-
Waves on Transmission Lines
4 Waves on Transmission Lines
After completing this section, students should be able to do the
following.
• Name several transmission lines.
• Explain the difference between phase shift and time delay of a
sinusoidalsignal.
• Evaluate whether the transmission line theory or circuit
theory has to beused based on the length of the line and the
frequency
• Students will explain the difference between lumped and
distributed circuitelements.
• Derive the voltage on a transmission line from a consideration
of a time-delay due to the finite speed of signals in a
transmission-line circuit.
• Explain parts of propagation constant and what they
represent.
• Calculate the phase and attenuation constant for specific
transmissionlines.
• Identify whether the wave travels in the positive or negative
direction fromthe equation of a wave.
• Describe how signal flows on a transmission line
• Recognize and explain transmission line equivalent circuit
model
• Derive the equations for voltage and current waves on a
transmission linefrom the equivalent circuit model.
• Describe forward and reflected wave on a transmission
line.
• Sketch forward and reflected wave as a function of distance,
and explainhow the graph changes as time passes.
• Derive phasor form of voltage and current wave from/to the
time-domainform.
• Describe what wavelength represents on a graph of a wave vs.
distance
• Explain how the wavelength is similar and different to waves
period?
• Derive and calculate the transmission line impedance and
reflection coef-ficient.
• Relate reflection coefficient to impedance.
51
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Waves on Transmission Lines
• Derive and calculate the input impedance of a transmission
line
• Calculate and visualize phasors of forward going voltage and
current wavesat various points on a transmission line.
52
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Types of Transmission Lines
4.1 Types of Transmission Lines
Any wire, cable, or line that guides energy from one point to
another is atransmission line. Whenever we make a circuit on a
breadboard, every wireattached forms a transmission line with the
ground wire. Whether we see thepropagation (transmission line)
effects on the line depends on the line length,and the frequency of
the signals used. At lower frequencies or short line lengths,we do
not see any difference between the signal’s phase at the generator
and theload, whereas we do at higher frequencies.
Types of transmission lines
(a) Coaxial Cable, Figure 19
(b) Microstrip, Figure 20
(c) Stripline, Figure 21
(d) Coplanar Waveguide, Figure 22
(e) Two-wire line, Figure 23
(f) Parallel Plate Waveguide, Figure 24
(g) Rectangular Waveguide, Figure 25
(h) Optical fiber, Figure 26
Figure 19: Coaxial Cable
53
-
Types of Transmission Lines
Figure 20: Microstrip
Figure 21: Stripline.
Figure 22: Coplanar Waveguide.
Figure 23: Two-wire line.
Figure 24: Parallel-Plate Waveguide.
Propagation modes on a transmission line
Coax, two-wire line, transmission line support TEM waves. Waves
on microstriplines can be approximated as TEM up to the 30-40 GHz
(unshielded), up to140 GHz shielded.
54
-
Types of Transmission Lines
Figure 25: Rectangular Waveguide.
Figure 26: Optical Fiber.
(a) Transversal Electro-Magnetic Field (TEM). Electric (E), and
Magnetic(M) fields are entirely transversal to the direction of
propagation
(b) Transversal Electric field (TE), Transversal Magnetic Field
(TM), M or Efield is in the direction of propagation
Transmission lines we will discuss in this course carry TEM
fields.
55
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Wave Equation
4.2 Wave Equation
Wave equation on a transmission line
In this section, we will derive the expression for voltage and
current on a trans-mission line. This expression will have two
variables, time t, and space z. So far,we have only seen voltages
and currents as a function of time, because all circuitelements
seen so far are lumped elements. In distributed systems, we want
toderive the equations for voltage and current for the case when
the transmissionline is longer than the fraction of a wavelength.
To make sure that we do notencounter any transmission line effects
to start with, we can look at the pieceof a transmission line that
is much smaller than the fraction of a wavelength.In other words,
we cut the transmission line into small pieces to make surethere
are no transmission line effects, as the pieces are shorter than
the fractionof a wavelength. We then represent each piece with an
equivalent circuit, asshown in Figure 29 (a). To derive expressions
for current and voltage on thetransmission line, we will use the
following five-step plan
(a) Look at an infinitesimal length of a transmission line
∆z.
(b) Represent that piece with an equivalent circuit.
(c) Write KCL, KVL for the piece in the time domain (we get
differentialequations)
(d) Apply phasors (equations become linear)
(e) Solve the linear system of equations to get the expression
for the voltageand current on the transmission line as a function
of z.
Look at a small piece of a transmission line and represented it
with an equivalentcircuit. What is modeled by the circuit
elements?
Write KVL and KCL equations for the circuit above.
KVL
−v(z, t) +R∆z i(z, t) + L∆z ∂i(z, t)∂t
+ v(z + ∆z, t) = 0
KCL
i(z, t) = i(z + ∆z) + iCG(z + ∆z, t)
i(z, t) = i(z + ∆z) +G∆z v(z + ∆z, t) + C ∆z∂v(z + ∆z, t)
∂t
56
-
Wave Equation
Figure 27: Coaxial cable is cut in short pieces.
Figure 28: Equivalent circuit of a section of transmission
line.
Figure 29: Equivalent circuit of transmission line.
57
-
Wave Equation
Rearrange the KCL and KVL Equations 79, 83 and divide them with
∆z. Equa-tions 80, 84. let ∆z → 0 and recognize the definition of
the derivative Equations,81, 85.
