Notes 12 ECE 6340 Intermediate EM Waves Fall 2016 Prof. David R. Jackson Dept. of ECE Notes 12

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

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

Wave Impedance (cont.) TEz wave: Note:

Transverse Equivalent Network (TEN) Denote where Note: V(z) behaves as a voltage function. Also, write (assume TMz wave) 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 Note: There is still flexibility here, since there can be any real scaling constant in the definition of t. We have not uniquely determined t yet.

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

Transverse Equivalent Network (cont.) Define and Note: I(z) behaves as a transmission-line current function. We then have Note: We could also introduce a different characteristic impedance (Z0  ZTM), and then have a different t function for the magnetic field.

Transverse Equivalent Network (cont.) Summary

Transverse Equivalent Network (cont.) +z -z z Waveguide Z0 = ZTM or ZTE + V (z) - I (z) z kz = kz of waveguide mode TEN

Transverse Equivalent Network (cont.) Power flowing in waveguide TEN: WG: Hence

Transverse Equivalent Network (cont.) The total complex power flowing down the waveguide is then or If we choose Note: This uniquely defines t, and hence voltage and current. then

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

Example: Rectangular Waveguide b a z z = 0 x y #0 #1 TE10 mode incident at a junction The TEN method gives us an exact solution since the two WGs have the same cross sectional shape (the same t function). TEN:

Example: Rectangular Waveguide (cont.) Similarly,

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

(This is because the impedances of the two guides are different.) Example (cont.) (conservation of energy and orthogonality) conservation of energy othogonality (in Notes 13) Note: (This is because the impedances of the two guides are different.)

Example (cont.) Alternative calculation:

Fields in the waveguides Example (cont.) Fields in the waveguides TE10 mode: This choice of t does not correspond to equal powers for the WG and the TEN. V(z) = Ey at the center of the WG

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

A couple of commonly used matching elements: Inductive post (narrow) x y TEN Z0TE Lp Capacitive diaphragm (iris) x y TEN Z0TE Cd

Matching Elements (cont. Note: Planar discontinuities are modeled as purely shunt elements. End view Inductive iris Capacitive iris Resonant iris Þ The equivalent circuit gives us the correct reflection and transmission for the dominant TE10 mode.

Discontinuity TE10 Top view: Rectangular waveguide with a post: Post ap = radius of post TE10 z x TE10 a Top view: Assumption: Only the TE10 mode can propagate. Higher-order mode region

Discontinuity (cont.) TEN : Top view: Lp Z0TE The element is chosen to give the same reflected and transmitted TE10 waves as in the actual waveguide. TEN : Note: The discontinuity is 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).

Discontinuity (cont.) Top view: Flat strip model (strip of width w) x z x TE10 w z = 0- z = 0+ x0 Top view: Narrow strip model for post (w = 4ap) Assume center of post is at x = x0 In this model (flat strip model) the equivalent circuit is exactly a shunt inductor (proof given later).

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

Discontinuity (cont.) Transverse fields radiated by the post current (no y variation): By symmetry, the scattered field should have no y variation. We assume a field representation in terms of TEm0 waveguide modes (therefore, there is only a y component of the electric field). Note: TMm0 modes do not exist.

Discontinuity (cont.) Top view: x z From boundary conditions: Hence TE10 w z = 0- z = 0+ x0 Top view: Note: The incident field is continuous at z = 0. From boundary conditions: Hence

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

Discontinuity (cont.) This establishes that the circuit model for the flat strip must be a parallel (shunt) element. Z0TE jXp Z0TE

Discontinuity (cont.) The fields inside the waveguide are then: There is one set of unknown coefficients Am.

Discontinuity (cont.) Magnetic field: From the Fourier series for the magnetic field, we then have: Note: We can use the scattered field here, since he incident field is continuous. or Represent the strip current as

Discontinuity (cont.) Therefore Hence In order to solve for jm, we need to enforce the condition that Ey = 0 on the strip.

(I0 is the total current on the strip.) Discontinuity (cont.) Assume that the strip is narrow, so that a single “Maxwell” function describes accurately the shape of the current on the strip: I0 is now the unknown. (I0 is the total current on the strip.) Basis function B(x) Hence

Discontinuity (cont.) Hence where (Please see the Appendix.) Note: cm can be evaluated in closed form (in terms of the Bessel function J0). (Please see the Appendix.)

Discontinuity (cont.) The fields inside the waveguide are now given by: We still need to solve for the unknown post current I0, by enforcing that the electric field vanish on the strip.

To solve for I0, enforce the electric field integral equation (EFIE): Discontinuity (cont.) To solve for I0, enforce the electric field integral equation (EFIE): on strip (unit-amplitude incident mode) Hence

Discontinuity (cont.)

Discontinuity (cont.) This is in the form To solve for the unknown I0, we can use the idea of a “testing function.” We multiply both sides by a testing function and then integrate over the strip.

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

Discontinuity (cont.) Hence This integral is cm.

Discontinuity (cont.) Hence so or

Discontinuity (cont.) Summary

Discontinuity (cont.) Reflection coefficient: so Post reactance: where jXp TEN From these two equations, Xp may be found in terms of .

Discontinuity (cont.) Result: where

Discontinuity (cont.) Results for many different types of waveguide discontinuities may be found in: N. Marcuvitz, The Waveguide Handbook, IET (The Institution of Engineering and Technology), IEE Electromagnetic Wave Series, 1985. (The book was originally published in 1951 as vol. 10 of the MIT Radiation Laboratory series.)

In this appendix we evaluate the cm coefficients . Use

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

Appendix (cont.) or

Appendix (cont.) Next, use the transformation

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

Appendix (cont.) Hence