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Antenna & Propagation Impedance Matching
1
Chapter 4 – Impedance Matching
1 – Quarter-wave transformer, series section transformer
2 – Stub matching, lumped element networks, feed point location
3 – Gamma match
4 – Delta- and T-match, Baluns
Antenna & Propagation Impedance Matching
2
REVISION
1 – 2-port network
2 – Smith Chart
Antenna & Propagation Impedance Matching
3
2-Port Network Representation
• Six ways to represent a two-port network in terms of V and I at each port:
– Z-matrix open-circuit impedance
– Y-matrix short-circuit admittance
– ABCD-matrix chain or transmission parameters
– B-matrix inverse transmission parameters
– H-matrix hybrid parameters
– G-matrix inverse hybrid parameters
V2V1
I1 I2
Two-Port
Network
Antenna & Propagation Impedance Matching
4
2 -Port Network Representation
• 7th way to represent a two-port network in terms of waves entering and leaving each port:
– S-matrix scattering parameters
• In general, there could be n ports in a network:
V2V1
I1 I2
Two-Port
Network
1
2
3
n
n-port
network
Antenna & Propagation Impedance Matching
5
[S] for Two-Port Network
Definition of normalized voltage waves:
1
1
11 i
o
i PZ
Va ==
1
1
11 r
o
r PZ
Vb ==
Incident wave:
Reflected wave:
~
a1
b1 Two-Port
NetworkZo1
Scattering parameters are then defined as:
ZLZo2
a2
b2
2
2
22 i
o
i PZ
Va ==
2
2
22 r
o
r PZ
Vb ==
2121111 aSaSb +=
2221212 aSaSb +=
=
2
1
2221
1211
2
1
a
a
SS
SS
b
bor
outputs inputs
aSb =or
1 2
Antenna & Propagation Impedance Matching
6
S11 for a One-Port Network
Measured S11 of an 800 MHz square microstrip
patch antenna on a 1.6 mm FR4 substrate
Antenna
Bandwidth
( RL=13 dB)
Antenna & Propagation Impedance Matching
7
Measured S21 of a 542 MHz hair-pin bandpass
filter on a 1.6 mm FR4 substrate
S21 for a One-Port Network
Antenna & Propagation Impedance Matching
8
Useful Reference• VSWR
– Visualization on VSWR• http://www.youtube.com/watch?v=mh3o8gUu4AE (10:30)
– Finding VSWR using Smith Chart • http://www.youtube.com/watch?v=YbZ9RBw7-js
• Smith Chart
– Finding values on Smith Chart• http://www.youtube.com/watch?v=hmqM8PnUkmo
– Reflection Coefficient using Smith Chart • http://www.youtube.com/watch?v=9KlIgae0ad8
– Matching using smith chart• http://www.youtube.com/watch?v=YdYFfxnB7fg
Antenna & Propagation Impedance Matching
9
Smith Chart Impedance or Admittance
Antenna & Propagation Impedance Matching
10
Smith Chart: Exercises
1. Use as a Z-chart, locate s/c, o/c, Zo , 1+ j1
2. Use as a Y-chart, locate s/c, o/c, Yo , 1+ j1
3. Move towards generator l/l
4. Move towards load l/l
5. Read VSWR, |G |, Vmax , Vmin
6. Read attenuation
Antenna & Propagation Impedance Matching
11
IMPEDANCE MATCHING
1 – Introduction
2 – Importance of Impedance Matching
Antenna & Propagation Impedance Matching
12
INTRODUCTION
• The operation of an antenna system over a frequency range is not completely dependent upon the frequency response of the antenna element itself but rather on the frequency characteristics of the transmission line-antenna element combination.
• In practice, the characteristic impedance of the transmission line is usually real whereas that of the antenna element is complex.
• Also the variation of each as a function of frequency is not the same.
• The efficient coupling-matching networks must be designed which attempt to couple-match the characteristics of the two devices over the desired frequency range.
Antenna & Propagation Impedance Matching
13
Feed-Point Impedance: Za
• Za = antenna impedance at its feed-point.
