signales in processes VHDL

Oussama Guest 2017-10-06 14:32 #5166516

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1 entity klausurCircuit is
2 end klausurCircuit;
3
4 architecture behavior of klausurCircuit is
5 signal a,b,c,d,e,f : bit;
6 begin
7
8 process (a,b)
9 begin
10 c <= not (a and b) after 6 ns;
11 d <= not (b or c) after 5 ns;
12 end process;
13
14 process (a,c)
15 begin
16 e <= transport c xor a after 5 ns;
17 end process;
18
19 end architecture behavior;

what I don't understand is when you have in process(a,b) two signales 
what does that mean, when will the d change. Is what I made correct.

Thank you for the explanation, I am new to Vhdl just trying to 
understand how it works.

Best Regards

Attached files:

im1.png (100 KB)

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Michael W. Guest 2017-10-19 13:34 #5180624

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I think this is the inner meaning of this "Klausur". you will have to 
find out, when the signals operate and when not.

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Lothar M. (lkmiller) Moderator (Company: Titel) 2017-10-19 14:59 #5180736

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Oussama wrote:
> 8 process (a,b)
> 9 begin
> 10 c <= not (a and b) after 6 ns;
> 11 d <= not (b or c) after 5 ns;
> 12 end process;
Just as a hint: this process has an incomplete sensitivity list. 'c' is 
missing. So the solution is fairly tricky, because c changes 6ns after a 
or b has changed. But because c is missing in the sensitivity list the 
process is not evaluated again and d does not change.
I recommend to ask your teacher about that sensitivity list...

For the signal e you should calculate a xor-signal e' without delay. And 
then simply shift that e' for 5ns.

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Michael W. Guest 2017-10-19 17:56 #5180924

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Lothar M. wrote:
> Just as a hint: this process has an incomplete sensitivity list.

Maybe this is intended since sampling is exactely simulated this way:

The result is taken over only the point of time when the gated clock 
becomes active.

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Lothar M. (lkmiller) Moderator (Company: Titel) 2017-10-19 18:21 #5180960

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Mar. W. wrote:
> the gated clock
I can't see no clock of any kind in the whole design...

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Michael W. Guest 2017-10-21 11:07 #5182610

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process (a,b)
d <= not (b or c) after 5 ns;

is a method to describe a gate because c is progressed into d only when 
b is active so any changes of c at others poits of time are not 
relevant.

this way also clocks are simulated finally by converting the rising edge 
command into a signal which is active for one simulation step only

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Klakx Guest 2017-10-21 12:12 #5182681

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.. bad design practice

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Lothar M. (lkmiller) Moderator (Company: Titel) 2017-10-21 12:20 #5182687

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Klakx wrote:
> .. bad design practice
Because simulation result will not be the same like the behaviour of the 
synthesized design.
So this "cute insider trick" is ugly coding no one will understand a 
half year later. Even the coder himself. If this kind of coding is 
teached anywhere then the school and the teacher is not worth the money!

Lothar M. wrote:
> I recommend to ask your teacher about that sensitivity list...
I strongly recommend...

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