Chapter 4, Problem 60E

Program Plan Intro

Circuit:

• The circuit is known as the combination of gates that is used to achieve a difficult logical operation.
• It has two general categories; they are:
  • Combinational circuit
  • Sequential circuit

Explanation of Solution

Given circuit diagram:

Behavior of the circuit:

• From the circuit diagram:
  • First, the input A is passed to NOT gate to perform the inverse of the A and produces the output as $\overline{\text{A}}$.
  • Next, the input A and input B is passed in the AND gate to perform the product of A and B, to produce the output as $\text{A}\cdot \text{B}$.
  • Finally, the output of NOT gate and the output of AND gate is passed as the input of XOR gate.
    • Note: when both the inputs are the same, then the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
    • That is, “$\overline{\text{A}}$” and “$\text{A}\cdot \text{B}$” are passed as the input for XOR gate and produce as the output $\overline{\text{A}}\oplus \text{(A}\cdot \text{B)}$.

Truth table for the given circuit diagram:

Step 1:

• The inputs are A and B for the above circuit diagram:

A	B	$\overline{A}$	$A\cdot B$	$\overline{A}\oplus \text{(}A\cdot B\text{)}$
0	0
0	1
1	0
1	1

Step 2:

• When the inputs are A as 0 and B as 0:

A	B	$\overline{A}$	$A\cdot B$	$\overline{A}\oplus \text{(}A\cdot B\text{)}$
0	0	1	0	1
0	1
1	0
1	1

• First, the input A as 0 is passed to NOT gate to perform the inverse of the A and produces the output as $1\oplus 0=1$ .
• Next, the input A as 0 and input B as 0 is passed to the AND gate to perform the product of A and B and produce the output as  $s$.
• Finally, the output of NOT gate and the output of AND gate is passed as the input of XOR gate.
  • Note: when both the inputs are the same, then the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
  • That is, “1” and “0” are passed as inputs to XOR gate and produces output as  $1\oplus 0=1$.

Step 3:

• When the inputs are A as 0 and B as 1:

A	B	$\overline{A}$	$A\cdot B$	$\overline{A}\oplus \text{(}A\cdot B\text{)}$
0	0	1	0	1
0	1	1	0	1
1	0
1	1

• First, the input A as 0 is passed to NOT gate to perform the inverse of the A and produces the output as $\overline{0}=1$.
• Next, the input A as 0 and input B as 1 is passed to the AND gate to perform the product of the A and B and produce the output as  $0\cdot 1=$ .
• Finally, the output of NOT gate and the output of AND gate is passed as the input of XOR gate.
  • Note: when both the inputs are the same, then the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
  • That is, “1” and “0” are passed as input for XOR gate and produce output as $1\oplus 0=1$.

Step 4:

• When the inputs are A as 1 and B as 0:

A	B	$\overline{A}$	$A\cdot B$	$\overline{A}\oplus \text{(}A\cdot B\text{)}$
0	0	1	0	1
0	1	1	0	1
1	0	0	0	0
1	1

• First, the input A as 1 is passed to NOT gate to perform the inverse of A and produces the output as $\overline{\text{ 1 }}=0$ .
• Next, the input A as 1 and input B as 0 is passed in the AND gate to perform the product of the A and B and produce the output as $1\cdot 0=0$.
• Finally, the output of NOT gate and the output of AND gate is passed as the input to XOR gate.
  • Note: when both the inputs of XOR gate are the same, the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
  • That is, “0” and “0” are passed as input for XOR gate and produce as the output $0\oplus 0=0$.

Step 5:

• When the inputs are A as 1 and B as 1:

A	B	$\overline{A}$	$A\cdot B$	$\overline{A}\oplus \text{(}A\cdot B\text{)}$
0	0	1	0	1
0	1	1	0	1
1	0	0	0	0
1	1	0	1	1

• First, the input A as 1 is passed to NOT gate to perform the inverse of the A and produces the output as $\overline{\text{ 1 }}=0$.
• Next, the input A and input B as 1 is passed in the AND gate to perform the product of the A and B and produce the output as $1\cdot 1=1$.
• Finally, the output of NOT gate and the output of AND gate is passed as the input of XOR gate.
  • Note: when both the inputs of XOR gate are the same, the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
  • That is, “0” and “1” are passed as input for XOR gate and produce as the output $0\oplus 1=1$.

Final truth table:

Therefore, truth table of the given circuit diagram is:

A	B	$\overline{A}$	$A\cdot B$	$\overline{A}\oplus \text{(}A\cdot B\text{)}$
0	0	1	0	1
0	1	1	0	1
1	0	0	0	0
1	1	0	1	1