• Za is complex generally.
aaa jXRZ +=Za
Antenna & Propagation Impedance Matching
14
Importance of Impedance Matching
• Increased power throughput (e.g. maximum power transfer)
• Increased power handling capability in a transmission line (due to reduced VSWR)
• Reduced effects on impedance matching sensitive circuits (e.g. frequency pulling effect on the signal source)
• With "controlled mismatch", an RF amplifier can operate with minimum noise generation (i.e. minimum noise figure)
Antenna & Propagation Impedance Matching
15
Concept of maximum power transfer
Zo
ZLV
iI V
L
LoL
iLLL Z
ZZ
VZIIVP
2
2
2
1
2
1
2
1
+===
Power maximum whence ZL = Zo
Power deliver at ZL is
In lump circuitP
L
ZL
Zo
Antenna & Propagation Impedance Matching
16
continueIn transmission line
oL
oL
ZZ
ZZ
+
−=
No reflection whence ZL = Zo , hence 0=
The load ZL can be matched as long as ZL not equal to zero (short-
circuit) or infinity (open-circuit)
The important parameter is reflection coefficient
Antenna & Propagation Impedance Matching
17
Features of Impedance Matching Networks
• Complexity
– use the simplest --> cheap, low loss
• Bandwidth
– perfect match usually at a single frequency
– wider bandwidth increases complexity
• Implementation
– select matching components to suit an application:
– lumped-element (L, C, R)
– distributed-element (transmission line, waveguides)
• Adjustability
– for loads with variable impedance
Antenna & Propagation Impedance Matching
18
MATCHING METHODS
1 – calculations
2 – smith chart
Antenna & Propagation Impedance Matching
19
Matching with lumped elements
The simplest matching network is an L-section using two reactive elements
jX
jB ZL
Zo
jX
jB ZL
Zo
Configuration 1
Whence RL>Zo
Configuration 2
Whence RL<Zo
ZL=RL+jXL
Antenna & Propagation Impedance Matching
20
continue
Configuration 1Configuration 2
If the load impedance(normalized) lies in unitycircle, configuration 1 isused.Otherwiseconfiguration 2 is used.
The reactive elements are either inductors or capacitors. So there are 8 possibilities for matching circuit for various load impedances.
Matching by lumped elements are possible for frequency below 1 GHz or for higher frequency in integrated circuit(MIC, MEM).
Antenna & Propagation Impedance Matching
21
Impedances for serial lumped elements
Serial circuit
Reactance relationship values
+ve X=2pfL L=X/(2pf)
-ve X=1/(2pfC) C=1/(2pfX)
L C
RR
Antenna & Propagation Impedance Matching
22
Impedances for parallel lumped elements
Parallel circuit
Susceptance relationship values
+ve B=2pfC C=B/(2pf)
-ve B=1/(2pfL) L=1/(2pfB)
LC
RR
Antenna & Propagation Impedance Matching
23
Lumped elements for microwave integrated circuit
Spiral inductorLoop inductor
Interdigital
gap capacitor
Planar resistor Chip resistor
Lossy film
Metal-insulator-
metal capacitor Chip capacitor
r
Dielectric
r
Antenna & Propagation Impedance Matching
24
Matching by calculation for configuration 1
jX
jB ZL
Zo
( )LLo
jXRjBjXZ
+++=
/1
1
For matching, the total impedance of L-section plus ZL should
equal to Zo,thus
Rearranging and separating into real and imaginary parts gives us
( ) oLoLL ZRZXXRB −=−
( ) LLoL XRBZBXX −=−1
*
**
Antenna & Propagation Impedance Matching
25
continue
22
22/
LL
LoLLoLL
XR
RZXRZRXB
+
−+=
Since RL>Zo, then argument of the second root is always positive,
the series reactance can be found as
L
o
L
oL
BR
Z
R
ZX
BX −+=
1
Note that two solution for B are possible either positive or negative
+ve inductor
-ve capacitor
+ve capacitor
-ve inductor
Solving for X from simultaneous equations (*) and (**) and substitute X
in (**) for B, we obtain
Antenna & Propagation Impedance Matching
26
Matching by calculation for configuration 2
( )LLo XXjRjB
Z +++=
11
For matching, the total impedance of L-section plus ZL should equal to
1/Zo, thus
Rearranging and separating into real and imaginary parts gives us
( ) LoLo RZXXBZ −=+
( ) LoL RBZXX =+
*
**
jX
jB ZL
Zo
Antenna & Propagation Impedance Matching
27
continueSolving for X and B from simultaneous equations (*) and (**) , we obtain
( )
o
LLo
Z
RRZB
/−=
Since RL<Zo,the argument of the square roots are always positive,
again two solution for X and B are possible either positive or
negative
( ) LLoL XRZRX −−=+ve inductor
-ve capacitor
+ve capacitor
-ve inductor
Antenna & Propagation Impedance Matching
28
parallel capacitance
parallel inductanceserial capacitance
serial inductance
(-ve)
(-ve)
(+ve)
(+ve)
Antenna & Propagation Impedance Matching
29
C1
L1
C2
L2
50C
2L
!10
C1L
2
10 50
Serial C
Parallel
L
Serial
L
Parallel
C
Matching using lumped
components
Antenna & Propagation Impedance Matching
30
ExampleDesign an L-section matching network to match a series RC load with an impedance ZL=200-j100 W, to a 100 W line, at a frequency of 500 MHz.