KVL
−(v(z + ∆z, t)− v(z, t)) = R∆z i(z, t) + L∆z ∂i(z, t)∂t
(79)
−v(z + ∆z, t)− v(z, t)∆z
= R i(z, t) + L∂i(z, t)
∂t(80)
lim∆z→0
{−v(z + ∆z, t)− v(z, t)∆z
} = lim∆z→0
{R i(z, t) + L ∂i(z, t)∂t
} (81)
−v(z, t)∂z
= Ri(z, t) + L∂i(z, t)
∂t(82)
KCL
−(i(z + ∆z, t)− i(z, t)) = G∆z v(z + ∆z, t) + C ∆z ∂v(z + ∆z,
t)∂t
(83)
− i(z + ∆z, t)− i(z, t)∆z
= Gv(z + ∆z, t) + C∂v(z + ∆z, t)
∂t(84)
lim∆z→0
{− i(z + ∆z, t)− i(z, t)∆z
} = lim∆z→0
{Gv(z + ∆z, t) + C ∂v(z + ∆z, t)∂t
} (85)
− i(z, t)∂z
= Gv(z + ∆z, t) + C∂v(z + ∆z, t)
∂t(86)
We just derived Telegrapher’s equations in time-domain:
−v(z, t)∂z
= R i(z, t) + L∂i(z, t)
∂t
− i(z, t)∂z
= Gv(z + ∆z, t) + C∂v(z + ∆z, t)
∂t
Telegrapher’s equations are two differential equations with two
unknowns, i(z, t),v(z, t). It is not impossible to solve them;
however, we would prefer to have lin-ear algebraic equations. We
then express time-domain variables as phasors.
v(z, t) = Re{Ṽ (z)ejωt}i(z, t) = Re{Ĩ(z)ejωt}
Ṽ (z), Ĩ(z) are the voltage, and current anywhere on the line,
and they dependon the position on the line z. The Telegrapher’s
equations in phasor form are
58
-
Wave Equation
−∂Ṽ (z)∂z
= (R+ jωL)Ĩ(z) (87)
−∂Ĩ(z)∂z
= (G+ jωC)Ṽ (z) (88)
Two equations, two unknowns. To solve these equations, we first
take a deriva-tive of both equations with respect to z.
−∂2Ṽ (z)
∂z2= (R+ jωL)
∂Ĩ(z)
∂z(89)
−∂2Ĩ(z)
∂z2= (G+ jωC)
∂Ṽ (z)
∂z(90)
Rearange the previous equations:
− 1(R+ jωL)
∂Ĩ(z)
∂z=∂2Ṽ (z)
∂z2(91)
− 1(G+ jωC)
∂Ṽ (z)
∂z=∂2Ĩ(z)
∂z2(92)
Substitute Eq.91 into Eq.88 and Eq.92 into Eq.87 and we get
−∂2Ṽ (z)
∂z2= (G+ jωC)(R+ jωL)Ṽ (z) (93)
−∂2Ĩ(z)
∂z2= (G+ jωC)(R+ jωL)Ĩ(z) (94)
Or if we rearrange
∂2Ṽ (z)
∂z2− (G+ jωC)(R+ jωL)Ṽ (z) = 0 (95)
∂2I(z)
∂z2− (G+ jωC)(R+ jωL)I(z) = 0 (96)
The above Equations 95-96 are called wave equations, and they
represent currentand voltage wave on a transmission line. γ =
(G+jωC)(R+jωL) is the complexpropagation constant. This constant
has a real and an imaginary part.
γ = α+ jβ
59
-
Wave Equation
where α is the attenuation constant and β is the phase
constant.
α = Re{√
(G+ jωC)(R+ jωL)}β = Im{
√(G+ jωC)(R+ jωL)}
We can now write wave equations as:
∂2Ṽ (z)
∂z2− γṼ (z) = 0 (97)
∂2I(z)
∂z2− γI(z) = 0 (98)
The general solution of the second order differential equations
with constantcoefficients Equations 97 - 98 is:
Ṽ (z) = Ṽ +0 e−γz + Ṽ −0 e
γz
Ĩ(z) = Ĩ+0 e−γz + Ĩ−0 e
γz
Where Ṽ (z)+ = Ṽ +0 e−γz represents the forward-going voltage
wave, Ṽ (z)− =
Ṽ −0 eγz represents the reflected voltage wave, Ĩ(z)+ = Ĩ+0
e
−γz represents theforward going current wave, and Ĩ(z)− = Ĩ−0
e
γz represents the reflected currentwave. We will see later that
Ṽ +0 is the forward-going voltage wave at the load,Ṽ −0 is the
reflected voltage wave at the load, Ĩ
+0 is the forward going current at
the load, and Ĩ−0 reflected current at the load.
In the next several sections, we will look at how to find the
constants β, Ṽ +0 ,Ṽ −0 , Ĩ
+0 , Ĩ
−0 . In order to find the constants, we will introduce the
concepts
of transmission line impedance Z0, reflection coefficient Γ(z),
input impedanceZin.
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Visualization of waves on lossless transmission lines
4.3 Visualization of waves on lossless transmissionlines
Ṽ (z) = Ṽ +0 e−γz + Ṽ −0 e
γz
Ĩ(z) = Ĩ+0 e−γz + Ĩ−0 e
γz
In this equation Ṽ (z) is the total voltage anywhere on the
line (at any point z),
Ĩ(z) is the total current anywhere on the line (at any point
z), Ṽ0+
and Ṽ0−
arethe phasors of forward and reflected voltage waves at the
load (where z=0), and
Ĩ0+
and Ĩ0−
are the phasors of forward and reflected current wave at the
load(where z=0).These voltages and currents are also phasors and
have a constant
magnitude and phase in a specific circuit, for example Ṽ0+
= |Ṽ0+|eΦ = 4e25
0
,
and Ĩ0+
= |Ĩ0+|eΦ = 5e−40
0
. We can get the time-domain expression for thecurrent and
voltage on the transmission line by multiplying the phasor of
thevoltage and current with ejωt and taking the real part of
it.
v(t) = Re{(Ṽ +0 e(−α−jβ)z + Ṽ−0 e
(α+jβ)z)ejωt}v(t) = |Ṽ +0 |e−αz cos(ωt− βz + ∠Ṽ
+0 ) + |Ṽ
−0 |eαz cos(ωt+ βz + ∠Ṽ
−0 ) (99)
If the signs of the ωt and βz terms are oposite the wave moves
in the forward+z direction. If the signs of ωt and βz are the same,
the wave moves in the −zdirection.
