Transmission Lines
Complex Numbers, Phasors and Circuits
Complex numbers are defined by points or vectors in the complex
plane, and can be represented in Cartesian coordinates
z a jb
j 1
or in polar (exponential) form
z A exp( j) A cos jA sin a A cos real part
b A sin imaginary part
where
2
A a b
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2
tan
1 b a
1
Transmission Lines
Im
z
b
A
a
Note :
Re
z A exp( j) A exp( j j 2 n)
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Transmission Lines
Every complex number has a complex conjugate
z* a jb * a jb
so that
z z* ( a jb) ( a jb)
a2 b2 z
2
A2
In polar form we have
z* A exp( j) * A exp( j)
A exp j 2 j A cos jA sin © Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
The polar form is more useful in some cases. For instance, when
raising a complex number to a power, the Cartesian form
z n ( a jb) ( a jb) ( a jb)
is cumbersome, and impractical for noninteger exponents.
polar form, instead, the result is immediate
In
n
z n A exp( j) An exp jn In the case of roots, one should remember to consider + 2k as
argument of the exponential, with k = integer, otherwise possible
roots are skipped:
n z n A exp j j 2 k n A exp j j 2 k n
n The results corresponding to angles up to 2 are solutions of the
root operation.
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Transmission Lines
In electromagnetic problems it is often convenient to keep in mind
the following simple identities
j exp j 2
j exp j 2
It is also useful to remember the following expressions for
trigonometric functions
exp( jz) exp( jz)
exp( jz) exp( jz)
cos z ; sin z 2
2j
resulting from Euler’s identity
exp( jz) cos( z) j sin( z)
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Transmission Lines
Complex representation is very useful for time-harmonic functions
of the form
A cos t Re A exp j t j Re A exp j exp j t Re A exp j t The complex quantity
A A exp j contains all the information about amplitude and phase of the
signal and is called the phasor of
A cos t If it is known that the signal is time-harmonic with frequency , the
phasor completely characterizes its behavior.
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Transmission Lines
Often, a time-harmonic signal may be of the form:
A sin t and we have the following complex representation
A sin t Re jA cos t j sin t Re jA exp j t j Re A exp j / 2 exp j exp j t Re A exp j / 2 exp j t Re A exp j t with phasor
A A exp j / 2 This result is not surprising, since
cos( t / 2) sin( t )
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Transmission Lines
Time differentiation can be greatly simplified by the use of phasors.
Consider for instance the signal
V ( t ) V0 cos t with phasor
V V0 exp j The time derivative can be expressed as
V ( t)
V0 sin t t
Re jV0 exp j exp j t jV0 exp j j V
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is the phasor of
V ( t)
t
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Transmission Lines
With phasors, time-differential equations for time harmonic signals
can be transformed into algebraic equations. Consider the simple
circuit below, realized with lumped elements
R
L
i (t)
v (t)
C
This circuit is described by the integro-differential equation
d it
1 t
v( t ) L
Ri i( t ) dt
dt
C © Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
Upon time-differentiation we can eliminate the integral as
d2 i t
d v( t )
di 1
L
R i( t )
dt
dt C
dt 2
If we assume a time-harmonic excitation, we know that voltage and
current should have the form
v( t ) V0 cos( t V )
phasor V V0 exp( jV )
i( t ) I0 cos( t I )
phasor I I0 exp( j I )
If V0 and V are given,
I0 and I are the unknowns of the problem.
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Transmission Lines
The differential equation can be rewritten using phasors
L Re 2 I exp j t R Re j I exp j t 1
Re I exp j t Re j V exp j t C
Finally, the transform phasor equation is obtained as
1 V R j L j
IZI
C where
Z
Impedance
R
Resistance
1 j L C
Reactance
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Transmission Lines
The result for the phasor current is simply obtained as
V
V
I0 exp j I I 1 Z R j L j C which readily yields the unknowns I0 and I .
The time dependent current is then obtained from
i( t ) Re I0 exp j I exp j t I0 cos t I © Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
The phasor formalism provides a convenient way to solve timeharmonic problems in steady state, without having to solve directly
a differential equation. The key to the success of phasors is that
with the exponential representation one can immediately separate
frequency and phase information. Direct solution of the timedependent differential equation is only necessary for transients.
Integro-differential
equations
Transform
Algebraic equations
based on phasors
I=?
i(t)=?
Direct Solution
( Transients )
i(t)
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Solution
AntiTransform
I
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Transmission Lines
The phasor representation of the circuit example above has
introduced the concept of impedance. Note that the resistance is
not explicitly a function of frequency. The reactance components
are instead linear functions of frequency:
Inductive component
proportional to Capacitive component inversely proportional to Because of this frequency dependence, for specified values of L
and C , one can always find a frequency at which the magnitudes of
the inductive and capacitive terms are equal
r L 1
r C
r 1
LC
This is a resonance condition. The reactance cancels out and the
impedance becomes purely resistive.
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Transmission Lines
The peak value of the current phasor is maximum at resonance
I0 | I0 |
V0
1 2 R L C
2
IM
r
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Transmission Lines
Consider now the circuit below where an inductor and a capacitor
are in parallel
I
L
R
C
V
The input impedance of the circuit is
1
j C Zin R j L
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1
R
j L
1 2 LC
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Transmission Lines
When
0
Zin R
1
LC
Zin Zin R
At the resonance condition
r 1
LC
the part of the circuit containing the reactance components
behaves like an open circuit, and no current can flow. The voltage
at the terminals of the parallel circuit is the same as the input
voltage V.
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Transmission Lines
Power in Circuits
Consider the input impedance of a transmission line circuit, with an
applied voltage v(t) inducing an input current i(t).
i(t)
v(t)
Zin
For sinusoidal excitation, we can write
v ( t ) V0 cos( t )
i ( t ) I0 cos( t )
/2 , /2 where is the phase difference between voltage and current. Note
that = 0 only when the input impedance is real (purely resistive).
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Transmission Lines
The time-dependent input power is given by
P( t ) v ( t ) i ( t ) V0 I0 cos( t ) cos( t )
V0 I0
cos() cos(2 t )
2
The power has two (Fourier) components:
(A) an average value
V0 I0
cos()
2
(B) an oscillatory component with frequency 2f
V0 I0
cos(2 t )
2
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Transmission Lines
The power flow changes periodically in time with an oscillation like
(B) about the average value (A). Note that only when = 0 we have
cos() = 1, implying that for a resistive impedance the power is
always positive (flowing from generator to load).
When voltage and current are out of phase, the average value of the
power has lower magnitude than the peak value of the oscillatory
component. Therefore, during portions of the period of oscillation
the power can be negative (flowing from load to generator). This
means that when the power flow is positive, the reactive component
of the input impedance stores energy, which is reflected back to the
generator side when the power flow becomes negative.
For an oscillatory excitation, we are interested in finding the
behavior of the power during one full period, because from this we
can easily obtain the average behavior in time. From the point of
view of power consumption, we are also interested in knowing the
power dissipated by the resistive component of the impedance.
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Transmission Lines
We can also rewrite the time-dependent current as
i ( t ) I0 cos( t ) I0 cos cos t I0 sin sin t
where we have used the trigonometry formula
cos A B cos A cos B sin A sin B
This result yields an equivalent expression for the power
P( t ) V0 cos( t ) I0 cos( t ) cos() V0 cos( t ) I0 sin( t ) sin()
V0 I0
1
2
sin() sin(2 t )
V0 I0 cos() cos ( t ) 1
2
Active (Real) Power
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Reactive Power
21
Transmission Lines
The active power corresponds to the power dissipated by the
resistive component of the impedance, and it is always positive.
The reactive power corresponds to the power stored and then
reflected by the reactive component of the impedance. It oscillates
from positive to negative during the period.
Until now we have discussed properties of instantaneous power.
Since we are considering time-harmonic periodic signals, it is very
convenient to consider the time-average power
1 T
P( t ) P( t ) dt
T 0
where T = 1 / f is the period of the oscillation.
We can use either the Fourier or the active/reactive power
formulation to determine the time-average power.
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Transmission Lines
Fourier representation
1 T V0 I0
P( t ) cos() dt
T 0
2
1 T V0 I0
cos(2 t ) dt
0
T
2
V0 I0
cos() 0
2
As one should expect, the time-average power flow is simply given
by the Fourier component corresponding to the average of the
original signal.
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Transmission Lines
Active/Reactive power representation
1 T
P( t ) V0 I0 cos() cos2 ( t ) dt
T 0
1 T V0 I0
sin() sin(2 t ) dt
0
T
2
V0 I0
cos() 0
2
This result tells us that the time-average power flow is the average
of the active power. The reactive power has zero time-average,
since power is stored and completely reflected by the reactive
component of the input impedance during the period of oscillation.
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Transmission Lines
The maximum of the reactive power is
V I
V I
max {Preac } = max{ 0 0 sin (φ ) sin ( 2ω t )} = 0 0 sin ( φ )
2
2
Since the time-average of the reactive power is zero, we often use
the maximum value above as an indication of the reactive power.
The sign of the phase φ tells us about the imaginary part of the
impedance (reactance)
φ > 0 The reactance is inductive
Current is lagging with respect to voltage
Voltage is leading with respect to current
φ < 0 The reactance is capacitive
Voltage is lagging with respect to current
Current is leading with respect to voltage
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Transmission Lines
If the net reactance is inductive
V Z I R I j L I
Im
j L I
V
Current lags
>0
I
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RI
Re
26
Transmission Lines
If the net reactance is capacitive
Im
1
VZI RI j
I
C
I
RI
Re
Voltage lags
<0
V
- j I / C
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Transmission Lines
In many engineering situations, we use the root-mean-square
(r.m.s.) values of quantities. For a given signal
v( t ) V0 cos( t )
the r.m.s. value is defined as
1 T 2
1 T
2
2
Vrms V
t
dt
V
cos
cos
t d t
0
0
T 0
T 0
1 2 2
1
V0
cos d V0
2 0
2
This result is valid for sinusoidal signals. Each signal shape
corresponds to a specific coefficient ( peak factor = V0 / Vrms ) that
allows one to convert directly from peak to r.m.s. values.
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Transmission Lines
The peak factor for sinusoidal signals is
V0
Vrms
2
1.4142
For a symmetric triangular signal the peak factor is
V0
3 1.732
Vrms
V0
t
For a symmetric square signal the peak factor is simply
V0
1
Vrms
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V0
t
29
Transmission Lines
For a non-sinusoidal periodic signal, we can determine the r.m.s.
value by using a very important theorem of vector spaces. If we
decompose the non-sinusoidal signal into its Fourier components
V ( t ) Vav V1 ( t ) V2 ( t ) V3 ( t ) k Vk t then
2
V t rms k Vk ( t)rms 2
2
V t
k k rms
So, the r.m.s. value of the signal is computed as
2
Vrms Vav
V1 rms V2 rms V3 rms 2
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2
2
30
Transmission Lines
So far, we have used peak values for the amplitude of voltage and
current. In terms of r.m.s. values, the time-average power for a
sinusoidal signal is
V0 I0
cos() Vrms Irms cos()
P( t ) 2 2
