NETWORK ANALYSIS
AND SYNTHESIS
FRANKLIN
F.
KUO
SECOND EDITION
£1.50
i-
->
PETER CLARKE
SECOND HAND BOOKS
T«l.
Rochdale 50514
Network
Anal/sis and
Synthesis
Second Edition
Wiley International Edition
Network
Analysis and
Synthesis
Second Edition
by Franklin
Bell
F.
Kuo
Telephone Laboratories,
John Wiley
&
Sons,
Inc.,
New York
Toppan Company. Ltd.
,
|
London
|
Inc.
Sydney
Tokyo, Japan
©
1962, 1966 by John Wiley
Copyright
All Rights Reserved
& Sons, Inc.
This book or any part thereof
must not be reproduced in any form
without the written permission of the publisher
Wiley International Edition
This book
to
which
it
not to be sold outside the country
is consigned by the publisher.
is
Libray of Congress Catalog Card
Number
:
66-16127
Printed in Singapore by Toppan Printing Co. (S) Pte. Ltd.
To My
Father and Mother
Preface
In the second edition, I have tried to keep the organization
of the first
Most of the new material are additions aimed at strengthening
the weaknesses of the original edition. Some specific
changes deserve
mention. The most important of these is a new chapter
edition.
on computer
computers have
brought about many significant changes in the content of
engineering
subject matter concerned with both analysis and
design. In analysis,
computation has become an important adjunct to theory. Theory
estate
lishes the foundation of the subject matter; computation
provides clarity,
depth, and insight. In design, the computer has not
only contributed
precision and speed to existing procedures but has
made practicable
design methods that employ iteration and simulation. The
importance of
computer-aided design cannot be overemphasized. In Chapter 15
I have
attempted to survey some digital computer applications in the
areas of
network analysis and design. I strongly encourage all students
applications (Chapter 15).
In the past
five years, digital
to read
this chapter for cultural interest, if
It
not for survival.
Another new section contains a rigorous treatment of the unit impulse.
was difficult to decide whether to incorporate this material in Chapter
2
in the discussion of signals or in a separate appendix.
ized functions in
By putting
general-
an appendix, I have left the decision of whether to teach
the rigorous treatment up to the individual instructor.
Other changes worth mentioning are: (1) two new sections
on the
Fourier integral in Chapter 3 (2) a section on initial and
final conditions
in Chapter 5; (3) a section on Bode plots in Chapter
8; (4) revised
material on two-port parameters in Chapter 9; and
(5) new sections on
frequency and transient responses of filters in Chapter 13. Major
or
minor changes may be found in every chapter, with the exception
of
Chapters 11 and 12. In addition, many new problems are included
at the
end of each chapter.
;
vii
Preface
viii
Chapters 1 and 2
brief description of the subject matter follows.
characteristics of
general
certain
deal with signal representation and
and includes the
analysis,
Fourier
linear systems. Chapter 3 deals with
A
and 5
impulse method for evaluating Fourier coefficients. Chapters 4
domain.
time
the
in
equations
differential
discuss solutions of network
two preceding
In Chapters 6 and 7, the goals are the same as those of the
domain.
frequency
the
of
that
here
is
viewpoint
chapters, except that the
system function.
Chapter 8 deals with the amplitude, phase, and delay of a
Chapter
synthesis.
network
with
concerned
are
chapters
The final seven
realizability theory
of
elements
the
10,
Chapter
In
two-ports.
9 deals with
elementary drivingare presented. Chapters 11 and 12 are concerned with
Chapter 13, some
In
procedures.
synthesis
function
transfer
and
point
Chapter 14
fundamental concepts in modern filter design are introduced.
design.
and
analysis
deals with the use of scattering matrices in network
digital
of
survey
brief
And, as mentioned earlier, Chapter 15 contains a
five
are
there
addition,
computer techniques in system analysis. In
functions,
appendices covering the rudiments of matrix algebra, generalized
complex variables, proofs of Brune's theorems, and a visual aid to
filter
approximation.
intended for a two-semester course in network theory.
or
Chapters 1 through 8 may be used in a one-semester undergraduate
analysis.
system
linear
or
analysis
transient
beginning graduate course in
The book
is
network
Chapters 9 and 15 are to be used in a subsequent course on
synthesis.
The second edition is largely a result of the feedback from the professors
who have used this book and from their students, Who have discovered
express my sincere
errors and weaknesses in the original edition. I wish to
who provided this feedback.
due to the following people: Robert Barnard of
are
thanks
Special
Belove and Peter Dorato of the Polytechnic
Charles
State,
Wayne
Sherman of the Bell
Institute of Brooklyn, James Kaiser and Philip
Telephone Laboratories, Evan Moustakas of San Jose State College,
Welch of the University of Texas, and David Landgrebe of Purdue.
appreciation to those
A.
I
J.
am particularly indebted to Mac Van Valkenburg of Princeton University
and Robert Tracey of Illinois for editorial advice and to Donald Ford
of Wiley for help and encouragement.
Zicchino, and
addition, I wish to thank Elizabeth Jenkins, Lynn
In
Joanne Mangione of the Bell Telephone Laboratories for their
and careful typing of the manuscript.
F. F.
»
tj
u.
Heights,
Berkeley
•
i
New Jersey,
December, 1965
efficient
Kuo
Preface to the First Edition
This book is an introduction to the study of electric networks based upon
a system theoretic approach. In contrast to many present textbooks,
the emphasis is not on the form and structure of a network but rather on
excitation-response properties. In other words, the major theme is
concerned with how a linear network behaves as a signal processor.
Special emphasis is given to the descriptions of a linear network by its
system function in the frequency domain and its impulse response in the
time domain. With the use of the system function as a unifying link, the
its
transition
from network
analysis to synthesis can
be accomplished with
relative ease.
The book was
originally conceived as a set of notes for a second course
in network analysis at the Polytechnic Institute of Brooklyn. It assumes
had a course in steady-state circuit analysis.
should be familiar with Kirchhoff's laws, mesh and node equations,
standard network theorems, and, preferably, he should have an elementary
that the student has already
He
understanding of network topology.
A
brief description of the subject matter follows.
Chapters
1
and 2
deal with signal representation and certain characteristics of linear net3, 4, 5, and 6 discuss transient analysis from both a time
domain viewpoint, i.e., in terms of differential equations and the impulse
response, and a frequency domain viewpoint using Fourier and Laplace
transforms. Chapter 7 is concerned with the use of poles and zeros in
both transient and steady-state analysis. Chapter 8 contains a classical
works. Chapters
treatment of network functions.
The final five chapters deal with network synthesis. In Chapter 9, the
elements of realizability theory are presented. Chapters 10 and 1 1 are
concerned with elementary driving-point and transfer function synthesis
procedures. In Chapter 12, some fundamental concepts in modern filter
design are introduced. Chapter 13 deals with the use of scattering matrices
ix
Preface to the First Edition
x
in network analysis
and
synthesis.
