Prof. D. R. Wilton

1. ECE 3317 Prof. D. R. Wilton Notes 20 Rectangular Waveguides [Chapter 5]

2. Rectangular Waveguide y ,  b x a cross section Rectangular Waveguide • 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: Note that kc is an unknown at this point. We then have: Dividing by the exp(-jkzz) 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 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

8. TMz Modes (cont.) 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 kcis given by Next, recall that Hence

12. TMz Modes (cont.) Summary of TMz Solution for (m,n) Mode (Hz= 0) Other (transverse) field components obtained by differentiation of the above component.

13. TMz Modes (cont.) Cutoff frequency Note: The number kcis 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 cross section Boundary conditions: 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 Same formula for both cases Same formula for both cases TMz TEz

19. Wavenumber General formula for the wavenumber TMzor 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 Hence

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

22. Wavenumber Plot General behavior of the wavenumber

23. Dominant Mode y ,  b x a cross section 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.) 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 (note no y-variation):

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

28. Example Standard X-band* waveguide (air-filled): a= 0.900 inches (2.286 cm) b= 0.400 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= 0.900 inches (2.286 cm) b= 0.400 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 =188.62 [rad/m] At 9.00 GHz: =129.13 [rad/m] At 5.00 GHz: =88.91 [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 gis 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): vp > c ! Hence: (This does not violate relativity.)

33. Phase and Group Velocity (cont.) t 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.) + - A pulse consists of a "group" of frequencies (according to the Fourier transform). waveguiding system Vi (t)

34. Phase and Group Velocity (cont.) waveguiding system Vi (t) If the phase velocity is a function of frequency, the pulse will be distorted as it travels down the 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 nulls of waves  k1  z g a x k2  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).