I have a basic question about the path of a signal traveling on a transmission line with a short circuit as termination.

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Suppose that this line is driven by another line (at left) with Zc = 50 Ohm characteristic impedance. The line in my scheme is matched with it but its termination is not, so there will be reflection (in this case total reflection because it is a short circuit load).

I'd say that the signal which comes from left travels along al the line in the scheme, arrives at the short circuit, and then it is reflected back.

But if you evaluate the input impedance of the transmission line and then find the correspondent reflection coefficient, you see that there is total reflection also at the beginning of the line (as shown in the scheme with the equation |Ґ(0)| = 1). From this consideration, it seems that the signal is reflected at the beginning of the line.

It's seems not correct for me since the signal knows the presence of the short circuit only when it arrives at it, but math says that |Ґ(0)| = 1. Which is the solution? And why do people often evaluates the input reflection coefficient, since the reflection will be in other positions (I think so)?

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    \$\begingroup\$ The whole idea why we call a piece of line "matched" is that it's like it wasn't there (aside from delay). So, this is to be expected. \$\endgroup\$ Commented Jan 17, 2020 at 19:56
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    \$\begingroup\$ How did you evaluate the input reflection coefficient. Show the math because that is possibly where you are mistaken. \$\endgroup\$
    – Andy aka
    Commented Jan 17, 2020 at 20:04
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    \$\begingroup\$ Why can't both be true? If it is reflected completely, then anywhere you look at the magnitude of the reflection, it'll be the same! The phase of course changes as you move down the line, which is the same thing as saying the reflection happens further away. \$\endgroup\$
    – tomnexus
    Commented Jan 17, 2020 at 20:20

2 Answers 2


There are two useful models for this situation:

  1. The input wave enters from the left, travels rightward, is reflected by the short, travels back to the left and exits via the cable.
  2. There’s a standing wave with a voltage zero at the short which extends through the entire cable.

In the linear, no loss limit these are exactly the same both mathematically and physically.

Both are useful for calculation, but you can only use one at a time. Combining them leads to confusion.

If you want to think about the conditions at different points on the line, the standing wave model is often simpler. In this case, it shows that the conditions of the short repeat every half-wavelength.

  • \$\begingroup\$ Thank you very much, what does it happen if there are losses? \$\endgroup\$
    – Kinka-Byo
    Commented Jan 18, 2020 at 8:31
  • \$\begingroup\$ Cable losses mean (1) in the waves model, the amplitude drops along both waves and therefore (2) the zeros of the standing wave are less-zero as you move away from the short. At some number of wavelengths to the left, the return wave and resulting standing wave are negligible. \$\endgroup\$ Commented Jan 18, 2020 at 16:52

You have not given any data how far from each other your drawn lines are placed - are they parallel and what are the geometric forms of the lines. Thus we have no idea what the structure between points O and L actually is.

If one assumes the lines are straight, they are parallel, they are made of lossless 50 Ohm coaxial cable and they are so near each other that the resulted structure is a shorted transmission line he can guess the characteristic impedance of the coaxial cable pair is more than 50 Ohm. The guess is based on reduced capacitance per unit length. It's a guess because I cannot calculate what's caused by the dispersion - the wave propagates partially in the insulation of the coaxial cables and partially in the free space around the coaxial cables with higher speed.

But the input impedance seen by the feeding cable (not drawn) must be zero or infinite or a pure reactance which depends on the frequency and the length L. Nothing else is possible because these are the options which can create 100% reflection.

The feeded wave reflects partially at point O and the rest is reflected from the short circuit. Because the joint at O is not matched, the reflected wave partially continues to the feeding line and partially returns to the right to be re-reflected by the short.

Your drawn structure is a resonator and at resonance frequencies your feeding line sees zero or infinite Ohm, at other frequencies there's a reactance which is non-predictable without geometric & material data and frequency. If the data were given, the characteristic impedance of your coaxial cable pair could be calculated but the calculation is beyond the scope of this answer.


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