In addition, there are three appendices
and proofs
covering the rudiments of matrix algebra, complex variables,
of Brune's realizability theorems.
The book is intended for a two-semester course in network theory.
or
Chapters 1 through 7 can be used in a one-semester undergraduate
analysis.
system
beginning graduate course in transient analysis or linear
Chapters 8 through 13 are to be used in a subsequent course on network
synthesis.
was my very good fortune to have studied under Professor M. E. Van
Valkenburg at the University of Illinois. I have been profoundly influenced
emphasis
by his philosophy of teaching and writing, which places strong
It
of exposition. In keeping with this philosophy, I have tried
viewpoint, and I have
to present complicated material from a simple
exercises. In addition,
and
examples
illustrative
included a large number of
rigor and
mathematical
between
ground
middle
I have tried to take a
upon
clarity
Unless a proof contributes materially to the
it is omitted in favor of an intuitive argument.
theorem,
of
a
understanding
of unit impulses, a development in terms
treatment
in
the
example,
For
intuitive understanding.
unit
of a generalized function is first introduced. It is stressed that the
whose
functions
of
sequence
a
actually
but
function
a
not
really
is
impulse
notion of an
limit point is undefined. Then, the less rigorous, intuitive
the
impulse "function" is presented. The treatment then proceeds along
nonrigorous path.
There are a number of topics which have been omitted. One of these
at present. I have puris network topology, which seems to be in vogue
that
posely omitted topology because it seems out of place in a book
analysis.
network
de-emphasizes the form and structure approach to
In an expository book of this nature, it is almost impossible to reference
field of
adequately all the original contributors in the vast and fertile
network theory. I apologize to those whose names were omitted either
through oversight or ignorance. At the end of the book, some supplementary textbooks are listed for the student who either wishes to fill in
some gaps in his training or wants to obtain a different point of view.
given to me by my
I acknowledge with gratitude the help and advice
former colleagues
by
my
and
Laboratories
Telephone
Bell
colleagues at the
my sincere
express
wish
to
I
Brooklyn.
of
Institute
Polytechnic
at the
were
appreciation to the many reviewers whose advice and criticism
invaluable in revising preliminary drafts of the manuscript. Professors
Stanford
R. D. Barnard of Wayne State University and R. W. Newcomb of
entire
of
the
University deserve specific thanks for their critical reading
manuscript and numerous helpful suggestions and comments.
Miss Elizabeth
In addition, I wish to thank Mrs. Elizabeth Jenkins and
Preface to the First Edition
La
xi
Jeunesse of the Bell Telephone Laboratories for their efficient and
careful typing of the manuscript.
Finally, to my wife Dora, I owe a special debt of gratitude. Her
encouragement and cooperation made the writing of this book an
enjoyable undertaking.
Murray
Hill,
January, 1962
New Jersey,
F. F.
KUO
Contents
Chapter
I:
Signals and Systems
I
1.1
Signal Analysis
1
1.2
Complex Frequency
Network Analysis
Network Synthesis
4
1.3
1.4
Chapter 2:
Signals
7
14
and Waveforms
20
General Characteristics of Signals
General Descriptions of Signals
The Step Function and Associated Waveforms
The Unit Impulse
20
24
28
33
The Frequency Domain: Fourier Analysis
46
3.1
Introduction
3.2
Orthogonal Functions
Approximation Using Orthogonal Functions
46
47
48
50
52
2.1
2.2
2.3
2.4
Chapter 3:
3.3
3.4
3.5
3.6
Fourier Series
Evaluation of Fourier Coefficients
Evaluation of Fourier Coefficients Using Unit
Impulses
3.7
The Fourier
3.8
Properties of Fourier Transforms
Integral
xiii
58
63
67
xiv
Contents
Chapter 4:
Differential Equations
75
4.1
Introduction
75
4.2
Homogeneous Linear Differential Equations
Nonhomogeneous Equations
76
4.4
Step and Impulse Response
82
85
4.5
Integrodifferential Equations
91
4.6
Simultaneous Differential Equations
93
4.3
Chapter 5:
Network
Analysis:
100
I
100
103
106
5.1
Introduction
5.2
5.5
Network Elements
Initial and Final Conditions
Step and Impulse Response
Solution of Network Equations
5.6
Analysis of Transformers
114
122
The Laplace Transform
134
6.2
The Philosophy of Transform Methods
The Laplace Transform
6.3
Properties of Laplace Transforms
134
135
137
6.4
Uses of Laplace Transforms
Partial-Fraction Expansions
Poles and Zeros
Evaluation of Residues
The Initial and Final Value Theorems
5.3
5.4
Chapter 6:
6.1
6.5
6.6
6.7
6.8
Chapter
7:
7.1
7.2
7.3
7.4
7.5
7.6
Transform Methods
in
Network Analysis
The Transformed Circuit
Thevenin's and Norton's Theorems
The System Function
The Step and Impulse Responses
The Convolution Integral
The Duhamel Superposition Integral
HI
144
148
155
162
165
175
175
180
187
194
197
201
Contents
Chapter 8:
212
8.1
Amplitude and Phase Response
212
8.2
Bode
221
Plots
8.3
Single-Tuned Circuits
8.4
Double-Tuned Circuits
On Poles and Zeros and Time Delay
229
238
245
Network
253
8.5
Chapter 9:
Analysis:
II
9.1
Network Functions
9.2
Relationships Between Two-Port Parameters
Transfer Functions Using Two-Port Parameters
9.4
Interconnection of Two-Ports
9.5
Incidental Dissipation
9.6
Analysis of Ladder Networks
253
264
266
271
276
279
10:
Elements of Readability Theory
290
9.3
Chapter
Amplitude, Phase, and Delay
xv
10.1
Causality and Stability
10.2
Hurwitz Polynomials
Positive Real Functions
Elementary Synthesis Procedures
290
294
299
308
Synthesis of One-Port Networks with Two Kinds
of Elements
315
10.3
10.4
Chapter
11:
11.1
11.2
11.3
11.4
11.5
11.6
Properties of L-C Immittance Functions
Synthesis of L-C Driving-Point Immittances
Properties of R-C Driving-Point Impedances
Synthesis of R-C Impedances or R-L. Admittances
Properties of R-L Impedances and R-C Admittances
Synthesis of Certain
R-L-C Functions
315
319
325
329
331
333
xvi
Contents
Elements of Transfer Function Synthesis
341
12.1
Properties of Transfer Functions
341
12.2
Zeros of Transmission
Synthesis of Yn and ZS1 with a 1-Q Termination
Synthesis of Constant-Resistance Networks
345
347
352
Topics in Filter Design
365
The Filter Design Problem
The Approximation Problem in Network Theory
The Maximally Flat Low-Pass Filter Approximation
365
365
368
373
388
392
Chapter
12:
12.3
12.4
Chapter
13:
13.1
13.2
13.3
13.4
