ECE 6345
Spring 2015
Prof. David R. Jackson
ECE Dept.
Notes 2
1
Overview
This set of notes treats circular polarization, obtained by
using a single feed.
y
L» W
y0 » x0
W
(x0, y0)
x
L
2
Overview
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
3
Amplitude of Patch Currents
y
L
W
( x0 , y0 )
I
h
w
r
x
I  1 [ A]
First Step: Find the patch currents (x and y directions), and relate them
to the input impedance of the patch.
4
Amplitude of Patch Currents (cont.)
y
L
W
( x0 , y0 )
I
h
w
r
x
I  1 [ A]
x 
ˆ
x-directed current mode (1,0): J  x Ax sin 

L


x
s
æπy ÷
ö
ç
y-directed current mode (0,1): J = ˆy Ay sin ç ÷
÷
çèW ø
y
s
5
Amplitude of Patch Currents (cont.)
The x mode (TM10):
J s  nˆ  H   zˆ  H
so
J sx  H y
æπx ÷
ö
H = ˆy Ax sin çç ÷
çè L ÷
ø
To find E, use
  H  j E
 H y H x 
1
 
x
Ez 

Ax   cos 



j 0 r  x
y  j 0 r  L 
 L 
1
6
Amplitude of Patch Currents (cont.)
V
Z in = = V = - hEz = - h (Ezx + Ezy ) = Z inx + Z iny
I
For the (1,0) mode we have
or
jh   
  x0 
Z 
  Ax cos 

 0 r  L 
L


x
in
   0 r L 
Zinx
Ax 
  x0   j h 
cos 

L


A similar derivation holds for the y mode.
7
Amplitude of Patch Currents (cont.)
The y mode (TM10):
   0 rW 
Ziny
Ay 
  y0   j h 
cos 

W


The patch current amplitudes
can then be written as:
Ax  A1 x  Z inx
where
 x
A1
 y
Ay  A1 Z iny
   0 r L 
1

  x0   j h 
cos 

 L 
A1  
y
   0 rW 
  y0   j h 
cos 

W


1
8
Amplitude of Patch Currents (cont.)
 x
A1
   0 r L 
1

  x0   j h 
cos 

L


 y
A1
   0 rW 

  y0   j h 
cos 

W


1
y
Assume
L W
x0  y0
Then
 x
1
A
 y
1
A
W
 A1
(x0, y0)
x
L
9
Amplitude of Patch Currents (cont.)
Because of the nearly equal dimensions and the feed along the
diagonal, we also have
Rx  R y  R
(The resistance in the circuit model for either mode is the same.)
Ri = resonant input resistance of the mode i, when excited by itself
(Rx only depends on x0, Ry only depends on y0).
We then have:
where
A2  A1R
Ax  A2 Z inx
Ay  A2 Z
y
in
 A2 
   0 r L 
R
  x0   j h 
cos 

 L 
Reminder: The bar denotes impedances that are normalized by R (either Rx or Ry).
10
Amplitude of Patch Currents (cont.)
The A2 coefficient can be written as
   0 r L 
R
A2 
  x0   j h 
cos 

L


2   x0 
Redge cos 
    L 
L


0 r

  x0   j h 
cos 

L


so
  x0     0 r L 
A2  Redge cos 


L
j

h



11
Circular Polarization Condition
y
Let
L = W (1+ δ)
y0 = x0
W
(x0, y0)
x
L
The CP condition is
Ay
j
Ax = amplitude of x mode
Ax
Ay = amplitude of y mode
- for RHCP
+ for LHCP
12
Circular Polarization Condition (cont.)
The frequency fCP is defined as the frequency for which we get CP
at broadside.
f  fCP
f
We have:
fx
A2
Ax 
1  j 2Q  f rx  1
A2
Ay 
1  j 2Q  f ry  1
fy
where
f rx 
f
fx
f ry 
f
fy
fx = resonance frequency of (1,0) mode
fy = resonance frequency of (0,1) mode
13
Circular Polarization Condition (cont.)
LHCP:
Ay
Ax
Choose:
1
f rx  1 
2Q
1
f ry  1  
2Q
j
at f  fCP
Then we have
A
Ax 
1 j
A
Ay 
1 j
1 j


