How to explain TE20 mode in a rectangular waveguide from reflection

In summary, Feynman explains the ##TE_{10}## mode of waveguide by considering a line source in the middle of the waveguide. The adjacent sources are out-of-phase, resulting in interference with an optical path that is half of the wavelength. The relationship between wavelength in the waveguide and free-space is shown in a picture. Feynman also mentions that at high frequencies, there can be two or more possible directions for the waves to appear. This happens when ##\lambda_0<\frac{2}{3}a##, corresponding to ##TE_{30}## modes. However, it could also happen when ##\lambda_0<a##, corresponding to additional waves and higher guide modes. Feyn
  • #1
Dale12
19
1
In Feynman's lectures, he explained the ##TE_{10}## mode of waveguide by considering a line source in the middle of waveguide as below:
Snipaste_2020-06-29_17-06-35.png

since the adjacent sources are all out-of-phase, which means to have interference, the adjacent optical path would be about half of wavelength as below:
Snipaste_2020-06-29_17-07-16.png

where
$$\sin\theta = \frac{\lambda_0}{2a}$$ (24.33)

and the relationship between wavelength in waveguide and free-space is show in this picture:
Snipaste_2020-06-29_17-13-35.png


and we have:
$$\cos\theta = \frac{\lambda_0}{\lambda_g}$$,

combing with ##\sin\theta = \frac{\lambda_0}{2a}##, we have:

$$\lambda_g = \frac{\lambda_0}{\cos\theta} = \frac{\lambda_0}{\sqrt{1-(\lambda_0/2a)^2}}$$

Feynman then summarized with:
“ If the frequency is high enough, there can be two or more possible directions in which the waves will appear. For our case, this will happen if ##\lambda_0<\frac{2}{3}a##. In general, however, it could also happen when ##\lambda+0<a##. These additional waves correspond to the higher guide modes we have mentioned. ”

However, ##\lambda_0<\frac{2}{3}a## is corresponding to ##TE_{30}## modes and the differential phase between adjacent source is about 3/2 wavelength, but how to explain ##\lambda+0<a## which corresponding to one wavelength or ##TE_{20}## mode?

Thanks!
 
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  • #2
And also it seems work for TE modes of parallel plate waveguide, but it seems hard to explain TM modes of parallel plate waveguide.
 
  • #3
I think I have make it clear now.
Q1: I think ##TE_{20}## mode can't be generated by putting a line source in the middle of wavguide, because the distribution of electric field for the ##TE_{20}## mode at a/2 is zero, in order to generate ##TE_{20}## mode, we can put the line source in the place of a/4, and mirror the source in the same way as Feynman does.

Q2: I think the line source can't generate the TM mode for the parallel plate waveguide, however, we can explain TE and TM both in view of incident and reflection from the PEC border of waveguide, and then for E components parallel to PEC, the reflection should have an -1, so that it has a form of sin function to be standing wave, and it's similar for H components perpendicular to PEC. for E components perpendicular and H components parallel to PEC, we have the same phase of reflection and then a standing wave of form cos function.
 

Related to How to explain TE20 mode in a rectangular waveguide from reflection

1. What is TE20 mode in a rectangular waveguide?

The TE20 mode is a type of electromagnetic wave that propagates through a rectangular waveguide. It is the second-order transverse electric mode, meaning that the electric field is perpendicular to the direction of propagation and has no component in the direction of propagation.

2. How is the TE20 mode produced in a rectangular waveguide?

The TE20 mode is produced when the dimensions of the rectangular waveguide are such that the ratio of the width to the height is equal to 2. This allows for the electric field to have a node in the width direction and an antinode in the height direction, resulting in the TE20 mode.

3. What is the significance of the TE20 mode in a rectangular waveguide?

The TE20 mode is important because it is the first higher-order mode in a rectangular waveguide and has a higher cutoff frequency than the dominant TE10 mode. It also has a different propagation constant and field distribution, making it useful for certain applications such as higher frequency transmission.

4. How can the TE20 mode be explained through reflection in a rectangular waveguide?

The TE20 mode can be explained through reflection by considering the boundary conditions at the walls of the rectangular waveguide. When the electric field is reflected off the walls, it must satisfy the boundary conditions, resulting in the formation of the TE20 mode.

5. What are some practical applications of the TE20 mode in a rectangular waveguide?

The TE20 mode has various applications in microwave and radio frequency systems, such as in satellite communication, radar systems, and microwave heating. It is also used in waveguide filters and couplers for higher frequency transmissions.

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