Lecture 09 em transmission lines

ELEN 3371 Electromagnetics Fall 2008
1
Lecture 9: EM Transmission Lines
and Smith Chart
Instructor:
Dr. Gleb V. Tcheslavski
Contact:
gleb@ee.lamar.edu
Office Hours:
Room 2030
Class web site:
www.ee.lamar.edu/gleb/em/Index.htm

2
Equivalent electrical circuits
In this topic, we model three electrical transmission systems that can be used to
transmit power: a coaxial cable, a strip line, and two parallel wires (twin lead).
Each structure (including the twin lead) may have a dielectric between two
conductors used to keep the separation between the metallic elements
constant, so that the electrical properties would be constant.

3
Coaxial cable

4
Microstrip line

5
Twin lead

6
Instead of examining the EM field distribution within these transmission
lines, we will simplify our discussion by using a simple model
consisting of distributed inductors and capacitors. This model is valid if
any dimension of the line transverse to the direction of propagation is
much less than the wavelength in a free space.
The transmission lines considered here support the propagation of
waves having both electric and magnetic field intensities transverse to
the direction of wave propagation. This setup is sometimes called a
transverse electromagnetic (TEM) mode of propagation. We assume
no loss in the lines.

7
Distributed
transmission line
Its equivalent
circuit
∆z is a short distance containing the distributed circuit parameter.
and are distributed inductance and distributed capacitance.
Therefore, each section has inductance and capacitance
ˆˆL C
ˆˆL L z C C z= ∆ = ∆ (9.7.1)

8
(9.8.1)
(9.8.2)
(9.8.3)
Note: the equations for a microstrip line are simplified and do not include
effects of fringing.
We can model the transmission line with an equivalent circuit consisting of an
infinite number of distributed inductors and capacitors.

9
The following simplifications were used:
1)No energy loss (resistance) was incorporated;
2)We neglected parasitic capacitances between the wires that
constitute the distributed inductances. We will see later that these
parasitic capacitances will lead to changes in phase velocity of the
wave (dispersion);
3)Parameters of the line are constant.
We can analyze EM transmission lines either as a large number of distributed
two-port networks or as a coupled set of first-order PDEs that are called the
telegraphers’ equations.

10
Transmission line equations
While analyzing the equivalent circuit of the lossless transmission line, it is
simpler to use Kirchhoff’s laws rather than Maxwell’s equations.
Therefore, we will
consider the
equivalent circuit of
this form:
For simplicity, we define the inductance and capacitance per unit length:
ˆˆ ;
L C
L C
z z
= =
V V
(9.10.1)
which have units of Henries per unit length and Farads per unit length, respectively.

11
The current entering the node at the location z is I(z). The part of this current
will flow through the capacitor, and the rest flows into the section. Therefore:
( , )ˆ( , ) ( , )
V z t
I z t C z I z z t
t
∂
= + +
∂
V V
( , ) ( , ) ( , )ˆI z z t I z t V z t
C
z t
+ − ∂
= −
∂
V
V
(9.11.1)
(9.11.2)
, the LHS of (9.11.2) is a spatial derivative. Therefore:0If z →V
( , ) ( , )ˆI z t V z t
C
z t
∂ ∂
= −
∂ ∂
(9.11.3)

12
Similarly, the sum of the voltage drops in this section can be calculated via the
Kirchhoff’s law also:
( , )ˆ( , ) ( , )
I z t
V z z t L z V z t
t
∂
− = +
∂
V V
( , ) ( , ) ( , )ˆV z t V z z t I z t
L
z t
− − ∂
= −
∂
V
V
(9.12.1)
(9.12.2)
, the LHS of (9.12.2) is a spatial derivative. Therefore:0If z →V
( , ) ( , )ˆV z t I z t
L
z t
∂ ∂
= −
∂ ∂
(9.12.3)

13
The equations (9.11.3) and (9.12.3) are two linear coupled first-order PDEs
called the telegrapher’s (Heaviside) equations. They can be composed in a
second-order PDE:
2 2
2 2
( , ) ( , )ˆˆ 0
I z t I z t
LC
z t
∂ ∂
− =
∂ ∂
2 2
2 2
( , ) ( , )ˆˆ 0
V z t V z t
LC
z t
∂ ∂
− =
∂ ∂
(9.13.1)
(9.13.2)
We may recognize that both (9.13.1) and (9.13.2) are wave equations with
the velocity of propagation:
1
ˆˆ
v
LC
= (9.13.3)

14
Example 9.1: Show that a transmission line consisting of distributed linear
resistors and capacitors in the given configuration can be used to model diffusion.
We assume that the
resistance and the
capacitance per unit length
are defined as
ˆˆ ;
R C
R C
z z
= =
V V
(9.14.1)
Potential drop over the resistor R and the current through the capacitor C are:
ˆ( , ) ( , )
( , )ˆ( , )
V z t I z t R z
V z t
I z t C z
t
∆ = ∆
∂
∆ = ∆
∂
(9.14.2)
(9.14.3)

15
( , ) ˆ0 ( , )
( , ) ( , )ˆ
V z t
If z I z t R
z
I z t V z t
C
z t
∂
∆ → ⇒ =
∂
∂ ∂
=
∂ ∂
The corresponding second-order PDE for the potential is:
2
2
( , )( , ) ( , )ˆ( ,ˆ ˆ)ˆˆI z t V z t
R R
V z t V z t
RC
z
C
z t t
∂ ∂ 
= = ÷
∂ ∂
∂ ∂
 
=
∂ ∂
(9.15.2)
(9.15.1)
(9.15.3)
Which is a form of a diffusion equation with a diffusion coefficient:
1
ˆˆ
D
RC
= (9.15.4)

16
Example 9.2: Show that a particular solution for the diffusion equation is given by
2
4
1
( , )
2
z
Dt
V z t e
D tπ
−
=
Differentiating the solution with respect to z:
2
4
3 2
( , ) 1
22
z
Dt
V z t z
e
z DtDπ
−∂  
= − ÷
∂  
2
2 2
4
2 3 2 2 5 2
( , ) 1
2 42
z
Dt
V z t z z
e
z Dt D tDπ
− ∂
= − + ÷
∂  
Differentiating the solution with respect to t:
2
2
4
3 2 5 2
1 ( , ) 1 1
2 42
z
Dt
V z t t z
e
D t D t DtDπ
− ∂
= − + ÷
∂  
(9.16.1)
(9.16.2)
(9.16.3)
(9.16.4)

17
Since the RHSs of (9.16.3) and (9.16.4) are equal, the diffusion equation is
satisfied.
The voltages at different times
are shown. The total area under
each curve equals 1.
This solution would be valid if a
certain amount of charge is
placed at z = 0 at some moment
in the past.
Note: the diffusion is significantly
different from the wave
propagation.

