Logic Gates Questions and Answers

Answer: D

Correct answer: All of these

All three statements describe adders: they add bits, a full adder can be built from two half adders and an OR gate, and a half adder takes two inputs and produces two outputs, the sum and the carry.

Answer: A

Correct answer: words, high

The gates are called decision-making elements because they respond to particular input words - whole combinations of input bits - and ignore the rest. The gate signals that it has recognised its word by driving its output high.

Answer: B

Correct answer: EX-NOR or EX-OR gates

XOR gives 1 when an odd number of inputs is 1, so a chain of XOR gates across a word computes its parity directly. That single property serves both purposes: generating the parity bit to be sent, and checking it on arrival. XNOR does the same with the opposite sense.

Answer: B

Correct answer: gives an output when all input signals are present simultaneously

An AND gate outputs 1 only when every input is 1 at the same instant. It holds no state, so it is not a memory circuit, and it is a negative-logic OR rather than a negative OR in the sense of the option offered.

Answer: C

Correct answer: Carry-save-adder

A carry-save adder reduces three numbers to two - a partial sum and a carry vector - without propagating carries, so several numbers can be added at once and the carries resolved only at the final step. The other adders take two operands at a time.

Answer: B

Correct answer: 1 ⊕ 1 ⊕ 0 = 1

XOR gives 1 when an odd number of its arguments is 1. Here two of the three are 1, an even number, so the result is 0, not 1. Checking the others: \(1\oplus 0 = 1\) (one 1), \(1\oplus 1\oplus 1 = 1\) (three 1s), and \(1\oplus 1 = 0\) (two 1s).

Answer: D

Correct answer: Both (a) & (b)

XOR is its own inverse, so an equation may be rearranged freely: from \( A\oplus B=C \), XOR both sides with \(B\) to get \( A = B\oplus C \), or with \(A\) to get \( B = A\oplus C \). Both rearrangements hold, which is why the third option - \( A\oplus B\oplus C = 0 \) - is also true but is not the only one.

Answer: D

Correct answer: Pin 1

The dot, and the notch beside it, mark pin 1; the pins are then numbered anticlockwise from it when the chip is viewed from above. Power and ground pins differ from one device to another, so no fixed pin can be assumed for them.

Answer: D

Correct answer: 1023

Building a \(2^n\) input multiplexer from two-input multiplexers gives a binary tree: \(2^{n-1}\) at the first level, \(2^{n-2}\) at the next, and so on down to one. The total is

\[ 2^{9}+2^{8}+\cdots+2^{0}=2^{10}-1=1023 \]

Answer: C

Correct answer: carry from ( K - 1 )TH bit

A full adder at bit position k adds the two operand bits of that position together with the carry produced by position \(k-1\). It is that third input that distinguishes it from a half adder.