Spring 2015 - University of Houstoncourses.egr.uh.edu/ECE/ECE6345/Class Notes/Topic 4... · Spring 2015. Notes 5 . ECE 6345 . Prof. David R. Jackson . ECE Dept. 1 . Overview This

Spring 2015

Notes 5

ECE 6345

Prof. David R. Jackson ECE Dept.

1

Overview

This set of notes discusses improved models of the probe inductance of a coaxially-fed patch (accurate for thicker

substrates). A parallel-plate waveguide model is initially assumed (at the end of the notes we will also look at the actual finite patch).

z

h ,r rε µ2a

x2b

2

Overview (cont.)

The following models are investigated:

Cosine-current model

Gap-source model

Frill model

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.

Reference:

3

Derivations are given in the Appendix. Even more details may be found in the reference below.

Cosine Current Model

cosI z k z h

Note: The derivative of the current is zero at the top conductor (PEC).

2

2

0c

inPZ

I

Pc = complex power radiated by probe current

z

h ,r rε µ2a

x

I (z)

We assume a tube of current (as in Notes 4) but with a z variation.

4

*12

p

c z szS

P E J dS= ∫

5

( ) 222 (2)0 0 0 0 0 0

0

1 1( ) sec (1 ) ( ) ( )8

min r m m m mmr

Z k h k h I k H k a J k aρ ρ ρη ε δε

∞

=

= +

∑

Cosine Current Model (cont.)

Final result:

where

0

1, 00, 0m

mm

δ=

= ≠

2 2

2 ( ) sin( )1 ( ) ( )m

mo

khI khkh mδ π

= + −

1/2221m

mk khρπ = −

0/m mk k kρ ρ=

Note: The wavenumber kρm is chosen to be a positive real number or a

negative imaginary number.

Gap Source Model

An ideal gap voltage source of height ∆ is assumed at the bottom of the probe.

z

h ,r rε µ2a

x1V + - ∆

1

inZI

6

Gap Source Model (cont.)

7

(2)0

(2)00 0 0

( )1 24 sinc(1 ) ( )

min r

m m m m

H k aa mY jh k H k a h

ρ

ρ ρ

ππ εη δ

∞

=

′ ∆ = + ∑

Final result:

where 1/22

21m

mk khρπ = −



0

1, 00, 0m

mm

δ=

= ≠

0/m mk k kρ ρ=

A magnetic frill of radius b is assumed on the mouth of the coax.

s ρˆˆ ˆM z E z ρ E

z

h ,r rε µ2a

xb

1 1

lnρEρ b / a

sM Eφ ρ= −

(TEM mode of coax, assuming 1 V)

10inZ

I

Frill Model

8

Choose:

Frill Model (cont.)

9

( )

( ) ( )

( )

2 20 0

2200 0 0 0

( ) ( )1 1 1 4ln / ( ) (1 ) ( )

m min r

m m m m

H k b H k aY j

k h b a k H k aρ ρ

ρ ρ

πεη δ

∞

=

− = +

∑

Final result:

where

0

1, 00, 0m

mm

δ=

= ≠



1/2221m

mk khρπ = −

0/m mk k kρ ρ=

Comparison of Models

Next, we show results that compare the various models, especially as the substrate thickness increases.

z

h ,r rε µ2a

x2b

10

Comparison of Models

0 0.005 0.01 0.015 0.02 0.025h (m)

-300

-200

-100

0

100

200

300

Xin

(Ω)

Uniform current modelFrill modelCosine current modelHFFS simulation

Models are compared for changing substrate thickness.

εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)

11

z

h ,r rε µ2a

x2b

(λ0 = 15 cm)

Comparison of Models (cont.)

0 0.005 0.01 0.015 0.02 0.025h (m)

0

50

100

150

200

250

300

350

400

450

500

Rin

(Ω)

Uniform current modelFrill modelCosine current modelHFFS simulation

εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)

/ 4 0.025 [m]dλ =Note:

Models are compared for varying substrate thickness.

12

z

h ,r rε µ2a

x2b

(λ0 = 15 cm)

For the gap-source model, the results depend on ∆.

0 0.005 0.01 0.015 0.02 0.025h (m)

-300

-200

-100

0

100

200

300

Xin

(Ω)

Delta=0.01mmDelta=0.1mmDelta=1mm


εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)

13

z

h ,r rε µ2a

x1V+-∆

(λ0 = 15 cm)

The gap-source model is compared with the frill model,

for varying ∆, for a fixed h.