Finally, we can relate the time-average power to the phasors of
voltage and current. Since
v ( t ) V0 cos( t ) Re V0 exp( j t )
i ( t ) I0 cos( t ) Re I0 exp( j) exp( j t )
we have phasors
V V0
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I I0 exp( j)
31
Transmission Lines
The time-average power in terms of phasors is given by
1
1
*
P( t ) Re V I Re V0 I0 exp( j)
2
2
V0 I0
cos()
2
Note that one must always use the complex conjugate of the phasor
current to obtain the time-average power. It is important to
remember this when voltage and current are expressed as
functions of each other. Only when the impedance is purely
resistive, I = I* = I0 since = 0.
Also, note that the time-average power is always a real positive
quantity and that it is not the phasor of the time-dependent power.
It is a common mistake to think so.
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Transmission Lines
Now, we consider power flow including explicitly the generator, to
understand in which conditions maximum power transfer to a load
can take place.
ZR
Vin VG
ZG ZR
1
Iin VG
ZG ZR
1
*
Pin Re Vin Iin
2
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Iin
ZG
VG
ZR
Vin
Generator
Load
33
Transmission Lines
As a first case, we examine resistive impedances
ZG RG
ZR RR
Voltage and current are in phase at the input. The time-average
power dissipated by the load is
RR
1
1
*
VG
P( t ) VG
RG RR
RG RR
2
RR
1 2
VG
2
( RG RR )2
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Transmission Lines
To find the load resistance that maximizes power transfer to the
load for a given generator we impose
d P( t )
0
dRR
from which we obtain
d RR
0
dRR ( RG RR )2 ( RG RR )2 2 RR ( RG RR )
( RG RR )
4
( RG RR ) 2 RR 0
0
RR RG
We conclude that for maximum power transfer the load resistance
must be identical to the generator resistance.
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Transmission Lines
Let’s consider now complex impedances
ZR RR jX R
ZG RG jXG
For maximum power transfer, generator and load impedances must
be complex conjugate of each other
*
ZR ZG
RR RG
X R XG
This can be easily understood by considering that, to maximize the
active power supplied to the load, voltage and current of the
generator should remain in phase. If the reactances of generator
and load are opposite and cancel each other along the path of the
current, the generator will only see a resistance. Voltage and
current will be in phase with maximum power delivered to the load.
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Transmission Lines
The total time-average power supplied by the generator in
conditions of maximum power transfer is
1
1 2 1
1 2 1
*
Ptot Re VG Iin VG
VG
RR
2
2
2 RR
4
The time-average power supplied to the load is
*
ZR
1
1 1
*
*
VG Pin Re Vin Iin Re VG
2
2 ZG ZR
ZG ZR !
RR jX R 1 2
1 2 1
VG Re VG
RR
2
4 RR !
8
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37
Transmission Lines
The power dissipated by the internal generator impedance is
1
PG Re
2
1 2
VG
4
*
(VG Vin ) Iin
1
1 2 1
1 2 1
VG
VG
RR
RR
RR
8
8
We conclude that, in conditions of maximum power transfer, only
half of the total active power supplied by the generator is actually
used by the load.
The generator impedance dissipates the
remaining half of the available active power.
This may seem a disappointing result, but it is the best one can do
for a real generator with a given internal impedance!
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38
Transmission Lines
Transmission Line Equations
A typical engineering problem involves the transmission of a signal
from a generator to a load. A transmission line is the part of the
circuit that provides the direct link between generator and load.
Transmission lines can be realized in a number of ways. Common
examples are the parallel-wire line and the coaxial cable. For
simplicity, we use in most diagrams the parallel-wire line to
represent circuit connections, but the theory applies to all types of
transmission lines.
Load
Generator
ZG
VG
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Transmission line
ZR
39
Transmission Lines
Examples of transmission lines
d
d
D
d
Two-wire line
D
Coaxial cable
W
t
D
Microstrip
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Transmission Lines
If you are only familiar with low frequency circuits, you are used to
treat all lines connecting the various circuit elements as perfect
wires, with no voltage drop and no impedance associated to them
(lumped impedance circuits). This is a reasonable procedure as
long as the length of the wires is much smaller than the wavelength
of the signal. At any given time, the measured voltage and current
are the same for each location on the same wire.
Load
Generator
ZG
VG
ZR
VR VG
ZG ZR
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VR
VR
VR
ZR
L << 41
Transmission Lines
Let’s look at some examples. The electricity supplied to households
consists of high power sinusoidal signals, with frequency of 60Hz
or 50Hz, depending on the country. Assuming that the insulator
between wires is air ( 0), the wavelength for 60Hz is:
c 2.999 108
5.0 106 m 5, 000 km
60
f
which is the about the distance between S. Francisco and Boston!
Let’s compare to a frequency in the microwave range, for instance
60 GHz. The wavelength is given by
c 2.999 108
5.0 10 3 m 5.0 mm
f
60 109
which is comparable to the size of a microprocessor chip.
Which conclusions do you draw?
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Transmission Lines
For sufficiently high frequencies the wavelength is comparable with
the length of conductors in a transmission line. The signal
propagates as a wave of voltage and current along the line, because
it cannot change instantaneously at all locations. Therefore, we
cannot neglect the impedance properties of the wires (distributed
impedance circuits).
V (z) V e j z V e j z
Generator
Load
ZG
VG
V( 0 )
V( z )
V( L )
ZR
L
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Transmission Lines
Note that the equivalent circuit of a generator consists of an ideal
alternating voltage generator in series with its actual internal
impedance. When the generator is open ( ZR ) we have:
Iin 0 and Vin VG
If the generator is connected to a load ZR
VG
Iin ZG ZR VG ZR
Vin ZG ZR If the load is a short ( ZR
VG
Iin ZG
and
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Iin
ZG
VG
Vin
ZR
0)
Vin 0
Generator
Load
44
Transmission Lines
The simplest circuit problem that we can study consists of a
voltage generator connected to a load through a uniform
transmission line. In general, the impedance seen by the generator
is not the same as the impedance of the load, because of the
presence of the transmission line, except for some very particular
cases.
Zin ZR
Zin
Transmission line
ZR
only if
Ln
2
[ n integer ]
L
Our goal is to determine the equivalent impedance seen by the
generator, that is, the input impedance of the line terminated by the
load. Once that is known, standard circuit theory can be used.
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45
Transmission Lines
Load
Generator
ZG
Transmission line
VG
ZR
Equivalent Load
Generator
ZG
VG
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Zin
46
Transmission Lines
A uniform transmission line is a “distributed circuit” that we can
describe as a cascade of identical cells with infinitesimal length.
The conductors used to realize the line possess a certain series
inductance and resistance. In addition, there is a shunt capacitance
between the conductors, and even a shunt conductance if the
medium insulating the wires is not perfect. We use the concept of
shunt conductance, rather than resistance, because it is more
convenient for adding the parallel elements of the shunt. We can
represent the uniform transmission line with the distributed circuit
below (general lossy line)
L dz
R dz
C dz
dz
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L dz
G dz
R dz
C dz
G dz
dz
47
Transmission Lines
The impedance parameters L, R, C, and G represent:
L = series inductance per unit length
R = series resistance per unit length
C = shunt capacitance per unit length
G = shunt conductance per unit length.
Each cell of the distributed circuit will have impedance elements
with values: Ldz, Rdz, Cdz, and Gdz, where dz is the infinitesimal
length of the cells.
If we can determine the differential behavior of an elementary cell of
the distributed circuit, in terms of voltage and current, we can find a
global differential equation that describes the entire transmission
line. We can do so, because we assume the line to be uniform
along its length.
So, all we need to do is to study how voltage and current vary in a
single elementary cell of the distributed circuit.
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48
Transmission Lines
Loss-less Transmission Line
In many cases, it is possible to neglect resistive effects in the line.
In this approximation there is no Joule effect loss because only
reactive elements are present. The equivalent circuit for the
elementary cell of a loss-less transmission line is shown in the
figure below.
L dz
I (z)+dI
I (z)
V (z)
V (z)+dV
C dz
dz
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49
Transmission Lines
The series inductance determines the variation of the voltage from
input to output of the cell, according to the sub-circuit below
L dz
I (z)
V (z)
V (z)+dV
dz
The corresponding circuit equation is
(V dV ) V j L dz I
which gives a first order differential equation for the voltage
dV
j L I
dz
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50
Transmission Lines
The current flowing through the shunt capacitance determines the
variation of the current from input to output of the cell.
I (z)
dI
I (z)+dI
C dz
V (z)+dV
The circuit equation for the sub-circuit above is
dI j Cdz(V dV ) j CVdz j C dV dz
The second term (including dV dz) tends to zero very rapidly for
infinitesimal length dz and can be ignored, giving a first order
differential equation for the current
dI
j C V
dz
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51
Transmission Lines
We have obtained a system of two coupled first order differential
equations that describe the behavior of voltage and current on the
uniform loss-less transmission line. The equations must be solved
simultaneously.
dV
j L I
dz
dI
j C V
dz
These are often called “telegraphers’ equations” of the loss-less
transmission line.
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52
Transmission Lines
One can easily obtain a set of uncoupled equations by
differentiating with respect to the space coordinate. The first order
differential terms are eliminated by using the corresponding
telegraphers’ equation
dI
j C V
dz
d 2V
dI
j L
j L j CV 2 LC V
dz
dz 2
d2 I
dV
j C
j C j L I 2 LC I
dz
dz 2
dV
j L I
dz
These are often called “telephonists’ equations”.
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53
Transmission Lines
We have now two uncoupled second order differential equations for
voltage and current, which give an equivalent description of the
loss-less transmission line.
Mathematically, these are wave
equations and can be solved independently.
The general solution for the voltage equation is
V (z) V e j z V e j z
where the wave propagation constant is
LC
Note that the complex exponential terms including have unitary
magnitude and purely “imaginary” argument, therefore they only
affect the “phase” of the wave in space.
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54
Transmission Lines
We have the following useful relations:
2 2 f vp
vp
r r
0 0 r r c
Here, v p f is the wavelength of the dielectric medium
surrounding the conductors of the transmission line and
1
1
vp 0 r 0 r
is the phase velocity of an electromagnetic wave in the dielectric.
As you can see, the propagation constant can be written in many
different, equivalent ways.
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Transmission Lines
The current distribution on transmission line can be readily
obtained by differentiation of the result for the voltage
dV
j z
j z
j
V
e
j
V
e
j L I
dz
which gives
1
C j z
j z
j z
j z
I (z) V e
V
e
V
e
V
e
L
Z0
The real quantity
Z0 L
C
is the “characteristic impedance” of the loss-less transmission line.
© Amanogawa, 2000 – Digital Maestro Series
56
Transmission Lines
Lossy Transmission Line
The solution for a uniform lossy transmission line can be obtained
with a very similar procedure, using the equivalent circuit for the
elementary cell shown in the figure below.
L dz
R dz
I (z)+dI
I (z)
V (z)
C dz
G dz
V (z)+dV
dz
© Amanogawa, 2000 – Digital Maestro Series
57
Transmission Lines
The series impedance determines the variation of the voltage from
