Prof. David R. Jackson ECE Dept.

Spring 2016

Notes 36

ECE 6341

1

Radiation Physics in Layered Media

Note: TMz and also TEy (since ) 0z∂=

∂

( ) 00 0

0

1 14

y x

z

jk y jk xTEx x

y

AI

k e e dkj k

ψµ

ψπ

+∞ − −

−∞

=

= + Γ ∫

For y > 0:

( )ˆ ,zE z E x y=

y

0Ix

rεh

2

Line source on grounded substrate

Reflection Coefficient

( ) ( ) ( )( ) ( )

0

0

TE TEin x xTE

x TE TEin x x

Z k Z kk

Z k Z k−

Γ =+

( ) ( )1 1tanTE TEin x yZ k jZ k h=

01

1

TE

y

Zkωµ

= ( )1/ 22 21 1y xk k k= −

where

00

0

TE

y

Zkωµ

= ( )1/ 22 20 0y xk k k= −

3

00TEZ

01TEZ

I V+ -

z

Poles

Poles:

( ) ( ) ( )( ) ( )

0

0

TE TEin x xTE

x TE TEin x x

Z k Z kk

Z k Z k−

Γ =+

( ) ( )0TE TEin xp xpZ k Z k= −

This is the same equation as the TRE for finding the wavenumber of a surface wave:

kxp = roots of TRE = kxSW

x xpk k=

4

( ) ( )0TE SW TE SWin x xZ k Z k= − 00

TEZ

01TEZ

I V+ -

z

If a slight loss is added, the SW poles are shifted off the real axis as shown.

Poles (cont.) Complex kx plane

Cxrk

xik

SW0k1k0k−1k−

SW

x xr xik k jk= +

5

Poles (cont.)

6

Cxrk

xik

0k1k

0k−1k−

For the lossless case, two possible paths are shown here.

Cxrk

xik

0k1k

0k−1k−

Review of Branch Cuts and Branch Points

In the next few slides we review the basic concepts of branch points and branch cuts.

7

Consider ( ) 1/ 2f z z= jz x jy r e θ= + =

( )1/21/2 /2j jz r e r eθ θ= =

1z = 0 :θ = 1/2 1z =

2 :θ π= 1/2 1z = −

4 :θ π= 1/2 1z =

There are two possible values.

Choose

Branch Cuts and Points (cont.)

8

1r⇒ =

1z =x

y

The concept is illustrated for ( ) 1/ 2f z z= jz r e θ=

1/ 2 / 2jz r e θ=

x

y

AB C

r = 1

Consider what happens if we encircle the origin:

Branch Cuts and Points (cont.)

9

Branch Cuts and Points (cont.)

1/ 2 / 2jz r e θ=

1/2

A 0 1B +C 2 -1

z

j

θ

ππ

point

We don’t get back the same result!

10

x

y

AB C

r = 1

Now consider encircling the origin twice:

1/ 2 / 2jz r e θ=

1/2

A 0 1B +C 2 -1D 3 -E 4 1

z

j

j

θ

ππππ +

point

We now get back the same result!

Hence the square-root function is a double-valued function.

Branch Cuts and Points (cont.)

x

y

AB CD

E

r = 1 r

11

In order to make the square-root function single-valued, we must put a “barrier” or “branch cut”.

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.

x

Branch cut

y

Here the branch cut was chosen to lie on the negative real axis (an arbitrary choice).

Branch Cuts and Points (cont.)

12

We must now choose what “branch” of the function we want.

jz r e θ= 1/ 2 / 2jz r e θ=

π θ π− < <

Branch Cuts and Points (cont.)

This is the "principle" branch, denoted by . z

13

x

Branch cut

y

1z =1/ 2 1z =MATLAB : π θ π− < ≤

Here is the other choice of branch.

jz r e θ= 1/ 2 / 2jz r e θ=

3π θ π< <

x

Branch cut

y

1z =1/ 2 1z = −

Branch Cuts and Points (cont.)

14

Note that the function is discontinuous across the branch cut.

jz r e θ= 1/ 2 / 2jz r e θ=

π θ π− < <

x

Branch cut

y

1z =1/2 1z =

1,z θ π −= − =

1,z θ π −= − = −

1/ 2z j=

1/ 2z j= −

Branch Cuts and Points (cont.)

15

The shape of the branch cut is arbitrary.

jz r e θ=1/ 2 / 2jz r e θ=

/ 2 3 / 2π θ π− < <

x

Branch cut

y

1z =1/2 1z =

Branch Cuts and Points (cont.)