Solution
Normalized ZL we
have : ZL= 2-j1
ZL= 2-j1Parallel C
(+j0.3)Serial L
(j1.2)
Serial C
(-j1.2)
Parallel L
(-j0.7)
Solution 1
Solution 2
Antenna & Propagation Impedance Matching
31
continue
ZL=200-j100
2.61pF
46.1nH
200-j1000.92pF
38.8nHpF
Zf
bC
o
92.02
==p
nHf
ZxL o 8.38
2==
p
pFZfx
Co
61.22
1=
−=
p
nHbf
ZL o 1.46
2=
−=
p
Reflection coefficient
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5
freq (GHz)
refl
ec
tio
n c
oe
ffic
ien
t
solution 1
solution 2Solution 2 seems to be
better matched at higher
frequency
Antenna & Propagation Impedance Matching
32
MATCHING TECHNIQUES
1 – Discrete lumped-element
2 – Transmission Line
Antenna & Propagation Impedance Matching
33
Impedances for serial lumped elements
Serial circuit
Reactance relationship values
+ve X=2pfL L=X/(2pf)
-ve X=1/(2pfC) C=1/(2pfX)
L C
RR
Antenna & Propagation Impedance Matching
34
Impedances for parallel lumped elements
Parallel circuit
Susceptance relationship values
+ve B=2pfC C=B/(2pf)
-ve B=1/(2pfL) L=1/(2pfB)
LC
RR
Antenna & Propagation Impedance Matching
35
Lumped elements for microwave integrated circuit
Spiral inductorLoop inductor
Interdigital
gap capacitor
Planar resistor Chip resistor
Lossy film
Metal-insulator-metal
capacitor Chip capacitor
r
Dielectric
r
Antenna & Propagation Impedance Matching
36
Resistive L-Section (Using Resistors)
• Advantage: Broadband
• Disadvantage: Very lossy
36
Matched conditions:
(a)
(b)
Solving (a) and (b) to give R1 and R2 :
R1
R2 Ro2Ro1
21 oo RR
(a) (b)
)//( 2211 oo RRRR +=
)//( 1122 oo RRRR +=
)( 2111 ooo RRRR −=
21
122
oo
oo
RR
RRR
−=
Antenna & Propagation Impedance Matching
37
Resistive L-Section (Using Resistors)
• Advantage: Broadband
• Disadvantage: Very lossy
Attenuation:R1
R2 Ro2
Ro1
21 oo RR
(a) (b)
)//(
//
2211
22
oo
o
in
out
RRRR
RR
V
V
++=
Vout~Vin
Antenna & Propagation Impedance Matching
38
Resistive L-Section: Example
38
Design a broadband resistive L-section matching network to match a 75
TV antenna output to a 50 transmission line. Calculate the attenuation of
the matching network.
Solution:
Ro1 = 75 , Ro2 = 50
Substituting these values into previous equations, we get:
R1 = 43.3 , R2 = 86.6
= 0.2113 = –13.5 dB
Balun75 -50
matching
network
75
txn line
50
txn line
7.313.4375
7.31
++=
in
out
V
V
Antenna & Propagation Impedance Matching
39
Reactive L-Section (Using L & C)
• Advantages: Low loss, simplicity in design
• Disadvantages: Narrow-band, fixed Q
Matched conditions:
(a)
(b)
Solving (a) and (b) to give Xs and Xp :
jXs
jXp RpRs
sp RR
(a) (b)
)//( ppss RjXjXR +=
)//( sspp jXRjXR +=
1−=s
p
R
RQ
1. −==s
p
sssR
RRRQX
Xs & Xp: opposite sign
1−
==
s
p
pp
p
R
R
R
Q
RX Xs = +ve for L, -ve for C
Xp = -ve for C, +ve for L
39
Antenna & Propagation Impedance Matching
40
Reactive LC L-Section: Example
Design an L-section matching network with L and C to match a 50
source to a 600 load for maximum power transfer at 400 MHz.
