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( jφ) = A cos ( φ ) + jA sin ( φ )
a = A cos ( φ ) real part
b = A sin ( φ ) imaginary part
where
2
A= a + b
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2
φ = tan
−1  b 
 
 a
1
Transmission Lines
Im
z
b
A
φ
a
Note :
Re
z = A exp( jφ) = A exp( jφ ± j 2 nπ)
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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)
2
2
=a +b = z
2
= A2
In polar form we have
z* = ( A exp( jφ) ) * = A exp(− jφ)
= A exp ( j 2 π − jφ )
= A cos ( φ ) − jA sin ( φ )
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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 non−integer exponents.
polar form, instead, the result is immediate
In
n
z = [ A exp( jφ)] = An exp ( jnφ )
n
In the case of roots, one should remember to consider φ + 2kπ as
argument of the exponential, with k = integer, otherwise possible
roots are skipped:
n z = n A exp jφ + j 2 kπ = n A exp  j φ + j 2 kπ 
(
)


n 
 n
The results corresponding to angles up to 2π 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
 π
j = exp  j 
 2
 π
− 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 ( ω t + φ ) = Re [ A exp ( jω t + jφ )]
= Re [ A exp ( jφ ) exp ( jω t )]
= Re [ A exp ( jω t )]
The complex quantity
A = A exp ( jφ )
contains all the information about amplitude and phase of the
signal and is called the phasor of
A cos ( ω t + φ )
If it is known that the signal is time-harmonic with frequency
phasor completely characterizes its behavior.
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ω, the
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Transmission Lines
Often, a time-harmonic signal may be of the form:
A sin ( ω t + φ )
and we have the following complex representation
A sin ( ω t + φ ) = Re  − jA ( cos ( ω t + φ ) + j sin ( ω t + φ ) ) 
= Re [ − jA exp ( jω t + jφ )]
= Re [ A exp ( − jπ / 2 ) exp ( jφ ) exp ( jω t )]
= Re  A exp ( j ( φ − π / 2 ) ) exp ( jω t )
= Re [ A exp ( jω t )]
with phasor
A = A exp ( j ( φ − π / 2 ) )
This result is not surprising, since
cos(ω t + φ − π / 2) = sin(ω t + φ)
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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 ( ω t + φ )
with phasor
V = V0 exp ( jφ )
The time derivative can be expressed as
∂ V ( t)
= −ωV0 sin ( ω t + φ )
∂t
= Re { jωV0 exp ( jφ ) exp ( jω t )}
⇒
jωV0 exp ( jφ ) = jω 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 i(t)
1 t
v( t ) = L
+ R i + ∫ i( t ) dt
dt
C −∞
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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(ω t + αV )
phasor ⇒ V = V0 exp( jαV )
i( t ) = I0 cos(ω t + α I )
phasor ⇒
If
I = I0 exp( jα I )
V0 and αV are given,
⇒
I0 and αI are the unknowns of the problem.
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Transmission Lines
The differential equation can be rewritten using phasors
{
}
L Re −ω2 I exp ( jω t ) + R Re { jω I exp ( jω t )}
1
+ Re { I exp ( jω t )} = Re { jωV exp ( jω t )}
C
Finally, the transform phasor equation is obtained as
1 

V =  R + jω L − j
I=ZI

ωC 

where
Z
=
Impedance
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R
Resistance
+
1 

j ωL −

C
ω


Reactance
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Transmission Lines
The result for the phasor current is simply obtained as
V
V
I= =
= I0 exp ( jα I )
1 
Z 
 R + jω L − j ω C 


which readily yields the unknowns I0 and αI .
The time dependent current is then obtained from
i( t ) = Re { I0 exp ( jα I ) exp ( jω t )}
= I0 cos ( ω t + α I )
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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 ω
Capacitive component ⇒ inversely proportional to ω
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
ωr L =
1
ωr C
⇒
ωr =
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 

R + ωL −

ω
C


2
2
IM
ωr
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ω
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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

Zin = R + 
+ jω C 
 jω L

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−1
= R+
jω L
1 − ω2 LC
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Transmission Lines
When
ω=0
Zin = R
1
ω=
LC
ω→∞
Zin → ∞
Zin = R
At the resonance condition
ωr =
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(ω t + φ)
φ ∈ [ −π/2 , π/2 ]
i (t ) = I 0 cos(ω t )
where V0 and I0 are peak values and φ is the phase difference
between voltage and current. Note that φ = 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 I 0 cos(ω t + φ) cos(ω t )
V0 I 0
=
[ cos (φ) + cos(2ω t + φ )]
2
The power has two (Fourier) components:
(A) an average value
V0 I 0
cos(φ)
2
(B) an oscillatory component with frequency 2f
V0 I 0
cos(2ω t + φ)
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 φ = 0 we have
cos(φ) = 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
Using cos ( A + B ) = cos A cos B − sin A sin B one can write
v (t ) = V0 cos(ω t + φ) = V0 cos φ cos ω t − V0 sin φ sin ω t
in phase with
current
in quadrature
with current
This gives an alternative expression for power:
P (t ) = V0 cos( ω t ) I 0 cos( ω t ) cos( φ ) − V0 cos( ω t ) I 0 sin( ω t ) sin( φ )
2
= V0 I 0 cos( φ ) cos ( ω t ) −
Real Power
V0 I 0
sin( φ ) sin(2ω t )
2
Reactive Power
V0 I 0
V0 I 0
cos( φ ) +
cos( φ ) cos(2ω t ) −
=
2
2
V0 I 0
sin( φ ) sin(2ω t )
2
Real Power
Reactive Power
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Transmission Lines
The real power corresponds to the power dissipated by the resistive
component of the impedance, and it is always positive.
The reactive power corresponds to 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.
To determine the time-average power, we can use either the Fourier
or the real/reactive power formulation.
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Transmission Lines
Fourier representation
1 T V0 I 0
1 T V0 I 0
cos(φ) dt + ∫
cos(2ω t − φ) dt
⟨ P (t ) ⟩ = ∫
0
0
2
2
T
T
=0
V0 I 0
=
cos(φ)
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
Real/Reactive power representation
1 T
T
⟨ P (t ) ⟩ =
( ∫ V0 I 0 cos(φ) dt + ∫ V0 I 0 cos(φ) cos(2ω t ) dt )
0
2T 0
=0
1 T
−
V0 I 0 sin(φ) sin(2ω t ) dt
∫
0
2T
=0
=
V0 I 0
cos(φ)
2
This result tells us that the time-average power flow is the average
of the real 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.
© Amanogawa, 2006 – Digital Maestro Series
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Transmission Lines
The maximum of the reactive power is
V0 I 0
V0 I 0
max{ Preac } = max{
sin ( φ ) sin ( 2ω t ) } =
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 or 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 total reactance is inductive
V = Z I = R I + jω L I
Im
jωL I
V
Current lags
φ>0
φ
I
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RI
Re
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Transmission Lines
If the total reactance is capacitive
1
V =ZI=RI− j
I
ωC
Im
I
RI
Re
φ
Voltage lags
φ<0
V
- j I /ω C
© Amanogawa, 2006 – Digital Maestro Series
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Transmission Lines
In many situations, we may use the root-mean-square (r.m.s.)
values of quantities, instead of the peak value. For a given signal
v(t ) = V0 cos(ω t )
the r.m.s. value is defined as
1 T 2
1 2π 2
2
Vrms =
V0 cos ( ω t ) dt = V0
cos ( ω t ) d ω t
∫
∫
T 0
ωT 0
1 2π 2
1
= V0
V0
cos ( ϑ ) d ϑ =
∫
2π 0
2
This result is valid for sinusoidal signals. Any given signal shape
corresponds to a specific coefficient ( peak factor = V0 / Vrms ) that
allows one to convert directly from peak value to r.m.s. value.
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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
© Amanogawa, 2006 – Digital Maestro Series
V0
t
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Transmission Lines
For a non-sinusoidal periodic signal, we can use a decomposition
into orthogonal Fourier components to obtain the r.m.s. value:
V (t ) = Vav + V1 (t ) + V2 (t ) + V3 (t ) … =
[Vrms ]
=
1
2
T
∫
T 0
=
[∑
1
T
∫
T 0
2
V (t ) dt =
2
(t )]
V
k k
dt +
1
1
T
∫
T 0
∑ k Vk ( t )
[ ∑ k Vk (t )] dt =
2
T
[ ∑ i ≠ j 2 Vi (t ) V j (t )] dt =
∫
0
T
orthogonal
1 T 2
1 T
2