16

The branch cut does not have to be a straight line.

jz r e θ=1/ 2 / 2jz r e θ=

In this case the branch is determined by requiring that the square-root function (and hence the angle θ ) change continuously as we start from a specified value (e.g., z = 1).

x

Branch cut

y

1z =1/2 1z =

1z = −1/ 2z j=

z j=( )1/ 2 / 4 1 / 2jz e jπ= = +

z j= −

( )1/ 2 / 4 1 / 2jz e jπ−= = −

Branch Cuts and Points (cont.)

17

Consider this function:

( )1/ 22( ) 1f z z= −

Branch Cuts and Points (cont.)

What do the branch points and branch cuts look like for this function?

(similar to our wavenumber function)

18

( ) ( ) ( ) ( ) ( )( )1/ 2 1/ 21/ 2 1/ 2 1/ 22( ) 1 1 1 1 1f z z z z z z= − = − + = − − −

Branch Cuts and Points (cont.)

x

y

1−

1

There are two branch cuts: we are not allowed to encircle either branch point.

19

( ) ( )( )1/ 21/ 2 1/ 2 1/ 21 2( ) 1 1f z z z w w= − − − =

Branch Cuts and Points (cont.) Geometric interpretation

( )( )1 2/ 2 / 21 2( ) j jf z r e r eθ θ=

1

2

1 1

2 2

1

( 1)

j

j

w z r e

w z r e

θ

θ

= − =

= − − =

The function f (z) is unique once we specify its value at any point. (The function must change continuously away from this point.)

20

x

y

1−

11w2w

1θ2θ

z

Riemann Surface

21

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.

The Riemann surface is a set of multiple complex planes connected together.

The function z1/2 has a surface with two sheets.

Riemann Surface

The concept of the Riemann surface is illustrated for

( ) 1/ 2f z z=

( 1 1)

3 ( 1 1)

π θ π

π θ π

− < < =

< < = −

Top sheet:

Bottom sheet:

Consider this choice:

jz r e θ=

22

For a single complex plane, this would correspond to a branch cut on the negative real axis.

Riemann Surface (cont.) x

y

BD

Top

Bottom

BD

D

B

B

Dside view

x

y

top view 23

Riemann Surface (cont.)

24

Bottom sheet

Top sheet

Branch cut

(where it used to be)

Branch point (where it used to be)

xy

Riemann Surface (cont.)

x

y

AB CD

EBD

Connection between Sheets (“escalator”)

1/2

A 0 1B +C 2 -1D 3 -E 4 1

z

j

j

θ

ππππ +

point

r = 1

25

Branch Cuts in Radiation Problem

Now we return to the problem (line source over grounded slab):

( )1/22 20 0y xk k k= −

( ) 00 0

0

1 14

y x

z

jk y jk xTEx x

y

AI

k e e dkj k

ψµ

ψπ

+∞ − −

−∞

=

= + Γ ∫

26

Note: There are no branch points from ky1:

( )1/22 21 1y xk k k= − ( ) ( )1 1tanTE TE

in x yZ k jZ k h= 01

1

TE

y

Zkωµ

=

(The integrand is an even function of ky1.)

Branch Cuts

( ) ( ) ( )

( ) ( )

1/2 1/2 1/22 20 0 0 0

1/2 1/20 0

y x x x

x x

k k k k k k k

j k k k k

= − = + −

= − − +

Branch points appear at 0xk k= ±

No branch cuts appear at 1xk k= ± (The integrand is an even function of ky1.)

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.

27

Branch Cuts (cont.)

( ) ( )1/2 1/20 0 0y x xk j k k k k= − − +

Cxrk

xik

0k−

0k

1k1k−

Branch cuts are lines we are not allowed to cross. 28

For 0

0 0

,x

y y

k k

k j k

= >

= −

real

xrk

xik

0k−

0k

Choose ( )( )

0

0

arg 0

arg 0x

x

k k

k k

− =

+ =

This choice then uniquely defines ky0 everywhere in the complex plane.

