Realization of a Boolean function:1. For the function Y= ABCD+ Ã BC D+ ABC D+ ABC D+ AB C D+ ABC D+AB CDa. Write the truth table.b. Using Karnaugh maps, find a minimal sum-of-products (minterms)expression that could be implemented by the circuit:74087406143274097406Fig.2: Simplified circuit for implementing the function Yc. To implement this circuit:Check the components to see that they work properlyInsert the appropriate IC onto the breadboard• Make connections as shown in Fig.2Provide the input data via the input switches and observe theoutput on LEDS• Verify the Truth table2. For the given truth table, realize a logical circuit on the breadboard toproduce the output Y using basic gates (Hint: Use K-maps). You must verifythe circuit by referring to the truth table.InputsOutputABDY11111111111111111111111111111

Answered: Realization of a Boolean function: 1.…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Related questions

Question

Realization of a Boolean function:
1. For the function Y= ABCD+ Ã BC D+ ABC D+ ABC D+ AB C D+ ABC D+AB CD
a. Write the truth table.
b. Using Karnaugh maps, find a minimal sum-of-products (minterms)
expression that could be implemented by the circuit:
7408
7406
1432
7409
7406
Fig.2: Simplified circuit for implementing the function Y
c. To implement this circuit:
Check the components to see that they work properly
Insert the appropriate IC onto the breadboard
• Make connections as shown in Fig.2
Provide the input data via the input switches and observe the
output on LEDS
• Verify the Truth table
2. For the given truth table, realize a logical circuit on the breadboard to
produce the output Y using basic gates (Hint: Use K-maps). You must verify
the circuit by referring to the truth table.
Inputs
Output
A
B
D
Y
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1

Expert Solution

Step 1

Write the expression of the function and simplify it.

$\begin{array}{rcl} Y & = & \bar{A} \bar{B} C \bar{D} + \bar{A} B C \bar{D} + A B C \bar{D} + A \bar{B} C \bar{D} + A \bar{B} \bar{C} \bar{D} + A \bar{B} \bar{C} D + A \bar{B} C D \\ = & \bar{A} C \bar{D} (\bar{B} + B) + A C \bar{D} (B + \bar{B}) + A \bar{B} \bar{C} (\bar{D} + D) + A \bar{B} C D \\ = & \bar{A} C \bar{D} + A C \bar{D} + A \bar{B} \bar{C} + A \bar{B} C D \\ = & C \bar{D} (\bar{A} + A) + A \bar{B} \bar{C} + A \bar{B} C D \\ = & C \bar{D} + A \bar{B} \bar{C} + A \bar{B} C D \\ = & C \bar{D} + A \bar{B} \end{array}$