# TM and TE waves in coaxial waveguide

Gold Member

## Homework Statement

1)Find the EM fields that go through a coaxial waveguide (inner radius equals a, outer radius equals b) filled with a dielectric material where both TE and TM modes propagates.
2)Find the cutoff frequency.
3)Find the cutoff frequency if we close both ends of the waveguide with a conducting material, separated by a distance L.

## Homework Equations

Helmholtz equation for both TE and TM modes. Each one of these have a different boundary conditions.

## The Attempt at a Solution

1)First I want to get the explicit form for both TE and TM modes, so that I can write both ##\vec E _{\text{TM}}## and both ##\vec H _{\text{TM}}## as well as both ##\vec E _{\text{TE}}## and both ##\vec H_{\text{TE}}##, the sum of this all is the total EM field they ask for, I suppose.
So I tackled the problem by trying to find the TE modes, in which case I must solve ##(\nabla_\perp ^2 + \gamma ^2)H_z(\rho, \theta)=0## where I use cylindrical coordinates. The boundary conditions are ##\frac{\partial H_z(\rho, \theta)}{\partial \rho} \big |_S=0##. Notice that there are 2 surfaces, one for when ##\rho=a## and one for when ##\rho=b##. So I solve this Helmholtz equation via separation of variables and I got to solve the Bessel equation for the radial part/function.
Now my "problem" appears when solving this Bessel equation and applying the boundary conditions above. Namely the radial part is ##R(\rho)=A_{mn}J_m(\rho n \gamma)+B_{mn}Y_m(\rho n \gamma)##. Unfortunately when applying the boundary conditions none of ##A_{mn}## nor ##B_{mn}## must be 0... so this complicates things a lot.
Applying the boundary conditions indeed yield 2 equations from which (after eliminating both ##A_{mn}## and ##B_{mn}##), ##J'_m(an\gamma) Y'_m(bn\gamma)-Y'_m(an\gamma ) J'_m(bn\gamma )=0##. I hoped to obtain an expression for the eigenvalues of the modes, ##\gamma _{mn}## but how can I get an explicit form for them? So far I can only give an implicit form for ##\gamma_{mn}##.
Had I reached an explicit form, I would have obtained the cutoff frequency using the relation ##\omega _c=\frac{\gamma _c}{\sqrt{\mu \varepsilon}}## where this gamma_c is the smallest possible I believe.

So I'd like any comment, especially if I'm missing something or if I'm on the right track. Thanks a lot!

## Answers and Replies

You are on the right track, although I'm not sure where the "n" comes from in the arguments of your Bessel function expressions. The gamma values are entirely numerical in nature and can be plotted and found.

EDIT: Clarity

• fluidistic
Gold Member
You are on the right track, although I'm not sure where the "n" comes from in the arguments of your Bessel function expressions. The gamma values are entirely numerical in nature and can be plotted and found.

EDIT: Clarity
Whoops indeed. I get ##R(\rho )=A_nJ_n(\gamma \rho ) + B_n Y_n(\gamma \rho )##. This yields the equation ##J'_n(\gamma a ) Y'_n (\gamma b)-J'_n (\gamma b ) Y'_n (\gamma a)=0##. I realize this will hold for infinitely many positive values for gamma that I could probably determine numerically, but the problem expect me not to use any computer. So how would I continue from there?

I forgot to point out that this assumes gamma is nonzero. In the case where gamma is zero there is a nontrivial solution where both the z components of H and E are zero, this solution corresponds to what is known as the TEM mode for coaxial waveguides. There is no cutoff frequency. As you try to launch power down a coaxial waveguide this TEM mode will be the dominant excitation for this reason. The reason that the TEM mode appears is that there are multiple conductors (and hence multiple boundary conditions to be met). They are not possible, for instance, in a single conductor cylindrical waveguide with a homogeneously filled core.

For the part where the coaxial waveguide becomes enclosed on both ends (making it a cavity) you may no longer assume propagating solutions but rather standing waves. There are separate TM or TE solutions which have similar but not identical resonant frequencies (no longer cutoff). For instance now instead of exponential propagating solutions in the lateral direction you should assume sine functions.