Thought experiment: Waveguide with lossless materials - Page 1
Forum section: electronics / rf-microwave

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: October 02, 2018, 06:18:32 am
content: So here is a little though experiment that is going around the office here to 'challenge' your understanding of EM.Imagine we build a rectangular waveguide with perfect electric conductors as walls, and lossless dielectric inside. This waveguide, as all rectangular waveguides, has a lower cut-off frequency below which no propagating modes exist.Now imagine I launch a wave into this waveguide (how does not matter - just know we manage to get energy into the waveguide). What happens to this wave? We know that the math says that we can't have propagation below cut-off, but where does this power we put in go?

author: IconicPCB
author_rank: Super Contributor
timestamp: October 02, 2018, 06:59:58 am
content: holly structural return loss Batman!!!

author: MiDi
author_rank: Frequent Contributor
timestamp: October 02, 2018, 08:54:44 am
content: Nice thought experiment.I am not able to translate what IconicPCB means, but I try with an analogy:There is a glass front and you can look through (transmissive for light).But you cannot go through (non transmissive for matter) and when you push hard enough the glass breaks and you can go through.Same applies for the waveguide: you would not get energy into this waveguide until it breaks.

author: richard.cs
author_rank: Super Contributor
timestamp: October 02, 2018, 11:03:03 am
content: Any energy you launch at it reflects back to the source / you can't launch energy down it, are essentially equivalent answers.If you want to get into the real detail what you actually get in a cutoff waveguide is an evanescent mode, and if the waveguide is not infinitely long you get non-zero energy coupled out of the far end. A cutoff waveguide isn't a brick wall, the field decays exponentially with length and if the waveguide opens out again such that a propagating wave can be supported then some energy is always coupled into the propagating mode the other side of the cutoff section, but usually a negligible amount by the time the cutoff section is a few wavelengths long. It's essentially the same maths and physics as quantum tunnelling.

author: coppercone2
author_rank: Super Contributor
timestamp: October 02, 2018, 01:10:49 pm
content: in a normal waveguide, if there is energy going through it, does the wave exert some kind of miniscule force on the waveguide walls? Like do the H/E fields cause pressure on the waveguide walls?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 02, 2018, 04:55:04 pm
content: You always have a source and a load. Yet your problem neglects to define them. So which is it?The QM version of this problem is, a "particle in a box" has a wave function that decays exponentially in the (finite) potential of the box walls; and if another box is beside the one, the wave "tunnels" through, having nonzero probability on the other side. Which, I don't remember offhand but presume -- if the boxes are slightly mismatched in size or potential, the particle would tend to oscillate back and forth between them, just as energy "sloshes" back and forth between two coupled, differently-tuned oscillators (of whatever sort, LC, pendulum, etc.).The practical version of this problem is, superconducting waveguides and resonators still have finite Q, of course; it's quite high, 10^7 or thereabouts (for properly prepared Nb at 4 K), but not infinite. And there is still the matter of coupling the source and load to such a waveguide, which in their application (particle accelerators), is probably pretty weak (a few microamperes isn't much space charge, and therefore beam-induced conductance either), hence the high Q requirement, but also the high power demand (usually ~10^5 W from pulsed klystrons, AFAIK).Come to think of it, due to exactly that factor -- the high resonator Q and the light loading from the beam -- they probably make extensive use of below-cutoff waveguide, as a means of setting that coupling factor just as needed.Tim

author: Wolfgang
author_rank: Super Contributor
timestamp: October 02, 2018, 05:14:52 pm
content: I try to give a "pictorial" answer.Say you have waveguide and with an exciting antenna right at the entry hole.You assume that the waveguide losses are zero.Now you excite this antenna with a signal of a given frequency and solve Maxwells equations for all this arrangement.Two possibilities exist:- The excitation frequency is below the cutoff frequency of the waveguide. Then you have an asymptotic damping of the field inside the waveguide, surprisingly *not* dominated by finite wall conductivity and or other losses.If you want, this is just a "near field" of the antenna, but no propagation. All energy the antenna radiates into the waveguide is reflected.- If your frequency is above the cutoff, Maxwells equations have a solution that propagate losslessly (we assumed a lossless waveguide) along the waveguide forever.

author: coppercone2
author_rank: Super Contributor
timestamp: October 02, 2018, 06:24:38 pm
content: I am kind of lost in these explanations.Is shining a laser light, or just general light, into a waveguide some how related to this?is this like a zone of operation that has some kind of frequency range between the highest specified mode of operation of the waveguide and some kind of frequency thats mathematically related (i.e. 100 times the waveguide cutoff?)Like if you shoot a laser beam into a waveguide, or even a really really high THz beam, doesn't it act just like a laser beam in a mirrored chamber? I thought this had something to do with debrogalie wavelength. If the wavelength is short enough doesn't it just ignore the waveguide and treat everything as far field reflection?Like if its a long strait waveguide, a laser will go right through it and not care about anything, so long it does not bend, and if it does bend, its gonna depend on the surface finish and angles at which it hits (so you can have it propagate like a periscope if the angle is right, or if its some weird internal geometry it could start bouncing all over the placed and essentially have a 'mirrored room' effect? and how about really high frequency, like a x-ray, that will just barely even interact with the waveguide material because of the frequency, so it will mainly continue through on its initial vector.

