The Gateway to Computer Science Excellence
+7 votes
4.4k views

Minimum number of $2 \times 1$ multiplexers required to realize the following function, $f = \bar{A} \bar{B} C + \bar{A} \bar{B} \bar{C}$

Assume that inputs are available only in true form and Boolean a constant 1 and 0 are available.

  1. 1
  2. 2
  3. 3
  4. 7
in Digital Logic by Veteran (105k points)
recategorized by | 4.4k views
+1
Answer is 2 Right
first Simplify the equation and get the equation A(bar)B(bar)
than at the first mux take the input I(zero)=1 and I(one) =0 and take the select line A than output will be A (bar) and At the second Mux take the input A(bar) at I(zero) than and take select line as B at Second Mux and u can take Input Line I(1) as Zero than output Will be A(bar)B(bar)
0
Is it not possible to implement this using only one mux? That is, give A as selection line, B' to 0 input and 0 to 1 input. Then if A is 0, choose B'. B is 0 means output is 1. B is 1 means output is 0. In case if A is 1, then 1 input has 0.

3 Answers

+18 votes
Best answer

Answer  : 2

f= A'B'C + A'B'C' ===> A'B'( C + C' ) ===> A'B' ====> (A+B)' Final function represents XOR .So with the help of 2*1 mux we can implement it

by Boss (45.4k points)
selected by
+1

shekhar chauhan u meant to say NOR not XOR as (A+B)' is NOR not XOR and in either case for making NOR or XOR we would need 2 2*1 MUX (one mux to complement either A or B and second mux for NOR) because imputs are available in true form only not in their complement forms.

So the answer would be 3 MUX needed.

0
it is along with a NOT Gate .otherwise we need 3        2*1 mux
0
can u please create the circuit and show us. it will be really greatful. I am not able to understand it
0
plz explain why 2 ?
+6
For $1^{st}$ MUX take $I_0 = 1, I_1 = B$ and give $A$ as select line. So output will be $A + \overline{A}B = A + B$

Give $A+B$ as select line to second MUX and take $I_0 = 0$ and $I_1 = 1$ for second MUX. You get $\overline{A+B}$ as output. So, $2$ MUX required.
0
okk.. thankew i got the answer but i think for first mux aslo we cant take B as input as its given we take only 1 or 0 na ?
0
Read question again. We can take inputs in true form means $A,B$ are allowed but not $\overline{A} , \;\overline{B}$.
+1
ohh.. sorry .. thanku again
+1

mcjoshi Veteran aren't u writing the reverse

For 1st1st MUX I0=B,I1=B=1,... Similarly inputs for second mux should also be reverse according to me

+4 votes
by Loyal (9.9k points)
0
Inputs available in true form means??
0 votes
Ans is 2. {i.e. 2(2x1) MUX or 1(2x1) + 1 NOT}

Explanation : f= A'B'C + A'B'C' = A'B'(C+C') = A'B' = (A+B)'   which is NOR so if we take A as a select line then the value of A either 0 or 1,   if is A is 0 then output of nor gate is  B'  and  if A is1 then output of nor gate is 0

Now input for the multiplexer is B' and 0 and select line is A then output is A'B' = (A+B)'
by (11 points)
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,324 answers
198,404 comments
105,169 users