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The circuit shown below implements a 2-input NOR gate using two $2-4$ 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
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Why does question says

2−4 MUX,it uses 1-2 MUX ?

$f = Az+B\bar z$ (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

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Yes. @john-carter is correct.

shoudn't the characterstic eq. be (A.z' + B.z) .

so, f= x( A.z'+ B.z )' + y( A.z'+ B.z )then, put option (a)

x=1, y=0 & z=B

=>( A.B' + B.B )'

=> ( A.B' + B )'

=> ( A + B )' is correct, which is given as 'C'

So option (a) is correct.

The PDF there is taking top input as select 0, which is usually common. But here select 1 and 0 are given in question and hence we must use that.

thanks, my fault,

came across a similar problem when solving another similar question from gate 2007

+1 vote
ans should b D.
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)'
Z = B, means Z should be selecting B input rt?
I dont know much about mux ,i just did a solution using the function of a mux.
Probably you are right..
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.