Chapter 4, Problem 61E

Program Plan Intro

Circuit:

• The circuit is known as the combination of gates that is used to achieve a difficult logical operation.
• It contains 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 A and produces the output as $\overline{\text{A}}$.
  • Next, the inputs B and C are passed into XOR gate to perform the XOR operation of the B and C to produce the output as $\text{B}\oplus \text{C}$.
    • Note: XOR operation - when both the inputs are the same, then the output of XOR gate is 0. Otherwise, the output of XOR gate is 1.
  • Finally, the output of NOT gate and output of XOR gate is passed as the input of AND gate.
    • That is, “$\overline{\text{A}}$” and “$\text{B}\oplus \text{C}$” are passed as input for AND gate and produces the output $\overline{\text{A}}\cdot \text{(B}\oplus \text{C)}$.

Truth table:

Step 1:

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

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

Step 2:

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

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

Step 3:

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

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

Step 4:

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

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

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

Step 5:

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

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

Step 6:

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

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

Step 7:

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

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

Step 8:

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

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

Step 9:

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

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

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

Therefore, Truth table for the given circuit is:

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