Download III. Eliminating transmission line mismatch

Chapter 2
Part 1: Transmission Lines Basics
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Definition: A transmission line is a mechanical structure that guides EM energy in a
desired direction.
Basic properties of transmission lines:

TEM mode transmission lines: E and H are in the transverse plane, a plane
that is perpendicular to the direction of the EM wave propagation.
Commonly used lines consisting a pair of parallel conductors usually
belong to this class.

Higher-order transmission lines: One of the field components is in the
direction of the EM wave propagation. Waveguide types consisting a
hollow conductor usually belong to this class (next semester).

Dispersion in most commonly used transmission lines such as coaxial
lines is negligible. However, the line loss is usually high especially at
microwave frequencies.

On the other hand, waveguides suffer much lower losses at microwave
frequencies but dispersion can be a problem.
The most popular TEM lines are (Fig. 2-4, P. 38):

Coaxial cable: Used mainly in the high-frequency regime (MHz and up) at
low- to medium -power levels.

Parallel-wire: Used mainly for lower frequency transmission (AC power
line) at high-power levels.

Microstrip: Used mainly for microwave IC and PC board fabrications.
The most popular higher order transmission lines are (Fig. 2-4, P. 38):

Waveguide: Hollow conductors used at high frequencies and high-power.

Optical fiber: Concentric dielectric layers used at optical frequencies.
Differences and similarities between a transmission line and a waveguide:

They both are mechanical structure used to carry EM energy.

Transmission line has two parallel conductors; waveguide has only one,
usually hollow, conductor.

Waves inside transmission line are TEM; waveguide waves are TE or TM.

Transmission line has higher loss, especially at high frequencies.

Waveguide waves may be distorted due to dispersion.

Transmission line has no low-frequency limit; waveguide acts like a highpass filter.
In general, the impact on the signal integrity is much severe from the effect of
distortion than from the effect of losses.
Beside guiding EM energy, short sections of transmission line can sometimes be used
as high-frequency discrete circuit components such as Z transformer, filter, …
Simplifications (Fig. 2-1, P. 35):

Source: A voltage source and a series resistor are used to represent the
generator (Thevenin’s theorem)

Load: A resistance (impedance) is used to represent the load.

Transmission line: A two-port network is used to describe a general
transmission line.

The effect of short  (transmission line effect):

Guideline: l > 0.01  (1% of the wavelength)

Phase shift: The voltage and current along a transmission line generally
vary as a function of position.
Short 
Long 



Reflection: If the line is not terminated properly (with a match load)
portion of the energy will be reflected back to the generator. This is
usually bad because:
1) Reflected energy is wasted
2) Generator may be damaged if the reflection is significant.
Power loss: Energy is dissipated due to conduction loss and dielectric loss.
Distortion: A signal containing more than one frequency component may
be distorted due to dispersion—different frequency components traveling
at different speeds in a waveguide.
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Lumped-element model

We will first consider the general properties of a TEM mode line.

Similar to your calculus classes, we are going to cut a thin slice
(differential section) of the line and examine its electrical properties.
Transmission line parameters (Formula table—P. 41):

Series L’ and R’ per unit length

Parallel C’ and G’ per unit length
Transmission line equations (Telegrapher’s equations): We can derive a set of circuit
equations for a differential slice of the above model and from them a set differential
equations can be developed to describe the voltage and current in a transmission line.
~
dV
~ ~
~ ~
~
V  (V  dV )  I ( R ' j L' )dz  
 ( R ' j L' ) I
dz
~
dI
~ ~ ~
~
~
~
I  ( I  dI )  (V  dV )(G ' j C ' )dz  
 (G ' j C ' )V
dz
Wave equations: Combining the two 1st order/2-unknown equations we have a pair of
2nd order/1-unknown equations.
~
~
d 2V
d 2I
~
2 ~
 V  0
  2 I  0     j  ( R' j L' )(G ' j C ' )
2
2
dz
dz
~
~
  z
  z
V  V0 e  V0 e
I  I 0 e  z  I 0 e  z
V0 V0
R' j L'
   Z0 

I0
I0
G ' j C '
u p  f 


 – propagation constant
 – attenuation constant = Re{},  – phase constant = Im{},

up – phase velocity, Z0 – characteristic impedance = v/i for a single wave
The time domain solution to the voltage wave equation due to sinusoidal excitation is:
v( z, t )  V0 e  z cos(t  z    )  V0 e  z cos(t  z    )
 
Travels in the  z direction Travels in the  z direction

Electrical length:

The electrical length of a line is the ratio of the physical length to the
wavelength of the line signal. As a result, a physically long line may have
a short electrical length if a low-frequency signal (long ) is being
transmitted and vice versa.