(
)
[
2
(
)
(
)
]
=
(
V
)
= ∑k
+
V
t
dt
V
t
V
t
dt
∑ k krms
j
 T ∫0 k
 ∑ i ≠ j T ∫0 i
⇒ Vrms =
∑ k (Vkrms )2
=
2
Vav + (V1
rms
2
) + (V2
2
rms
) +…
The final result holds for any decomposition into orthogonal
functions and it is known in mathemics as Parseval’s identity.
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Transmission Lines
In terms of r.m.s. values, the time-average power for a sinusoidal
signal is then
V0 I 0
⟨ P (t ) ⟩ =
cos(φ) = Vrms I rms cos(φ)
2 2
Finally, we can relate the time-average power to the phasors of
voltage and current. Since
v (t ) = V0 cos(ω t ) = Re{ V0 exp( jω t ) }
i (t ) = I 0 cos(ω t − φ) = Re{ I 0 exp(− jφ) exp( jω t ) }
we have phasors
V = V0
I = I 0 exp(− jφ)
© Amanogawa, 2006 – Digital Maestro Series
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Transmission Lines
The time-average power in terms of phasors is given by
1
1
*
⟨ P (t ) ⟩ = Re{ V I } = Re{ V0 I 0 exp( jφ) }
2
2
V0 I 0
=
cos(φ)
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 φ = 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.
© Amanogawa, 2006 – Digital Maestro Series
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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 + Z R
1
Iin = VG
ZG + Z R
1
*
⟨ Pin ⟩ = Re{ Vin Iin
}
2
© Amanogawa, 2006 – Digital Maestro Series
Iin
ZG
VG
ZR
Vin
Generator
Load
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Transmission Lines
As a first case, we examine resistive impedances
ZG = RG
Z R = RR
Voltage and current are in phase at the input. The time-average
power dissipated by the load is
1
RR
1
*
⟨ P (t ) ⟩ = VG
VG
2
RG + RR
RG + RR
1
RR
2
= VG
2
( RG + RR )2
© Amanogawa, 2006 – Digital Maestro Series
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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
d RR
from which we obtain

d 
RR

=0
2
dRR  ( RG + RR ) 
( 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.
© Amanogawa, 2006 – Digital Maestro Series
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Transmission Lines
Let’s consider now complex impedances
Z R = RR + jX R
ZG = RG + jX G
For maximum power transfer, generator and load impedances must
be complex conjugate of each other:
*
Z R = ZG
⇒
RR = RG
X R = − XG
RG
VG
jXG
−jXG
RG
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.
© Amanogawa, 2006 – Digital Maestro Series
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Transmission Lines
The total time-average power supplied by the generator in
conditions of maximum power transfer is
1
1
1
2 1
2 1
*
⟨ Ptot ⟩ = Re{ VG Iin } = VG
= VG
RR
2
2
2 RR
4
The time-average power supplied to the load is
*