input to output of the cell, according to the sub-circuit
L dz
V (z)
R dz
I (z)
V (z)+dV
dz
The corresponding circuit equation is
(V dV ) V ( j Ldz Rdz) I
from which we obtain a first order differential equation for the
voltage
dV
( j L R) I
dz
© Amanogawa, 2000 – Digital Maestro Series
58
Transmission Lines
The current flowing through the shunt admittance determines the
input-output variation of the current, according to the sub-circuit
I (z)
I (z)+dI
dI
C dz
G dz
V (z)+dV
The corresponding circuit equation is
dI ( j Cdz Gdz)(V dV )
( j C G)Vdz ( j C G)dV dz
The second term (including dV dz) can be ignored, giving a first
order differential equation for the current
dI
( j C G)V
dz
© Amanogawa, 2000 – Digital Maestro Series
59
Transmission Lines
We have again a system of coupled first order differential equations
that describe the behavior of voltage and current on the lossy
transmission line
dV
( j L R) I
dz
dI
( j C G)V
dz
These are the “telegraphers’ equations” for the lossy transmission
line case.
© Amanogawa, 2000 – Digital Maestro Series
60
Transmission Lines
One can easily obtain a set of uncoupled equations
differentiating with respect to the coordinate z as done earlier
by
dI
( j C G)V
dz
d 2V
dI
( j L R) ( j L R)( j C G)V
dz
dz2
d2 I
dV
( j C G )
( j C G)( j L R) I
dz
dz2
dV
( j L R) I
dz
These are the “telephonists’ equations” for the lossy line.
© Amanogawa, 2000 – Digital Maestro Series
61
Transmission Lines
The telephonists’ equations for the lossy transmission line are
uncoupled second order differential equations and are again wave
equations. The general solution for the voltage equation is
V (z) V e z V e z V e z e j z V e z e j z
where the wave propagation constant is now the complex quantity
( j L R)( j C G) j
The real part of the propagation constant describes the
attenuation of the signal due to resistive losses. The imaginary part
describes the propagation properties of the signal waves as in
loss-less lines.
The exponential terms including are “real”, therefore, they only
affect the “magnitude” of the voltage phasor. The exponential
terms including have unitary magnitude and purely “imaginary”
argument, affecting only the “phase” of the waves in space.
© Amanogawa, 2000 – Digital Maestro Series
62
Transmission Lines
The current distribution on a lossy transmission line can be readily
obtained by differentiation of the result for the voltage
dV
( j L R) I V e z V e z
dz
which gives
( j C G)
I (z) V ez V ez
( j L R)
1
V ez V ez
Z0
with the “characteristic impedance” of the lossy transmission line
( j L R)
Z0 ( j C G)
© Amanogawa, 2000 – Digital Maestro Series
Note: the characteristic
impedance is now complex !
63
Transmission Lines
For both loss-less and lossy transmission lines
the characteristic impedance does not depend on the line length
but only on the metal of the conductors, the dielectric material
surrounding the conductors and the geometry of the line crosssection, which determine L, R, C, and G.
One must be careful not to interpret the characteristic impedance
as some lumped impedance that can replace the transmission line
in an equivalent circuit.
This is a very common mistake!
Z0
© Amanogawa, 2000 – Digital Maestro Series
ZR
Z0
ZR
64
Transmission Lines
We have obtained the following solutions for the steady-state
voltage and current phasors in a transmission line:
Loss-less line
V (z) V e jz V e j z
1
I (z) V e jz V e j z
Z0
Lossy line
V (z) V ez V ez
1
I (z) V ez V ez
Z0
Since V (z) and I (z) are the solutions of second order differential
(wave) equations, we must determine two unknowns, V+ and V ,
which represent the amplitudes of steady-state voltage waves,
travelling in the positive and in the negative direction, respectively.
Therefore, we need two boundary conditions to determine these
unknowns, by considering the effect of the load and of the
generator connected to the transmission line.
© Amanogawa, 2000 – Digital Maestro Series
65
Transmission Lines
Before we consider the boundary conditions, it is very convenient
to shift the reference of the space coordinate so that the zero
reference is at the location of the load instead of the generator.
Since the analysis of the transmission line normally starts from the
load itself, this will simplify considerably the problem later.
ZR
New Space Coordinate
z
d
0
We will also change the positive direction of the space coordinate,
so that it increases when moving from load to generator along the
transmission line.
© Amanogawa, 2000 – Digital Maestro Series
66
Transmission Lines
We adopt a new coordinate d = z, with zero reference at the load
location. The new equations for voltage and current along the lossy
transmission line are
Loss-less line
jd
jd
V (d) V e
Lossy line
d
d
V e
1
I (d) V e jd V e jd
Z0
V (d) V e
V e
1
I (d) V ed V e d
Z0
At the load (d = 0) we have, for both cases,
V (0) V V 1
I (0) V V
Z0
© Amanogawa, 2000 – Digital Maestro Series
67
Transmission Lines
For a given load impedance ZR , the load boundary condition is
V (0) ZR I (0)
Therefore, we have
ZR
V V V V
Z0
from which we obtain the voltage load reflection coefficient
V
ZR Z0
R Z Z
V
R
0
© Amanogawa, 2000 – Digital Maestro Series
68
Transmission Lines
We can introduce this result into the transmission line equations as
Loss-less line
Lossy line
V e jd
2 j d
I (d) e
1
R Z0
V (d) V e jd 1 R e2 j d
V e d
2 d
I (d) e
1
R Z0
V (d) V ed 1 R e2 d
At each line location we define a Generalized Reflection Coefficient
(d) R e2 j d
(d) R e2 d
and the line equations become
V (d) V e jd 1 (d) V (d) V ed 1 (d) V e jd
I (d) 1 (d) Z0
V ed
I (d) 1 (d) Z0
© Amanogawa, 2000 – Digital Maestro Series
69
Transmission Lines
We define the line impedance as
1 (d )
V (d )
Z0
Z( d ) 1 (d )
I (d )
A simple circuit diagram can illustrate the significance of line
impedance and generalized reflection coefficient:
Req = (d)
d
© Amanogawa, 2000 – Digital Maestro Series
ZR
Zeq=Z(d)
0
70
Transmission Lines
If you imagine to cut the line at location d, the input impedance of
the portion of line terminated by the load is the same as the line
impedance at that location “before the cut”. The behavior of the line
on the left of location d is the same if an equivalent impedance with
value Z(d) replaces the cut out portion. The reflection coefficient of
the new load is equal to (d)
Req ( d ) ZReq Z0
ZReq Z0
If the total length of the line is L, the input impedance is obtained
from the formula for the line impedance as
1 ( L)
Vin V ( L)
Z0
Zin 1 ( L)
Iin
I ( L)
The input impedance is the equivalent impedance representing the
entire line terminated by the load.
© Amanogawa, 2000 – Digital Maestro Series
71
Transmission Lines
An important practical case is the low-loss transmission line, where
the reactive elements still dominate but R and G cannot be
neglected as in a loss-less line. We have the following conditions:
L R
C G
so that
( j L R)( j C G)
R G 1
j L j C 1 j
L
j
C
R
G
RG
j LC 1 j L j C 2 LC
The last term under the square root can be neglected, because it is
the product of two very small quantities.
© Amanogawa, 2000 – Digital Maestro Series
72
Transmission Lines
What remains of the square root can be expanded into a truncated
Taylor series
1 R
G j LC 1 2
j
L
j
C
1
C
L
R
G
j LC
2
L
C
so that
1
C
L
R
G
2
L
C
© Amanogawa, 2000 – Digital Maestro Series
LC
73
Transmission Lines
The characteristic impedance of the low-loss line is a real quantity
for all practical purposes and it is approximately the same as in a
corresponding loss-less line
R j L
L
Z0 G j C
C
and the phase velocity associated to the wave propagation is
1
vp LC
BUT NOTE:
In the case of the low-loss line, the equations for voltage and
current retain the same form obtained for general lossy lines.
© Amanogawa, 2000 – Digital Maestro Series
74
Transmission Lines
Again, we obtain the loss-less transmission line if we assume
R0
G0
This is often acceptable in relatively short transmission lines, where
the overall attenuation is small.
As shown earlier, the characteristic impedance in a loss-less line is
exactly real
L
Z0 C
while the propagation constant has no attenuation term
( j L)( j C ) j LC j
The loss-less line does not dissipate power, because = 0.
© Amanogawa, 2000 – Digital Maestro Series
75
Transmission Lines
For all cases, the line impedance was defined as
1 (d)
V (d)
Z0
Z(d) 1 (d)
I (d)
By including the appropriate generalized reflection coefficient, we
can derive alternative expressions of the line impedance:
A) Loss-less line
1 Re2 jd
ZR jZ0 tan( d)
Z0
Z(d) Z0
2 jd
jZR tan( d) Z0
1 Re
B) Lossy line (including low-loss)
1 R e2 d
ZR Z0 tanh( d)
Z0
Z(d) Z0
2 d
ZR tanh( d) Z0
1 Re
© Amanogawa, 2000 – Digital Maestro Series
76
Transmission Lines
Let’s now consider power flow in a transmission line, limiting the
discussion to the time-average power, which accounts for the
active power dissipated by the resistive elements in the circuit.
The time-average power at any transmission line location is
1
P(d , t ) Re V (d) I * (d)
2
This quantity indicates the time-average power that flows through
the line cross-section at location d. In other words, this is the
power that, given a certain input, is able to reach location d and
then flows into the remaining portion of the line beyond this point.
It is a common mistake to think that the quantity above is the power
dissipated at location d !
© Amanogawa, 2000 – Digital Maestro Series
77
Transmission Lines
The generator, the input impedance, the input voltage and the input
current determine the power injected at the transmission line input.
Iin
Zin
Vin VG
ZG Zin
ZG
VG
Vin
Generator
© Amanogawa, 2000 – Digital Maestro Series
Zin
Line
1
Iin VG
ZG Zin
1
*
Pin Re Vin Iin
2
78
Transmission Lines
The time-average power reaching the load of the transmission line
is given by
1
P(d=0 , t ) Re V (0) I * (0)
2
*
1
1
Re V 1 R V 1 R 2
Z0*
This represents the power dissipated by the load.
The time-average power absorbed by the line is simply the
difference between the input power and the power absorbed by the
load
P line Pin P(d 0 , t ) Remember that the internal impedance of the generator dissipates
part of the total power generated.
© Amanogawa, 2000 – Digital Maestro Series
79
Transmission Lines
The time-average power injected into the input of the transmission
line is maximized when the input impedance of the transmission
line and the internal generator impedance are complex conjugate of
each other.
Zin
Load
Generator
ZG
VG
ZG Z*in
© Amanogawa, 2000 – Digital Maestro Series
Transmission line
ZR
for maximum power transfer
80
Transmission Lines
In a loss-less transmission line no power is absorbed by the line, so
the input time-average power is the same as the time-average
power absorbed by the load. The characteristic impedance of the
loss-less line is real and we can express the power flow as
1
P(d , t ) Re V (d) I* (d)
2
1
Re V e j d 1 Re j 2 d
2
1
* j d
j 2 d * (V ) e
1 Re
Z0
1
1
2
2
2
V
V
R
2 Z0
2 Z0
Incident wave
© Amanogawa, 2000 – Digital Maestro Series
Reflected wave
81
Transmission Lines
In the case of low-loss lines, the characteristic impedance is again
real, and the time-average power flow along the line is given by
1
P(d , t ) Re V (d) I * (d)
2
1
Re V e d e j d 1 Re2 d
2
1
* d j d
2 d * (V ) e e
1 Re
Z0
1
1
2
2 2 d
2 2 d
V
e
V
e
R
2 Z0
2 Z0
Incident wave
© Amanogawa, 2000 – Digital Maestro Series
Reflected wave
82
Transmission Lines
Note that in a lossy line the reference for the amplitude of the
incident voltage wave is at the load and that the amplitude grows
exponentially moving towards the input. The amplitude of the
incident wave behaves in the following way
V e L
V e d
V
input
inside the line
load
The reflected voltage wave has maximum amplitude at the load, and
it decays exponentially moving back towards the generator. The
amplitude of the reflected wave behaves in the following way
V R e L
V R e d
V R
input
inside the line
load
© Amanogawa, 2000 – Digital Maestro Series
83
Transmission Lines
For a general lossy line the characteristic impedance is complex,
and the time-average power is
1
P(d , t ) Re V (d) I * (d)
2
1
Re V e d e j d 1 (d) 2
*
* * d j d
Y0 (V ) e e
1 (d) G0 2 2 d G0 2 2 d
2
R
V
e
V
e
2
2
B0 V
© Amanogawa, 2000 – Digital Maestro Series
2
e2 d Im((d))
84
Transmission Lines
We have introduced for convenience the characteristic admittance
of the line
Y0 1
G0 jB0
Z0
since a complex characteristic impedance would appear at
denominator in the expression for the power.
Note that for a low-loss transmission line the characteristic
impedance is approximately real and
B0 0
The previous result for the low-loss line can be readily recovered
from the time-average power for the general lossy line.
