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I have a question concerning reflections due to impedance mismatching. Now, assuming a lossless TL, the latter is modelled by sections of lumped L-C segments where L in a series (sub)element of the transmission line and the capacitors connect to GND. I understand that this models the two wires of the TL and mathematically leads to the characteristic impedance Sqrt(L/C) used to calculate reflections once source and load impedances are also known.. Now to my question.. I know how to compute the reflection coefficient when source and load impedances are given. But how do i calculate the reflection introduced by a series capacitance in the TL, right between the source and the TL? (Or between TL and the load). I assume i have to include the series C into source or load, respectively...but how to deal then (for example) when there are two TLs in series and the series C is right between those?

Thanks a lot for explanations ;)

Removed EDIT to a new question

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  • \$\begingroup\$ This is exactly the kind of question a Smith chart is meant to solve. \$\endgroup\$ – The Photon Apr 12 '16 at 21:00
  • \$\begingroup\$ Oh... That's right.. I will try this one tomorrow. \$\endgroup\$ – Junius Apr 12 '16 at 21:05
  • \$\begingroup\$ Ask your new question as a separate question, it's a different and interesting one, you might get some good answers. \$\endgroup\$ – tomnexus Apr 28 '16 at 1:23
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You can solve any simple TL problem like this :

  1. start at the load, and work towards the generator.
  2. take the end load Z (if it's a reactance, convert to Z using frequency)
  3. Transform the impedance down the line, either using a Smith chart, the TL equation or the lossless TL equation.
  4. If you have a series capacitor in the middle of a line, treat it as two lines. Calculate Z up to the cap, then add 1/jwC, then carry on.
  5. If you're looking for reflection coefficient of the whole complex circuit, Load, TL, capacitor, TL, then just convert back from impedance.
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  • \$\begingroup\$ Thanks for your comment, too! Assume i start from from the load and go until the point right before the source... This is the reflecrion coefficient assuming a 50ohm source, right(or a source of the value to which i normalised the end load when starting in the smith chart, right?) So how o get the reflection coefficient for another source impedance? How to "re-normalize then?) \$\endgroup\$ – Junius Apr 13 '16 at 5:22
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    \$\begingroup\$ In your problem, the source impedance is not part of the circuit, it's to the left of the accumulated calculation. Γ = (ZL - ZS)/(ZL + ZS) where ZL is the load impedance and ZS is the source impedance. So you can calculate Γ for any source impedance. \$\endgroup\$ – tomnexus Apr 13 '16 at 5:26
  • \$\begingroup\$ Ok, but the reflection coefficient cannot be independant of the source impedance? So if i am at that point where the source is to the left and i transformed anything from the end load to this point... This is then the reflection coefficient of.. what? ;-) \$\endgroup\$ – Junius Apr 13 '16 at 5:32
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    \$\begingroup\$ Source impedance is ZS in the equation, it's vital for calculating reflection coefficient. I just meant that it's not part of the right hand side of the circuit. So you compute the load impedance, as transformed by the lines, and then test it with the source impedance. So you draw a line through the circuit, and calculate reflection coefficient there. \$\endgroup\$ – tomnexus Apr 13 '16 at 10:47
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The easiest way is to use any Spice simulator. It has "ideal transmission line" element. You can learn everything about this circuit using "frequency domain analysis".

Alternatively, you need to count for reactive impedance of the capacitor (1/(i*2*pifC)). There are two waves for each frequency: one running from the source to the load and another one running opposite direction. Make the amplitude and phase of the second (opposite) way variables and solve for correct voltage and current on the reactive load.

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  • \$\begingroup\$ Well i know i could simulate but i wanted an analytical solution... And i think it should be solvable without considering the two "partial" waves (forward and reflected)... Also, to be correct, the incident wave is not only once reflected at the load (or any impedance discontinity), but multiple times... Until it has been absorbed "completely" by the system \$\endgroup\$ – Junius Apr 12 '16 at 20:52
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The real transmission line (50 Ohm cable, or PCB line...) does not have discrete capacitors and inductors. So the waves are reflected from the ends only.

The line made of discrete components is a low pass filter with multiple degrees of freedom. The analytic solution becomes a nightmare in this case.

The approach of two waves is the simplest one, unless you do not find a ready solution for reactive load.

Oh, may be I misunderstood your sentence about multiple reflections. Yes, sure the return wave reflects from "source" until it is not terminated exactly to RL (50 Ohms...). However, the system of equations takes care of these reflections.

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  • \$\begingroup\$ Yeah, you are right of course. Never mind my sentence about "multiple reflections". This was both, a thinking mistake and also a bad verbal explanation of the thought i had :D EDIT: i know that the lumped element version is not easy to consider, i did not want to do this anyways. I just thought about an " easy" solution for the case where a series capacitance is in between two transmission lines... \$\endgroup\$ – Junius Apr 12 '16 at 21:16
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Your particular question is begging to be solved using two port network theory and admittance and/or scattering parameters. An ideal lossless transmission line typically is analyzed analytically like this.

I must note that the partial differential equations render to the wave equation in the case of a distributed LRC circuit, so you see wave like propagation. This matches an ideal transmission line. Those same PDE's when using lumped values gives a very different result.

In SPICE an ideal transmission line is modelled very differently than a lossy transmission line. The lossy is modelled as lumped elements.

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