1. Applied Electromagnetic Waves Rectangular Waveguides
ECE 3317
Applied Electromagnetic Waves
Prof. David R. Jackson
Fall 2018
Notes 20
Rectangular Waveguides

2. Rectangular Waveguide
Cross section
We assume that the boundary is a perfect electric conductor (PEC).
We analyze the problem to solve for Ez or Hz (all other fields come from these).
TMz: Ez only
TEz: Hz only

3. TMz Modes
(Helmholtz equation)
(PEC walls)
Guided-wave assumption:

4. Note that kc is an unknown at this point.
TMz Modes (cont.)
Define:
We then have:
Note that kc is an unknown at this point.
Dividing by the exp(-j kz z) term, we have:
We solve the above equation by using the method of separation of variables.
We assume:

5. Both sides of the equation must be a constant!
TMz Modes (cont.)
Hence
Divide by XY :
Hence
Both sides of the equation must be a constant!
This has the form

6. TMz Modes (cont.) Denote General solution: Boundary conditions: (1)
(2)

7. TMz Modes (cont.) From the last slide:
This gives us the following result:
Hence
Now we turn our attention to the Y (y) function.

8. TMz Modes (cont.)
We have
Hence
Denote
Then we have
General solution:

9. TMz Modes (cont.) Boundary conditions: (3) (4)
Equation (4) gives us the following result:

10. TMz Modes (cont.) The Y(y) function is then Therefore, we have
New notation:
The TMz field inside the waveguide thus has the following form:

11. TMz Modes (cont.) Recall that Hence
Therefore the solution for kc is given by
Next, recall that
Hence

12. TMz Modes (cont.) Summary of TMz Solution
Note: If either m or n is zero, the entire field is zero.

13. TMz Modes (cont.) Cutoff Frequency for Lossless Waveguide
We start with
where
Set
Observation:
The number kc is the value of k for which the wavenumber kz is zero.

14. The mode will propagate only above the cutoff frequency.
TMz Modes (cont.)
Hence or
The cutoff frequency fc of the TMm,n mode is then
The mode will propagate only above the cutoff frequency.
This may be written as

15. TEz Modes We now start with
Using the separation of variables method again, we have
where
and

16. TEz Modes (cont.) Boundary conditions: The result is
This can be shown by using the following equations:

17. TEz Modes (cont.) Summary of TEz Solution
Same formula for cutoff frequency as the TEz case!
Note: The (0,0) TEz mode is not valid, since it violates the magnetic Gauss law:

18. Summary TMz TEz TMz TEz same formula for both cases

19. Note: The (m,n) notation is suppressed here.
Wavenumber
TMz or TEz mode:
with
Note: The (m,n) notation is suppressed here.
Above cutoff:
Recall:
Hence

20. Wavenumber (cont.)
Below cutoff:
Hence
Hence

21. Wavenumber (cont.)
Recall that
Hence we have

22. Wavenumber Plot
“Light line”

23. Dominant Mode
The "dominant" mode is the one with the lowest cutoff frequency.
Assume b < a
Lowest TMz mode: TM11
Lowest TEz mode: TE10
TEz
TMz
The dominant mode is the TE10 mode.

24. Dominant Mode (cont.)
Summary (TE10 Mode)

25. Dominant Mode (cont.)
Here we develop a simple rule for how big a waveguide has to be to propagate the dominant mode.
Start with
At the cutoff frequency (f = fc):
Example: Air-filled waveguide at 1 GHz:

26. Dominant Mode (cont.)
What is the mode with the next highest cutoff frequency?
Assume b < a / 2
Then the next highest is the TE20 mode. fc
TE10
TE20
Useful operating region
TE01

27. Dominant Mode (cont.)
What is the mode with the next highest cutoff frequency?
If b > a / 2
Then the next highest is the TE01 mode.
The useable bandwidth is now lower. fc
TE10
TE20
Useful operating region
TE01

28. Dominant Mode (cont.) Fields of the Dominant TE10 Mode
Find the other fields from these equations:

29. Dominant Mode (cont.)
From these, we find the fields to be:
where

30. This is the same formula as for a TEz plane wave!
Dominant Mode (cont.)
Wave impedance:
Note:
This is the same formula as for a TEz plane wave!
Define the wave impedance:

31. Example Find the single-mode operating frequency region
for air-filled X-band waveguide.
Standard X-band* waveguide:
a = inches (2.286 cm)
b = inches (1.016 cm)
Note: b < a / 2
Use
Hence, we have:
* X-band: from 8.0 to 12 GHz.

32. Example (cont.) Find the phase constant of the TE10 mode at 9.00 GHz.
Find the attenuation in dB/m at 5.00 GHz
Recall:
At GHz:  = [rad/m]
At GHz:  = [nepers/m]

33. Example (cont.)
At 5.0 GHz:
Recall:
This is a very rapid attenuation!

34. (This assumes that we are above the cutoff frequency.)
Guide Wavelength
Recall: The guided wavelength g is the distance z that it takes for the wave to repeat itself.
(This assumes that we are above the cutoff frequency.)
From this we have
Hence we have the result

35. Phase and Group Velocity
Recall that the phase velocity is given by
(This assumes that we are above the cutoff frequency.)
Hence
We then have
For a hollow waveguide (cd = c):
Hence:
(This does not violate relativity.)

36. Phase and Group Velocity (cont.)
The group velocity is the velocity at which a pulse travels on a structure.
The group velocity is given by:
(derivation omitted)
(This assumes that we are above the cutoff frequency.) + -
Waveguiding system

37. Phase and Group Velocity (cont.)
To calculate the group velocity for a waveguide, we use
Hence we have:
We then have the following final result:
For a hollow waveguide:
Note:

38. Phase and Group Velocity (cont.)
For a lossless transmission line or a lossless plane wave (lossless TEMz wave):
Recall:
We then have
(a constant)
For a lossless transmission line or a lossless plane wave there is no distortion, since the phase velocity is constant with frequency.
Hence we have

39. Pulse Distortion
If the phase velocity is a function of frequency (the system has “dispersion”), the pulse will be distorted as it travels down the system. + -
Waveguiding system
A pulse will get distorted in a rectangular waveguide!
(very much frequency dependent!)

40. Plane Wave Interpretation of Dominant Mode
Consider the electric field of the dominant TE10 mode:
where
Separating the terms, we have: #1 #2
This form is the sum of two plane waves:

41. Plane Wave Interpretation (cont.)
Top view
TE10 mode
At the cutoff frequency, the angle  is 90o. At high frequencies (well above cutoff) the angle  approaches zero.

42. Plane Wave Interpretation (cont.)
Crests of waves
Waves cancel out
Snapshot of two plane waves
The two plane waves add to give an electric field that is zero on the side walls of the waveguide (x = 0 and x = a).

43. Waveguide Modes in Transmission Lines
A transmission line normally operates in the TEMz mode, where the two conductors have equal and opposite currents.
At high frequencies, waveguide modes can also propagate on transmission lines.
This is undesirable, and limits the high-frequency range of operation for the transmission line.

44. Waveguide Modes in Coax
Dominant waveguide mode in coax (derivation omitted):
TE11 mode
Example: RG 142 coax