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

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

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

4. TE z wave: Wave Impedance (cont.) Note: 4

5. Transverse Equivalent Network (TEN) Denote (assume TM z wave) Also, write where Note: V ( z ) behaves as a voltage function. 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). Hence we have There is still flexibility here, since there can be any real scaling constant in the definition of  t. 5

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

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

8. Transverse Equivalent Network (cont.) Summary 8

9. TEN +z -z z Waveguide Z 0 = Z TM or Z TE +V (z)-+V (z)- I (z)I (z) z k z = k z of waveguide mode Transverse Equivalent Network (cont.) 9

10. Note on Power Hence TEN: WG: 10

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

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

13. Example: Rectangular Waveguide TE 10 mode incident TEN: b a z a z = 0 x y #0 #1 13

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

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

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

17. Alternative calculation: Example (cont.) 17

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

19. Matching Elements A couple of commonly used matching elements: Capacitive diaphragm x y Inductive post (narrow) x y TEN Z 0 TE LpLp TEN Z 0 TE CdCd 19

20. Discontinuity Rectangular waveguide with a post: Top view: Higher-order mode region z x TE 10 Post TE 10 a p = radius of post 20 Assumption: only the TE 10 mode can propagate.

21. Discontinuity (cont.) TEN : Z 0 TE LpLp 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). Top view: The TL discontinuity is chosen to give the same reflected and transmitted TE 10 waves as in the actual waveguide. 21

22. Discontinuity (cont.) Top view: Narrow strip model for post ( w = 4a p ) Assume center of post is at x = x 0 Flat strip model (strip of width w ) In this model (flat strip model) the equivalent circuit is exactly a shunt inductor (proof given later). z x TE 10 w z = 0 - z = 0 + x0x0 22

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

24. Discontinuity (cont.) We assume a field representation in terms of TE m 0 waveguide modes (therefore, there is only a y component of the electric field). Note that TM m0 modes do not exist. Transverse fields radiated by the post current (no y variation): By symmetry, the scattered field should have no y variation. 24

25. Discontinuity (cont.) Top view: From boundary conditions: Hence z x TE 10 w z = 0 - z = 0 + x0x0 25

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

27. Discontinuity (cont.) Z 0 TE jX p This establishes that the circuit model for the flat strip must be a parallel element. 27

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

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

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: Hence I 0 is the unknown ( I 0 is the total current on the strip.) 30

31. Discontinuity (cont.) Hence where Note: c m can be evaluated in closed form (in terms of the Bessel function J 0 ). (Please see the Appendix.) 31

32. Discontinuity (cont.) To solve for I 0, enforce the electric field integral equation (EFIE): on strip Hence (unit-amplitude incident mode) 32

33. Discontinuity (cont.) or 33

34. Discontinuity (cont.) This is in the form To solve for the unknown I 0, we can use the idea of a “testing function.” We multiply both sides by a testing function and then integrate over the strip. 34

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

36. Discontinuity (cont.) Hence 36

37. Discontinuity (cont.) Hence so or 37

38. Discontinuity (cont.) Summary 38

39. Discontinuity (cont.) Reflection Coefficient: From these equations, X p may be found. Post Reactance: where jX p 39

40. Discontinuity (cont.) Result: 40

41. Appendix In this appendix we evaluate the c m coefficients. Use 41

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

43. Appendix (cont.) or 43

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

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

46. Hence Appendix (cont.) 46