 
1
1 
ZR
1
*
*
VG 
⟨ Pin ⟩ = Re{ Vin Iin } = Re VG
 
2
2  ZG + Z R
Z
Z
+
 G
R 


 RR + jX R 
1
1
2
2 1
= VG Re 
 = VG
2
2
8
RR
 4 RR 
© Amanogawa, 2006 – Digital Maestro Series
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Transmission Lines
The power dissipated by the internal generator impedance is
1
*
}
⟨ PG ⟩ = Re { (VG − Vin ) Iin
2
1
1
1
2 1
2 1
2 1
= VG
− VG
= VG
4
8
8
RR
RR
RR
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!
© Amanogawa, 2006 – Digital Maestro Series
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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
© Amanogawa, 2006 – Digital Maestro Series
Transmission line
ZR
39
Transmission Lines
Examples of transmission lines
d
d
D
d
Two-wire line
D
Coaxial cable
w
t
h
Microstrip
© Amanogawa, 2006 – Digital Maestro Series
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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
© Amanogawa, 2006 – Digital Maestro Series
VR
VR
VR
ZR
L << λ
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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 (ε ≈ ε0), the wavelength for 60Hz is:
c 2.999 × 10 8
λ= =
≈ 5.0 × 106 m = 5, 000 km
f
60
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 × 10 8
λ= =
≈ 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?
© Amanogawa, 2006 – Digital Maestro Series
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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− jβ z + V − e jβ z Load
Generator
ZG
VG
V( 0 )
V( z )
V( L )
ZR
L
© Amanogawa, 2006 – Digital Maestro Series
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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
ZG
VG
Vin
ZR
= 0)
and Vin = 0
© Amanogawa, 2006 – Digital Maestro Series
Iin
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
only if
λ
L=n
2
[ n = integer ]
ZR
L
Our first goal is to determine the equivalent impedance seen by the
generator, that is, the input impedance of a line terminated by the
load. Once that is known, standard circuit theory can be used.
© Amanogawa, 2006 – Digital Maestro Series
45
Transmission Lines
Generator
Load
ZG
Transmission line
VG
Generator
VG
© Amanogawa, 2006 – Digital Maestro Series
ZR
Equivalent Load
ZG
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
© Amanogawa, 2006 – Digital Maestro Series
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.
© Amanogawa, 2006 – Digital Maestro Series
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
© Amanogawa, 2006 – Digital Maestro Series
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 = − jω L dz I
which gives a first order differential equation for the voltage
dV
= − jω L I
dz
© Amanogawa, 2006 – Digital Maestro Series
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 = − jω Cdz(V + dV ) = − jω CVdz − jω C dV dz
The second term (including dV dz) tends to zero very rapidly in the
limit of infinitesimal length dz leaving a first order differential
equation for the current
dI
= − jω C V
dz
© Amanogawa, 2006 – Digital Maestro Series
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
= − jω L I
dz
dI
= − jω C V
dz
These are often called “telegraphers’ equations” of the loss-less
transmission line.
© Amanogawa, 2006 – Digital Maestro Series
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
= − jω C V
dz
d 2V
dI
= − jω L
= jω L jω CV = − ω2 LC V
dz
dz2
d2 I
dV
= − jω C
= jω C jω L I = − ω2 LC I
dz
dz2
dV
= − jω L I
dz
These are often called “telephonists’ equations”.
© Amanogawa, 2006 – Digital Maestro Series
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− jβ z + V − e jβ z
where the wave propagation constant is
β = ω LC
Note that the complex exponential terms including β have unitary
magnitude and purely “imaginary” argument, therefore they only
affect the “phase” of the wave in space.
© Amanogawa, 2006 – Digital Maestro Series
54
Transmission Lines
We have the following useful relations:
2π 2π f ω
β=
=
=
λ
vp
vp
ω εr µr
=
= ω ε0 µ0 ε r µ r = ω ε µ
c
Here, λ = v p f is the wavelength of the dielectric medium
surrounding the conductors of the transmission line and
1
1
vp =
=
ε0 ε r µ0 µ r
εµ
is the phase velocity of an electromagnetic wave in the dielectric.
As you can see, the propagation constant β can be written in many
different, equivalent ways.
© Amanogawa, 2006 – Digital Maestro Series
55
Transmission Lines
The current distribution on the transmission line can be readily
obtained by differentiation of the result for the voltage
dV
= − jβ V + e− jβ z + jβ V − e jβ z = − jω L I
dz
which gives
(
)
(
C + − jβ z
1
− jβ z
−V e
=
I (z) =
V e
V + e− jβ z − V − e jβ z
L
Z0
)
The real quantity
Z0 =
L
C
is the “characteristic impedance” of the loss-less transmission line.
© Amanogawa, 2006 – 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, 2006 – 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 = − ( jω Ldz + Rdz) I
from which we obtain a first order differential equation for the
voltage
dV
= − ( jω L + R) I
dz
© Amanogawa, 2006 – 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 = − ( jω Cdz + Gdz)(V + dV )
= − ( jω C + G)Vdz − ( jω C + G)dV dz
The second term (including dV dz) can be ignored, giving a first
order differential equation for the current
dI
= − ( jω C + G)V
dz
© Amanogawa, 2006 – 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
= − ( jω L + R) I
dz
dI
= − ( jω C + G)V
dz
These are the “telegraphers’ equations” for the lossy transmission
line case.
© Amanogawa, 2006 – 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
= − ( jω C + G)V
dz
d 2V
dI
= − ( jω L + R) = ( jω L + R)( jω C + G)V
2
dz
dz
d2 I
dV
= − ( jω C + G)
= ( jω C + G)( jω L + R) I
dz
dz2
dV
= − ( jω L + R) I
dz
These are the “telephonists’ equations” for the lossy line.
© Amanogawa, 2006 – 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−γ z + V − eγ z = V + e−α z e− jβ z + V − eα z e jβ z
where the wave propagation constant is now the complex quantity
γ = ( jω L + R)( jω C + G) = α + jβ
The real part α of the propagation constant γ describes the
attenuation of the signal due to resistive losses. The imaginary part
β describes the propagation properties of the signal waves as in
loss-less lines.
The exponential terms including α are “real”, therefore, they only
affect the “magnitude” of the voltage phasor. The exponential
terms including β have unitary magnitude and purely “imaginary”
argument, affecting only the “phase” of the waves in space.
© Amanogawa, 2006 – 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
= − ( jω L + R) I = −γ V + e−γ z + γ V − eγ z
dz
which gives
( jω C + G )
I (z) =
(V + e−γz − V − e γ z )
( jω L + R )
1
(V + e−γz − V − e γ z )
=
Z0
with the “characteristic impedance” of the lossy transmission line
( jω L + R)
Z0 =
( jω C + G)
© Amanogawa, 2006 – 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, 2006 – 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
+ − jβ z
− jβ z
V (z) = V e
Lossy line
+ −γz
− γz
+V e
(
1
I (z) =
V + e − jβ z − V − e jβ z
Z0
V (z) = V e
)
(
+V e
1
I (z) =
V + e− γ z − V −e γ z
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, 2006 – 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, 2006 – 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
+ jβd
− − jβ d
V (d) = V e
I (d) =
1
Z0
(
Lossy line
+ γd
− −γ d
+V e
+ jβd
V e
− − jβ d
−V e
V (d) = V e
)
I (d) =
1
Z0
(
+V e
+ γd
V e
− −γd
−V e
)
At the load (d = 0) we have, for both cases,
V (0) = V + + V −
(
1
I (0) =
V + −V −
Z0
© Amanogawa, 2006 – Digital Maestro Series
)
67
Transmission Lines
For a given load impedance ZR , the load boundary condition is
V (0) = Z R I (0)
Therefore, we have
(
ZR
V +V =
V + −V −
Z0
+
−
)
from which we obtain the voltage load reflection coefficient
V−
Z R − Z0
ΓR = + =
Z R + Z0
V
© Amanogawa, 2006 – Digital Maestro Series
68
Transmission Lines
We can introduce this result into the transmission line equations as
Loss-less line
Lossy line
+ jβ d
V (d) = V e
+ jβ d
I (d) =
V e
Z0
(
1+ ΓR e
(
1 − Γ Re
−2 jβ d
−2 jβ d
)
)
+ γd
V (d) = V e
+ γd
I (d) =
V e
Z0
(
1+ ΓR e
(
1 − Γ Re
−2 γ d
−2 γ d
)
)
At each line location we define a Generalized Reflection Coefficient
Γ (d) = Γ R e−2 jβ d
Γ (d) = Γ R e−2 γ d
and the line equations become
V (d) = V + e jβ d (1 + Γ (d) )
V (d) = V + e γ d (1 + Γ (d) )
V + e jβ d
I (d) =
(1 − Γ (d) )
Z0
V +e γd
I (d) =
(1 − Γ (d) )
Z0
© Amanogawa, 2006 – Digital Maestro Series
69
Transmission Lines
We define the line impedance as
V (d) 1+ Γ (d)
Z (d) =
=
I (d) 1 − Γ (d)
A simple circuit diagram can illustrate the significance of line
impedance and generalized reflection coefficient:
ΓReq = Γ(d)
d
© Amanogawa, 2006 – 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 Γ(d)
Γ Req = Γ ( d ) =
Z Req − Z 0
Z Req + Z 0
If the total length of the line is L, the input impedance is obtained
from the formula for the line impedance as
Vin V ( L ) 1 + Γ ( L )
Zin =
=
=
Iin I ( L ) 1 − Γ ( L )
The input impedance is the equivalent impedance representing the
entire line terminated by the load.
© Amanogawa, 2006 – 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:
ω L >> R
ω C >> G
so that
γ = ( jω L + R )( jω C + G )
=

R 
G 
jω L jω C  1 +
1+


j
ω
L
j
ω
C



R
G
RG
≈ jω LC 1 +
+
−
jω L jω C ω2 LC
The last term under the square root can be neglected, because it is
the product of two very small quantities.
© Amanogawa, 2006 – Digital Maestro Series
72
Transmission Lines
What remains of the square root can be expanded into a truncated
Taylor series
 1 R
G 
+
γ ≈ jω LC 1 + 

2
j
L
j
C
ω
ω



1
C
L
= R
+G
 + jω LC
2
L
C
so that
1
C
L
α = R
+G

L
C
2
© Amanogawa, 2006 – Digital Maestro Series
β = ω 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 + jω L
L
Z0 =
≈
G + jωC
C
and the phase velocity associated to the wave propagation is
ω
vp = ≈
β
1
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, 2006 – Digital Maestro Series
74
Transmission Lines
Again, we obtain the loss-less transmission line if we assume
R=0
G=0
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
γ = ( jω L)( jω C ) = jω LC = jβ
The loss-less line does not dissipate power, because α = 0.
© Amanogawa, 2006 – Digital Maestro Series
75
Transmission Lines
For all cases, the line impedance was defined as
V (d)
1 + Γ (d)
Z (d) =
= Z0
I (d)
1 − Γ (d)
By including the appropriate generalized reflection coefficient, we
can derive alternative expressions of the line impedance:
A) Loss-less line
1 + Γ R e−2 jβd
Z R + jZ 0 tan(β d)
Z (d) = Z 0
= Z0
−2 jβd
jZ R tan(β d) + Z 0
1 − Γ Re
B) Lossy line (including low-loss)
1 + Γ R e −2 γ d
Z R + Z 0 tanh( γ d)
Z (d) = Z 0
= Z0
−2 γ d
Z R tanh( γ d) + Z 0
1 − Γ Re
© Amanogawa, 2006 – 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, 2006 – 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
ZG
VG
Vin
Generator
© Amanogawa, 2006 – Digital Maestro Series
Zin
Line
Zin
Vin = VG
ZG + Zin
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 the general expression
{
}
1
⟨ P(d=0 , t ) ⟩ = Re V (0) I * (0)
2
 +
1
1
= Re V (1 + Γ R ) * V + (1 − Γ 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 ⟩ = ⟨ P in ⟩ − ⟨ P (d = 0 , t ) ⟩
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. Remember that the internal impedance
of the generator dissipates part of the total power generated.
© Amanogawa, 2006 – Digital Maestro Series
79
Transmission Lines
It is instructive to develop further the general expression for the
time-average power at the load, using Z0=R0+jX0 for the
characteristic impedance, so that
R0 + jX 0 R0 + jX 0
= *
=
= 2
*
2
Z0 Z0 Z0
R0 + X 02
Z0
1
Z0
Alternatively, one may simplify the analysis by introducing the line
characteristic admittance
1
Y0 =
= G0 + jB0
Z0
It may be more convenient to deal with the complex admittance at
the numerator of the power expression, rather than the complex
characteristic impedance at the denominator.
© Amanogawa, 2006 – Digital Maestro Series
80
Transmission Lines









*
0
⟨ P(d=0, t ) ⟩ = 1 Re V + 1+Γ R  1 V + 1−Γ

Z

2
=
V+
2 Z0
=
V+
2 Z0
=
V+
2 Z0
=
V+
2 Z0
2
2





Re R0 + jX
2


0








R
+
jX
Re
0
0 
2
2












1+ Re Γ + j Im Γ





R



2
2





2






2

R0 − R0 Γ R − 2 X 0 Im Γ
© Amanogawa, 2006 – Digital Maestro Series
 
R  






=




1− Re Γ


2
2






1−   Re Γ   +  Im Γ    +






 R  
 R   










Re
+
+
2Im
R
jX
j
 1− Γ
Γ
R
0
0
2

* 
 
R  






 

R   

 





R
+ j Im Γ




  
R   



j 2Im Γ R  

 

 
=


R 

81
Transmission Lines
Equivalently, using the complex characteristic admittance:








*
0

⟨ P(d=0, t ) ⟩ = 1 Re V + 1+Γ R  Y V + 1−Γ



2
=
=
=
=
V+
2
2
2
2












1+ Re Γ + j Im Γ





R



2






 
R  






=




1− Re Γ


2
2




 




1−   Re Γ
 R   +  Im Γ R    +




 
 













2
Re G0 − jB0  1− Γ R + j 2Im Γ
2
V+







Re G0 − jB0 
2
V+


0
Re G0 − jB
2
V+





* 
 
R  

2

G0 − G0 Γ R + 2B0 Im Γ
© Amanogawa, 2006 – Digital Maestro Series





 

R   

 





R
+ j Im Γ




  
R   




 
j 2Im Γ R  

 

 
=


R 

82
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
Generator
Load
ZG
VG
ZG = Z*in
© Amanogawa, 2006 – Digital Maestro Series
Transmission line
ZR
for maximum power transfer
83
Transmission Lines
The characteristic impedance of the loss-less line is real and we
can express the power flow, anywhere on the line, as
1
⟨ P (d , t ) ⟩ = Re{ V (d) I * (d) }
2
1
 + jβ d
1 + Γ R e− j 2β d
= Re V e
2

