Why Gray and Johnson state encoding avoid metastable state?

Hi Sir/Madam,

Can anyone share why Gray and Johnson state encoding avoid metastable state?

Thanks.

Re: Why Gray and Johnson state encoding avoid metastable sta

The idea is that with that code only 1 bit change each time and so also you're sure that when one change the other still be the same and so you don't have problem about the 2 commutations.

Example
only 4 state = stateA, stateB, stateC, stateD

stateA = "00"
stateB = "01"
stateC = "11"
stateD = "10"


In this example you can see that if the state are coded in that way and you allow only transition by one state to the only one near it only 1 bit change, so if your flip flop of hte state change in 1 clk or in the next you remain in the old state till you go in the new one.

If you code the state as

stateA = "00"
stateB = "01"
stateC = "10"
stateD = "11"

You can see that on the transition from stateB to stateC change 2 bit, but if you've metastability problem on theese flip flop they can change in different time and so you can have that from 01 you go in 11 or 00 and not in the state you wanna.

Re: Why Gray and Johnson state encoding avoid metastable sta

Can any one suggest a good book to study and learn State Machines Concept that I can use in Embedded programming using C..

It would be really helpful if this book can show some example codes so that i can understand better..


Many Thanks,

Re: Why Gray and Johnson state encoding avoid metastable sta

Here is a doc to description the SM, hope it will be helpful to you

Re: Why Gray and Johnson state encoding avoid metastable sta

Some time ago I've found this paper about Gray code and VHDL, it might be useful because point also some interesting issue about synthesis trouble.

Hope it help.

Bye
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