author: Wolfgang
author_rank: Super Contributor
timestamp: October 02, 2018, 06:45:01 pm
content: Yes, shining a laser into a waveguide is related. Laser ist just electromagnet radiation, as radio waves.Now think of a laser of a given wavelength, and a waveguide. Now you make the waveguide smaller and smaller.Light can still be coupled (i.e., you can shine thru) in the waveguide *until* the waveguide gets so small that no half wavelengths of the light fits into the waveguide anymore. Then the transparency is gone.Read this:https://www.radio-electronics.com/info/antennas/waveguide/cutoff-frequency.php

author: Marco
author_rank: Super Contributor
timestamp: October 02, 2018, 07:04:13 pm
content: Quote from: TheUnnamedNewbie on October 02, 2018, 06:18:32 amIt stays put, it can't move but it can still oscillate as a standing\evanescent wave.

author: coppercone2
author_rank: Super Contributor
timestamp: October 02, 2018, 07:26:24 pm
content: lol, I now understand, the thread is asking the lower cutoff frequency, not the specified upper cut off frequency (which is specified by the manufacturer as the practical limit of propagation in a mode useful towards the couplers). Thats why it did not make sense, when its too small rather then too big you are talking about the stuff I am just coming to understand with evanescent waves.They are specified as having a band pass response, but described in literature mostly as having a high pass response. So the question is what happens to a photon when its put into an area smaller then its dimensions or something like that. Conceptually I initially imagine it as being some how squished into the tube.Can you still think of the wave, if time is frozen, as having some kind of dimensions, when its stuck in the small waveguide? You can't define it as a electromagnetic wave anymore, its something else that fits in that structure some how? is there some kind of 'strain force'? in its being squished ? It's no longer in its natural state and it took some kind of weird effect to get it in there in the first place, probably with alot of energy, so it must have a desire to revert back to normal...?is it different if the waveguide is really long or really short (i.e. its being asked to go through a several atom thin rectangular hole in a sheet structure rather then a long rectangular tube?)Is there some mechanism by which they change wavelength to escape? what would be a theoretical way to get it into the wave guide in the first place, some kind of warping of space time to make some kinda tunnel that leads it in there in the first place?I am having a hard time imagining it not requiring energy expenditure to squish it in there, since there must be some kind of exclusion principle (like pauli exclusion).Is there a mathematical function to describe its topology as its squeezed in there some how? i think the word would be convolution ? Like if you squeeze an egg into a narrow bottle it kinda deforms in this mushroom shape as it goes through. When I think of a tunnel I think of rounded edges and some kind of nonlinear transition into a linear tunnel zone. Like in terms of a Cartesian coordinate grid shrinking, otherwise there is some kind of cusp., if you conserve the line count, otherwise you need more lines that branch out from other lines or just appear.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: October 03, 2018, 05:28:56 am
content: Quote from: coppercone2 on October 02, 2018, 07:26:24 pmI seem to recall explaining to you before that waveguide modes are all high-pass, but that the specified band is the single-mode band. Here:https://www.eevblog.com/forum/rf-microwave/waveguide-high-frequency-(past-rating)-behavior/msg1610530/#msg1610530I'm still not confident as to anyone having the right answer (I don't know myself, we are arguing about this in the office still). The reason I am confused is the following: Say we have a transition from a waveguide with lower cutoff \$f_1\$ to a waveguide with cuttoff \$f_2 > f_1\$. I now launch a nice propagating mode into waveguide 1 at a frequency of \$f_w\$, with \$f_2 > f_w > f_1\$. There will be a very large amount of reflection, but still, some energy will couple into waveguide 2. So we are putting energy into the waveuigde, but where does this energy go? Since the waveguide is lossless, it can't be absorbed by the dielectric is dissipated as resistance.The best I can come up with right now is that you don't get a strict propagating mode anymore, but your wave guide just starts to act like a (hollow) single conductor, where you have all the energy inside some kind of surface mode? Even if you then have a finite-length wave guide, only part of the energy 'comes out' the other end. So where does the rest of this energy go?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: October 03, 2018, 07:38:14 am
content: 1. What is the source? Are we assuming an ordinary, well-behaved, impedance-matched source?2. What is the load? Are we assuming an infinite waveguide, or anything that looks like that (e.g., a very long, lossy (but nonreflective) waveguide, or a matched terminator)?3. What is the length of the below-cutoff section (in terms of electrical length)?4. Are we speaking in terms of transient events (in which case, we're talking about the energy of a wavefront launched into the structure, but we're also at the same time necessarily talking about a wideband event), or AC steady state (in which case we should be speaking of power and transmission/reflection only, not energy)?I'm not entirely sure that length matters, actually. (You probably know better than I do, you've had higher level E&M and more recently than I!)If 1 and 2 are true, then the answer is simple: what doesn't "tunnel" through the below-cutoff section, is transmitted; the rest is reflected back at the source. What is transmitted, is just power like any other, but it won't be at the same level as if the source were tied to it directly.So, obviously, the impedance seen by the source is wrong.If we insert an antenna tuner (lossless as well), forcing the source to a match, then necessarily the full source power must be transmitted. Can't go anywhere else, it's getting to the output somehow or another. The fields in the front end may be... interesting, depending on how much we've blocked by the sub-cutoff section, and depending on how exactly the coupling element is made, and probably having an interesting frequency response (i.e., we're basically constructing a part of a filter here).The conclusion should be easy to figure out, given suitably specific setup to the problem; it sounds to me more like a matter of lacking definitions, so perhaps your office should discuss what information is missing, first?Tim

