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

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A $4:2$ priority encoder has active-high inputs $\mathrm{D}_0,\mathrm{D}_1,\mathrm{D}_2,\mathrm{D}_3$, where $\mathrm{D}_3$ has the highest priority and $\mathrm{D}_0$ has the lowest priority. Outputs are $\mathrm{Y}_1,\mathrm{Y}_0$ and valid output $\mathrm{V}$. Which set of equations is correct?

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

1 Answer

1 1 vote

In a priority encoder, if more than one input is $1$, the highest priority input is encoded.

Priority order is:

$\mathrm{D}_3>\mathrm{D}_2>\mathrm{D}_1>\mathrm{D}_0$.

Now check $\mathrm{Y}_1$.

The output code is $10$ when $\mathrm{D}_2$ is the highest active input.

The output code is $11$ when $\mathrm{D}_3$ is active.

So, $\mathrm{Y}_1=1$ whenever $\mathrm{D}_2=1$ or $\mathrm{D}_3=1$.

$\boxed{\therefore \mathrm{Y}_1=\mathrm{D}_2+\mathrm{D}_3}$

Now check $\mathrm{Y}_0$.

$\mathrm{Y}_0=1$ when $\mathrm{D}_3=1$.

Also, $\mathrm{Y}_0=1$ when $\mathrm{D}_1=1$ and no higher input $\mathrm{D}_2$ is active.

$\boxed{\therefore \mathrm{Y}_0=\mathrm{D}_3+\mathrm{D}_2'\mathrm{D}_1}$

Valid output $\mathrm{V}$ is $1$ if at least one input is active.

$\boxed{\therefore \mathrm{V}=\mathrm{D}_0+\mathrm{D}_1+\mathrm{D}_2+\mathrm{D}_3}$
 

Answer: C

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