GATE CSE 1990 | Question: 5-b

Question

Show with the help of a block diagram how the Boolean function :

$f=AB+BC+CA$

can be realised using only a $4:1$ multiplexer.

Comments

$AB + BC + CA = AB + \bar ABC + A\bar BC$

Answer 1

$AB+BC+CA=AB+A'BC+AB'C$

Comments

@kirti how u take c is a input or B and C as a select line......but here it is not given any information?????

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C+C' IS 1. SO NO NEED OF OR AND NOT GATE

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I0 = 0

I1 = I2 = C

I3 = 1

S0 =A

S1 = B

will give the answer

Answer 2

here people directly given answer and not procedure

 

AB+BC+CA

can be written as

AB(C+C') + (A+A')BC + A(B+B')C

Now if we choose A&B as selector then

for first term i.e. for AB(C+C') (C+C') will become 1. And AB indicate input 11 it should be connected to I4

second term gives us ABC and A'BC

third term gives us ABC and AB'C

A'BC means A=0 and B=1 hence C should be connected to I2

AB'C means A=1 and B=0 hence C should be connected to I3

We don't get any term like A'B' which means A=0 and B=0 hence you can connect antthing to I1

Answer 3

This is Simplified form:

$f=AB+BC+CA$

Actual form:

$f=ABC'+A'BC+AB'C+ABC....................(1)$

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Two select lines:

$A\ and\ B$

Let Inputs be:

$I_{0},I_{1},I_{2},I_{3}$

$f=A'B'I_{0}+A'BI_{1}+AB'I_{2}+ABI_{3}....................(2)$

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Compare $(1)$ and $(2)$

$I_{0}=0$

$I_{1}=C$

$I_{2}=C$

$I_{3}=1$

Answer 4

$\text{without using NOT gate}$

Comments

Add this inside explaination:

F= AB + BC +AC

 = ABC+ A'BC + ABC'+​​​​​​​ AB'C  = ABC' + AB'C  + BC

Means :

when B'C' connect it to 0

when BC' connect it to A ->ABC'

when B'C connect it to A ->AB'C

when BC connect it to 1 ->BC

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@Anu Thanks.

it is understandable .I  only added answer to construct mux without using NOT.