In the next several sections, we will look at how to find the
constants β, Ṽ +0 ,Ṽ −0 , Ĩ
+0 , Ĩ
−0 . In order to find the constants, we will introduce the
concepts
of transmission line impedance Z0, reflection coefficient Γ(z),
input impedanceZin.
Example 9. We will show next that if the signs of the ωt and βz
have theopposite sign, as in Equation 100, the wave moves in the
forward +z direction.If the signs of ωt and βz are the same, as in
Equation 101, the wave moves inthe −z direction. In order to see
this, we will visualize Equations 100 and 101using Matlab code
below.
vf (t) = |Ṽ +0 |e−αz cos(ωt− βz + ∠Ṽ+0 ) (100)
vr(t) = |Ṽ −0 |eαz cos(ωt+ βz + ∠Ṽ−0 ) (101)
Explanation. Figure 30 shows forward and reflected waves on a
transmissionline. z represents the length of the line, and on the
y-axis is the magnitude of
61
-
Visualization of waves on lossless transmission lines
the voltage. The red line on both graphs is the voltage signal
at a time .1 ns. Wewould obtain Figure 30 if we had a camera that
can take a picture of the voltage,and we took the first picture at
t1 = .1 ns on the entire transmission line. Theblue dotted line on
both graphs is the same signal .1 ns later, at time t2 =.2 ns.We
see that the signal has moved to the right in 1 ns, from the
generator to theload. On the bottom graph, we see that at a time .1
ns, the red line representsthe reflected signal. The dashed blue
line shows the signal at a time .2 ns. Wesee that the signal has
moved to the left, from the load to the generator.
Figure 30: Forward (top) and reflected (bottom) waves on a
transmission line.
clear all
clc
f = 10^9;
w = 2*pi*f
c=3*10^8;
beta=2*pi*f/c;
lambda=c/f;
t1=0.1*10^(-9)
t2=0.2*10^(-9)
x=0:lambda/20:2*lambda;
y1=sin(w*t1 - beta.*x);
y2=sin(w*t2 - beta.*x);
y3=sin(w*t1 + beta.*x);
y4=sin(w*t2 + beta.*x);
62
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Visualization of waves on lossless transmission lines
subplot (2,1,1),
plot(x,y1,’r’),...
hold on
plot(x,y2,’--b’),...
hold off
subplot (2,1,2),
plot(x,y3,’r’)
hold on
plot(x,y4,’--b’)
hold off
Using Matlab code above, repeat the visualization of signals in
the previous sec-tion for a lossy transmission line. Assume that α
= 0.1 Np, and all other vari-ables are the same as in the previous
section. How do the voltages compare inthe lossy and lossless
cases?
Question 13 In the following simulation, we have three waves as
a functionof distance z. One is fixed cos(βz+ 00) with a constant
phase of 00, and for theother two signals the phase can be changed
manually by changing the slider tthat represents time. In the
simulation, β = 1 and ω = 1. This simulation isrealistic only if
time moves forward from 0 to 5. Observe how phase change ωtas the
time increases from 0 to 5, then answer the question below.
Geogebra link: https://tube.geogebra.org/m/x5q7p7jx
The sign in front of βz and ωt is opposite for the forward going
wave.
Multiple Choice:
(a) True
(b) False
63
https://tube.geogebra.org/m/x5q7p7jx
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Propagation constant and loss
4.4 Propagation constant and loss
Lossless transmission line
In many practical applications, conductor loss is low R → 0, and
dielectricleakage is low G → 0. These two conditions describe a
lossless transmissionline.
In this case, the transmission line parameters are
• Propagation constant
γ =√
(R+ jωL)(G+ jωC)
γ =√jωLjωC
γ = jω√LC = jβ
• Transmission line impedance will be defined in the next
section, but it isalso here for completeness.
Z0 =
√R+ jωL
G+ jωC
Z0 =
√jωL
jωC
Z0 =
√L
C
• Wave velocity
v =ω
β
v =ω
ω√LC
v =1√LC
• Wavelength
64
-
Propagation constant and loss
λ =2π
β
λ =2π
ω√LC
λ =2π
√ε0µ0εr
λ =c
f√εr
λ =λ0√εr
Voltage and current on lossless transmission line
On a lossless transmission line, where γ = jβ current and
voltage simplify to
Ṽ (z) = Ṽ +0 e−jβz + Ṽ −0 e
jβz
Ĩ(z) = Ĩ+0 e−jβz + Ĩ−0 e
jβz
What does it mean when we say a medium is lossyor lossless?
In a lossless medium, electromagnetic energy is not turning into
heat; there is noamplitude loss. An electromagnetic wave is heating
a lossy material; therefore,the wave’s amplitude decreases as
e−αx.
medium attenuation constant α [dB/km]
coax 60
waveguide 2
fiber-optic 0.5
In guided wave systems such as transmission lines and
waveguides, the attenu-ation of power with distance follows
approximately e−2αx. The power radiatedby an antenna falls off as
1/r2. As the distance between the source and loadincreases, there
is a specific distance at which the cable transmission is
lossierthan antenna transmission.