author: ejeffrey
author_rank: Super Contributor
timestamp: October 03, 2018, 04:25:11 pm
content: Quote from: TheUnnamedNewbie on October 02, 2018, 06:18:32 amThis is basically "what happens when an unstoppable force meets an immovable object" problem. You can't launch a wave into a waveguide below cutoff, so you are basically saying "suppose I set up initial conditions that violate my specified boundary conditions, what happens next".If you ask a well posed question, you can get a well posed answer.In most cases, the answer will be that whatever source you try to put in the cavity, whether a feedline or an atom in an excited state (think like in a laser or maser), it won't be able to radiate. There will be a local evanescent wave in the vicinity of the source, but it won't have net power transfer. The field amplitudes will fall of exponentially from the source. From the perspective of a feedline, the antenna impedance will be purely imaginary. In the case of an excited atom you get what is called the Purcell effect: suppression of spontaneous emission by eliminating possible decay modes.Stepping down from a larger waveguide is just a special case of a feedline. The reflection coefficient from the discontinuity is magnitude 1. It will have some phase shift caused by the evanescent wave (i.e., stored energy in the neighborhood of the junction), but the power transfer is zero.

author: rfeecs
author_rank: Frequent Contributor
timestamp: October 03, 2018, 09:40:30 pm
content: Quote from: TheUnnamedNewbie on October 03, 2018, 05:28:56 amNowhere. It just sits there.You are putting no net energy into the waveguide. The same amount of energy that you are putting in is reflected back out the input.You have a purely reactive load. If you have a generator that drives a purely reactive load, where does the energy go? Same thing.I'm saying what has already been said. You got a good answer by reply #3.You can think of the lowest mode of a rectangular waveguide as a plane wave that is bouncing back and forth between two opposite walls of the waveguide. As the angle of reflection goes away from perpendicular to the walls, the resulting wave has a net velocity down the waveguide. As the frequency increases above cutoff, the angle has to go up to maintain the boundary condition of a short at each wall. The cutoff frequency is where the angle of reflection is perpendicular, and the wave just bounces back and forth and does not propagate.Of course when you say perfect conductors and no loss, if you can launch a wave into the cutoff condition exactly, then the wave will just build up larger and larger to infinity.You can make some analogy to a generator and/or transmission line that is terminated in an ideal capacitor or inductor or perfectly conducting lossless resonator.

author: richard.cs
author_rank: Super Contributor
timestamp: October 04, 2018, 08:50:21 am
content: If you had a magic box that could force energy into the waveguide, then what it would be really doing is forcing an ever-increasing E field at the entrance, and therefore storing an increasing amount of energy in the evanescent fields inside the waveguide, these extend down the whole length but decay exponentially so the vast majority of that energy is stored in the first few wavelengths. If you stopped increasing the external E field then no more energy would flow in, and if you took it away that stored energy would radiate from the open end.If you keep increasing the external E field forever with your magic box you will be storing more energy in the evanescent field, until one of the following happens:1) Subject to this enormous field strength something (air in the waveguide?) starts behaving non-linearly, coupling energy to higher frequencies which are above cutoff, energy then propogates away and some equilibrium is reached.2) As the field builds up it increases proportionally down the whole length so if the length is finite the "tunnelling" energy eventually becomes significant, and when that equals the energy input then again your reach equilibrium.3) Anything else is basically the question "what happens if I attempt to store near-infinite energy in the first few wavelengths of the cutoff guide?" Most likely the E field gets high enough to destroy everything and you may discover exciting new physics whilst being blasted with exotic particles.