Electrically short lines: For example, AC power lines less than 50 miles
long are usually considered electrically short. For such cables we can
consider them as discrete passive components.
R + jX

Medium-length lines: In general, power lines between 50 miles and 150
miles long are considered to be medium-length lines. We can also consider
them as discrete passive components with a more complex structure.
Z/2
Z/2
Y


Short and medium lines are important in low-frequency applications such
as power transmission. They are handled with lumped-circuit theory.

Electrically long lines: A line is considered to be electrically long if its
physical length is comparable to the wavelength. For example, a 15 cm
coaxial cable would be considered electrically long at 3 GHz ( = c/f =
3E8 / 3E9 = 0.1 m = 10 cm).

Because at high frequencies the physical length of an electrically long line
is often very short, the series resistance becomes negligible.

Since low-loss materials are usually required to fabricate high-frequency
cables, the shunt conductance can also be neglected.
Lossless (ideal) line (R’ = G’ = 0   = 0):
~
~
d 2V
d 2I
~
2 ~
 V 0
  2 I  0    L'C '
2
2
dz
dz
~
~
  j z
  j z
V  V0 e
 V0 e
I  I 0 e  j z  I 0 e  j z
V0 V0
   Z0 
I 0
I0
L'
C'
u p  f 



For all TEM lines:
G' 
L' C '   and

C' 
    
 
0
r
and
up 
where
0 
1

c
f

c
r
(free space wavelengt h)


Dispersion refers to the spreading of a signal due to the differences in
speed between different frequency components. It is often detrimental to
digital communication since digital pulses contain multiple frequency
components.

Since the phase velocity is independent of frequency, ideal TEM lines are
nondispersive.
Terminating an ideal line

The characteristic impedance Z0 of a transmission line is not a real
resistance even though it has a unit of ohm. Z0 of a line merely represents
the ratio of the voltage and current of one traveling wave along the line. If
more than one wave (for example, incident and reflected waves) are
present in the line, the ratio of the total voltage and total current would not
equal to Z0.
I-,V- I+,V+
Z0
ZL
l
z
VL  V0  V0

0
I L  I 0  I 0
V   V0 V0  V0
VL V0 V0


 0

Z L Z0 Z0
ZL
Z0
ZL
1
V
I
Z L  Z0 Z0
j r
   e 
 

Z L  Z0 Z L
V
I
1
Z0
If a transmission line is terminated with a load resistance ZL equal to its
characteristic impedance Z0 there will be no energy reflected from the
load. In other words, a matched load acts like an infinitely long line with
the same Z0.
For a mismatch load, part of the energy will be reflected back to the
generator. The amplitude of the reflected wave is according to the above
formula.
The portion of power reflected back is 2.
 = 1 for o/c and  = -1 for s/c
|| = 1 for purely reactive loads.

0

0






0

0


1.
2.
3.
4.
5.
Examples (P. 50-51)
More examples: Z0 = 50 
ZL = 55 
 = (ZL – Z0) / (ZL + Z0) = (55 – 50) / (55 + 50) = 5/105 = 0.048
2 = 0.0482 =0.0023 = 0.23%
ZL = 45 
 = (ZL – Z0) / (ZL + Z0) = (45 – 50) / (45 + 50) = -5/95 = -0.053
2 = 0.0532 =0.0028 = 0.28%
ZL = 5 
 = (ZL – Z0) / (ZL + Z0) = (5 – 50) / (5 + 50) = -45/55 = -0.82
2 = 0.822 =0.67 = 67%
ZL = 5 K
 = (ZL – Z0) / (ZL + Z0) = (5000 – 50) / (5000 + 50) =
4950/5050 = 0.98
2 = 0.982 =0.96 = 96%
ZL = 75 + j25 
Z0  50
 
ZL  75  25j 
ZL  Z0
ZL  Z0  25  25i 
ZL  Z0
ZL  Z0  125  25i 
ZL  Z0  35.355 
arg  ZL  Z0  45 deg
ZL  Z0  127.475 
arg  ZL  Z0  11.31 deg
35.355
127.475
 0.277
  0.231  0.154i
  0.277

45  11.31  33.69
arg     33.69 deg


 2  7.692 %
Impedance transformation:

The input impedance is the ratio of the total line voltage to the total line
current. This ratio in general varies along the line and thus the input
impedance is a function of the distance between the measurement point
and the load.

