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Consider the following multiplexer where 10, 11, 12, 13 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
asked in Digital Logic by Veteran (68.8k points)
edited by | 1.8k views
Incomplete figure.Please someone edit it.

Here is full image.

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

1 Answer

+33 votes
Best answer

As X connected to I0  I1 .Y connected to I2, Y' connected to I& 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 ]

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

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

=XYZ'

answered by Veteran (49.2k points)
selected by

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



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