1

Prof. David R. JacksonECE Dept.

Spring 2014

Notes 12

ECE 6341

2

Tube of Surface Current

Top view:

An infinite hollow tube of surface current, flowing

in the z direction:

1 2

x

y

a ( )szJ

From superposition, we know we can solve the problem using

ˆA z (TMz)

y

x

z

( )szJ a

1 2

3

Tube of Current (cont.)

Represent Jsz(f) as a (complex exponential) Fourier series:

( ) jnsz n

n

J c e

2

0

1( )

2jn

n szc J e d

The coefficients cn are assumed to be known.

4

r < a 1 ( )jnn n

n

a e J k

(2)2 ( )jn

n nn

b e H k

Apply B.C.’s (at r = a) :

1 2

2 1 ( )z z

sz

E E

H H J

r > a

Tube of Current (cont.)

5

The first B.C. gives

1 2

2 11( )szJ

The second B.C. gives

Hence we have, from the first BC,

2( ) ( )n n n na J ka b H ka

Tube of Current (cont.)

1z H

TM Table :

6

From the second BC we have

Substituting for an from the first equation, we have

Next, look at the term

21( ) ( )n n n n nk b H ka a J ka c

2 ( )( ) ( )

( )

nn

n n n nn

H kab H ka J ka c

J ka k

2 2( ) ( ) ( ) ( )n n n nJ ka H ka J ka H ka

Tube of Current (cont.)

2 2( ) ( ) ( ) ( ) ( )n n n n n n nb J ka H ka J ka H ka c J kak

7

This may be written as (using x = ka)

Hence,

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2

n n n n n n

n n n n

J x J x x jY x J x J x jY x

j J x Y x J x Y x

jx

2( )n n n

jb c J ka

ka k

(This follows from the Wronskian identity below.)

Tube of Current (cont.)

2( ) ( ) ( ) ( )n n n nJ x Y x J x Y x

x

8

or ( )2n n n

ab j c J ka

Using

(2) ( )2n n n

aa j c H ka

(2)( ) ( )n n n na J ka b H ka

we have

Tube of Current (cont.)

9

r < a 1 ( )zjk z jnn n

n

e a e J k

(2)2 ( )zjk z jn

n nn

e b e H k

r > a

Tube of Current (cont.)Generalization: a phased tube of current

( ) zjk z jnsz n

n

J e c e

(2) ( )2n n n

aa j c H k a

( )

2n n n

ab j c J k a

10

Inductive Post

Equivalence principle: remove coax aperture and probe metal

2a ,r r hI0

0

2sz

IJ

a

The magnetic-current frill is ignored in calculating the fields.

Coax feed

2b The probe current is assumed to be uniform

(no z or variation).

Parallel-plate waveguide

sM

I0

11

Inductive Post (cont.)

Tube problem:

I0 ˆ zE z E

0

0

E

E

For r a: 22 ( )jn

z n nn

A b e H k

( )2n n n

ab j c J ka

The probe problem is thus equivalent to an infinite tube of current in free space.

Tangential electric field is zero

where

(There is no z or variation of the fields.)

21E

j z

21E

j z

22

2

1zE k

j z

12

Hence

20 0 ( )zA b H k

and

Inductive Post (cont.)

0

0

1

2

1

2 2

, 020, 0

jnn sz

jn

c J e d

Ie d

a

In

an

0 0 0 ( )2

ab j c J ka

0

0 2

Ic

a

13

so

200 0 ( )

2 2z

IaA j J ka H k

a

22

2

1z z zE k A j A

j z

200 0( ) ( )

4z

IE J ka H k

Inductive Post (cont.)

We then have

14

Complex source power radiated by tube of current (height h):

*

*

1

21

22

ss s

sz z a

P E J dS

a h J E

or

*0

*20 0

0 0

1(2 )

2 2

1(2 ) ( ) ( )

2 2 4

s z a

IP a h E

a

I Ia h J ka H ka

a

Inductive Post (cont.)

15

or

or

2 20 0 0( ) ( )

8s

hP I J ka H ka

2 (2)0 0 0

( )( ) ( )

8s

khP I J ka H ka

0 0r

r rr

k k

where

Inductive Post (cont.)

16

2*0 0 0

1 1

2 2s inP V I Z I

2

0

2 s

in

PZ

I

so

Inductive Post (cont.)

Equivalent circuit seen by coax:

+I0

ZinCoaxV0

-

17

For ka << 1

20 0

( )( ) ( )

4in

khZ J ka H ka

( ) 21 ln

4 2in

kh kaZ j

Therefore

We thus have the “probe reactance”:

( ) 2ln

4 2p

kh kaX

Inductive Post (cont.)

0.5772156

Imp inX Z

18

or

2( ) ln

2pX khka

Xin

0.5772156

Inductive Post (cont.)

1 2( ) ln

2p

p

XL kh

ka

Probe inductance:

19

Summary

00

0

2( ) ln

2p r

r r

X k hk a

Inductive Post (cont.)

0

0

2ln

2p r

r r

hL

k a

Probe inductance:

Probe reactance:

20

Inductive Post (cont.)

Comments:

The probe inductance increases as the length of the probe increases.

The probe inductance increases as the radius of the probe decreases.

The probe inductance is usually positive since there is a net stored magnetic energy in the vicinity of the probe.

This problem provides a good approximation for the probe reactance of a microstrip antenna (as long as kh << 1).

21

Example

0

2.94

0.1524 [cm] 60mils

0.0635 [cm] 50[ ]

2.0 [GHz] 14.9896 [cm]

r

h

a

f

SMA Connector

12.3 [ ]pX

0.979 [nH]pL

Practical patch antenna case:

22

RLCpin ZjXZ

R L CZ0

Lp

resf

fpX

R

0f X

0resf f f

Microstrip Antenna Model

R Xor

inZ R jX

f0 = resonance frequency of RLC circuit

The frequency where R is maximum is different from the frequency where the reactance is zero, due to

the probe reactance.

23

Lossy Post

If the post has loss (finite conductivity), then we need to account for the skin effect, which accounts for the magnetic energy stored inside the post.

00

0

2( ) ln

2extp r

r r

X k hk a

2intp s

hX X

a

1

2s sX R

24

Example (Revisited)

0

2.94

0.1524 [cm] 60mils

0.0635 [cm] 50[ ]

2.0 [GHz] 14.9896 [cm]

r

h

a

f

SMA Connector

12.3 [ ]extpX

0.0062 [ ]intpX

Practical patch antenna case:

73.0 10 S / m Assume:

25

Accuracy of Probe Formula

2.2

2.0 GHz

0.635 mm (SMA connector)

2.188 mm (50 coax)

r

f

a

b

H. Xu, D. R. Jackson, and J. T. Williams, “Comparison of Models for the Probe Inductance for a Parallel Plate Waveguide and a Microstrip Patch,” IEEE Trans. Antennas and Propagation, vol. 53, pp. 3229-3235, Oct. 2005.

0 0.005 0.01 0.015 0.02 0.025

h (m)

-300

-200

-100

0

100

200

300

Xin

(

)

Uniform current modelFrill modelCosine current modelHFFS simulation

0/ 0.1h

HFSS