1. Rectangular Waveguides
ECE 3317
Prof. David R. Jackson
Notes 20
Rectangular Waveguides
[Chapter 5]

2. Rectangular Waveguideb x y
, 
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
TMz
(Helmholtz equation)
(PEC walls)
Guided-wave assumption:

4. TMz Modes (cont.) Define: We then have:
Note that kc is an unknown at this point.
We then have:
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. TMz Modes (cont.) where Hence, Divide by XY : Hence This has the form
Both sides of the equation must be a constant!
This has the form

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

7. TMz Modes (cont.) The second boundary condition is:
This gives us the following result:
Hence
Now we turn our attention to the Y (y) function.

8. TMz Modes (cont.) General solution:
Now we turn our attention to the Y function:
Hence
Define
Then we have
General solution:

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

10. TMz Modes (cont.) Hence 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 for (m,n) Mode (Hz = 0)

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

14. TMz Modes (cont.) Hence which gives us
The cutoff frequency fc of the TMm,n mode is then
This may be written as

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

16. TEz Modes (cont.) y b ,  x a Boundary conditions: cross section
The result is
This can be shown by using the following equations:

17. TEz Modes (cont.) Summary of TEz Solution for (m,n) Mode (Ez = 0)
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. Wavenumber General formula for the wavenumber TMz or TEz mode: with
Note: The (m,n) notation is suppressed here.
Recall:
Hence

20. Wavenumber When we are below cutoff it is convenient to write Hence

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

22. Wavenumber Plot
General behavior of the wavenumber

23. Dominant Mode
The "dominant" mode is the one with the lowest cutoff frequency.
Assume b < a a b x y
, 
cross section
Lowest TMz mode: TM11
Lowest TEz mode: TE10
TEz
TMz
The dominant mode is the TE10 mode.

24. Dominant Mode (cont.) Formulas for the dominant TE10 mode
At the cutoff frequency:

25. Dominant Mode (cont.)
What is the mode with the next highest cutoff frequency?
Assume b < a / 2
The next highest is the TE20 mode.
useful operating region fc
TE10
TE20

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

27. Dominant Mode (cont.) E H b a
From these, we find the other fields to be:
where x y a b E H

28. Example Standard X-band* waveguide (air-filled):
a = inches (2.286 cm)
b = inches (1.016 cm)
Find the operating frequency region.
Use
Hence, we have
* X-band: from 8.0 to 12 GHz.

29. Example (cont.) Standard X-band* waveguide (air-filled):
a = inches (2.286 cm)
b = inches (1.016 cm)
Find the phase constant of the TE10 mode at 9.00 GHz.
Find the attenuation in dB/m at 5.00 GHz
Recall:
At 9.00 GHz: k = [rad/m]
At 9.00 GHz:  = [rad/m]
At 5.00 GHz:  = [nepers/m]

30. Example (cont.) At 5.00 GHz:  = 88.91 [nepers/m] Therefore,
A very rapid attenuation!
Note: We could have also used

31. Guide Wavelength
The guide 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

32. Phase and Group Velocity
Recall that the phase velocity is given by
Hence
We then have
For a hollow waveguide (cd = c):
Hence:
vp > c !
(This does not violate relativity.)

33. 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
(The derivation of this is omitted.) t
A pulse consists of a "group" of frequencies (according to the Fourier transform). + -
Vi (t)
waveguiding system

34. Phase and Group Velocity (cont.)
If the phase velocity is a function of frequency, the pulse will be distorting as it travels down the system.
Vi (t) + -
waveguiding system
A pulse will get distorted in a rectangular waveguide!

35. 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:
vg < c

36. Phase and Group Velocity (cont.)
For a lossless transmission line or a plane wave (TEMz waves):
We then have
For a lossless transmission line there is no distortion.
Hence we have

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

38. Plane Wave Interpretation (cont.)
Picture (top view): z k1  a
TE10 mode k2 x
At the cutoff frequency, the angle  is 90o. At high frequencies (well above cutoff) the angle  approaches zero.

39. Plane Wave Interpretation (cont.)
Picture of two plane waves a k2 k1 z x 
crests of waves g
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).