Radiation Physics in Layered Media Notes

ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 36 1

Radiation Physics in Layered Media y Line source on grounded substrate 0I x r h = ( ) = z E , E x y 0 Note: TMz and also TEy (since ) z z For y>0: = A z 0 0 I 1 + ( ) jk y = + jk x TE 1 k e e dk 0 y x x x 4 j k 0 y 2

Reflection Coefficient ( ( ) ) ( ( ) ) + TE in TE in TE Z Z k k Z Z k k ( ) 0 = x x TE k x TE 0 x x TE Z 00 I where V - + ( ) ( ) = TE in TE tan Z k jZ k h z 1 1 x y TE Z 01 ( ) 1/2 = 2 0 2 x = TE k k k 0 Z 0 y 0 k 0 y ( ) 1/2 = 2 2 x k k k = TE 0 Z 1 1 y 1 k 1 y 3

Poles ( ) ( ) x k ( ( ) ) + TE in TE in TE Z Z k Z Z k k ( ) 0 = x x TE k x TE 0 x = k k Poles: x xp ( ) ( ) = TE in TE Z k Z k 0 xp xp This is the same equation as the TRE for finding the wavenumber of a surface wave: ( 0 in x x Z k Z k = ) ( ) TE SW TE SW TE Z 00 I V - + z kxp= roots of TRE =kxSW TE Z 01 4

Poles (cont.) Complex kx plane k xi = + k k jk x xr xi SW C k xr 1k 0k 0k 1k SW If a slight loss is added, the SW poles are shifted off the real axis as shown. 5

Poles (cont.) k xi C 0k 1k k xr 0k 1k k xi C 0k 1k k xr 0k 1k For the lossless case, two possible paths are shown here. 6

Review of Branch Cuts and Branch Points In the next few slides we review the basic concepts of branch points and branch cuts. 7

Branch Cuts and Points (cont.) ( ) f z = 1/2 = + = j z z x jy re Consider ( ) 1/2 = = 1/2 /2 j j z re r e z = 1 = = = Choose 1/2 0: 1 z 1 r = 1/2 = 2 : 1 z y = 4 : = 1/2 z 1 x z = 1 There are two possible values. 8

Branch Cuts and Points (cont.) ( ) f z = = 1/2 j z z re The concept is illustrated for = 1/2 /2 j z r e y Consider what happens if we encircle the origin: r= 1 C B x A 9

Branch Cuts and Points (cont.) y = 1/2 /2 j z r e r= 1 C B x A 1/2 point z A B C 2 0 1 + We don t get back the same result! j -1 10

Branch Cuts and Points (cont.) = 1/2 /2 j z r e y Now consider encircling the origin twice: r = 1 r D C B x A E 1/2 point z A B C 2 D 3 E 4 0 1 + j -1 - + We now get back the same result! j Hence the square-root function is a double-valued function. 1 11

Branch Cuts and Points (cont.) The origin is called a branch point: we are not allowed to encircle it if we wish to make the square-root function single-valued. In order to make the square-root function single-valued, we must put a barrier or branch cut . y Branch cut x Here the branch cut was chosen to lie on the negative real axis (an arbitrary choice). 12

Branch Cuts and Points (cont.) We must now choose what branch of the function we want. = = j 1/2 /2 j z re z r e z This is the "principle" branch, denoted by . y Branch cut x z = 1 MATLAB: = 1/2 1 z 13

Branch Cuts and Points (cont.) Here is the other choice of branch. re = = j 1/2 /2 j z z r e 3 y Branch cut x z = 1 = 1/2 1 z 14

Branch Cuts and Points (cont.) Note that the function is discontinuous across the branch cut. re = = j 1/2 /2 j z z r e y = z = = 1, z Branch cut 1/2 j x z = 1 = z = = 1, z = 1/2 1 1/2 z j 15

Branch Cuts and Points (cont.) The shape of the branch cut is arbitrary. 3 /2 re = j /2 z y = 1/2 /2 j z r e x z = 1 = 1/2 Branch cut 1 z 16

Branch Cuts and Points (cont.) The branch cut does not have to be a straight line. re = j In this case the branch is determined by requiring that the square- root function (and hence the angle ) change continuously as we start from a specified value (e.g., z= 1). z = 1/2 /2 j z r e y = z j e z = 1 ( ) = = + 1/2 /4 j 1 / 2 z j = 1/2 z j x z = 1 Branch cut = 1/2 1 z = z j ( ) = = 1/2 /4 j 1 / 2 z e j 17

