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I'm facing a doubt in transient simulation in ADS.
In many setups I've seen placing a resistor in series to the transient voltage step generator. Placing the series resistor, and simulating the below circuit, I obtain the plots which follow. enter image description here enter image description here

My goal is to simulate just the reflection between a step generator and a matched transmission line. I don't understand if the step generator already includes a series resistor in the ADS component (I guess not, and this is the reason why we place the 50 ohm resistor). However, what I don't understand is that at the node 'bbb' the voltage is 0.5 V (as it can be seen from the plot). Since no reflection should occur, should not the node 'bbb' be equal to 1 V? I'm thinking about a practical setup: if we set the signal generator with impedance 50 ohm and next in series we place a matched load (50 ohm), should not be the node 'bbb' at the same voltage of the signal generator (no reflections i.e. no variations)?

Thanks in advance!

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However, what I don't understand is that at the node 'bbb' the voltage is 0.5 V (as it can be seen from the plot).

NB: the step generator in the simulator does "never" include internal R.

There is no reflection in your system because all is matched.

And you have 1/2 volt (Vg = 1 V) because the emission coefficient, at the first input line, is \$Ke=0.5= Vg * Zo/(Zo+50)\$.

NB: the Ke (coefficient emission), as its name says, is the factor that says what part of the voltage generator is "injected" into the transmission line which follows.
So, one must consider that the first line is "matched" at output ... and it follows that its input impedance is considered as Zo= 50 Ohm. You replace the line and all at the right with 50 Ohm. Finally you have only the generator, the internal impedance of the generator, and at last ... 50 Ohm load.
The voltage (which will go into the line ...) on this load is \$ Vg * 50/(50+50)=0.5 V \$ which is also Ke (by definition).

Some other coefficients can also be defined:
Ke (emission coefficient at the generator)
Kr,g (reflection coefficient at the generator)
Kr,l (reflection coefficient at the load)
Kt 1,2 (transmission coefficient from line 1 to line 2, different impedances)
Kt 2,1 (transmission coefficient from line 2 to line 1, different impedances)

Here is an example of the point (check) in the middle of the two lines.
Check the voltage of 1.6 V (NB: Vg = 2 V).

enter image description here

The real generator does "always" include a 50 Ohm (sometimes, 75 ... 600 Ohm), and it must be taken into account (even if you don't materially see it, don't forget it) when calculating Ke.

Here is also 2 examples of waves when the impedances are not matched.
Note that the voltage can be as higher as 2 x Vg (=2 V) = ~ 4 V.

enter image description here

enter image description here

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  • \$\begingroup\$ So, if I've unerstood correctly, it's correct to place a resistor in series to the step generator in order to emulate a real signal generator (please correct me if I'm wrong). Now, my doubt is, if we've emulated correctly a real signal generator, should not be the node 'bbb' equal to 1 V since everything is matched? Could you give me more reference for the emission coefficient concept? \$\endgroup\$
    – Anice41
    Jun 4, 2022 at 11:33
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    \$\begingroup\$ Correct for real generator. No, the voltage is not 1 V because it must match the DC voltage when reflections are ended ... and this circuit ends in a generator with 50 Ohm AND a 50 Ohm load ... (lines are disappeared in this DC case). So voltage is half the true generator voltage (DC voltage). \$\endgroup\$
    – Antonio51
    Jun 4, 2022 at 11:44
  • \$\begingroup\$ I've still one doubt. Imagine in a practical setup to place in series to the signal generator (already including internally the 50 ohm resistor) a matched resistor (50 ohm). If we measure the voltage at the node between the signal generator and the matched resistor, will not be 1 V? I agree that this doesn't agree with the DC analysis, where the voltage between 2 identical resistor will be equal to half of the supply voltage. \$\endgroup\$
    – Anice41
    Jun 4, 2022 at 11:58
  • \$\begingroup\$ If you place another 50 Ohm in series with the generator (the generator has already its own 50 Ohm inside), the Ke will be 50 (Zo of the line) divided by (50 generator+50 added +Zo line) -> Ke=50/150=0.333........ If Vg=1 V -> Ke*Vg=0.333... V. \$\endgroup\$
    – Antonio51
    Jun 4, 2022 at 12:02
  • \$\begingroup\$ @Anice41 you have the simulator set up. Try some things out. For example, increase the load termination to 1000 ohms, or decrease the source termination to 1 ohm. What happens in those cases? Can you get close to the full 1 V across bbb? But what is the trade-off? \$\endgroup\$
    – The Photon
    Jun 4, 2022 at 14:53

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