18
Sinusoidal waves
We are looking for the solutions of wave equations (9.13.1) and
(9.13.2) for the time-harmonic (AC) case. We must emphasize that –
unlike the solution for a static DC case or quasi-static low-frequency
case (ones considered in the circuit theory) – these solutions will be in
form of traveling waves of voltage and current, propagating in either
direction on the transmission line with the velocity specified by (9.13.3).
We assume here that the transmission line is connected to a distant generator
that produces a sinusoidal signal at fixed frequency ω = 2πf. Moreover, the
generator has been turned on some time ago to ensure that transient response
decayed to zero; therefore, the line is in a steady-state mode.
The most important (and traditional) simplification for the time-harmonic case
is the use of phasors. We emphasize that while in the AC circuits analysis
phasors are just complex numbers, for the transmission lines, phasors are
complex functions of the position z on the line.

19
Sinusoidal waves
{ } { }( , ) Re ( ) ; ( , ) Re ( )j t j t
V z t V z e I z t I z eω ω
= =
2
2
2
2
2
2
( )
( ) 0
( )
( ) 0
d V z
k V z
dz
d I z
k I z
dz
+ =
+ =
Therefore, the wave equations will become:
Here, as previously, k is the wave number:
2
k
v
ω π
λ
= =
Velocity of propagation
Wavelength of the voltage or current wave
(9.19.1)
(9.19.2)
(9.19.3)
(9.19.4)

20
Sinusoidal waves
A solution for the wave equation (9.13.3) can be found, for instance, in one of
these forms:
1 1
2 2
( ) cos sin
( ) jkz jkz
V z A kz B kz
V z A e B e− +
= +
= +
(9.20.1)
(9.20.2)
We select the exponential form (9.20.2) since it is easier to interpret in terms of
propagating waves of voltage on the transmission line.
Example 9.3: The voltage of a wave propagating through a transmission line was
continuously measured by a set of detectors placed at different locations along
the transmission line. The measured values are plotted. Write an expression for
the wave for the given data.

21
Sinusoidal waves
Slopeofthetrajectory…
Thedata

22
Sinusoidal waves
We assume that the peak-to peak amplitude of the wave is 2V0. We also
conclude that the wave propagates in the +z direction.
The period of the wave is 2s, therefore, the frequency of oscillations is ½ Hz.
The velocity of propagation can be found from the slope as:
5 1
4
1 0
v m s
−
= =
−
The wave number is:
12 0.5
4 4
k m
v
ω π π −×
= = =
The wavelength is:
2 4 2
8 m
k
π π
λ
π
×
= = =
Therefore, the wave is:
4
0( , )
z
j t
V z t V e
π
π
 
− ÷
 
=
(9.22.2)
(9.22.3)
(9.22.4)
(9.22.5)
Not in vacuum!
2 fω π π= = (9.22.1)

23
Sinusoidal waves
Assuming next that the source is located far from the observation point (say, at z
= -∞) and that the transmission line is infinitely long, there would be only a
forward traveling wave of voltage on the transmission line. In this case, the
voltage on the transmission line is:
0( ) jkz
V z V e−
=
The phasor form of (9.12.3) in this case is
ˆ(
(
)
)
) (jkV z j LI z
dV z
dz
ω− = −=
0( ) ( )
ˆ ˆ
jkzk k
I z V z V e
L Lω ω
−
= =
Which may be rearranged as:
(9.23.1)
(9.23.2)
(9.23.3)

24
Sinusoidal waves
The ratio of the voltage to the current is a very important transmission line
parameter called the characteristic impedance:
ˆ( )
( )
c
V z L
Z
I z k
ω
= =
1
ˆˆ
k and v
v LC
ω
= =Since
Then: [ ]
ˆ
ˆc
L
Z
C
Ω=
(9.24.1)
(9.24.2)
(9.24.3)
We emphasize that (9.24.3) is valid for the case when only one wave (traveling
either forwards or backwards) exists. In a general case, more complicated
expression must be used. If the transmission line was lossy, the characteristic
impedance would be complex.

25
Sinusoidal waves
If we know the characteristics of the transmission lane and the forward voltage
wave, we may find the forward current wave by dividing voltage by the
characteristic impedance.
Another important parameter of a transmission line is its length L, which is often
normalized by the wavelength of the propagation wave.
Assuming that the dielectric between conductors has ε and µ

26
Sinusoidal waves
The velocity of propagation does not depend on the dimensions of
the transmission line and is only a function of the parameters of the
material that separates two conductors. However, the characteristic
impedance DOES depend upon the geometry and physical
dimensions of the transmission line.

27
Sinusoidal waves
Example 9.4: Evaluate the velocity of propagation and the characteristic impedance
of an air-filled coaxial cable with radii of the conductors of 3 mm and 6 mm.
The inductance and capacitance per unit length are:
The velocity of propagation is:
The characteristic impedance of the cable is:
Both the v and the Zc may be decreased by insertion of a dielectric between leads.
7
0 4 10 6ˆ ln ln 0.14
2 2 3
H m
b
L
a
µ
µ π
π π
−
×   
= = = ÷  ÷
   
( ) ( )
12
02 2 8.854 10ˆ 80
ln ln 6 3
pF mC
b a
πε π −
× ×
= = =
8
6 12
1 1
3 10
ˆˆ 0.14 10 80 10
m sv
LC
− −
= = = ×
× × ×
6 12ˆˆ 0.14 10 80 10 42cZ L C − −
Ω= = × × ≈
(9.27.1)
(9.27.2)
(9.27.3)
(9.27.4)

28
Terminators
So far, we assumed that the
transmission line was infinite. In
the reality, however,
transmission lines have both the
beginning and the end.
The line has a real characteristic
impedance Zc. We assume that the
source of the wave is at z = -∞ and
the termination (the end of the line)
is at z = 0. The termination may be
either an impedance or another
transmission line with different
parameters. We also assume no
transients.

29
Terminators
The phasor voltage at any point on the line is:
2 2( ) jkz jkz
V z A e B e− +
= +
The phasor current is:
2 2
( )
jkz jkz
c
A e B e
I z
Z
− +
−
=
At the load location (z = 0), the ratio of voltage to current must be equal ZL:
2 2
2 2
( 0)
( 0)
L c
A BV z
Z Z
I z A B
+=
= =
= −
Note: the ratio B2 to A2 represents the magnitude of the wave incident on the
load ZL.
(9.29.1)
(9.29.2)
(9.29.3)

30
Terminators
(9.30.1)
We introduce the reflection coefficient for the transmission line with a load as:
2
2
L c
L c
Z ZB
A Z Z
−
Γ ≡ =
+
Often, the normalized impedance is used:
L
L
c
Z
z
Z
=
The reflection coefficient then becomes:
2
2
1
1
L
L
B z
A z
−
Γ ≡ =
+
(9.30.2)
(9.30.3)

31
Terminators
Therefore, the phasor representations for the voltage and the current are:
0
0
( )
( )
jkz jkz
jkz jkz
c
V z V e e
V
I z e e
Z
− +
− +
 = + Γ 
 = − Γ 
(9.31.1)
(9.31.2)
The total impedance is:
( )
( )
( )
V z
Z z
I z
=
generally a complicated function of the position and NOT equal to Zc. However,
a special case of matched load exists when:
L cZ Z=
In this situation: ( ) cZ z Z=
(9.31.3)
(9.31.4)
(9.31.5)

32
Terminators
Example 9.5: Evaluate the reflection coefficient for a wave that is incident from
z = -∞ in an infinitely long coaxial cable that has εr = 2 for z < 0 and εr = 3 for z > 0.
The characteristic impedance is:
( )ˆ ln
ˆ 2
c
b aL
Z
C
µ
ε π
= =
The load impedance of a line is the characteristic impedance of the line for z > 0.

