Transmission Lines
Ranga Rodrigo
January 27, 2009
Antennas and Propagation: Transmission Lines
1/72
1
Standing Waves
2
Smith Chart
3
Impedance Matching
Series Reactive Matching
Shunt Reactive Matching
Antennas and Propagation: Transmission Lines
Outline
2/72
Transmission Line Model
Z0
G
R
L
C G
R
L
C G
R
C
L
Z0
R : series loss resistance (copper losses).
G : shunt loss conductance (losses in dielectric).
L : series inductance representing energy
storage within the line.
C : shunt capacitance representing energy
storage within the line.
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
3/72
I1
I3
I2
V1
V3
V2
l
l
If an infinitely long pair of wires were considered
and the voltage and current were somehow
measured at uniform spaced points along the
line, then
V1 V2
Vk
=
= ··· =
= constant = Z0 Ω.
I1
I2
Ik
(1)
This is termed the characteristic impedance of
the line and is denoted by Z0 .
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
4/72
Characteristic Impedance
R + j ωL
Z0 =
G + j ωC
µ
¶1/2
Ω.
(2)
At very low frequencies:
µ ¶1/2
R
Z0 =
Ω.
G
(3)
For high frequencies (ωL À R and ωC À G ):
µ ¶1/2
L
Z0 =
Ω
C
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
(4)
5/72
Propagation Constant
d 2V
= (R + j ωL)(G + j ωC )V.
d x2
(5)
Equation 5 can be written as
d 2V
= γ2V,
2
dx
(6)
where γ = (R + j ωL)(G + j ωC ) is termed the
propagation constant. This is usually expressed as
p
γ = α + j β,
(7)
where α represents the attenuation per unit length
and β the phase shift per unit length.
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
6/72
Equation 6 has a solution of the form
V (x) = Ae −γx + B e γx .
(8)
This suggests that the line will contain two waves,
one traveling in the positive x -direction (e −γx ) and the
other traveling in the negative x -direction (e γx ).
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
7/72
Sending-End Impedance
I inc
I ref
Z0
ZT
l
ZS
x =0
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
8/72
−2γl ZT −Z0
ZS 1 + e
ZT (1 + e −2γl ) + Z0 (1 − e −2γl )
ZT +Z0
=
. (9)
=
Z0 1 − e −2γl ZT −Z0 ZT (1 − e −2γl ) + Z0 (1 + e −2γl )
ZT +Z0
ZS ZT cosh(γl ) + Z0 sinh(γl )
=
.
Z0 ZT sinh(γl ) + Z0 cosh(γl )
(10)
ZT
+ tanh(γl )
ZS
Z0
=
.
Z0 1 + ZT tanh(γl )
Z0
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
(11)
9/72
Short Circuit Termination
If a short circuit is used as the load termination, then
the sending-end impedance becomes,
ZS |SC = Z0 tanh(γl ),
(12)
and, neglecting line losses for short lengths we get
ZS |SC = Z0 tanh( j βl ) = j Z0 tan(βl ).
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
(13)
10/72
Open Circuit Termination
When the termination is open circuit, then the
sending-end impedance becomes,
ZS |OC =
Z0
,
tanh(γl )
(14)
and, neglecting line losses for short lengths we get
ZS |OC =
− j Z0
= − j Z0 cot(βl ).
tanh( j βl )
Antennas and Propagation: Transmission Lines
Summary of last Week’s Lecture
(15)
11/72
− tan(βl )
6
4
2
0
π
2
π
3π
2
2π
5π
2
βl = 2πl
λ
−2
−4
−6
Normalized input resistance versus βl for a
short-circuited,
lossless, transmission
Antennas and Propagation:
Transmission Lines
Summary of last Week’s line.
Lecture
12/72
− cot(βl )
6
4
2
0
π
2
π
3π
2
2π
5π
2
βl = 2πl
λ
−2
−4
−6
Normalized input resistance versus βl for a
open-circuited,
lossless, transmission
Antennas and Propagation:
Transmission Lines
Summary of last Week’s line.
