Chapter 4, Problem 62E

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:

Behavior of the circuit:

• From the above circuit diagram,
  • First, the inputs A and B are passed to AND gate to perform the product of A and B and produce the output as $\text{A}\cdot \text{B}$.
  • Next, pass the same input B from the AND gate, and input C is passed in the NAND gate to perform the inverse of product of B and C and produce the output as $\overline{\text{B}\cdot \text{C}}$.
  • Next, pass the same input C from the NAND gate to NOT gate, to perform the inverse of C to produce the output as $\overline{\text{C}}$.
  • Next, pass the output of AND gate and output of NOT gate as the input to NOR gate to perform the inverse of sum of “$\text{A}\cdot \text{B}$” and “$\overline{\text{C}}$” and produce the output as $\overline{\text{A}\cdot \text{B +}\overline{\text{ C}}}$.
  • Finally, the output of NOR gate and output of NAND gate is passed as the input of OR gate.
    • Therefore, “$\overline{\text{B}\cdot \text{C}}$” and “$\overline{\text{A}\cdot \text{B +}\overline{\text{ C}}}$” are passed as input to OR gate and produce the output as $\overline{\text{BC}}\text{ + }\overline{\text{(AB + }\overline{\text{C}}\text{)}}$.

Truth table for the above circuit diagram:

Step 1:

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

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

• First, the inputs A as 0 and B as 0 are passed to AND gate to perform the product of 0 and 0 and produce the output as $0\cdot 0=0$.
• Next, pass the same input B as 0 from the AND gate $\text{A}\cdot \text{B}$, and input C as 0 is passed in the NAND gate to perform the inverse of product of 0 and 0, and produces the output as $\overline{\text{0}\cdot \text{0}}\text{ =}\overline{\text{ 0 }}=1$.
• Next, pass the same input C as 0 from the NAND gate $\overline{\text{B}\cdot \text{C}}$ in the NOT gate to perform the inverse of 0 to produce the output as $\overline{\text{0}}=1$.
• Next, pass the output of AND gate and output of NOT gate as the input for NOR gate to perform the inverse of sum of “0” and “1”, and produce the output as $\overline{\text{0 +1}}=\overline{1}=0$.
• Finally, the output of NOR gate and output of NAND gate is passed as the input of OR gate.
  • Therefore, “0” and “1” are passed as input to OR gate and produce the output as $\text{0 + 1 = 1}$.

Step 3:

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

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

• First, the inputs A as 0 and B as 0 are passed to AND gate to perform the product of 0 and 0 and produce the output as $0\cdot 0=0$.
• Next, pass the same input B as 0 from the AND gate $\text{A}\cdot \text{B}$, and input C as 1 is passed in the NAND gate to perform the inverse of product of 0 and 1 and produces the output as $\overline{\text{0}\cdot 1}\text{ =}\overline{\text{ 0 }}=1$.
• Next, pass the same input C as 1 from the NAND gate $\overline{\text{B}\cdot \text{C}}$ in the NOT gate to perform the inverse of 1 to produce the output as $\overline{\text{1}}=0$.
• Next, pass the output of AND gate and output of NOT gate as the input for NOR gate to perform the inverse of sum of “0” and “0” and produces the output as $\overline{\text{0 +0}}=\overline{\text{ 0 }}=1$.
• Finally, the output of NOR gate and output of NAND gate is passed as the input of OR gate.
  • Therefore, “1” and “1” are passed as input to OR gate and produce the output as $\text{1 + 1 = 1}$.

Step 4:

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

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

• First, the inputs A as 0 and B as 1 are passed to AND gate to perform the product of 0 and 1 and produce the output as $0\cdot 1=0$.
• Next, pass the same input B as 1 from the AND gate $\text{A}\cdot \text{B}$, and input C as 0 is passed in the NAND gate to perform the inverse of product of 1 and 0 and produces the output as $\overline{1\cdot 0}\text{ =}\overline{\text{ 0 }}=1$.
• Next, pass the same input C as 0 from the NAND gate $\overline{\text{B}\cdot \text{C}}$ in the NOT gate to perform the inverse of 0 to produce the output as $\overline{\text{0}}=1$.
• Next, pass the output of AND gate and output of NOT gate as the input for NOR gate to perform the inverse of sum of “0” and “1” and produces the output as $\overline{\text{0 +1}}=\overline{\text{ 1 }}=0$.
• Finally, the output of NOR gate and output of NAND gate are passed as the input of OR gate.
  • Therefore, “0” and “1” are passed as input to OR gate and produce the output as $\text{0 + 1 = 1}$.

Step 5:

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

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

• First, the inputs A as 0 and B as 1 are passed to AND gate to perform the product of 0 and 1 and produce the output as $0\cdot 1=0$.
• Next, pass the same input B as 1 from the AND gate $\text{A}\cdot \text{B}$, and input C as 1 is passed in the NAND gate to perform the inverse of product of 1 and 1 and produces the output as $\overline{1\cdot 1}\text{ =}\overline{\text{ 1 }}=0$.
• Next, pass the same input C as 1 from the NAND gate $\overline{\text{B}\cdot \text{C}}$ in the NOT gate to perform the inverse of 1 to produce the output as $\overline{\text{1}}=0$.
• Next, pass the output of AND gate and output of NOT gate as the input for NOR gate to perform the inverse of sum of “0” and “0” and produces the output as $\overline{\text{0 + 0}}=\overline{\text{ 0 }}=1$.
• Finally, the output of NOR gate and output of NAND gate are passed as the input of OR gate.
  • Therefore, “1” and “0” are passed as input to OR gate and produce the output as $\text{1 + 0 = 1}$.

