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47 votes
47 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. $f=X_1+X_2'$
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5 Answers

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

$= X_1'X_2' + X_1X_2$

So, (C).
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9 votes
9 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).
4 votes
4 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.

2 votes
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)

Answer:

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