Lecture
13/72
Standing Waves
For any transmission line, a sinusoidal signal of
appropriate frequency introduced at the sending
end by a generator will propagate along the
length of the transmission line.
If the line has infinite length, then the signal
never reaches the end of the line.
If the signal is viewed at some distance down
the line away from the generator, then it will
appear to have the same frequency, but will
exhibit smaller peak to peak voltage swing than
at the generator.
The signal, therefore, has been attenuated by
the line losses due to the conductor resistance
and
dielectric
Antennas and
Propagation:
Transmission Linesimperfections.
Standing Waves
15/72
Reflection
If the line is lossless, then the signal viewed at
some remote point will be identical to that at the
generator but time delayed by an amount
dependant on the position.
When a sinusoidal signal reaches the open end
of a section of a lossless transmission line, it
can dissipate no energy.
This means that all the energy propagating
along the line in the forward direction (incident)
will be reflected completely, on reaching the
open circuit termination.
The reflected wave (backward wave) must be
such that the total current at the open circuit is
zero.
Antennas and
Propagation: Transmission Lines
Standing Waves
16/72
Reflection
As the reflected signal travels back along the
line toward the generator, it reinforces the
incident waveform at certain points forming
maxima (nodes).
Similarly, it can cancel the incident waveform at
certain other points producing minima
(antinodes).
In an open circuit line, node voltage points will
occur at the same position as the antinode
current points.
These waves do not represent traveling waves.
They are standing waves, implying that there is
no net power flow from generator to the
(open-circuit)
load.
Antennas and
Propagation: Transmission Lines
Standing Waves
17/72
A point one-quarter wavelength away from a
short circuit will have voltage and current
magnitudes equivalent to those obtained for an
open circuit.
For a lossless line the peak value of the standing
wave envelope is twice that of the incident wave.
There are nulls as well due to the complete
cancelation.
For lossy (practical) line, the peak will be less
than twice that of the incident wave, and
complete cancelation rarely results.
Antennas and Propagation: Transmission Lines
Standing Waves
18/72
Voltage Standing Wave Ratio
Node to antinode voltage ratio is infinity for the
ideal case.
Voltage standing wave ratio (VSWR) or simply
standing wave ratio is defined as
Vmax
Vmin
|Vinc | + |Vref |
=
|Vinc | − |Vref |
VSWR =
(16)
|Vinc | is the peak value of the incident voltage
wave, similarly the others. Here the voltage
values are peak values. RMS values can be
used as well.
Antennas and Propagation: Transmission Lines
Standing Waves
19/72
Reflection Coefficient
I inc A
I ref
I trans
ZT
Z0
A0
Considering the current continuity at the boundary
A A 0 , we get
Vinc Vref Vtrans
−
=
.
Z0
Z0
ZT
(17)
For voltage continuity across the boundary
(18)
Vinc + Vref = Vtrans .
Antennas and Propagation: Transmission Lines
Standing Waves
20/72
From Equations 17 and 18 we can obtain the
reflection coefficient Γ.
Γ=
Vref ZT − Z0
=
.
Vinc ZT + Z0
(19)
An alternative symbol for the reflection coefficient is
ρ . Sometimes ρ is used to represent |Γ|. From
Equation 16
¯ ¯
¯V ¯
1 + ¯ Vinc ¯ 1 + |Γ|
¯ ref ¯ =
VSWR =
.
¯V ¯
1 − ¯ Vinc ¯ 1 − |Γ|
(20)
¯
¯
¯ Vinc ¯ VSWR − 1
¯=
|Γ| = ¯¯
.
Vref ¯ VSWR + 1
(21)
ref
Therefore,
Antennas and Propagation: Transmission Lines
Standing Waves
21/72
Example
A certain transmission line with a characteristic
impedance of 50 Ω is required to deliver 1 kW over a
short distance to a load. The cable can withstand a
maximum of 250 V. Determine
1
the maximum VSWR, and
2
amount of power that must be provided by the
generator.
Antennas and Propagation: Transmission Lines
Standing Waves
22/72
Solution
We can assume that the line is lossless. Power
dissipated by the load P L is
P L = P inc − P ref .