33
Terminators
The reflection coefficient can be expressed as:
[ ] [ ]
[ ] [ ]
0 0
2 0 1 02 1
2 1 0 0
2 0 1 0
2 1
2 1
ln ln
2 2
ln ln
2 2
1 1 1 1
3 2 0.1
1 1 1 1
3 2
r r
r r
r r
r r
b a b a
Z Z
Z Z b a b a
µ µ
ε ε π ε ε π
µ µ
ε ε π ε ε π
ε ε
ε ε
   
− ÷  ÷
−    Γ = =
+    
+ ÷  ÷
   
− −
= = ≈ −
+ +

34
Terminators
The reflection coefficient Γ is completely determined by the value of the
impedance of the load and the characteristic impedance of the transmission
line. The reflection coefficient for a lossless transmission line can have any
complex value with magnitude less or equal to one.
If the load is a short circuit (ZL = 0), the reflection coefficient Γ = -1. The voltage
at the load is a sum of voltages of the incident and the reflected components
and must be equal to zero since the voltage across the short circuit is zero.
If the load is an open circuit (ZL = ∞), the reflection coefficient Γ = +1. The
voltage at the load can be arbitrary but the total current must be zero.
If the load impedance is equal to the characteristic impedance (ZL = Zc), the
reflection coefficient Γ = 0 – line is matched. In this case, all energy of
generator will be absorbed by the load.

35
Terminators
For the shorten transmission line:
{ }0
0
1
( , ) Re
2 sin cos
2
jkz jkz j t
V z t V e e e
V kz t
ω
π
ω
− +
Γ = −
 = − 
 
= × − ÷
 
(9.35.1)
(9.35.2)
For the open transmission line:
{ }
( )
0
0
1
( , ) Re
2 cos cos
jkz jkz j t
V z t V e e e
V kz t
ω
ω
− +
Γ = +
 = + 
= ×
(9.35.3)
(9.35.4)
In both cases, a standing wave is created.
The signal does not appear to propagate.

36
Terminators
Since the current can be found as 0 0 cI V Z=
ZL = ∞
ZL = 0
Note that the current wave differs from the voltage wave by 900

37
Terminators
Another important quantity is the ratio of the maximum voltage to the minimum
voltage called the voltage standing wave ratio:
max
min
1
1
V
VSWR
V
+ Γ
= =
− Γ
Which leads to
1
1
VSWR
VSWR
−
Γ =
+
VSWR, the reflection coefficient, the load impedance, and the characteristic
impedance are related.
Even when the amplitude of the incident wave V0 does not exceed the maximally
allowed value for the transmission line, reflection may lead to the voltage V0(1+|Γ|)
exceeding the maximally allowed.
Therefore, the load and the line must be matched.
(9.37.1)
(9.37.2)

38
Terminators
Example 9.6: Evaluate the VSWR for the coaxial cable described in the Example
9.5. The reflection coefficient was evaluated as Γ = -0.1.
1 1 0.1
1.2
1 1 0.1
VSWR
+ Γ + −
= = =
− Γ − −
Note: if two cables were matched, the VSWR would be 1.

39
Terminators

40
Impedance and line matching
The ratio of total phasor voltage to total phasor current on a transmission line
has units of impedance. However, since both voltage and current consist of the
incident and reflected waves, this impedance varies with location along the line.
0
0
( )
( )
( )
jkz jkz jkz jkz
c jkz jkz
jkz jkz
c
V e eV z e e
Z z Z
VI z e ee e
Z
− + − +
− +
− +
 + Γ + Γ = = =
− Γ − Γ 
Incorporating (9.30.1), we obtain:
2 co tan
(
s 2 sin
2 cos 2 si
)
tan n
jkz jkzL c
L c L c
c c
jkz
L c
jkzL c c L
L
L
c
c
c
Z Z
e e
Z Z Z kz j Z kz
Z Z
Z Z
Z jZ kz
Z
Z kz
z Z
Z jZ kzj Z kze e
Z Z
− +
− +
−
+
+ −
= =
−
−−
−
=
−
+
(9.40.1)
(9.40.2)
The last formula is most often used to find the impedance at the line terminals.

41
Considering the transmission line shown, we assume that a load with the
impedance ZL is connected to a transmission line of length L having the
characteristic impedance Zc and the wave number k.
The input impedance can be found as an impedance at z = -L:
tan
( ) ( )
tan
L c
in c
c L
Z jZ kL
Z L Z z L Z
Z jZ kL
+
= = − =
+
(9.41.1)
Or as the normalized input impedance: tan
( )
1 tan
L
in
L
z j kL
z L
jz kL
+
=
+
(9.41.2)

42
Impedance and line matching (Ex)
Example 9.7: A signal generator whose frequency f = 100 MHz is connected to a
coaxial cable of characteristic impedance 100 Ω and length of 100 m. The
velocity of propagation is 2⋅108
m/s. The line is terminated with a load whose
impedance is 50 Ω. Calculate the impedance at a distance 50 m from the load.
The normalized load impedance is 50 100 0.5L L cz Z Z= = =
The wave number is
The normalized input impedance is
8
1
8
2 2 2 1 10
2 10
f
k m
v
π π π
π
λ
−× ×
= = = =
×
( )
( )
0.5 tan 50tan 1
( 50 )
1 tan 1 0.5tan 50 2
L
in
L
jz j kL
z z m
jz kL j
π
π
++
= − = = =
+ +
Therefore, the input impedance is
0.5 100 50in in cZ z Z= = × = Ω

43
The wave number can be expressed in terms of wavelength:
2
k
π
λ
=
Therefore:
2
kL L
π
λ
=
and ( ) ( )
2
tan tan tan ; integerkL L kL n n
π
π
λ
 
= = + − ÷
 
If the length of the transmission line is one quarter of a wavelength,
2
4 2
kL
π λ π
λ
= =
tangent (9.42.3) approaches infinity and
( )
2
4
4
L c c
in c in
c L L
Z jZ Z
Z z Z Z
Z jZ Zλ
λ + ∞ 
= − = → = ÷
+ ∞ 
(9.43.1)
(9.43.2)
(9.43.3)
(9.43.4)
(9.43.5)
Implying that the normalized input impedance zin of a λ/4 line terminated with the
load ZL is numerically equal to the normalized load admittance yL = 1/zL.