(
)
(
1
(V + )* e− jβ d 1 − Γ R e− j 2β d
Z0
1
+2
=
V
2Z0
Incident wave
) 
*
1
+2
−
V
ΓR 2
2Z0
Reflected wave
This result is valid for any location, including the input and the load,
since the transmission line does not absorb any power.
© Amanogawa, 2006 – Digital Maestro Series
84
Transmission Lines
In the case of low-loss lines, the characteristic impedance is again
real, but the time-average power flow is position dependent
because the line absorbs power.
{
{
}
(
1
⟨ P (d , t ) ⟩ = Re V (d) I * (d)
2
1
= Re V + eα d e jβ d 1 + Γ R e−2 γ d
2
1
(V + )* eα d e− jβ d 1 − Γ R e−2 γ d
Z0
)
(
) 
*
1
1
+ 2 2α d
+ 2 −2α d
=
V
e
−
V
e
ΓR 2
2Z0
2Z0
Incident wave
© Amanogawa, 2006 – Digital Maestro Series
Reflected wave
85
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 + eα L
⇔
input
V + eα d
inside the line
⇔
V+
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 + Γ R e−α L
⇔
input
© Amanogawa, 2006 – Digital Maestro Series
V + Γ R e−α d
⇔
inside the line
V +ΓR
load
86
Transmission Lines
For a general lossy line the power flow is again position dependent.
Since the characteristic impedance is complex, the result has an
additional term involving the imaginary part of the characteristic
admittance, B0, as
{
{
}
1
⟨ P (d , t ) ⟩ = Re V (d) I * (d)
2
1
= Re V + eα d e jβ d (1 + Γ (d) )
2
}
Y0* (V + )* eα d e− jβ d (1 − Γ (d) )*
G0 + 2 2α d G0 + 2 −2α d
=
V
e
−
V
e
ΓR 2
2
2
+ B0 V
© Amanogawa, 2006 – Digital Maestro Series
+2
e2α d Im(Γ (d))
87
Transmission Lines
For the general lossy line, keep in mind that
Z 0*
R0 − jX 0 R0 − jX 0
1
Y0 =
=
=
= 2
= G0 + jB0
*
2
2
Z0 Z0 Z0
R0 + X 0
Z0
−X0
R0
=
G0 = 2
B
0
R0 + X 02
R02 + X 02
Recall that for a low-loss transmission line the characteristic
impedance is approximately real, so that
B0 ≈ 0
and
Z 0 ≈ 1 G0 ≈ R0 .
The previous result for the low-loss line can be readily recovered
from the time-average power for the general lossy line.
© Amanogawa, 2006 – Digital Maestro Series
88
Transmission Lines
To completely specify the transmission line problem, we still have
to determine the value of V+ from the input boundary condition.
¾ The load boundary condition imposes the shape of the
interference pattern of voltage and current along the line.
¾ 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 + Γ (L)
Zin = Z 0
1 − Γ (L)
Z R + jZ 0 tan(β L)
Zin = Z 0
jZ R tan(β L) + Z 0
loss - less line
Z R + Z 0 tanh( γ L)
Zin = Z 0
Z R tanh( γ L) + Z 0
lossy line
© Amanogawa, 2006 – Digital Maestro Series
89
Transmission Lines
For a loss-less transmission line:
V (L) = V + e jβ L [1 + Γ (L) ] = V + e jβ L (1 + Γ R e− j 2β L )
⇒
Zin
1
V = VG
ZG + Zin e jβ L (1 + Γ R e− j 2β L )
+
For a lossy transmission line:
(
V (L) = V + e γ L [1 + Γ (L) ] = V + e γ L 1 + Γ R e−2 γ d
⇒
)
Zin
1
V = VG
ZG + Zin e γ L (1 + Γ R e−2 γ L )
+
© Amanogawa, 2006 – Digital Maestro Series
90
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
Γ1
Z01
© Amanogawa, 2006 – Digital Maestro Series
Zin
Zin − Z01
Γ1 =
Zin + Z01
91
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 jβ 0 (1 + Γ R e− j 2β 0 )
= V + (1 + Γ R ) = 0
⇒
© Amanogawa, 2006 – Digital Maestro Series
Γ R = −1
92
Transmission Lines
Since
ΓR =
V−
V+
⇒ V − = −V +
We can write the line voltage phasor as
V (d) = V + e jβ d + V − e− jβ d
= V + e jβ d − V + e− jβ d
= V + ( e jβ d − e− jβ d )
= 2 jV + sin(β d)
© Amanogawa, 2006 – Digital Maestro Series
93
Transmission Lines
For the line current phasor we have
1
(V + e jβ d − V − e− jβ d )
I (d) =
Z0
1
(V + e jβ d + V + e− jβ d )
=
Z0
V + jβ d
(e
=
+ e− jβ d )
Z0
2V +
cos(β d)
=
Z0
The line impedance is given by
V (d)
2 jV + sin(β d)
=
= jZ0 tan(β d)
Z(d) =
+
I (d) 2V cos(β d) / Z0
© Amanogawa, 2006 – Digital Maestro Series
94
Transmission Lines
The time-dependent values of voltage and current are obtained as
V (d, t ) = Re[V (d) e jω t ] = Re[2 j | V + | e jθ sin(β d) e jω t ]
= 2| V + |sin(β d) ⋅ Re[ j e j (ω t +θ) ]
= 2| V + |sin(β d) ⋅ Re[ j cos(ω t + θ) − sin(ω t + θ)]
= −2| V + |sin(β d) sin(ω t + θ)
I (d, t ) = Re[ I (d) e jω t ] = Re[2 | V + | e jθ cos(β d) e jω t ] / Z0
= 2| V + |cos(β d) ⋅ Re[ e j (ω t +θ) ] / Z0
= 2| V + |cos(β d) ⋅ Re[ (cos(ω t + θ) + j sin(ω t + θ)] / Z0
| V+ |
=2
cos(β d) cos(ω t + θ)
Z0
© Amanogawa, 2006 – Digital Maestro Series
95
Transmission Lines
The time-dependent power is given by
P(d, t ) = V (d, t ) ⋅ I (d, t )
| V + |2
=−4
sin(β d) cos(β d) sin(ω t + θ) cos(ω t + θ)
Z0
| V + |2
=−
sin(2β d) sin (2ω t + 2θ)
Z0
and the corresponding time-average power is
1 T
< P(d, t ) > = ∫ P(d, t ) dt
T 0
| V + |2
1 T
=−
sin(2β d) ∫ sin (2ω t + 2θ) = 0
Z0
T 0
© Amanogawa, 2006 – Digital Maestro Series
96
Transmission Lines
ZR → ∞ (OPEN CIRCUIT)
ZR → ∞
Z0
The load boundary condition due to the open circuit is I (0) = 0
V + jβ 0
⇒ I (d = 0) =
e (1 − Γ R e− j 2β 0 )
Z0
V+
=
(1 − Γ R ) = 0
Z0
⇒
© Amanogawa, 2006 – Digital Maestro Series
ΓR = 1
97
Transmission Lines
Since
ΓR =
V−
V+
⇒ V− = V+
We can write the line current phasor as
1
I (d) =
(V + e jβ d − V − e− jβ d )
Z0
1
=
(V + e jβ d − V + e− jβ d )
Z0
V + jβ d − jβ d
2 jV +
=
−e
(e
)=
sin(β d)
Z0
Z0
© Amanogawa, 2006 – Digital Maestro Series
98
Transmission Lines
For the line voltage phasor we have
V (d) = (V + e jβ d + V − e− jβ d )
= (V + e jβ d + V + e− jβ d )
= V + ( e jβ d + e− jβ d )
= 2V + cos(β d)
The line impedance is given by
V (d)
2V + cos(β d)
Z0
Z(d) =
=
=−j
I (d) 2 jV + sin(β d) / Z0
tan(β d)
© Amanogawa, 2006 – Digital Maestro Series
99
Transmission Lines
The time-dependent values of voltage and current are obtained as
V (d, t ) = Re[V (d) e jω t ] = Re[2 | V + | e jθ cos(β d) e jω t ]
= 2| V + |cos(β d) ⋅ Re[ e j(ω t +θ) ]
= 2| V + |cos(β d) ⋅ Re[ (cos(ω t + θ) + j sin(ω t + θ)]
= 2| V + |cos(β d) cos(ω t + θ)
I (d, t ) = Re[ I (d) e jω t ] = Re[2 j | V + | e jθ sin(β d) e jω t ] / Z0
= 2| V + |sin(β d) ⋅ Re[ j e j(ω t +θ) ] / Z0
= 2| V + |sin(β d) ⋅ Re[ j cos(ω t + θ) − sin(ω t + θ)] / Z0
| V+ |
= −2
sin(β d) sin(ω t + θ)
Z0
© Amanogawa, 2006 – Digital Maestro Series
100
Transmission Lines
The time-dependent power is given by
P(d, t ) = V (d, t ) ⋅ I (d, t ) =
| V + |2
=−4
cos(β d) sin(β d) cos(ω t + θ) sin(ω t + θ)
Z0
| V + |2
=−
sin(2β d) sin (2ω t + 2θ)
Z0
and the corresponding time-average power is
1 T
< P(d, t ) > = ∫ P(d, t ) dt
T 0
| V + |2
1 T
=−
sin(2β d) ∫ sin (2ω t + 2θ) = 0
Z0
T 0
© Amanogawa, 2006 – Digital Maestro Series
101
Transmission Lines
ZR = Z0 (MATCHED LOAD)
Z0
ZR = Z0
The reflection coefficient for a matched load is
ZR − Z0 Z0 − Z0
ΓR =
=
=0
ZR + Z0 Z0 + Z0
no reflection!
The line voltage and line current phasors are
V (d) = V + e jβ d (1 + Γ R e−2 jβ d ) = V + e jβ d
+
V + jβ d
V
(1 − Γ R e−2 jβ d ) =
I( d ) =
e
e jβ d
Z0
Z0
© Amanogawa, 2006 – Digital Maestro Series
102
Transmission Lines
The line impedance is independent of position and equal to the
characteristic impedance of the line
V (d) V + e jβ d
Z(d) =
=
= Z0
+
I (d) V jβ d
e
Z0
The time-dependent voltage and current are
V (d, t ) = Re[| V + | e jθ e jβ d e jω t ]
= | V + | ⋅ Re[e j(ω t +β d +θ) ] =| V + | cos(ω t + β d + θ)
I (d, t ) = Re[| V + | e jθ e jβ d e jω t ] / Z0
+
| V+ |
|
V
|
j (ω t +β d +θ)
=
⋅ Re[e
]=
cos(ω t + β d + θ)
Z0
Z0
© Amanogawa, 2006 – Digital Maestro Series
103
Transmission Lines
The time-dependent power is
+
|
|
V
P(d, t ) = | V + | cos(ω t + β d + θ)
cos(ω t + β d + θ)
Z0
| V + |2
cos2 (ω t + β d + θ)
=
Z0
and the time average power absorbed by the load is
1 t | V + |2
< P(d) > = ∫
cos2 (ω t + β d + θ) dt
T 0 Z0
| V + |2
=
2 Z0
© Amanogawa, 2006 – Digital Maestro Series
104
Transmission Lines
ZR = jX (PURE REACTANCE)
ZR = j X
Z0
The reflection coefficient for a purely reactive load is
ZR − Z0 jX − Z0
ΓR =
=
=
ZR + Z0 jX + Z0
( jX − Z0 )( jX − Z0 )
=
=
( jX + Z0 )( jX − Z0 )
© Amanogawa, 2006 – Digital Maestro Series
X 2 − Z02
Z02 + X 2
+2j
XZ0
Z02 + X 2
105
Transmission Lines
In polar form
Γ R = Γ R exp( jθ)
where
ΓR =
(
(
)
) (
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 
θ = 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 power and
then returns it to the line without dissipation.
© Amanogawa, 2006 – Digital Maestro Series
106
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
 2π f 
 2π 
Zin = j Z0 tan ( β L ) = j Z0 tan  L  = j Z0 tan 
L