PL =
2
Vinc
2
Vref
−
,
Z0
Z0
1
= (Vinc − Vref )(Vinc + Vref ),
Z0
= 1000 W.
Antennas and Propagation: Transmission Lines
Standing Waves
23/72
(Vinc − Vref )(Vinc + Vref ) = 50000.
Because RMS voltage in the line must be less than
250 V,
(Vinc + Vref )|max < 250.
(Vinc − Vref ) > 200.
|Vinc | + |Vref |
,
VSWR =
|Vinc | − |Vref |
250
VSWR <
= 1.25.
200
Antennas and Propagation: Transmission Lines
Standing Waves
24/72
We can show that
µ
¶
PL
VSWR − 1 2
= 1−
.
P inc
VSWR + 1
For this value of VSWR, 98.8% of the incident power
is delivered to the load. Therefore, the generator
must be capable of supplying at least 1012 W and
must be matched to the line. In normal design,
VSWR is kept below 1.1.
Antennas and Propagation: Transmission Lines
Standing Waves
25/72
Sending-End Reflection Coefficient
l
Γin =
ZS −Z0
ZS +Z0
Z0 , γ
Γin
ZT
ΓL =
ZT −Z0
ZT +Z0
ΓL
Rearranging Equation 9,
−2γl ZT −Z0
ZS 1 + e
ZT +Z0
=
,
Z0 1 − e −2γl ZT −Z0
ZT +Z0
we can obtain
Γin = ΓL e −2γl .
Antennas and Propagation: Transmission Lines
(22)
Standing Waves
26/72
Impedance Matching
One of the major tasks in transmission line
circuit design is impedance matching, and Smith
chart is a graphical procedure for solving
impedance transformation problems.
A mismatched transmission line (a line
terminated in an impedance other than its
characteristic impedance) would reflect some of
the incident wave back along the line toward the
generator.
This interaction of the forward waves and
reflected waves give rise to a resultant waveform
with nodes and antinodes at fixed points along
length of the transmission line.
This
way,
standing
waves are formed.
Antennas and
Propagation:
Transmission
Lines
Smith Chart
28/72
Effects of Mismatch
Adverse Effects
Energy loss resulting in reduced system
performance.
Frequency pulling of the signal generator.
Possible transmitter (and other circuitry)
damage.
Desirable Effect
Carefully controlled mismatch can improve
system performance by improving the noise
characteristics.
Antennas and Propagation: Transmission Lines
Smith Chart
29/72
Controlling Impedance Mismatch
There are two requirements:
1
A methodical scheme for determining the
degree of mismatch between the source and the
load impedance.
2
A method whereby the degree of mismatch can
be varied in a known manner.
Antennas and Propagation: Transmission Lines
Smith Chart
30/72
Smith Chart
Matching requires a way of handling the
expressions governing the impedance
transformation that occurs along a length of
transmission line with known termination.
In 1939, P. H. Smith, an engineer with the Bell
Telephone Laboratories, developed a graphical
presentation of transmission line data.
He found that for passive loads (0 < |Γ| < 1) the
graphical representation is bounded.
Smith chart is a plot of the voltage reflection.
Antennas and Propagation: Transmission Lines
Smith Chart
31/72
All possible values of impedance and
admittance can be can be plotted on the chart.
An easy method of converting impedances to
admittances is available.
Smith chart provides a simple graphical method
for determining the impedance transformations
due to a length of transmission line.
Antennas and Propagation: Transmission Lines
Smith Chart
32/72
Normalized Resistance Loci
Γ=
ZL − Z0
ZL + Z0
(23)
Let
Γ = u + j v and
ZL
= Zn = R n + j X n .
Z0
(R n − 1) + j X n
.
u+ jv =
(R n + 1) + j X n
(24)
Equating real and imaginary parts gives
R n (u − 1) − X n v = −(u + 1),
R n v + X n (u − 1) = −v.
