# TM and TE waves in coaxial waveguide

• fluidistic
In summary, the EM fields that go through a coaxial waveguide filled with a dielectric material are determined by solving the Helmholtz equation for both TE and TM modes. These modes have different boundary conditions and can be expressed as the sum of the electric and magnetic fields. The cutoff frequency for these modes can be found using the relation ωc=γc√με, where γc is the smallest possible value for the numerical parameter γ. The TEM mode, which corresponds to a zero value for γ, is also possible in coaxial waveguides. However, when the waveguide is enclosed on both ends, standing waves are formed and the resonant frequencies are different for TM and TE modes.
fluidistic
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!

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
deskswirl said:
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.

## 1. What are TM and TE waves in a coaxial waveguide?

TM and TE waves, also known as transverse magnetic and transverse electric waves, are types of electromagnetic waves that propagate through a coaxial waveguide. They are characterized by their electric and magnetic fields being perpendicular to the direction of propagation.

## 2. How do TM and TE waves differ from each other?

The main difference between TM and TE waves is the orientation of their electric and magnetic fields. In TM waves, the electric field is perpendicular to the direction of propagation, while in TE waves, the magnetic field is perpendicular to the direction of propagation.

## 3. What are the main applications of TM and TE waves in coaxial waveguides?

TM and TE waves are commonly used in microwave and radio frequency communication systems. They are also used in radar systems, satellite communications, and medical imaging devices.

## 4. How are TM and TE waves generated in a coaxial waveguide?

TM and TE waves can be generated in a coaxial waveguide by using a signal source, such as a microwave generator, and coupling it to the waveguide. The signal will then propagate through the waveguide as TM or TE waves, depending on the orientation of the electric and magnetic fields.

## 5. What are the advantages of using TM and TE waves in a coaxial waveguide?

TM and TE waves have low attenuation, meaning they can travel long distances without losing much energy. They also have a high power-carrying capacity and can be easily guided through the waveguide without interference from external sources. Additionally, they have a low noise level, making them ideal for high-frequency applications.

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