Step 6:

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

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

• First, the inputs A as 1 and B as 0 are passed to AND gate to perform the product of 0 and 1 and produce the output as $1\cdot 0=0$.
• Next, pass the same input B as 0 from the AND gate $\text{A}\cdot \text{B}$, and input C as 0 is passed in the NAND gate to perform the inverse of product of 0 and 0 and produces the output as $\boxed{\overline{0\cdot 0}\text{ =}\overline{\text{ 0 }}=1}$.
• Next, pass the same input C as 0 from the NAND gate $\overline{\text{B}\cdot \text{C}}$ in the NOT gate to perform the inverse of 0 to produce the output as $\boxed{\overline{\text{0}}=1}$.
• Next, pass the output of AND gate and output of NOT gate as the input for NOR gate to perform the inverse of sum of “0” and “1” and produces the output as $\boxed{\overline{\text{0 +1}}=\overline{\text{ 1 }}=0}$.
• Finally, the output of NOR gate and output of NAND gate are passed as the input of OR gate.
  • Therefore, “0” and “1” are passed as input to OR gate and produce the output as $\boxed{\text{0 + 1 = 1}}$.

Step 7:

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

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

• First, the inputs A as 1 and B as 0 are passed to AND gate to perform the product of 0 and 1 and produce the output as $\boxed{1\cdot 0=0}$.
• Next, pass the same input B as 0 from the AND gate $\text{A}\cdot \text{B}$, and input C as 1 is passed in the NAND gate to perform the inverse of product of 0 and 1 and produces the output as $\boxed{\overline{0\cdot 1}\text{ =}\overline{\text{ 0 }}=1}$.
• Next, pass the same input C as 1 from the NAND gate $\overline{\text{B}\cdot \text{C}}$ in the NOT gate to perform the inverse of 1 to produce the output as $\boxed{\overline{\text{1}}=0}$.
• Next, pass the output of AND gate and output of NOT gate as the input for NOR gate to perform the inverse of sum of “0” and “0” and produces the output as $\boxed{\overline{\text{0 + 0}}=\overline{\text{ 0 }}=1}$.
• Finally, the output of NOR gate and output of NAND gate are passed as the input of OR gate.
  • Therefore, “1” and “1” are passed as input to OR gate and produce the output as $\boxed{\text{1 + 1 = 1}}$.

Step 8:

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

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

• First, the inputs A as 1 and B as 1 are passed to AND gate to perform the product of 1 and 1 and produce the output as $\boxed{1\cdot 1=1}$.
• Next, pass the same input B as 1 from the AND gate $\text{A}\cdot \text{B}$, and input C as 0 is passed in the NAND gate to perform the inverse of product of 1 and 0 and produces the output as $\boxed{\overline{1\cdot 0}\text{ =}\overline{\text{ 0 }}=1}$.
• Next, pass the same input C as 0 from the NAND gate $\overline{\text{B}\cdot \text{C}}$ in the NOT gate to perform the inverse of 0 to produce the output as $\boxed{\overline{\text{0}}=1}$.
• Next, pass the output of AND gate and output of NOT gate as the input for NOR gate to perform the inverse of sum of “1” and “1” and produces the output as $\boxed{\overline{\text{1 + 1}}=\overline{\text{ 1 }}=0}$.
• Finally, the output of NOR gate and output of NAND gate are passed as the input of OR gate.
  • That is, “0” and “1” are passed as input to OR gate and produce the output as $\boxed{\text{0 + 1 = 1}}$.

Step 9:

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

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

• First, the inputs A as 1 and B as 1 are passed to AND gate to perform the product of 1 and 1 and produce the output as $\boxed{1\cdot 1=1}$.
• Next, pass the same input B as 1 from the AND gate $\text{A}\cdot \text{B}$, and input C as 1 is passed in the NAND gate to perform the inverse of product of 1 and 1 and produces the output as $\boxed{\overline{1\cdot 1}\text{ =}\overline{\text{ 1 }}=0}$.
• Next, pass the same input C as 1 from the NAND gate $\overline{\text{B}\cdot \text{C}}$ in the NOT gate to perform the inverse of 1 to produce the output as $\boxed{\overline{\text{1}}=0}$.
• Next, pass the output of AND gate and output of NOT gate as the input for NOR gate to perform the inverse of sum of “1” and “0” and produces the output as $\boxed{\overline{\text{1 + 0}}=\overline{\text{ 1 }}=0}$.
• Finally, the output of NOR gate and output of NAND gate are passed as the input of OR gate.
  • Thus, “0” and “0” are passed as input to OR gate and produce the output as $\boxed{\text{0 + 0 = 0}}$.

Therefore, the truth table for the given circuit is:

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