1. Prof. David R. Jackson Dept. of ECE Notes 8 ECE 5317-6351 Microwave Engineering Fall 2015 Waveguides Part 5: Transverse Equivalent Network (TEN) 1

2. Waveguide Transmission Line Model Our goal is to come up with a transmission line model for a waveguide mode. + - z The waveguide mode is not a TEM mode, but it can be modeled as a wave on a transmission line. z a b x y 2

3. For a waveguide mode, voltage and current are not uniquely defined. The voltage depends on x ! x y a b A B TE 10 Mode z a b x y 3 Waveguide Transmission Line Model (cont.)

4. For a waveguide mode, voltage and current are not uniquely defined. The current depends on the length of the interval! TE 10 Mode x y a b x1x1 x2x2 Current on top wall: Note: If we integrate around the entire boundary, we get zero current (the top and bottom walls have opposite current). 4 Waveguide Transmission Line Model (cont.) z a b x y

5. Wave impedance Examine the transverse ( x, y ) fields: Note: The minus sign above arises from: 5 Modal amplitudes Waveguide Transmission Line Model (cont.)

6. Introduce a defined voltage into the field equations: 6 We may use whatever definition of voltage we wish here. (In other words, C 1 is arbitrary). Waveguide Transmission Line Model (cont.) We then have:

7. Introduce a characteristic impedance into the equations: 7 The constant C 2 is arbitrary here. Waveguide Transmission Line Model (cont.)

8. Summary: 8 Waveguide Transmission Line Model (cont.) The z dependence of the transverse fields behaves like voltage and current on a transmission line.

9. The transmission-line model is called the Transverse Equivalent Network (TEN) model of the waveguide. + - z 9 Waveguide Transmission Line Model (cont.) TEN Waveguide

10. Power flow down the waveguide (complex power): 10 Waveguide Transmission Line Model (cont.) Complex power flowing down the transmission line.

11. Assume we choose to have: Then we have: 11 Waveguide Transmission Line Model (cont.) It is not necessary to make this assumption of equal powers, but it is a possible choice that can be made.

12. Summary of Constants (assuming equal powers) The most common choice: Once we pick Z 0, the constants are determined. 12 Waveguide Transmission Line Model (cont.)

13. We have two constants ( C 1 and C 2 ). Here are possible constraints we can choose to determine the constants:  We can define the voltage  We can define the current  We can define the characteristic impedance  We can impose the power equality condition Any two of these are sufficient to determine the constants. 13 Waveguide Transmission Line Model (cont.) Possible Choices

14. Method 1:  Define voltage  Define current (This determines Z 0 ) Example: TE 10 Mode of Rectangular Waveguide Method 2:  Choose Z 0 = Z TE  Assume power equality z a b x y 14

15. Define: Example: TE 10 Mode (cont.) Method 1 15 z a b x y

16. Since we have defined both voltage and current, the characteristic impedance is not arbitrary, but is determined. Hence 16 Example: TE 10 Mode (cont.) z a b x y

17. 17 Example: TE 10 Mode (cont.) z a b x y

18. 18 z a b x y Example: TE 10 Mode (cont.)

19. Method 2 19 Example: TE 10 Mode (cont.) z a b x y

20. Solution: Take the conjugate of the second one and then multiply the two equations together. The solution is unique to within a common phase term. 20 z a b x y Example: TE 10 Mode (cont.)

21. (as expected) 21 z a b x y Example: TE 10 Mode (cont.)

22. Example: Waveguide Discontinuity a = 2.2856 cm b = 1.016 cm  r = 2.54 f = 10 GHz b a z a z = 0 x y A B For a 1 [ V/m ] (field at the center of the guide) incident TE 10 mode E-field in guide A, find the TE 10 mode fields in both guides, and the reflected and transmitted powers. 22

23. Example (cont.) Convention:  Choose Z 0 = Z TE  Assume power equality TEN 23 (“Method 2”)

24. Equivalent reflection problem: 24 TEN Note: The above TEN enforces the continuity of voltage and current at the junction, and hence the tangential electric and magnetic fields are continuous in the WG problem. Example (cont.)

25. Recall that for the TE 10 mode: 25 TEN Example (cont.)

26. Hence, for the waveguide problem we have the fields as: 26 Example (cont.)

27. Substituting in, we have (guide A): 27 Example (cont.)

28. Substituting in, we have (guide A): 28 (characteristic impedance appears here) (wave impedance appears here) Example (cont.)

29. Substituting in, we have (guide B): 29 Example (cont.)

30. Substituting in, we have (guide B): 30 Example (cont.) (wave impedance appears here) (characteristic impedance appears here)

31. Summary of Fields 31 Example (cont.)

32. Power Calculations: Alternative: Note: In this problem, Z 0 and  are real. 32 Example (cont.) (watts) (no VARS)

33. Final Results: For a 1 [ V/m ] incident TE 10 mode E-field in guide A (field at the center of the guide), find the TE 10 mode fields in both guide, and the reflected and transmitted powers. b a z a z = 0 x y A B 33 Example (cont.) a = 2.2856 cm b = 1.016 cm  r = 2.54 f = 10 GHz

34. Discontinuities in Waveguide 34 Rectangular Waveguide (end view) Note: Planar discontinuities are modeled as purely shunt elements. The equivalent circuit gives us the correct reflection and transmission of the TE 10 mode. Inductive iris Capacitive iris Resonant iris   

35. Discontinuities in Waveguide (cont.) 35 Inductive iris in air-filled waveguide Top view: z x TE 10 Higher-order mode region Z 0 TE LpLp TEN Model Because the element is a shunt discontinuity, we have  T 1

36. Discontinuities in Waveguide (cont.) 36 Much more information can be found in the following reference: N. Marcuvitz, Waveguide Handbook, Peter Perigrinus, Ltd. (on behalf of the Institute of Electrical Engineers), 1986.  Equivalent circuits for many types of discontinuities  Accurate CAD formulas for many of the discontinuities  Graphical results for many of the cases  Sometimes, measured results