44
The last example represents a one quarter-wavelength transmission line that is
useful in joining two transmission lines with different characteristic impedances or
in matching a load. One of the simplest matching techniques is to use a quarter-
wave transformer – a section of a transmission line that has a particular
characteristic impedance Zc(λ/4).
This characteristic impedance Zc(λ/4)
must be chosen such that the
reflection coefficient at the input of
the matching transmission line
section is zero.
This happens when
( 4)c c LZ Z Zλ = (9.44.1)
One considerable disadvantage of this method is its frequency dependence since
the wavelength depends on the frequency.

45
When the load impedance equals the characteristic impedance, the load and the
line are matched and no reflection of the wave occurs.
For the short circuit and the open circuit:
0
( ) tan( )
( ) cot( )
tan( )
L
L
in cZ
c
in cZ
Z z L jZ kL
Z
Z z L jZ kL
j kL
=
=∞
= − =
= − = = −
In practice, it is easier to make short circuit
terminators since fringing effects may exist in open
circuits. In both cases, the input impedance will be a
reactance, Zin = jXin as shown for a short-circuited
(a) and an open-circuited (b) transmission lines. The
value of the impedance depends on the length of
the transmission line, which implies that we can
observe/have any possible value of reactance that is
either capacitive or inductive.
(9.45.1)
(9.45.2)

46
Types of input
impedance of short-
circuited and open-
circuited lossless
transmission lines:
We introduce the characteristic admittance of the transmission line:
1c cY Z= (9.46.1)
and the input susceptance of the transmission line:
1in inB X= − (9.46.2)

47
Assuming that a transmission line is terminated with
a load impedance ZL or load admittance YL that is not
equal to the line’s characteristic admittance Yc.
Let the input admittance of the line be Yc + jB at the
distance d1 from the load.
If, at this distance d1, we connect a susceptance –jB
in parallel to the line, the total admittance to the left
of this point (d1) will be Yc.
The transmission line is matched from the insertion
point (d1) back to the generator.
In practice, matching can be done by insertion of a
short-circuited transmission line of particular length.

48
Such transmission line used to match another (main) transmission line is called a
stub. The length of a stub is chosen to make its admittance be equal –jB. This
process of line matching is called single-stub matching.
The length of the stub can be made adjustable. Such adjustable-length
transmission lines are sometimes called a trombone line.
Note that single-stub matching requires two adjustable distances: location of the
stub d1 and the length of the stub d2. In some situations, only the stub’s length can
be adjusted. In these cases, additional stub(s) may be used.
The distances mentioned here are normalized to the wavelength. Therefore, this
method allows line matching at particular discrete frequencies.

49
Impedance and line matching (Ex)
Example 9.8: A lossless transmission line is
terminated with an impedance whose value is a
half of the characteristic impedance of the line.
What impedance should be inserted in parallel
with the load at the distance λ/4 from the load
to minimize the reflection from the load?
To minimize the reflection, the parallel combination of the ZQ and the input
impedance at that location should equal to the characteristic impedance of the line.
22
( 4)
2 2
( 4)
22
2
2
cc
QQ
Q in Q ccL
c
Q in Q cc c
Q Q
L c
ZZ
ZZ
Z Z Z ZZZ
Z
Z Z Z ZZ Z
Z Z
Z Z
λ
λ
  
 ÷ ÷
   = = = =
+ +   
+ + ÷  ÷
   
2
1 2
2
Q
Q c
Q c
Z
Z Z
Z Z
= ⇒ =
+

50
Smith chart
The input impedance of a transmission line depends on the impedance of the
load, the characteristic impedance of the line, and the distance between the load
and the observation point. The value of the input impedance also periodically
varies in space.
The input impedance can be found graphically via so called Smith chart.
tan
( )
1 tan
L
in
L
z j kL
z L
jz kL
−
=
+
The normalized impedance at any location is complex and can be found as:
An arbitrary normalized load impedance is:
Lz r jx= +
Where: ;L L
c c
R X
r x
Z Z
= =
Since the line is assumed to be lossless, its characteristic impedance is real.
(9.50.1)
(9.50.2)
(9.50.3)

51
Smith chart
The reflection coefficient will be complex:
1
1
L
r i
L
z
j
z
−
Γ = Γ + Γ =
+
( ) ( )
2 2
2 22 2
1 1 21
1 1 1 1
r i r i i
L
r i r i r i
j
z r jx j
j
+ Γ + Γ + Γ − Γ Γ+ Γ
⇒ = = = + = +
− Γ − Γ − Γ − Γ + Γ − Γ + Γ
Therefore, by equating the real and the imaginary parts:
2 2
2 1
1 1
r i
r
r r
   
Γ − + Γ = ÷  ÷
+ +   
( )
2 2
2 1 1
1r i
x x
   
Γ − + Γ − = ÷  ÷
   
(9.51.1)
(9.51.2)
(9.51.3)
(9.51.4)
(9.51.3) and (9.51.4) represent family of circles in a plane whose axes are Γr and Γi.
The center and radius of each circle are determined by the normalized resistance r
and reactance x. All circles are inside the unit circle since maximal Γ = 1.

52
Smith chart
For the constant r circles:
1.The centers of all the constant r
circles are on the horizontal axis – real
part of the reflection coefficient.
2.The radius of circles decreases when
r increases.
3.All constant r circles pass through the
point Γr =1, Γi = 0.
4.The normalized resistance r = ∞ is at
the point Γr =1, Γi = 0.
For the constant x (partial) circles:
1.The centers of all the constant x
circles are on the Γr =1 line. The circles
with x > 0 (inductive reactance) are
above the Γr axis; the circles with x < 0
(capacitive) are below the Γr axis.
2. The radius of circles decreases when absolute value of x increases.
3. The normalized reactances x = ±∞ are at the point Γr =1, Γi = 0

53
Smith chart
The constant r circles are orthogonal to the constant x circles at every intersection.
The actual (de-normalized) load impedance is:
( )L cZ Z r jx= +
The intersection of an r circle and an x circle specifies the normalized impedance.
The evenly spaced marks on the circumference indicate the fraction of a half-
wavelength, since the impedance repeats itself every half-wavelength.
Since the transmission line is lossless, the magnitude of the reflection coefficient is
constant at every point between the load and the signal generator.
The horizontal Γr and the vertical Γi axes have been removed from the chart.
Smith chart can be expressed in terms of either impedance or admittance.
(9.53.1)

54
Smith chart (Example)
Example 9.9: On the simplified Smith chart, locate the following normalized
impedances:
) 1 0;
) 0.5 0.5
) 0 0;
) 0 1;
) 1 2;
)
a z j
b z j
c z j
d z j
e z j
f z
= +
= −
= +
= −
= +
= ∞

55
Smith chart
The reflection coefficient can also be expressed in the following form:
1
1
Lj L
L
z
e
z
θ −
Γ = Γ =
+
The magnitude of the reflection coefficient
0 1≤ Γ ≤
is determined by the value of the normalized load impedance zL and is constant for
all locations along the lossless transmission line.
θL is a phase angle associated with the reflection coefficient Γ.
(9.55.1)
(9.55.2)
The input impedance at any arbitrary point (say, -z’) is
' ' 2 '
' ' 2 '
( ') 1
( ')
( ') 1
jkz jkz j kz
in c cjkz jkz j kz
V z e e e
Z z Z Z
I z e e e
− −
− −
− + Γ + Γ
− = = =
− −Γ − Γ
(9.55.3)