Branch Cuts and Points (cont.) Consider this function: ( ) 1/2 = 2 ( ) f z 1 z (similar to our wavenumber function) What do the branch points and branch cuts look like for this function? 18

Branch Cuts and Points (cont.) ( ) ( ) 1/2 ( ) ( ) ( ) ( ) 1/2 1/2 1/2 1/2 = = + = 2 ( ) f z 1 1 1 1 1 z z z z z y 1 x 1 There are two branch cuts: we are not allowed to encircle either branch point. 19

Branch Cuts and Points (cont.) Geometric interpretation ( ) ( ) ( ) 1/2 1/2 = = 1/2 1 1/2 2 ( ) f z 1 1 z z w w y = = = j z 1 w z r e 1 1 1 = j ( 1) w z r e 2 2 2 w 2 1 w 1 1 2 x 1 The function f (z) is unique once we specify its value at any point. (The function must change continuously away from this point.) ( )( ) = /2 /2 j j ( ) f z r e r e 1 2 1 2 20

Riemann Surface http://upload.wikimedia.org/wikipedia/commons/thumb/8/82/Georg_Friedrich_Bernhard_Riemann.jpeg/225px-Georg_Friedrich_Bernhard_Riemann.jpeg The Riemann surface is a set of multiple complex planes connected together. The function z1/2 has a surface with two sheets. Georg Friedrich Bernhard Riemann (September 17, 1826 July 20, 1866) was an influential German mathematician who made lasting contributions to analysis and differential geometry, some of them enabling the later development of general relativity. The function z1/2 is continuous everywhere on this surface (there are no branch cuts). It also assumes all possible values on the surface. 21

Riemann Surface The concept of the Riemann surface is illustrated for ( ) f z = 1/2 = j z z re Consider this choice: = = ( 1 1) Top sheet: 3 ( 1 1) Bottom sheet: For a single complex plane, this would correspond to a branch cut on the negative real axis. 22

Riemann Surface (cont.) x Top y D B Bottom B D y D B x D B top view side view 23

Riemann Surface (cont.) Branch point (where it used to be) Top sheet x y Branch cut (where it used to be) Bottom sheet 24

Riemann Surface (cont.) y Connection between Sheets ( escalator ) r = 1 D D C B B x A E 1/2 point z A B C 2 D 3 E 4 0 1 + j -1 - + j 1 25

Branch Cuts in Radiation Problem Now we return to the problem (line source over grounded slab): = A z 0 0 I 1 + ( ) jk y = + jk x TE 1 k e e dk 0 y x x x 4 j k 0 y ( ) 1/2 = 2 0 2 x k k k 0 y Note: There are no branch points from ky1: ( 1 1 y x k k k = ( ) ) 1/2 ( ) = 2 2 TE in TE tan Z k jZ k h = TE 0 Z 1 1 x y 1 k 1 y (The integrand is an even function of ky1.) 26

Branch Cuts ( ) 1/2 ( ) ( ) ) 1/2 1/2 = = + 2 2 x k k k k k k k 0 0 0 0 y x x ( ) ( 1/2 1/2 = + j k k k k 0 0 x x Note: It is arbitrary that we have factored out a j instead of a +j, since we have not yet determined the meaning of the square roots yet. = xk k Branch points appear at 0 = xk k No branch cuts appear at (The integrand is an even function of ky1.) 1 27

Branch Cuts (cont.) ( ) ( ) 1/2 1/2 = + k j k k k k 0 0 0 y x x k xi 0k C k xr 1k 1k 0k Branch cuts are lines we are not allowed to cross. 28

Branch Cuts (cont.) = real , k k ( ( ) ) + = = arg 0 k k 0 x For 0 x Choose = k j k arg 0 k k 0 0 y y 0 x at this point ( ) 1/2 = ( k 2 0 j k 2 x k k k 0 y xi ) ( ) 1/2 1/2 = + k k k k 0 0 0 y x x 0k k xr 0k = k j k 0 0 y y This choice then uniquely defines ky0 everywhere in the complex plane. 29

Branch Cuts (cont.) = real ( ( ) ) , k k + = = arg k k we have x For 0 x 0 k arg 0 k k 0 x 0 x ( ) 1/2 = 2 0 2 x k k k = + /2 0/2 j j k j k k e k k e 0 y 0 0 0 y x x ( ) ( ) 1/2 1/2 = + k j k k k k 0 0 0 y x x k xi = /2 j k je k Hence 0 0 y y = k k 0 0 y y k xr 0k 0k 30