Ax 1  j
Ay
2e
2e
j

4
j

e
j

2
 j
4
(LHCP)
14
Circular Polarization Condition (cont.)
The frequency conditions for f = fCP can be written as:
1
f rx  1 
2Q
fCP
1
 1
fx
2Q
fx
1
 1
fCP
2Q
1
f ry  1  
2Q
fCP
1
 1
fy
2Q
fy
fx  f y
so
fCP
or
fCP
fCP
1
 1
2Q
2
1
  fx  f y 
2
15
Circular Polarization Condition (cont.)
Also,
fy
fx
1
=
fCP fCP 2Q
Let
æ 1 ö
1
÷
çç÷
=
÷
çè 2Q ÷
ø Q
f  f y  f x
f
1

f CP Q
Then we have
fCP
2Q
fCP
fCP
2Q
f
fx
fy
16
Circular Polarization Condition (cont.)
Summary of frequencies

1 
f x  fCP 1 

2
Q



1 
f y  fCP 1 

2
Q


(LHCP)

1 
f x  fCP 1 

2
Q



1 
f y  fCP 1 

2
Q


(RHCP)
fCP = frequency for which we get CP at broadside.
17
Patch Dimensions for CP
W
L
r
L
PMC
L
r
PMC
L  f h,  r ,W 
(Hammerstad formula)
18
Physical Dimensions for CP (cont.)
L  f h,  r ,W 
Le  L  2L
k0  r L  
e
(resonance condition)
Let
k0 x  2 f x 0 0
k0 y  2 f y 0 0
(known wavenumbers)
19
Physical Dimensions for CP (cont.)
k0 x  L  2L   r  
k0 x Le  r  
Similarly, we have
k0 y W  2W   r  
Hence
L
W

k0 x  r

k0 y  r
 2 L  h,  r , W 
 2 W  h ,  r , L 
Note: For W, we use the
same formula as L , but
replace W  L.
20
Physical Dimensions for CP (cont.)
Since the patch is nearly square, the two fringing extensions are
nearly equal (L W). Hence we have
L
W

k0 x  r

k0 y  r
where
 2 L
 2 L
k0 x  2 f x 0 0
k0 y  2 f y 0 0
(known wavenumbers)
Note: The fringing length L depends on W, so these equations must be
solved numerically (using, e.g., iteration).
21
Hammerstad’s Formula
W


0.262
h
   re  0.300 
L  0.412h 


W


0.258

  0.813   re
h

æεr + 1ö
æεr ÷
ç
çç
εre = ç
+
÷
÷ èç 2
çè 2 ø
1ö÷
÷
ø÷
1
1 + 12
h
W
22
Input Impedance of CP Patch
R
R
Z in  f   Z  f   Z  f  

1  j 2Q  f rx  1 1  j 2Q  f ry  1
x
in
At fCP:
so
or
y
in
f rx  1  1/ (2Q) and f ry  1  1/ (2Q)
(LHCP)
R
R
Z in  f 0  

1 j 1 j
R (1  j )  R (1  j ) R (1  j )  R (1  j )


(1  j )(1  j )
2
Z in  R
Recall: R = Rx = R for TM10 mode
The CP frequency fCP is also the
resonance frequency where the
input impedance is real (if we
neglect the probe inductance).
23
Input Impedance of CP Patch
Hence, at the resonance (CP) frequency fCP we have
  x0 
Z in  Redge cos 

 L 
2
Redge = Redge for TM10 mode
y
Note: We have a CAD formula for Redge.
W
(x0, y0)
x
The fed position x0 can be chosen to give
the desired input resistance at the
resonance frequency fCP.
L
24
CP (Axial Ratio) Bandwidth
We now examine the frequency dependence of the term Ay / Ax .
Ax 
A2
1  j 2Q  f rx  1
Ay 
A2
1  j 2Q  f ry  1
1  j 2Q( f rx  1)

Ax 1  j 2Q( f ry  1)
Ay
where
f
f rx 

fx
f 
1 

1 
 (LHCP)
f
2
Q


1

CP 
fCP 1 

2
Q


f
25
CP Bandwidth (cont.)
Define
Then
Similarly,
fr 
f
f CP
This is the ratio of the
operating frequency to the
CP frequency.

