Quadrature Coupler and Rat-Race Coupler Notes 3

Appendix A (cont.) Adding even and odd mode cases together: + V= V + +V + 1 V1− S11 = + V1 ⇒ a2= a3= a4= 0 Hence S11 = 0 S11 = V1- e + V1-o V1- e + V1-o = = + + V +V 2V + 1  V1- e V1-o  + +  + 2V V  1 0 0 = ( S11e + S11o ) =+ 2 By symmetry: S= S= S= S= 0 11 22 33 44 23 Appendix A (cont.) + + + V= V + V 1 V2 − S21 = + V1 a2= a3= a4= 0 V2 − e + V2 − o V2 − e + V2 − o 1 e o S21 S21 + S21 = = = ⇒ ( ) + + + 2V 2 V +V 1   −1 − j   1 − j   = + 2  2   2   −j = 2 −j S= S= S= By symmetry and reciprocity: S= 21 12 43 34 2 24 Appendix A (cont.) + V= V + +V + 1 V3− S31 = + V1 a2= a3= a4= 0 V3− e + V3− o V3− e + V3− o V2 − e − V2 − o 1 e o S31 S S = = = − ⇒ = ( ) 21 21 2V + 2V + 2 V + +V + 1   −1 − j   1 − j   =  − 2  2   2   −1 = 2 By symmetry and reciprocity: S= S= S= S= 31 13 24 42 −1 2 25 Appendix A (cont.) + V= V + +V + 1 V4 − S41 = + V1 ⇒ a2= a3= a4= 0 S41= V4 − e + V4 − o V4 − e + V4 − o V1− e − V1− o = = = V + +V + 2V + 2V + By symmetry and reciprocity: 1 e S11 − S11o )= 0 ( 2 S= S= S= S= 0 41 14 23 32 26 Appendix B Here we analyze the Rat-Race Ring coupler. 2 1 2 3 4 1 Plane of symmmetry 3 4 Layout Schematic 27 Appendix B (cont.) Port 1 Excitation “even” problem 1 2 #1 #2 Ys1 = jY0 s1 tan ( β s ls1 ) ( ) = j Y0 / 2 tan (π / 4 ) 4 3 Y0 = 1/ Z 0 Y0 s = Y0 / 2 = jY0 / 2 Ys 2 = jY0 s 2 tan ( β s ls 2 ) ( ) = j Y0 / 2 tan ( 3π / 4 ) = − jY0 / 2 28 Appendix B (cont.) Port 1 Excitation “odd” problem #2 #1 1 2 Ys1 = − jY0 s1 cot ( β s ls ) ( ) = − j Y0 / 2 cot (π / 4 ) 4 3 = − jY0 / 2 Ys1 = − jY0 s 2 cot ( β s ls ) Y0 = 1/ Z 0 Y0 s = Y0 / 2 ( ) = − j Y0 / 2 cot ( 3π / 4 ) = jY0 / 2 29 Appendix B (cont.) Proceeding as for the 90o coupler, we have:  1 [ ABCD ]0e =  ± jY0  2 0  0  1 1  j   2 Z 0 j 2Z0   1  jY 0   0   2  1 =  ± jY0   2 0  ±1  1 1  j   2 Z 0 j 2Z0   0    ±1  = 2 j Z 0  j 2Z0    1   0  1  30 Appendix B (cont.) Converting from the ABCD matrix to the S matrix, we have:  ±1  = ABCD ]0e  2 [ j Z 0  j 2Z0    1   Note: We are describing a two-port device here, in the even and odd mode cases. This is a 2×2 matrix, not a 4×4 matrix. Use Table 4.2 in Pozar [ S ]0e −j = 2  ±1 1   1 1   31 Appendix B (cont.) For the S parameters coming from port 1 excitation, we then have: V1− S11 = + V1 V1− e + V1− o 1 e o = = + S11 S S ) ( 11 11 2V + 2 j  1 −j = +  2  2 2 =0 a2= a3= a4= 0 S= S= 0 11 33 (symmetry) V2 − S21 = + V1 a2= a3= a4= 0 S= S= S= S= 21 12 34 43 −j 2 (symmetry and reciprocity ) V2 − e + V2 − o S21 = = 2V + 1 −j −j  = +   2 2 2 −j = 2 1 e o S21 + S21 ( ) 2 32 Appendix B (cont.) S31 = − V3 V1+ a2= a3= a4= 0 S= S= 31 13 (symmetry) j 2 V3− e + V3− o 1 e o = = − S31 S S ( 11 11 ) + 2V 2 1 −j j  = −  2  2 2 −j = 2 Similarly, exciting port 2, and using symmetry and reciprocity, we have the following results (derivation omitted): S= S= 0 22 44 S= S= S= S= 0 23 32 14 41 S= S= 24 42 j 2 33