( )1/22 20 0y xk k k= −

( ) ( )1/2 1/20 0 0y x xk j k k k k= − − +

Branch Cuts (cont.)

at this point

29

0 0y yk j k= −

For 0

,0

x

x

kk k=

< <

real we have ( )( )

0

0

arg

arg 0x

x

k k

k k

π− =

+ =

/ 2 0/20 0 0

j jy x xk j k k e k k eπ = − − +

0 0y yk k=

Hence

Branch Cuts (cont.)

xik

xrk0k− 0k

30

/ 20 0

jy yk je kπ= −

( )1/22 20 0y xk k k= −

( ) ( )1/2 1/20 0 0y x xk j k k k k= − − +

Riemann Surface

( )1/22 20 0y xk k k= −

There are two sheets, joined at the blue lines.

xik

xrk0k−

0k

Top sheet

Bottom sheet

0 0y yk j k= −

0 0y yk j k= +

31 The path of integration is on the top sheet.

Proper / Improper Regions Let

“Proper” region:

0 0 0

x xr xik k jk

k k jk

= +

′ ′′= −

( )1/22 20 0y xk k k= −

“Improper” region:

0

0

Im 0

Im 0

<

>y

y

kk

Boundary: 0Im 0yk =

2 2 20 0 0= real real > 0y y xk k k k⇒ = − =

The goal is to figure out which regions of the complex plane are "proper" and "improper."

32

Proper / Improper Regions (cont.)

Hence

One point on curve:

( ) ( )2 2 2 20 0 0 02 2 real 0xr xi xr xik k k k j k k k k′ ′′ ′ ′′− − + + − − = >

0 0xr xik k k k′ ′′= −

0

0

xr

xi

k k

k k

′=

′′= −

xik

0k

0−k

xrk

( ) ( )2 2

0 0 real 0xr xik jk k jk′ ′′− − + = >

Therefore

0 0 0xk k k jk′ ′′= = −

(hyperbolas)

0 0 0k k jk′ ′′= −

33

Proper / Improper Regions (cont.)

Also 2 2 2 20 0 0xr xik k k k′ ′′− − + >

xik

0k

0−k

xrk

The solid curves satisfy this condition.

34

Complex plane: top sheet

xik

0k

0−k

xrk

Improper region

Proper

On the complex plane corresponding to the bottom sheet, the proper and improper regions are reversed from what is shown here.

Proper / Improper Regions (cont.)

35

0 0y yk j k≈ −

Sommerfeld Branch Cuts

Complex plane corresponding to top sheet: proper everywhere

Complex plane corresponding to bottom sheet: improper everywhere

xik

0k

0−k

xrk

Hyperbola

36

Sommerfeld Branch Cuts

Note: We can think of a two complex planes with branch cuts, or a Riemann surface with hyperbolic-shaped “ramps” connecting the two sheets.

xik

0k

0−k

xrk

Riemann surface

xik

0k

0−k

xrk

Complex plane

The Riemann surface allows us to show all possible poles, both proper (surface-wave) and improper (leaky-wave).

37

Sommerfeld Branch Cut

Let

xik

0k

0−kxrk

0 0k ′′ →

The branch cuts now lie along the imaginary axis, and part of the real axis.

38

Path of Integration xik

0k

0−kxrk

1k

1−k C

The path is on the complex plane corresponding to the top Riemann sheet.

39

Numerical Path of Integration

xik

0k

0−k

xrk1k

1−kC

40

Leaky-Mode Poles

(improper)

( ) ( )0

0Im 0

LW LWin x x

y

Z k Z k

k

= −

>

Note: TM0 never becomes improper.

Frequency behavior on the Riemann surface

41

TRE:

xik

0kxrk

1k

= cf fSW

ISW

LW

0f →

sf f=

Bottom sheet

Riemann Surface

xik

0k

0−k

xrk1k

1−kC

SWP LWP

BP

0 0LWk kβ− ≤ ≤

The LW pole is then “close” to the path on the Riemann surface (and it usually makes an important contribution).

We can now show the leaky-wave poles!

LW LW LWxpk jβ α= −

( )ReLW LWxpkβ =

42

SW and CS Fields

Total field = surface-wave (SW) field + continuous-spectrum (CS) field

Note: The CS field indirectly accounts for the LW pole.

43

xik

0kxrk

1k

SW LW bC

CS field SW field

pC

Leaky Waves LW poles may be important if

0 0LWk kβ− ≤ ≤

0LW kα

Physical Interpretation

Re( )LWLW xpkβ =

0θ

leaky wave

radiation

0 0sinLW kβ θ≈

The LW pole is then “close” to the path on the

Riemann surface.

44

Improper Nature of LWs

The rays are stronger near the beginning of the wave: this gives us exponential growth vertically.

Region of strong leakage fields

β α= −LWxpk j

“leakage rays”

45

Improper Nature (cont.) Mathematical explanation of exponential growth (improper behavior):

Equate imaginary parts:

( )( ) ( )

( ) ( )

1/22 20 0

2 220 0

2 220

LW LWy xp

LW LWy xp

y y

k k k

k k k

j k jβ α β α

= −

= −

− = − −

β α β α

βα αβ

= −

= −

y y

yy

(improper) 0 0yα α> → <46