Riemann Surface ( ) 1/2 = 2 0 2 x k k k Top sheet = 0 y k xi k j k 0 0 y y 0k k xr 0k = + k j k 0 0 y y Bottom sheet There are two sheets, joined at the blue lines. The path of integration is on the top sheet. 31

Proper / Improper Regions Let = + k k jk x xr xi The goal is to figure out which regions of the complex plane are "proper" and "improper." = k k jk 0 0 0 ( ) 1/2 = 2 0 2 x k k k 0 y Im 0 k Proper region: 0 y Im 0 k Improper region: 0 y = Im 0 yk Boundary: 0 = = 2 y 2 0 2 x = real real >0 k k k k 0 0 y 32

Proper / Improper Regions (cont.) ) ( 2 ( ) 2 + = real 0 k jk k jk Hence ( 0 0 xr xi ) ) ( + + = 2 2 2 xr 2 xi 2 2 real 0 k k k k j k k k k 0 0 0 0 xr xi k k = k k (hyperbolas) Therefore 0 0 xr xi One point on curve: k xi = k k 0 xr k 0 = k k 0 xi k xr 0k = = xk k k jk = k k jk 0 0 0 0 0 0 33

Proper / Improper Regions (cont.) + 2 2 2 xr 2 xi 0 k k k k Also 0 0 k xi The solid curves satisfy this condition. k 0 k xr 0k 34

Proper / Improper Regions (cont.) k xi Complex plane: top sheet k j k 0 0 y y k 0 k xr 0k Proper Improper region On the complex plane corresponding to the bottom sheet, the proper and improper regions are reversed from what is shown here. 35

Sommerfeld Branch Cuts k xi k 0 k xr 0k Hyperbola Complex plane corresponding to top sheet: proper everywhere Complex plane corresponding to bottom sheet: improper everywhere 36

Sommerfeld Branch Cuts k k xi xi Complex plane Riemann surface k k 0 0 k k xr xr 0k 0k Note: We can think of a two complex planes with branch cuts, or a Riemann surface with hyperbolic-shaped ramps connecting the two sheets. The Riemann surface allows us to show all possible poles, both proper (surface-wave) and improper (leaky-wave). 37

Sommerfeld Branch Cut k xi k 0 Let 0 k 0 k xr 0k The branch cuts now lie along the imaginary axis, and part of the real axis. 38

Path of Integration k xi C k k 0 1 k xr 0k 1k The path is on the complex plane corresponding to the top Riemann sheet. 39

Numerical Path of Integration k xi C k k 0 1 k 0k 1k xr 40

Leaky-Mode Poles TRE: Frequency behavior on the Riemann surface ( Im ) ( ) = LW x k LW x Z k Z k k 0 in xi 0 0 y f 0 (improper) = f f c 0k SW k xr 1k ISW Note: TM0 never becomes improper. LW = f f s Bottom sheet 41

Riemann Surface We can now show the leaky-wave poles! k xi C k k 1 0 BP k xr 0k 1k SWP LWP = ) LW xp LW LW k j ( = LW LW xp Re k The LW pole is then close to the path on the Riemann surface (and it usually makes an important contribution). LW k k 0 0 42

SW and CS Fields k xi SW field CS field 0k 1k k xr C p SW LW C b Total field = surface-wave (SW) field + continuous-spectrum (CS) field Note: The CS field indirectly accounts for the LW pole. 43

Leaky Waves LW poles may be important if LW k k The LW pole is then close to the path on the Riemann surface. 0 0 LW k 0 radiation Physical Interpretation 0 sin k 0 0 LW = LW xp Re( ) k LW leaky wave 44

Improper Nature of LWs Region of strong leakage fields leakage rays = j LW xp k The rays are stronger near the beginning of the wave: this gives us exponential growth vertically. 45

Improper Nature (cont.) Mathematical explanation of exponential growth (improper behavior): ( ) 1/2 = 2 2 LW y LW xp k k k 0 0 ( ) ) ( ( ) 2 2 = 2 LW y LW xp k k k 0 0 ( 2 ) 2 = 2 j k j 0 y y Equate imaginary parts: y = y = y y 0 0 (improper) y 46