Antennas and Propagation: Transmission Lines
Smith Chart
(25)
33/72
Xn =
R n (u − 1) + (u + 1)
.
v
(26)
Substituting 26 in 25,
Rn v +
[R n (u − 1) + (u + 1)] (u − 1)
= −v.
v
v 2 (R n + 1) − 2uR n + u 2 (R n + 1) = 1 − R n .
1 − Rn
2uR n
v2 −
+ u2 =
.
1 + Rn
1 + Rn
¶
µ
R n2
Rn 2
1 − Rn
2
−
v + u−
=
.
1 + Rn
(1 + R n )2 1 + R n
(27)
(28)
(29)
Finally, the desired relationship is
Rn
v + u−
1 + Rn
2
µ
Antennas and Propagation: Transmission Lines
¶2
=
1
.
(1 + R n )2
Smith Chart
(30)
34/72
When plotted on the u − v plane, in Cartesian
coordinates, Equation 30,
Rn
v2 + u −
1 + Rn
µ
¶2
=
1
,
(1 + R n )2
represents a circle with center
u=
and radius
Rn
and v = 0,
Rn + 1
1
.
Rn + 1
For different values of R n , we can generate a family
of circles. They represent normalized resistance loci.
Antennas and Propagation: Transmission Lines
Smith Chart
35/72
v
Normalized Resistance
u
Antennas and Propagation: Transmission Lines
Smith Chart
36/72
Normalized Reactance Loci
We can repeat the above process for the normalized
reactive element X n .
¶ µ ¶2
1
1
(u − 1) + v −
=
.
Xn
Xn
2
µ
(31)
This equation too results in a family of circles, this
time, centered at
u = 1 and v =
and with radius
Antennas and Propagation: Transmission Lines
1
,
Xn
1
.
Xn
Smith Chart
37/72
v
Normalized Reactance
u
Antennas and Propagation: Transmission Lines
Smith Chart
38/72
v
u
Antennas and Propagation: Transmission Lines
Smith Chart
39/72
For any passive impedance, Re(Z ) ≥ 0 and z 0
real, |Γ| ≤ 1. Thus all possible values of Γ can be
plotted on this polar chart having a maximum
radius value of unity.
It can accommodate reflection coefficient values
for all impedances from short (Γ = 1∠180◦ ) to an
open (Γ = 1∠0◦ )
Antennas and Propagation: Transmission Lines
Smith Chart
40/72
Constant VSWR Circles
Smith chart is really a plot of the reflection coefficient
Γ = u + j v . Constant VSWR circles are actually |Γ| =
constant circles. From Equation 20 we have the
relation
1 + |Γ|
VSWR =
.
1 − |Γ|
We can, therefore, relate |Γ| and VSWR to plot the
circles.
For a matched system, the reflection coefficient
will be zero and VSWR will be unity, the center
of the Smith chart.
For al reflection coefficient of one, the VSWR
will be infinity and will overlay the contour for
zero resistance.
Antennas and Propagation: Transmission Lines
Smith Chart
41/72
v
Constant VSWR
∞
7
3
1.67
u
Antennas and Propagation: Transmission Lines
Smith Chart
42/72
v
Normalized resistance
Normalized reactance
Constant VSWR
u
Antennas and Propagation: Transmission Lines
Smith Chart
43/72
Points and Properties of Smith Chart
At point u = 1, v = 0 normalized resistance and
reactance terms are large: position for an open
circuit.
At point u = 0, v = 1 the value is 0 + j 1: position
for perfect inductive reactance.
At point u = 0, v = −1 the value is 0 − j 1: position
for perfect capacitive reactance.
At point u = −1, v = 0 zero resistance and zero
reactance contours intersect: position for the
short circuit.
Point u = 0, v = 0, the center of the Smith chart,
represents the normalized impedance 1 + j 0,
which under normal circumstances, represents
the
transmission
line characteristic
Antennas and
Propagation:
Transmission Lines
Smithimpedance.
