it can be represented
(b) 16.125 = 10000.0010
Since 5 bits is consumed by 16 ∴ it cannot be executed
(c) 3.03125 = 0011.00001
Since 5 bits is consumed by 0.3125 ∴ it cannot be executed
(d) 1.6875 = 0001.1011
Therefore, it can be represented
Hence (b) and (c) cannot be represented

Question 8

2 / -0.33

Consider the function F = A + B̅C, where F is a three-variable Boolean function in which A is MSB and C is LSB. Which of the following is/are false among the given options?

F = ∑ (1, 4, 5, 6, 7)

F = ∑ (1, 3, 5, 6, 7)

F = ∏ (0, 2, 4)

F = ∏ (0, 2, 3)

Solution

Concept:

Sum of product (SOP) – The concept of sum of products mainly includes min-term. Min-term is a Boolean expression resulting in 1 for the output of a single cell and 0’s for all other cells in K-Map. Min-term is all the 1’s in the truth table.

Product of sum (POS) – Product of sum mainly includes max term. Max term is a Boolean expression resulting in 0 for the output of a single cell and 1 for all other cells in K-Map. Maxterm is all the 0’s in the truth table.

Explanation:

Function: X = A + B̅C

Construct the truth table for this function:

A	B	C	A + B̅C
0	0	0	0
0	0	1	1
0	1	0	0
0	1	1	0
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	1

Here F = ∑ (1, 4, 5, 6, 7) means sum of product which is all the 1’s in output of truth table.

F = ∏ (0, 2, 3) means product of sum which is all the 0’s in the output of truth table.

Option 2 and 3 are false.

Question 9
2 / -0.33

Consider the 64-bit register which stops floating point number in IEEE Double precision format
What is the value of the number if 64 bits are given below?
1
10000000111
010111000…

A
-348--348

B
-3484--3481

C
-3487--3486

D
-3489--3482

Solution

In IEEE 64-bit format,
signed(s)
Exponent(E)
Mantissa(M)
1 bit
11 bits
52 bits

(-1)^s × 1.M × 2^{E – 1023}
(-1)¹ × 1.010111 × 2^{1031 – 1023}
(-1)¹ × 1.010111 × 2⁸
-1×(1 + 2^-2 + 2^-4 + 2^-5 + 2^-6)×2⁸
-1(256 + 64 + 16 + 8 + 4) = -348
Question 10
2 / -0.33

P, Q, and R are the decimal integers corresponding to the 4-bit binary number 1100 considered in signed magnitude, 1’s complement, and 2’s complement representations, respectively. The 6-bit 2’s complement representation of (P + Q + R) is

A
110101

B
110010

C
111101

D
111001
Solution
Concept:
1. Signed magnitude representation uses the most significant bit (MSB) a sign bit.
- If the sign bit is ‘0’ then the number is positive.
- If the sign bit is ‘1’ then the number is negative.
The remaining bits represent the magnitude of the binary number.
2. 1’s complement representation:
It is a representation of a binary number obtained by toggling all bits in it i.e. transforming the 0 bit to 1 and the 1 bit to 0.
3. 2’s complement representation:
It is obtained by simply adding 1 to the 1’s complement of that binary number.
Calculation:
Given: 4-bit binary number = 1100
Signed magnitude representation (P): - 4
1’s complement representation (Q): - 3
2’s complement representation (R): - 4
So,
P + Q + R = (-4) + (-3) + (-4) = -11
(-11)₁₀ is represented in 2’s complement as:
-(11)₁₀ = 10101
Since the Options are in 6 bits so, we copy sign bit once towards left.
So 6-bit representation of (-11)₁₀ is 110101
Question 11
2 / -0.33

How many functions does f1.f2 and f1 + f2 represents respectively if the given function is
\({\rm{f}}1{\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)_{}} = \mathop \sum \nolimits_M \;\left( {0,{\rm{\;}}1,{\rm{\;}}4} \right) + {\rm{\;}}\mathop \sum \limits_\emptyset \left( {3,{\rm{\;}}5,{\rm{\;}}7} \right)\)
\({\rm{f}}2\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right) = \mathop \sum \nolimits_M \;\left( {2,3} \right) + {\rm{\;}}\mathop \sum \limits_\emptyset \left( {1,{\rm{\;}}6,{\rm{\;}}7} \right)\)
where ∑_m represents sum of min-terms and ∑_ϕ represents and don’t care

