in "Lessons in Electric Circuits", VOL III, ChPT 14, all the possible configurations of standing wave for a certain transmission line are shown. For instance, this is the case of a transmission line left open at the ends:

enter image description here

According to the operating frequency and the transmission line characteristics, there may be:

enter image description here

My question is: do these different configurations of standing waves exist simultaneusly (i.e. the total voltage and current waveforms are their superposition) or does each of them exclude each other?

From the author's analysis, it looks like there is only one of these configurations, at a certain frequency. But I've seen many similar situations in which there is a superposition of standing waves. For instance, I'm thinking at the analogy with a rectangular waveguide:

enter image description here

In this situation there are not V and I, but only E and H fields, but the situation is similar: the metallic surfaces of the shell's lateral wall set the condition $$E = 0$$ (as well as I = 0 or V = 0 are set respectively by the open circuit or the short circuit at the end of a transmission line), which may determine the standing waves shown in figure.

For this structure, I've always been told that all these modes exist simultaneously: when a source is put inside the waveguide, it excites all propagation modes, some of which are above cut-off (and will propagate) and some other which are below cut-off (and will be attenuated). So, according to the operating frequence, multiple standing waves may exist.

Is there a similar situation, with cut-off standing waves frequencies, for transmission lines?

  • \$\begingroup\$ You can clearly see in your waveguide drawings that the waves satisfying the criteria are of different wavelength and thus different frequency. Absent a non-linearity they are present or not wholly independently of each other. \$\endgroup\$ Jan 2, 2021 at 17:52

1 Answer 1


Exciting each resonance requires a different signal frequency.

Suppose the first resonance occurs at 50 MHz. Then the 2nd one will be at 100 MHz, the 3rd at 150 MHz, and so on.

If you excite the line with all these frequencies simultaneously, then you can produce a superposition of all the different standing wave patterns at the same time.

If you excite the line with only one frequency at a time, then you'll only excite one resonance at a time.

is it the same for the waveguide in the second picture? Or is it how I've always been told ("Given an excitation frequency, there will be all resonance modes above cut-off?")?

The two situations are different. In the rectangular waveguide, if you express the modes as a combination of plane waves, each plane wave component has a momentum vector

$$\vec{k} = k_x \hat{x} + k_y \hat{y} + k_z \hat{z}, $$


$$\left|\vec{k}\right| = \frac{2\pi f}{c}$$

and the boundary conditions for each mode are

$$ k_x = \frac{n}{2}\frac{2\pi}{w_x}, n=1,2,3,..$$ $$ k_y = \frac{m}{2}\frac{2\pi}{w_y}, m=1,2,3,..$$

where \$w_x\$ and \$w_y\$ are the transverse dimensions of the waveguide.

Notice there is no restriction on \$k_z\$. That means for any \$f\$ above a given mode's cut-off we can find a \$k_z\$ that allows \$k_x\$ and \$k_y\$ to "fit" in the transverse dimensions and thus allows the mode to propagate.

In the longitudinal resonance case, you're already assuming a particular transverse mode, so \$k_x\$ and \$k_y\$ are fixed for any particular \$f\$ and there's no such additional degree of freedom available to allow \$k_z\$ (aka \$\beta\$) to vary to obtain a resonance except at the specific wavelengths where

$$\frac{2\pi f}{c}=k_z=\frac{n}{2}\frac{2\pi}{\ell}$$

  • \$\begingroup\$ About your last statement, I seem to understand that if I excite the line with 150MHz, I'll see only the 3rd resonance, and not also the 1st and 2nd. Just a curiosity: is it the same for the waveguide in the second picture? Or is it how I've always been told ("Given an excitation frequency, there will be all resonance modes above cut-off?")? \$\endgroup\$
    – Kinka-Byo
    Jan 2, 2021 at 6:47
  • \$\begingroup\$ A last question: in case of an open transmission line, the open circuit sets the last condition you have written for Kz. This condition identifies all the modes we were talking about with specific frequencies (f1, f2, f3...). What if the source frequency f is between two of them (for instance f1 < f < f2)? We may see from the pictures I put in the question, that there will be the standing wave corresponding to f1. So, the voltage frequency at the source terminals is f, while that's on the transmission line is f1, correct? \$\endgroup\$
    – Kinka-Byo
    Jan 3, 2021 at 13:09
  • 1
    \$\begingroup\$ No, if the source generates frequency \$f\$, then \$f\$ is what will be on the line. There wil still be an antinode at the open termination (and a node of the current waveform) but the pattern on the rest of the line will be changing rapidly. The source will see an equivalent load that is capacitive or inductive rather than short or open, and the energy stored in the line will be fluctuating rapidly. \$\endgroup\$
    – The Photon
    Jan 3, 2021 at 16:38
  • \$\begingroup\$ Perfect, and will the voltage and current waveforms be always standing waves also in this case? \$\endgroup\$
    – Kinka-Byo
    Jan 3, 2021 at 17:02
  • 1
    \$\begingroup\$ @Kinka-Byo, they won't, they'll be messy. \$\endgroup\$
    – The Photon
    Jan 3, 2021 at 17:21

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