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
Here is full image.
how to simply this:
F = (XYZ' +XYZ + Y'ZY + ZY')Z'
= (XYZ' +XYZ + ZY')Z' //this step to final step XYZ'
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
enabled at 0 and disabled at 1?
As you mentioned - "Z is connected to A0 "
then how could you substitute A0' as 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.
@gari thank u
how we know "X connected to I0 & I1 .Y connected to I2 "
A bit more specific guide for GO can be found...