Sequential synchronous circuit exercise

Thread Starter

SpoNNw

Joined May 25, 2020
4
I need some help in understanding an exercise.

Here I have a State Diagram (Moore Machine), and the exercise asks me to write the table of transitions between states and machine outputs.

I can understand almost everything, but something is confusing me, the outputs expression, I have no idea how to determine what is the output value by those expressions. As I understand I have 2 Inputs A and B right? But I'm assuming that I can only have one value for my input like for example, in STATE S2 we can see (A⋅B') so for me That means (A )A=1 AND (B') B=0 so my OUTPUT Value should = 0 (1⋅0=0), in S1 we got A' that means my OUTPUT value should be 0, and so on... but I'm not shore if I am thinking correctly, if someone could explain me what I am doing wrong pls tell me.



1623340404500.png
* I hope I did design the table correctly, if I don't pls let me know, because other question is how can e represent the 3rd transition of S2 (A') with this table.

Normally in this type exercises the INPUTS are represented by 0's or 1's like this:
1623341059044.png

But as you can see in the exercise it is represented by expressions, while I don't realize it I can't determine the next state in the transition table.
 

Papabravo

Joined Feb 24, 2006
21,228
In a Moore Machine, the outputs are ONLY a function of the state you are in. In particular, they do not depend on how you got there.
In a Mealy Machine, the outputs are a function of how you get to a particular state, That is they are a function of which transition arc takes you from state A to state B and not on which of those two states you are in.

Does that help?
 

Thread Starter

SpoNNw

Joined May 25, 2020
4
[QUOTE = "Papabravo, postagem: 1639306, membro: 5202"]
Em uma Máquina Moore, as saídas são APENAS uma função do estado em que você se encontra. Em particular, elas não dependem de como você chegou lá.
Em uma Mealy Machine, as saídas são uma função de como você chega a um determinado estado. Ou seja, são uma função de qual arco de transição o leva do estado A para o estado B e não em qual desses dois estados você está.

Isso ajuda?
[/CITAR]
Obrigado pela resposta, mas não me ajudou muito, minhas dúvidas permanecem
In a Moore Machine, the outputs are ONLY a function of the state you are in. In particular, they do not depend on how you got there.
In a Mealy Machine, the outputs are a function of how you get to a particular state, That is they are a function of which transition arc takes you from state A to state B and not on which of those two states you are in.

Does that help?
Thanks for the answer and try to help me, but it didn't help me much, my doubts remain :(
 

Papabravo

Joined Feb 24, 2006
21,228
When I say that the outputs are ONLY a function of the state you are in: what that means is that I can take the bits which "define the state I am in", apply them to a combinatorial circuit that implements any arbitrary Boolean function and produce each and every output bit.

When I say that the outputs are a function of the inputs, AND the state I am coming from, AND the state I am going to (the arc that show a transition from one state to another), you can see that you have a much more complicated function with more inputs that are used to determine the outputs.

It does not matter which approach you take. They are both equivalent machines and can produce identical behaviors. the proof of this is complicated and involved and may be beyond your capabilities at the moment. What you need to do is be comfortable describing the behavior as it is represented in a state machine diagram or a state transition matrix.

Would it make more sense if I drew an actual state machine circuit with gates, and clocks, and registers?
 
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