Class Test (11-08-20)

Which of the following statements is not correct?

• Music on a CD is a digital signal

• Variation of resistance in Light Dependent Resistor is an analog signal

• Human voice is a digital signal

• Water tap is an analog signal

Which of the following statements is false?

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

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

• 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.

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

What is the Minterm equivalent of A’ + B’

• Σ(0, 1)

• Σ(0, 1, 2)

• Σ(1, 2)

• Σ(1, 2, 3)

Which of the following expression is dual of

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

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

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

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

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

• W’Y’Z + W’X’Z + W’X’Y’ + X’Y’Z’

• W’Y’Z + W’X’Y’ + WXY + X’Y’Z’

• W’Y’Z + W’X’Z + WXY + W’X’Y’

• W’Y’Z + W’X’Z + WXY + W’X’Y

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

• a. W’X’ +X’Y’ + X’Z

• b. W’X’ +WYZ

• c. W’X’ +X’Y’ + Z

• d. W’X’ +WYZ + X’Z

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

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

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

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

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

F =Σ(2,3,12,13,1,4,15) simplifies to

• a. AD + A’B’D’

• b. AB + CD

• c. AB + A’B’C

• d. AC + BD’

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

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

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

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

An active low output decoder can be implemented using

• a. AND gates and OR gates

• b. NAND gates and Inverters

• c. AND gates and Inverters

• d. OR gates and Inverters