Prof. David R. Jackson ECE Dept. Spring 2014 Notes 35 ECE 6341 1.
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Transcript of Prof. David R. Jackson ECE Dept. - ece.ee.uh.eduece.ee.uh.edu/ECE/ECE6341/Class Notes/Topic 6...
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Prof. David R. Jackson ECE Dept.
Spring 2016
Notes 36
ECE 6341
1
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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
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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
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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
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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= +
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Poles (cont.)
6
Cxrk
xik
0k1k
0k−1k−
For the lossless case, two possible paths are shown here.
Cxrk
xik
0k1k
0k−1k−
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Review of Branch Cuts and Branch Points
In the next few slides we review the basic concepts of branch points and branch cuts.
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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
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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.)
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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
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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
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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.)
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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− < ≤
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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.)
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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.)
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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.)
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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.)
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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)
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( ) ( ) ( ) ( ) ( )( )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.
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( ) ( )( )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
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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.
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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.
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Riemann Surface (cont.) x
y
BD
Top
Bottom
BD
D
B
B
Dside view
x
y
top view 23
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Riemann Surface (cont.)
24
Bottom sheet
Top sheet
Branch cut
(where it used to be)
Branch point (where it used to be)
xy
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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
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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.)
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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.
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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
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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
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0 0y yk j k= −
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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= − − +
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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.
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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."
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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′ ′′= −
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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.
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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.)
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0 0y yk j k≈ −
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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
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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).
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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.
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Path of Integration xik
0k
0−kxrk
1k
1−k C
The path is on the complex plane corresponding to the top Riemann sheet.
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Numerical Path of Integration
xik
0k
0−k
xrk1k
1−kC
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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
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TRE:
xik
0kxrk
1k
= cf fSW
ISW
LW
0f →
sf f=
Bottom sheet
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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 =
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SW and CS Fields
Total field = surface-wave (SW) field + continuous-spectrum (CS) field
Note: The CS field indirectly accounts for the LW pole.
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xik
0kxrk
1k
SW LW bC
CS field SW field
pC
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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.
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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”
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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