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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. $f=X_1+X_2'$

edited | 2.6k views
0
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

$g = X_1'$
So, $f = ac' + bc$

$= X_1'X_2' + X_1X_2$

So, (C).
by Veteran (432k points)
selected by
0
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?
+1
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.
+5
Why anyone is not noticing that a should be equal to 1 and b should be equal to 0 in the question
+1
This is so weird. The solution is different from what I'm getting. I think the diagram is wrong.
+1
Image was wrong. Fixed now.
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).
by (345 points)

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)
by Junior (987 points)

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

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

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 Boss (19.2k points)
by Loyal (8.6k points)