For an ideal line (lossless) terminated with a match load the input
impedance Zin measured at any distance always equals to Z0.

On the other hand, if ZL  Z0 then Zin will vary as a function of the distance
away from the load.
Zin
Z0
l
ZL

The input impedance Zin:
~
 e  j z   e  j z 
V ( z ) V0 e  j z  V0 e  j z
Z in ( z )  ~
   j z

Z
0   j z

 V0 e  j z
  e  j z 
I ( z ) V0 e
e
Z0
Let z  l
 e j l   e  j l 
1  e 2 j l 
Z in (l )  Z 0  j l

Z
0
 j l 
 2 j l 
 e  e

1   e

The input impedance Zin measured at any distance l from the load is:
Z cos  l  jZ 0 sin  l
Z  jZ 0 tan  l
Z in (l )  Z 0 L
 Z0 L
Z 0 cos  l  jZ L sin  l
Z 0  jZ L tan  l
As a result, the input impedance of an unmatched line can vary drastically
as a function of l.
l
l
0
0
90
/4
180
/2
270
3/4
360

Examples: Z0 = 50 , f = 3 GHz
1) ZL = 0  and  = 3E8 / 3E9 = 10 cm
Zin = jZ0 (sin  l / cos  l)
Zin = 0 for  l = 0, 180, 360, …
Zin =  for  l = 90, 270, …
A shorted line 2.5 cm long sometimes looks like an open.
2) ZL =   and  = 3E8 / 3E9 = 10 cm
Zin = jZ0 (cos  l / sin  l)
Zin =  for  l = 0, 180, 360, …
Zin = 0 for  l = 90, 270, …
An open line 2.5 cm long sometimes looks like a short.
3) For ZL = 30  and v = 0.8 c, find Zin for a 5.5 cm line



8
Z0  50
l  l
ZL  30
0.8 3 10 
9
f  3 10 Hz
2
l  5.5 10 m
360deg
Zin  Z0
l
l  247.5 deg

ZL cos  l  j  Z0 sin  l
360 deg
 
f
m
s
  0.08 m
 0.687
Z0 cos  l  j  ZL sin  l
ZL cos  l  11.481
j  Z0 sin  l  46.194i
ZL cos  l  j  Z0 sin  l  11.481  46.194i
11.481  46.194i  47.599
j  ZL sin  l  27.716i
Z0 cos  l  j  ZL sin  l  19.134  27.716i
19.134  27.716i  33.679
arg ( 11.481  46.194i)  103.957 deg
ZL cos  l  j  Z0 sin  l
Z0 cos  l  19.134
arg ( 19.134  27.716i)  124.62 deg

 1.413arg 
ZL cos  l  j  Z0 sin  l 
  20.663 deg
Z0 cos  l  j  ZL sin  l
Zin  66.119  24.935i
 Z0 cos  l  j ZL sin  l 
 24.935  
L 
9
L  1.323  10 H
Zin  70.665
arg  Zin  20.663 deg
 2  f 
4) For ZL = 30  and v = 0.8 c, find Zin for a 2.2 cm line
Z0  50
l 
l

 360deg
Zin  Z0
2
ZL  30
l  2.2 10
Z0 cos  l  j ZL sin  l
j  Z0 sin  l  49.384i
ZL cos  l  j Z0 sin  l
Z0 cos  l  j ZL sin  l
Zin  79.859  13.161i
9

C  2  3 10  13.161
j ZL sin  l  29.631i
Z0 cos  l  j ZL sin  l  7.822  29.631i
arg ( 19.134  27.716i)  124.62 deg
 ZL cos  l  j Z0 sin  l 
  9.359 deg
 Z0 cos  l  j ZL sin  l 
 1.619
arg 
Zin  80.936
1
Z0 cos  l  7.822
19.134  27.716i  33.679
arg ( 11.481  46.194i)  103.957 deg

  0.08
ZL cos  l  j Z0 sin  l
11.481  46.194i  47.599

9
3 10
ZL cos  l  j Z0 sin  l  4.693  49.384i

0.8 3 10
l  99 deg
ZL cos  l  4.693

8
 
arg  Zin  9.359 deg
 12
C  4.031  10
From the examples we conclude that a short section of transmission line
can transform a short to an open and vice versa. It can also transform a
capacitance to an inductance and vice versa.

