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fsm design in verilog

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arya.jagadeesh

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in moore type fsm

Code Verilog - [expand]
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always@(.....)
 
begin
writnig steates transition code
 
z= (state variable== some state)
end


i found there is difference if i keep assignment outside the always block.....how this is happening why signal is delayed in first case
 
Last edited by a moderator:

Hi arya

Would u plz elaborate

Thanks
 

Code Verilog - [expand]
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modulesimple (Clock, Resetn, w, z);
input Clock, Resetn, w;
outputz;
reg [2:1] y;
parameter[2:1] A = 2’b00, B = 2’b01, C = 2’b10;
// Define the sequential block
always@(negedgeResetn orposedgeClock)
begin
if (Resetn == 0) y<=A;
else
case(y)
A: if(w) y<=B;
else y<=A;
B: if(w) y<=C;
else y<=A;
C: if(w) y<=C;
else y<=A;
default:y<=2’bxx;
endcase
// z=(y==c); output at this is one clock cycle after y changes to c,z here is reg
end
// Define output
assignz=(y==C); output at this becomes one at same time as y changes to c
endmodule
 
Last edited by a moderator:

First assignment between endcase and end is a registered output.
Second assignment between end and endmodule is a combinatorial output.

What don't you understand? The behavior is correct as you stated in your original post.
 

    V

    Points: 2
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hi arya
first assignment is in always block and that's and u have used non blocking state ment that's why it is updating after one clock.. if u see synthesis it will come out from a flop
and one is continuous statement hence it will update as the transaction is done if u synthesis rtl it will be from combinational circuit..

I hope u will get ur all answer by this paragraph

Thanks and Regards
VIr_1602
 

    V

    Points: 2
    Helpful Answer Positive Rating
thank you for your reply
i have a question with registerd outputs

in general, all the assignments values are updated in the next clock?
 

Hi arya

Thats what the sequential circuit means.

Regd
Vir
 


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