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Prof. David R. Jackson ECE Dept. Spring 2016 Notes 15 ECE 6341 1
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### Transcript of ECE 1100 Introduction to Electrical and Computer Engineeringcourses.egr.uh.edu/ECE/ECE6341/Class...

• Prof. David R. Jackson ECE Dept.

Spring 2016

Notes 15

ECE 6341

1

• Arbitrary Line Current

TMz: ( , )ρzA z

Introduce Fourier Transform:

1( ) ( )2

zjk zz zI z I k e dkπ

+∞−

−∞

= ∫

( )( ) zjk zzI k I z e dz+∞

+

−∞

= ∫

2

y

z

x

( )I z

• Arbitrary Line Current (cont.) View this as a collection of phased line currents::

Then

0( ) zjk zdI z I e−=

01 ( )

2π= z zI I k dk

( ) ( )

( ) ( )

20 00

200

41 ( )

4 2

z

z

jk zz

jk zz z

IdA H k ej

I k dk H k ej

ρ

ρ

µ ρ

µ ρπ

=

=

(from Notes 11)

3

1( ) ( )2

zjk zz zI z I k e dkπ

+∞−

−∞

= ∫

• Hence, from superposition, the total magnetic vector potential is

( ) ( )20 0( )8zjk z

z z zA I k H k e dkj ρµ ρπ

+∞−

−∞

= ∫

Arbitrary Line Current (cont.)

4

( )1/22 2zk k kρ = −

2 2

2 2

,

,z z

z z

k k k kk

j k k k kρ

− ≤= − − ≥

• Example

Uniform phased line current

00( ) z

jk zI z I e−=

0

0

0

( )0

0 0

( )

2 ( )

z z

z z

jk z jk zz

j k k z

z z

I k I e e dz

I e dz

I k kπδ

+∞− +

−∞

+∞−

−∞

=

=

= −

( ) 12

12

x

x

jk xx

jk xx

x e dk

e dk

δπ

π

∞−

−∞

∞+

−∞

=

=

5

y

z

x

( )I z

Note:

• Example (cont.) Hence

( ) ( )

[ ] ( ) ( )

( ) ( )( ) ( )

0

0

200

200 0 0

200 0 0

20 00 0

( )8

2 ( )8

28

4

z

z

z

z

jk zz z z

jk zz z z

jk z

jk z

A I k H k e dkj

I k k H k e dkj

I H k ej

I H k ej

ρ

ρ

ρ

ρ

µ ρπ

µ πδ ρπµ π ρπµ ρ

+∞−

−∞

+∞−

−∞

=

= −

=

=

( )1/22 20 0zk k kρ = −where 6

• Example (cont.) Hence

( )1/22 20 0zk k kρ = −where

0(2)00 0( )4

zjk zz

IA H k ej ρ

µ ρ −

=

7

If kz0 is real, then

2 20 0

0 2 20 0

,

,z z

z z

k k k kk

j k k k kρ

− ≤= − − ≥

This is the correct choice of the wavenumber.

y

z

x

( )I z

• Example (cont.)

( )1/22 20 0zk k kρ = −

8

If kz0 is complex:

( )( )

0 0

0 0

0 : Im 0 ( )

: Im 0 ( )

z

z

k k

k kρ

ρ

β

β

< < >

> <

improper

proper

0 0 0z z zk jβ α= −

This is the “physical” choice of the wavenumber.

Note that the radiation condition at infinity is violated (z → - ∞), so we lose uniqueness. However, we can still talk about what choice of the square root is “physical.”

y

z

x

( )I z

• Example (cont.)

9

( )0 00 : Im 0 ( )z k kρβ< < > improper

A semi-infinite leaky-wave line source produces a cone of radiation in the near field.

y

z

x

( )I z

Cone of leakage Region of exponential growth

• Example

Dipole

( ) ( )I z Il zδ=

( )zI k Il=

10

y

z

x

Il

• Example (cont.)

Also,

( ) ( )20 0 ( )8zjk z

z zA I H k e dkj ρµ ρπ

+∞ −

−∞= ∫

( ) 04

jkr

zeA I

rµπ

= (from ECE 6340)

( ) ( )20 0( )8zjk z

z z zA I k H k e dkj ρµ ρπ

+∞−

−∞

= ∫

Hence

11

2 2

2 2

,

,z z

z z

k k k kk

j k k k kρ

− ≤= − − ≥

• Example (cont.) Hence,

A spherical wave is thus expressed as a collection of cylindrical waves.

( )20

1 ( )2

z

jkrjk z

ze H k e dk

r j ρρ

− +∞ −

−∞= ∫

12

• Example (cont.) Use a change of variables:

( ) ( )

( )

2 20 0

20

1 1( ) ( )2 2

1 ( )2

z z

CCW

z

CW

jk z jk zz

zC

jk z

zC

kH k e dk H k e dk

j j k

kH k e dk

j k

ρρ ρ ρ

ρρ ρ

ρ ρ

ρ

+∞ − −

−∞

−=

=

∫ ∫

13

( )

( )

1/22 20

1/22 20

z

zz

k k k

k kdk dk dk

kk k

ρ

ρ ρρ ρ

ρ

= −

= − = −−

( )20

1 ( )2

z

jkrjk z

z

ke H k e dkr j k

ρρ ρρ

− +∞ −

−∞

=

We then have

or

(The path has been deformed back to the real axis, please see the next two slides)

• Example (cont.)