0.0001 0.0002 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0009 0.001

∆ (m)

-300

-200

-100

0

100

200

300

400

500

Rin

or X

in (Ω

)

Gap source model reactance Frill model reactance Gap source model resistance Frill model resistance


εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)

0.52mm3

b a−=

h = 20 mm

R

X

14

z

h ,r rε µ2a

x1V+-∆

(λ0 = 15 cm)

These results suggest the “1/3” rule: The best ∆ is chosen as

3b a−

∆ =

This rule applies for a coax feed that has a 50 Ω impedance.


15

z

h ,r rε µ2a

x1V+-∆

z

h ,r rε µ2a

x2b


0 0.005 0.01 0.015 0.02 0.025h (m)

-300

-200

-100

0

100

200

Xin

(Ω)

Gap source modelFrill model

The gap-source model is compared with the frill model, using the

optimum gap height (1/3 rule).

εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)

16

z

h ,r rε µ2a

x2b

Reactance

(λ0 = 15 cm)


The gap-source model is compared with the frill model, using the

optimum gap height (1/3 rule).

0 0.005 0.01 0.015 0.02 0.025h (m)

0

100

200

300

400

500

Rin

(Ω)

Gap source modelFrill model

εr = 2.2 a = 0.635 mm f = 2 GHz Z0 = 50 Ω (b = 2.19 mm)

17

z

h ,r rε µ2a

x2b

Resistance

(λ0 = 15 cm)

Probe in Patch A probe in a patch does not see an infinite parallel-plate waveguide.

Exact calculation of probe reactance:

L

W

x

y

(x0, y0)

cavityin p inZ jX Z

0

Imp in fX Z

f0 = frequency at which Rin is maximum

Zin may be calculated by HFSS or any other software, or it may be measured.

0cavityinX

18

Probe in Patch (cont.)

Using the cavity model, we can derive an expression for the probe reactance (derivation given later)

This formula assumes that there is no z variation of the probe current or cavity fields (thin-substrate approximation), but it does accurately account for the actual patch dimensions.

Le

We

x

y

PMC

(x0, y0)

Cavity Model

19

2 2 20 0

0 2 21 0 20 0

cos cos sinc2

1 14

p

e ep r

m,n , e em n

e e

nπwmπx nπyL W

X ω μ μ hW L mπ nπδ δ k

L W

3 2/pw e a

a = probe radius

(x0, y0) = probe location


20

Final result:

Using image theory, we have an infinite set of “image probes.”


Image theory can be used to improve the simple parallel-plate waveguide model when the probe gets close to the patch edge.

Image Theory

21

A simple approximate formula is obtained by using two terms: the original probe current in a parallel-plate waveguide and one image. This should be an improvement when the probe is close to an edge.

( ) ( ) ( ) ( )two 1 10 0 0 04 4 2inX = khY ka J ka khY k J ks aη η− −

0

0 /r

r

k k ε

η η ε

=

=


Original

Image

two probe imagein in inX = X X+

22

As shown on the next plot, the best overall approximation in obtained by using the following formula:

( )probe twomax ,in in inX X X=


“modified CAD formula”

23

Probe in Patch (cont.) Results show that the simple formula (“modified CAD formula”) works fairly well.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1Xr

0

5

10

15

20

25

30

35

X f (Ω

)

Uniform current modelCavity modelUniform current model (with one image)Modified CAD formulaHFFS simulation

L

W

x

y

(x0, y0)

0 / 2/ 2r

x LxL−

=

24

( )probe twomax ,in in inX X X=

center edge

Next, we investigate each of the improved probe models in more detail:

Cosine-current model Gap-source model Frill model

25

Appendix

Cosine Current Model

Assume that ( ) cos ( )I z k z h= −

Note: (0) cos ( )I kh=

z

h ,r rε µ2a

x

I (z)

26

21 (0)2c inP Z I=


Circuit Model: (0)I

coax feed inZ

2

2

0c

inPZ

I

27

Represent the probe current as: 0

( ) cosmm

m zI z Ihπ∞

=

=

∑

2

2

0c

inPZ

I

This will allow us to find the fields and hence the power radiated by the probe current.

z

h ,r rε µ2a

x

I (z)

28


Hence

00 0

( ) cos cos cosh h

mm

m z m z m zI z dz I dzh h hπ π π∞

=

′ ′ =

∑∫ ∫

( )

( )

00

00

( )cos 12

2 ( )cos1

h

m m

h

mm

m z hI z dz Ih

m zI I z dzh h

π δ

πδ

= + ⇒ = +

∫

∫

Using Fourier-series theory:


The integral is zero unless m = m′.

