Updating implicit signal

[valtih1978]

In the
http://rti.etf.bg.ac.rs/rti/ri5rvl/tutorial/TUTORIAL/IEEE/HTML/1076_12.HTM#12.6.3
I read

<blockquote>
For any implicit signal S'Stable(T), the current value of the signal
(and likewise the current state of the corresponding driver) is modified
if and only if one of the following statements is true:

-- An event has occurred on S in this simulation cycle.

-- The driver of S'Stable(T) is active.

If an event has occurred on signal S, then S'Stable(T) is updated by
assigning the value FALSE to the variable representing the current value
of S'Stable(T), and the driver of S'Stable(T) is assigned the waveform
TRUE after T. Otherwise, if the driver of S'Stable(T) is active, then
S'Stable(T) is updated by assigning the current value of the driver to
the variable representing the current value of S'Stable(T). Otherwise,
neither the variable nor the driver is modified. </blockquote>

Suppose I have activity at S at 0 an 0.5 ns. The first activity makes
S'Stable(1) FALSE in (0, 1 ns) and TRUE after 1 ns on. Second activity
comes at 0.5 ns. I expect that S'Stable will be FALSE up to 1.5 ns.
However, there is S'Stable=>TRUE schedule at 1.0 ns. I do not see
anything in the LRM (or it was a Tutorial?) than cleans up this
schedule. So, when I am 1 ns since start, S'Stable(1 ns) will report
TRUE, despite event half second before.

 

[Andy]

Don't forget transport vs. inertial. Inertial is the default.

Andy

 

[valtih1978]

By default, inertial delay = transport delay = 0. I want to understand
this case first.

 

[Andy]

On Thursday, September 12, 2013 3:15:55 AM UTC-5, valtih1978 wrote:
> By default, inertial delay = transport delay = 0. I want to understand this case first.

No, the default pulse rejection limit for inertial delay is the time expression associated with the first waveform element.

Therefore:

S <= reject 0 ns inertial not S after 10 ns;

is equivalent to:

S <= transport not S after 10 ns;

but is NOT equivalent to:

S <= inertial not S after 10 ns; -- with or without "inertial"

Andy

 

[valtih1978]

I feel it is the time to unveil how is this is related with my question

 

[Andy]

On Friday, September 13, 2013 7:45:05 AM UTC-5, valtih1978 wrote:
> I feel it is the time to unveil how is this is related with my question

"You can lead a horse to water, but you cannot make it drink."

Andy

 

[valtih1978]

On 16.09.2013 22:15, Andy wrote:
> "You can lead a horse to water, but you cannot make it drink."

That is right. If you do not want to say what you mean, I cannot force you.