1 ECE 6345 Spring 2011 Prof. David R. Jackson ECE Dept. Notes 2.

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


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


( )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


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


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


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


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:

14

Qf

Qf

ry

rx

2

11

2

11

j

AA

j

AA

y

x

1

1

Choose

Then we have

(LHCP)


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


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


Then we have

f

17

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)


Summary of frequencies

f0 = frequency for which we get CP at broadside.

18

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.)

20

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


Hence

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

21


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

22

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+

23

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).

24

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.

25

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

26

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.)

27


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

28


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

29


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:

30


2 3 dBAR AR

348.0x

Set

From a numerical solution:

31


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

32

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= + =

+ -

33

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


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


36

Summary Summary

0.348AR CPBWQ

1.414imp CPBWQ

0.707imp LinBWQ

CP

Linear

1 ECE 6345 Spring 2011 Prof. David R. Jackson ECE Dept. Notes 2.

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