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GATE IT 2005 | Question: 48

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49 49 votes

The circuit shown below implements a $\text{2-input}$ NOR gate using two $2-1$ MUX (control signal $1$ selects the upper input). What are the values of signals $x, y$ and $z$?

  1. $1, 0, B$
  2. $1, 0, A$
  3. $0, 1, B$
  4. $0, 1, A$

7 Answers

Best answer
58 58 votes

$f = Az+B\bar z$ (As $A$ will be selected when $z$ is high) . 

So, next function will become $g = xf +y\bar f$ 

$ = x\left(Az+B \bar z \right) + y (\overline {Az+B \bar z})$

Putting $x = 0, y = 1, z = A$, we get $g = \overline {AA+B \bar A} = \overline {A + B} \: \left( \because A + B\bar A = A + B \right)$ and answer will become D

• edited by
3 3 votes
Instead of finding a function 'f' for the first MUX (in terms of z,x and y) and the second MUX (in terms of f , x and y), we can logically find out the answer.We can put in 4 values of A and B and analyse the output C. We can get the values appropriate values of x and y by drawing and comparing with a 2x1 MUX when A and B are given as inputs respectively.
1 1 vote
Let the lower mux function be f=Az'+Bz Upper mux g=xf'+yf Now put option a in function g it comes out to be a'b' which is (a+b)'
0 0 votes

In the MUX1, equation is : g = Az + Bz’ In MUX2, equation is : f = xg + yg’ = x(Az + Bz’) + y (Az + Bz’)’
Function ‘f’ should be equal to (A + B)’ .
Using hit and try method, put x = 0, y = 1 and z = A. f = 0(AA + BA’) + 1(AA + BA’)’ f = (A + A’B)’ f = (A + B)’
 So thats correct Answer is (D)

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