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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 I2Two-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 I2Two-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
Smith Chart Impedance or
Admittance
Antenna & Propagation Impedance Matching
9
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
10
IMPEDANCE MATCHING
1 – Introduction
2 – Importance of Impedance Matching
Antenna & Propagation Impedance Matching
11
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
12
Feed-Point Impedance: Za
• Za = antenna impedance at its feed-point.
• Za is complex generally.
aaa jXRZ +=Za
Antenna & Propagation Impedance Matching
13
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
14
Concept of maximum power
transferZo
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 circuitPL
ZL
Zo
Antenna & Propagation Impedance Matching
15
continueIn transmission line
oL
oL
ZZ
ZZ
+−
=ρ
No reflection whence ZL = Zo , hence 0=ρ
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
16
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
17
MATCHING METHODS
1 – calculations
2 – smith chart
Antenna & Propagation Impedance Matching
18
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
19
continue
Configuration 1Configuration 2
If the load impedance
(normalized) lies in unity
circle, configuration 1 is
used.Otherwise
configuration 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
20
Impedances for serial lumped elements
Serial circuit
Reactance relationship values
+ve X=2πfL L=X/(2πf)-ve X=1/(2πfC) C=1/(2πfX)
L C
RR
Antenna & Propagation Impedance Matching
21
Impedances for parallel lumped elements
Parallel circuit
Susceptance relationship values
+ve B=2πfC C=B/(2πf)-ve B=1/(2πfL) L=1/(2πfB)
LC
RR
Antenna & Propagation Impedance Matching
22
Lumped elements for microwave integrated
circuit
Spiral inductorLoop inductor
Interdigital
gap capacitor
Planar resistor Chip resistor
Lossy film
Metal-insulator-
metal capacitor Chip capacitor
εr
Dielectric
εr
Antenna & Propagation Impedance Matching
23
Matching by calculation for
configuration 1jX
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
24
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
25
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
26
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
27
parallel capacitance
parallel inductanceserial capacitance
serial inductance
(-ve)
(-ve)
(+ve)
(+ve)
Antenna & Propagation Impedance Matching
28
C1
L1
C2
L2
50ΩC2
L!
10Ω
C1L
2
10Ω 50Ω
Serial C
Parallel
L
Serial
L
Parallel
C
Matching using lumped
components
Antenna & Propagation Impedance Matching
29
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-j1 Parallel C
(+j0.3)Serial L
(j1.2)
Serial C
(-j1.2)
Parallel L
(-j0.7)
Solution 1
Solution 2
Antenna & Propagation Impedance Matching
30
continue
ZL=200-j100
2.61pF
46.1nH
200-j1000.92pF
38.8nHpF
Zf
bC
o
92.02
==π
nHf
ZxL o 8.38
2==
π
pFZfx
Co
61.22
1=
−=
π
nHbf
ZL o 1.46
2=
−=
πReflection coefficient
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5
freq (GHz)
reflection c
oeffic
ient
solution 1
solution 2Solution 2 seems to be
better matched at higher
frequency
Antenna & Propagation Impedance Matching
31
MATCHING TECHNIQUES
1 – Discrete lumped-element
2 – Transmission Line
Antenna & Propagation Impedance Matching
32
Impedances for serial lumped elements
Serial circuit
Reactance relationship values
+ve X=2πfL L=X/(2πf)-ve X=1/(2πfC) C=1/(2πfX)
L C
RR
Antenna & Propagation Impedance Matching
33
Impedances for parallel lumped elements
Parallel circuit
Susceptance relationship values
+ve B=2πfC C=B/(2πf)-ve B=1/(2πfL) L=1/(2πfB)
LC
RR
Antenna & Propagation Impedance Matching
34
Lumped elements for
microwave integrated circuit
Spiral inductorLoop inductor
Interdigital
gap capacitor
Planar resistor Chip resistor
Lossy film
Metal-insulator-metal
capacitor Chip capacitor
εr
Dielectric
εr
Antenna & Propagation Impedance Matching
35
Resistive L-Section (Using
Resistors)• Advantage: Broadband
• Disadvantage: Very lossy
35
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
36
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
37
Resistive L-Section: Example
37
Design a broadband resistive L-section matching network to match a 75 ΩTV antenna output to a 50 Ω transmission line. Calculate the attenuation of
the matching network.
