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?
Machine of Mealy and Moore
Page 1 of 14 Replies  4933 Views  Last Post: 28 June 2010  06:09 AM
Replies To: Machine of Mealy and Moore
#2
Re: Machine of Mealy and Moore
Posted 20 June 2010  11:18 AM
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.
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
#3
Re: Machine of Mealy and Moore
Posted 25 June 2010  12:29 PM
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 DFF 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
#5
Re: Machine of Mealy and Moore
Posted 28 June 2010  06:09 AM
No problem. Just click the little plus sign if it really helped. I hope it did anyways.
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