1

ECE 6345ECE 6345Spring 2011

Prof. David R. JacksonECE Dept.

Notes 2

2

OverviewOverview

This set of notes treats circular polarization, obtained by using a single feed.

x

y

L

W

(x0, y0)

L W»

0 0y x=

3

OverviewOverview

Goals:

Find the optimum dimensions of the CP patch Find the input impedance of the CP patch Find the pattern (axial-ratio) bandwidth Find the impedance bandwidth of CP patch

4

Amplitude of Patch CurrentsAmplitude of Patch Currents

w

r

Ly

x

0 0( , )x y

h

W

1 [ ]I A

First Step: Find the patch currents (x and y directions), and relate them to the input impedance of the patch.

5

Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)

ˆ sinxs x

xJ x A

L

x-directed current mode (1,0):

w

r

Ly

x

0 0( , )x y

h

W

1 [ ]I A

sinys y

πyˆJ y A

W

æ ö÷ç= ÷ç ÷÷çè øy-directed current mode (0,1):

6

ˆ ˆsJ n H z H

ysx HJ

Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)

x mode:

so

sinx

πxˆH y A

L

æ ö÷ç= ÷ç ÷÷çè ø

EjH To find E, use

0 0

1 1cosy x

z xr r

H H xE A

j x y j L L

7

Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)

( )x y x yin z z z in in

VZ V hE h E E Z Z

I= = =- =- + = +

0

0

cosxin x

r

xjhZ A

L L

0

0cos

xin r

x

Z LA

x j hL

For the (1,0) mode we have

or

A similar derivation holds for the y mode.

8

Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)

0

0cos

yin r

y

Z WA

y j hW

y mode:

1

1

x xx in

y yy in

A A Z

A A Z

01

0

1

cos

x rLAx j hL

01

0

1

cos

y rWAy j h

W

The patch current amplitudes can then be written as

where

9

Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)

0 0

L W

x y

Assume

01

0

1

cos

x rLAx j hL

01

0

1

cos

y rWAy j h

W

x

y

L

W

(x0, y0) 1 1 1x yA A A Then

10

Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)

x yR R R

2

2

xx in

yy in

A A Z

A A Z

Because of the nearly equal dimensions and the feed along the diagonal, we have

We then have

Reminder: The bar denotes impedances that are normalized by R (either Rx or Ry).

2 1A AR

Ri = resonant input resistance of the mode i, when excited by itself (e.g., by a feed along the centerline).

where

11

Amplitude of Patch Currents (cont.)Amplitude of Patch Currents (cont.)

02

0

2 0

0

0

0 0

cos

cos

cos

cos

r

edger

redge

LRA

x j hL

xR

LLx j hL

x LR

L j h

The A2 coefficient can be written as

12

Circular Polarization ConditionCircular Polarization Condition

xA

yA

( )1L W δ= +

amplitude of y mode

amplitude of x mode

x

y

L

W

(x0, y0)

0 0y x=

jA

A

x

y

The CP condition is

13

Circular Polarization Condition (cont.)Circular Polarization Condition (cont.)

2

2

1 2 1

1 2 1

xrx

y

ry

AA

j Q f

AA

j Q f

The frequency f0 is defined as the frequency for which we get CP at broadside.

xf

0f

yff

0 0rx ry

x y

f ff f

f f

where

frx = resonance frequency of (1,0) mode

fry = resonance frequency of (0,1) mode

At f = f0:

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Qf

Qf

ry

rx

2

11

2

11

j

AA

j

AA

y

x

1

1

Choose

Then we have

(LHCP)

Circular Polarization Condition (cont.)Circular Polarization Condition (cont.)

je

e

e

j

j

A

A j

j

j

x

y

2

4

4

2

2

1

1

15

0

0

0

0

1 1 11 1 1

2 2 2

1 1 11 1 1

2 2 2

xrx

x

yry

y

f ff

Q f Q f Q

fff

Q f Q f Q

0

2x yf f

f

yx fff 2

10

or

The frequency conditions can be written as

so

Circular Polarization Condition (cont.)Circular Polarization Condition (cont.)

16

0

2

f

Q

0 0

1 1 1

2 2y xf f

f f Q Q Q

æ ö÷ç ÷- = - - =ç ÷ç ÷çè ø

xy fff

0

1f

f Q

Also

Let

xf

0f

yf

0

2

f

Q

Circular Polarization Condition (cont.)Circular Polarization Condition (cont.)

Then we have

f

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0 0

1 11 1

2 2x yf f f fQ Q

(LHCP)

0 0

1 11 1

2 2x yf f f fQ Q

(RHCP)

Circular Polarization Condition (cont.)Circular Polarization Condition (cont.)

Summary of frequencies

f0 = frequency for which we get CP at broadside.

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Patch Dimensions for CPPatch Dimensions for CP

L

W

L

r PMCPMC

L

WhfL r ,,

r

19

2eL L L

WhfL r ,,

0e

rk L

0 0 0

0 0 0

2

2

x x

y y

k f

k f

Let

(resonance condition)

Physical Dimensions for CP (cont.)Physical Dimensions for CP (cont.)

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0 0 2ex r x rk L k L L

0 2y rk W W

0

0

2 , ,

2 , ,

r

x r

r

y r

L L h Wk

W W h Lk

Similarly, we have

Physical Dimensions for CP (cont.)Physical Dimensions for CP (cont.)

