prof. david r. jackson ece dept. spring 2014 notes 37 ece 6341 1
TRANSCRIPT
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Prof. David R. JacksonECE Dept.
Spring 2014
Notes 37
ECE 6341
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Line Source on a Half Space
1/2 1/22 2 2 20 0 1 1y x y xk k k k k k
There are branch points at 0 1,x xk k k k
r
y
x
00 0
0
11
4y x
z z
jk y jk xTEz x x
y
E j A
IA k e e dk
j k
01
1
TE
y
Zk
00
0
TE
y
Zk
1 0
1 0
TE TEx xTE
x TE TEx x
Z k Z kk
Z k Z k
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Steepest-Descent Transformation
0 0 0sin cosx yk k k k
There are no branch points in the plane from ky0 (cos is analytic).
There are still branch points in the plane from ky1:
Steepest-descent transformation:
1/22 2 21 1 0 sinyk k k
1sin n
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Line Source on a Grounded Substrate
1/2 1/22 2 2 20 0 1 1y x y xk k k k k k
There are branch points only at 0xk k
00 0
0
11
4y x
z z
jk y jk xTEz x x
y
E j A
IA k e e dk
j k
01
1
TE
y
Zk
00
0
TE
y
Zk
0
0
TE TEin x xTE
x TE TEin x x
Z k Z kk
Z k Z k
r
y
x
1 1tanTE TEin x yZ k jZ k h
(even function of ky1)
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Steepest-Descent Path Physics
0 0 0sin cosx yk k k k
There are no branch points in the plane (cos is analytic).
Both sheets of the kx plane get mapped into a single sheet of the plane.
Steepest-descent transformation:
We focus on the grounded substrate problem for the remaining discussion.
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Steepest-Descent Path Physics
Examine ky0 to see where the plane is proper and improper:
0 0
0
cos
cos cosh sin sinh
y r i
r i r i
k k j
k j
0 0Im sin sinhy r ik k
0
0
: Im 0
: Im 0
y
y
k
k
Proper
Improper
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SDP Physics (cont.)
0 0Im sin sinhy r ik k
P: properI: improper i
I
2
r
2
I
IIP P
P P
C
0
0
: Im 0
: Im 0
y
y
k
k
Proper
Improper
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SDP Physics (cont.)
0 0sin sin cosh cos sinhx r i r ik k k j Mapping of quadrants
in kx plane
Non-physical “growing” LW pole
(conjugate solution)
xrk
xik
12
3 4
0k0k
/ 2 / 2conjr r r
conji i
i
I
2
r
2
LWP
SWP
I
II
PP
PP
4
1
1
4
3
2
2
3
C
(symmetric about /2 line)
*LWxp xk k
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SDP Physics (cont.)
SDP: cos cosh 1 r i
A leaky-wave pole is considered to be physical if it is captured when
deforming to the SDP (otherwise, there is no direct residue contribution).
i
2
r
2
SDP
LWP
SWP
C
0
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SDP Physics (cont.)
LWP:
0
3/2
jk x
z
eE A
x
SDP:
2 ResLWxjk x
zE j e (exists if pole is captured)
The leaky-wave field is important if:
1) The pole is captured (the pole is said to be “physical”).2) The residue is strong enough.3) The attenuation constant is small.
Comparison of Fields:
LWxk j
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SDP Physics (cont.)
LWP captured: b
b rp
Note:
The angle b represents the
boundary for which the leaky-wave field exists.
i
r
SDP
b rp
p rp ipj LWP
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SDP Physics (cont.)
0
cos0 cos
y xjk y jk x
z x x
j k
C
E F k e e dk
F e k d
cos
02 Res cos pj kLWz p pE j F k e
Behavior of LW field:
0xp y pjk x jk yLWzE Ae e
In rectangular coordinates:
LWxp xk k j where
(It is an inhomogeneous plane-wave field.)
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SDP Physics (cont.)
Examine the exponential term:
cos cos
cos cosh sin sinh
p rp ip
rp ip rp ip
j
j
Hence
0
0
sin sinh
sinh sin
rp ip
ip rp
k
k
e
e
cos pj ke
0ip since
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Radially decaying:
rp
SDP Physics (cont.) 0 sinh sinip rpk
e
LW exists:
b
b rp
Also, recall that LW exists
r
rp
b
y
x
LW decays radially
Line source
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Power Flow
0
0 0
0
ˆ ˆRe Re( )
ˆ ˆRe sin cos
ˆ ˆsin cosh cos cosh
x y
rp ip rp ip
rp ip rp ip
k x k yk
x k j y k j
k x y
0
x
y
Power flows in the direction of the vector.
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Power Flow (cont.)
0 rp Hence
0tan tanxrp
y
Note that
0 ˆ ˆsin cosh cos coshrp ip rp ipk x y
Also, 0 0ˆ ˆsin cosx y
rp
r
y
x
b
0
Note: There is no
amplitude change along the rays ( is perpendicular to in a lossless region).