Properly terminating the output of microwave amplifiers and oscillators is
important.

Tune circuits at microwave frequencies usually require L and C with
extremely small values. Using small segments of transmission line is one
way to accomplish this requirement.
Complete solution of the wave equations (Example 2-5, P. 57-58):
~
 Vg Z in  
1




V0  g

j

l

j

l
 Z g  Z in   e  e



Standing waves:

Transmission line waves (incident and reflected) are traveling waves in
opposite directions. It is usually difficult to observe them separately.

The combined wave displays an alternating pattern (repeats at every ½ )
of constructive and destructive interference between the incident and
reflective waves.

The envelop of this pattern is called the standing wave.

The best way to visualize the standing wave pattern is to visit the
following web site: http://graphics.cs.ucdavis.edu/~schussma/xl.html
In this transmission line simulation you can specify the magnitude and
phase of the reflection coefficient. The program will then display the
incident, reflective, and total wave along the line as well as the envelop of
the total wave (standing wave pattern).
!!! Everybody should try this !!!

Match load: Since no reflection exists in the line, no standing wave pattern
is observed (0 V swing between maximum and minimum).

Open-circuit (o/c) load: Since V+ and V- are the same at ZL, we expect the
load voltage to be doubled at ZL and complete cancellation at ¼  away (2
V swing). This pattern repeats every ½ .

Short-circuited (s/c) load: Since V+ and V- are opposite at ZL, we expect
complete cancellation at ZL and the voltage to be doubled at ¼  away (2
V swing). This pattern repeats every ½ .

Larger load-100 : Similar to o/c with less voltage swing between
maximum (0.667V) and minimum (0.333 V).

Smaller load-25 : Similar to s/c with less voltage swing between
maximum (0.667V) and minimum (0.333 V).

In summary, voltage maxima are located at n  ½  (1/2 , 1 , 1 1/2 ,
…) for resistive loads larger than Z0 and at ¼  + n  ½  (1/4 , 3/4 , 1
1/4 , …) for resistive loads smaller than Z0.
Minima are located at ¼  away from maxima.
The locations on the line corresponding to voltage maxima also
correspond to current minima, and vice versa.
In general, voltage maxima occur at locations where the incident and
reflective waves are in phase (constructive interference). For a complex
load impedance, the locations of voltage maxima are given by:
~
V ( z )  V0 (e  j z  e  j z )  V0 (e  j z   e j r e  j z )



   z   r   z   r  2  z  0,  2n
 l max 
 r  2 n   r  n


2
4
2
n  1,2,... if

n  0,1,2,... if
r  0
r  0




r
In general, voltage minima occur at locations where the incident and
reflective waves are out of phase (destructive interference). For a complex
load impedance, the locations of the first voltage maximum and minimum
are given by:
 r  2

 r  
if  r  0
 4  2  4  r  2   2 

lmax  
 r        r
if  r  0
r
 4
4
2
l   / 4 if lmax   / 4

 r      r      max
lmin 
4
2
lmax   / 4 if lmax   / 4
If we measure the first maximum or minimum of the standing wave the
phase angle of the reflective coefficient can be computed:
  4 

if lmax  or 2  lmax  
lmax     2  lmax
4
  
r  

 l    4   2  l  2
if lmax  or 2  lmax  
max
 max 2   
4
 r  2  lmin  
Summary of r calculations:


0
-
2
2 lmin
r
0

2
2 lmax
-

VSWR (Voltage Standing Wave Ratio): A quantity that can be measured
easily. The reflective coefficient and input impedance can be computed
from the VSWR and the locations of the first minimum or first maximum.
~
Vmax 1  
S ~ 
Vmin 1  
S 1
S 1
1   
Z L  Z0 
1   
Example 2-5 (Slotted-line), P.54-55



Special cases:

