Question

(i) Write the truth table for logic gates marked $P$ and $Q$ in the given circuit.
(ii) Write the truth table for the circuit.

Answer 1

The gate marked $P$ is a $NOT$ gate.
The gate marked $Q$ is a $NOR$ gate.
$A$ when passed through $NOT$ gate returns the inverted output.
When $B$ and the output of $A$ through $NOT$ gate are passed through $NOR$ gate, the output returned is high if and only if both the inputs are low.

Complete step by step solution:
(i) Gate marked $P$ is a $NOT$ GATE. It inverts the input and returns the value as output. Therefore, if the input is high, output is low and when input is low, output is high.

The logic equation of $NOT$ gate is as follows:
$Y = \overline A$
Where, $Y$ is the output and $A$ is the input.
Truth table of $NOT$ GATE IS:

INPUT( $A$ )	OUTPUT( $Y$ )
$1$	$0$
$0$	$1$

The gate marked $Q$ is a $NOR$ gate. It gives a high output when both the inputs are low, and a low output when one or both the inputs are high.

The logic equation of $NOR$ gate is as follows:
$Y = \overline {A + B}$
Where, $Y$ is the output while $A$ and $B$ are the input.
Truth table of $NOR$ GATE IS:

Input 1 ( $A$ )	Input 2 ( $B$ )	$Y = \overline {A + B}$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$0$

(ii) The given circuit is a combinational circuit made out of basic logic gates:
The logic expression for the circuit is:
$Y = \overline {\overline A + B}$
Where, $Y$ is the output while both $A$ and $B$ are inputs.
Truth table for $NOR$ gate is:

INPUT FOR P ( $A$ )	INPUT1 ( $\overline A$ )	INPUT2 ( $B$ )	OUTPUT $Y = \overline {\overline A + B}$
$0$	$1$	$0$	$0$
$0$	$1$	$1$	$0$
$1$	$0$	$0$	$1$
$1$	$0$	$1$	$0$

Note: $NOT$ is a single input logic gate and is also known as the Inverter.
A $NOR$ gate can have two or more inputs and is the result of negation of $OR$ gate.
Remember that the final circuit takes inverted value of input $A$ as its own input.