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Class Test (11-08-20)

Total questions: 12

Worksheet time: 14mins

Name
Class
Date
1.

Which of the following statements is not correct?

a)

Music on a CD is a digital signal

b)

Variation of resistance in Light Dependent Resistor is an analog signal

c)

Human voice is a digital signal

d)

Water tap is an analog signal

2.

Which of the following statements is false?

a)

A maxterm is sum of all the literals within the Boolean expression.

b)

The maxterm should evaluate to zero for each combination of input variables.

c)

For any two input binary variable x and y, if x+y represents one of the maxterm, then x’+y’ would be its corresponding minterm.

d)

For any two input binary variable x and y, if xy represents one of the minterm, then x’+y’ would be its corresponding maxterm.

3.

What is the Minterm equivalent of A’ + B’

a)

Σ(0, 1)

b)

Σ(0, 1, 2)

c)

Σ(1, 2)

d)

Σ(1, 2, 3)

4.

Which of the following expression is dual of

a)

(AB + CD) = (A + C) (A + D) (B + C) (B + D)

b)

(AC + BD) = (A + C) (A + D) (B + C) (B + D)

c)

(AB + CD) = (A + B) (A + D) (B + C) (C + D)

d)

(AD + BC) = (A + B) (A + C) (B + C) (B + D)

5.

Find the SOP for the function F = Σ(0, 1, 3, 5,14) + Σd(8, 15)

a)

W’Y’Z + W’X’Z + W’X’Y’ + X’Y’Z’

b)

W’Y’Z + W’X’Y’ + WXY + X’Y’Z’

c)

W’Y’Z + W’X’Z + WXY + W’X’Y’

d)

W’Y’Z + W’X’Z + WXY + W’X’Y

6.

Find the SOP for the function F = Σ(0, 1, 2, 8,11) + Σd(3,9,15)

a)

a. W’X’ +X’Y’ + X’Z

b)

b. W’X’ +WYZ

c)

c. W’X’ +X’Y’ + Z

d)

d. W’X’ +WYZ + X’Z

7.

F = Σ(0, 1, 2,3,5,7,8,9,10,12,13) simplifies to

a)

a. (A + B’ + D) (A’ + C’ + D’) (A’ + B’ + C’)

b)

b. (A’ + B’ + D) (A’ + C + D’) (A’ + B + C’)

c)

c. (A + B’ + D) (A + C + D) (A + B’ + C’)

d)

d.(A + B’ + D') (A’ + C’ + D’) (A’ + B’ + C’)

8.

F =Σ(2,3,12,13,1,4,15) simplifies to

a)

a. AD + A’B’D’

b)

b. AB + CD

c)

c. AB + A’B’C

d)

d. AC + BD’

9.

For a 4-bit magnitude comparator, which compares two 4-bit numbers A and B, the Boolean expression for finding A > B (A greater than B) is

a)

A_GT_B = A3’ . B3’ + C3 . A2 . B2’+ C3 . C2 . A1 . B1’ + C3 . C2 . C1 . A0 . B0’

b)

A_GT_B = A3’ . B3’ + C3’ . A2 . B2’+ C3 . C2 . A1 . B1’ + C3 . C2 . C1 . A0 . B0’

c)

A_GT_B = A3 . B3’ + C3 . A2 . B2’+ C3 . C2 . A1 . B1’ + C3 . C2 . C1 . A0 . B0’

d)

A_GT_B = A3 . B3’ + C3 . A2 . B2’+ C3’ . C2’ . A1 . B1’ + C3 . C2 . C1 . A0 . B0’

10.

An active low output decoder can be implemented using

a)

a. AND gates and OR gates

b)

b. NAND gates and Inverters

c)

c. AND gates and Inverters

d)

d. OR gates and Inverters

11.

Which of the following logic circuit can be used to convert a Binary number to an Octal number

a)

a. A 2-to-1 Multiplexer

b)

b. A 8-to-3 Encoder

c)

c. A 3-to-8 Decoder

d)

d. A 4-to-2 Priority Encoder

12.

Consider the 1:4 demultiplexercircuit shown below. What would be the output bits for input condition S0 = 1, S1 = 1 and Din = 1?

a)

a. Y0 = 0, Y1 = 1, Y2= 1, Y3 = 1

b)

b. Y0 = 0, Y1 = 0, Y2= 0, Y3 = 1

c)

c. Y0 = 0, Y1 = 0, Y2= 1, Y3 = 0

d)

d. Y0 = 0, Y1 = 0, Y2= 1, Y3 = 1