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The following circuit compares two $2$-bit binary numbers, $X$ and $Y$ represented by $X_1X_0$ and $Y_1Y_0$ respectively. ($X_0$ and $Y_0$ represent Least Significant Bits)

Under what conditions $Z$ will be $1$?

  1. $X>Y$
  2. $X<Y$
  3. $X=Y$
  4. $X!=Y$
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3 Answers

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Ans: a) X > Y

The function Z can be represented as:

$Z = X_1.Y_1' + (X_1 \odot Y_1).(X_0.Y_0')$

The first term, $X_1.Y_1'$ would be 1 only when $X_1 > Y_1$

The second term would be 1 when $X_1 == Y_1$ and $X_0 > Y_0$
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The output equation will be $(X_1⊙Y_1).(X_0.\overline{Y}_0)+X_1.\overline{Y}_1$

Clearly, when $X=0$, ie, when $X_0$ and $X_1$ are both $0$, output can't be 1.

 

So, Options B, C and D are incorrect, because in them, $X=0$ is valid.

Option A
Answer:

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