Give two solutions.
Solution:
Rs = 50 , Rp = 600
Substituting these values into previous equations, we get:
Xs = Q.Rs = 166
Xp = Rp/Q = 181
317.311150
6001 ==−=−=
s
p
R
RQ
Antenna & Propagation Impedance Matching
41
Reactive LC L-Section: Example (cont'd)
Solution 1:
Xs = +166 (inductive), Xp = –181 (capacitive)
Solution 2:
Xs = –166 (capacitive), Xp = +181 (inductive)
nHf
XL s
s 6610*400*2
166
2 6===
pp
pFXf
Cp
p 2.2181*10*400*2
1
2
16
===pp
nHf
XL
p
p 7210*400*2
181
2 6===
pp
pFXf
Cs
s 4.2166*10*400*2
1
2
16
===pp
Antenna & Propagation Impedance Matching
42
Reactive LC L-Section: Example (cont'd)
Solution 1:
Xs = +166 (inductive), Xp = –181 (capacitive)
Solution 2:
Xs = –166 (capacitive), Xp = +181 (inductive)
Antenna & Propagation Impedance Matching
43
L-Section for Complex Impedances
• If Rs has a reactance jX', simply introduce a series with an
equal but opposite-sign reactance (–jX').
jXs
jXp RpRs
jX'–jX'
extra reactances
• The –jX' reactance is then combined with jXs to give its
final value.
• Similar treatment can be applied to Rp by introducing a
parallel jX' and a parallel –jX'.
43
Antenna & Propagation Impedance Matching
44
T-Section
• Advantages: R1 can be < or > R2
Variable Q (higher than L-section)
• Disadvantages: More L C elements
Method:(a) Introduce a hypothetical resistance level Rh at jX2, such that
Rh > R1 and Rh > R2
(Rh determines the Q of the matching network)
(b) Split jX2 into jX2' and jX2" (in parallel)
(c) Treat the T-section as two L-sections (L-1 and L-2)
NOTE: Q of L-1 section > Q of 2-element L-section (as Rh > R2)
jX1
jX2' R2R1
Rh
jX3
jX2 jX2"L-1 L-2
44
Antenna & Propagation Impedance Matching
45
T-Section: Q Values
45
Using the resistance values of the previous 2-element L-section:
(a) Q of original L-section:
(b) Q of T-section if Rh = 5050 (Note: Q1 > QL)
L-1 section:
L-2 section:
317.3150
6001
1
2 =−=−=R
RQL
10150
50501
1
1 =−=−=R
RQ h
723.21600
50501
2
2 =−=−=R
RQ h
50 600
Antenna & Propagation Impedance Matching
46
T-Section: Design Example
46
Design a T-section LC matching network to match R1 = 50
and R2 = 600 at 400 MHz, and Q associated with R1 is 10.
(a) For L-1 network, select Q1 = 10.
(b)
(c)
(d)
=+=+= 5050)110(50)1( 22
11 QRRh
50 600
=== 50510
5050
1
'
2Q
RX h
=== 50010*50111 QRX
–ve for C (arbitrarily selected)
+ve for L (opposite of X2' )
Antenna & Propagation Impedance Matching
47
T-Section: Design Example
47
(a) For L-2 network,
(b)
(c)
(d)
(e)
(f)
(g)
50 600
=== 1854723.2
5050"
2
2Q
RX h
=== 1634723.2*600223 QRX
723.21600
50501
2
2 =−=−=R
RQ h
−=+−
−=
+== 694
1854505
1854*505
"'
"'."//'
22
22222
XX
XXXXX
+ve for L (arbitrarily selected)
–ve for C (opposite of X2'' )
–ve means C
nHf
XLX 199
10*400*2
500
2 6
111 ===
pp
pFXf
CX 57.0694*10*400*2
1
2
16
2
22 ===pp
pFXf
CX 24.01634*10*400*2
1
2
16
3
33 ===pp
Antenna & Propagation Impedance Matching
48
T-Section: Design Example (cont'd)
48
50 600
nHL 1991 = pFC 57.02 = pFC 24.03 =
"AUTO-TRANSFORMER"
another possible solution
Antenna & Propagation Impedance Matching
49
p-Section
• Advantages: R1 can be < or > R2
Variable Q (higher than L-section)
• Disadvantage: More L C elements
Method:(a) Introduce a hypothetical resistance level Rh at jX2, such that
Rh < R1 and Rh < R2
(Rh determines the Q of the matching network)
(b) Split jX2 into jX2' and jX2" (in series)
(c) Treat the p-section as two L-sections (L-1 and L-2)
NOTE: Q of L-1 section > Q of 2-element L-section (as Rh < R2)
jX1
jX2'
R2R1
Rh
jX3
jX2
jX2"
L-1 L-2
49
Antenna & Propagation Impedance Matching
50
p-Section: Design Example
Design a p-section LC matching network to match R1 = 50
and R2 = 600 at 400 MHz, and Q associated with R1 is 10.