29

Result:

( ) 0

2 cos ( ) cos1

h

mmo

m zI k z h dzh h

πδ

= − + ∫or

2 2

2 ( ) sin( )1 ( ) ( )m

mo

khI khkh mδ π

= + −

(derivation omitted)


30

(Time-Harmonic Fields)

Note: We have both Ez and Eρ To see this:

so

0E∇⋅ =

1 1( ) 0zE EEz

φρρ

ρ ρ ρ φ∂ ∂∂

+ + =∂ ∂ ∂

0Eρ ≠


31

For Ez, we represent the field as follows:

where

( )00

cosz m mm

m zE A J kh ρπ ρ

∞− −

=

=

∑aρ <

aρ > ( )(2)0

0cosz m m

m

m zE A H kh ρπ ρ

∞+ +

=

=

∑

( )

1/ 222

1/ 22 2

m

zm

mk kh

k k

ρπ = −

= −


32

At

so

aρ =

z zE E+ −= (BC 1)

( ) ( )(2)0 0m m m mA H k a A J k aρ ρ

+ −=


33

Also we have

To solve for Eρ , use

(BC 2) 2 1 szH H Jφ φ− =

1 zE EHj z

ρφ ωµ ρ

∂ ∂−= − ∂ ∂

EjH ωε=×∇


where

34

so 1

1

z HHj Ez

HE

j z

φρ

φρ

ωερ φ

ωε

∂∂= −

∂ ∂∂

= −∂

Hence we have 2

2

1 1 zH EHj j z

φφ ωµ ωε ρ

∂ ∂= − − −

∂ ∂

For the mth Fourier term:

( )( )

( ) 2 ( )1 1 mm m z

zmEH k H

j jφ φωµ ωε ρ ∂

= − − − − ∂


35

Hence

( )2 ( ) 2 ( )

mm m z

zmEk H k H jφ φ ωερ

∂− = −

∂

2 2 2zm mk k kρ− =

so that

( )( )

2

mm z

m

EjHkφρ

ωερ

∂= −

∂

where


36

where ( )( )

2

mm z

m

EjHkφρ

ωερ

∂= −

∂

2 1 2szIH H J

aφ φ π− = =

( ) ( ) ( )2 1 2m m m m

szIH H J

aφ φ π− = =

For the mth Fourier term:


37

Hence

we have

( ) ( )(2)0 02 ( )

2m

m m m m mm

Ij k A H k a A J k ak aρ ρ ρρ

ωεπ

+ − − ′ ′− =

(2)0(2)

0 00

( )( ) ( )

( ) 2m mm

m m m pmm

H k a kIA H k a A J k aJ k a a j

ρ ρρ

ρ π ωε+ +

′ ′− = −


( ) ( )(2)0 0m m m mA H k a A J k aρ ρ

+ −=Using (BC 1)

38

(2) (2)0 0 0 0 0( ) ( ) ( ) ( ) ( )

2mm

m m m m m m

kIA J k a H k a H k a J k a J k aa j

ρρ ρ ρ ρ ρπ ωε

+ ′ ′− = −

02 ( )

2mm

m mm

kIA j J k ak a a j

ρρ

ρπ π ωε+ − = −

( )2

014

mm m m

kA I J k aρ

ρωε+

= −

Hence

or

or

(using the Wronskian identity)


39

*

2*

0 0

*

0

*

0

(2) *00

0 ' 0

12

1 ( )2

( )

( ) ( )21 ( )cos cos2

scs

h

z sz

h

z sz

h

z

h

m m mm m

P E J dS

E a J a dz d

a E a J dz

a E a I z dza

m z m zA H k a I dzh h

π

ρ

φ

π

ππ

π π∞ ∞+

′= =

−= ⋅

−=

= −

= −

′ = − ⋅

∫

∫ ∫

∫

∫

∑ ∑∫

We now find the complex power radiated by the probe:


40

( )

( )

* (2)0 0

0

2(2) *

0 0 00

1 ( ) 12 2

11 ( ) ( )4 4

c m m m mm

mm m m m m

m

hP A I H k a

kh H k a I J k a I

ρ

ρρ ρ

δ

δωε

∞+

=

∞

=

= − +

= − + −

∑

∑

Integrating in z and using orthogonality, we have:

Hence, we have:

