PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<
PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<
5th Edition
ISBN: 9781266481475
Author: Petruzella
Publisher: MCG
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Chapter 4, Problem 2P
Program Plan Intro

Logic gate:

  • Logic gate is an electronic circuit that is used to take logical decisions based on the input.
  • It contains one or more number of inputs and one output.
  • The working of logic gate is based on the binary principle that has two states either logic 0 or logic 1.
  • The output of logic gate is produced when it satisfies any of its logic conditions.
  • The logic condition depends upon the type of the gates and the number of inputs.
  • The primary logic gates include AND, OR and NOT and the combinations of these gates are used to implement any of the other logic gates.

AND gate:

  • The AND gate refers to a logic gate whose output will be HIGH only when all the inputs are HIGH.
  • The output of AND gate will be LOW when any one of its input is LOW.
  • The symbol to represent AND gate is given below.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  1

  • The truth table for AND gate is as follows.
INPUT AINPUT BOUTPUT Y
000
010
100
111

OR gate:

  • The OR gate refers to a logic gate whose output will be HIGH when any one of its inputs are HIGH.
  • The output of AND gate will be LOW when both the inputs are LOW.
  • The symbol to represent OR gate is given below.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  2

  • The truth table for OR gate is as follows.
INPUT AINPUT BOUTPUT Y
000
011
101
111

NOT gate:

  • The NOT gate refers to a logic gate whose output will be HIGH when it’s input is LOW and whose output will be LOW when it’s input is HIGH.
  • The symbol to represent NOT gate is given below.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  3

  • The truth table for NOT gate is as follows.
INPUT AOUTPUT Y
01
10

Expert Solution
Check Mark

Explanation of Solution

a.

Logic gate circuit:

The logic gate circuit is as follows.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  4

Explanation:

In the above given logic gate circuit,

  • The inputs “A” and “B” are connected to logic OR gate and the corresponding output will be (A+B).
  • Now, the resultant along with other inputs “C” and “D” are connected to logic AND gate whose output will be Y=(A+B)CD.

Therefore, the Boolean expression for the given logic circuit is Y=(A+B)CD.

Explanation of Solution

b.

Logic gate circuit:

The logic gate circuit is as follows.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  5

Explanation:

In the above given logic gate circuit,

  • The inputs “A” and “B” are connected to logic OR gate and the corresponding output will be (A+B).
  • The input “C” is connected to logic NOT gate and the corresponding output will be “ˉC”.
  • Then, the inputs “ˉC” and “D” are connected to another logic OR gate and the corresponding output will be (ˉC+D).
  • Now, the output (A+B) and (ˉC+D) are connected to a logic AND gate whose output will be Y=(A+B)(ˉC+D).

Therefore, the Boolean expression for the given logic circuit is Y=(A+B)(ˉC+D).

Explanation of Solution

c.

Logic gate circuit:

The logic gate circuit is as follows.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  6

Explanation:

In the above given logic gate circuit,

  • The inputs “A” and “B” are connected to logic AND gate and the corresponding output will be (AB).
  • Now, the resultant along with other input “C” is connected to logic OR gate whose output will be Y=(AB+C).

Therefore, the Boolean expression for the given logic circuit is Y=(AB+C).

Explanation of Solution

d.

Logic gate circuit:

The logic gate circuit is as follows.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  7

Explanation:

In the above given logic gate circuit,

  • The inputs “A” and “B” are connected to logic AND gate and the corresponding output will be (AB).
  • Now, the resultant along with other input “C” is connected to logic OR gate whose output will be (AB+C).
  • The input “E” is connected to logic NOT gate and the corresponding output will be “ˉE”.
  • Now, the outputs (AB+C) and ˉE along with the input “D” are connected to a logic AND gate whose output will be Y=[(AB+C)DˉE].

Therefore, the Boolean expression for the given logic circuit is Y=[(AB+C)DˉE].

Explanation of Solution

e.

Logic gate circuit:

The logic gate circuit is as follows.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  8

Explanation:

In the above given logic gate circuit,

  • The inputs “A” and “B” are connected to logic OR gate and the corresponding output will be (A+B).
  • Then, the inputs “C” and “D” are connected to logic OR gate and the corresponding output will be (C+D).
  • Now, the output (A+B) and (C+D) are connected to a logic AND gate whose output will be Y=(A+B)(C+D).

Therefore, the Boolean expression for the given logic circuit is Y=(A+B)(C+D).

Explanation of Solution

f.

Logic gate circuit:

The logic gate circuit is as follows.

PROGRAMMABLE LOGIC CONTROL.(LL)>CUSTOM<, Chapter 4, Problem 2P , additional homework tip  9

Explanation:

In the above given logic gate circuit,

  • The input “B” is connected to logic NOT gate and the corresponding output will be “ˉB”.
  • The inputs “A” and “ˉB” are connected to logic AND gate and the corresponding output will be (AˉB).
  • Similarly, the input “D” is connected to logic NOT gate and the corresponding output will be “ˉD”.
  • The inputs “C” and “ˉD” are connected to logic AND gate and the corresponding output will be (CˉD).
  • Then, the input “F” is connected to logic NOT gate and the corresponding output will be “ˉF”.
  • The inputs “E” and “ˉF” are connected to logic AND gate and the corresponding output will be (EˉF).
  • Finally, the outputs AˉB,  CˉD and EˉF are connected to a logic OR gate whose output will be Y=(AˉB+CˉD+EˉF).

Therefore, the Boolean expression for the given logic circuit is Y=(AˉB+CˉD+EˉF).

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