(a) For L-1 network, select Q1 = 10.
(b)
(c)
(d)
=+
=+
= 495.0110
50
122
1
1
Q
RRh
50 600
=== 0.510
50
1
11
Q
RX
=== 95.410*495.01
'
2 QRX h +ve for L (arbitrarily selected)
–ve for C (opposite of X2' )
50
Antenna & Propagation Impedance Matching
51
p-Section: Design Example (cont'd)
(a) For L-2 network,
(b)
(c)
(d)
(e)
(f)
(g)
50 600
=== 23.1780.34*495.0" 22 QRX h
=== 24.1780.34
600
2
23
Q
RX
80.341495.0
60012
2 =−=−=hR
RQ
=+=+= 18.2223.1795.4"' 222 XXX
+ve for L (arbitrarily selected)
–ve for C (opposite of X2'' )
+ve means L
nHf
XLX 83.8
10*400*2
18.22
2 6
222 ===
pp
pFXf
CX 58.795*10*400*2
1
2
16
1
11 ===pp
pFXf
CX 08.2324.17*10*400*2
1
2
16
3
33 ===pp
51
Antenna & Propagation Impedance Matching
52
p-Section: Design Example (cont'd)
50 600
pFC 58.791 = nHL 83.82 = pFC 08.233 =
This is just one of the four possible solutions.
52
Antenna & Propagation Impedance Matching
53
Common Transmission-Line Applications
• Inductor and capacitor
• Quarter-wave transformer
• Single-stub tuner
• Double-stub tuner
• Triple-stub tuner
• [A good reference for the Double- and Triple-stub tuners is: R.E.Collin, Foundations for Microwave Engineering, McGraw Hill]
53
Antenna & Propagation Impedance Matching
54
Transmission Line Inductor
• The feed-point impedance of a terminated transmission line is:
• If the load impedance is a short-circuit, ZL = 0, then
• Thus, the terminated transmission line behaves like an inductor (inductance = L) if bl < p/2.
54
)ltan(jZZ
)ltan(jZZZZ
Lo
oLoin
+
+=
Lj)ltan(jZZ oin ==
)ltan(ZL o=
Antenna & Propagation Impedance Matching
55
Example
What is the equivalent inductance at 1 GHz at the feed-point of a 50 W, l/8 transmission line which is terminated with a short-circuit?
nH.)(
.tan)ltan(Z
L o 067102
8
250
9=
==p
p
ZL
l
Zo,
Zin
LZin
Antenna & Propagation Impedance Matching
56
Transmission Line Capacitor
• The feed-point impedance of a terminated transmission line is:
• If the load impedance is an open-circuit, ZL = , then
• Thus, the terminated transmission line behaves like a capacitor (capacitance = C) if bl < p/2.
56
)ltan(jZZ
)ltan(jZZZZ
Lo
oLoin
+
+=
Cj)ltan(j
ZZ o
in
1==
oZ
)ltan(C
=
Antenna & Propagation Impedance Matching
57
Example
What is the equivalent capacitance at 1 GHz at the feed-point of a 50 W, l/8 transmission line which is terminated with an open-circuit?
pFZ
lC
o
18.350)10(2
8.