65
-
Propagation constant and loss
Low-Loss Transmission Line
This section is optional.
In some practical applications, losses are small, but not
negligible. R
-
Propagation constant and loss
of a modulated signal are attenuated the same amount, and there
is nodispersion on the line. When the phase constant is a linear
function of
frequency, β = const ω, then the phase velocity is a constant vp
=ω
β=
1
const, and the group velocity is also a constant, and equal to
the phase
velocity. In this case, all frequencies of the modulated signal
propagate atthe same speed, and there is no distortion of the
signal.
Transmission-line parameters R, G, C, and L
To find the complex propagation constant γ, we need the
transmission-line pa-rameters R, G, C, and L. Equations for R, G,
C, and L for a coaxial cable aregiven in the table below.
Transmission-line R G C L
Coaxial CableRsd2π
(1
a+
1
b
)2πσ
ln b/a
2πε
ln b/a
µ
2πln b/a
Where Rsd =√πfµm/σm is the resistance associated with
skin-depth. f is the
frequency of the signal, µm is the magentic permeability of
conductors, σc is theconductivity of conductors.
Example 10. Calculate capacitance per unit length of a coaxial
cable if theinner radius is 0.02 m, the outer radius is 0.06 m, the
dielectric constant isεr = 2. Use the applet below, Matlab,
Matematica, or other software that youuse.
Geogebra link: https: // tube. geogebra. org/ m/ whkrg2pu
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-
Transmission Line Impedance
4.5 Transmission Line Impedance
This section will relate the phasors of voltage and current
waves through thetransmission-line impedance.
In equations 109-110 Ṽ +0 e−γz and Ṽ −0 e
γz are the phasors of forward and reflectedgoing voltage waves
anywhere on the transmission line (for any z). Ĩ+0 e
−γz andĨ−0 e
γz are the phasors of forward and reflected current waves
anywhere on thetransmission line.
Ṽ (z) = Ṽ +0 e−γz + Ṽ −0 e
γz (109)
I(z) = Ĩ+0 e−γz + Ĩ−0 e
γz (110)
To find the transmission-line impedance, we first substitute the
voltage waveequation 109 into Telegrapher’s Equation Eq.111 to
obtain Equation 112.
−∂Ṽ (z)∂z
= (R+ jωL)I(z) (111)
γṼ +0 e−γz − γṼ −0 eγz = (R+ jωL)I(z) (112)
We now rearrange Equation 112 to find the current I(z) and
multiply throughto get Equation 113.
I(z) =γ
R+ jωL(Ṽ +0 e
−γz + Ṽ −0 eγz)
I(z) =γṼ +0
R+ jωLe−γz − γṼ
−0
R+ jωLeγz (113)
We can now compare Equation 110 for current, a solution of the
wave equation,with the Eq.113. Since both equations represent
current, and for two transcen-dental equations to be equal, the
coefficients next to exponential terms have tobe the same. When we
equate the coefficients, we get the equations below.
Ĩ+0 =γṼ +0
R+ jωL(114)
Ĩ−0 = −γṼ −0
R+ jωL(115)
Definition 13. We define the characteristic impedance of a
transmission line asthe ratio of the voltage to the current
amplitude of the forward wave as shown in
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-
Transmission Line Impedance
Equation 114, or the ratio of the voltage to the current
amplitude of the reflectedwave as shown in Equation 115.
Z0 =Ṽ +0Ĩ+0
=R+ jωL
γ(116)
Z0 = −Ṽ −0Ĩ−0
=R+ jωL
γ(117)
We can further simplify Equations 116-117 to obtain the final
Equation 118for the transmission line impedance. This equation is
valid for both lossy andlossless transmission lines.
Z0 =Ṽ +0Ĩ+0
Z0 =R+ jωL
γ
Z0 =
√R+ jωL
G+ jωC
For lossless transmission line, where R→ 0 and G→ 0, the
equation simplifiesto
Z0 =
√L
C(118)
Equations for voltage and current on a transmission-line
Using the definition of transmission-line impedance Z0, we can
now simplify theEquations 109-110 for voltage and current on the
transmission line, by replacingthe currents Ĩ+0 = Z0/Ṽ
+0 , and Ĩ
−0 = −Z0/Ṽ
−0 .
Ṽ (z) = Ṽ +0 e−jβz + Ṽ −0 e
jβz (119)
I(z) =Ṽ +0Z0
e−jβz − Ṽ−0
Z0ejβz (120)
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-
Reflection Coefficient
4.6 Reflection Coefficient
In this section, we will derive the equation for the reflection
coefficient. Thereflection coefficient relates the forward-going
voltage with reflected voltage.
Reflection coefficient at the load
Equations 121-122 represent the voltage and current on a
lossless transmissionline shown in Figure 31.
Ṽ (z) = Ṽ +0 e−jβz + Ṽ −0 e
jβz (121)
I(z) =Ṽ +0Z0
e−jβz − Ṽ−0
Z0ejβz (122)
Figure 31: Transmission Line connects generator and the
load.
We set up the z-axis so that the z = 0 is at the load, and the
generator is atz = −l. At z = 0, the load impedance is connected.
The definition of impedanceis Z = V/I, therefore at the z=0 end of
the transmission line, the voltage andcurrent on the transmission
line at that point have to obey boundary conditionthat the load
impedance imposes.
ZL =V (0)
I(0)
Substituting z=0, the boundary condition, in Equations 121-122,
we get Equa-tions 123-124.