56
Smith chart
The normalized input impedance at the point -z’ is
2 '
2 '
1( ') 1
( ')
1 1
jj kz
in
in j kz j
c
eZ z e
z z
Z e e
φ
φ
−
−
+ Γ− + Γ
− = = =
−Γ − Γ
Where 2 'L kzφ θ= −
(9.56.1)
(9.56.2)
Compared to (9.51.2), we conclude that the only difference is in a phase shift φ
that is linearly proportional to coordinate z’ and can be translated to the Smith
chart by rotating the initial value of the load impedance along a circle with the
radius equal to the magnitude of the reflection coefficient. A clockwise rotation is
equivalent to moving toward the signal generator; a counterclockwise rotation –
toward the load. The amount of rotation depends on the distance 2kz’ = 4πz’/λ.
If the distance z’ = λ/4, the rotation will be equal to π radians. In this situation, the
numerical value of an impedance will be converted into the numerical value of an
admittance (refer back to our quarter wavelength lines discussion).

57
Smith chart
For example, the impedance
z = 0.5 + j 0.5
corresponds to the admittance
1
1 1
0.5 0.5
y j
j
= = −
+
Surprisingly, this calculation can be
done graphically using the Smith
chart!
First, we locate the normalized
impedance on the chart.
Second, we draw a circle (or a
semicircle “to generator”) centered
at the center of the Smith chart and
passing through the impedance.
Third, we plot a straight line through the center of the Smith chart and through this
impedance. The intersection of the line with the semicircle yields the value for the
admittance.

58
Smith chart
This method suggests that the
normalized load impedance of 0
will yield a normalized load
admittance of ±∞.
The last observation implies that
the input impedance at the
location that is λ/4 from a short
circuit will be an input
impedance of an open circuit!
Therefore, we may avoid using
open circuits since they are not
very attractive due to fringing,
and use short circuits only…

59
Example 9.10: A load impedance 50 50LZ j= +
terminates a transmission line that is 5 m long and has Zc = 25 Ω. Using the Smith
chart, find the impedance at the signal generator if the frequency of oscillation f =
105
Hz. The phase velocity for this transmission line is v = 2 Mm/s.
The wavelength is 6
5
2 10
20
1 10
v
m
f
λ
×
= = =
×
Therefore, the distance between the load and the generator is: 5/20 = λ/4.
The normalized load impedance is
50 50
2 2
25
L
j
z j
+
= = +
We locate the impedance on the Smith chart, place a compass leg at the center of
the chart, and draw an arc a distance λ/4 (half a circle) in the clockwise direction,
which is “toward the generator”.

60
Finally, we draw a line
passing through the
impedance and the
center of the Smith chart.
The intersection of that
line with the arc
determines the
normalized load
impedance of
0.25 5inz j= − 0.2
Therefore, the load
impedance is
25(0.25 5)
6.25 6.25
in c inZ Z z
j
j
=
= − 0.2
= −

61
An interesting property of the Smith chart is that it can be used equally well in terms
of an impedance or an admittance. Any constant coefficient circle centered at the
center of the Smith chart will pass the “real part equals 1” circle at two locations.
The truth is that at any of these two locations, an admittance can be added in
parallel to the line to match it with the load!
Example 9.11: A load admittance is 0.2 5Ly j= − 0.
Find the locations where a matching admittance should
be placed. Also, find the value for the matching
admittance.
The input admittance will have the value 1iny jb= ±
The value of a reflection coefficient can be
determined from the load admittance…
Or we can simply use the Smith chart.
http://my.ece.ucsb.edu/sanabria/tattoo.html

62
We draw a circle centered
at the center of the Smith
chart and passing through
the input admittance we
have calculated.
Two intersections of that
circle with the “real part 1”
circle determine the two
values of the matching
admittances. The location
where these admittances
are to be inserted can be
determined by the angles
that correspond to each
arc.

63
Transient effects and Bounce diagram
Let us consider a transmission
line connected to the battery
with an internal impedance Zb
through a switch. The line has
a characteristic impedance Zc
and is connected to the load
ZL. We assume first that all
impedances are pure
resistances.
The signal propagates with the velocity v and will reach the load L/v seconds after it
was launched. The amplitude of the wave V1 that is launched on the line can be
determined by the voltage divider rule:
1
c
b
b c
Z
V V
Z Z
=
+
(9.63.1)
Note: we are discussing a DC potential propagating along the line.

64
Assuming that the switch was engaged at the time t = 0, at the time τ = L/v the
front of this propagating voltage step arrives at the load impedance ZL. At this
time, a portion of this incident voltage will be reflected from the load impedance,
and another portion will be absorbed by the load (“transmitted” to the load). The
reflection coefficient at the load is
2
1
L c
L
L c
Z Z V
Z Z V
−
Γ = =
+
(9.64.1)
The front of this propagating reflected wave (whose amplitude V2 can be either
positive or negative) reaches the battery impedance at a time 2τ = 2L/v. This front
will also be reflected from the battery impedance with the reflection coefficient
3
2
b c
b
b c
Z Z V
Z Z V
−
Γ = =
+
(9.64.2)
Therefore, the battery impedance acts like a load impedance for the wave V2 and
the new voltage step V3 will be reflected towards the load…

65
This process of wave reflections from two impedances may continue indefinitely
long. The front of the propagating voltage step “bounces” back and forth between
the load impedance and the battery impedance.
This graphical technique used to evaluate the voltage
at any location of the line as a function of time is
called a bounce diagram.
The horizontal axis represents the normalized
position; the vertical axis – the normalized time. The
prediction of the temporal response at a given
location is obtained by inserting a vertical line on the
diagram at that location. The intersection of the
trajectory with that line indicates that the voltage at
that location will change its value over time since
more and more components (reflected waves) will be
added.