2tan
)tan(9
=
==p
p
o/c
l
Zo,
Zin
CZin
Antenna & Propagation Impedance Matching
58
Quarter-Wave Transformer
• The feed-point impedance of a terminated transmission line is:
• When l = l/4, i.e. bl = p/2, then
• Thus, we can transform the load impedance ZL to another value Zin by designing the l/4 transmission line with a suitable characteristic impedance Zo:
58
)ltan(jZZ
)ltan(jZZZZ
Lo
oLoin
+
+=
L
oin
Z
ZZ
2
=
Lino ZZZ =
=== 2617550 .xZZZ Lino
Antenna & Propagation Impedance Matching
59
Example
Match a 75 W load to a 50 W source at 300 MHz using a l/4 transformer.
The physical length of the transmission line is determined by the wavelength, which depends on the velocity factor of the transmission line.
ZLZo,
Zin
l=/4
Antenna & Propagation Impedance Matching
60
STUB MATCHING
Antenna & Propagation Impedance Matching
61
Single-Stub Tuner
• In a single-stub tuner, a short transmission line (called "stub") is added in parallel with the main transmission line.
• The stub usually has the same Zo as the main transmission line.
• By choosing the appropriate stub length d and its distance l from the load, it is possible to achieve zero reflection from the source to the point where the stub is added.
• Therefore, the stub is always placed close to the load.
• The stub is usually a short-circuit stub to minimize radiation loss (which is more likely to occur with an open-circuit stub).
61
Antenna & Propagation Impedance Matching
62
Single-Stub Tuner
ZLZo
l
d
Short-
circuit stub
Zo
Source Load
Antenna & Propagation Impedance Matching
63
Single-Stub Tuner (cont'd)
• At point A where the stub is added, it is desired to have the impedance equal to ZA=Zin=Zo. It will be more convenient to use the normalized impedance zA=zin=1.
• Also, as the stub is added in parallel, it is more convenient to work in admittances:
• The matching process involves:
– using the distance l to transform yL to 1+jb, and
– using the stub to add –jb to cancel the +jb susceptance.
• Although the matching process can be carried out using equations, a Smith Chart is most conveniently used for this application.
63
11
==in
inz
y11
==o
oz
yL
Lz
y1
=
Antenna & Propagation Impedance Matching
64
Single-Stub Tuner Using the Smith Chart• Locate zL and yL. (yL is diagonally opposite zL.)
• Using a compass, draw a locus towards the Generator along a constant |G| circle until r = 1 circle is reached (Point A). The distance moved is l (in units of ).
• Read the jb value at point A. (Two possible solutions.)
• The stub is required to have –jb.
• Choose either a short-circuit or open-circuit stub.
• Determine the stub length d to produce the required susceptance.
EXAMPLE: (zL = 2)
• Locate zL = 2, and yL = 0.5.
• Move towards Generator until point PA or PB. Read yA = 1 + j0.71 and yB = 1 – j0.71
zLyL
PA
PB
Antenna & Propagation Impedance Matching
65
zLyL
PA
PB
ZLZo
l
dShort-
circuit
stub
Zo
Source Load
A
Antenna & Propagation Impedance Matching
66
Single-Stub Tuner Using the Smith Chart (cont’d)
EXAMPLE: (zL = 2)
• Locate zL = 2, and yL =0.5.
• Move towards Generator until point PA or PB.
• Read on the circumference: lA= 0.152 and lB= 0.348.
• Read at point PA and PB:
yA=1 + j0.71 yB=1 – j0.71
• At point PA, the stub must have a susceptance of –j0.7. If it is a short-circuit stub, its length dAS = 0.402 – 0.25 = 0.152. If it is an open-circuit stub, its length dAO = 0.402. (see next slide.)
• At point PB, the stub must have a susceptance of +j0.7. If it is a short-circuit stub, its length dBS = 0.098 + 0.25 = 0.348. If it is an open-circuit stub, its length dBO = 0.098.
ZLZo
l
d
Short-
circuit
stub
Zo
Source LoadA
zLyL
PA
PB
lA
lB
=stuby0=stuby
66
Antenna & Propagation Impedance Matching
67
For a stub with a
susceptance of –j0.7, if it
is a short-circuit stub, its
length dAS = 0.402 – 0.25
= 0.152. If it is an open-
circuit stub, its length dAO
= 0.402.
=stuby0=stuby
250.
4020.
00.
open-circuit short-circuit
Single-Stub Tuner Using the Smith Chart (cont’d)
67
Antenna & Propagation Impedance Matching
68
Single-Stub Tuner: Exercise
The impedance of a WiFi monopole antenna at 2.48 GHz was 20 – j25 . The antenna was connected to the transmitter via a 50 microstrip line with an effective dielectric constant of 2. Design a single-stub matching network to match the antenna to the transmission line. Use the shortest short-circuit stub.