Short-circuited line: A short-circuited line behaves like a discrete reactive
component with:
Z inSC  jZ 0 tan  l
For length between 0 and ¼  it is inductive with equivalent L:
 Leq 
Z tan  l
1

Leq  0
 l  tan 1 


 Z0 




For length between ¼  and ½  it is capacitive with equivalent C:
 1 
1
1

Ceq  
 l    tan 1 
 C Z 
Z 0 tan  l
 
 eq 0 
The reactive type repeats beyond l > ½  following the same pattern (Fig.
2-15d, P. 59).
Open-circuited line: An open-circuited line behaves like a discrete reactive
component with:
 jZ 0
Z inOC   jZ 0 cot  l 
tan  l
For length between 0 and ¼  it is capacitive with equivalent C:
tan  l
1
Ceq 
 l  tan 1 Ceq Z 0 
Z 0

For length between ¼  and ½  it is with inductive equivalent L:
 Z 
Z0
1 
Leq  
 l    tan 1   0 
 L 
 tan  l
  2
eq 

The reactive type repeats beyond l > ½  following the same pattern (Fig.
2-17d, P. 61).
In microwave and high-speed circuits small C and L are often needed. It is
much easier to fabricate them using transmission lines segments.
Transmission parameters (Z0 and ) can be determined from the results of
OC and SC measurements (Example 2-8, P. 61).
  Z SC 
1
in 
Z 0  Z inSC  Z inOC
  tan 1 
OC 

l
 Z in 
1/2 wave window: A 1/2  section of transmission line, regardless of
characteristic impedance, can be inserted into an existing transmission line
system without disturbing the electrical properties of the system.
Z cos  l  jZ1 sin  l
Z (1)  jZ1 (0)
Z in  Z1 L
 Z1 L
 ZL
Z1 cos  l  jZ L sin  l
Z1 (1)  jZ L (0)
Zin = ZL
Z0
Z1
l
/2
ZL

1/4 wave transformer: A 1/4  section of transmission line (Z02) can be
inserted into used to match any Z01 to ZL.
Z cos  l  jZ 02 sin  l
Z (0)  jZ02 (1)
jZ (1)
Z in  Z 02 L
 Z 02 L
 Z 02 02  Z 01
Z 02 cos  l  jZ L sin  l
Z 02 (0)  jZ L (1)
jZ L (1)
 Z 02  Z 01  Z L


Zin = Z01
Z01
Z02
l
/4
ZL
Example: match a 50- line to a 200- load at 100 MHz (v = c/8).
/4 = (c/8) / f / 4 = 0.094 m
Z1 = (50  200)1/2 = 100 
Summary of lossless transmission line properties (Table 2-3, P. 63):

Zin is real at maxima and minima.

At voltage maxima: Zin = Z0 [(1 + ||)/(1 – ||)]

At voltage minima: Zin = Z0 [(1 – ||)/(1 + ||)]
Power flow in ideal lines

Instantaneous power:
P i (t ) 
Z0
P r (t )  

2
V
2
cos 2 ( t    )
V
Z0
Average power:
Pavi 
V
cos 2 ( t      r )
2
2Z 0
Pavr   
2
V
2
Pav 
2
V
2
2Z 0
2Z 0
   Pavi
2
1   
2
Chapter 2
Part 2: Smith chart



Introduction

The primary objective of the Smith chart is to graphically compute the
input impedance at any position along a transmission line terminated with
an arbitrarily load.

Powerful graphical tool for computing the complex impedance of
transmission line circuits (Originally called transmission line calculator).

Useful CAD presentation tool for displaying the performance (noise, gain,
stability, …) of microwave circuits.

Still a good way to design matching networks.

Useful for lossless or lossy lines.
Components of the chart

Polar plot for the reflection coefficient at the load (Fig. 2-20, P. 67).

Lines of a pair of parametric equations, one for the real (resistive) part and
the other for the imaginary (reactive) part, are plotted on the chart to
indicate the normalized load impedance (zL = ZL / Z0) (Fig. 2-21, P. 69).

Three concentric scales are on the perimeter of the chart (Fig. 2-22, P. 70).

The innermost scale indicates the phase angle of the reflection coefficient.