66
Assuming that the voltage of the battery is constant, we may conclude that the
voltage of each individual component will also be constant over time and will
have the same value as that of the front. The voltage at any location of the
transmission line is a sum of individual components:
( )
( )( ) ( )( )
1 2 3
2 2
... 1 ...
1 ... 1 ...
c
L L b L b L b
b c
c
L b L b L L b L b b
b c
Z
V V V V V
Z Z
Z
V
Z Z
= + + + = + Γ + Γ Γ + Γ Γ Γ
+
 = + Γ Γ + Γ Γ + + Γ + Γ Γ + Γ Γ +
 +
It can be seen that every “next” component in the sum in (9.66.1) has lower
amplitude than the previous components. Consequently, voltage on the line will
converge to its steady-state value:
The absolute value of the reflection coefficient cannot exceed one. Therefore, the
quantity in the brackets can be expressed as the closed form summation:
2 1
1 ... 1
1
forε ε ε
ε
+ + + = <
−
(9.66.1)
(9.66.2)

67
Therefore, the steady-state (asymptotic) voltage will be:
1
1
c L
b
b c L b
Z
V V
Z Z
 + Γ
=  ÷
+ − Γ Γ 
or, using the definitions of reflection coefficients:
[ ] [ ]
[ ] [ ]{ } [ ] [ ]{ }
1
1
L c L cc
b
b c L c L c b c b c
Z Z Z ZZ
V V
Z Z Z Z Z Z Z Z Z Z
 + − +
=  ÷
 ÷+ − − + − + 
L
b
b L
Z
V V
Z Z
=
+
Which simplifies to
The current flowing through the load is
b
L b L
VV
I
Z Z Z
= =
+
(9.67.1)
(9.67.2)
(9.67.3)
(9.67.4)

68
Example 9.12: A 12 V battery is connected via a switch to a transmission line that
is 6 m long. The characteristic impedance of the line is 50 Ω, the battery
impedance is 25 Ω, and the load impedance is 25 Ω. The velocity of propagation is
2 ⋅ 106
m/s. Find and sketch the voltage at the midpoint of this line during the time
interval 0 < t < 9 µs.
The amplitude of the wave that is launched on the transmission line is
1
50
12 8
25 50
c
b
b c
Z
V V V
Z Z
= = =
+ +
The reflection coefficient at the load is
25 50 1
25 50 3
L c
L
L c
Z Z
Z Z
− −
Γ = = = −
+ +
The reflection coefficient at the battery is
25 50 1
25 50 3
b c
b
b c
Z Z
Z Z
− −
Γ = = = −
+ +
(9.68.1)
(9.68.2)
(9.68.3)

69
We will use the bounce diagram to evaluate the
voltage at the midpoint of the line. We will identify
the amplitudes of the waves. The normalized time
is t/τ, where
3L v sτ µ= =
The voltage at the midpoint is zero until the first
wave arrives. At this moment, the amplitude
becomes 8. later on, the wave, reflected from the
load arrives and the voltage will be changed…
The steady-state voltage found from (9.67.3) is
25
12 6
25 25
L
b
b L
Z
V V V
Z Z
= = =
+ +
(9.69.1)
(9.69.2)

70

71
Example 9.13: A battery with no internal impedance has an open circuit voltage of
100 V. At a time t = 0, this battery is connected to an air-felled 50 Ω coaxial cable
via a 150 Ω resistor. The cable is 300 m long and is terminated in a load of 33.3 Ω.
a)Sketch a bounce diagram for the first 4 µs after the switch is closed;
b)Plot the voltage on the load as a function of time;
c)Find the asymptotic value of the load voltage.
The reflection coefficients and the
voltage step propagating on the
line are:
150 50 1
150 50 2
b c
b
b c
Z Z
Z Z
− −
Γ = = =
+ +
33.3 50 1
33.3 50 5
L c
L
L c
Z Z
Z Z
− −
Γ = = = −
+ +
1
50
100 25
150 50
c
b
b c
Z
V V V
Z Z
= = =
+ +
(9.71.1)
(9.71.2)

72
Since the coaxial line is filled with air, the velocity of propagation
is 3⋅108
m/s. Therefore, the signal takes 1 µs to travel from one
end to another.
The asymptotic voltage and current computed from (9.67.3) and
(9.67.4) are
18.2
0.55
t
t
V V
I A
→∞
→∞
=
=

73
Pulse propagation
Up to this point, we were discussing propagation of a step voltage along the
transmission line. We learned that the front propagates with a certain velocity and
it takes specific time (proportional to L/v) for the signal to travel through the line of
length L.
Highly important is the case when a voltage pulse is propagating through the line.
We will consider the line
connecting a pulse
generator to the load.
As previously, we can
estimate the amplitude of the
pulse launched on the line:
1
c
g
g c
Z
V V
Z Z
=
+
(9.73.1)

74
Pulse propagation
We assume next that the temporal width of the pulse is much less than the time
needed for the pulse to travel through the line:
t L vV =
For instance, if the length of the transmission line is 3 m and the velocity of
propagation is c:
9
8
3
10 10
3 10
L
s
v
−
≈ ≈ ×
×
the pulse duration must be less than 10 ns.
Note that the velocity of propagation along the transmission line is usually less than
that for vacuum and, therefore, the critical pulse duration will be slightly increased.
The pulse of the voltage V1 is launched from the generator towards the load. A
portion of the pulse is absorbed by the load, and a portion of the pulse is reflected
back toward the generator. The amount of the energy that is absorbed (and
reflected) is determined by the ratio of the load impedance to the line’s
characteristic impedance.
(9.74.1)
(9.74.2)

75
Pulse propagation
We can conclude that the energy of the incident pulse will be divided into the
energy of the reflected wave and the energy absorbed by the load:
{
.
inc ref abs
inc energy
P t P t P t= +V V V
Or, in terms of impedances and reflection coefficients:
( )
22 2
11 L
c c L
VV V
Z Z Z
Γ
= +
Here, VL is the voltage that appears on the load. Incorporating the reflection
coefficient, we arrive at:
{ } { }( )
2
2 2
11 L c L c L
c c L
Z Z Z Z VV V
Z Z Z
− +  
= +
(9.75.1)
(9.75.2)
(9.75.3)

76
Pulse propagation
The last equation can be solved for the load voltage:
{ } { }
2
1 1
2L c L c LL L
L
c c L c
Z Z Z Z ZZ Z
V V V
Z Z Z Z
− +  = − =
+
At this point, we re-introduce the transmission coefficient as:
1
2
1L L
L c
V Z
V Z Z
Τ = = = + Γ
+
(9.76.1)
(9.76.2)

77
Pulse propagation (Example)
Example 9.14: Evaluate the transmission coefficient for a wave that propagates in
the +z direction through a coaxial cable. The dielectric constant in the region z < 0
is 2 and in the region z > 0 is 3. the physical dimensions of the cable are constant.
The characteristic impedance of the coax cable can be found as:
( )ln
2
c
b a
Z
µ
ε π
=
We will consider the region 2 (z > 0) as a load for the line in the region 1. Thus, the
transmission coefficient given by (9.76.2) is:
( )
( ) ( )
0
0 0
2 3 ln 2 2 3
0.9
3 ln 2 2 ln 2 1 3 1 2
b a
b a b a
µ ε π
µ ε π µ ε π
Τ = = =
+ +

78
Pulse propagation
Let us consider the joining of two
transmission lines having
different dimensions. Apparently,
the characteristic impedances in
two regions will differ and a
portion of the signal will be
reflected back from the second
transmission line.
If we send a pulse to propagate through such a system of
two joined lines, we may observe a pulse reflected back
from a boundary (the second line in our case). If the
velocity of propagation on a transmission line v is known,
and the time ∆t needed for a pulse to travel to and back
from the reflective boundary is measured, the distance to
the boundary is:
2d v t= V (9.78.1)
Can be used for fault locations: time domain reflectometry.

79
Pulse propagation
Example 9.15: Using the reflection and transmission coefficients, show that
energy is conserved at the junction between two lossless transmission lines.
The conservation of energy
implies that (9.75.1) or
(9.75.2) must be satisfied:
( ) ( )
2 22
1 1 2
inc incinc
c c c
V VV
Z Z Z
Γ Τ
= +
Therefore:
( ) ( ) ( )
2 2
2 1 2 1
2 22 2 2 2
2 1 2 1 2 2 1 2 1 2 1
1 1 2 1 2 1
2 2
2 2
2 1 22 2 2 2
2 1 2 1 1 2 1 2 1
2 2 2
2 2 1 2 1 1 2 1 2 1
2
2 21
4
2 4
2 21
2 2
c c c c
c c c c c c c c c c c
c c c c c c
c c
c c c
c c c c c c c c c
c c c c c c c c c c
Z Z Z Z
Z Z Z Z Z Z Z Z Z Z Z
Z Z Z Z Z Z
Z Z
Z Z Z
Z Z Z Z Z Z Z Z Z
Z Z Z Z Z Z Z Z Z Z
− +
− + +   Γ Τ + +   = + = + =
− + +
+ + + +
+ = =
+ + + + 2
1
1
cZ
=

80
Example 9.16: A 1 V pulse propagates on a transmission line terminated in an
open circuit at z = 0. Four oscilloscopes are triggered by the same pulse generator
and are located at za = -6, zb = -4, zc = -2, zd = 0 m. Find the velocity of propagation
and interpret the signals on the oscilloscopes. Sketch the signals if the line is
terminated with a short circuit.
From the traces on the oscilloscopes A and B:
6
6
2
2 10
1 10
z
v m s
t −
= = = ×
×
V
V
The oscilloscope D is at the location of the open circuit, and the incident and the
reflected pulses add together.

81
The signals detected after t = 4 µs are the reflected pulses propagating backwards
to the generator. If the line was terminated with a short circuit, the oscilloscopes
would detect something like:
Note that the voltage across the
short circuit must be zero, as it is
depicted by the oscilloscope D.
Also, the reflected pulses will be
opposite to the incident ones.

82
Lossy transmission lines
Up to this point, our discussion was limited to lossless transmission lines
consisting of equivalent inductors and capacitors only. As a result, the
characteristic impedance for such lines is real. Let us incorporate ohmic
losses within the conductors and leakage currents between conductors.
In this case, the characteristic impedance becomes complex and the new
model will be:

83
The new first-order PDEs (telegrapher’s equations) are
( , ) ( , )ˆ ˆ ( , )
( , ) ( , )ˆ ˆ ( , )
I z t V z t
C GV z t
z t
V z t I z t
L RI z t
z t
∂ ∂
= − −
∂ ∂
∂ ∂
= − −
∂ ∂
(9.83.1)
(9.83.2)
Where the circuit elements are defines as
ˆ ˆˆ ˆ; ; ;
L C R G
L C R G
z z z z
= = = =
V V V V
A time-harmonic excitation of the transmission line leads to a phasor notation:
( ) ˆ ˆ ( )
( ) ˆ ˆ ( )
I z
G j C V z
z
V z
R j L I z
z
ω
ω
∂  = − + ∂
∂
 = − + ∂
(9.83.3)
(9.83.4)
(9.83.5)

84
The quantities in square brackets are denoted by distributed admittance and
distributed impedance leading to
ˆY
ˆZ
( ) ˆ ( )
( ) ˆ ( )
dI z
YV z
dz
dV z
ZI z
dz
= −
= −
(9.84.1)
(9.84.2)
Second-order ODEs can be derived for current and voltage:
2
2
2
2
( ) ˆ ˆ ( )
( ) ˆ ˆ ( )
d I z
ZYI z
dz
d V z
ZYV z
dz
=
=
(9.84.3)
(9.84.4)
The phasor form solutions:
1 2
1 2
( )
( )
z z
z z
V z V e V e
I z I e I e
γ γ
γ γ
− +
− +
= +
= +
(9.84.5)
(9.84.6)

85
Here, γ is the complex propagation constant:
( )( )ˆ ˆˆ ˆ ˆ ˆj ZY R j L G j Cγ α β ω ω= + = = + + (9.85.1)
As previously, we denote by V1 and I1 the amplitudes of the forward (in the +z
direction) propagating voltage and current waves; and V2 and I2 are the amplitudes
of the backward (in the -z direction) propagating voltage and current waves.
Therefore, the time varying waves will be in the form:
( ) ( )
( ) ( )
1 2
1 2
( , ) cos cos
( , ) cos cos
z z
z z
V z t V e t z V e t z
I z t I e t z I e t z
α α
α α
ω β ω β
ω β ω β
− +
− +
= − + +
= − + +
(9.85.2)
(9.85.3)
We recognize that (9.85.2) and (9.85.3) are exponentially decaying propagating
waves: the forward wave (V1, I1) propagates and decays in the +z direction, while
the backward wave (V2, I2) propagates and decays in the –z direction.

86
An example of a forward
propagating wave: it is
possible to determine the
values of α and β from the
graph as shown.

87
For small losses, employing a binomial approximation (1 – x)n
≅ 1 – nx for x << 1,
the complex propagating constant is approximately
ˆ ˆˆ ˆ
ˆ ˆˆ ˆ1 1 1
ˆ ˆ ˆ ˆ2 2
R G R G
j j L j C j LC j
j L Lj C C
γ α β ω ω ω
ω ωω ω
     
= + = + + ≈ − +  ÷  ÷ ÷  ÷  ÷
       
The attenuation constant can be approximated as
ˆ ˆˆ ˆ1ˆ ˆˆ ˆ
ˆ ˆ ˆ ˆ22 2
R G R G
LC LC
L LC C
α ω
ω ω
   
≈ + = + ÷  ÷ ÷  ÷
   
The attenuation constant does not depend on frequency.
(9.87.1)
(9.87.2)

88
Lossy transmission lines (Ex)
Example 9.17: Find the complex propagation constant if the circuit elements satisfy
the ratio . Interpret this situation.ˆ ˆˆ ˆR L G C=
The complex propagation constant is:
( )( ) ( ) ( )
ˆ ˆˆ
ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ
RC C
j R j L G j C R j L j C R j L
L L
γ α β ω ω ω ω ω
 
= + = + + = + + = + ÷ ÷
 
The attenuation constant is independent on frequency, which implies no distortion
of a signal as it propagates on this transmission line and a constant attenuation.
The characteristic impedance of this transmission line
( )
ˆ ˆ ˆ ˆ ˆ ˆ
ˆ ˆ ˆ ˆˆ ˆˆ ˆc
Z R j L R j L L
Z
Y G j C CRC L j C
ω ω
ω ω
+ +
= = = =
+ +
does not depend on frequency. Therefore, this transmission line is distortionless.