Ṽ (0) = Ṽ +0 e−jβ0 + Ṽ −0 e
jβ0 = Ṽ +0 + Ṽ−0 (123)
I(0) =Ṽ +0Z0
e−jβ0 − Ṽ−0
Z0ejβ0 =
Ṽ +0Z0− Ṽ
−0
Z0(124)
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-
Reflection Coefficient
Dividing the two above equations, we get the impedance at the
load.
ZL = Z0Ṽ +0 + Ṽ
−0
Ṽ +0 − Ṽ−0
(125)
We can now solve the above equation for Ṽ −0
ZLZ0
(Ṽ +0 − Ṽ−0 ) = Ṽ
+0 + Ṽ
−0
(ZLZ0− 1)Ṽ +0 = (
ZLZ0
+ 1)Ṽ −0
Ṽ −0Ṽ +0
=ZLZ0− 1
ZLZ0
+ 1
Ṽ −0Ṽ +0
=ZL − Z0ZL + Z0
(126)
Definition 14. ΓL =Ṽ −0Ṽ +0
=ZL − Z0ZL + Z0
is the voltage reflection coefficient at the
load. ΓL relates the reflected and incident voltage phasor and
the load ZL andtransmission line impedance Z0. The voltage
reflection coefficient at the load is,in general, a complex number,
it has a magnitude and a phase ΓL = |ΓL|ej∠ΓL .
Example
(a) 100 Ω transmission line is terminated in a series connection
of a 50 Ω re-sistor and 10 pF capacitor. The frequency of operation
is 100 MHz. Findthe voltage reflection coefficient.
(b) For purely reactive load ZL = j50Ω, find the reflection
coefficient.
Voltage and Current on a transmission line
Now that we related forward and reflected voltage on a
transmission line withthe reflection coefficient at the load, we
can re-write the equations for the currentand voltage on a
transmission line as:
Ṽ (z) = Ṽ +0 (e−jβz + ΓLe
jβz) (127)
I(z) =Ṽ +0Z0
(e−jβz − Γejβz) (128)
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Reflection Coefficient
We see that if we know the length of the line, line type, the
load impedance, andthe transmission line impedance, we can
calculate all variables above, except forṼ +0 . In the following
chapters, we will derive the equation for the forward goingvoltage
at the load, but first, we will look at little more at the various
reflectioncoefficients on a transmission line.
Reflection coefficient anywhere on the line
Equations 121-122 can be concisely written as
Ṽ (z) = Ṽ (z)+ + Ṽ (z)− (129)
Ĩ(z) = Ĩ(z)+ + Ĩ(z)− (130)
Where Ṽ (z)+ is the forward voltage anywhere on the line, Ṽ
(z)− is reflectedvoltage anywhere on the line, Ĩ(z)+ is the
forward current anywhere on the line,and Ĩ(z)− is the reflected
current anywhere on the line.
We can then define a reflection coefficient anywhere on the line
as
Definition 15. Γ(z) =Ṽ (z)−
Ṽ (z)+=
Ṽ −0 ejβz
Ṽ +0 e−jβz
=Ṽ −0Ṽ +0
e2jβz is a voltage reflection
coefficient anywhere on the line. Γ(z) relates the reflected and
incident voltagephasor at any z.
Since we already defined ΓL =Ṽ −0Ṽ +0
as the reflection coefficient at the load, we
can now simplify the general reflection coefficient as
Γ(z) = ΓLe2jβz (131)
It is important to remember that we defined points between the
generator andthe load as the negative z-axis. If the line length
is, for example, l m long, thegenerator is then at z=-l m, and the
load at z=0. To find the reflection coefficientat some distance l/2
m away from the load, at z = −l/2 m, the equation for thereflection
coefficient will be
Γ(z = −l/2) = ΓLe−2jβl/2 (132)
Since we already defined the reflection coefficient at the load,
the reflection atany point on the line z = −l is
Γ(z = −l) = ΓLe−2jβl (133)Γ(z = −l) = |ΓL|ej(∠ΓL−2βl) (134)
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Reflection Coefficient
Reflection coefficient at the input of the trans-mission
line
Using the reasoning above, the reflection coefficient at the
input of the linewhose length is l is
Γ(z = −l) = Γin = ΓLe−2jβl (135)
Example 11. The reflection coefficient at the load is ΓL =
0.5ej600 . Find the
input reflection coefficient if the electrical length of the
line is βl = 450.
Explanation. The reflection coefficient at the input of the line
is Γ(z = −l) =Γin = |ΓL|ej(∠ΓL−2βl).
We substitute the expression for ΓL = 0.5ej600 and βl = 900, we
get the reflec-
tion coefficient at the input of the line Γin = 0.5e−j30.
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Input impedance of a transmission line
4.7 Input impedance of a transmission line
Again, we will look at a transmission line circuit in Figure 32
to find the inputimpedance on a transmission line.
Figure 32: Transmission Line connects generator and the
load.
The equations for the voltage and current anywhere (any z) on a
transmissionline are
Ṽ (z) = Ṽ +0 (ejβz + ΓLe
−jβz) (136)
I(z) =Ṽ +0Z0
(ejβz − Γe−jβz) (137)
The voltage and current equations at the generator z = −l
are:
Ṽin = Ṽ (z = −l) = Ṽ +0 (ejβl + ΓLe−jβl) (138)
Ĩin = I(z = −l) =Ṽ +0Z0
(ejβl − Γe−jβl) (139)
Input impedance as a function of reflection coef-ficient
The input impedance is defined as Zin =VinIin
. Since the line length is l, the
input impedance is
Zin =Ṽ +0 (e
−jβz + ΓLejβz)
Ṽ +0Z0
(e−jβz − ΓLejβz)(140)
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-
Input impedance of a transmission line
If we cancel common terms, we get
Zin = Z0(e−jβz + ΓLe
jβz)
(e−jβz − ΓLejβz)(141)
Now we can take e−jβz in front of parenthesis from both
numerator and denom-inator and then cancel it.
