1. 1 ECE 6345 Spring 2011 Prof. David R. Jackson ECE Dept. Notes 2

2. 2 Overview This set of notes treats circular polarization, obtained by using a single feed. x y L W (x 0, y 0 )

3. 3 Overview Goals:  Find the optimum dimensions of the CP patch  Find the input impedance of the CP patch  Find the pattern (axial-ratio) bandwidth  Find the impedance bandwidth of CP patch

4. 4 Amplitude of Patch Currents First Step: Find the patch currents ( x and y directions), and relate them to the input impedance of the patch.

5. 5 Amplitude of Patch Currents (cont.) x- directed current mode (1,0): y- directed current mode (0,1):

6. 6 Amplitude of Patch Currents (cont.) x mode : so To find E, use

7. 7 Amplitude of Patch Currents (cont.) For the (1,0) mode we have or A similar derivation holds for the y mode.

8. 8 Amplitude of Patch Currents (cont.) y mode: The patch current amplitudes can then be written as where

9. 9 Amplitude of Patch Currents (cont.) Assume x y L W (x 0, y 0 ) Then

10. 10 Amplitude of Patch Currents (cont.) Because of the nearly equal dimensions and the feed along the diagonal, we have We then have Reminder: The bar denotes impedances that are normalized by R (either R x or R y ). R i = resonant input resistance of the mode i, when excited by itself (e.g., by a feed along the centerline). where

11. 11 Amplitude of Patch Currents (cont.) The A 2 coefficient can be written as

12. 12 Circular Polarization Condition amplitude of y mode amplitude of x mode x y L W (x 0, y 0 ) The CP condition is

13. 13 Circular Polarization Condition (cont.) The frequency f 0 is defined as the frequency for which we get CP at broadside. where f rx = resonance frequency of (1,0) mode f ry = resonance frequency of (0,1) mode At f = f 0 :

14. 14 Choose Then we have (LHCP) Circular Polarization Condition (cont.)

15. 15 or The frequency conditions can be written as so Circular Polarization Condition (cont.)

16. 16 Also Let Circular Polarization Condition (cont.) Then we have

17. 17 (LHCP) (RHCP) Circular Polarization Condition (cont.) Summary of frequencies f 0 = frequency for which we get CP at broadside.

18. 18 Patch Dimensions for CP PMC

19. 19 Let (resonance condition) Physical Dimensions for CP (cont.)

20. 20 Similarly, we have Physical Dimensions for CP (cont.) Hence Note: For  W, we use the same formula as  L, but replace W  L.

21. 21 Physical Dimensions for CP (cont.) where Note: For the calculation of  L it is probably accurate enough to use the patch dimensions that come from neglecting fringing. Since the patch is nearly square, the two fringing extensions are nearly equal. Hence we have

22. 22 Hammerstad’s Formula

23. 23 Input Impedance of CP Patch At f 0 so or (LHCP) The CP frequency f 0 is also the resonance frequency where the input impedance is real (if we neglect the probe inductance).

24. 24 Input Impedance of CP Patch Hence, at the resonance (CP) frequency f 0 we have Note: We have a CAD formula for R edge. The fed position x 0 can be chosen to give the desired input resistance at the resonance frequency f 0.

25. 25 CP (Axial Ratio) Bandwidth We now examine the frequency dependence of the term (LHCP) where

26. 26 CP Bandwidth (cont.) Similarly, Then Define (This is the ratio of the operating frequency to the CP (resonance) frequency.)

27. 27 CP Bandwidth (cont.) Hence Note: Then Let

28. 28 CP Bandwidth (cont.) x y

29. 29 CP Bandwidth (cont.) where In our case, From ECE 6340:

30. 30 CP Bandwidth (cont.) Set From a numerical solution:

31. 31 CP Bandwidth (cont.) Therefore Hence so Hence

32. 32 Impedance Bandwidth where Note: At x = 0 we have

33. 33 Set (derivation omitted) Impedance Bandwidth (cont.) (bandwidth limits)

34. 34 Hence Impedance Bandwidth (cont.) so The band edges (in normalized frequency) are then

35. 35 Hence Impedance Bandwidth (cont.)

36. 36 Summary CP Linear