Chart
44/72
Point
u = 1,
v =0
u = 0,
v =1
u = 0,
v = −1
u = −1,
v =0
u = 0,
v =0
Comment
Representation
normalized
resis- open circuit
tance and reactance
terms are large
0+ j1
perfect inductive
reactance
0− j1
perfect capacitive reactance
zero resistance and short circuit
zero reactance contours intersect
center of the Smith characteristic
chart,
normalized impedance
impedance 1 + j 0
Antennas and Propagation: Transmission Lines
Smith Chart
45/72
Example
Consider a load impedance ZL = 17.7 + j 11.8 Ω
connected to a 50 Ω line of length l = λ/6. Plot ΓL ,
and graphically obtain Γin . Compare the result with
the computed value.
Antennas and Propagation: Transmission Lines
Smith Chart
46/72
Solution
Form Equation 18
ΓL =
ΓV
= ΓL eZ−2γl
inref
T − .Z 0
=
.
Γ=
Vinc ZT + Z0
17.7 + j 11.8 − 50
= 0.5∠150◦ .
17.7 + j 11.8 + 50
We notice that if the line is lossless α = 0, and γ = j β.
So, 2βl = 2 × 2π/λ × (λ/6) = 2π/3 = 120◦ . We can use
Equation 22 to obtain the input reflection coefficient.
Γin = 0.5∠150◦ − 120◦ = 0.5∠30◦ .
The same result can be graphically obtained.
Because |ΓL | = |Γin | for a lossless line, we merely
have to rotate clockwise from ΓL an angle 2βl (120◦ ).
Antennas and Propagation: Transmission Lines
Smith Chart
47/72
v
Normalized resistance
Normalized reactance
Constant Γ
0.75
+30◦
+150◦
0.5
0.25
u
Antennas and Propagation: Transmission Lines
Smith Chart
48/72
Impedance Transformation Using
Smith Chart
The above example showed us that the
graphical solution of the reflection coefficient
transformation is simple.
However, we had to compute ΓL using the
impedances. It is possible to solve impedance
transformations completely graphically.
We just need to replace every point on the polar
reflection coefficient chart by it normalized
equivalent impedance.
We can do this by using the following equations:
Z
1+Γ
= Zn =
,
Z0
1−Γ
Antennas and Propagation: Transmission Lines
and
Y
1−Γ
= Yn =
Y0
1+Γ
Smith Chart
(32)
49/72
Example
Plot the following on the Smith chart.
1
Γ1 = 1∠180◦
2
Γ2 = 1∠0◦
3
Γ3 = 0.5∠180◦
4
Γ4 = 0.5∠0◦
5
Γ5 = 0.5∠90◦
6
Γ6 = 0.5∠−90◦
Antennas and Propagation: Transmission Lines
Smith Chart
50/72
Fortunately, we do not haveNormalized
to useresistance
the
Normalized reactance
equations as, on the Smith chart,
theVSWR
norConstant
malized impedance grid has been superimposed.
Γ5 : Z = 0.6 + j 0.8
+j1
+j2
7
+ j 0.4
+j4
3
+ j 0.2
∞ = Z : Γ2
3
1
0
Γ1 : Z = 0
1
3
1.67
Γ4 : Z = 3
Γ3 : Z = 1/3
− j.2
−j4
Γ6 : Z = 0.6 − j 0.8
− j.4
−j2
−j1
Antennas and Propagation: Transmission Lines
Smith Chart
51/72
Example
A 5.2 cm length of lossless 100 Ω line is terminated
in a load impedance ZL = 30 + j 50 Ω.
1
Calculate ΓL and SWR along the line.
2
Determine the impedance and admittance at the
input and at a point 2.0 cm from the load end.
The signal frequency is 750 MHz.
Solution
ZL = 30 + j 50, and Z0 = 100. Therefore, ZLn = 0.3 + j 0.5.
Antennas and Propagation: Transmission Lines
Smith Chart
52/72
+j1
123.42◦
Normalized resistance
Normalized reactance
Constant VSWR
+ j 0.5
+j2
7
+ j 0.4
4.23
ZLn
+j4
3
+ j 0.2
3
1
0
0.3
1
3
1.67
Z1n = R n = 4.23
− j.2
−j4
− j.4
−j2
|Γ| = 0.6176
−j1
Antennas and Propagation: Transmission Lines
Smith Chart
53/72
1
Γ = 0.617∠123.42◦ . The constant |Γ| circle is the
SWR circle. We obtain 4.23 as the SWR value
from the point of intersection between the SWR
circle and the right half of the resistance axis.