Zin = Z01 + ΓLe
2jβz
1− ΓLe2jβz(142)
We have previously defined the reflection coefficient at the
transmission line’sinput as Γin = ΓLe
2jβz. The final equation for the input impedance is
therefore
Zin = Z01 + Γin1− Γin
(143)
Input impedance as a function of load impedance
If we now look back at the Equation 141, here we can also use
Euler’s formulaejβz = cos(βz) + j sin(βz), and the equation for the
reflection coefficient at the
load ΓL =ZL − Z0ZL + Z0
we find the input impedance of the line as shown below.
Zin = Z0ZL + jZ0 tanβl
Z0 + jZL tanβl(144)
This equation will be soon become obsolete when we learn how to
use the SmithChart.
Example 12. Find the input impedance if the load impedance is ZL
= 0Ω, andthe electrical length of the line is βl = 900.
Explanation. Since the load impedance is a short circuit, and
the angle is 900
the equation simplifies to Zin = jZ0 tanβl =∞.
When we find the input impedance, we can replace the
transmission line and theload, as shown in Figure 33. In the next
section, we will use input impedanceto find the forward going
voltage on a transmission line.
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-
Input impedance of a transmission line
Figure 33: Transmission Line connects generator and the
load.
76
-
Forward voltage on a transmission line
4.8 Forward voltage on a transmission line
Again, we will look at a transmission line circuit in Figure 34
to find the inputimpedance on a transmission line.
Figure 34: Transmission Line connects generator and the
load.
The equations for the voltage and current anywhere (any z) on a
transmissionline are
Ṽ (z) = Ṽ +0 (e−jβz − ΓLejβz) (145)
I(z) =Ṽ +0Z0
(e−jβz − ΓLejβz) (146)
Using the equations from the previous section, we can replace
the transmissionline with its input impedance, Figure 35.
Figure 35: Transmission Line connects generator and the
load.
Forward voltage phasor as a function of load impedance
From Figure 35, we can find the input voltage on a transmission
line using thevoltage divider.
77
-
Forward voltage on a transmission line
Ṽin =Zin
Zin + ZgṼg (147)
Using Equation 146, we can also find the input voltage. The
input voltageequation at the generator z = −l is:
Ṽin = Ṽ (z = −l) = Ṽ +0 (ejβl + ΓLe−jβl) (148)
Since these two equations represent the same input voltage we
can make themequal.
Ṽ +0 (ejβl + ΓLe
−jβl) =Zin
Zin + ZgṼg (149)
Rearranging the equation, we find Ṽ +0 .
Ṽ +0 =ṼgZin
Zg + Zin
1
ejβl + ΓLe−jβl(150)
(151)
Forward voltage phasor as a function of input re-flection
coefficient
There is another way to find the input impedance as a function
of the inputreflection coefficient.
We write KVL for the circuit in Figure 35.
Ṽg = Zg Ĩin + Ṽin (152)
Using Equations 146-146, we can also find the input voltage and
current. Inputvoltage and current equation at the generator z = −l
are:
Ṽin = Ṽ (z = −l) = Ṽ +0 (ejβl + ΓLe−jβl) (153)
Ĩin = I(z = −l) =Ṽ +0Z0
(ejβl − Γe−jβl) (154)
Substituting these two equations in Equation 152 we get
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Forward voltage on a transmission line
Ṽg = ZgṼ +0Z0
(ejβl − ΓLe−jβl) + Ṽ +0 (ejβl + ΓLe−jβl) (155)
We can re-write this equation as follows.
Ṽg = Zgejβl Ṽ
+0
Z0(1− ΓLe−2jβl) + Ṽ +0 ejβl(1 + ΓLe−2jβl) (156)
Using that Γin = ΓLe−2jβl is the input reflection coefficient,
and multiplying
through with Z0.
ṼgZ0 = Zgejβl Ṽ
+0
1(1− Γin) + Ṽ +0 Z0ejβl(1 + Γin) (157)
Rearranging the equation, we get Ṽ +0
Ṽ +0 = Ṽge−jβl Z0
Z0(1 + Γin) + Zg(1− Γin)(158)
Γin = ΓLe−2jβl is the input reflection coefficient.
Special case - forward voltage when the generatorand
transmission-line impedance are equal
Because the generator’s impedance is equal to the transmission
line impedance,we will use the second equation. When Zg = Z0 we see
that the denominatorsimplifies into Z0(1 + Γin) + Zg(1− Γin) = Z0(1
+ Γin+ 1− 1 + Γg ∈) and we
can further simplify the fraction to get the final value of Ṽ
+0 =Vg2e−jβl.
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Traveling and Standing Waves
4.9 Traveling and Standing Waves
Standing Waves
In the previous section, we introduced the voltage reflection
coefficient thatrelates the forward to reflected voltage
phasor.
Ṽ (z) = Ṽ +0 (e−jβz + ΓLe
jβz) (159)
I(z) =Ṽ +0Z0
(e−jβz − ΓLejβz) (160)
Let us look at the physical meaning of these variables.
(a) ΓL is the voltage reflection coefficient at the load
(z=0),
(b) z is the axis in the direction of wave propagation,
(c) β is the phase constant,
(d) Z0 is the impedance of the transmission line.