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ESDP (Extreme SDP)
/ 2
ESDP
i
r
2
Fast
Slow
Set
We can show that the ESDP divides the LW region into slow-wave and fast-wave regions.
The ESDP is important for evaluating the fields on the interface (which determines the far-field pattern).
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ESDP (cont.)
(SDP)
(ESDP)
cos cosh 1
sin cosh 1
r i
r i
Recall that
0
0
sin
sin
xp p
rp ip
k k
k j
0
Re
sin cosh
xp
rp ip
k
k
To see this:
Hence
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ESDP (cont.)
Fast-wave region:
Slow-wave region:
0
sin cosh rp ipk
0
1
k
0
1
k
Hence
sin cosh 1r i
sin cosh 1rp ip
sin cosh 1rp ip
Compare with ESDP:
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ESDP (cont.)
/ 2
i
r
ESDP
2
Fast
Slow
The ESDP thus establishes that for fields on the interface, a leaky-wave pole is physical (captured) if it is a fast wave.
LWP captured
LWP not captured
SWP
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SDP in kx Plane
0
0
sin
sin
x
r i
k
k j
k
0
0
sin cosh
cos sinh
xr r i
xi r i
k k
k k
cos cosh 1r i SDP:
We now examine the shape of the SDP in the kx plane.
The above equations allow us to numerically plot the shape of
the SDP in the kx plane.
so that
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SDP in kx Plane (cont.)
2
(Please see the appendix for a proof.)
LW
SDP
xik
0kxrk
0 sinxk k
C
SW
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Fields on Interface
2
The leaky-wave pole is captured if it is in the fast-wave region.
LW
ESDP
xik
0kxrk
SW
fast-wave region
The SDP is now a lot simpler!
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Fields on Interface (cont.)
SW CS
z z z
SW LW RWz z z
E E E
E E E
The contribution from the ESDP is called the “space-wave” field or the “residual-wave” (RW) field.
(It is similar to the lateral wave in the half-space problem.)
2
LW
ESDP
xik
0kxrk
SW
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Asymptotic Evaluation of “Residual-Wave” Field
( 0)xjk xRWz x x
EDSP
E F k e dk y
Use
0x
x
k k js
dk j ds
xik
0k
xrk
- + s
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Define
0
0
00
00
jk xRW sxz
jk x sx
E je F k js e ds
je F k js e ds
H s F s F s
0
0
00
0
0
jk xRW sxz
jk x sx
E je F k js e ds
je F k js e ds
Asymptotic Evaluation of “Residual-Wave” Field (cont.)
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Then
0
0
jk xRW sxzE je H s e ds
x x for
Assume ~ 0H s As s as
0
1
1~ jk xRW
z
AE je
x
Watson’s lemma (alternative form):
We then have
Asymptotic Evaluation of “Residual-Wave” Field (cont.)
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It turns out that for the line-source problem at an interface,
1
2
Hence0
3/2
3~
2
jk xRWz
eE j A
x
Note: For a dipole source we have0
1 2
jkRWz
eE A
Asymptotic Evaluation of “Residual-Wave” Field (cont.)
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Discussion of Asymptotic Methods
We have now seen two ways to asymptotically evaluate the fields on
an interface as x for a line source on a grounded substrate:
1) Steepest-descent ( ) plane
2) Wavenumber (kx) plane
There are no branch points in the steepest-descent plane. The function f ( ) is analytic at the saddle point 0 = = /2, but is zero there. The fields on the interface correspond to a higher-order saddle-point evaluation.
The SDP becomes an integration along a vertical path that descends from the branch point at kx = k0. The integrand is not
analytic at the endpoint of integration (branch point) since there is a square-root behavior at the branch point. Watson’s lemma is used to asymptotically evaluate the integral.
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Summary of Waves
0
3/2~
jk xRWz RW
eE A
x
LWxjk xLW
z LWE A eSWxjk xSW
z SWE A e
x
y
LW
SWRW
Continuous spectrum
LW LW LWxk j SW SW
xk
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Interpretation of RW Field
The residual-wave (RW) field is actually a sum of lateral-wave fields.
x
y
c
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Proof of angle property:
0tan
tan
xr xr
xi xi
r i
k k k
k k
Hence ~ r
Appendix: Proof of Angle Property
0
0
sin cosh
cos sinh
xr r i
xi r i
k k
k k
The last identity follows from
or ~ r
2
tanxrr
xi
k
k
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As
Hence
2
i
r
~2
2
Proof (cont.)
On SDP:
~2
or
(the asymptote)
2
0u
r
2
0u
0u
0u
0u
i
SDP
SAP
2
2
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2
ESDP:2
0
Hence
To see which choice is correct:
In the kx plane, this corresponds to a vertical line for which
Proof (cont.)