© Amanogawa, 2000 – Digital Maestro Series
85
Transmission Lines
To completely specify the transmission line problem, we still have
to determine the value of V+ from the input boundary condition.
The load boundary condition imposes the shape of the
interference pattern of voltage and current along the line.
The input boundary condition, linked to the generator, imposes
the scaling for the interference patterns.
We have
Zin
Vin V ( L) VG
ZG Zin
or
with
1 (L)
Zin Z0
1 (L)
ZR jZ0 tan( L)
Zin Z0
jZR tan( L) Z0
loss - less line
ZR Z0 tanh( L)
Zin Z0
ZR tanh( L) Z0
lossy line
© Amanogawa, 2000 – Digital Maestro Series
86
Transmission Lines
For a loss-less transmission line:
V (L) V e j L 1 (L) V e j L (1 Re j 2 L )
!
1
Zin
V VG
ZG Zin e j L (1 Re j 2 L )
For a lossy transmission line:
V (L) V e L 1 (L) V e L 1 Re2 d
!
1
Zin
V VG
ZG Zin e L (1 Re2 L )
© Amanogawa, 2000 – Digital Maestro Series
87
Transmission Lines
In order to have good control on the behavior of a high frequency
circuit, it is very important to realize transmission lines as uniform
as possible along their length, so that the impedance behavior of
the line does not vary and can be easily characterized.
A change in transmission line properties, wanted or unwanted,
entails a change in the characteristic impedance, which causes a
reflection. Example:
Z01
Z02
ZR
1
Z01
© Amanogawa, 2000 – Digital Maestro Series
Zin
Zin Z01
1 Zin Z01
88
Transmission Lines
Special Cases
ZR 0 (SHORT CIRCUIT)
ZR = 0
Z0
The load boundary condition due to the short circuit is V (0) = 0
V (d 0) V e j 0 (1 R e j 2 0 )
V (1 R ) 0
© Amanogawa, 2000 – Digital Maestro Series
R 1
89
Transmission Lines
Since
R V
V
V V We can write the line voltage phasor as
V (d) V e j d V e j d
V e j d V e j d
V ( e j d e j d )
2 jV sin( d)
© Amanogawa, 2000 – Digital Maestro Series
90
Transmission Lines
For the line current phasor we have
1
(V e j d V e j d )
I (d) Z0
1
(V e j d V e j d )
Z0
V j d
(e
e j d )
Z0
2V cos( d)
Z0
The line impedance is given by
2 jV sin( d)
V (d)
jZ0 tan( d)
Z(d) I (d) 2V cos( d) / Z0
© Amanogawa, 2000 – Digital Maestro Series
91
Transmission Lines
The time-dependent values of voltage and current are obtained as
V (d, t ) Re[V (d) e j t ] Re[2 j | V | e j sin( d) e j t ]
2| V |sin( d) Re[ j e j( t ) ]
2| V |sin( d) Re[ j cos(t ) sin(t )]
2| V |sin( d) sin(t )
I (d, t ) Re[ I (d) e j t ] Re[2 | V | e j cos( d) e j t ] / Z0
2| V |cos( d) Re[ e j ( t ) ] / Z0
2| V |cos( d) Re[ (cos(t ) j sin(t )] / Z0
| V |
cos( d) cos(t )
2
Z0
© Amanogawa, 2000 – Digital Maestro Series
92
Transmission Lines
The time-dependent power is given by
P(d, t ) V (d, t ) I (d, t )
| V |2
sin( d) cos( d) sin(t ) cos(t )
4
Z0
| V |2
sin(2 d) sin (2t 2)
Z0
and the corresponding time-average power is
1 T
P(d, t ) P(d, t ) dt
T 0
| V |2
1 T
sin(2 d) sin (2t 2) 0
Z0
T 0
© Amanogawa, 2000 – Digital Maestro Series
93
Transmission Lines
ZR (OPEN CIRCUIT)
ZR Z0
The load boundary condition due to the open circuit is I (0) = 0
V j 0
I (d 0) e (1 R e j 2 0 )
Z0
V
(1 R ) 0
Z0
© Amanogawa, 2000 – Digital Maestro Series
R 1
94
Transmission Lines
Since
R V
V
V V
We can write the line current phasor as
1
(V e j d V e j d )
I (d) Z0
1
(V e j d V e j d )
Z0
2 jV V j d j d
(e
)
sin( d)
e
Z0
Z0
© Amanogawa, 2000 – Digital Maestro Series
95
Transmission Lines
For the line voltage phasor we have
V (d) (V e j d V e j d )
(V e j d V e j d )
V ( e j d e j d )
2V cos( d)
The line impedance is given by
2V cos( d)
V (d)
Z0
j
Z(d) tan( d)
I (d) 2 jV sin( d) / Z0
© Amanogawa, 2000 – Digital Maestro Series
96
Transmission Lines
The time-dependent values of voltage and current are obtained as
V (d, t ) Re[V (d) e j t ] Re[2 | V | e j cos( d) e j t ]
2| V |cos( d) Re[ e j( t ) ]
2| V |cos( d) Re[ (cos(t ) j sin(t )]
2| V |cos( d) cos( t )
I (d, t ) Re[ I (d) e j t ] Re[2 j | V | e j sin( d) e j t ] / Z0
2| V |sin( d) Re[ j e j ( t ) ] / Z0
2| V |sin( d) Re[ j cos( t ) sin(t )] / Z0
| V |
sin( d) sin(t )
2
Z0
© Amanogawa, 2000 – Digital Maestro Series
97
Transmission Lines
The time-dependent power is given by
P(d, t ) V (d, t ) I (d, t ) | V |2
cos( d) sin( d) cos(t ) sin(t )
4
Z0
| V |2
sin(2 d) sin (2t 2)
Z0
and the corresponding time-average power is
1 T
P(d, t ) P(d, t ) dt
T 0
| V |2
1 T
sin(2 d) sin (2t 2) 0
Z0
T 0
© Amanogawa, 2000 – Digital Maestro Series
98
Transmission Lines
ZR Z0 (MATCHED LOAD)
Z0
ZR = Z0
The reflection coefficient for a matched load is
ZR Z0 Z0 Z0
R 0
ZR Z0 Z0 Z0
no reflection!
The line voltage and line current phasors are
V (d) V e j d (1 R e2 j d ) V e j d
V j d
V
(1 R e2 j d ) I( d ) e
e j d
Z0
Z0
© Amanogawa, 2000 – Digital Maestro Series
99
Transmission Lines
The line impedance is independent of position and equal to the
characteristic impedance of the line
V (d) V e j d
Z0
Z(d) I (d) V j d
e
Z0
The time-dependent voltage and current are
V (d, t ) Re[| V | e j e j d e j t ]
| V | Re[e j( t d ) ] | V | cos(t d )
I (d, t ) Re[| V | e j e j d e j t ] / Z0
| V |
|
|
V
j ( t d )
]
cos(t d )
Re[e
Z0
Z0
© Amanogawa, 2000 – Digital Maestro Series
100
Transmission Lines
The time-dependent power is
|
|
V
cos(t d )
P(d, t ) | V | cos(t d )
Z0
| V |2
cos2 (t d )
Z0
and the time average power absorbed by the load is
1 t | V |2
cos2 (t d ) dt
P(d) T 0 Z0
| V |2
2 Z0
© Amanogawa, 2000 – Digital Maestro Series
101
Transmission Lines
ZR jX (PURE REACTANCE)
ZR = j X
Z0
The reflection coefficient for a purely reactive load is
ZR Z0 jX Z0
R ZR Z0 jX Z0
( jX Z0 )( jX Z0 )
( jX Z0 )( jX Z0 )
© Amanogawa, 2000 – Digital Maestro Series
X 2 Z02
Z02 X 2
2j
XZ0
Z02 X 2
102
Transmission Lines
In polar form
R R exp( j)
where
R 2 2
X Z0
4 X 2 Z02
2
2
Z02 X 2
Z02 X 2
2
2
2 2
Z0 X
1
2
Z02 X 2
1 2 XZ0 tan X 2 Z2 0
The reflection coefficient has unitary magnitude, as in the case of
short and open circuit load, with zero time average power absorbed
by the load. Both voltage and current are finite at the load, and the
time-dependent power oscillates between positive and negative
values. This means that the load periodically stores and returns
powers to the line without dissipation.
© Amanogawa, 2000 – Digital Maestro Series
103
Transmission Lines
Reactive impedances can be realized with transmission lines
terminated by a short or by an open circuit. The input impedance of
a loss-less transmission line of length L terminated by a short
circuit is purely imaginary
2 f 2 L
Zin j Z0 tan L j Z0 tan L j Z0 tan vp At a specified frequency f, any reactance value can be obtained by
changing the length of the line from 0 to /2. Since the tangent
function is periodic, the behavior of the impedance will repeat
identically for each line increment of length /2.
Note that a similar periodic behavior is obtained when the length of
the transmission line is fixed and the frequency of operation is
changed.
© Amanogawa, 2000 – Digital Maestro Series
104
Transmission Lines
Shorted transmission line – Fixed frequency
L
L 0
0 L
L
4
4
L
4
2
L
2
3
L
2
4
3
L
4
3
L 4
Z in 0
s h o rt c irc u it
Im Z in 0
in d u c ta n c e
Z in o p e n c irc u it
c a p a c ita n c e
Im Z in 0
Z in 0
s h o rt c irc u it
Im Z in 0
in d u c ta n c e
Z in o p e n c irc u it
Im Z in 0
c a p a c ita n c e
© Amanogawa, 2000 – Digital Maestro Series
105
Z(L)/Zo = j tan( L)
Transmission Lines
Impedance of a short circuited transmission line
(fixed frequency, variable length)
40
30
20
inductive
Normalized Input Impedance
10
inductive
inductive
0
-10
capacitive
cap.
-20
-30
-40
0
100
200
300
400
500
L
[deg]
Line Length L
© Amanogawa, 2000 – Digital Maestro Series
106
Z(L)/Zo = j tan( L)
Transmission Lines
Impedance of a short circuited transmission line
(fixed length, variable frequency)
40
30
20
inductive
Normalized Input Impedance
10
inductive
inductive
0
-10
capacitive
cap.
-20
-30
-40
0
100
vp / (4L)
200
vp / (2L)
300
3vp / (4L)
400
vp / L
500
5vp / (4L)
[deg]
f
Frequency of operation
© Amanogawa, 2000 – Digital Maestro Series
107
Transmission Lines
For a transmission line of length L terminated by an open circuit,
the input impedance is again purely imaginary
Z0
j
Zin j
tan L Z0
Z0
j
2 2 f tan L tan L
v
p We can use the open circuited line to realize any reactance, but
starting from a capacitive value when the line length is very short.
The reactance varies linearly with frequency in the case of lumped
elements since
X L (inductance)
or
1
(capacitance)
X
C
This is not the case when the reactance is realized with a section of
transmission line.
© Amanogawa, 2000 – Digital Maestro Series
108
Transmission Lines
Open transmission line – Fixed frequency
L
L 0
0 L
L
4
4
L
4
2
L
2
3
L
2
4
3
L
4
3
L 4
Z in o p e n c irc u it
Im Z in 0
c a p a c ita n c e
Z in 0
s h o rt c irc u it
Im Z in 0
in d u c ta n c e
Z in o p e n c irc u it
Im Z in 0
c a p a c ita n c e
Z in 0
s h o rt c irc u it
Im Z in 0
in d u c ta n c e
© Amanogawa, 2000 – Digital Maestro Series
109
Normalized Input Impedance
Z(L)/ Zo = - j cotan( L)
Transmission Lines
Impedance of an open circuited transmission line
(fixed frequency, variable length)
40
30
20
inductive
inductive
inductive
10
0
-10
capacitive
capacitive
capacitive
-20
-30
-40
0
100
200
300
400
500
[deg]
L
Line Length L
© Amanogawa, 2000 – Digital Maestro Series
110
Normalized Input Impedance
Z(L)/ Zo = - j cotan( L)
Transmission Lines
Impedance of an open circuited transmission line
(fixed length, variable frequency)
40
30
20
inductive
inductive
inductive
10
0
-10
capacitive
capacitive
capacitive
-20
-30
-40
0
100
vp / (4L)
200
vp / (2L)
300
3vp / (4L)
400
vp / L
500
5vp / (4L)
[deg]
f
Frequency of operation
© Amanogawa, 2000 – Digital Maestro Series
111
Transmission Lines
It is possible to realize resonant circuits by using transmission
lines as reactive elements. For instance, consider the circuit below
realized with lines having the same characteristic impedance:
I
L1
short
circuit
Z0
L2
V
Zin1
Zin1 j Z0 tan L 1
© Amanogawa, 2000 – Digital Maestro Series
Z0
short
circuit
Zin2
Zin2 j Z0 tan L 2
112
Transmission Lines
The circuit is resonant if L1 and L2 are chosen such that an
inductance and a capacitance are realized.
A resonance condition is established when the total input
impedance of the parallel circuit is infinite (or, equivalently, when
the input admittance of the parallel circuit is zero)
1
1
0
j Z0 tan r L1 j Z0 tan r L 2 or
r
r
tan L1 tan L 2 vp
vp
with
2 r
r r vp
Since the tangent is a periodic function, there is a multiplicity of
possible resonant angular frequencies r that satisfy the condition
above. The solutions can be found by using a numerical procedure.
© Amanogawa, 2000 – Digital Maestro Series
113
Transmission Lines
Transient and Steady-State on a Transmission Line
We need to give now a physical interpretation of the mathematical
results obtained for transmission lines. First of all, note that we are
considering a steady-state regime where the wave propagation
along the transmission line is perfectly periodic in time. This
means that all the transient phenomena have already decayed.
To give a feeling of what the steady-state regime is, consider a
transmission line that is connected to the generator by closing a
switch at a reference time t = 0. For simplicity we assume that all
impedances, including the line characteristic impedance, are real.
Switch
Generator
Load
RG
t=0
VG
© Amanogawa, 2000 – Digital Maestro Series
Transmission line
RR
114
Transmission Lines
After the switch is closed, the voltage at the input of the
transmission line will vary nearly instantaneously from the open
voltage of the generator VG to a value V+ , with a current I+
beginning to flow into the line.
A transient takes place in the
transmission line, as charges in the conductors move, transporting
the current towards the load. Until the front first reaches the end of
the transmission line, the load voltage remains zero.
Initially, the input impedance of the transmission line appears to be
the same as the line characteristic impedance, because the current
cannot yet sense the value of the load impedance. Therefore, a
voltage front V+ propagates with the current front I+ , where
Z0
V I Z0 VG
RG Z0
V
V0
I Z0 RG Z0
© Amanogawa, 2000 – Digital Maestro Series
115
Transmission Lines
If the load does not match exactly the characteristic impedance of
the line, the voltage V+ and the current I+ cannot be established
across the load RR , when the front reaches the end of the line,
because
V I RR
Therefore, voltage and current adjust themselves at the load by
–
reflecting back a wave front with voltage V and current I such
that
V V ( I I ) RR
© Amanogawa, 2000 – Digital Maestro Series
116
Transmission Lines
Since also the reflected front will see an impedance Z0 we have