(e) Ṽ +0 is the phasor of the forward-going wave at the
load,
(f) ΓLṼ+0 is the phasor of the reflected going wave at the
load,
(g) Ṽ +0 e−jβz is a forward voltage anywhere on the line,
(h) ΓLṼ+0 e
jβz is a reflected voltage anywhere on the line,
(i) Ṽ (z) is the total voltage phasor on the line, the sum of
forward and re-flected voltage.
(j) Ĩ+0 is the phasor of the forward-going current at the
load,
(k) −ΓLṼ +0Z0
is the phasor of the reflected current at the load,
(l) Ĩ+0 e−jβz is the phasor of a forward current anywhere on
the line,
(m) −ΓLṼ +0Z0
ejβz is the phasor of a reflected current anywhere on the
line,
(n) Ĩ(z) is the phasor of the total current on the line, the
sum of forward andreflected voltage.
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Traveling and Standing Waves
The magnitude of a complex number can be found as |z|
=√zz∗.Therefore the
magnitude of the voltage anywhere on the line is |Ṽ (z)| =√Ṽ
(z)Ṽ (z)∗. We
can simplify this equation as shown in Figure 161.
|Ṽ (z)| =√Ṽ (z)Ṽ (z)∗
|Ṽ (z)| =√
(Ṽ +0 )2(e−jβz − |ΓL|ejβz+Θr )Ṽ +0 (ejβz −
|ΓL|e−(jβz+∠ΓL))
|Ṽ (z)| = Ṽ +0√
(e−jβz − |ΓL|ejβz+|∠ΓL)(ejβz − |ΓL|e−(jβz+∠ΓL))
|Ṽ (z)| = Ṽ +0√
1 + |ΓL|e−(2jβz+|∠ΓL) + |ΓL|ej2βz+∠ΓL + |ΓL|2)
|Ṽ (z)| = Ṽ +0√
1 + |ΓL|2 + |ΓL|(e−(2jβz+∠ΓL)∠ΓL + e(j2βz+∠ΓL))
|Ṽ (z)| = Ṽ +0√
1 + |ΓL|2 + 2|ΓL| cos(2βz + ∠ΓL) (161)
The Equation 161 is written in terms of z. We set up the load at
z = 0, and thegenerator at z = −l. The positions of the maximums
and minimum total voltageon the line will be at some position at
the negative part of z-axis zmax = −lmax,and the minimums will be
at zmax = −lmin.The magnitude of the total voltage on the
transmission line is given by Eq.161.We will now visualize how the
magnitude of the voltage looks on the transmissionline.
Example 13. Find the magnitude of the total voltage anywhere on
the trans-mission line if ΓL = 0.
Explanation. Let us start from a simple case when the voltage
reflection coef-ficient on the transmission line is ΓL = 0 and draw
the magnitude of the totalvoltage as a function of z. Equation 161
shows the magnitude of the total volt-age anywhere on the line is
equal to the magnitude of the voltage at the load|Ṽ (z)| = |Ṽ +0
|. The magnitude of the voltage is constant everywhere on
thetransmission line, and so the line is called ”flat,” and it
represents a single, for-ward traveling wave from the generator to
the load. The magnitude is the greenline in Figure 36. To see the
movie of this transmission line, go to the classweb page under
Instructional Videos. Forward voltage is shown in red,
reflectedvoltage in pink, and the magnitude of the voltage is
green.
Example 14. Find the magnitude of the total voltage anywhere on
the trans-mission line if ΓL = 0.5e
j0.
Explanation.
Let’s look at another case, ΓL = 0.5 and ∠ΓL = 0. Equation 162
represents themagnitude of the voltage on the transmission line,
and Figure 37 shows in greenhow this function looks on a
transmission line. This case is shown in Figure 37.
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Traveling and Standing Waves
Figure 36: Flat line.
|Ṽ (z)| = Ṽ +0
√5
4+ cos2βz (162)
Figure 37: Voltage on a transmission line with reflection
coefficient magnitude0.5, and zero phase.
The function in Equation 162 is at its maximum when cos(2βz +
∠ΓL) = 1 or
z =k
2λ, and the function value is Ṽ (z) = 1.5Ṽ +0 . It is at its
minimum when
cos(2βz + ∠ΓL) = −1 or z =2k + 1
4λ and the function value is Ṽ (z) = 0.5Ṽ +0
The function that we see looks like a cosine with an average
value of Ṽ +0 , but it isnot a cosine. The minimums of the
function are sharper than the maximums.
Question 14 Observe waves in the app below.
Geogebra link: https://tube.geogebra.org/m/bmr8euzu
82
https://tube.geogebra.org/m/bmr8euzu
-
Traveling and Standing Waves
Is the black wave moving left or right?
Multiple Choice:
(a) left
(b) right
Is the black wave a forward, reflected or the sum of the
two?
Multiple Choice:
(a) Forward
(b) Reflected
(c) The sum of the two
Example 15. Find the magnitude of the total voltage anywhere on
the trans-mission line if ΓL = 1.
Explanation. Another case we will look at is when the reflection
coefficient isat its maximum of ΓL = 1. The function is shown if
Figure 38. In this case,we have a pure standing wave on a
transmission line.
Figure 38: Shorted Transmission Line.
Question 15 Observe waves in the app below.
Geogebra link: https://tube.geogebra.org/m/fwebfheh
83
https://tube.geogebra.org/m/fwebfheh
-
Traveling and Standing Waves
Is the black wave moving left or right? The wave is not moving
left or right, itis standing in place, so it is called a standing
wave. Is the black wave a forward,reflected or the sum of the
two?
