I am trying to figure out if I implemented a 2-1 Mux correctly based on a 3 variable K-map.

my function states F(x,y,z)= Σm(0,1,3,6), and I have created a kmap as follows

  00  01  11 10

0] 1 , 0, 1 , 0

1] 1 , 1 , 0 , 0

Note: the most sig bit is x,and xy are the top inputs and z is the side inputs

from this I found that a 2-1 Mux could be implemented by using z as my selector and ~b=0 and ~a=1, as my inputs to be pushed. Is this correct or have I done this wrong?


Yes,it's right.You can use z as your selector pin.Here's how finally it's look like


simulate this circuit – Schematic created using CircuitLab

If you give z=0,the top portion of K-map will be active.That is 00 and 11,for z=0,the output is 1.

    Similarly when z=1,the output F=1 for x=0 (after some simplifications)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.