1 

f rx  f r 1 
 2Q 
f ry

1 

 f r 1 
 2Q 
26
CP Bandwidth (cont.)
Hence
æ æ
ö
1 ö
÷
ç
÷
ç
1 + j 2Q çç f r ç1 +
- 1÷
÷
÷
÷
ç
÷
÷
ç
2
Q
Ay 1 + j 2Q ( f rx - 1)
è
ø
è
ø
=
=
æ æ 1 ö ÷
ö
Ax 1 + j 2Q ( f ry - 1)
ç
÷
ç
1 + j 2Q çç f r ç1- 1÷
÷
÷
÷
ç
÷ ø
÷
çè è 2Q ø
Note:
Let
Then
fr  1 
Ay
Ax

1 j
 j
1 j
x  2 Q  f r  1
1  j (f r  x) 1  j (1  x)


Ax 1  j (f r  x) 1  j (1  x)
Ay
27
CP Bandwidth (cont.)
1  j (1  x)

Ax 1  j (1  x)
Ay
x  2 Q  f r  1
y
B
fr 
f
f CP
E(t )

x
A
A
AR 
B
28
CP Bandwidth (cont.)
From ECE 6340:
AR  cot 
where
sin 2  sin(2 ) sin 
  tan
1
Ay
Ax
 Ay 
  arg  
 Ax 
In our case,
  tan
1
1  (1  x) 2
1  (1  x) 2
  tan 1 (1  x)  tan 1 (1  x)
29
CP Bandwidth (cont.)
Set
AR  2
From a numerical solution:
 AR  3 dB
x  0.348
30
CP Bandwidth (cont.)
Hence
2Q  f r  1  0.348
0.348
fr  1 
2Q
so
0.348
2Q
0.348
 1
2Q
f r  1 
f r
0.348
Therefore, f r  f  f 
Q

r

r
Hence, we have
BW
AR CP
f f

 f r  f r
fCP
BW
AR CP
0.348

Q
31
Impedance Bandwidth
Z in  f  

R
R

1  j 2Q  f rx  1 1  j 2Q  f ry  1
R
R

 
 
1  
1  
1  j 2Q  f r 1 
 1 1  j 2Q  f r 1 
 1


  2Q  
  2Q  
R
R


1  j (f r  x) 1  j (f r  x)

R
R

1  j (1  x) 1  j (1  x)
where
Note: At x = 0 we have Zin =
 for
f r  1
x  2Q f r  1
R
R
+
= R
1 + j 1- j
32
Impedance Bandwidth (cont.)
Zin
1
1
Zin 


R 1  j (1  x) 1  j (1  x)
Zin  Z0 Zin  R Zin  1



Zin  Z0 Zin  R Zin  1
S  SWR 
Set
1 
1 
S  S0  2 (bandwidth limits)
x0   2
(derivation omitted)
33
Impedance Bandwidth (cont.)
Hence
2Q  f r  1   2
so
2
fr  1 
2Q
The band edges (in normalized frequency) are then
1
f  1
2Q
1

fr  1 
2Q

r
34
Impedance Bandwidth (cont.)
BW
Hence
Hence
imp CP
f f

 f r  f r
fCP
 1 
 2

 2Q 
BW
imp CP
BW
imp CP
2

Q
35
Summary
BW
AR CP
0.348

Q
CP antenna
BW
Linear antenna
BW
imp CP
imp Lin
2 1.414


Q
Q
1
0.707


Q
2Q
36