Multiple Choice:
(a) Forward
(b) Reflected
(c) The sum of the two
Example 16. Find the magnitude of the total voltage anywhere on
the trans-mission line for any ΓL, and the position of voltage
maximums on the line.
Explanation. The magnitude of the total voltage on the line is
given in Equa-tion161. In general the voltage maximums will occur
when the cosine functionis at its maximum cos(2βz+∠ΓL) = 1. In this
case, the maximum value of themagnitude of total voltage on the
line is shown in Equation 163.
|Ṽ (z)max| = |Ṽ +0 |√
1 + |ΓL|2 + 2|ΓL||Ṽ (z)max| = |Ṽ +0 |
√(1 + |ΓL|)2
|Ṽ (z)max| = |Ṽ +0 |(1 + |ΓL|) (163)
The Equation 164 shows position of voltage maximums zmax on the
line.
cos(−2βlmax + ∠ΓL) = 1cos(2βlmax − ∠ΓL) = 1
2βzmax − ∠ΓL = 2nπ
zmax =2nπ + ∠ΓL
2β
zmax = λ2nπ + ∠ΓL
4π(164)
In general the voltage minimums will occur when cos(2βz) =
−1.
|Ṽ (z)min| = |Ṽ +0 |√
1 + |ΓL|2 − 2|ΓL||Ṽ (z)min| = |Ṽ +0 |
√(1− |ΓL|)2
|Ṽ (z)min| = |Ṽ +0 |(1− |ΓL|) (165)
The Equation 166 shows the position of voltage minimums on the
line.
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Traveling and Standing Waves
cos(2βzmax − ∠ΓL) = −12βzmax + ∠ΓL = (2n+ 1)π
zmax =(2n+ 1)π + ∠ΓL
2β
zmax = λ(2n+ 1)π + ∠ΓL
4π(166)
Voltage Standing Wave Ratio (VSWR) - pron:”vee-s-uh-are”
The ratio of voltage minimum on the line over the voltage
maximum is calledthe Voltage Standing Wave Ratio (VSWR) or just
Standing Wave Ratio (SWR).
SWR =Ṽ (z)max
Ṽ (z)min
SWR =1 + |ΓL|1− |ΓL|
(167)
Note that the SWR is always equal or greater than 1.
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Example Transmission Line Problem
4.10 Example Transmission Line Problem
Example 17. A transmitter operated at 20MHz, Vg=100V with Zg =
50Ωinternal impedance is connected to an antenna load through
l=6.33m of the line.The line is a lossless Z0 = 50Ω, β =
0.595rad/m. The antenna impedance at20MHz measures ZL = 36 + j20Ω.
Set the beginning of the z-axis at the load,as shown in Figure
39.
(a) What is the electrical length of the line?
(b) What is the input impedance of the line Zin?
(c) What is the forward going voltage at the load Ṽ +0 ?
(d) Find the expression for forward voltage anywhere on the
line.
(e) Find the expression for reflected voltage anywhere on the
line.
(f) Find the total voltage anywhere on the line.
(g) Find the expression for forward current anywhere on the
line.
(h) Find the expression for reflected current anywhere on the
line.
(i) Find the total current anywhere on the line.
(j) Instead of the antenna, a load impedance Zl = 50Ω is
connected to this 50Ohm line. How will that change the equations
above?
Figure 39: Transmission Line connects generator and the
load.
Explanation. The equations for the voltage and current anywhere
(any z) ona transmission line are
Ṽ (z) = Ṽ +0 (e−jβz − |Γ|ejβz+Θr ) (168)
I(z) =Ṽ +0Z0
(e−jβz − Γejβz) (169)
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Example Transmission Line Problem
We are given phase constant β, and we have to find the other
unknowns: phasorof voltage at the load Ṽ +0 , and the reflection
coefficient Γ.
Since we know the load impedance ZL, and the transmission-line
impedance Z0,we can find the reflection coefficient Γ using
Equation 170.
Γ =ZL − Z0ZL + Z0
= 0.279 ej112o
= 0.279 ej1.95 rad (170)
To find the input impedance of the line, we use the equation
Zin = Z0ZL + jZ0 tanβl
Z0 + jZL tanβl= 70.8 + j27.1Ω (171)
We can use one of the following two equations to find the
forward going voltageat the load:
Ṽ +0 =ṼgZin
Zg + Zin
1
ejβl + Γe−jβl(172)
Ṽ +0 = Vge−jβl Z0
Z0(1 + Γin) + Zg(1− Γin)(173)
Because the generator’s impedance is equal to the transmission
line impedance,we will use the second equation. When Zg = Z0 we see
that the denominatorsimplifies into Z0(1 + Γin) + Zg(1 − Γin) =
Z0(1 + Γin + 1 − Γin) and we can
further simplify the fraction to get the final value of Ṽ +0
=Vg2e−jβl. Since
βl =2π
λ0.6λ, the forward going voltage at the load is Ṽ +0 = 50 e
−j1.2π =
50 e−j3.768 = 50 e−j216o
.
The equations of the voltage and current anywhere on the line
are therefore
Ṽ (z) = 50e−j3.768(e−j0.595z − 0.279ej0.595z+1.95) (174)I(z) =
e−j3.768(e−j0.595z − 0.279ej0.595z+1.95) (175)
Suppose we replace the antenna with another load of impedance
50Ω. In thatcase, the reflection coefficient from the load will be
zero, and the reflected voltageswill disappear, so the voltage and
current will be equal to the forward-goingvoltage on the
transmission line.
Ṽ (z) = Ṽ +0 e−jβz (176)
I(z) =Ṽ +0Z0
e−jβz (177)
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