This point is marked as Z1n = R n . At this point
.
R n = 1+|Γ|
1−|Γ|
2
λ = 3 × 108 /750 × 106 = 40 cm. In order to find the
impedance at the input, we must travel
5.2/40 = 0.130λ toward the generator, along a
constant VSWR circle.
Antennas and Propagation: Transmission Lines
Smith Chart
54/72
Normalized resistance
Normalized reactance
Constant VSWR
+j1
123.42◦ , 0.0786λ
+ j 0.5
+j2
7
+ j 0.4
4.23
ZLn
+j4
3
+ j 0.2
Zinn
3
2
1
0
0.3
1
3
1.67
Z1n = R n = 4.23
− j.2
−j4
YLn
− j.4
−j2
|Γ| = 0.6176
−j1
Antennas and Propagation: Transmission Lines
Smith Chart
55/72
1
2
3
We read off the chart that Zin n = 2 + j 2.
Therefore, Zin = 200 + j 200.
Admittance points are directly opposite (across
(1 + j 0)) in the Smith chart. We read off the chart
that YLn = 0.88 − j 1.47.
Do the other calculations on your own.
Antennas and Propagation: Transmission Lines
Smith Chart
56/72
Some Impedance Matching
Techniques
There are two common types of impedance
matching:
Conjugate Matching: The matching of a load
impedance to a generator for
maximum transfer of power.
Z0 Matching: The matching of a load impedance
to a transmission line to eliminate
wave reflection at the load.
Antennas and Propagation: Transmission Lines
Impedance Matching
58/72
Conjugate Matching
Maximum power is delivered to a load when ZL
is set equal to the complex conjugate of the
generator impedance ZG . That is
ZL = ZG∗ = RG − j X G .
In situations where the load impedance is not
adjustable, a matching network may be placed
between the generator and the fixed load.
Antennas and Propagation: Transmission Lines
Impedance Matching
59/72
Z0 Matching
This type matches a load impedance to the
characteristic impedance of a transmission line,
i.e., ZL = Z0 .
In this case ΓL = 0 and hence SWR along the
line is unity.
If ZL 6= Z0 , a matching network may be used to
eliminate the standing waves on the line.
Vref ZTline
− Z0
With ZL connected to the transmission
Γ=
=
through the matching network, V
Zinc
ZT + Z0
0 matching
|Γ|the network
1 + of
requires that the input impedance
VSWR =
equal to Z0 or Zinn = Zin /Z0 = 1 +1j −
0.|Γ|
For a Smith
chart normalized to Z0 , this means that Zinn
must be at the center of the chart.
Antennas and Propagation: Transmission Lines
Impedance Matching
60/72
For a well-designed source, ZG = Z0 , the
characteristic impedance of its output line.
With Z0 real, matching the load to the line
( ZL = Z0 ) results in a conjugate match between
the generator and the load.
Antennas and Propagation: Transmission Lines
Impedance Matching
61/72
Series Reactive Matching
Example
Find the reactance X in the following figure to
achieve matching.
Matching network
jX
To generator
ZL =
Z0 = 50 Ω
Z01 = Z0
25 + j 30
l
in
Antennas and Propagation: Transmission Lines
A
L
Impedance Matching
62/72
108◦
Normalized resistance
Normalized reactance
61.09◦
Constant VSWR
+j1
0.065λ
+j2
7
+ j 0.4
ZLn
+ j 0.2
3
2.87
ZE
+j4
0.235λ
Effect of adding j X n = − j 1.1045
3
1
0
1
3
1.67
Effect of adding j X n = j 1.1045
ZA
− j.2
−j4
− j.4
−j2
−j1
Antennas and Propagation: Transmission Lines
−61.09◦
Impedance Matching
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If we assume that the design frequency is 2000
MHz and the wavelength within the line is equal
to the free space wavelength, we find that
l = 0.065 × λ = 0.065 × 15 cm. l = 0.975 cm. This is
if we pick the point E , the first intersection point
between the constant VSWR circle and R n = 1
circle.