V Z0 I ;
V
V Z0 I RR Z0
V
RR Z0
The quantity
V
RR Z0
R R Z
V
R
0
is the load reflection coefficient.
© Amanogawa, 2000 – Digital Maestro Series
117
Transmission Lines
The wave reflected by the load propagates in the negative direction
and interferes with the oscillating values of voltage and current
found along the transmission line, which continue to be injected by
the generator.
When the reflected wave reaches the input of the transmission line,
it terminates on the generator impedance RG and if this does not
match the line characteristic impedance, reflection back into the
line again occurs, generating now a forward wave with associated
voltage
RG Z0
V2 V
RG Z0
Remember that the ideal voltage source of the generator will
behave simply as a short for the reflected wave attempting to exit
the line from the input.
© Amanogawa, 2000 – Digital Maestro Series
118
Transmission Lines
The front reflected by the generator side will again reach the load,
and waves of ever decreasing amplitude will keep bouncing back
and forth along the line until the process associated to that initial
wave front dies out.
Every subsequent wave front injected over time by the oscillating
generator undergoes an identical phenomenon of multiple
reflections. If we assume a sinusoidal generator, voltage and
current injected into the line repeat periodically, according to the
period of the oscillation. Therefore, successive reflections at the
ends of the line obey the same reflection coefficients, but involving
in time different values of amplitude and phase.
If the generator continues to supply the line with a stable
oscillation, after a sufficient time the combined interference of
forward and backward waves becomes stable, and one can identify
two well-defined incoming and reflected steady-state waves, arising
from the superposition of the infinite transient components
travelling on the line.
© Amanogawa, 2000 – Digital Maestro Series
119
Transmission Lines
Note that the wave fronts travel with a phase velocity equal to the
speed of light in the medium surrounding the wires. Also, note that
the length of the line will affect the interference pattern of the wave
superposition, so that lines of different length will result in different
voltage and current distributions along the line.
When we study the steady-state voltages and currents in a
transmission line, we only need to know the phasors that represent
the stable steady-state oscillations at each line location. The
phasors provide a snapshot of how values of voltage and current
relate to each other in space, at a reference time, in terms of
amplitude and phase. The actual time oscillation can be easily
recovered, because in steady-state we know that voltage and
current are perfectly periodic at each line location, according to the
period of the generator.
If the generator provides more than one frequency of oscillation, at
steady-state the behavior of each frequency in the spectrum can be
studied independently and the total result is obtained by
superposition.
© Amanogawa, 2000 – Digital Maestro Series
120
Transmission Lines
Standing Wave Patterns
In practical applications it is very convenient to plot the magnitude
of phasor voltage and phasor current along the transmission line.
These are the standing wave patterns:
V (d) V 1 (d) Loss - less line V
1 (d) I (d) Z0
V (d) V e
d 1 d Lossy line V e
d
1 d I (d) Z0
© Amanogawa, 2000 -–Digital Maestro Series
121
Transmission Lines
The standing wave patterns provide the top envelopes that bound
the time-oscillations of voltage and current along the line. In other
words, the standing wave patterns provide the maximum values
that voltage and current can ever establish at each location of the
transmission line for given load and generator.
The standing wave pattern gives a clear representation of wave
interference in a transmission line.
The patterns present a
succession of maxima and minima which repeat in space with a
period of length /2, due to constructive or destructive interference
between forward and reflected waves. The patterns for a loss-less
line are periodic in space, repeating exactly with a /2 period.
Again, note that although we talk about maxima and minima of the
standing wave pattern we are always examining a maximum of
voltage or current that can be achieved at a transmission line
location during any period of oscillation.
© Amanogawa, 2000 -–Digital Maestro Series
122
Transmission Lines
We limit now our discussion to the loss-less transmission line case
where the generalized reflection coefficient varies as
(d) R exp j 2 d R exp j exp j 2 d Note that the magnitude of an exponential with imaginary argument
is always unity, so
exp j exp j 2 d 1
and that in a loss-less line it is always true that, at any line location,
(d) R
When d increases, moving from load to generator, the generalized
reflection coefficient on the complex plane moves clockwise on a
circle with radius |R| and is identified by the angle - 2 d .
© Amanogawa, 2000 -–Digital Maestro Series
123
Transmission Lines
The voltage standing wave pattern has a maximum at locations
where the generalized reflection coefficient is real and positive
(d) R
exp j exp j 2 d 1
2 d 2 n
At these locations we have
1 (d) 1 R
Vmax V (d max ) V 1 R The phase angle - 2 d changes by an amount 2, when moving
from one maximum to the next. This corresponds to a distance
between successive maxima of /2.
© Amanogawa, 2000 -–Digital Maestro Series
124
Transmission Lines
The voltage standing wave pattern has a minimum at locations
where the generalized reflection coefficient is real and negative
(d) R
exp j exp j 2 d 1
2 d 2 n 1 At these locations we have
1 (d) 1 R
Vmin V (d min ) V 1 R Also when moving from one minimum to the next, the phase angle
- 2 d changes by an amount 2. This again corresponds to a
distance between successive minima of /2.
© Amanogawa, 2000 -–Digital Maestro Series
125
Transmission Lines
The voltage standing wave pattern provides immediate information
on the transmission line circuit
If the load is matched to the transmission line ( ZR = Z0 ) the
voltage standing wave pattern is flat, with value | V+ |.
If the load is real and ZR > Z0 , the voltage standing wave
pattern starts with a maximum at the load.
If the load is real and ZR < Z0 , the voltage standing wave
pattern starts with a minimum at the load.
If the load is complex and Im(ZR ) > 0 (inductive reactance),
the voltage standing wave pattern initially increases when
moving from load to generator and reaches a maximum first.
If the load is complex and Im(ZR ) < 0 (capacitive reactance),
the voltage standing wave pattern initially decreases when
moving from load to generator and reaches a minimum first.
© Amanogawa, 2000 -–Digital Maestro Series
126
Transmission Lines
Since in all possible cases
d 1
the voltage standing wave pattern
V (d) V 1 (d) cannot exceed the value 2 | V+ | in a loss-less transmission line.
If the load is a short circuit, an open circuit, or a pure reactance,
there is total reflection with
d 1
since the load cannot consume any power. The voltage standing
wave pattern in these cases is characterized by
Vmax 2 V © Amanogawa, 2000 -–Digital Maestro Series
and
Vmin 0 .
127
Transmission Lines
The quantity 1 + (d) is in general a complex number, that can be
constructed as a vector on the complex plane. The number 1 is
represented as 1 + j0 on the complex plane, and it is just a vector
with coordinates (1,0) positioned on the Real axis. The reflection
coefficient (d) is a complex number such that |(d)| 1.
Im
1+
1
© Amanogawa, 2000 -–Digital Maestro Series
Re
128
Transmission Lines
We can use a geometric construction to visualize the behavior of
the voltage standing wave pattern
V (d) V 1 (d) simply by looking at a vector plot of |(1 + (d))| . |V+| is just a
scaling factor, fixed by the generator. For convenience, we place
the reference of the complex plane representing the reflection
coefficient in correspondence of the tip of the vector (1, 0).
Example: Load
with inductive
reactance
Im( )
1+R
R
1
© Amanogawa, 2000 -–Digital Maestro Series
Re ( )
129
Transmission Lines
Im( )
Maximum of
voltage standing
wave pattern
R
1+(d)
(d)
1
2 dmax
Re ( )
d 2 d max 0
Im( )
Minimum of
voltage standing
wave pattern
1+(d)
1
d 2 d min © Amanogawa, 2000 -–Digital Maestro Series
(d)
R
Re ( )
2 dmin
130
Transmission Lines
The voltage standing wave ratio (VSWR) is an indicator of load
matching which is widely used in engineering applications
Vmax 1 R
VSWR Vmin 1 R
When the load is perfectly matched to the transmission line
R 0
VSWR 1
When the load is a short circuit, an open circuit or a pure reactance
R 1
VSWR We have the following useful relation
VSWR 1
R VSWR 1
© Amanogawa, 2000 -–Digital Maestro Series
131
Transmission Lines
Maxima and minima of the voltage standing wave pattern.
Load with inductive reactance
Im ZR 0
The load reflection coefficient
is in this part of the domain
ZR Z0 Im R Im 0
ZR Z0 Im()
Re()
1
The first maximum of the voltage standing wave pattern is closest
to the load, at location
d 2 d max 0
© Amanogawa, 2000 -–Digital Maestro Series
d max 4
132
Transmission Lines
Load with capacitive reactance
Im ZR 0
ZR Z0 Im R Im 0
ZR Z0 The load reflection coefficient
is in this part of the domain
Im()
1
Re()
The first minimum of the voltage standing wave pattern is closest to
the load, at location
d 2 d min © Amanogawa, 2000 -–Digital Maestro Series
d min 4
133
Transmission Lines
A measurement of the voltage standing wave pattern provides the
locations of the first voltage maximum and of the first voltage
minimum with respect to the load.
The ratio of the voltage magnitude at these points gives directly the
voltage standing wave ratio (VSWR).
This information is sufficient to determine the load impedance ZR ,
if the characteristic impedance of the transmission line Z0 is
known.
STEP 1: The VSWR provides the magnitude of the load
reflection coefficient
VSWR 1
R VSWR 1
© Amanogawa, 2000 -–Digital Maestro Series
134
Transmission Lines
STEP 2: The distance from the load of the first maximum or
minimum gives the phase of the load reflection coefficient.
Vmax
|V|
For an inductive reactance, a voltage
maximum is closest to the load and
4
d max
2 d max Vmin
dmax
0
|V|
Vmax
For a capacitive reactance, a voltage
minimum is closest to the load and
4
2 d min d min
Vmin
dmin
0
© Amanogawa, 2000 -–Digital Maestro Series
135
Transmission Lines
STEP 3: The load impedance is obtained by inverting the
expression for the reflection coefficient
ZR Z0
R R exp j ZR Z0
© Amanogawa, 2000 -–Digital Maestro Series
1 R exp j ZR Z0
1 R exp j 136
Transmission Lines
Smith Chart
The Smith chart is one of the most useful graphical tools for high
frequency circuit applications. The chart provides a clever way to
visualize complex functions and it continues to endure popularity
decades after its original conception.
From a mathematical point of view, the Smith chart is simply a
representation of all possible complex impedances with respect to
coordinates defined by the reflection coefficient.
Im(Γ )
1
Re(Γ )
© Amanogawa, 2000 - Digital Maestro Series
The domain of definition of the
reflection coefficient is a circle of
radius 1 in the complex plane. This
is also the domain of the Smith chart.
137
Transmission Lines
The goal of the Smith chart is to identify all possible impedances on
the domain of existence of the reflection coefficient. To do so, we
start from the general definition of line impedance (which is equally
applicable to the load impedance)
1 + Γ (d )
V (d )
= Z0
Z( d ) =
1 − Γ (d )
I (d )
This provides the complex function Z( d ) = f {Re ( Γ ) , Im ( Γ )} that
we want to graph. It is obvious that the result would be applicable
only to lines with exactly characteristic impedance Z0.
In order to obtain universal curves, we introduce the concept of
normalized impedance
Z(d ) 1 + Γ (d )
z( d ) =
=
1 − Γ (d )
Z0
© Amanogawa, 2000 - Digital Maestro Series
138
Transmission Lines
The normalized impedance is represented on the Smith chart by
using families of curves that identify the normalized resistance r
(real part) and the normalized reactance x (imaginary part)
z ( d ) = Re ( z ) + j Im ( z ) = r + jx
Let’s represent the reflection coefficient in terms of its coordinates
Γ ( d ) = Re (Γ ) + j Im ( Γ )
Now we can write
1 + Re ( Γ ) + j Im ( Γ )
r + jx =
1 − Re ( Γ ) − j Im ( Γ )
=
1 − Re 2 ( Γ ) − Im2 ( Γ ) + j 2 Im ( Γ )
© Amanogawa, 2000 - Digital Maestro Series
2
1
Re
Im
−
Γ
+
( ))
(Γ )
(
2
139
Transmission Lines
The real part gives
r=
1 − Re
2
Add a quantity equal to zero
( Γ ) − Im 2 ( Γ )
(1 − Re ( Γ ))2 + Im2 ( Γ )
(
2
 r ( Re ( Γ ) − 1 )2 +

