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Consider the following multiplexer where $I0, I1, I2, I3$ are four data input lines selected by two address line combinations $A1A0=00,01,10,11$ respectively and $f$ is the output of the multiplexor. EN is the Enable input.

The function $f(x,y,z)$ implemented by the above circuit is

1. $xyz'$
2. $xy + z$
3. $x + y$
4. None of the above

### 1 comment

It could be done via Table and K-map but that will be more time consuming.

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As $x$ is connected to $I0$  & $I1$, $y$ connected to $I2,$ $y'$ connected to $I3$ & $A1,$ $z$ connected to $A0$ and $z'$ connected to ENABLE $(EN),$

$f= ( {\overline{A1}}.{\overline{A0}}.I0+ {\overline{A1}}.A0.I1 +A1.{\overline{A0}}.I2 + A1.A0.I3 ).EN$

$\implies f = (xyz' +xyz + y'z'y + zy')z'$
$\qquad= (xyz' +xyz + zy')z' =xyz'$

Correct Answer: $A$

how to simply this:

F = (XYZ' +XYZ + Y'ZY + ZY')Z'

= (XYZ' +XYZ + ZY')Z'  //this step to final step XYZ'

=XYZ'

z*z' =0
@karthik

(XYZ' +XYZ + Y'ZY + ZY')Z'

=XYZ'Z'+XYZZ'+Y'ZYZ' + ZY'Z'

=XYZ'+0+0+0

=XYZ'
Where is X,Y,Z in the circuit diagram?

why is EN  multiplied in characteristic equation? Kindly explain . Thanks

yes it will multiply

When EN = 1, themultiplexer performs its operation depending on the selection line

Thanks :)

in the above question enable will be in complemented form

enabled at  0 and disabled at 1?

yes for the above circuit because inverter is present there

As you mentioned - "Z is connected to A"

then how could you substitute A0as Z for the third min-term.

It should be Y'Z'Y instead of Y'ZY.

Correct me if I am wrong. Although I know that it won't affect the final answer by doing so.

how you write this function in xyz form from A1, A0??????????
@raj see in circuit...  A1 is getting y' and AO is getting z..

@gari thank u

how we know "X connected to I0  I1 .Y connected to I2  "

I think u should look at the diagram carefully
This is totally due to depression .. thanks @akanshadewangan24
I want to know, why you multiply enable with the all the input ??

What the reason for that ?
edited

how come ZY' for A1.A0.I3  why not Y'ZY'

ZY' equals to ZY'Y' @sandeep verma

Option A