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GO Classes DPP | GATE CS | Digital Logic | MUX with Enable

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6 6 votes

A $2:1$ MUX has inputs $\mathrm{I}_0,\mathrm{I}_1$, select line $\mathrm{S}$, and active-low enable input $\mathrm{E}$. When $\mathrm{E}=1$, the output is $0$. When $\mathrm{E}=0$, the MUX works normally. Which expression represents the output $\mathrm{F}$?

  1. $\mathrm{F}=\mathrm{E}(\mathrm{S}'\mathrm{I}_0+\mathrm{S}\mathrm{I}_1)$
     
  2. $\mathrm{F}=\mathrm{E}'+\mathrm{S}'\mathrm{I}_0+\mathrm{S}\mathrm{I}_1$
     
  3. $\mathrm{F}=\mathrm{E}'(\mathrm{S}'\mathrm{I}_0+\mathrm{S}\mathrm{I}_1)$
     
  4. $\mathrm{F}=\mathrm{S}'\mathrm{I}_0+\mathrm{S}\mathrm{I}_1$

2 Answers

2 2 votes

For a normal $2:1$ MUX,

$\mathrm{F}=\mathrm{S}'\mathrm{I}_0+\mathrm{S}\mathrm{I}_1$.

But here enable is active-low.

That means the MUX works only when $\mathrm{E}=0$.

So, the working condition is represented by $\mathrm{E}'$.

Therefore, the output becomes:

$\mathrm{F}=\mathrm{E}'(\mathrm{S}'\mathrm{I}_0+\mathrm{S}\mathrm{I}_1)$.

When $\mathrm{E}=1$, $\mathrm{E}'=0$, so $\mathrm{F}=0$.

When $\mathrm{E}=0$, $\mathrm{E}'=1$, so the MUX works normally.

Answer: C

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