Analysis of Two-Layer Stripline and Parallel-Plate Waveguide Structures

ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 7 1

Two-Layer Stripline Structure x , h2 2 2 r r , h1 1 1 r r z Goal: Derive a transcendental equation for the wavenumber kzof theTMx modes of propagation. Assumption: There is no variation of the fields in the y direction, and propagation is along the z direction. 2

Two-Layer Stripline Structure (cont.) R TEN: Z01 Z02 x h2 h1 TMx impedances: Note: Because this is a lossless closed structure, kz is either real (above cutoff) or imaginary (below cutoff). xk xk = = 2 1 Z Z 02 01 2 1 ( ) ( ) 1/2 1/2 = 2 2 2 z k k k = 2 2 z k k k 2 x 1 1 x = Z Z Transverse Resonance Equation (TRE): in in 3

Two-Layer Stripline Structure (cont.) Z Z in in Z02 Z01 x h1 h2 ( ) ( ) = tan Z jZ 1 1 x k h = tan Z jZ k h 01 in 02 2 2 in x ( ) ( ) = tan tan Z 1 1 x k h Z k h TRE: 01 02 2 2 x or k k ( ) ( ) = tan tan 1 2 x x 1 1 x k h k h 2 2 x 1 2 4

TEN (cont.) Hence ( ) ( ) = tan tan k 1 1 x k h k k h 2 1 1 2 2 2 r x r x x Assuming nonmagnetic materials, ( ) ( ) 1/2 1/2 = 2 0 2 z k k k = 2 0 2 z k k k 2 2 x r 1 1 x r A similar analysis could be applied for the TEx modes. 5

Parallel-Plate Waveguide Special case: parallel-plate waveguide x = + h h h 1 2 , h r r z = = k k k 1 2 x x x 6

Parallel-Plate Waveguide (cont.) ( ) ( ) = tan tan k k h k k h 1 2 r x x r x x or ( ) ( ) + = tan tan 0 k h k h 1 2 x x Helpful identity: + tan( ) 1 tan( )tan( ) a tan( ) a b b ( ) a b + = tan We then have ( ) ( ) ( ) + 1 tan = tan tan 0 k h k h k h k h 1 2 1 2 x x x x This will be satisfied if ( ) + = tan 0 k h k h 1 2 x x 7

Parallel-Plate Waveguide (cont.) Hence we have ( ) xk h = tan 0 so = = , 0,1,2 xk h m m We have (from the separation equation) = 2 0 2 x k k k z r Note: TMx and TEx modes have the same wavenumber, but only the TMx mode can exist for m = 0. (the electric field is perpendicular to the metal walls). Hence 2 m = 2 0 k k z r h 8

Waveguide With Slab y TExmn modes TMxmn modes b x w a ( ) / 2 = L a w TEN : w L L Z Z Z 01 00 00 9

Waveguide With Slab (cont.) k = TM = TE 0 x Z 0 Z 00 00 k 0 0 x k = TM = TE 1 x Z Z 1 01 01 k 1 1 x Wavenumbers: Note: n y b n y b 1/2 ( ) 2 = jk z sin A F x e n z = 2 0 2 z x k k k 0 x b ( ) = jk z cos F G x e z 1/2 2 x n = 2 0 2 z k k k 1 x r b Note: In the BCs, Ax acts like Ex while Fx acts like Hx. 10

Waveguide With Slab (cont.) First try reference plane at x = 0: = Z Z = 0 Z R Z Z Z Z 01 00 00 x 11

Waveguide With Slab (cont.) Z (1) in (2) in Z Z Z Z Z 01 00 00 w L L x + + tan tan Z Z jZ jZ l l = 0 L Z Z General formula: 0 in 0 L = (1) in tan( ) Z jZ k L 00 0 x Apply this twice: + (1) in tan( tan( ) ) Z Z jZ jZ k w k w = (2) in 01 (1) in 1 x Z Z 01 + 01 1 x 12

Waveguide With Slab (cont.) Z (1) in (2) in Z Z Z Z Z 01 00 00 w L L x Apply again: ( ) 2 in + tan( tan( ) ) Z Z jZ jZ k L k L = 00 0 x Z Z ( ) 2 in 00 + 00 0 x A mess ! 13

Waveguide With Slab (cont.) Now try a reference plane at the center of the structure (the origin is now re-defined here). R w L L x Z01 Z00 Z00 Z Z TRE: = Z Z = 0 Z Hence (This is the only finite number that will work.) But (from symmetry): = Z Z 14

Waveguide With Slab (cont.) Now use an admittance formulation w L L Z01 x Z00 Z00 Y Y TRE: = 0 Y Hence = Y Y (only finite number that will work) But (from symmetry): = = Z Hence Y Y 15

Waveguide With Slab (cont.) 0 Hence, there are two valid solutions: = Z = 0 Z PEC wall = Z PMC wall PEC wall PMC wall 16

Waveguide With Slab (cont.) From symmetry: ( ) ( ) x = E x AE z z Even mode: A = +1 Odd mode: A = -1 z E x = 0 z E = 0 Plots of Ez Odd mode: PEC wall Even mode: PMC wall 17

Waveguide With Slab (cont.) H n E n = = = PEC: 0, 0, 0 t E H PEC wall: odd mode t n = = = PMC wall: even mode PMC: 0, 0, 0 t H E t n Note: even/odd is classification is based on the Ez field. z E x = 0 z E = 0 Plots of Ez Odd mode: PEC wall Even mode: PMC wall 18

Example: TEx Even Modes Z = Z ( ) 1 in Z R w + (1) in tan Z jZ k 01 1 x 2 w = Z Z Z Z Z 01 00 01 00 + (1) in tan Z jZ k 01 1 x 2 w L L Set w + = (1) in tan 0 Z jZ k 01 1 x 2 w + = tan( ) tan 0 Z j jZ k L k 01 00 0 1 x x 2 w = tan( )tan 0 Z Z k L k 01 00 0 1 x x 2 19

TEx Even Modes (cont.) w ( ) = tan tan 0 0 0 k L k 0 1 x x 2 k k 1 0 x x or w = tan( )tan 0 k k k L k 0 1 0 1 x x x x 2 TEx even 1/2 1/2 2 2 n n where = = 2 2 z 2 2 z k k k k k k 0 0 1 0 x x r b b Hence, the transcendental equation is in the form = ( ) F k 0 z 20

Four Possible Cases (We have only done one of them: TEx even.) TEx even TMx even TMx odd TEx odd 21

Cutoff Frequency (TEx Even) zk = 0 (closed structure) Set Then 1 2 1 2 2 2 n n = = 2 0 2 z 2 0 2 z k k k k k k 0 1 x x r b b Hence, the transcendental equation is now in the form = ( ) F k 0 0 = = cutoff frequency 2 k f f 0 0 0 c where c 22

LSE/LSM Terminology y This terminology is often used in the microwave community. x LSE: Longitudinal Section Electric . This means the same thing as TEx. The electric field vector of the mode has no x component, and hence it lies within the yz plane (This is called the longitudinal plane, which means the plane parallel to the slab face. LSM: Longitudinal Section Magnetic . This means the same thing as TMx. The magnetic field vector of the mode has no x component, and hence it lies within the yz plane (the longitudinal plane). 23