Hey guys,

I have a little problem. I've got a circuit with 3 flip/flops (A, B & C) and one input (X), using the following equations:

D(a) = (BC' + B'C)X + (BC + B'C')X'
D(b) = A
D(c) = B

Now I think I've got the circuit diagram for this (http://i41.tinypic.com/1fzfm.jpg) all right, but I'm having trouble working out the next state table.

This is because I don't really understand HOW to work it out.

Could someone give me a brief explanation of how I'd go about working this out? I don't want/need the answer, as I do need to learn this myself, but if someone could just say "oh - you need to add this to this etc." that would be massively appreciated.

Thanks. Tom.

 

You need to know what type of flip-flop is being used. We will assume they are D-type flip-flops.

Begin with state 000.
Go down sequentially through all eight states (including X) - ABCX.
Write down the results for each function D(a), D(b) and D(c). Of course D(b) and D(c) are trivial.
You do not need a circuit diagram. Use the equations given.

Edit: all 16 states.

 

Okay. Thanks Mr. Chips. So just to make sure I'm doing it right, the first few lines of the truth table would look like this:

Present Next
A B C X A B C
0 0 0 0 1 0 0
0 0 1 0 0 0 0
0 1 0 0 0 0 1
1 0 0 0 1 1 0

Thanks.

 

Yes, I have not gone through the logic but that is the right idea, except do all states in sequential order

ABCX
0000
0001
0010
0011
:
:

etc, or

ABCX
0000
0010
0100
0110
:
:
:
0001
0011
0101
0110
:
:
:
this way you don't miss any state. There should be 16 lines total for 4 variables.

 

Yeah got 16 lines. Thanks very much for your help!

 

Note that:

(BC' + B'C)X + (BC + B'C')X'

Is the expression for a 3-input XOR gate, which can be built from two 2-input XOR gates by XORing any two of the inputs and then XORing the output of that with the third input.