1

Prof. David R. JacksonECE Dept.

Spring 2014

Notes 34

ECE 6341

2

Example

By using a Fourier-transform method, the solution is

ˆ ,zA x ywhere, for y > 0,

0 14

y xjk y jk xx

y

Ie e dk

j k

2

(Please see the appendix.)

y

0Ix

0zJ x y I

1k

,

Line current

1

2 2 2 y xk k k

3

Example (cont.)

The vertical wavenumber is

The wavenumber ky is interpreted as

1

2 2 2 y xk k k

y x

y

y x

k k kk

j k k k

A convenient change of variables is the “steepest-descent transformation”

sinxk k ( ) r ij

(This follows from the radiation condition at infinity.)

4

Example (cont.)

Then

1

2 2 2 2sin cos yk k k k

The path C in the complex -plane is not unique until we choose either + or – here.

This is because the path is not uniquely determined by only sinxk k

To see this in more detail, write

sin

sin cosh cos sinh

x r i

r i r i

k k j

k j

5

Example (cont.)

Because kx is real,

Hence

cos sinh 0 r i

0

3, ,...

2 2

i

r

or

i

r

0xk

0xk

0xk0xk

0xk

0xk

sin coshx r ik k

/ 2/ 2

6

Example (cont.)There are four possible paths.

i

r

/ 2/ 2

r

i

/ 2/ 2

r

i

/ 2/ 2

r

i

/ 2/ 2

7

Example (cont.)

kx will vary from - to along each of these paths.

The path must be chosen so that along the path

Re 0

Im 0

y

y

k

k

Assume we choose the + sign (an arbitrary choice):

cos cos

cos cosh sin sinh

y r i

r i r i

k k k j

k j

8

Example (cont.)

Correct path C :

cos cosh sin sinhy r i r ik k j

y yk j k

y yk j k

y yk j k

y yk j k

y yk k y yk k

i

r/ 2 / 2

C

9

Example (cont.)

Now proceed with the change of variables:

sin

cosx

x

k k

dk k d

0 1

4y x

jk y jk xx

y

e e dkk

Ij

sin cos0 1cos

4 cosjk x y

C

Ie k d

j k

Hence, we have

cosyk k

10

Example (cont.)

Next, let

sin cos0

4jk x y

C

Ie d

j

sin

cos

x

y

x

y

2

sin cos sin sin cos cos cosx y

11

Example (cont.)

The integral then becomes

cos0

4jk

C

Ie d

j

Ignoring the constant in front, we can identify

1f

0 Hence

cos

sin

cos

g j

g j

g j

k

0

0

g j

g j

12

Example (cont.)

SDP:

so

0Im Im Im 1v g g j constant

cos

cos

cos cosh sin sinh

r i

r i r i

g j

j j

j j

cos cosh 1r i (SDP or SAP)

cos coshr iv

Hence

13

Example (cont.)

cos cosh sin sinhr i r ig j j

sin sinhr iu

Using

Using

Reu g

we also see that

This will help us determine which curve is the SDP and which is the SAP.

14

Example (cont.)

sin sinhr iu

cos cosh 1r i (SDP or SAP)

2

0u

r

2

0u

0u

0u

0u

i

SDP

SAP

2

2

15

Example (cont.)

Examination of the original path allows us to determine the direction of integration along the SDP.

2

0u

r

2

0u

0u

0u

0u

i

SDP

SAP

2

2

2

SDP

i

r

2

C

16

Example (cont.)

Calculate :

From the figure we see that the correct choice is

so

SDP

0

2 2

arg arg2

4 2

SDP

SDP

g j

4SDP

or4SDP

3

4SDP

17

Example (cont.)

We then have

0 4

2~

4

jjkIe e

j k j

0

00

2~ SDPg z jI f z e e

g z

0 4

2~

4

jjkIe e

j k

or

Recipe:

4SDP

0

1f

0g j

k

0g j

18

Example (cont.)

It can easily be verified that

The exact solution is: 2004

ex IH k

j

~ ex 1 kfor

0 4

2~

4

jjkIe e

j k

( ) ( )2 22 4 40

2 2( ) ~ ( ) ~

j x j xH x e H x e

x x

19

AppendixDerivation of formula

,zA x y

2 2

2 22

2 2

0

0

k

kx y

TMz :

y

0Ix

0zJ x y I

1k

,

20

Appendix (cont.)

, ,

1, ,

2

x

x

jk xx

jk xx x

k y x y e dx

x y k y e dk

2

2 22

0xk ky

, ,0 ,yj k y

x x xk y k e k y

We then have

1

2 2 2 y xk k kDefine:

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Appendix (cont.)

Choose - sign for

Boundary Conditions at y = 0:

(satisfied automatically)

0y

, ,0 yjk y

x xk y k e

0

0

z z

x x sx

x x

E E

H H J I x

H H I

1

xHy

22

Appendix (cont.)Hence

1,0

1,0

x y x

x y x

H jk k

H jk k

0

0

2,0

,02

y x

xy

jk k I

Ik

j k

We then have

23

Appendix (cont.)

Hence

And then

01,0

2 2xjk x

xy

Ix e dk

jk

0 1,

4y x

jk y jk xx

y

Ix y e e dk

j k