v
 λ 
p


For a specified frequency f, any reactance value (positive or
negative!) can be obtained by changing the length of the line from 0
to λ/2. An inductance is realized for L < λ/4 (positive tangent)
while a capacitance is realized for λ/4 < L < λ/2 (negative tangent).
When L = 0 and L = λ/2 the tangent is zero, and the input
impedance corresponds to a short circuit. However, when L = λ/4
the tangent is infinite and the input impedance corresponds to an
open circuit.
© Amarcord, 2006 – Digital Maestro Series
107
Transmission Lines
Since the tangent function is periodic, the same impedance
behavior of the impedance will repeat identically for each additional
line increment of length λ/2. A similar periodic behavior is also
obtained when the length of the line is fixed and the frequency of
operation is changed.
At zero frequency (infinite wavelength), the short circuited line
behaves as a short circuit for any line length. When the frequency
is increased, the wavelength shortens and one obtains an
inductance for L < λ/4 and a capacitance for λ/4 < L < λ/2, with
an open circuit at L = λ/4 and a short circuit again at L = λ/2.
Note that the frequency behavior of lumped elements is very
different. Consider an ideal inductor with inductance L assumed to
be constant with frequency, for simplicity. At zero frequency the
inductor also behaves as a short circuit, but the reactance varies
monotonically and linearly with frequency as
X = ωL (always an inductance)
© Amarcord, 2006 – Digital Maestro Series
108
Transmission Lines
Short circuited transmission line – Fixed frequency
L
Z in = 0
L= 0
0< L<
L=
λ
4
λ
2
4
λ
2
λ
k p
Im Z in > 0
in d u c ta n c e
Z in → ∞
o p e n c irc u it
k p
c a p a c ita n c e
Im Z in < 0
Z in = 0
2
< L<
L=
4
λ
< L<
L=
λ
3λ
4
3λ
4
3λ
< L< λ
4
s h o rt c irc u it
k p
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
k p
Im Z in < 0
c a p a c ita n c e
…
© Amarcord, 2006 – Digital Maestro Series
109
Z(L)/Zo = j tan(β 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
π/(2β)= λ/4
200
π/β =λ/2
300
3π/(2β)= 3λ/4
400
2π/β= λ
500
5π/(2β)= 5λ/4
θ
L
[deg]
Line Length L
© Amarcord, 2006 – Digital Maestro Series
110
Z(L)/Zo = j tan(β 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)
θ
[deg]
f
Frequency of operation
© Amarcord, 2006 – Digital Maestro Series
111
Transmission Lines
For a transmission line of length L terminated by an open circuit,
the input impedance is again purely imaginary
Z0
Zin = − j
=−j
tan ( β L )
Z0
Z0
=−j
 2π 
 2π f 
tan  L 
tan 
L
λ




v
 p 
We can also use the open circuited line to realize any reactance, but
starting from a capacitive value when the line length is very short.
Note once again that the frequency behavior of a corresponding
lumped element is different. Consider an ideal capacitor with
capacitance C assumed to be constant with frequency. At zero
frequency the capacitor behaves as an open circuit, but the
reactance varies monotonically and linearly with frequency as
1
X=
(always a capacitance)
ωC
© Amarcord, 2006 – Digital Maestro Series
112
Transmission Lines
Open circuit transmission line – Fixed frequency
L
L= 0
0< L<
L=
λ
4
λ
2
4
λ
2
λ
2
< L<
L=
4
λ
< L<
L=
λ
3λ
4
3λ
4
3λ
< L< λ
4
Z in → ∞
o p e n c irc u it
Im Z in < 0
k p
c a p a c ita n c e
Z in = 0
s h o rt c irc u it
k p
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
k p
c a p a c ita n c e
Z in = 0
s h o rt c irc u it
k p
Im Z in > 0
in d u c ta n c e
…
© Amarcord, 2006 – Digital Maestro Series
113
Normalized Input Impedance
Z(L)/ Zo = - j cotan(β L)
Transmission Lines
Impedance of an open circuited transmission line
(fixed frequency, variable length)
40
30
20
inductive
10
inductive
inductive
0
-10
capacitive
capacitive
capacitive
-20
-30
-40
0
0
100
π/(2β) = λ/4
200
π/β = λ/2
300
3π/(2β) = 3λ/4
400
2π/β = 2λ
500
5π/(2β) = 5λ/4
θ
[deg]
L
Line Length L
© Amarcord, 2006 – Digital Maestro Series
114
Normalized Input Impedance
Z(L)/ Zo = - j cotan(β L)
Transmission Lines
Impedance of an open circuited transmission line
(fixed length, variable frequency)
40
30
20
inductive
10
inductive
inductive
0
-10
capacitive
capacitive
capacitive
-20
-30
-40
0
0
100
vp / (4L)
200
vp / (2L)
300
3vp / (4L)
400
vp / L
500
5vp / (4L)
θ
[deg]
f
Frequency of operation
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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
V
Zin1
Zin1 = j Z0 tan ( β L 1 )
© Amarcord, 2006 – Digital Maestro Series
L2
Z0
short
circuit
Zin2
Zin2 = j Z0 tan ( β L 2 )
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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 ( β r L1 ) j Z0 tan ( β r L 2 )
or
 ωr

 ωr

tan  L1  = − tan  L 2 
 vp

 vp





with
2π ωr
βr =
=
λ r vp
Since the tangent is a periodic function, there is a multiplicity of
possible resonant angular frequencies ωr that satisfy the condition
above. The values can be found by using a numerical procedure to
solve the trascendental equation above.
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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.
Generator
Switch
Load
RG
t=0
VG
© Amanogawa, 2006 – Digital Maestro Series
Transmission line
RR
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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
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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
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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
ΓR =
=
+ R +Z
V
R
0
is the load reflection coefficient.
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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 an additional forward wave with
associated voltage
+
− RG − Z0
V2 = V
RG + Z0
Remember that the ideal voltage source part of the generator will
behave simply as a short for the reflected wave attempting to exit
the line from the input.
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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.
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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.
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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 + Γ(d) )

