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ECE 6345ECE 6345Spring 2011
Prof. David R. JacksonECE Dept.
Notes 2
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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=
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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
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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.
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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):
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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or
The frequency conditions can be written as
so
Circular Polarization Condition (cont.)Circular Polarization Condition (cont.)
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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
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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)
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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
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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