2 2 (2)0 0 0

0

1 (1 ) ( ) ( )16c m pm m m m

m

hP I k H k a J k aρ ρδωε

∞

=

= + +

∑

Am+ coefficient


41

Therefore,

2

2cos ( )

cin

PZkh

=

( )22 2 (2)0 0 0

0

1 sec ( ) 1 ( ) ( )8in m m m m m

m

hZ kh I k H k a J k aρ ρ ρδωε

∞

=

= +

∑

Define: 0

2

0

mm

r r

kk

k

mk h

ρρ

πε µ

=

= −

42


We then have

Also, use r

kεη

ωεεµω

ωε0000 ==

( ) 222 (2)0 0 0 0 0 0

0

1 1( ) sec (1 ) ( ) ( )8

min r m m m mmr


∞

=

= +

∑

The probe reactance is: Im( )p inX Z=

43


Keep only the m = 0 term

( ) 222 (2)0 0 0 0 0 0

0

1 1( ) sec (1 ) ( ) ( )8

min r m m m mmr


∞

=

= +

∑

0 1:k h <<

(2)0 0 0 0

1 ( ) ( ) ( )4in rZ k h J ka H kaη µ≈

(same as previous result using uniform model)

Thin substrate approximation

2 2

2 ( ) sin( )1 ( ) ( )m

mo

khI khkh mδ π

= + −

The result is

44


Gap Model z

h ,r rε µ2a

x1V + - ∆

( ) 1/ , 0,

0, otherwise.z

zE z a

− ∆ < < ∆=

( ) ( ) ( )20

0, cosz m m

m

m zE z B H khρπρ ρ

∞

=

=

∑

Note: It is not clear how best to choose ∆, but this will be re-visited later.

45

Gap Model (cont.) z

h ,r rε µ2a

x1V + - ∆

( ) ( ) ( )20 0

2 sinc1m

m m

mBhh H k aρ

πδ

− ∆ = +

From Fourier series analysis (details omitted):

( ) ( ) ( )20

0

1 / , 0, cos

0, otherwise.z m mm

zm zE z a B H k ahρπ∞

=

− ∆ < < ∆ = =

∑

At ρ = a:

46

Gap Model (cont.)

( )2in szY a Jπ= ∆ ( ) ( )szJ z H zφ=where

The magnetic field is found from Ez , with the help of the magnetic vector potential Az (the field is TMz):

1 zAHφ µ ρ∂

= −∂

22

2

1z zE k A

j zωµε ∂

= + ∂

( ) ( ) ( )20

0, cosz m m

m

m zA z A H khρπρ ρ

∞

=

=

∑

where

z

h ,r rε µ2a

x1V + - ∆

Use:

Setting ρ = a allows us to solve for the coefficients Am.

47

Gap Model (cont.)

(2)0

(2)0 0 0

( )1 24 sinc(1 ) ( )

min

m m m m

H k aa mY j kh k H k a h

ρ

ρ ρ

ππη δ

∞

=

′ ∆ = + ∑

Final result:

z

h ,r rε µ2a

x1V + - ∆

48

z

h ,r rε µ2a

xb

Frill Model

To find the current I (z) , use reciprocity.

10inZ

I

Introduce a ring of magnetic current K = 1 in the φ direction at z (the testing current “B”).

( )1 1

ln /sMb aφ ρ

= −

F

a b

V

b a

V

bs

S

I z H M dV A,B B,A

H M dV

H M dS

A

B SF

z

1V frill

49

Frill Model (cont.)

( )

( )

( )

( ) ( )

( ) ( )

2

0

,0

2 ,0

1 12 ,0ln /

2 ,0ln /

F

bs

S

bb

sa

bb

sa

bb

a

bb

a

I z H M dS

H M d d

H M d

H db a

H db a

π

φ φ

φ φ

φ

φ

ρ ρ ρ φ

π ρ ρ ρ

π ρ ρ ρρ

π ρ ρ

= ⋅

=

=

= −

= −

∫

∫ ∫

∫

∫

∫

A B

SF z

50

( ) ( ) ( )2 ,0ln /

bgap

a

I z H db a φπ ρ ρ= − ∫

z

b

The magnetic current ring B may be replaced by a 1V gap source of zero height (by the equivalence principle).

Let z → 0:

The field of the gap source is then calculated as was done in the gap-source model, using ∆ = 0.

Frill Model (cont.)

z

b

51

Final result:

( )

( ) ( )

( )

2 20 0

220 0 0

( ) ( )1 1 4ln / ( ) (1 ) ( )

m min

m m m m

H k b H k aY j k

h b a k H k aρ ρ

ρ ρ

πη δ

∞

=

− = +

∑

Frill Model (cont.)

z

h ,r rε µ2a

xb

52

Spring 2015 - University of Houstoncourses.egr.uh.edu/ECE/ECE6345/Class Notes/Topic 4... · Spring 2015. Notes 5 . ECE 6345 . Prof. David R. Jackson . ECE Dept. 1 . Overview This

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