EBK COMPUTER SCIENCE ILLUMINATED
EBK COMPUTER SCIENCE ILLUMINATED
7th Edition
ISBN: 9781284174755
Author: Dale
Publisher: JONES+B CO
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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

Expert Solution & Answer
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Explanation of Solution

Given circuit:

EBK COMPUTER SCIENCE ILLUMINATED, Chapter 4, Problem 62E

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 AB.
    • 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 BC¯.
    • Next, pass the same input C from the NAND gate to NOT gate, to perform the inverse of C to produce the output as 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 “AB” and “C¯” and produce the output as AB + C¯¯.
    • Finally, the output of NOR gate and output of NAND gate is passed as the input of OR gate.
      • Therefore, “BC¯” and “AB + C¯¯” are passed as input to OR gate and produce the output as BC¯ + (AB + C¯)¯.

Truth table for the above circuit diagram:

Step 1:

  • The inputs are A, B, and C for the circuit diagram:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
000     
001     
010     
011     
100     
101     
110     
111     

Step 2:

  • When the inputs are A as 0, B as 0, and C as 0:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
001     
010     
011     
100     
101     
110     
111     
  • 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 00=0.
  • Next, pass the same input B as 0 from the AND gate AB, 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 00¯ = 0 ¯=1.
  • Next, pass the same input C as 0 from the NAND gate BC¯ in the NOT gate to perform the inverse of 0 to produce the output as 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 0 +1¯= 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 0 + 1 = 1.

Step 3:

  • When the inputs are A as 0, B as 0, and C as 1:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
00101011
010     
011     
100     
101     
110     
111     
  • 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 00=0.
  • Next, pass the same input B as 0 from the AND gate AB, 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 01¯ = 0 ¯=1.
  • Next, pass the same input C as 1 from the NAND gate BC¯ in the NOT gate to perform the inverse of 1 to produce the output as 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 0 +0¯= 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 1 + 1 = 1.

Step 4:

  • When the inputs are A as 0, B as 1, and C as 0:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
00101011
01001101
011     
100     
101     
110     
111     
  • 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 01=0.
  • Next, pass the same input B as 1 from the AND gate AB, 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 10¯ = 0 ¯=1.
  • Next, pass the same input C as 0 from the NAND gate BC¯ in the NOT gate to perform the inverse of 0 to produce the output as 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 0 +1¯= 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 0 + 1 = 1.

Step 5:

  • When the inputs are A as 0, B as 1, and C as 1:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
00101011
01001101
01100011
100     
101     
110     
111     
  • 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 01=0.
  • Next, pass the same input B as 1 from the AND gate AB, 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 11¯ = 1 ¯=0.
  • Next, pass the same input C as 1 from the NAND gate BC¯ in the NOT gate to perform the inverse of 1 to produce the output as 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 0 + 0¯= 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 1 + 0 = 1.

Step 6:

  • When the inputs are A as 1, B as 0, and C as 0:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
00101011
01001101
01100011
10001101
101     
110     
111     
  • 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 10=0.
  • Next, pass the same input B as 0 from the AND gate AB, 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 00¯ = 0 ¯=1.
  • Next, pass the same input C as 0 from the NAND gate BC¯ in the NOT gate to perform the inverse of 0 to produce the output as 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 0 +1¯= 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 0 + 1 = 1.

Step 7:

  • When the inputs are A as 1, B as 0, and C as 1:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
00101011
01001101
01100011
10001101
10101011
110     
111     
  • 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 10=0.
  • Next, pass the same input B as 0 from the AND gate AB, 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 01¯ = 0 ¯=1.
  • Next, pass the same input C as 1 from the NAND gate BC¯ in the NOT gate to perform the inverse of 1 to produce the output as 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 0 + 0¯= 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 1 + 1 = 1.

Step 8:

  • When the inputs are A as 1, B as 1, and C as 0:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
00101011
01001101
01100011
10001101
10101011
11011101
111     
  • 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 11=1.
  • Next, pass the same input B as 1 from the AND gate AB, 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 10¯ = 0 ¯=1.
  • Next, pass the same input C as 0 from the NAND gate BC¯ in the NOT gate to perform the inverse of 0 to produce the output as 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 1 + 1¯= 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 0 + 1 = 1.

Step 9:

  • When the inputs are A as 1, B as 1, and C as 1:
ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
00101011
01001101
01100011
10001101
10101011
11011101
11110000
  • 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 11=1.
  • Next, pass the same input B as 1 from the AND gate AB, 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 11¯ = 1 ¯=0.
  • Next, pass the same input C as 1 from the NAND gate BC¯ in the NOT gate to perform the inverse of 1 to produce the output as 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 1 + 0¯= 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 0 + 0 = 0.

Therefore, the truth table for the given circuit is:

ABCABBC¯C¯AB + C¯¯BC¯ + (AB + C¯)¯
00001101
00101011
01001101
01100011
10001101
10101011
11011101
11110000

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