in Digital Logic edited by
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39 votes

Consider the circuit shown below. The output of a $2:1$ MUX is given by the function $(ac' + bc)$.

Which of the following is true?

  1. $f=X_1'+X_2$
  2. $f=X_1'X_2+X_1X_2'$
  3. $f=X_1X_2+X_1'X_2'$
  4. $f=X_1+X_2'$
in Digital Logic edited by
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3 Comments

g = (a and x1′) or (b and x1)
g = (1 and x1’) or (0 and x1)
g = x1’

f = ac’ + bc
f = (a and x2′) or (b and x2)
f = (g and x2′) or (x1 and x2)
f = x1’x2’ + x1x2

Ref: Geeksforgeeks.org
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the diagram doesn't match with GO 2020 soft copy
12
this question diagram is still wrong in Go pdf 2020.please correct in the next issue
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5 Answers

32 votes
 
Best answer
$g = X_1'$
So, $f = ac' + bc$

$= X_1'X_2' + X_1X_2$

So, (C).
selected by
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8 Comments

What is the significance of ac'+bc here?

We don't need it to get the answer. So why it is mentioned in the question?
0
here , f = ac'+bc where for first mux c = x1 and for second mux c = x2.

Now substitute for g = ac' + bc => a x1' + b x , now for f , f= ac'+bc => g x2' + x1 x2 => x1' x2' + x1 x2.
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Why anyone is not noticing that a should be equal to 1 and b should be equal to 0 in the question
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This is so weird. The solution is different from what I'm getting. I think the diagram is wrong.
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Image was wrong. Fixed now.
2

Out of the two inputs how come we assume a is I0(Input 0) and b is I1 (Input 1). Unless we assume this is there anyway to get into the answer? 

0
The image is wrong in GO PDF.
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image is still wrong in Hard copy of GO PYQ, (page no – 471), Digital logic part.
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8 votes
Given output of the mux is $a{c}' + bc$

For the first mux, $g$ is the output and $a=0,b=1,c = x_{1}$

Hence,

$g=1.{x_{1}}'+0.x_{1} = {x_{1}}'$  $........(1)$

For the second mux,

$a=g={x_{1}}' [from (1)],b=x_{1},c=x_{2}$

$f={x_{1}}'.{x_{2}}'+x_{1}.x_{2}$

Hence, option (c).
2 votes

$\underline{\textbf{Answer:}\Rightarrow}$

$\underline{\textbf{Explanation:}\Rightarrow}$

A multiplexer has $\mathbf 3$ things:

  1. Selection Line.
  2. Input Line.
  3. Output Lines. 
  4. For $\mathbf{2:1}$ Multiplexer, there are $\mathbf 2$ inputs and $\mathbf 1$ output and $\mathbf 1$ selection line.

So, when Selection Line, $\mathbf  {X_1}$ is set $\mathbf 0$, then $\mathbf {a}$ will be selected and when $\mathbf {X_1 = 1}$, then $\mathbf b$ will be selected.

$\therefore $ Equation will be: $\mathbf{g = 1.X_1' + 0.X_1 = X_1'}$

Now, this $\mathbf{g = X_1}$ will be given as input to second multiplexer.

Following the same procedure for second multiplexer and taking one input as $\mathbf{X_1'}$, we will get:

$\mathrm{f = X_2'X_1' + X_2X_1}$

$\therefore \mathbf C$ is the correct answer.

by

2 Comments

edited by
A silly doubt: How do we consider multiplying the select line inputs like, why X1’ *1 and not X1’*0?

As input is given as 0 in a and 1 in b. Corresponding to that the select line should be taken, right? Please help me with this doubt.

Just I need to know that why didn’t you multiplied X1’ with 0. As input is zero so select line has to be complemented, please correct me if I’m wrong.
0

same doubt!!

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2 votes

$f=ac'+bc$

Now, solve without considering $X1,X2,1,0$ and try to get function $f.$

Output of $mux_{1}:$ $g=c'a+cb$ and

Output of $mux_{2}:$ $c'g+cb\Rightarrow c'(c'a+cb)+cb=c'a+cb=$ $ac'+bc$

With this we are clear that with $X1'$ we have to consider $a's$ input and with $X1$ we have to consider $b's$ input.

So,

Output of $mux_{1}:$ $g=X1'.1+X1.0=X1'$

Output of $mux_{2}:$ $X2'g+X2.X1=X2'(X1')+X2.X1=$ $X1X2+X1'X2'$

 

Correct Answer is (C)

1 comment

which among a and b is msb/lsb? how to know when I0/I1 is not given...
0
0 votes
Answer : C

g = (X1' * 1) + (X1 * 0) =  X1'

a = g from 2nd MUX

f = (ac' + bc) = (X2' * a)+ ( X2 * b)

X2' = c' , b= X1 (given)

 so, f = (X1' * X2')+(X1*X2)
Answer:

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