Loss - less line 
V+
⋅ (1 − Γ(d) )
 I (d) =
Z0

 V (d) = V + eα d ⋅ (1 + Γ ( d ) )

Lossy line 
V + eα d
⋅ (1 − Γ ( d ) )
 I (d) =
Z0

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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, due to the
interference of incident and refelected wave.
The patterns present a succession of maxima and minima which
repeat in space with a period of length λ/2, due to constructive or
destructive interference between forward and reflected waves. The
patterns for a loss-less line are exactly periodic in space, repeating
with a λ/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.
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Transmission Lines
We limit now our discussion to the loss-less transmission line case
where the generalized reflection coefficient varies as
Γ(d) = Γ R exp ( − j 2β d ) = Γ R exp ( jφ ) exp ( − j 2β d )
Note that the magnitude of an exponential with imaginary argument
is always unity
exp ( jφ ) exp ( − j 2β d ) = 1
In a loss-less line it is always true that, for any line location,
Γ(d) = Γ R
When d increases, moving from load to generator, the generalized
reflection coefficient on the complex plane moves clockwise on a
circle with radius |ΓR| and is identified by the angle φ - 2β d .
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Transmission Lines
The voltage standing wave pattern has a maximum at locations
where the generalized reflection coefficient is real and positive
Γ(d) = Γ R
exp ( jφ ) exp ( − j 2β d ) = 1
⇒
φ − 2β d = 2 nπ
At these locations we have
1 + Γ(d) = 1 + Γ R
⇒ Vmax = V (d max ) = V + ⋅ (1 + Γ R )
The phase angle φ - 2β d changes by an amount 2π, when moving
from one maximum to the next. This corresponds to a distance
between successive maxima of λ/2.
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Transmission Lines
The voltage standing wave pattern has a minimum at locations
where the generalized reflection coefficient is real and negative
Γ(d) = − Γ R
exp ( jφ ) exp ( − j 2β d ) = −1
⇒
φ − 2β d = ( 2 n + 1 ) π
At these locations we have
1 + Γ(d) = 1 − Γ R
⇒ Vmin = V (d min ) = V + ⋅ ( 1 − Γ R )
Also when moving from one minimum to the next, the phase angle
φ - 2β d changes by an amount 2π. This again corresponds to a
distance between successive minima of λ/2.
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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.
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Transmission Lines
Since in all possible cases
Γ (d) ≤ 1
the voltage standing wave pattern
V (d) = V + ⋅ (1 + Γ(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
Γ (d) = 1
since the load cannot consume any power. The voltage standing
wave pattern in these cases is characterized by
Vmax = 2 V +
© Amanogawa, 2006 -–Digital Maestro Series
and
Vmin = 0 .
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Transmission Lines
The quantity 1 + Γ(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 Γ(d) is a complex number such that |Γ(d)| ≤ 1.
Im
Γ
1+Γ
1
© Amanogawa, 2006 -–Digital Maestro Series
Re
132
Transmission Lines
We can use a geometric construction to visualize the behavior of
the voltage standing wave pattern
V (d) = V + ⋅ (1 + Γ(d) )
simply by looking at a vector plot of |(1 + Γ (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( Γ )
1+ΓR
φ
ΓR
1
© Amanogawa, 2006 -–Digital Maestro Series
Re ( Γ )
133
Transmission Lines
Im( Γ )
Maximum of
voltage standing
wave pattern
φ
1+Γ(d)
ΓR
1
Γ(d)
2 β dmax
Re ( Γ )
∠ Γ ( d ) = φ − 2β d max = 0
Im( Γ )
Minimum of
voltage standing
wave pattern
φ
1+Γ(d)
1
∠ Γ ( d ) = φ − 2β d min = −π
© Amanogawa, 2006 -–Digital Maestro Series
Γ(d)
ΓR
Re ( Γ )
2 β dmin
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Transmission Lines
The voltage standing wave ratio (VSWR) is an indicator of load
matching which is widely used in engineering applications
Vmax 1 + Γ R
VSWR =
=
Vmin 1 − Γ R
When the load is perfectly matched to the transmission line
ΓR = 0
⇒
VSWR = 1
When the load is a short circuit, an open circuit or a pure reactance
ΓR = 1
⇒
VSWR → ∞
We have the following useful relation
VSWR − 1
ΓR =
VSWR + 1
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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 ( Γ R ) = Im 
>0

 ZR + Z0 
Im Γ
Re Γ
1
The first maximum of the voltage standing wave pattern is closest
to the load, at location
∠ Γ ( d ) = φ − 2β d max = 0
© Amanogawa, 2006 -–Digital Maestro Series
⇒
φ
d max =
λ
4π
136
Transmission Lines
‰
Load with capacitive reactance
Im ( ZR ) < 0
⇒
 ZR − Z0 
Im ( Γ R ) = Im 
<0

 ZR + Z0 
The load reflection coefficient
is in this part of the domain
Im(Γ)
1
Re(Γ)
The first minimum of the voltage standing wave pattern is closest to
the load, at location
∠ Γ ( d ) = φ − 2β d min = π
© Amanogawa, 2006 -–Digital Maestro Series
⇒
π−φ
λ
d min =
4π
137
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
ΓR =
VSWR + 1
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Transmission Lines
‰
STEP 2: The distance from the load of the first maximum or
minimum gives the phase φ of the load reflection coefficient.
Vmax
|V|
For an inductive reactance, a voltage
maximum is closest to the load and
4π
d max
φ = 2β d max =
λ
Vmin
dmax
0
|V|
Vmax
For a capacitive reactance, a voltage
minimum is closest to the load and
4π
φ = −π + 2β d min = −π + d min
λ
Vmin
dmin
0
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Transmission Lines
‰
STEP 3: The load impedance is obtained by inverting the
expression for the reflection coefficient
ZR − Z0
Γ R = Γ R exp ( jφ ) =
ZR + Z0
⇒
© Amanogawa, 2006 -–Digital Maestro Series
1 + Γ R exp ( jφ )
ZR = Z0
1 − Γ R exp ( jφ )
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