[Apprentice123]

As from a Mealy machine to determine an equivalent Moore machine? And unlike?

The Mealy machine:

q0 -> (b,0) -> q0
q0 -> (a,1) -> q1
q1 -> (b,0) -> q1
q1 -> (a,1) -> q2
q2 -> (a,1) -> q1
q2 -> (b,0) -> q2

q2 is final state

How do I find an equivalent Moore machine?

[mojo666]

I think you can just make copies of states for each transition. For example, we have these two groups.

q0 -> (a,1) -> q1
q2 -> (a,1) -> q1

q1 -> (b,0) -> q1

These are all transitions to q1. The top group outputs 1 and the bottom group outputs 0. So we turn q1 into two states (q11 and q10). Thus, our transitions are now

q0 -> (a) -> q11
q2 -> (a) -> q11

q1 -> (b ) -> q10

However, q1 is no longer a valid source state so we have to make two copies of each transition with q1 as its source and replace q1 with q11 and q10.

q0 -> (b,0) -> q0
q0 -> (a) -> q11
q10-> (b ) -> q11
q11-> (b ) -> q11
q11-> (a,1) -> q2
q10-> (a,1) -> q2
q2 -> (a) -> q10
q2 -> (b,0) -> q2

Now do the same with q2 and q0 and you should be close to the solution. I'm not sure how to handle start and final states.

This post has been edited by mojo666: 20 June 2010 - 11:19 AM

[NeoTifa]

A Mealy machine I do believe is a function of input and states based on the flip flops being used and the way they're wired. A Moore machine depends solely on states, so in order to get a sort of equivilent machine you'd have to rethink your function equation and put them in terms with your projected characteristic table. A D-FF usually has the S'-R' latch attached to initialize the states to zero or such, so the use of flip flops and logic gates through your circuit should be arranged to reflect your task. Wait, you're talking about registers right? XD

[Apprentice123]

Thanks

[NeoTifa]

No problem. Just click the little plus sign if it really helped. I hope it did anyways.