Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1

Transcript of Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

Page 1: Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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Prof. David R. JacksonECE Dept.

Spring 2014

Notes 37

ECE 6341

Page 2: Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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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

Page 7: Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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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

Page 8: Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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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

Page 19: Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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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

Page 31: Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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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

Page 33: Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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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

Page 34: Prof. David R. Jackson ECE Dept. Spring 2014 Notes 37 ECE 6341 1.

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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.)