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GO Classes DPP | GATE CS | Digital Logic | Binary Encoder

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For a $4:2$ binary encoder with active-high inputs $\mathrm{I}_0,\mathrm{I}_1,\mathrm{I}_2,\mathrm{I}_3$ and outputs $\mathrm{Y}_1,\mathrm{Y}_0$, assume exactly one input is $1$ at a time. Which output equations are correct?

  1. $\mathrm{Y}_1=\mathrm{I}_1+\mathrm{I}_3,\ \mathrm{Y}_0=\mathrm{I}_2+\mathrm{I}_3$
     
  2. $\mathrm{Y}_1=\mathrm{I}_2+\mathrm{I}_3,\ \mathrm{Y}_0=\mathrm{I}_1+\mathrm{I}_3$
     
  3. $\mathrm{Y}_1=\mathrm{I}_0+\mathrm{I}_1,\ \mathrm{Y}_0=\mathrm{I}_2+\mathrm{I}_3$
     
  4. $\mathrm{Y}_1=\mathrm{I}_3,\ \mathrm{Y}_0=\mathrm{I}_1$

2 Answers

2 2 votes

A $4:2$ encoder gives the binary code of the active input.

If $\mathrm{I}_0=1$, output is $00$.

If $\mathrm{I}_1=1$, output is $01$.

If $\mathrm{I}_2=1$, output is $10$.

If $\mathrm{I}_3=1$, output is $11$.

Now check $\mathrm{Y}_1$.

$\mathrm{Y}_1=1$ for inputs $\mathrm{I}_2$ and $\mathrm{I}_3$.

So, $\mathrm{Y}_1=\mathrm{I}_2+\mathrm{I}_3$.

Now check $\mathrm{Y}_0$.

$\mathrm{Y}_0=1$ for inputs $\mathrm{I}_1$ and $\mathrm{I}_3$.

So, $\mathrm{Y}_0=\mathrm{I}_1+\mathrm{I}_3$.

Answer: B

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