(Re
r ( Re ( Γ ) − 1 ) + Re
2
(Γ ) − 1
2
=0
) + r Im
(Γ ) − 1
)
2
( Γ ) + Im ( Γ ) +
2
1
1+ r
−
1
1+ r
=0
1
2

+
+ (1 + r ) Im ( Γ ) =
1 + r 
1+ r
1
2
 2

1
r
r
2
(1 + r )  Re ( Γ ) − 2 Re ( Γ )
+
+ (1 + r ) Im ( Γ ) =

2
1 + r (1 + r ) 
1+ r

⇒
 Re ( Γ ) −


1 + r 
r
2
© Amanogawa, 2000 - Digital Maestro Series
+ Im
2
(Γ ) =
( )
1
2
Equation of a circle
1+ r
140
Transmission Lines
The imaginary part gives
x=
x
2
Multiply by x and add a
quantity equal to zero
2 Im ( Γ )
(1 − Re ( Γ ))2 + Im 2 ( Γ )
=0
(1 − Re ( Γ ))2 + Im 2 ( Γ ) − 2 x Im ( Γ ) + 1 − 1 = 0


(1 − Re ( Γ ))2 + Im 2 ( Γ ) − 2 Im ( Γ ) + 1 = 1

 x
2
2
x
x
2
1 
1
2 
2
(1 − Re ( Γ )) +  Im ( Γ ) − Im ( Γ ) + 2  = 2
x

x  x
⇒
2
1

=
( Re ( Γ ) − 1 ) + Im ( Γ ) −
2


x
x
2
© Amanogawa, 2000 - Digital Maestro Series
1
Equation of a circle
141
Transmission Lines
The result for the real part indicates that on the complex plane with
coordinates (Re(Γ), Im(Γ)) all the possible impedances with a given
normalized resistance r are found on a circle with
{
}
1
Radius =
1+ r
r
,0
Center =
1+ r
As the normalized resistance r varies from 0 to ∞ , we obtain a
family of circles completely contained inside the domain of the
reflection coefficient | Γ | ≤ 1 .
Im(Γ )
r=1
r=5
r=0
Re(Γ )
r = 0.5
© Amanogawa, 2000 - Digital Maestro Series
r →∞
142
Transmission Lines
The result for the imaginary part indicates that on the complex
plane with coordinates (Re(Γ), Im(Γ)) all the possible impedances
with a given normalized reactance x are found on a circle with
{ }
1
Center = 1 ,
x
1
Radius =
x
As the normalized reactance x varies from -∞ to ∞ , we obtain a
family of arcs contained inside the domain of the reflection
coefficient | Γ | ≤ 1 .
Im(Γ )
x = 0.5
x=1
x →±∞
Re(Γ )
x=0
x = - 0.5
© Amanogawa, 2000 - Digital Maestro Series
x = -1
143
Transmission Lines
Basic Smith Chart techniques for loss-less transmission lines
Given Z(d) ⇒ Find Γ(d)
Given ΓR and ZR
Find dmax and dmin (maximum and minimum locations for the
voltage standing wave pattern)
Find the Voltage Standing Wave Ratio (VSWR)
Given Z(d) ⇒ Find Y(d)
Given Γ(d) ⇒ Find Z(d)
⇒ Find Γ(d) and Z(d)
Given Γ(d) and Z(d) ⇒ Find ΓR and ZR
Given Y(d) ⇒ Find Z(d)
© Amanogawa, 2000 - Digital Maestro Series
144
Transmission Lines
Given Z(d) ⇒ Find Γ(d)
1.
Normalize the impedance
Z (d ) R
X
=
+ j
= r+ j x
z (d ) =
Z0
Z0
Z0
2.
3.
4.
Find the circle of constant normalized resistance r
Find the arc of constant normalized reactance x
The intersection of the two curves indicates the reflection
coefficient in the complex plane. The chart provides
directly the magnitude and the phase angle of Γ(d)
Example: Find Γ(d), given
Z ( d ) = 25 + j 100 Ω
© Amanogawa, 2000 - Digital Maestro Series
with Z0 = 50 Ω
145
Transmission Lines
1. Normalization
z (d) = (25 + j 100)/50
= 0.5 + j 2.0
2. Find normalized
resistance circle
3. Find normalized
reactance arc
1
x = 2.0
05
2
3
r = 0.5
50.906 °
0.2
0
0.2
0.5
1
2
5
4. This vector represents
the reflection coefficient
-0.2
Γ (d) = 0.52 + j0.64
-3
|Γ (d)| = 0.8246
∠ Γ (d) = 0.8885 rad
-0 5
= 50.906 °
-2
-1
1.
0.8246
© Amanogawa, 2000 - Digital Maestro Series
146
Transmission Lines
Given Γ(d) ⇒ Find Z(d)
1.
2.
3.
Determine the complex point representing the given
reflection coefficient Γ(d) on the chart.
Read the values of the normalized resistance r and of the
normalized reactance x that correspond to the reflection
coefficient point.
The normalized impedance is
z (d ) = r + j x
and the actual impedance is
Z(d) = Z0 z ( d ) = Z0 ( r + j x ) = Z0 r + j Z0 x
© Amanogawa, 2000 - Digital Maestro Series
147
Transmission Lines
Given ΓR and ZR
⇐⇒ Find Γ(d) and Z(d)
NOTE: the magnitude of the reflection coefficient is constant along
a loss-less transmission line terminated by a specified load, since
Γ (d ) = Γ R exp ( − j 2β d ) = Γ R
Therefore, on the complex plane, a circle with center at the origin
and radius | ΓR | represents all possible reflection coefficients
found along the transmission line. When the circle of constant
magnitude of the reflection coefficient is drawn on the Smith chart,
one can determine the values of the line impedance at any location.
The graphical step-by-step procedure is:
1.
Identify the load reflection coefficient ΓR and the
normalized load impedance ZR on the Smith chart.
© Amanogawa, 2000 - Digital Maestro Series
148
Transmission Lines
2.
3.
Draw the circle of constant reflection coefficient
amplitude |Γ(d)| =|ΓR|.
Starting from the point representing the load, travel on
the circle in the clockwise direction, by an angle
2π
d
θ= 2βd = 2
λ
4.
The new location on the chart corresponds to location d
on the transmission line. Here, the values of Γ(d) and
Z(d) can be read from the chart as before.
Example: Given
ZR = 25 + j 100 Ω
with
Z0 = 50 Ω
find
Z( d ) and Γ( d )
© Amanogawa, 2000 - Digital Maestro Series
for
d = 0.18λ
149
Transmission Lines
Circle with constant | Γ |
1
zR
05
ΓR
2
3
0.2
∠ ΓR
0
0.2
0.5
-0.2
Γ(d) = 0.8246 ∠-78.7°
= 0.161 – j 0.809
1
2
= 2 (2π/λ) 0.18 λ
= 2.262 rad
= 129.6°
θ
Γ (d)
-3
z(d) -2= 0.236 – j1.192
Z(d) = z(d) × Z0 = 11.79 – j59.6 Ω
-0 5
-1
© Amanogawa, 2000 - Digital Maestro Series
5
θ=2βd
z(d)
150
Transmission Lines
Given ΓR and ZR
1.
2.
3.
⇒ Find dmax and dmin
Identify on the Smith chart the load reflection coefficient
ΓR or the normalized load impedance ZR .
Draw the circle of constant reflection coefficient
amplitude |Γ(d)| =|ΓR|. The circle intersects the real axis
of the reflection coefficient at two points which identify
dmax (when Γ(d) = Real positive) and dmin (when Γ(d) =
Real negative)
A commercial Smith chart provides an outer graduation
where the distances normalized to the wavelength can be
read directly. The angles, between the vector ΓR and the
real axis, also provide a way to compute dmax and dmin .
Example: Find dmax and dmin for
ZR = 25 + j 100 Ω ; ZR = 25 − j100Ω
© Amanogawa, 2000 - Digital Maestro Series
(Z0 = 50 Ω )
151
ZR 25 j 100 Im(ZR) > 0
( Z0 50 )
Transmission Lines
2β dmax = 50.9°
dmax = 0.0707λ
1
ZR
05
ΓR
2
3
0.2
∠ ΓR
0
0.2
0.5
1
2
5
-0.2
-3
-2
-0 5
-1
© Amanogawa, 2000 - Digital Maestro Series
2β dmin = 230.9°
dmin = 0.3207λ
152
ZR 25 j 100 Im(ZR) < 0
( Z0 50 )
Transmission Lines
2β dmax = 309.1°
dmax = 0.4293 λ
1
05
2
3
0.2
0
0.2
0.5
1
2
5
∠ ΓR
-0.2
ΓR
-3
-2
-0 5
2β dmin = 129.1°
dmin = 0.1793 λ
© Amanogawa, 2000 - Digital Maestro Series
ZR
-1
153
Transmission Lines
Given ΓR and ZR ⇒ Find the Voltage Standing Wave Ratio (VSWR)
The Voltage standing Wave Ratio or VSWR is defined as
Vmax 1 + Γ R
=
VSWR =
Vmin 1 − Γ R
The normalized impedance at a maximum location of the standing
wave pattern is given by
1 + Γ ( dmax ) 1 + Γ R
z ( dmax ) =
=
= VSWR!!!
1 − Γ ( dmax ) 1 − Γ R
This quantity is always real and ≥ 1. The VSWR is simply obtained
on the Smith chart, by reading the value of the (real) normalized
impedance, at the location dmax where Γ is real and positive.
© Amanogawa, 2000 - Digital Maestro Series
154
Transmission Lines
The graphical step-by-step procedure is:
1.
2.
3.
4.
Identify the load reflection coefficient ΓR and the
normalized load impedance ZR on the Smith chart.
Draw the circle of constant reflection coefficient
amplitude |Γ(d)| =|ΓR|.
Find the intersection of this circle with the real positive
axis for the reflection coefficient (corresponding to the
transmission line location dmax).
A circle of constant normalized resistance will also
intersect this point. Read or interpolate the value of the
normalized resistance to determine the VSWR.
Example: Find the VSWR for
ZR1 = 25 + j 100 Ω ; ZR2 = 25 − j100Ω
© Amanogawa, 2000 - Digital Maestro Series
(Z0 = 50 Ω )
155
Transmission Lines
Circle with constant | Γ |
1
zR1
05
2
3
ΓR1
0.2
0
0.2
0.5
1
2
5
ΓR2
-0.2
Circle of constant
conductance r = 10.4
z(dmax )=10.4
-3
-2
-0 5
zR2
-1
© Amanogawa, 2000 - Digital Maestro Series
For both loads
VSWR = 10.4
156
Transmission Lines
Given Z(d) ⇐⇒ Find Y(d)
Note:
The normalized impedance and admittance are defined as
1 + Γ (d )
z( d ) =
1 − Γ (d )
1 − Γ (d )
y( d ) =
1 + Γ (d )
Since
λ

Γ  d +  = −Γ ( d )
4

⇒
λ
1 + Γ  d + 
λ
4  1 − Γ (d )


z d +  =
=
= y(d )
λ  1 + Γ (d )
4


1− Γ d + 
4

© Amanogawa, 2000 - Digital Maestro Series
157
Transmission Lines
Keep in mind that the equality
λ

z  d +  = y( d )
4

is only valid for normalized impedance and admittance. The actual
values are given by
λ
λ