Other Low-Pass Filter Approximations
13.5
Transient Response of Low-Pass Filters
13.6
13.7
A Method to Reduce Overshoot in Filters
A Maximally Flat Delay and Controllable Magnitude
Approximation
Synthesis of Low-Pass Filters
13.9 Magnitude and Frequency Normalization
13.10 Frequency Transformations
13.8
Chapter
14:
395
397
402
404
413
The Scattering Matrix
14.1
Incident and Reflected
14.2
14.3
The
The
14.4
Properties of the Scattering Matrix
14.5
Insertion Loss
413
415
419
426
429
14.6
Darlington's Insertion Loss Filter Synthesis
431
Chapter
15:
Power Flow
Network
a Two-Port Network
Scattering Parameters for a One-Port
Scattering Matrix for
Computer Techniques
in Circuit Analysis
in Circuit Analysis
15.1
The Uses of Digital Computers
15.2
Amplitude and Phase Subroutine
A Fortran Program for the Analysis of Ladder
Networks
Programs that Aid in Darlington Filter Synthesis
15.3
15.4
438
438
450
453
457
Contents
Appendix A:
xvii
Introduction to Matrix Algebra
461
Fundamental Operations
Elementary Concepts
Operations on Matrices
Solutions of Linear Equations
461
A.2
A.3
A.4
A. 5
References
A.1
Appendix B:
462
464
468
469
on Matrix Algebra
Generalized Functions and the Unit Impulse
470
B.l
Generalized Functions
B.2
Properties of the Unit Impulse
470
476
Elements of Complex Variables
481
C.l
Elementary Definitions and Operations
C.2
C.3
C.4
Analysis
481
483
486
487
Appendix C:
Appendix D:
Appendix E:
Singularities and Residues
Contour Integration
Proofs of
Functions
An Aid
Some Theorems on
Positive
Real
490
to the Improvement of Filter Approxi-
mation
493
493
494
495
496
498
502
504
E.l
Introduction
E.2
E.3
E.4
Constant Logarithmic Gain Contours
Constant Phase Contours
Contour Drawings
E.5
Correction Procedure
E.6
Correction Network Design
E.7
Conclusion
Bibliography
£05
Name Index
$09
Subject Index
si I
chapter
I
Signals and systems
This book is an introduction to electric network theory. The first half
of the book is devoted to network analysis and the remainder to network
synthesis and design. What are network analysis and synthesis? In a
generally accepted definition of network analysis
and
synthesis, there are
three key words: the excitation, the network, and the response as depicted
in Fig. 1.1. Network analysis is concerned with determining the response,
given the excitation and the network. In network synthesis, the problem
to design the network given the excitation and the desired response.
In this chapter we will outline some of the problems to be encountered
in this book without going into the actual details of the problems.
is
We
will also discuss
I.I
some
basic definitions.
SIGNAL ANALYSIS
networks, the excitation and response are given in terms of
and currents which are functions of time, t. In general, these
functions of time are called signals. In describing signals, we use the two
universal languages of electrical engineering— time and frequency. Strictly
speaking, a signal is a function of time. However, the signal can be
For
electric
voltages
described equally well in terms of spectral or frequency information. As
between any two languages, such as French and German, translation is
needed to render information given in one language comprehensible in the
Response
Excitation
Network
>-
FIG.
I.I.
The
>
objects of our concern.
Network
analysis and synthesis
m
Ao-
FIG.
Sinusoidal signal.
1.2.
is effected by the
and the Laplace transform. We shall
have ample opportunity to define and study these terms later in the book.
At the moment, let us examine how a signal can be described in terms of
both frequency and time. Consider the sinusoidal signal
other.
Between time and frequency, the translation
Fourier series, the Fourier integral,
s(t)
where
A
is
the amplitude,
=A
O is
sin (a>
t
+
the phase
0.1)
O)
shift,
and
<w
is
the angular
frequency as given by the equation
coo
where
=
Y
<
L2>
T is the period of the sinusoid. The signal is plotted aguinst time
An equally complete description of the signal is obtained if we
in Fig. 1.2.
aD Ao
io.
E
W
b>0
Angular frequency
FIG.
1.3a. Plot
of amplitude
A
versus angular frequency
o>o
to
Angular frequency
FIG. 1.3b. Plot of phase 6 versus angular frequency
<o.
<o.
Signals and systems
— «a
—6>4 —6)3
FIG.
±
— a>2 — 6>i
&>0
1.4a. Discrete
— fc>4 — W3 — ft)2 —
FIG.
let
the angular frequency
signal
is
is
is
+<<>
ft>4
0)2
0>3
ft>4
+0)
phase spectrum.
be the independent variable. In
described in terms of ^4
amplitude
T
ft>i
1.4b. Discrete
co
"3
amplitude spectrum.
b)i
—ft)
»2
«">1
>
w o>
an£l
plotted against frequency,
^0.
as
and in
shown in
Fig. 1.36,
this case, the
where
where phase shift
Fig. 1.3a,
plotted.
Now suppose that the signal is made up of 2n +
s(t)
=%A
<—
i
sin
(<v
+
1
sinusoidal components
(1.3)
fy)
The spectral description of the signal would then contain 2n + l lines at
±(»„ as given in Figs. 1.4a and b. These discrete spectra
±<*>i, ±«>2
of amplitude A versus co and phase shift 6 versus co are sometimes called
line spectra. Consider the case when the number of these spectral lines
become infinite and the intervals oo i+1 — w f between the lines approach
zero. Then there is no longer any discrimination between one frequency
and another, so that the discrete line spectra fuse into a continuous spectra,
as shown by the example in Figs. 1.5a and b. In the continuous case, the
sum in Eq. 1.3 becomes an integral
xo
where A{m)
is
known
£
A(o>) sin [mt
+
0(to)]
as the amplitude spectrum
dm
and
(1.4)
0(co)
as the phase
spectrum.
As we
shall see later, periodic signals
such as the sine wave in Fig. 1.2
can be described in terms of discrete spectra through the use of Fourier
series. On the other hand, a nonperiodic signal such as the triangular
Network
4
anal/sis
and synthesis
Mo>)
—a
<a
FIG.
1.5a.
FIG.
Continuous amplitude spectrum.
1.5b.
Continuous phase spectrum.
pulse in Fig. 1.6 can only be described in terms of continuous spectra
through the Fourier integral transform.
1.2
COMPLEX FREQUENCY
In this section,
we
shall see, the
we
will consider the concept of
complex frequency. As
complex frequency variable
s
= a + ja>
(1.5)
a generalized frequency variable whose real part a describes growth and
decay of the amplitudes of signals, and whose imaginary part/cu is angular
frequency in the usual sense. The idea of complex frequency is developed
by examining the cisoidal signal
is
S(f)
FIG.
1.6.
=
aJ»*
Ae
Triangular signal.
(1.6)
Signals and systems
5
ReS
FIG.
1.7.