If we pick point A , we get l = 0.212 × 15 = 3.525
cm. Since ωL/Z0 = 1.1, we have ωL = 55 Ω. The
we get L = 55/2π f = 4.38 nH.
With the matching network, Zinn = 1 + j 0 and thus
the SWR on the 50 Ω line to the left of Zin is unity.
Antennas and Propagation: Transmission Lines
Impedance Matching
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The matching network that we designed will not
perfectly match at frequencies other than the design
frequency. This is because both l /λ0 and ωL/Z0 are
frequency dependant. λ0 is the free space
wavelength.
Freq. λ0 (cm)
(MHz)
1800
16.67
2000
15.00
2200
13.64
l /λ0
ωl
Z0
Z An
0.212 0.99 1.4 − j 1.25
0.235 1.10 1.0 − j 1.10
0.259 1.21 0.7 − j 0.95
Antennas and Propagation: Transmission Lines
Zin
SWR
1.4 − j 0.26
1.00
0.7 + j 0.26
1.5
1.00
1.65
Impedance Matching
65/72
Shunt Reactive Matching
Example
Find the susceptance B in the following figure to
achieve matching.
Matching network
To generator
Z0 = 50 Ω
jB
ZL = 20
Z01 = Z0
l
in
Antennas and Propagation: Transmission Lines
A
L
Impedance Matching
66/72
Smith chart solution must be carried out in
admittances.
ZLn = 50/Z0 = .40 Ω.
We can find YLn by rotating the ZLn point halfway
around the chart on its SWR circle.
From the chart, we get YLn = 2.5 + j 0.
Since matching requires Yinn = 1, the length l is
chosen so that the real part of Y An is equal to
unity.
As before, there are two possibilities since the
SWR circle intersects the G n = 1 circle at two
points, Y An = 1 + j 0.95 and Y An = 1 − j 0.95.
Antennas and Propagation: Transmission Lines
Impedance Matching
67/72
+j1
Normalized resistance
Normalized reactance
Constant VSWR
+j2
7
+ j 0.4
+j4
3
+ j 0.2
2.5
YLn = 2.5 + j 0
3
0
ZLn
1
1
3
1.67
0◦
Effect of adding j X n = j 0.95
− j.2
Y An = 1 − j 0.95
−j4
0.09λ
− j.4
−j2
−j1
Antennas and Propagation: Transmission Lines
−64.623◦
Impedance Matching
68/72
For l = 0.09λ, Y An = 1 − j 0.95, and with Y0 = 0.01 S,
Y A = 0.02 − j 0.19 S.
This means that a shunt capacitance of
admittance value + j 0.019 S is required to cancel
the inductive effect of Y A .
Thus adding a capacitive susceptance
B = ωC = 0.019 S at plane A results in Yin = 0.02 S
or Yinn = 1.0, a matched condition.
Short-circuit or open-circuit stubs may be used
to obtain the required susceptance.
It is best to use a high impedance shorted line
for inductive susceptance and a low impedance
open-circuited line for capacitive susceptance.
Antennas and Propagation: Transmission Lines
Impedance Matching
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Shunt Reactive Matching with a
Shorted Stub
l
Z0
Z0
ZL
Z0s
short
ls
Antennas and Propagation: Transmission Lines
Impedance Matching
70/72
Shunt Reactive Matching with a
Open-Circuited Stub
l
Z0
Z0
ZL
Z0s
open
ls
Antennas and Propagation: Transmission Lines
Impedance Matching
71/72
Stub Tuners
We can do stub tuning using either series or
shunt stubs.
However, series and shunt reactance matching
techniques usually result in low SWR values
over narrow range (typically, less than 20%).
Improved broadband performance can often be
realized by utilizing series and parallel resonant
circuits.
Antennas and Propagation: Transmission Lines
Impedance Matching
72/72