Z  d +  = Z0 ⋅ z  d + 
4
4


y( d )
Y ( d ) = Y0 ⋅ y( d ) =
Z0
where Y0=1 /Z0 is the characteristic admittance of the transmission
© Amanogawa, 2000 - Digital Maestro Series
158
Transmission Lines
line.
The graphical step-by-step procedure is:
1.
2.
3.
Identify the load reflection coefficient ΓR and the
normalized load impedance ZR on the Smith chart.
Draw the circle of constant reflection coefficient
amplitude |Γ(d)| =|ΓR|.
The normalized admittance is located at a point on the
circle of constant |Γ| which is diametrically opposite to the
normalized impedance.
Example: Given
ZR = 25 + j 100 Ω
with Z0 = 50 Ω
find YR .
© Amanogawa, 2000 - Digital Maestro Series
159
Transmission Lines
Circle with constant | Γ |
1
z(d) = 0.5 + j 2.0
Z(d) = 25 + j100 [ Ω ]
05
2
3
0.2
0
0.2
0.5
1
2
5
θ = 180°
= 2β⋅λ/4
-0.2
-3
-2
-0 5
© Amanogawa, 2000 - Digital Maestro Series
y(d) = 0.11765 – j 0.4706
Y(d) = 0.002353 – j 0.009412 [ S ]
z(d+λ/4)-1= 0.11765 – j 0.4706
Z(d+λ/4) = 5.8824 – j 23.5294 [ Ω ]
160
Transmission Lines
The Smith chart can be used for line admittances, by shifting the
space reference to the admittance location. After that, one can
move on the chart just reading the numerical values as
representing admittances.
Let’s review the impedance-admittance terminology:
Impedance = Resistance + j Reactance
Z
=
R
+
jX
Admittance = Conductance + j Susceptance
Y
=
G
+
jB
On the impedance chart, the correct reflection coefficient is always
represented by the vector corresponding to the normalized
impedance.
Charts specifically prepared for admittances are
modified to give the correct reflection coefficient in correspondence
of admittance.
© Amanogawa, 2000 - Digital Maestro Series
161
Transmission Lines
Smith Chart for
Admittances
y(d) = 0.11765 – j 0.4706
-1
-0 5
Negative
(inductive)
susceptance
Γ
-2
-3
-0.2
2
5
1
0.5
0.2
0
0.2
3
Positive
(capacitive)
susceptance
2
05
1
z(d) = 0.5 + 2.0
© Amanogawa, 2000 - Digital Maestro Series
162
Transmission Lines
Since related impedance and admittance are on opposite sides of
the same Smith chart, the imaginary parts always have different
sign.
Therefore, a positive (inductive) reactance corresponds to a
negative (inductive) susceptance, while a negative (capacitive)
reactance corresponds to a positive (capacitive) susceptance.
Numerically, we have
1
z= r+ jx
y= g+ jb=
r+ jx
r− jx
r − jx
y=
= 2
( r + j x )( r − j x ) r + x2
r
x
g= 2
b=− 2
⇒
2
r +x
r + x2
© Amanogawa, 2000 - Digital Maestro Series
163
Transmission Lines
Impedance Matching
A number of techniques can be used to eliminate reflections when
line characteristic impedance and load impedance are mismatched.
Impedance matching techniques can be designed to be effective for
a specific frequency of operation (narrow band techniques) or for a
given frequency spectrum (broadband techniques).
One method of impedance matching involves the insertion of an
impedance transformer between line and load
Z0
Impedance
Transformer
ZR
In the following, we neglect effects of loss in the lines.
©Amanogawa, 2000 – Digital Maestro Series
164
Transmission Lines
A simple narrow band impedance transformer consists of a
transmission line section of length /4
ZA
ZB
Z01
Z02
Z01
ZR
/4
dmax or dmin
The impedance transformer is positioned so that it is connected to
a real impedance ZA. This is always possible if a location of
maximum or minimum voltage standing wave pattern is selected.
©Amanogawa, 2000 – Digital Maestro Series
165
Transmission Lines
Consider a general load impedance with its corresponding load
reflection coefficient
ZR RR jX R ;
ZR Z01
R R exp j ZR Z01
If the transformer is inserted at a location of voltage maximum dmax
1 d
1 R
ZA Z01
Z01
1 d
1 R
If it is inserted instead at a location of voltage minimum dmin
1 d
1 R
ZA Z01
Z01
1 d
1 R
©Amanogawa, 2000 – Digital Maestro Series
166
Transmission Lines
Consider now the input impedance of a line of length /4
Zin
Z0
ZA
L = /4
Since:
1 d
1 R
ZA Z01
Z01
1 d
1 R
we have
Zin lim
tan L ©Amanogawa, 2000 – Digital Maestro Series
ZA jZ0 tan( L)
Z0
jZA tan( L) Z0
Z02
ZA
167
Transmission Lines
Note that if the load is real, the voltage standing wave pattern at the
load is maximum when ZR > Z01 or minimum when ZR < Z01 . The
transformer can be connected directly at the load location or at a
distance from the load corresponding to a multiple of /4 .
ZA=Real
ZB
Z01
Z02
Z01
ZR=Real
d1
/4
n /4 ; n=0,1,2…
©Amanogawa, 2000 – Digital Maestro Series
168
Transmission Lines
If the load impedance is real and the transformer is inserted at a
distance from the load equal to an even multiple of /4 then
ZA ZR ;
d1 2 n n
4
2
but if the distance from the load is an odd multiple of /4
2
Z01
ZA ZR
©Amanogawa, 2000 – Digital Maestro Series
;
d1 (2 n 1)
n
4
2 4
169
Transmission Lines
The input impedance of the impedance transformer after inclusion
in the circuit is given by
2
Z02
ZB ZA
For impedance matching we need
2
Z02
Z01 ZA
Z02 Z01 ZA
The characteristic impedance of the transformer is simply the
geometric average between the characteristic impedance of the
original line and the load seen by the transformer.
Let’s now review some simple examples.
©Amanogawa, 2000 – Digital Maestro Series
170
Transmission Lines
Real Load Impedance
ZA
ZB
Z01 = 50 Z02 = ?
RR = 100 /4
2
Z02
ZB Z01 Z02 Z01 RR 50 100 70.71 RR
©Amanogawa, 2000 – Digital Maestro Series
171
Transmission Lines
Note that an identical result is obtained by switching Z01 and RR
ZA
ZB
Z01 = 100 Z02 = ?
RR = 50 /4
2
Z02
ZB Z01 Z02 Z01 RR 100 50 70.71 RR
©Amanogawa, 2000 – Digital Maestro Series
172
Transmission Lines
Another real load case
ZA
ZB
Z01 = 75 Z02 = ?
RR = 300 /4
2
Z02
ZB Z01 Z02 Z01 RR 75 300 150 RR
©Amanogawa, 2000 – Digital Maestro Series
173
Transmission Lines
Same impedances as before, but now the transformer is inserted at
a distance /4 from the load (voltage minimum in this case)
ZB
Z01 = 75 2
752
Z01
ZA 18.75 RR 300
ZA
Z02
/4
Z01
RR = 300 /4
2
Z02
ZB Z01 Z02 Z01 ZA 75 18.75 37.5 ZA
©Amanogawa, 2000 – Digital Maestro Series
174
Transmission Lines
Complex Load Impedance – Transformer at voltage maximum
ZA
ZB
Z01 = 50 Z02
Z01
/4
dmax
ZR = 100 + j 100
100 j100 50
R 0.62
100 j100 50
1 R
ZA Z0
213.28
1 R
Z02 Z01 ZA 50 213.28 103.27 ©Amanogawa, 2000 – Digital Maestro Series
175
Transmission Lines
Complex Load Impedance – Transformer at voltage minimum
ZA
ZB
Z01 = 50 Z02
Z01
/4
dmin
ZR = 100 + j 100
100 j100 50
R 0.62
100 j100 50
1 R
ZA Z0
11.72 1 R
Z02 Z01 ZA 50 11.72 24.21 ©Amanogawa, 2000 – Digital Maestro Series
176
Transmission Lines
If it is not important to realize the impedance transformer with a
quarter wavelength line, we can try to select a transmission line
with appropriate length and characteristic impedance, such that the
input impedance is the required real value
ZA
Z01
Z02
ZR = RR + jXR
L
RR jX R jZ02 tan( L)
Z01 ZA Z02
Z02 j RR jX R tan( L)
©Amanogawa, 2000 – Digital Maestro Series
177
Transmission Lines
After separation of real and imaginary parts we obtain the equations
Z02 ( Z01 RR ) Z01 X R tan L tan L Z02 X R
2
Z01 RR Z02
with final solution
2
2
Z01 RR RR
XR
Z02 1 RR / Z01
tan L 1 RR / Z01 Z01 RR RR2 X R2 XR
The transformer can be realized as long as the result for Z02 is real.
Note that this is also a narrow band approach.
©Amanogawa, 2000 – Digital Maestro Series
178
Transmission Lines
Single stub impedance matching
Impedance matching can be achieved by inserting another
transmission line (stub) as shown in the diagram below
ZA = Z0
ZR
Z0
Z0S
dstub
Lstub
© Amanogawa, 2000 – Digital Maestro Series
179
Transmission Lines
There are two design parameters for single stub matching:
The location of the stub with reference to the load dstub
The length of the stub line Lstub
Any load impedance can be matched to the line by using single
stub technique. The drawback of this approach is that if the load is
changed, the location of insertion may have to be moved.
The transmission line realizing the stub is normally terminated by a
short or by an open circuit. In many cases it is also convenient to
select the same characteristic impedance used for the main line,
although this is not necessary. The choice of open or shorted stub
may depend in practice on a number of factors. A short circuited
stub is less prone to leakage of electromagnetic radiation and is
somewhat easier to realize. On the other hand, an open circuited
stub may be more practical for certain types of transmission lines,
for example microstrips where one would have to drill the insulating
substrate to short circuit the two conductors of the line.
© Amanogawa, 2000 – Digital Maestro Series
180
Transmission Lines
Since the circuit is based on insertion of a parallel stub, it is more
convenient to work with admittances, rather than impedances.
YA = Y0
YR = 1/ZR
Y0 = 1/Z0
dstub
Y0S
Lstub
© Amanogawa, 2000 – Digital Maestro Series
181
Transmission Lines
For proper impedance match:
1
YA = Ystub + Y ( d stub ) = Y0 =
Z0
Line admittance at location
dstub before the stub is applied
Input admittance
of the stub line
Ystub
+
Y (dstub)
YR = 1/ZR
Y0S
Lstub
dstub
© Amanogawa, 2000 – Digital Maestro Series
182
Transmission Lines
In order to complete the design, we have to find an appropriate
location for the stub. Note that the input admittance of a stub is
always imaginary (inductance if negative, or capacitance if positive)
Ystub = jBstub
A stub should be placed at a location where the line admittance has
real part equal to Y0
Y (d stub ) = Y0 + jB (d stub)
For matching, we need to have
Bstub = − B ( d stub )
Depending on the length of the transmission line, there may be a
number of possible locations where a stub can be inserted for
impedance matching. It is very convenient to analyze the possible
solutions on a Smith chart.
© Amanogawa, 2000 – Digital Maestro Series
183
Transmission Lines
First location
suitable for
stub insertion
1
y(dstub1)
05
2
3
θ2
θ1
0.2
0
0.2
0.5
zR
1
2
5
Constant
|Γ(d)| circle
-0.2
-3
yR
Load
location
Unitary
conductance
circle
-2
-0 5
-1
© Amanogawa, 2000 – Digital Maestro Series
y(dstub2)
Second location
suitable for stub
insertion
184
Transmission Lines
The red arrow on the example indicates the load admittance. This
provides on the “admittance chart” the physical reference for the
load location on the transmission line. Notice that in this case the
load admittance falls outside the unitary conductance circle. If one
moves from load to generator on the line, the corresponding chart
location moves from the reference point, in clockwise motion,
according to an angle θ (indicated by the light green arc)
4π
θ = 2β d =
d
λ
The value of the admittance rides on the red circle which
corresponds to constant magnitude of the line reflection coefficient,
|Γ(d)|=|ΓR |, imposed by the load.
Every circle of constant |Γ(d)| intersects the circle Re { y } = 1
(unitary normalized conductance), in correspondence of two points.
Within the first revolution, the two intersections provide the
locations closest to the load for possible stub insertion.
© Amanogawa, 2000 – Digital Maestro Series
185
Transmission Lines
The first solution corresponds to an admittance value with positive
imaginary part, in the upper portion of the chart
(
Y ( d stub1 ) = Y0 + j B d stub1
Line Admittance - Actual :
y ( d stub1 ) = 1 + j b ( d stub1 )
Normalized :
θ1
Stub Location : d stub1 =
λ
4π
(
)
− j b ( d stub1 )
− j B d stub1
Stub Admittance - Actual :
Normalized :

λ
Stub Length : Lstub =
tan −1 
 Z0 s B
2π

λ
Lstub =
tan −1 Z0s B
2π
(
© Amanogawa, 2000 – Digital Maestro Series
)
1
d stub1

 ( short )


(
)
(d stub1 ))
(open)
186
Transmission Lines
The second solution corresponds to an admittance value with
negative imaginary part, in the lower portion of the chart
(
Y (d stub2 ) = Y0 − j B d stub2
Line Admittance - Actual :
y (d stub2 ) = 1 − j b ( d stub2 )
Normalized :
θ
Stub Location : d stub2 = 2 λ
4π
(
)
j b ( d stub2 )
j B d stub2
Stub Admittance - Actual :
Normalized :

λ
1
−1 
Stub Length : Lstub =
−
tan
 Z0 s B d stub
2π
2

λ
Lstub =
tan −1 − Z0s B d stub2
2π
(
© Amanogawa, 2000 – Digital Maestro Series
)
(
(

 ( short )