Rotating phasor.
when S(0 is represented as a rotating phasor, 1 as shown in Fig. 1.7.
The angular frequency m of the phasor can then be thought of as a velocity
end of the phasor. In particular the velocity to is always at right
shown in Fig. 1.7. However, consider the general
case when the velocity is inclined at any arbitrary angle y> as given in
Figs. 1.8a and 1.86. In this case, if the velocity is given by the symbol s,
we see that s is composed of a component m at right angle to the phasor S
as well as a component a, which is parallel to S. In Fig. 1.8a, s has a
component — a toward the origin. As the phasor S spins in a counterclockwise fashion, the phasor decreases in amplitude. The resulting wave
for the real and imaginary parts of S(r) are damped sinusoids as given by
at the
angles to the phasor, as
ReS(f)
ImS(0
M
= Ae~" cos cot
= X<r*'sina>f
(1.7)
]<b)
FIG. 1.8. (a) Rotating phasor with exponentially decreasing amplitude,
phasor with exponentially increasing amplitude.
1
(b)
Rotating
A phasor S is a complex number characterized by a magnitude and a phase angle
(see
Appendix C).
6
Network
analysis
and synthesis
/Envelope = Ae~
FIG.
1.9.
Damped
sinusoids.
/ ^~ Envelope = Ae**
FIG.
1. 10.
Exponentially increasing sinusoid.
Signals and systems
7
t
FIG.
which are shown in
I.I I.
Fig. 1.9.
Exponential signals.
Note that the damped sinusoid has an
In Fig. 1 .84, the phasor is shown with
exponential envelope decay, Ae~" %
.
component of velocity +o. Therefore, as the phasor spins,
the amplitudes of the real and imaginary parts increase exponentially
with an envelope Ae+at as shown by Im S(f) in Fig. 1.10.
a positive
real
,
From
this discussion, it is
S(0
apparent that the generalized cisoidal signal
= Ae" = Ae {a+Mt
(1.8)
and decay of the amplitudes in addition to angular
When a = 0, the sinusoid is undamped, and
the signal is an exponential signal
describes the growth
frequency in the usual sense.
wheny'a>
=
0,
= Ae*'
(1.9)
Finally, if a = ja> = 0, then the signal is a constant
S(0
as
shown in Fig.
A. Thus
1.3
we
1.11.
see the versatility of a
complex frequency
description.
NETWORK ANALYSIS
As mentioned before, the characterization of the excitation and response
and frequency makes up only part of the analysis problem.
The other part consists of characterizing the network itself in terms of
signals in time
time and frequency, and determining how the network behaves as a signal
processer. Let us turn our attention now to a brief study of the properties
of linear networks and the general characteristics of signal processing by
a linear system.
Network
8
analysis
and synthesis
BASIC DEFINITIONS
Linear
A
system (network) is linear if (a) the principle of superposition and
of proportionality hold.
(b) the principle
By
the superposition principle,
[eg(0, r 2(f)l
e[t)
if,
for a given network, [ex (t), r^t)]
and
are excitation-response pairs, then if the excitation were
= ex(t) + e2(t),
would be
the response
r(t)
= r,(f) + r£i).
By
the
C
proportionality principle, if the excitation were C^it), where
x is a
constant, then the response would be C^r^t), i.e., the constant of propor-
Cx
preserved by the linear network. The two conditions of
summarized in Fig. 1.12.
Another definition of a linear network is that the excitation and response
of the network are related by a linear differential equation. We shall
discuss this definition in Chapter 4 on differential equations.
tionality
is
linearity are
Passive
A linear network is passive* if (a) the energy delivered to the network is
nonnegative for any arbitrary excitation, and (b) if no voltages or currents
appear between any two terminals before an excitation is applied.
Reciprocal
A network is said to be reciprocal if when the points of excitation and
measurement of response are interchanged, the relationship between
Cieift)
C2 e2(t)
..
System
System
CuiM
.
C2 r^t)
_
Cm(t) *
Ci*t(0+Ck*gt)
System
FIG.
1.12.
CW<)
linear system.
1
G. Raisbeck, "A Definition of Passive Linear Networks in Terms of Time and
Energy," /. Appl. Phys., 25 (Dec. 1954), 1510-1514.
9
Signals and systems
tit)
r(t)
A
B
>-
System
-
— —.M.
t
t
3<•
Ht-Tit
-Ti)
A
7\
TihTi
Tl
t
FIG.
excitation
B
^
System
1.13.
(
Time-invariant system.
and response remains the same, Thus must be true
for any
choice of points of excitation and response.
Causal
We say a system is causal if its response is nonanticipatory, i.e., if
=
KO = °
«(0
then
In other words, a system is causal
oo
/
T, the response is zero for
—
=
Time
t
<T
'
<r
(1.10)
if
before an excitation
< <
t
is
applied at
T.
invariant
A system is time invariant ife(t) -* rtf) implies that e(t ±
T)-*-r(t± T),
where the symbol -» means "gives rise to." To understand the concept
of time invariance in a linear system, let us suppose that initially the
excitation is introduced at f = 0, which gives rise to a response r(t). If
the excitation were introduced at / = T, and if the shape of the response
waveform were the same as in the first case, but delayed by a time T
(Fig. 1.13), then we could say the system is time invariant. Another way
of looking at this concept is through the fact that time-invariant systems
contain only elements that do not vary with time. It should be mentioned
here that linear systems need not be time invariant.
Derivative property
From the time-invariant property we can show that,
if e{f)
at the input
gives rise to r(i) at the output (Fig. 1.14), then, if the input were
e'(f),
Network
10
analysis
and synthesis
System
de(t)
System
>
*
dr(t)
System
,
d*Ht)
'
dt'
it
d*e(t)
dt*
—
FIG.
1.14.
Some
k
>
System
e(T)dr
dt
C'.{,\
l'(T)
J
1
implications of linear time-invariant systems.
r'(t).
The proof is quite
where e is a real quantity. By the
time-invariant property, the response would be r(t + e). Now suppose the
i.e.,
the derivative of e(t), the response would be
simple. Consider an excitation e(*
+
e)
excitation were
e1 (r)--W*
+
«)-««)]
(1-11)
€
then according to the linearity and time-invariant properties, the response
would be
= - [r(I + e) -
ri (0
Taking the limit as
e ->- 0,
we
K0]
(112)
see that
limc1(0
e-»0
= 7-c(0
at
(1.13)
lim rx(0
«-»o
= ^ K0
at
We
can extend this idea to higher derivatives as well as for the integrals of
e(t)
and
Ideal
r(f),
as
shown
in Fig. 1.14.
models
Let us now examine some idealized models of linear systems. The
systems given in the following all have properties which make them very
useful in signal processing.
1
Signals and systems
Km.
e(t)
Amplifier
FIG.
1.15. Amplifier.
fit)
K dtd
K
FIG.
.
1
>K i
FIG.
Amplifier:
1.
An
1.18.
Time-delay network.
2
1.19.
2. Differentiator:
amplifier scales
The input
t
Excitation function.
up
= Ke(t), where Kis a constant (Fig.
up or down
ff VaT
/Tt-JX
uyif
1
lit)
dt
1.17. Integ[rat or.
m
FIG.
Kdfltt
.16. Differentiator.
m
FIG.
1
the magnitude of the input,
i.e.,
1.15).
signal is differentiated
and possibly scaled
(Fig. 1.16).
3. Integrator:
The output
is
the integral of the input, as
shown
in
Fig. 1.17.
4. Time delayer: The output is delayed by an amount
same wave shape as the input (Fig. 1.18).
Suppose we take the triangular pulse in
T,
but retains the
Fig. 1.19 as the input signal.
the outputs for each of the four systems just described are shown
in Figs. 1.200-1 .20(2.
Then
12
Network analysis and
synthesis
m
1
2
i
t
-1
<b)
tit)
_L
Ti
Ti
+
1
Ti
+
2
(d)
(c)
FIG. I JO. la) Amplifier output,
(ft)
Differentiator output, (c) Integrator output,
(rf)
Delayed output.
Ideal elements
In the analysis of electric networks, we use idealized linear mathematical
models of physical circuit elements. The elements most often encountered
are the resistor R, given in ohms, the capacitor C, given in farads, and
the inductor L, expressed in henrys. The endpoints of the elements are
port is defined as any pair of two terminals into which
called terminals.
energy is supplied or withdrawn or where network variables may be
measured or observed. In Fig. 1.21 we have an example of a two-port
A
network.
The energy sources that
are ideal
current or voltage sources,
polarities
make up the excitation functions
as shown in Figs. 1.22a and b. The
Two-port
network
FIG.
1.21.
Two-port network.
*a Response
'/ measurement
Signals and systems
13
o
+
w Q)
««p
FIG. 1.22a. Voltage source.
FIG. 1.22b. Current source.
indicated for the voltage source
and the direction of flow for the current
source are arbitrarily assumed for reference purposes only. An ideal
voltage source is an energy source that provides, at a given port, a voltage
signal that
is
independent of the current at that port. If we interchange the
words "current" and "voltage" in the
last definition,
we then
an
define
ideal current source.
In network analysis, the principal problem is to find the relationships
that exist between the currents and voltages at the ports of the network.
Certain simple voltage-current relationships for the network elements also
serve as defining equations for the elements themselves. For example,
when the currents and voltages are expressed as functions of time, then the
R, L, and
t>(0
C elements,
shown
in Fig. 1.23, are defined
= L -J9
or
i(t)
= -v(t)
or
iQ)
=i
»(0
=7
I'
CJo
v(x)
L *o
J
at
«(*)
dx
+ v(0)
or
i(t)
=C
by the equations
dx
+
j(0)
(1.14)
^
dt
where the constants of integration
discussed in detail later.
i'(0)
and
»(0) are initial conditions to
be
Expressed as a function of the complex frequency variable s, the equations
—m*v(t)
m
v(t)
M
FIG.
—Ut)
*—
v(t)
(b)
1.23. (a) Resistor,
(b) Inductor,
^=C
(e)
(c) Capacitor.
Network
14
analysis and synthesis
—Ks)»—
1(3)
»—
>
V(s)
]sL
V(s)
1.24. (a) Resistor,
defining the R, L,
initial
(c)
(b)
(a)
FIG.
sC
V(s)
and
conditions for the
C
(b) Inductor,
shown
elements,
(c)
Capacitor.
in Fig. 1.24, are (ignoring
moment)
V(s)
=
V(s)
= sLI(s)
RI(s)
or
I(s)
= ^V(s)
or
I(s)
=
R
—
V(s)
(1.15)
sL
7(s)
= -^/(s)
or
/(s)
=
sCV(s)
sC
We see that in the time domain, i.e., where the independent variable is
t,
the voltage-current relationships are given in terms of differential equations.
On the other hand, in the complex-frequency domain, the voltage-current
elements are expressed in algebraic equations.
Algebraic equations are, in most cases, more easily solved than differential
relationships for the
equations. Herein lies the ration d'itre for describing signals
in the frequency domain as well as in the time domain.
and networks
When
a network is made up of an interconnection of linear circuit
elements, the network is described by its system or transfer function H(s).
The response R(s) and the excitation E(s) are related by the equation
R(s)
= H(s) E(s).
(1.16)
In network analysis, we are given E(s), and we can obtain H(s) directly
from the network. Our task is to determine R(s).
1.4
NETWORK SYNTHESIS
We will now briefly introduce some of the problems germane to network
synthesis.
In network synthesis,
excitation E(s),
we
and the
from the
are given the response R(s)
and we are required to
synthesize the network
Signals and systems
m <b
FIG.
Z{s)
VM
1.25. Driving-point
=
15
FIG. I.M. Black box.
impedance
R.
system function
aw-™
(1.17)
£(s)
Since R(s) and £(5) are voltages or currents, then H(s) is denoted generally
as an immittance if R(s) is a voltage and E(s) is a current, or vice versa.
9
driving-point immittance is defined to be a function for which the
A
variables are measured at the
Z(s) at a given port
is
same
Thus a
port.
driving-point impedance
the function
TO-™
(1.18)
I(s)
where the excitation
shown
is
in Fig. 1.25.
a current
and the response is a voltage V(s), as
interchange the words "current" and
I(s)
When we
"voltage" in the last definition, we then have a driving-point admittance.
An example of a driving-point impedance is the network in Fig. 1.25,
where
Z(s)
= ^i = R
(1.19)
I(s)
Now suppose the resistor in Fig. 1.25 were enclosed in a "black box."
We have no access to this black box, except at the terminals 1-1' in Fig.
1.26. Our task is to determine the network in the black box. Suppose we
are given the information that, for a given excitation /(*), the voltage
response V(s) is proportional to I(s) by the equation
V(s)
= KI(s)
An obvious solution, though not unique,
is
(1.20)
that the network consists of a
=
KCl. Suppose next that the excitation
of value R
V(s), the response is a current I(s), and that
resistor
ns)
•IRE
_M.
3
+ 4j
is
a voltage
(1.21)
Standards on Circuits "Linear Passive Networks," Proe. IRE, 48, No. 9
(Sept. 1960), 1608-1610.
Network
16
analysis
and synthesis
—Us)
hM
'
1
I
lo-
+
>i
=r 4
Two-port
network
Vi(s)
o
+
V&)
l'o—
FIG.
1.27.
Network
FIG.
realiza-
1.28.
Two-port network.
tion for Y(s).
a network equivalent to the network in the black
box. From a close scrutiny of the driving-point admittance Y(s), we see
that a possible solution might consist of a resistor of J O in parallel with
a capacitor of 4 farads, as seen in Fig. 1.27.
The problem of driving-point synthesis, as shown from the examples
just given, consists of decomposing a given immittance function into
Our task is to
synthesize
basic recognizable parts (such as 3
+ 4s).
Before
we proceed with
the
mechanics of decomposition, we must first determine whether the function
is realizable, i.e., can it be synthesized in terms of positive resistances,
inductances, and capacitances? It will be shown that realizable drivingpoint immittances belong to a class of functions known as positive real or,
simply, p.r. functions. From the properties of p.r. functions, we can test a
given driving-point function for realizability. (The Appendices present a
short introduction to complex variables as well as the proofs of some
theorems on positive real functions.) With a knowledge of p.r. functions,
we then go on to examine special driving-point functions. These include
functions which can be realized with two kinds of elements only the L-C,
—
R-C, and R-L immittances.
Next we proceed to the synthesis of transfer functions. According to
4
the IRE Standards on passive linear networks, a transfer function or
transmittance is a system function for which the variables are measured at
different ports. There are many different forms which a transfer function
might take. For example, consider the two-port network in Fig. 1.28.
If the excitation is I^s) and the response Vjs), the transfer function is a
transfer impedance
Ztt(s)
=
(1.22)
On the other hand, if Vl(s) were the excitation and
we would
)=™
tf(S
Fi(»)
4
Loe.
eit.
V^s) the response, then
have a voltage-ratio transfer function
(1.23)
Signals and systems
17
\H(ju)\
FIG.
As
1.2?. Ideal
amplitude spectrum for low-pass
filter.
for driving-point functions, there are certain properties which a
transfer function
must
satisfy in
these realizability conditions
order to be realizable.
We
shall study
and then proceed to the synthesis of some
simple transfer functions.
The most important aspect of transfer function
synthesis
is filter
design.
A filter is defined as a network which passes
a certain portion of a frequency spectrum and blocks the remainder of the spectrum. By the term
"blocking," we imply that the magnitude response \H(ja>)\ of the filter is
approximately zero for that frequency range. Thus, an ideal low-pass
a network which passes all frequencies up to a cutoff frequency
and blocks all frequencies above w c as shown in Fig. 1.29.
One aspect of filter design is to synthesize the network from the transfer
function H(s). The other aspect deals with the problem of obtaining a
filter is
mc
,
,
realizable transmittance H(s) given the specification of, for example, the
magnitude characteristic in Fig.
1.29.
This part of the synthesis is generally
referred to as the approximation problem.
mation?" Because frequency response
Why
the
word "approxiand C
characteristics of the R, L,
elements are continuous (with the exception of isolated points called
made to
we can realize low-pass filters
resonance points), a network containing these elements cannot be
cut off abruptly at
<o c
in Fig. 1.29. Instead,
which have the magnitude
characteristics
of Fig.
1.30.
In connection with
\H(J<*)\
FIG.
1.30. Realizable
low-pass filter characteristics.
Network
18
analysis
and synthesis
we
problems in magnitude
a filter, we deal with
element values such as R — 0.5 ohm and C = 2 farads instead of "practical"
element values of, for example, R — 500,000 ohms and C = 2 picofarads
(pico = 10-12). Also we will study a method whereby low-pass filter
designs might be transformed into high-pass, band-pass, and bandelimination filters. The mathematical basis of this method is called
the
filter
design problems,
will discuss certain
and frequency normalization so
that, in designing
frequency transformation.
We next discuss some aspects of analysis and synthesis in which the
excitation and response functions are given in terms of power rather than
of voltage and current.
We will examine the power-transfer properties of
which describe the incident
power of the network at its ports.
Finally, in Chapter 15, we will examine some of the many uses of highspeed digital computers in circuit analysis and design. In addition to a
general survey of the field, we will also study some specific computer
programs in circuit analysis.
linear networks, using scattering parameters,
and
reflected
Problems
1.1
Draw the line
s(t)
1.2
JtTir/4
-
spectra for the signal
3 sin (t
+ ^1 + 4sin
(it
- 1| +
6 sin it
Find the response to the excitation sin / into a sampler that closes every
»= 0, 1, 2,
seconds where
Draw the response for ^ / ^ 2*.
K
13 Find the response to the excitation shown in the figure when the network
is (a)
an
ideal differentiator;
(ft)
an ideal integrator.
Mt)
_L
PROS.
1.4
If the system function of
2
1.5
1
1.3
a network
is
given as
1
H{s)~
(s
+ 2X* +
3)
find the response R(s) if the excitation
Signals and systems
19
network
realiz-
is
s
1.5
ations.
Given the driving-point functions find
their simplest
Z(s)
(«)
3*
(*)
Y(s)
= 2s
—
H
s
Z(s)
id)
r(s)
-3i+
+
2
*
1
_
s*+2
2,
1
3s
+
2
s*
+4
1.6 For the network shown, write the mesh equation in terms of (a) differential
equations and (b) the complex-frequency variable s.
R
l
——wv—
npfir*—
-0
PROB.
1.7
1.6
For the network shown, write the node equation in terms of (a) differential
equations and (6) complex-frequency form.
PROB.
1.8
r(t)
1.7
Suppose the response of a linear system to an excitation
«(/)
= 3e-*'. What would the response be to an excitation of t(t — 2)1
were
chapter
Signals and
Our main concern
in this chapter
is
2
waveforms
the characterization of signals as
signals
functions of time. In previous studies we have dealt with d-c
of
sinusoids
were
which
signals
a-c
or
time,
with
that were constant
practice,
engineering
In
sin
(cot
A
0).
s(t)
as
such
constant amplitude,
than
the class of signals encountered is substantially broader in scope
=
+
of
simple a-c or d-c signals. To attempt to characterize each member
signals
of
variety
infinite
almost
the
of
view
in
foolhardy
is
the class
be
encountered. Instead, we will deal only with those signals that can
building
characterized in simple mathematical terms and which serve as
will concentrate on formublocks for a large number of other signals.
than deal with
lating analytical tools to aid us in describing signals, rather
We
limitathe representation of specific signals. Because of time and space
behavior,
random
exhibit
tions, we will cover only signals which do not
characterized as functions of time.
i.e., signals which can be explicitly
These signals are often referred to as deterministic
signals.
Let us
first
discuss certain qualitative aspects of signals in general.
2.1
GENERAL CHARACTERISTICS OF SIGNALS
In this section we will examine certain behavior patterns of signals.
Once these patterns are established, signals can be classified accordingly,
and some simplifications result. The adjectives which give a general
continuous.
qualitative description of a signal axe periodic, symmetrical, and
Let us discuss these terms in the given order.
signal
First, signals are either periodic or aperiodic. If a
it is
is
periodic, then
described by the equation
s(t)
= s(t±kT)
k
20
=
0,1,2,...
(2.1)
Signals
and waveforms
21
1
-2T -r
T 2T 3T AT 5T
-l
FIG.
where
2.1.
Square wave.
T is the period of the signal. The sine wave, sin is periodic with
T = lit. Another example of a periodic signal is the square wane
in Fig. 2.1. On the other hand, the signals given in Fig. 2.2 are
/,
period
given
aperiodic, because the pulse patterns
do not repeat
interval T. Alternatively, these signals
may be considered "periodic" with
an
after
a certain
finite
infinite period.
Next, consider the symmetry properties of a signal. The key adjectives
signal function can be even or odd or neither.
here aie even and odd.
A
An even function obeys the relation
j(0-*(-0
For an odd function
For example, the function
pulse in Fig. 2.2a
is
=
s(i)
sin
t is
(2.2)
-s{-t)
(2.3)
odd, whereas cos
t is
even, whereas the triangular pulse
even.
is
odd
The square
(Fig. 2.26).
Observe that a signal need not be even or odd. Two examples of signals
of this type are shown in Figs. 2.3a and 2.4a. It is significant to note,
however, that any signal s(t) can be resolved into an even component 5,(0
and an odd component s9(t) such that
*(0
= *.(0 + *o(0
(2.4)
For example, the signals in Figs. 2.3a and 2.4a can be decomposed into
odd and even components, as indicated in Figs. 2.3b, 2.3c, 2.4b, and 2.4c.
m
o
t
2
M
FIG. 12.
(b)
(a)
Even
function, (b)
Odd function.
22
Network
analysis
and synthesis
s(t)
1
i
i
-1
-t
(a)
(a)
sjt)
sM
-1
o
i
1
t
(b)
(b)
soft)
«oW
-1
-i
(0
(O
FIG.
2.3.
Decomposition into
odd and even components,
(a)
Original
Even
From
part,
Eq. 2.4
(c)
function,
(6)
Odd part.
we observe
FIG. 2.4. Decomposition into
even and odd components,
(a) Unit step function.
(6)
Even part of unit step, (c)
Odd part of unit step.
that
s(-t)
= st(-t) + * (-0
- sXO - *o(0
(2.5)
Consequently, the odd and even parts of the signal can be expressed as
J#)- MK0 + *(-')]
*(o = two - K-m
(2 .6;
Consider the signal s(t), shown in Fig. 2.5a. The function s(-t) is equal
axis and is given in Fig. 2.5ft. We then
to s(t) reflected about the t
=
obtain st(t) and j„(0 as
shown
in Figs. 2.5c
and
d, respectively.
j
Signals
and waveforms
13
rf-tf
i
1
-1l
t
1
t
m
(a)
*Jt)
lift)
i
-]I
1
L
*
I
t
-i
(d)
(e)
FIG. US. Decomposition into odd and even components from
s(t)
and s(—t).
Now let us turn our attention to the continuity property of signals.
Consider the signal shown in Fig. 2.6. At t
T, the signal is discontinuous.
=
The
height of the discontinuity
is
f(T+) -f(T-
/(T+)
where
)
=A
(2.7)
= lim/(T+c)
€-»0
f(T-)
and
€ is
(2.8)
- lim/(T- «)
a real positive quantity. In particular, we are concerned with
neighborhood of f = 0. From Eq. 2.8, the points
discontinuities in the
M
{
T
FIG. 24. Signal with discontinuity.
t
Network
24
analysis
and synthesis
r«
Tl
FIG.
/(0+) and/(0-)
2.7. Signal
with two discontinuities.
are
/(0+)
- lim/(e)
€-•0
/(0-)
(2.9)
= lim/(-e)
For example, the square pulse in Fig. 2.7 has two discontinuities, at
Tt The height of the discontinuity at Tx is
Tx and
.
s(Tx +)
- s(T -) = K
Similarly, the height of the discontinuity at
2.1
(2.10)
x
Ta is — K.
GENERAL DESCRIPTIONS OF SIGNALS
we consider various time domain descriptions of signals.
In particular, we examine the meanings of the following terms: time
constant, rms value, d-c value, duty cycle, and crest factor. The term, time
In this section
constant, refers only to exponential waveforms;
the remaining four
terms describe only periodic waveforms.
Time constant
important to know how quickly a
of
the decay of an exponential is the
useful measure
waveform decays.
waveform described by
exponential
the
time constant T. Consider
In
many
physical problems,
it is
A
r(t)
From a plot
Also
of r(t) in Fig.
2.8,
= Ke~*IT u(t)
we
see that
when
(2.11)
t
=
T,
r(T)
= 0.37KO)
(2.12)
r(4T)
= 0.02r(0)
(2.13)
Signals and
waveforms
25
1.00
075
r(t)
0.50
037
-
025
O02
5
FIG.
2.8.
Normalized curve for time constant
T
1.
Observe that the larger the time constant, the longer it requires for the
waveform to reach 37% of its peak value. In circuit analysis, common
time constants are the factors RC and RjL.
RMS Value
The rms or root mean square value of a periodic waveform e(t) is defined
as
Crms
-
i?f>*r
(2.14)
where T is the period. If the waveform is not periodic, the term rms does
not apply. As an example, let us calculate the rms voltage for the periodic
waveform in
Fig. 2.9.
Inns
4^r(T-)M>]r
(2.15)
-fair
= J2J3Av
D-C Value
The
d-c value of a
waveform has meaning only when the waveform
waveform over one period
is
periodic. It is the average value of the
«oc
= -1
fT
TJ
e{t)dt
p
(2.16)
26
Network
analysis
and synthesis
2T
¥
-A
FIG.
The square wave
2.9. Periodic
waveform.
in Fig. 2.1 has zero d-c value, whereas the
waveform in
Fig. 2.9 has a d-c value of
1
[AT
A
ATI
(2.17)
Duty cycle
The term duty cycle, D, is defined as the ratio of the time duration of the
positive cycle t9 of a periodic waveform to the period, T, that is,
(2.18)
T
becomes important in dealing with waveforms of the type shown in Fig. 2.10, where most of the energy is conThe rms voltage of the waveform
centrated in a narrow pulse of width f
The duty
cycle of a pulse train
.
in Fig. 2.10
is
,h
*--(iM'
(2.19)
= AJ7JT
An
A
(o
FIG. 2.10. Periodic waveform with small duty
cycle.
Signals and
r
waveforms
27
T
TT
FIG.
2.1 1. Periodic
waveform with zero d-c and small duty
cycle.
Wc see that the smaller the duty cycle, the smaller the rms voltage. The
square wave in Fig. 2.1 has a 50% duty cycle.
Crest factor
Crest factor1
is defined as the ratio of the peak voltage of a periodic
waveform to the rms value (with the d-c component removed). Explicitly,
for any waveform with zero d-c such as the one shown in Fig. 2.1 1
crest
factor, CF, is denned as
—
CF = -^-
-^-
or
£rms
whichever
voltage
is
is greater.
(2.20)
£rms
For the waveform
in Fig. 2.11, the peak-to-peak
defined as
=
e, p
*„
+
(2.21)
<?»