The other two scales indicate the distance, in terms of electrical length (),
of the movement of the measurement plane.
o The outermost scale WTG ( toward generator) indicates the
movement of the measurement plane toward the generator (away from
the load).
o The middle scale WTL ( toward load) indicates the movement of the
measurement plane toward the load (away from the generator).
Graphical calculations

The reflection coefficient  along the line: The magnitude of  is constant
along the line. The only other unknown is the phase angle r.
1) Plot the normalized load impedance (zL = ZL / Z0) on the chart.
2) Draw a radial line through zL until it intersects the r scale.
3) r is at the position where the radial line intersects the r scale.

VSWR:
1) Plot the normalized load impedance (zL = ZL / Z0) on the chart.
2) Starting at zL, draw a complete circle centering at the origin. This is
called the constant-|| circle, or more commonly known as the SWR
circle since S is a function of || only.
3) S equals to the value where the SWR circle intersects the  = 0
radial line.

The first voltage minimum equals to the angular displacement between zL
and the  = 180 radial line using the WTG scale (Fig. 2-24, P. 74).

The first voltage maximum equals to the angular displacement between zL
and the  = 0 radial line using the WTG scale. You can also use lmin 
¼  to obtain lmax.


The normalized input impedance (zin = Zin / Z0) along the line:
1) Plot the normalized load impedance (zL = ZL / Z0) on the chart.
2) Draw a radial line through zL until it intersects the WTG scale. This is
the starting position WTGstart.
3) From the physical distance between the measurement plane and the
load, compute the equivalent electrical length  (in ).
4) From the WTG scale compute the equivalent angular movement.
WTGfinal = WTGstart + 
5) Draw a radial line through WTGfinal.
6) Starting at zL, draw a complete circle centering at the origin. This is
called the constant-|| circle, or more commonly known as the SWR
circle since S is a function of || only.
7) zin is at the position where the WTGfinal radial line intersects the the
SWR circle.

Zin is real at  = 0 and is maximum (= Z0 S); Zin is real at  = 180
and is minimum (= Z0 / S)

Z-Y conversion: It is often more convenient to work with Y since parallel
components are easier to insert into a transmission line circuit (???) and
the result can be obtained by simply adding the Y’s.

In a Smith chart z(= Z/Z0) and y(= Y/Y0 = Y Z0) are 180 apart.

Example 2-10, P. 76-78 and 2-11, P. 78.

Examples 1-5
Impedance matching

Whenever a transmission line is not terminated with a matched load (ZL =
Z0, portion of the incident energy will be reflected back. This usually
causes two problems: 1) energy is wasted; 2) the reflected energy could
upset/damage the transmitter.

Components are added between the generator and the load to achieve
matching. This is called an impedance-matching network.

The match point in a Smith chart is at the origin where zin = yin = 1 + j 0

The single-stub matching technique:
1) Since the stub, which is a section of transmission line with a sliding
short at the end, is connected in parallel to the line, it is more
convenient to work with admittance than impedance. First transform zL
into yL by rotating the zL 180 along the SWR circle.
2) Determine where to attach the stub by rotating yL in the CW (WTG)
direction along the SWR circle until it intersects the real circle = 1. At
this location the yin has a real part = 1.
3) There will be two points on the chart that satisfy the above condition
(real{yin} = 1).
4) Measure the electrical length (in ) for the above rotation and convert
it into physical distance d. This should be the distance between the
load and the stub.
5) Determine the imaginary part (susceptance) of yin (= j B) from the
chart.
6) Starting from the short circuit point (where and why), rotate in the CW
(WTG) direction along the SWR =  circle (outer rim) until it
intersects the imaginary line = -j B. At this location the yin of the stub
has a susceptance = -j B.
7) Measure the electrical length (in ) for the above rotation and convert
it into physical distance l. This should be the length of the stub.

Example 2-12, P. 80-84.

Example 6
Discrete conjugate matching: Handout
Chapter 2
Part 3: Transient (Pulse) Transmission Line Responses




Introduction

So far we have studied the high-frequency behavior of transmission lines
under harmonic excitations, in this part we are going to examine the pulse
responses of transmission lines.

Because of reflections due to mismatch, an electrically long transmission
line may respond irregularly to pulsed signals.

Transient response of a line is a time record of the pulses bouncing
between the transmitting and receiving end.
Transmission line responses

The steady-state response can be calculated using ordinary dc circuit
analysis techniques.

The transient response calculations must include the reflected waves
between the generator and load.

