Digital Logic

The AND gate receives \(X\) directly and \(\overline{X}\) through the inverter, so ideally \(Y = X \cdot \overline{X} = 0\) for every input: A is true and B is false.

Now let the NOT gate take time \(t\) to respond.

• \(X\) rises from 0 to 1: the direct input becomes 1 at once, but for time \(t\) the inverter still outputs its old value 1. Both AND inputs are briefly 1, so \(Y\) shows a short glitch to logic One (a static hazard). C is true.
• \(X\) falls from 1 to 0: the direct input drops to 0 at once, while the inverter output is still 0 and rises only after \(t\). The two inputs are never 1 together, so \(Y\) simply stays at 0. With no 1 to dip from, no transient Zero can appear, so D is false.

Answer: A, C

• A is true. The maxterms of a function are exactly the minterms it does not contain; out of 0 to 7 these are 0, 1, 2 and 4.
• B is true. \(XY + YZ + XZ\) is 1 exactly when at least two inputs are 1, that is for \(XYZ\) = 011, 101, 110 and 111, which are minterms 3, 5, 6 and 7. \(F\) is the 3-input majority function.
• C and D are false. \(F(X, 0, Z) = XZ\) while \(F(X, 1, Z) = X + Z\), so \(F\) depends on \(Y\). Likewise \(F(0, Y, Z) = YZ\) and \(F(1, Y, Z) = Y + Z\) differ, so \(F\) depends on \(X\).

Answer: A, B

A 2-to-1 multiplexer passes its input 0 when the select line is 0 and its input 1 when the select line is 1.

• M1 chooses between X1 = 1 and X2 = 1, so Q1 = 1 whatever A is.
• M2 chooses between X3 = 0 and X4 = 0, so Q2 = 0 whatever B is.
• M3 passes Q1 when C = 0 and Q2 when C = 1, so \(Y = \overline{C}\).

Y = 1 exactly when C = 0, while A and B can take any values: \(2 \times 2 = 4\) of the 8 combinations.

Answer: 4

All four have sign bit 0, so compare their values using the exponent bias of 127.

• A: exponent 0111 1111 is 127, so the value is \((2 - 2^{-23}) \times 2^{0}\), just under 2.
• B: exponent 1111 1110 is 254, so the value is \((2 - 2^{-23}) \times 2^{127}\), about \(3.4 \times 10^{38}\).
• C: an all-ones exponent is reserved. With a non-zero mantissa the pattern means NaN (not a number), which has no value to compare.
• D: exponent 127 with a zero mantissa is exactly 1.

B is in fact the largest finite single-precision number, so it is the largest option.

Answer: B

Check each statement for \(x = 0\) and \(x = 1\):

• A: \(x.1 = x\) is the identity law for AND, so it is true.
• B: \(x + 1 = 1\) for every \(x\), so \(x + 1 = x\) fails when \(x = 0\). False.
• C: \(x.x = x\) (idempotent law), so \(x.x = 0\) fails when \(x = 1\). False.
• D: \(x + \overline{x} = 1\) is the complement law, so it is true.

The false statements are B and C.

Answer: B, C

Convert every numeral to decimal using its place values.

• \(224_5 = 2 \times 5^{2} + 2 \times 5 + 4 = 50 + 10 + 4 = 64\).
• (A) \(64_{10} = 64\): equal.
• (B) \(100_8 = 1 \times 8^{2} = 64\): equal.
• (C) \(50_{16} = 5 \times 16 = 80\): not equal.
• (D) \(121_7 = 1 \times 7^{2} + 2 \times 7 + 1 = 49 + 14 + 1 = 64\): equal.

So three of the four numerals name the same value, 64; only the hexadecimal one differs.

Answer: A, B, D

Work with minterm sets: AND keeps the minterms common to both inputs, OR takes their union, and XOR keeps the minterms that appear in exactly one input.

• \(f1 \cdot f2\): common minterms \(\{3, 5, 7, 11, 13\}\).
• \(f3 + f4\): \(\{0, 1, 2, 4, 6, 11, 13\}\).
• XOR: 11 and 13 are in both sets and cancel, leaving \(\{0, 1, 2, 3, 4, 5, 6, 7\}\).

So \(Y = \sum(0,1,2,3,4,5,6,7)\), which is (C). Its maxterms are the remaining minterms 8 to 15, so (D) is the same function. Options (A) and (B) both describe \(\sum(0,1,2,11,13)\), which is not \(Y\).

Answer: C, D