Question

if one input of an and gate is low while the other is a clock signal, the output is
low.
high.
a clock signal.
cannot be determined

Explanation:

Step1: Recall AND gate operation

The AND gate has the property that \( Y = A\cdot B\). In digital logic, LOW is represented as \( 0\) and HIGH as \( 1\).

Step2: Substitute values

Let \( A = 0\) (LOW) and \( B\) be the clock - signal (which can be either \( 0\) or \( 1\)). Then \( Y=0\cdot B\).

Step3: Calculate result

Using the rule of multiplication in boolean algebra \( 0\cdot x = 0\) for any \( x\in\{0,1\}\). So the output \( Y\) is LOW.

Answer:

LOW.