The reflections may cause severe distortion to the wave shape of the
signal.

They may even damage the receiver or transmitter because of EOS
(electric overstress) or polarity reversal.
Computing the transient response

The transient response is usually difficult to compute. To simplify the
analysis, a rectangular pulse is decomposed into two step functions. The
step response is then computed and the pulse response is computed by
adding the two step responses (Fig. 2-32, P. 84).

Bounce diagram (Fig. 2-35, P. 89): The bounce diagram is used to keep
track of the multiple reflections along the line.

TDR: Measures the round-trip time as well as the shape of the reflected
pulse to determine the distance and the type of discontinuity. Used for
fault detection, location, and identification. It can also be used to measure
the length of long transmission lines (Example 2-13, P. 89).
Computer simulation results:

RG = RL = 10

RG = RL = 100

RG = 10 RL = 100

RG = 100 RL = 10
Chapter 2
Appendix
I.


Transient Response
The response of a circuit can be extremely complex. To simplify the analysis we
often separate the response into two parts:

Steady-state response: This response represents the behaviors of the circuit
after a long period of time (until the effect of initial conditions has
diminished to a negligible level). The steady-state response can usually be
computed using basic AC/DC circuit laws.

Transient response: This response represents the behaviors of the circuit
immediately after the input is applied. It is usually more difficult to solve
since it involves both the input and the initial conditions.

There are other classifications of the circuit responses such as frequency
response vs. time response, force response vs. natural response, and zerostate response vs. zero-input response.
Properties of transient response:

Preshoot: A change of amplitude in the opposite direction before the pulse.

Overshoot: The overshooting of the amplitude after the initial transition.

Rounding: The rounding of the amplitude after the initial transition.

Ringing: The peak-to-peak amplitude following the overshooting.
Note: The above parameters are expressed as % of the steady-state amplitude.

Period: The duration required for the waveform to repeat itself.

Pulse width: The time duration between the 50% point of the leading and
trailing edges of the pulse.

Rise time: The duration between the 10 and 90% of the steady-state value.

Settling time: The duration required to reach the steady-state amplitude
after the pulse has reached the 100% point.

Droop: The decrease of amplitude with time.

Nonlinearity: The deviation from the line between the 10 and 90% points.
II.





Electrical Component Parameters
Conductance (G) / resistance (R): The basic electrical property of a resistor is how
well (poorly) it conducts electricity (AC and DC).

R – The resistance of a resistive component.

G – The conductance of a resistive component.

R=1/G
Reactance (X) / susceptance (B): The basic electrical property of a capacitor or an
inductor (reactive component) is how much it shifts the phase angle of an AC
signal at a particular frequency.

X – The reactance of a reactive component.
XC = -1/2fC
XL = 2fL

B – The susceptance of a reactive component.
BC = 2fC
BL = -1/2fL

X=1/B
Inductance (L) / capacitance (C):

L: XL > 0, BL < 0

C: XC < 0, BC > 0
Impedance (Z) / admittance (Y): The combined property of a component or circuit
with both resistance and reactance.

Z = R + jX

Y = G + jB

Z=1/Y
Z – Y Conversion:

Z to Y:
Z  R  jX
1
Y 

Z
1
R2  X 2
G  Y cos Y

  1  X 
  tan  
 R 
 
 Y   Z  
   tan 1  X 
 
 
 R 

B  Y sin  Y
if R  0
if R  0
Y  G  jB
Y to Z:
Y  G  jB
1
1
Z  
2
Y
G  B2
R  Z cos Z
Z  R  jX
  1  B 
  tan  
 G 
 
 Z   Y  
   tan 1  B 
 
 
 G 

X  Z sin  Z
if G  0
if G  0
III.


Eliminating transmission line mismatch
Reference: David Royle; Designer’s Guide to Transmission Lines and
Interconnections; EDN, June 1988.
Criteria:

Time domain: The time required to travel the length of the interconnection
is greater than the 1/8th of the signal rise time.
Note: Rise time refers to the rise or fall time, whichever is smaller, and it corresponds to the
0% to 100% time duration. To convert from the 10% – 90% rise time to the 00% – 100%
rise time multiply by 1.25.


Frequency: The length of the interconnection is greater than 1/15th of the
wavelength of the highest frequency component of the signal.
Remedies: