inertial delay understanding

Thread Starter

anhnha

Joined Apr 19, 2012
905
This is a definition of Inertial delay in the web: http://fpgafreak.blogspot.com/2010/01/transport-and-inertial-delay.html
Inertial delay.inertial delay is the one which gate ( Component ) have,that is if a gate is modeling then in real situation it has some delay to model that inertial delay is used. For example If you model an inertial delay of, say 20 ns, and then put a pulse of, say, 10ns, through the model, it will be "swallowed" and will not appear at the output. Because I/p pulse that do not exceed the propagation delay of the gate do not propagate to the O/P.
Inertial delay is the time it takes for a signal to change its value.
This is usually representative of capacitance.The continuous-assignment will create an inertial delay.
By default delay is inertial.
Can anyone explain more about the bold sentence?
For example, when a pulse is transmitted from A to B, inertial delay is 20ns and the pulse is 10ns. Is the conclude bellow right?
1. In order to a pulse can be transmitted to B, the pulse at A has to be remain for a certain time?
What is theory related to this problem?
2. The transmission of pulse is different from that of moving a car from A to B.
When a car move from A to B, it doesn't have any connection to A?
 

WBahn

Joined Mar 31, 2012
30,238
How did a car get involved in this?

The processing path from the input to the output always has some type of parasitics (capacitance, inductance, and resistance) associated with it and before a pulse can appear at the output, the parasitics must charge up/down and that takes time. If the pulse goes away before enough time has elapsed, the parasitics never finish changing and decay back to what they were before without the output ever changing appreciably.
 

Thread Starter

anhnha

Joined Apr 19, 2012
905
Thank you, WBahn:
I think my problem is that I don't understand how a wave(a pulse) is propagated in a wire from A to B.
Do you mean that in order to proagated a pulse in a transmission, it need to charge capacitors in the way it go along?
Does transmission wire have many capacitor in parallel and the voltage will be transmit from A to B by charging these capacitors?
If so, why a short pulse cannot transmit to output? How this related to capacitors at the
A end?
Could you explain again? I still don't really get it.
 

WBahn

Joined Mar 31, 2012
30,238
We aren't talking about a transmission wire, but rather a circuit. But let's take a simple RC circuit and see what happens.

Rich (BB code):
o-------/\/\/\/\/------+---------o
            R          |
                     __|__
Vin                  _____ C    Vout
                       |
                       |
0----------------------+---------o
This has a time constant of RC, meaning that a step change in voltage will result in an exponentially decaying change that will move ~63% of the way to the new value in one time-constant. So if you put a pulse into the input that is 1 time-constant wide, the output pulse will rise to 63% of the pulse height and then die away. If the input pulse width is only 10% of the time constant, then the output will only rise to 10% of the pulse height.

This factor is just one of several factors that is at play. Which factor dominates depends on the type of circuit involved. In the case of a digital circuit, the signal at certain points in the circuit must reach a minimum threshold, which may be more than one-half of the final signal height, before anything happens from that point on.
 
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