)
))
(open)
187
Transmission Lines
If the normalized load admittance falls inside the unitary
conductance circle (see next figure), the first possible stub location
corresponds to a line admittance with negative imaginary part. The
second possible location has line admittance with positive
imaginary part. In this case, the formulae given above for first and
second solution exchange place.
If one moves further away from the load, other suitable locations for
stub insertion are found by moving toward the generator, at
distances multiple of half a wavelength from the original solutions.
These locations correspond to the same points on the Smith chart.
First set of locations
Second set of locations
© Amanogawa, 2000 – Digital Maestro Series
λ
= d stub1 + n
2
λ
= d stub2 + n
2
188
Transmission Lines
Second location
suitable for stub
insertion
1
y(dstub2)
05
2
3
0
yR
θ2
0.2
0.2
0.5
Load
location
1
θ1
2
5
Unitary
conductance
circle
Constant
|Γ(d)| circle
-0.2
zR
-3
-2
-0 5
-1
© Amanogawa, 2000 – Digital Maestro Series
y(dstub1)
First location
suitable for stub
insertion
189
Transmission Lines
Single stub matching problems can be solved on the Smith chart
graphically, using a compass and a ruler. This is a step-by-step
summary of the procedure:
(a) Find the normalized load impedance and determine the
corresponding location on the chart.
(b) Draw the circle of constant magnitude of the reflection
coefficient |Γ| for the given load.
(c) Determine the normalized load admittance on the chart. This is
obtained by rotating 180° on the constant |Γ| circle, from the
load impedance point. From now on, all values read on the chart
are normalized admittances.
© Amanogawa, 2000 – Digital Maestro Series
190
Transmission Lines
(a) Obtain the normalized load
1
impedance zR=ZR /Z0 and find
its location on the Smith chart
05
2
3
zR
0.2
0
0.2
0.5
-0.2
1
2
5
180° = λ /4
-3
yR
(b) Draw the
constant |Γ(d)|
circle
-2
-0 5
(c) Find the normalized load
admittance knowing that
yR = z(d=λ /4 )
From now on the
represents admittances.
chart
© Amanogawa, 2000 – Digital Maestro Series
-1
191
Transmission Lines
(d) Move from load admittance toward generator by riding on the
constant |Γ| circle, until the intersections with the unitary
normalized conductance circle are found. These intersections
correspond to possible locations for stub insertion. Commercial
Smith charts provide graduations to determine the angles of
rotation as well as the distances from the load in units of
wavelength.
(e) Read the line normalized admittance in correspondence of the
stub insertion locations determined in (d). These values will
always be of the form
y (d stub ) = 1 + jb
y (d stub ) = 1 − jb
© Amanogawa, 2000 – Digital Maestro Series
top half of chart
bottom half of chart
192
Transmission Lines
First Solution
(d) Move from load toward
generator and stop at a
location where the real
part of the normalized line
admittance is 1.
05
First location suitable for
stub insertion
1
dstub1=(θ1/4π)λ
2
3
zR
0.2
(e) Read here the
value of the
normalized line
admittance
y(dstub1) = 1+jb
θ1
0
0.2
0.5
1
2
5
-0.2
-3
yR
Load
location
-2
Unitary
conductance
circle
-0 5
-1
© Amanogawa, 2000 – Digital Maestro Series
193
Transmission Lines
Second Solution
(d) Move from load
toward generator and
stop at a location
where the real part of
the normalized line
05
admittance is 1.
1
2
3
θ2
0.2
0
0.2
0.5
zR
1
2
Unitary
conductance
circle
5
-0.2
-3
yR
Load
location
-2
-0 5
(e) Read here the
value of the
normalized line
admittance
y(dstub2) = 1 - jb
-1
Second location suitable
for stub insertion
dstub2=(θ2/4π)λ
© Amanogawa, 2000 – Digital Maestro Series
194
Transmission Lines
(f) Select the input normalized admittance of the stubs, by taking
the opposite of the corresponding imaginary part of the line
admittance
line: y ( d stub ) = 1 + jb
line: y ( d stub ) = 1 − jb
→
stub: ystub = − jb
→
stub: ystub = + jb
(g) Use the chart to determine the length of the stub. The
imaginary normalized admittance values are found on the circle
of zero conductance on the chart. On a commercial Smith chart
one can use a printed scale to read the stub length in terms of
wavelength. We assume here that the stub line has
characteristic impedance Z0 as the main line. If the stub has
characteristic impedance Z0S
must be renormalized as
≠ Z0 the values on the Smith chart
Y0
Z0 s
± jb' = ± jb
= ± jb
Y0 s
Z0
© Amanogawa, 2000 – Digital Maestro Series
195
Transmission Lines
1
05
2
3
Short circuit
0.2
0
0.2
-0.2
y=∞
0.5
1
2
5
(g) Arc to determine the length of a
short circuited stub with normalized
input admittance - jb
-3
(f) Normalized input
admittance of stub
-0 5
ystub = 0 - jb
-2
-1
© Amanogawa, 2000 – Digital Maestro Series
196
Transmission Lines
1
05
2
3
(g) Arc to determine the length of an
open circuited stub with normalized
input admittance - jb
0.2
0
0.2
0.5
1
2
5
-0.2
y=0
-3
(f) Normalized input
admittance of stub
Open circuit
-0 5
ystub = 0 - jb
-2
-1
© Amanogawa, 2000 – Digital Maestro Series
197
Transmission Lines
1
(f) Normalized input
admittance of stub
ystub = 0 + jb
05
2
3
(g) Arc to determine the length of a
short circuited stub with normalized
input admittance + jb
0.2
0
0.2
0.5
1
2
y=∞
Short circuit
5
-0.2
-3
-2
-0 5
-1
© Amanogawa, 2000 – Digital Maestro Series
198
Transmission Lines
1
(f) Normalized input
admittance of stub
ystub = 0 + jb
05
2
3
0.2
0
(g) Arc to determine the length of an
open circuited stub with normalized
input admittance + jb
0.2
0.5
1
2
5
-0.2
y=0
-3
Open circuit
-2
-0 5
-1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
First Solution
1
05
2
3
0.2
0
matching
condition
0.2
0.5
1
2
5
-0.2
-3
yR
-2
-0 5
After the stub is inserted,
the admittance at the stub
location is moved to the
center of the Smith chart,
which
corresponds
to
normalized admittance 1
and reflection coefficient 0
(exact matching condition).
If you imagine to add
gradually
the
negative
imaginary admittance of
the inserted stub, the total
admittance would follow
the yellow arrow, reaching
the match point when the
complete stub admittance
is added.
-1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
First Solution
If the stub does not have
the proper normalized input
admittance, the matching
condition is not reached
1
05
2
Effect of a stub with
negative susceptance of
insufficient magnitude
3
Effect of a stub with
positive susceptance
0.2
0
0.2
0.5
1
2
5
-0.2
-3
yR
-2
-0 5
Effect of a stub with
negative susceptance of
excessive magnitude
-1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
Double stub impedance matching
Impedance matching can be achieved by inserting two stubs at
specified locations along transmission line as shown below
YA = Y01
dstub2
Y01 = 1/Z01
dstub1
YR = 1/ZR
Y0S2
Y0S1
Lstub2
© Amanogawa, 2000 – Digital Maestro Series
Lstub1
202
Transmission Lines
There are two design parameters for double stub matching:
The length of the first stub line Lstub1
The length of the second stub line Lstub2
In the double stub configuration, the stubs are inserted at predetermined locations. In this way, if the load impedance is
changed, one simply has to replace the stubs with another set of
different length.
The drawback of double stub tuning is that a certain range of load
admittances cannot be matched once the stub locations are fixed.
Three stubs are necessary to guarantee that match is always
possible.
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
The length of the first stub is selected so that the admittance at the
location of the second stub (before the second stub is inserted) has
real part equal to the characteristic admittance of the line
Y’A = Y01 + jB
dstub2
Y01 = 1/Z01
dstub1
YR = 1/ZR
Y0S1
Lstub1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
YA = Y01 + jB – jB = Y01
dstub2
Y01 = 1/Z01
dstub1
YR = 1/ZR
Ystub = -jB
Y0S1
Lstub1
Y0S2
Lstub2
The length of the second stub is
selected to eliminate the imaginary
part of the admittance at the location
of insertion.
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Transmission Lines
1
05
2
3
0.2
0
0.2
0.5
1
2
At the location where
the second stub is
inserted, the possible
normalized admittances
that can give matching
are found on the circle
of unitary conductance
on the Smith chart.
5
-0.2
-3
The normalized admittance
that we
want at location
-0 5
dstub2
-2
is on this circle
-1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
dstub
Think of stub matching in a unified way.
Single stub
YR
YA = Y01
The two approaches solve the same problem
Y01 = 1/Z01
dstub2
Y0S2
Double stub
YR
Lstub2
Y0S1
Lstub1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
If one moves from the location of the second stub back to the load,
the circle of the allowed normalized admittances is mapped into
another circle, obtained by pivoting the original circle about the
center of the chart.
At the location of the first stub, the allowed normalized admittances
are found on an auxiliary circle which is obtained by rotating the
unitary conductance circle counterclockwise, by an angle
4
4
aux d stub2 d stub1 d 21
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
This angle of rotation
corresponds to a distance
1
d12 = dstub2 -dstub1
05
2
3
0.2
0
aux
0.2
0.5
1
2
5
Pivot here
-0.2
The normalized admittance
that we want at location dstub1
is on this auxiliary
circle.
-0 5
-3
-2
-1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
1
05
2
3
0.2
0
aux
0.2
0.5
1
2
5
This is-0.2
the auxiliary circle for
distance between the stubs
d21 = /8 + n /2.
-3
-2
-0 5
-1
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Transmission Lines
1
05
2
3
0.2
0
aux
0.2
0.5
1
2
5
-0.2
-3
-2
-0 5
This is the auxiliary circle for
distance between the stubs
-1
d21 = /4 + n /2.
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Transmission Lines
1
05
2
3
0.2
0
aux
0.2
0.5
1
2
5
-0.2
-3
-2
-0 5
-1
© Amanogawa, 2000 – Digital Maestro Series
This is the auxiliary circle for
distance between the stubs
d21 = 3 /8 + n /2.
212
Transmission Lines
1
05
2
3
0.2
0
aux
0.2
0.5
1
2
5
-0.2
-3
This is the auxiliary circle for
distance -0between
the stubs
5
d21 = n /2.
-2
NOTE: this is not a good
-1
choice for double stub design!
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
Given the load impedance, we need to follow these steps to
complete the double stub design:
(a) Find the normalized load impedance and determine the
corresponding location on the chart.
(b) Draw the circle of constant magnitude of the reflection
coefficient || for the given load.
(c) Determine the normalized load admittance on the chart. This is
obtained by rotating -180 on the constant || circle, from the
load impedance point. From now on, all values read on the chart
are normalized admittances.
(d) Find the normalized admittance at location dstub1 by moving
clockwise on the constant || circle.
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Transmission Lines
(e) Draw the auxiliary circle
(f) Add the first stub admittance so that the normalized admittance
point on the Smith chart reaches the auxiliary circle (two
possible solutions). The admittance point will move on the
corresponding conductance circle, since the stub does not alter
the real part of the admittance
(g) Map the normalized admittance obtained on the auxiliary circle
to the location of the second stub dstub2. The point must be on
the unitary conductance circle
(h) Add the second stub admittance so that the total parallel
admittance equals the characteristic admittance of the line to
achieve exact matching condition
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
(d) Move to the
first stub location
(a) Obtain the normalized load
1
impedance zR=ZR /Z0 and find
its location on the Smith chart
05
2
3
zR
0.2
0
0.2
0.5
-0.2
1
2
5
180 = /4
-3
yR
(b) Draw the
constant |(d)|
circle
-2
-0 5
(c) Find the normalized load
admittance knowing that
yR = z(d= /4 )
From now on the chart
represents admittances.
© Amanogawa, 2000 – Digital Maestro Series
-1
216
Transmission Lines
1
05
2
3
(f) First solution: Add
admittance of first stub to
reach auxiliary circle
0.2
0
0.2
0.5
1
2
5
-0.2
-3
yR
-2
-0 5
-1
(e) Draw the auxiliary circle
© Amanogawa, 2000 – Digital Maestro Series
(f) Second solution: Add
admittance of first stub to
reach auxiliary circle
217
Transmission Lines
1
(g) First solution: Map the normalized admittance
from the auxiliary circle to the location of the
05
second stub dstub2.
2
3
0.2
0
(h) Add second
stub admittance
0.2
0.5
1
2
5
-0.2
-3
-2
-0 5
First solution: Admittance at
location dstub2 before insertion
of second stub
© Amanogawa, 2000 – Digital Maestro Series
-1
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Transmission Lines
(g) Second solution: Map the normalized
admittance from the auxiliary circle to the location
of the second stub dstub2.
1
05
2
3
Second solution: Admittance at
0.2 Add second
(h)
stub admittance
0
0.2
0.5
location dstub2 before insertion
of second stub
1
2
5
-0.2
-3
-2
-0 5
-1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
As mentioned earlier, a double stub configuration with fixed stub
location may not be able to match a certain range of load
impedances.
This is easily seen on the Smith chart. If the normalized admittance
of the line, at the first stub location, falls inside a certain forbidden
conductance circle tangent to the auxiliary circle (and always
contained inside the unitary conductance circle), it is not possible
to find a value for the first stub that can bring the normalized
admittance to the auxiliary circle. Therefore, it is impossible to
position the normalized admittance of the second stub location on
the unitary conductance circle.
When this condition occurs, the location of one of the stubs must
be changed appropriately. Alternatively, a third stub could be
added.
Examples of forbidden regions follow.
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
1
05
2
Forbidden conductance
circle. If the admittance
at the first stub location
falls inside this circle,
match is not possible
with the given two stub
configuration.
3
0.2
0
aux
0.2
0.5
1
2
5
This is-0.2
the auxiliary circle for
distance between the stubs
d21 = /8 + n /2.
-3
-2
-0 5
The normalized conductance circle
for the normalized admittance does
not intersect the auxiliary circle.
-1
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
1
05
2
3
aux
0.2
0
0.2
0.5
1
2
Forbidden
5
conductance
circle
-0.2
-3
-2
-0 5
This is the auxiliary circle for
distance between the stubs
-1
d21 = /4 + n /2.
© Amanogawa, 2000 – Digital Maestro Series
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Transmission Lines
1
Forbidden
2
conductance
circle
05
aux
0.2
0
3
0.2
0.5
1
2
5
-0.2
-3
-2
-0 5
-1
© Amanogawa, 2000 – Digital Maestro Series
This is the auxiliary circle for
distance between the stubs
d21 = 3 /8 + n /2.
223