virtualgate's picture

Boolean Algebra, Minimisation, Number System

K map introduction
Content covered: 

K map introduction

More Less
0Comment
K map simplification
Content covered: 

K map simplification

More Less
0Comment

examples on KMap

Example on Boolen Expression

Consider the Boolean Operator @ with the following properties:

\(\text{x @ 1} = \overline{x}\)\(\text{x @ 0} = \text{x}\)\(\text{x @ x} = \text{0}\),  \(\text{x @ }\overline{\text{x}} = 1\). Then \(\text{x @ y}\) is equivalent to

A) \(\text{x}\overline{y} + \bar{\text{x}}y\)

B) \(\text{x}\overline{y} + \bar{\text{x}}\overline{y}\)

C) \(\bar{\text{x}}y + \text{x}y\)

D) \(\text{x}y + \bar{\text{x}}\bar{y}\)

Let us approach this question as follows:

We need to find \(\text{x @ y}\)

Given, 

\(\text{x @ 1} = \overline{x} ​​\)

\(\text{x @ 0} = \text{x}\)

\(\text{x @ x} = \text{0}\)

\(\text{x @ }\overline{\text{x}} = 1\)

 

Now, to find \(\text{x @ y}\), we replace \(y\) by \(1, 0, \text{x},\ \ and \ \ \ \bar{\text{x}}\), and find that the expression of option (A) holds True.

\(\therefore\)   Ans (A)

0Comment

  • This quiz contains 10 questions on the topic Boolean Algebra,Minimisation, Number System​
  • Lean well before you attempt the quiz
  • You can attempt the quiz unlimited number of times.

Difficulty Level:  intermediate
Sourabh V Bhalekar @sourabhbhalekar 22 Sep 2017 07:42 pm

Completed

 

Pages