14

( )1/22 20zk k kρ= −

Branch cut

( )Re kρCCCW

( )Im kρ

0k

Mapping equation for C:

Note: The value of kz is opposite across the branch cut.

CCCW: counterclockwise around branch cut

0k−

( ),CCW

z

k C

kρ ∈

∈ −∞ ∞

The square root is defined so that when kρ is on the real axis we have:

2 20 0,zk j k k k kρ ρ= − − >

• Example (cont.)

15

( )Re kρCCW

( )Im kρ

0k

CCW: clockwise around branch cut

Deform path

0k−

( )Re kρ

( )Im kρ

0k

Branch cut

0k−

“Sommerfeld path”

• Example (cont.)

16

Alternative form:

( )

( ) ( )

( ) ( )

20

0 2 20 00

0 2 20 00

0

( )2

( ) ( )2 2

( ) ( )2 2

2

z

z z

z z

jkrjk z

z

jk z jk z

z z

jk z jk z

z z

z

ke j H k e dkr k

k kj jH k e dk H k e dkk k

k kj jH k e dk H k e dkk k

kj Hk

ρρ ρ

ρ ρρ ρ ρ ρ

ρ ρρ ρ ρ ρ

ρ

ρ

ρ ρ

ρ ρ

− +∞ −

−∞

+∞− −

−∞

+∞− −

= −

= − − ′

′ ′= − − −

′ = +

∫ ∫

∫ ∫

( ) ( )

( ) ( )

( )

0 1 200

1 20 00 0

00

( ) ( )2

( ) ( )2 2

z z

z z

z

jk z jk z

z

jk z jk z

z z

jk z

z

kjk e dk H k e dkk

k kj jH k e dk H k e dkk k

kj J k e dk

k

ρρ ρ ρ ρ

ρ ρρ ρ ρ ρ

ρρ ρ

ρ ρ

ρ ρ

ρ

+∞− −

∞ +∞− −

∞ −

′ ′ −

′ ′= − −

= −

∫ ∫

∫ ∫

k kρ ρ′ = −

( ) ( ) ( ) ( )2 10 0H z H z− = − +

( ) ( ) ( ) ( ) ( )( )1 20 0 012J z H z H z= +

• Example (cont.)

17

00

1 ( ) zjkr

jk z

z

ke J k e dkr j k

ρρ ρρ

− +∞ − =

Sommerfeld Identities

Hence we have:

( )20

1 ( )2

z

jkrjk z

z

ke H k e dkr j k

ρρ ρρ

− +∞ −

−∞

=

The integrals are along a “Sommerfeld path” that stays slightly above or below the branch cuts.

• Far-Field Identity This identity is useful for calculating the far-field of finite (3-D) sources in cylindrical coordinates.

Note: We assume that the current decays at z = ± ∞ fast enough so that a 3-D far field exists.

18

y

z

x

( )I zr θ

• Far-Field Identity (cont.)

( )200( ) ( )8

zjk zz z zA I k H k e dkj ρ

µ ρπ

+∞ −

−∞= ∫

Exact solution:

19

y

z

x

( )I zr θ

• Far-Field Identity (cont.) From ECE 6340, as →∞r

0~ ( , )4

jkreA ar

µ θ φπ

( ) ( )sin cos sin sin cos

cos0

( , )

ˆ ( )4

jk x y z

V

jkrj kz

a J r e dx dy dz

ez I z e dzr

θ φ θ φ θ

θ

θ φ

µπ

′ ′ ′+ +

− +∞ ′

−∞

′ ′ ′ ′=

′ ′=

Hence

0~ ( cos )4

jkr

zeA I k

rµ θπ

20

• Far-Field Identity (cont.)

Hence, comparing these two,

( )20 00( ) ( ) ~ ( cos )8 4

z

jkrjk z

z zeI k H k e dk I k

j rρµ µρ θπ π

−+∞ −

−∞

r →∞as

( )20( ) ( ) ~ 2 ( cos )z

jkrjk z

z zeI k H k e dk j I k

rρρ θ

−+∞ −

−∞

or

21

• Far-Field Identity (cont.)

To generalize this identity, use

( ) ( )2 2 42( ) ~nj x

nH x ex

π π

π− − −

Hence

( ) ( )2 20( ) ~ ( )

nnH x j H x

22

• Far-Field Identity (cont.)

Therefore, we have

( )2 1( ) ( ) ~ 2 ( cos )zjkr

jk z nz n z

eI k H k e dk j I krρ

ρ θ−+∞ − +

−∞

Note: This is valid for ρ →∞

, 0, 180r θ→∞ ≠ Hence, this is valid for

23

• Far-Field Identity (cont.)

Since the current function is arbitrary, we can write

( )2 1( ) ( ) ~ 2 ( cos )zjkr

jk z nz n z

ef k H k e dk j f krρ

ρ θ−+∞ − +

−∞

for , 0, 180r θ→∞ ≠

24

Slide Number 1Slide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24