Digital Design - decoders and multiplexer help

[Saterial]

Homework Statement

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)

Homework Equations

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:

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?

 

[lewando]

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

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.

 

[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?

 

[lewando]

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

 

[rezeew]

I would approach this problem by first understanding the basic concepts of decoders and multiplexers. Decoders are digital circuits that convert coded inputs into a set of outputs, while multiplexers are circuits that select one of several inputs and allow it to pass through to the output.

In this problem, we are given a function f that takes in four inputs (x3, x2, x1, x0) and produces two outputs (y1, y0). From the truth table, we can see that the output y1 is 1 when the number of 1's in the input is even, and 0 otherwise. Similarly, y0 is 1 when the number of 1's in the input is odd, and 0 otherwise.

To implement this function using four 2-to-4 decoders, we can use the inputs x3, x2, x1, x0 as the select inputs for the decoders. Each decoder will have four outputs (A, B, C, D) which correspond to the binary representation of the input. For example, if the input is 0011, the decoder with select inputs x3 and x2 will produce outputs A=0, B=0, C=1, D=1. We can then use these outputs to construct the outputs y1 and y0 according to the given function.

To implement this function using a 4-to-1 multiplexer, we can use the inputs x0 and x1 as the select inputs for the multiplexer. The four inputs of the multiplexer will be x3, x2, x1, x0. The output of the multiplexer will be y1 and y0, which can be constructed using the same logic as before.

Overall, both methods will result in the same outputs y1 and y0 for a given set of inputs. It is up to the designer to choose which method is more efficient or suitable for their specific application.

I would also recommend researching and practicing with some basic examples of decoder and multiplexer circuits to gain a better understanding of their functionality and applications.