Elementary Question of cutoff frequency in waveguide

[electro_nemo]

Got a little confused by the cutoff phenomena happened in waveguides, need your help please.
For example a rectangular waveguide has its first two cutoff freq to be TE10 ~ 6.56GHz and TE20 ~ 13.12GHz, and I feed in a wave of freq 14GHz (does this wave have to be a TE or TM wave so that it can actually propagate?) , then there should have at least two modes(TE10&TE20) co-exist inside the waveguide right? and, although the wave bears two different modes for propagation, it still maintains its own frequency, i.e. 14GHz, right? if I'm right so far, then the thing bothers me a lot is that, how can this monochromatic wave having two different wavelengths inside the waveguide? Thank you so much for your patient explanation!
:|

 

[electro_nemo]

Is this question too naive or what?
somebody help please~~
Really appreciate any of your voice

 

[biff44]

You can send frequencies that are much higher than the cuttoff frequency of the physical waveguide you are using. There can be TE and TM modes propagating. There are a number of good reasons you would not want to do that, though. For instance, lets say you actually cared about the signal you were sending through the waveguide. Lets say the waveguide cross section dimensions were big enough so that 5 different modes could propagate. So the engergy at the input to the waveguide can take 5 different electrical length paths. Lets say 40% of the engergy took mode I, and 40% took mode II, and the remaining 20% took up modes III, IV, and V. And lets say the waveguide was long enough so that the electrical length in mode I was 180 degrees different than in mode II. So at the output (assuming you could launch and capture these different modes) you would have 80% of the input energy cancelling out. You waveguide would look like it had a lot of insertion loss. If you moved the frequency a little, the mode I and mode II energy might now be in-phase, and you would then have very little insertion loss. So you would have a large amount of insertion loss ripple vs frequency.

You would have trouble figuring out how to launch the higher order modes. You would need something like a double E field probe to launch a TE20, for instance.

And if there were any internal obstruction in the waveguide, like a right angle bend, all heck would break loose as energy in one mode might hop to another mode.

 

[electro_nemo]

Thank you biff44 for your reply, however, I'm still confused about my original questions, let me put them this way and see if they now make any more sense:
1), In order to form TE and TM modes in a rectangular waveguide, do I have to feed in TE/TM waves? (since rectangular waveguides cannot support TEM wave...)
2), If multiple modes of the transmitting wave formed inside the rectangular waveguide, does it mean these different modes take different transmitting wavelengths however maintains the same frequency???
3), this question arises from your reply, do different modes cancel each other if 180 out of phase?
Thank you so much for explaining!

You can send frequencies that are much higher than the cuttoff frequency of the physical waveguide you are using. There can be TE and TM modes propagating. There are a number of good reasons you would not want to do that, though. For instance, lets say you actually cared about the signal you were sending through the waveguide. Lets say the waveguide cross section dimensions were big enough so that 5 different modes could propagate. So the engergy at the input to the waveguide can take 5 different electrical length paths. Lets say 40% of the engergy took mode I, and 40% took mode II, and the remaining 20% took up modes III, IV, and V. And lets say the waveguide was long enough so that the electrical length in mode I was 180 degrees different than in mode II. So at the output (assuming you could launch and capture these different modes) you would have 80% of the input energy cancelling out. You waveguide would look like it had a lot of insertion loss. If you moved the frequency a little, the mode I and mode II energy might now be in-phase, and you would then have very little insertion loss. So you would have a large amount of insertion loss ripple vs frequency.

You would have trouble figuring out how to launch the higher order modes. You would need something like a double E field probe to launch a TE20, for instance.

And if there were any internal obstruction in the waveguide, like a right angle bend, all heck would break loose as energy in one mode might hop to another mode.

 

[biff44]

If you WANT to launch a specific mode, you have to go out of your way to launch that mode, and only that mode, at the front end of the waveguide. At the other end of the waveguide, you do the same thing again trying to capture ONLY that one mode.

If you look at a coaxial to waveguide transition, which are normally designed to only launch the TE10 mode, there is usually an Efield probe right in the middle of the widest wall, where the TE10 Efield would be at a maximum. Such a launch would not work well for a TE20 mode, or a TM mode, etc.

Once a single mode is launched in a waveguide, if all other modes are below their cuttoff frequency, then pretty much only that one mode propagates. If there is some discontinuity, like a waveguide post, then the higher order modes generated by that perturbation die out very quickly with distance from that discontinuity. So the waveguide is "self healing", in that IF any higher order modes are generated inside, they are attenuated many tens of dB/wavelength and die out quickly.

However, IF you have an overmoded waveguide, then these higher order modes do not die out. Once they are generated, they continue to propagate with low loss/length.

The Electric fields of the various modes all combine linearly at whatever point in the waveguide you are measuring. So, yes, you can have one mode cancelling out another mode at some point in a waveguide. Just like if at low frequency you hooked up one sine wave signal to a load resistor, and then hooked another sine wave of opposite magnitude to that same load. The two signals will interfere with each other, and if they were of the same amplitude, they would exactly cancel each other at the load.

I am trying to think of an analogy. Maybe an echo is close enough. If you were standing in a big hall, and someone shouted to you, you would probably hear the shout a couple of times. First, you would hear the most direct path of the shout. A little later, you might hear that shout that traveled to you by bouncing off of the roof of the hall, which is time delayed. You hear them both where you are standing--in other words you hear them linearly superimposed on each other. But the phase (or time) shift for each path is different.