ZLZo
l
dShort-
circuit
stub
Zo
Source Load
68
Antenna & Propagation Impedance Matching
69
GAMMA MATCH
Antenna & Propagation Impedance Matching
70
70
GAMMA MATCH
• Gamma Match– Unbalanced transmission lines.
– Equivalent circuit• The gamma match is equivalent to half of the T-match
• Requires a capacitor in series with the gamma rod
• The input impedance is:
( ) ( ) ag
ag
cinZZ
ZZjXZ
2
2
12
1
++
++−=
Za is the center point free space input impedance of the antenna in the
absence of gamma-match connection.
[3.21]
Antenna & Propagation Impedance Matching
71
71
GAMMA MATCH
• Design procedure– Determine the current division factor α by using Eq.
[3.13]
– Find the free space impedance (in the absence of the gamma match) of the driven element at the center point. Designate it as Za
– Divide Za by 2 and multiply by the step-up ratio (1+α)2. Designate the result as Z2
– Determine the characteristic impedance Z0 of the transmission line form by the driven element and the gamma rod using Eq. [ 3.15a]
( )2
12
222aZ
jXRZ +=+= [3.22]
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72
GAMMA MATCH
– Normalized Z2 of Eq. [3.22] by Z0 and designate it as z2
– Invert z2 of Eq. [3.23] and obtain its equivalent admittance y2=g2+jb2
– The normalized value Eq. [3.24]
=
+=+
==
2
'tan
22
0
22
0
22
lkjz
jxrZ
jXR
Z
Zz
g
[3.23]
[3.24]
Antenna & Propagation Impedance Matching
73
73
GAMMA MATCH
– Equivalent admittance yg=gg+jbg
– Add two parallel admittances to obtain the total input admittance at the gamma feed.
– Invert the normalize input admittance yin to obtain the equivalent normalized input impedance
( ) ( )
ininin
gggin
jxrz
bbjggyyy
+=
+++=+= 222[3.25]
[3.26]
Antenna & Propagation Impedance Matching
74
74
GAMMA MATCH
– Obtain the unnormalized input impedance by multiplying zin by Z0
– Select the capacitor C so that its reactance is equal in magnitude to Xin
in
inininin
XfCC
zZjXRZ
==
=+=
p 2
11
0 [3.27]
[3.28]
Antenna & Propagation Impedance Matching
75
BALUNS
How they work
How they are made
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76
What is a balun?
• A Balun is special type of transformer that performs two functions:
– Impedance transformation
– Balanced to unbalanced transformation
• The word balun is a contraction of “balanced to unbalanced transformer”
Antenna & Propagation Impedance Matching
77
Why do we need a balun?
• Baluns are important because many types of antennas (dipoles, yagis, loops) are balanced loads, which are fed with an unbalanced transmission line (coax).
• Baluns are required for proper connection of parallel line to a transceiver with a 50 ohm unbalanced output
• The antenna’s radiation pattern changes if the currents in the driven element of a balanced antenna are not equal and opposite.
• Baluns prevent unwanted RF currents from flowing in the “third” conductor of a coaxial cable.
Antenna & Propagation Impedance Matching
78
Balanced vs Unbalanced Transmission Lines
• A balanced transmission line is one whose currents are currents are symmetric with respect to ground so that all current flows through the transmission line and the load and none through ground.
• Note that line balance depends on the current through the line, not the voltage across the line.
Antenna & Propagation Impedance Matching
79
An example of a Balanced Line
• Here is an example of a balanced line. DC rather than AC is used to simplify the analysis:
• Notice that the currents are equal and opposite and the that the total current flowing through ground = 25mA-25mA = 0
I = 25 mA
V = +6 VDC
6 V
6 V
V = -6 VDC
I = -25 mA
Antenna & Propagation Impedance Matching
80
FAQ’s
• Do I really need a balun?
– Not necessarily. If you feed a balanced antenna with unbalanced line and you don’t want feed line radiation, use a balun!
• What kind of balun is best?
– There is no best balun for all applications. The choice of balun depends on the type of antenna and the frequency range.
• Will you make a Balun for me?
– No. However, I will be happy to show how to make your own.