A
2, 4

B
8, 8

C
4, 8

D
4, 4

Solution

Given:
\(f1.f2\left( {a,b,c} \right) = \mathop \sum \limits_\emptyset \left( {1,3,7} \right)\)
\(f1 + f2\left( {a,b,c} \right) = \sum \left( {0,1,2,3,4} \right) + \mathop \sum \limits_\emptyset \left( {5,6,7} \right)\)
It is given in min-term form with the don’t care.
In table form it is represented as :
a
b
c
f1
f2
f1.f2
f1+f2
0
0
0
1
0
0
1
0
0
1
1
Ø
Ø
1
0
1
0
0
1
0
1
0
1
1
Ø
1
Ø
1
1
0
0
1
0
0
1
1
0
1
Ø
0
0
Ø
1
1
0
0
Ø
0
Ø
1
1
1
Ø
Ø
Ø
Ø

Now,
\(f1\left( {a,b,c} \right) = \sum \left( {0,\;1,\;4} \right) + \;\mathop \sum \limits_\emptyset \left( {3,\;5,\;7} \right)\)
\(f2\left( {a,b,c} \right) = \sum \left( {2,3} \right) + \;\mathop \sum \limits_\emptyset \left( {1,\;6,7} \right)\)
As, in f1.f2(a ,b ,c), there are 3 don’t cares .So, number of functions possible in this are : 2³ = 8
Similarly, in f1 + f2(a, b, c), there are 3 don’t cares. So, number of functions possible are : 2³= 8
Question 12
2 / -0.33

D input of a clocked D-flip-flop receives and input A ⊕ Q_n where A is an external logic input and Q_n is the output of the n^th D-FF before the clock appears. The circuit works as

A
EXOR gate

B
T-FF

C
D-FF

D
JK-FF

Solution

Characteristic equation of a D flip flop is, Q (n + 1) = D
Characteristic equation of a T flip flop is, Q (n + 1) = T ⊕ Qn
The input of the D flip flop = A ⊕ Q_n
The output of the D flip flop will be, Q (n + 1) = A ⊕ Q_n
It represents, the characteristic equation of T flip flop.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Signed (S)	Mantissa (M)	Exponent (E)
1 bit	23 bits	8 bits
1 (negative)	11000110	131 (decimal)

Inputs				Outputs
D₀	D₁	D₂	D₃	X₀	X₁	V
0	0	0	0	x	x	0
1	0	0	0	0	0	1
x	1	0	0	0	1	1
x	x	1	0	1	0	1
x	x	x	1	1	1	1

a	b	c	f1	f2	f1.f2	f1+f2
0	0	0	1	0	0	1
0	0	1	1	Ø	Ø	1
0	1	0	0	1	0	1
0	1	1	Ø	1	Ø	1
1	0	0	1	0	0	1
1	0	1	Ø	0	0	Ø
1	1	0	0	Ø	0	Ø
1	1	1	Ø	Ø	Ø	Ø

signed(s)	Exponent(E)	Mantissa(M)
1 bit	11 bits	52 bits

Digital Logic Test 2

Result

Digital Logic Test 2

Score

-

Rank

-

SHARING IS CARING

Question 1

7

8

9

10

Solution

Question 2

F = (A + B + C) . (A + B̅ + C) . (A̅ + B + C)

F = (A̅ + B̅ + C̅) . (A̅ + B + C̅) + (A + B̅ + C̅)

F = (A + B + C̅) . (A + B̅ + C̅) . (A̅ + B + C̅) . (A̅ + B̅ + C) . (A̅ + B̅ + C̅)

F = (A̅ + B̅ + C) . (A̅ + B + C) . (A + B̅ + C) . (A + B + C̅) . (A + B + C)

Solution

Question 3

Mantissa = 11000110

Exponent = 132

Mantissa = ​11100011

Exponent = 131

Solution

Question 4

Priority encoder

Decoder

Multiplexer

Demultiplexer

Solution

Question 5

The overall modulus of the cascaded counter is 60

The lowest output frequency of the cascaded counter is 100 kHz

The overall modulus of the cascaded counter is 120

The lowest output frequency of the cascaded counter is 200 kHz

Solution

Question 6

\(\bar P Q+R\)

\(P+Q\bar R \)

\(P\bar Q +R\)

\(PQ+ R\)

Solution

Question 7

Only a and c

Only b and c

Only b and d

All (a), (b), (c) and (d)

Solution

Question 8

F = ∑ (1, 4, 5, 6, 7)

F = ∑ (1, 3, 5, 6, 7)

F = ∏ (0, 2, 4)

F = ∏ (0, 2, 3)

Solution

Question 9

-348--348

-3484--3481

-3487--3486

-3489--3482

Solution

Question 10

110101

110010

111101

111001

Solution

Question 11

2, 4

8, 8

4, 8

4, 4

Solution

Question 12

EXOR gate

T-FF

D-FF

JK-FF

Solution

Mantissa = 11100011