# Homework Help: Digital Design - decoders and multiplexer help

1. Jun 18, 2013

### Saterial

1. The problem statement, all variables and given/known data
Let f(x3; x2; x1; x0) = (y1; y0) such that y1 = 1 if the number of 1's in x3x2x1x0 is even, 0 otherwise, and, output y0 = 1 if the number of 1's in x3x2x1x0 is odd, 0 otherwise.

1. Implement f with four 2-to-4 decoders.
2. Implement f with a 4-to-1 multiplexer with selection inputs x0; x1 (in this order)

2. Relevant equations

3. The attempt at a solution

How do I even start solving this? This was a bonus as it was not really taught to me so I have to learn myself.

(IN THE CASE OF 0, 0, 0, 0, both y1 and y0 are 0 right? because there are no 1's at all)

I begin by making a truth table of:
Code (Text):

x3 x2 x1 x0   y1 y0
0  0  0  0    0  0
0  0  0  1    0  1
0  0  1  0    0  1
0  0  1  1    1  0
0  1  0  0    0  1
0  1  0  1    1  0
0  1  1  0    1  0
0  1  1  1    0  1
1  0  0  0    0  1
1  0  0  1    1  0
1  0  1  0    1  0
1  0  1  1    0  1
1  1  0  0    1  0
1  1  0  1    0  1
1  1  1  0    0  1
1  1  1  1    1  0
Where do I go from here?

Last edited by a moderator: Jun 18, 2013
2. Jun 18, 2013

### lewando

Zero is an even number.

If x3..x0 | y1,y0 = 0,0,0,0 | 0,0 then y1 would indicate: "number of ones is not even" y0 would indicate "number of ones is not odd" --a contradiction.

3. Jun 18, 2013

### Saterial

So that would mean the output of y1 and y0 in my case is 1, 0 for 0, 0 ,0 0 in my case because the number of 1's is even and not odd?

4. Jun 18, 2013

### lewando

Yes. And it should make your job somewhat easier as well.