Solution:
Ro1 = 75 Ω, Ro2 = 50 Ω
Substituting these values into previous equations, we get:
R1 = 43.3 Ω, R2 = 86.6 Ω
= 0.2113 = –13.5 dB
Balun75 Ω-50 Ω matching
network
75 Ωtxn line
50 Ωtxn line
7.313.4375
7.31
++=
in
out
V
V
Antenna & Propagation Impedance Matching
38
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
38
Antenna & Propagation Impedance Matching
39
Reactive LC L-Section: Example
Design an L-section matching network with L and C to match a 50 Ωsource to a 600 Ω load for maximum power transfer at 400 MHz.
Give two solutions.
Solution:
Rs = 50 Ω, Rp = 600 Ω
Substituting these values into previous equations, we get:
Xs = Q.Rs = 166 Ω
Xp = Rp/Q = 181 Ω
317.311150
6001 ==−=−=
s
p
R
RQ
Antenna & Propagation Impedance Matching
40
Reactive LC L-Section: Example (cont'd)
Solution 1:
Xs = +166 Ω (inductive), Xp = –181 Ω (capacitive)
Solution 2:
Xs = –166 Ω (capacitive), Xp = +181 Ω (inductive)
nHf
XL ss 66
10*400*2
166
2 6===
ππ
pFXf
Cp
p 2.2181*10*400*2
1
2
16
===ππ
nHf
XL
p
p 7210*400*2
181
2 6===
ππ
pFXf
Cs
s 4.2166*10*400*2
1
2
16
===ππ
Antenna & Propagation Impedance Matching
41
Reactive LC L-Section: Example (cont'd)
Solution 1:
Xs = +166 Ω (inductive), Xp = –181 Ω (capacitive)
Solution 2:
Xs = –166 Ω (capacitive), Xp = +181 Ω (inductive)
Antenna & Propagation Impedance Matching
42
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'.
42
Antenna & Propagation Impedance Matching
43
T-Section
• Advantages: R1 can be < or > R2Variable 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
43
Antenna & Propagation Impedance Matching
44
T-Section: Q Values
44
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 Ω (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 Ω 600 Ω
Antenna & Propagation Impedance Matching
45
T-Section: Design Example
45
Design a T-section LC matching network to match R1 = 50 Ωand R2 = 600 Ω at 400 MHz, and Q associated with R1 is 10.
(a) For L-1 network, select Q1 = 10.
(b)
(c)
(d)
Ω=+=+= 5050)110(50)1( 22
11 QRRh
50 Ω 600 Ω
Ω=== 50510
5050
1
'
2Q
RX h
Ω=== 50010*50111 QRX
–ve for C (arbitrarily selected)
+ve for L (opposite of X2' )
Antenna & Propagation Impedance Matching
46
T-Section: Design Example
46
(a) For L-2 network,
(b)
(c)
(d)
(e)
(f)
(g)
50 Ω 600 Ω
Ω=== 1854723.2
5050"
2
2Q
RX h
Ω=== 1634723.2*600223 QRX
723.21600
50501
2
2 =−=−=R
RQ h
Ω−=+−
−=
+== 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 ===⇒
ππpF
XfCX 57.0
694*10*400*2
1
2
16
2
22 ===⇒ππ
pFXf
CX 24.01634*10*400*2
1
2
16
3
33 ===⇒ππ
Antenna & Propagation Impedance Matching
47
T-Section: Design Example
(cont'd)
47
50 Ω 600 Ω
nHL 1991 = pFC 57.02 = pFC 24.03 =
"AUTO-TRANSFORMER"
another possible solution
Antenna & Propagation Impedance Matching
48
ππππ-Section
• Advantages: R1 can be < or > R2Variable 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 π-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
48
Antenna & Propagation Impedance Matching
49
ππππ-Section: Design Example
Design a π-section LC matching network to match R1 = 50 Ωand R2 = 600 Ω at 400 MHz, and Q associated with R1 is 10.
(a) For L-1 network, select Q1 = 10.