Hence

Note: For W, we use the same formula as L , but replace W L.

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Physical Dimensions for CP (cont.)Physical Dimensions for CP (cont.)

where

0

0

2

2

x r

y r

L Lk

W Lk

Note: For the calculation of L it is probably accurate enough to use the patch dimensions that come from neglecting fringing.

0 0 0

0 0 0

2

2

x x

y y

k f

k f

Since the patch is nearly square, the two fringing extensions are nearly equal. Hence we have

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Hammerstad’s FormulaHammerstad’s Formula

0.262 0.3000.412

0.2580.813

re

re

WhL hWh

1 1 1

2 21 12

r rre

ε εε

hW

æ ö æ ö+ -÷ ÷ç ç= +÷ ÷ç ç÷ ÷ç çè ø è ø+

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Input Impedance of CP PatchInput Impedance of CP Patch

1 2 1 1 2 1x y

in in inrx ry

R RZ f Z f Z f

j Q f j Q f

1 1/ (2 ) 1 1/ (2 )rx ryf Q f Q and

0 1 1

(1 ) (1 ) (1 ) (1 )

(1 )(1 ) 2

in

R RZ f

j j

R j R j R j R j

j j

RZ in

At f0

so

or

(LHCP)

The CP frequency f0 is also the resonance frequency where the input impedance is real (if we neglect the probe inductance).

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Input Impedance of CP PatchInput Impedance of CP Patch

2 0cosin edge

xZ R

L

Hence, at the resonance (CP) frequency f0 we have

Note: We have a CAD formula for Redge.

The fed position x0 can be chosen to give the desired input resistance at the resonance frequency f0.

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CP (Axial Ratio) BandwidthCP (Axial Ratio) Bandwidth

)1(21

)1(21

ry

rx

x

y

fQj

fQj

A

A

00

00

2

11

211

2

rxx

f ff

ff fQ

f f

f Qf

Q

We now examine the frequency dependence of the term

(LHCP)

where

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CP Bandwidth (cont.)CP Bandwidth (cont.)

0f

ff r

Qff rrx 2

11

Qff rry 2

11Similarly,

Then

Define(This is the ratio of the operating frequency to the CP (resonance) frequency.)

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CP Bandwidth (cont.)CP Bandwidth (cont.)

1 2 12

1 2 12

rr

y

x rr

fj Q f

A Q

A fj Q f

Q

æ ö÷ç+ + - ÷ç ÷ç ÷çè ø=

æ ö÷ç+ - - ÷ç ÷ç ÷çè ø

1 yr

x

Af j

A

2 1rx Q f

1 ( ) 1 (1 )

1 ( ) 1 (1 )y r

x r

A j f x j x

A j f x j x

Hence

Note:

Then

Let

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CP Bandwidth (cont.)CP Bandwidth (cont.)

1 (1 )

1 (1 )y

x

A j x

A j x

A

B

( )tE

x

y

AAR

B

2 1rx Q f 0f

ff r

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CP Bandwidth (cont.)CP Bandwidth (cont.)

1

sin 2 sin(2 )sin

tan

arg

y

x

y

x

A

A

A

A

21

2

1 1

1 (1 )tan

1 (1 )

tan (1 ) tan (1 )

x

x

x x

where

In our case,

cotAR From ECE 6340:

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CP Bandwidth (cont.)CP Bandwidth (cont.)

2 3 dBAR AR

348.0x

Set

From a numerical solution:

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CP Bandwidth (cont.)CP Bandwidth (cont.)

2 1 0.348

0.3481

2

r

r

Q f

fQ

Qf

Qf

r

r

2

348.01

2

348.01

0.348AR CPBWQ

Therefore

Hence

so

Hence

0.348r r rf f f

Q

0 0

AR CPr

f f fBW f

f f

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Impedance BandwidthImpedance Bandwidth

1 2 1 1 2 1

1 2 1 1 2 12 2

1 ( ) 1 ( )

11 (1 ) 1 (1 )

inrx ry

r rr r

r r

r

R RZ f

j Q f j Q f

R R

f fj Q f j Q f

Q Q

R R

j f x j f x

R Rf

j x j x

for

12 rfQxwhere

Note: At x = 0 we have1 1in

R RZ R

j j= + =

+ -

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1 1

1 (1 ) 1 (1 )in

in

ZZ

R j x j x

0

0

1

1in in in

in in in

Z Z Z R Z

Z Z Z R Z

1

1SWRS

2x

Set 0 2S S

(derivation omitted)

Impedance Bandwidth (cont.)Impedance Bandwidth (cont.)

(bandwidth limits)

34

2 1 2

21

2

r

r

Q f

fQ

Hence

Impedance Bandwidth (cont.)Impedance Bandwidth (cont.)

11

2

11

2

r

r

fQ

fQ

so

The band edges (in normalized frequency) are then

35

12

2imp CPBW

Q

2imp CPBWQ

Hence

Hence

imp CPr r rBW f f f

Impedance Bandwidth (cont.)Impedance Bandwidth (cont.)

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Summary Summary

0.348AR CPBWQ

1.414imp CPBWQ

0.707imp LinBWQ

CP

Linear