# Wave guide modes: what are they, how are they used

1. Oct 13, 2012

I posted this in the EE forum but I think the physics forum may be more helpful.

Title pretty much says it all. In EM waveguides (rectangular, circular, what have you), I understand there are various "modes." I'm trying to get a better physical understanding of all the various parameters and their effects.

I'm gathering that these represent specific, discrete solutions to the Hemholtz equation, and therefore there are only specific frequencies that will propagate?

What happens between these frequencies?

What determines which mode you would use in a waveguide and why?

Finally, the cutoff frequency - modes with frequencies above this number are the ones that don't propagate, correct (i.e. a HPF vs an LPF)?

Thank you in advance - additionally, any general explanation of modes and their physical meaning and relevance would be greatly appreciated. My material seems to just jump right into how to determine these without bothering to explain what they are, where they come from or how they're used.

2. Oct 14, 2012

### cmos

Yes.

Sort of, but I think you might have the wrong idea. Specific frequencies will propagate, but these frequencies form a continuum ABOVE a certain cut-off frequency. The eigenvalues of the Helmholtz equation is the axial component of the wavevector (in engineering contexts this component is usually termed the propagation constant and has the symbol β). So, for a given frequency above cutoff, there exists a discrete number of possible values of β. Each of those β's correspond to the same frequency, but their transverse field profile will be different. These are your waveguide modes.

Like I said, there is a discrete number of β's for a given frequency. Guided modes simply do not exist in between those β's. If you happen to input a wave that should have a forbidden β, that wave would suffer a high reflection at the input and also scatter away.

Usually the mode shape (i.e. the transverse field profile) because that determines (via diffraction) how the field will enter/exit. I should also add that the vast majority of applications will use the fundamental mode (i.e. the lowest order mode) because it gives the most uniform profile -- it essentially gives a "spot."

No, modes with frequencies below that number do not propagate. Think of it this way: Smaller frequencies correspond to longer wavelengths. At some point, the wavelength becomes to long to "fit" inside the waveguide.

3. Aug 18, 2013

### Jakecohen7

Not clear!

I am still not clear on this subject.

What determines the specific mode that propagates thru a fiber. It must have only to do with the geometry and the difference in index of refraction between the core and cladding, but I have never found a suitable explanation!

I can imagine that the angle at which total internal reflection takes place and also the entrance Na of the beam plays a role, but I have nothing quantitative (like a formula) that explains for a given geometry this spatial mode will exit.

Any help is appreciated.

4. Aug 18, 2013

### Born2bwire

I'd like to qualify cmos' comments because his talk about the \beta s may be a bit confusing in light of what he states regarding the frequencies.

When you solve for the modes of a waveguide, you are solving for a set of field profiles that match the boundary conditions associated with the waveguide. Take the case of two infinite parallel plate waveguides. At the plates, the tangential electric field must be zero. Assuming that the direction of the guided mode is z and the plates lie in the x-z plane, then the for the TE modes, the electric field, which is only in the x direction, must be zero at the plates. So we can imagine that the profile of the electric field in the y-direction will be sinusoidal so that the field is zero at y=0 and y=a where the plates lie. But a sine function has an infinite, but discrete number of periods that can satisfy this boundary condition that will be set by the propagation constant in the y direction, our \beta_y.

Each mode is going to be associated with one unique profile dictated by the \beta_y. One mode is going to be half a sinewave (one peak). The next mode will be a full period (one peak, one trough), and the next mode will be 1.5 periods, etc. Each of these modes has a minimum frequency that you have to meet to have a wave meet these boundary conditions. However, let's say that we have a mode, which may be the TE_{10} mode. This mode will support a wave that has any frequency above its cutoff. What happens is that the \beta_y component (which in this case defines the mode) still stays the same but the \beta_z component, the wavenumber in the guided direction, will lengthen so that the total wavenumber corresponds to the wavenumber at your given frequency. So the angle of the wave as it strikes the wall changes as we change the frequency in such a way that allows us to still satisfy the boundary conditions. You can imagine that as the frequency decreases, the wave will have to strike the wall closer and closer to the normal to satisfy the boundary conditions. At some point, the we hit a frequency where the wave has to travel normal to the wall and this is the cutoff frequency (since we no longer have any propagation in the guided direction). So in this way we have a continuum of solutions above a cut off frequency.

However, at a given frequency, there will only be a discrete number of modes that will satisfy the boundary conditions (provided we are above the lowest cutoff frequency of any mode). So at some frequency y, we may be able to have the TE_{10}, TM_{10} and TE_{20} modes if we are at a frequency that is above all their cutoffs but below the cutoffs of the higher order modes.

5. Aug 19, 2013

### ZapperZ

Staff Emeritus
The waveguide modes are dictated by what they will be used for. When you know what you want to use it for, you then design a waveguide with the correct geometry for the boundary conditions.

In many accelerating structures in a particle accelerator, the TM01 mode is often used for a cylindrical/pillbox cavity. This is because the E-field is pointed axially along the direction where you want to propagate the charged particles, and that the peak field is exactly along the axis.

Zz.