Computer Science Illuminated
Computer Science Illuminated
7th Edition
ISBN: 9781284155617
Author: Nell Dale, John Lewis
Publisher: Jones & Bartlett Learning
bartleby

Videos

Question
Book Icon
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
Check Mark

Explanation of Solution

Given circuit:

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Knowledge Booster
Background pattern image
Computer Science
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Database System Concepts
Computer Science
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:McGraw-Hill Education
Text book image
Starting Out with Python (4th Edition)
Computer Science
ISBN:9780134444321
Author:Tony Gaddis
Publisher:PEARSON
Text book image
Digital Fundamentals (11th Edition)
Computer Science
ISBN:9780132737968
Author:Thomas L. Floyd
Publisher:PEARSON
Text book image
C How to Program (8th Edition)
Computer Science
ISBN:9780133976892
Author:Paul J. Deitel, Harvey Deitel
Publisher:PEARSON
Text book image
Database Systems: Design, Implementation, & Manag...
Computer Science
ISBN:9781337627900
Author:Carlos Coronel, Steven Morris
Publisher:Cengage Learning
Text book image
Programmable Logic Controllers
Computer Science
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education
Boolean Algebra - Digital Logic and Logic Families - Industrial Electronics; Author: Ekeeda;https://www.youtube.com/watch?v=u7XnJos-_Hs;License: Standard YouTube License, CC-BY
Boolean Algebra 1 – The Laws of Boolean Algebra; Author: Computer Science;https://www.youtube.com/watch?v=EPJf4owqwdA;License: Standard Youtube License