89
Example 9.18: The attenuation on a 50 Ω distortionless transmission line is 0.01
dB/m. The line has a capacitance of 0.1⋅10-9
F/m. Find:
a)Transmission line parameters: distributed inductance, resistance, conductance;
b)The velocity of wave propagation.
Since the line is distortionless, the characteristic impedance is
2 2 9 7
ˆ
ˆˆ50 50 0.1 10 2.5 10
ˆc c
L
Z L Z C H m
C
− −
= = Ω ⇒ = = × × = ×
The attenuation constant is:
{ }
ˆ 0.01ˆ 0.01 1 20lg( ) 8.686 0.0012
ˆ 8.686
C
R dB m Np m e dB m Np m
L
α = = = = = = =
Therefore:
ˆ
ˆ 0.0012 50 0.0575
ˆ c
L
R Z m
C
α α= = = × = Ω

90
Lossy transmission lines (Ex)
The distortionless line criterium is:
5
2 2
ˆ ˆˆ ˆ ˆ 0.0575ˆ 2.3 10
ˆ ˆ ˆ 50c
R G RC R
G S m
ZL LC
−
= ⇒ = = = = ×
The phase velocity is:
8
7 9
1 1
2 10
ˆˆ 2.5 10 0.1 10
v m s
LC
− −
= = = ×
× × ×

91
Dispersion and group velocity
The losses of types considered previously (finite conductivity of conductors and
nonzero conductivity of real dielectrics) lead to attenuation of wave amplitude as
it propagates through the line.
When the wavelength is comparable with the physical dimensions of the line or
when the permittivity of the dielectric depends on the frequency, another
phenomenon called dispersion occurs.
We will model dispersion by the
insertion of a distributed parasitic
capacitance in parallel to the distributed
inductance.
While developing the telegrapher’s
equation for this case, we note that the
current entering the node will split into
two portions:
( , ) ( , ) ( , )L cI z t I z t I z t= + (9.91.1)

92
The voltage drop across the capacitor is:
( , )( , ) ˆ LI z tV z t
L
z t
∂∂
= −
∂ ∂
The voltage drop across the parasitic (shunt) capacitor is:
( , ) 1
ˆ c
s
V z t
I dt
z C
∂
= −
∂ ∫
Note: the units of this additional shunt capacitance are F⋅m rather than F/m.
The current passing through the shunt capacitor:
( , ) ( , )ˆI z t V z t
C
z t
∂ ∂
= −
∂ ∂
The wave equation will be:
2 2 4
2 2 2 2
( , ) ( , ) ( , )ˆ ˆˆ ˆ 0s
V z t V z t V z t
LC LC
z t z t
∂ ∂ ∂
− + =
∂ ∂ ∂ ∂
(9.92.1)
(9.92.2)
(9.92.3)
(9.92.4)

93
Assume that there is a time-harmonic signal generator connected to the infinitely
long transmission line. The complex time-varying wave propagating through the
line will be
( )
0( , ) j t z
V z t V e ω β−
=
Combining (9.92.4) and (9.93.1) leads to the dispersion relation (the terms in the
square brackets) relating the propagation constant β to the frequency of the wave:
( ) ( ) ( ) ( )
2 2 2 2( )
0
ˆ ˆˆ ˆ 0j t z
sV e j LC j LC j jω β
β ω ω β−  − − + − =
 
Therefore, the propagation constant is a nonlinear function of frequency:
2
ˆˆ
ˆˆ1 s
LC
LC
ω
β
ω
= ±
−
(9.93.1)
(9.93.2)
(9.93.3)

94
dispersion
no dispersion
The propagation constant
depends on frequency – this
phenomenon is called
dispersion.
The phase velocity is a function
of frequency also.

95
We can consider dispersion as a low pass filter acting on a signal. The
propagation constant will be a real number for frequencies less than the particular
cutoff frequency ω0.
0
1
ˆˆ
sLC
ω =
This cutoff frequency is equal to the resonant frequency of the “tank” circuit. Above
this frequency, the propagation constant will be imaginary, and the wave will not
propagate.
In the non-dispersive frequency range (below the cutoff frequency), the velocity of
propagation is
0
1
ˆˆ
V
LC
=
The wave number below the cutoff frequency is:
0
0
0v
ω
β =
(9.95.1)
(9.95.2)
(9.95.3)

96
Dispersion implies that
the propagation
constant depends on
the frequency.
There are positive and
negative dispersions.
Note that in the case of
negative dispersion,
waves of frequencies
less than a certain cutoff
frequency will propagate.
While for the positive dispersion, only the waves whose frequency exceeds the
cutoff frequency will propagate.

97
If two signals of different frequencies propagate through the same linear
dispersive medium, we must employ the concept of group velocity…
if a narrow pulse propagates in a dispersive region, according to Fourier
analysis, such a pulse consists of a number of high frequency components.
Each of them will propagate with different phase velocity.
Let us assume that two waves of the same amplitude but slightly different
frequencies propagate through the same dispersive medium. The frequencies are:
1 0 2 0;ω ω ω ω ω ω= + = −V V
The corresponding propagation constants are:
1 0 2 0;β β β β β β= + = −V V
The total signal will be a sum of two waves:
( ) ( )
( ) ( )
0 1 1 2 2
0 0 0
( , ) cos cos
2 cos cos
V z t V t z t z
V t z t z
ω β ω β
ω β ω β
= − + −  
= ∆ − ∆ −
(9.97.1)
(9.97.2)
(9.97.3)

98
Summation of two time-harmonic
waves of slightly different
frequencies leads to constructive
and destructive interference.
voltage
By detecting signals at two locations, we
can track a point of constant phase
propagating with the phase velocity:
0 0pv ω β=
and the peak of the envelope propagating with the group velocity: gv ω β= ∆ ∆
(9.98.1)
(9.98.2)
In dispersive media, phase and group velocities can be considerably different!

99
Dispersion and group velocity (Ex)
Example 9.19: Find the phase and group velocities for:
a)Normal transmission line;
b)A line in which the elements are interchanged.
a) The propagation
constant computed
according to (9.87.1)
will be:
( )( )ˆ ˆˆ ˆ ˆ ˆj YZ j L j C j LCγ α β ω ω ω= + = = =
The phase velocity is 1
ˆˆ
pv
LC
ω
β
= =
The group velocity is 1
1
ˆˆ
gv
LC
ω β
ωβ
∂ ∂
= = =
∂∂

100
Dispersion and group velocity (Ex)
The two velocities are equal in this case and both are independent on frequency.
b) The propagation constant computed according to (9.87.1) will be:
1 1 1ˆ ˆ
ˆ ˆ ˆˆ
j YZ
j Lj C j LC
γ α β
ωω ω
= + = = = −
The phase velocity is
2 ˆˆ
pv LC
ω
ω
β
= = −
The group velocity is
2 ˆˆ1gv LC
ω β
ω
ωβ
∂ ∂
= = =
∂∂
The phase and group velocities both depend on frequency and are in the
opposite directions.
?? QUESTIONS ??

Lecture 09 em transmission lines

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