Prof. David R. Jackson Dept. of ECE

1. ECE 6340 Intermediate EM Waves Fall 2013 Prof. David R. Jackson Dept. of ECE Notes 12

2. Wave Impedance Assume TMz wave traveling in the +z direction From previous notes: so or

3. Wave Impedance (cont.) -z wave: so we can simply substitute In this formula the wavenumber and the wave impedance are taken to be the same for negative and positive traveling waves. + sign: +z wave - sign: - z wave Summary:

4. Wave Impedance (cont.) TEzwave: Note:

5. Transverse Equivalent Network (TEN) Denote Note: V(z) behaves as a voltage function. where Also, write (assume TMzwave) Note: We can assume that both the unit vector and the transverse amplitude function are both real (because of the corollary to the “real” theorem). There is still flexibility here, since there can be any real scaling constant in the definition of t. Hence we have

6. Transverse Equivalent Network (cont.) For the magnetic field we have Now use The result for the transverse magnetic field is then

7. Transverse Equivalent Network (cont.) Define and We then have Note: I(z) behaves as a transmission-line current function.

8. Transverse Equivalent Network (cont.) Summary

9. Transverse Equivalent Network (cont.) +z z -z Waveguide I(z) + V(z) - Z0= ZTMor ZTE z TEN kz = kz of waveguide mode

10. Transverse Equivalent Network (cont.) Note on Power TEN: WG: Hence

11. Transverse Equivalent Network (cont.) The total complex power flowing down the waveguide is then or If we choose then

12. TEN: Junction + - + - z = 0 Junction: z same cross sectional shape At z = 0: (EM boundary conditions) Hence These conditions are also true at a TL junction.

13. Example: Rectangular Waveguide y #1 TE10mode incident z x #0 z = 0 b a a TEN:

14. Example: Rectangular Waveguide (cont.) Similarly,

15. Example (Numerical Results) X-band waveguide f = 10GHz Choose The results are:

16. Example (cont.) (conservation of energy and orthogonality) conservation of energy othogonality Note: This is because the impedances of the two guides are different.

17. Example (cont.) Alternative calculation:

18. Example (cont.) Fields TL part Air region: Dielectric region:

19. Matching Elements y TEN Lp Z0TE Z0TE x Inductive post (narrow) y TEN Cd Z0TE Z0TE x Capacitive diaphragm A couple of commonly used matching elements:

20. Discontinuity x TE10 TE10 z Higher-order mode region Rectangular waveguide with a post: Post ap= radius of post TE10 Top view: Assumption: only the TE10 mode can propagate.

21. Discontinuity (cont.) Lp Z0TE Z0TE Top view: The TL discontinuity is chosen to give the same reflected and transmitted TE10 waves as in the actual waveguide. TEN: Note: The discontinuity is a approximately a shunt load because the tangential electric field of the dominant mode is approximately continuous, while the tangential magnetic field of the dominant mode is not (shown later).

22. Discontinuity (cont.) x z= 0+ z= 0- TE10 TE10 w x0 z Flat strip model (strip of width w) Top view: Narrow strip model for post (w=4ap) Assume center of post is at x=x0 In this model (flat strip model) the equivalent circuit is exactly a shunt inductor (proof given later).

23. Discontinuity (cont.) Top view: (that is, the field scattered by the strip)

24. Discontinuity (cont.) Transverse fields radiated by the post current (no y variation): By symmetry, the scattered field should have no y variation. We assume a field representation in terms of TEm0 waveguide modes (therefore, there is only a y component of the electric field). Note that TMm0 modes do not exist.

25. Discontinuity (cont.) x z= 0+ z= 0- TE10 TE10 w x0 z Top view: From boundary conditions: Hence

26. Discontinuity (cont.) Equate terms of the Fourier series: Modeling equation: so

27. Discontinuity (cont.) This establishes that the circuit model for the flat strip must be a parallel element. jXp Z0TE Z0TE

28. Discontinuity (cont.) Magnetic field: From the Fourier series for the magnetic field, we then have: or Represent the strip current as

29. Discontinuity (cont.) Therefore Hence In order to solve for jm, we need to enforce the condition that Ey= 0 on the strip.

30. Discontinuity (cont.) Assume that the strip is narrow, so that a single “Maxwell” function describes accurately the shape of the current on the strip: I0 is the unknown (I0 is the total current on the strip.) Hence

31. Discontinuity (cont.) Hence where Note: cm can be evaluated in closed form (in terms of the Bessel function J0). (Please see the Appendix.)

32. Discontinuity (cont.) To solve for I0, enforce the electric field integral equation (EFIE): on strip (unit-amplitude incident mode) Hence

33. Discontinuity (cont.) or or

34. Discontinuity (cont.) This is in the form To solve for the unknown I0, we can use the idea of a “testing function.” We multiply both sides by a testing function and then integrate over the strip.

35. Discontinuity (cont.) Galerkin’s method: The testing function is the same as the basis function:

36. Discontinuity (cont.) Hence

37. Discontinuity (cont.) Hence so or

38. Discontinuity (cont.) Summary

39. Discontinuity (cont.) Reflection Coefficient: Post Reactance: where jXp From these equations, Xp may be found.

40. Discontinuity (cont.) Result:

41. Appendix In this appendix we evaluate the cm coefficients. Use

42. Appendix (cont.) or This term integrates to zero (odd function).

43. Appendix (cont.) or

44. Appendix (cont.) Next, use the transformation

45. Appendix (cont.) or Next, use the following integral identify for the Bessel function: so that

46. Appendix (cont.) Hence