(b)
(c)
(d)
Ω=+
=+
= 495.0110
50
122
1
1
Q
RRh
50 Ω 600 Ω
Ω=== 0.510
50
1
11
Q
RX
Ω=== 95.410*495.01
'
2 QRX h +ve for L (arbitrarily selected)
–ve for C (opposite of X2' )
49
Antenna & Propagation Impedance Matching
50
ππππ-Section: Design Example (cont'd)
(a) For L-2 network,
(b)
(c)
(d)
(e)
(f)
(g)
50 Ω 600 Ω
Ω=== 23.1780.34*495.0" 22 QRX h
Ω=== 24.1780.34
600
2
23
Q
RX
80.341495.0
60012
2 =−=−=hR
RQ
Ω=+=+= 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 ===⇒
ππ
pFXf
CX 58.795*10*400*2
1
2
16
1
11 ===⇒ππ
pFXf
CX 08.2324.17*10*400*2
1
2
16
3
33 ===⇒ππ
50
Antenna & Propagation Impedance Matching
51
ππππ-Section: Design Example (cont'd)
50 Ω 600 Ω
pFC 58.791 = nHL 83.82 = pFC 08.233 =
This is just one of the four possible solutions.
51
Antenna & Propagation Impedance Matching
52
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]
52
Antenna & Propagation Impedance Matching
53
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.
53
)ltan(jZZ
)ltan(jZZZZ
Lo
oLoin β
β++
=
Lj)ltan(jZZ oin ωβ ==
ωβ )ltan(Z
L o=
Antenna & Propagation Impedance Matching
54
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=
==π
λλπ
ωβ ZL
l
Zo, β
Zin
LZin
Antenna & Propagation Impedance Matching
55
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.
55
)ltan(jZZ
)ltan(jZZZZ
Lo
oLoin β
β++
=
Cj)ltan(j
ZZ oin ωβ
1==
oZ
)ltan(C
ωβ
=
∞
Antenna & Propagation Impedance Matching
56
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
=
==π
λλπ
ωβ
o/c
l
Zo, β
Zin
CZin
Antenna & Propagation Impedance Matching
57
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:
57
)ltan(jZZ
)ltan(jZZZZ
Lo
oLoin β
β++
=
L
oin
Z
ZZ
2
=
Lino ZZZ =
Ω=== 2617550 .xZZZ Lino
Antenna & Propagation Impedance Matching
58
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, β
Zin
l=λ/4
Antenna & Propagation Impedance Matching
59
STUB MATCHING
Antenna & Propagation Impedance Matching
60
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).
60
Antenna & Propagation Impedance Matching
61
Single-Stub Tuner
ZLZo
l
d
Short-
circuit stub
Zo
Source Load
Antenna & Propagation Impedance Matching
62
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.
62
11
==in
inz
y11
==o
oz
yL
Lz
y1
=
Antenna & Propagation Impedance Matching
63
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 |Γ| circle until r = 1
circle is reached (Point A). The distance moved is l
(in units of λ).
• 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
64
zLyL
PA
PB
ZLZo
l
dShort-
circuit
stub
Zo
Source Load
A
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65
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= 0.152λ and
lB= 0.348λ.
• 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 = 0.402λ – 0.25λ =
0.152λ. If it is an open-circuit stub, its
length dAO = 0.402λ. (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 = 0.098λ + 0.25λ =
0.348λ. If it is an open-circuit stub, its
length dBO = 0.098λ.
ZLZo
l
d
Short-
circuit
stub
Zo
Source LoadA
zLyL
PA
PB
lA
lB
∞=stuby0=stuby
65
Antenna & Propagation Impedance Matching
66
For a stub with a
susceptance of –j0.7, if it
is a short-circuit stub, its
length dAS = 0.402λ – 0.25λ= 0.152λ. If it is an open-circuit stub, its length dAO= 0.402λ.
∞=stuby0=stuby
λ250.
λ4020.
λ00.
open-circuit short-circuit
Single-Stub Tuner Using the Smith Chart (cont’d)
66
Antenna & Propagation Impedance Matching
67
Single-Stub Tuner: Exercise
The impedance of a WiFi
monopole antenna at 2.48
GHz was 20 – j25 Ω. The
antenna was connected to
the transmitter via a 50 Ωmicrostrip 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
67
Antenna & Propagation Impedance Matching
68
GAMMA MATCH
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69
69
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
α
α
++
++−=
Za is the center point free space input impedance of the antenna in the
absence of gamma-match connection.
[3.21]
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70
70
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
222aZjXRZ α+=+= [3.22]
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71
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]
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72
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]
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73
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
==
=+=
πω 2
11
0 [3.27]
[3.28]
Antenna & Propagation Impedance Matching
74
BALUNS
How they work
How they are made
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75
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”
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76
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.
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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.
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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
240 Ω
240 Ω
Antenna & Propagation Impedance Matching
79
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.