Since the waveform has zero d-c value
eJ*
= et(T-t )
(2.22)
(2.23)
Also,
eb
= e„D
and
ea
= ePP(l -
The rms value of the waveform
emg
D)
(2.24)
is
= ( e»( l ~ D? ° +
f
enWr-
t )\
*
(2.25)
1
G.
Justice,
'The
(Jan., 1964), 4-5.
eppy/D(l
-
D)
Significance of Crest Factor," Hewlett-Packard Journal, 15,
No. 5
Network
28
analysis and synthesis
Since crest factor
CF =
ejerm*,
we have
CF =
e„(\
-
D)
ep jD(l -
D)
(2.26)
-Vi/J>-i
For example,
if
D=
—
CF
(2.27)
= ^100 -
UD =
1
10
10,000'
CF=VlO,000- 1~100
(2.28)
A voltmeter with high crest factor is able to read accurately rms values of
whose waveforms differ from sinusoids, in particular, signals with
low duty factor. Note that the smallest value of crest factor occurs for
the maximum value of D, that is, X>m»x = 0.5,
signals
CFnto
= Vl/iW =1
1
(2.29)
THE STEP FUNCTION AND ASSOCIATED WAVEFORMS
Z3
The
unit step function w(f)
shown
„(/)
=
=
in Fig. 2.12
is
defined as
<
f£0
t
1
(2.30)
physical analogy of a unit step excitation corresponds to a switch S,
and connects a d-c battery of 1 volt to a given circuit,
t
the
as shown in Fig. 2.13. Note that the unit step is zero whenever
The
which closes at
=
u(t)
^o
Switch
lv
Network
T.
FIG. 2.12. Unit step function.
FIG. 2.13. Network analog of unit
step.
Signals
and waveforms
29
m
*<t-a)
l
4
a
FIG. 2.14. Shifted step function.
argument
argument
a
(/)
greater than zero.
> 0, is defined by
vit
and
is
occurs
FIG. 2.15. Square pulse.
within the parentheses
(t) is
2
1
(
- a) =
=
is
and is unity when the
the function u(t
a), where
negative,
Thus
—
/
1
<a
{Zil)
t^a
shown in Fig. 2.14. Note that the jump discontinuity of the step
when the argument within the parentheses is zero. This forms the
basis of the shifting property of the step function. Also, the height of the
jump discontinuity of the step can be scaled up or down by the multiplication of a constant K.
With the use of the change of amplitude and the shifting properties of
the step function, we can proceed to construct a family of pulse waveforms.
For example, the square pulse in Fig. 2.15 can be constructed by the sum
of two step functions
sif)
= 4u(t -
1)
+
The "staircase"
by the equation
as given in Fig. 2.16.
characterized
<
s(t)
=>J,u(t
(-4)
u(t
- 2)
function,
shown
(2.32)
in Fig. 2.17,
- kT)
*t)
4
-4
FIG.
2.16. Construction
of square pulse by step function.
is
(2.33)
Network
30
analysis
and synthesis
3
2
—
1
1
T
ZT
FIG.
t
2.17. Staircase function.
wave in Fig. 2.1. Using the
that the square wave is given by (for t ^ 0)
shifting
Finally, let us construct the square
property,
s(t)
A
we
see
-
m(0
simpler
way
-
2u(t
-
T)
+
2u(t
-
2T)
to represent the square
-
wave
2k(*
is
- 3D +
•
'
•
(2.34)
by using the property
zero whenever its argument is negative. Restricting
ourselves to the interval f ;> 0, the function
that the step function
is
s(0
is
by the waveform in Fig.
2. 1 can be represented
Fig.
in
wave
It is now apparent that the square
zero whenever sin (irtIT)
2.18.
(2.35)
-«(*»?)
S (0
is
negative, as seen
= «(sin^)-u(-sin^)
(2.36)
Another method of describing the square wave is to consider a generalization of the step function known as the sgn function (pronounced signum).
The sgn function
is
defined as
sgn [/(f)]
=1
=
>o
fit) < o
fit)
fit)
s(t)
T
FIG.
2. IS.
2T
The
3T
AT
signal u(sin7rr/r).
(2.37)
Signals
and waveforms
31
•ft)
ZT
3T
FIG. 2.19. Sine pulse.
Thus the square wave in
Fig. 2.1
s(0
is
=
simply expressed as
sgn
I
sin
—
(2.38)
1
Returning to the shifting property of the step function,
can be represented as
we
see that the
single sine pulse in Fig. 2.19
s(r)
- sin ^ [u(t - IT) - u(t -
3T)]
(2.39)
The
step function is also extremely useful in representing the shifted or
delayed version of any given signal. For example, consider the unit ramp
p(t)
=
tu(t)
(2.40)
shown in Fig. 2.20. Suppose the ramp is delayed by an amount t = a, as
shown in Fig. 2.21. How do we represent the delayed version of ramp?
l
FIG.2JM.
t
Ramp function with zero time shift.
Network
32
analysis
and synthesis
s(t)
Ramp function with
FIG. 2.21.
First, let
us replace the variable
t
When
p(t') is plotted against
of 5(0 versus
we
t
in Fig. 2.20.
t',
If,
When we plot p(t') against
t,
= a.
t'
— —
t
f u(t')
a.
Then
(2.41)
the resulting curve is identical to the plot
t ' in p(t'),
however, we substitute t — a
=
then have
p(f)
shift
by a new variable
=
p(r')
time
=
(t
- a) u(t -
(2.42)
a)
we have the delayed version of pit) shown in
Fig. 2.21.
From the preceding discussion, it is clear that if any signal /(/) u(t)
delayed by a time T, the delayed or shifted signal is given by
f{t')=f{t-T)u{t-T)
For example,
let
is
(2.43)
us delay the function (sin irtfT) u(i) by a period T. Then
the delayed function
s(t'),
s(t')
shown
=
[sin
in Fig. 2.22,
- (t -
is
T)] u(t
-
FIG. 2.22. Shifted sine wave.
T)
(2.44)
Signals
and waveforms
33
m2-
2
1
FIG. 2.23. Triangular pulse.
As a final example, consider the waveform in Fig. 2.23, whose component parts are given in Fig. 2.24. For increasing t, the first nonzero
component is
the function 2{t
=
—
1) u(t
—
1),
which represents the
straight
At ? = 2, the rise of the straight line is to be
arrested, so we add to the first component a term equal to — 2(t — 2)u(t — 2)
with a slope of —2. The sum is then a constant equal to 2. We then add a
line
of slope 2 at
— 2u(t —
term
s(t)
/
1.
2) to bring the level
= 2{t -
1) u(t
-
1)
-
down
2(t
-
to zero. Thus,
2) u(t
-
2)
-
2u(t
-
2)
(2.45)
1A THE UNIT IMPULSE
The unit impulse, or deltafunction, is a mathematical anomaly. P. A. M.
first used it in his writings on quantum mechanics. 2 He defined the
Dirac
2(t-\)u(t-\)
FIG. 2.24. Decomposition of the triangular pulse in Fig. 2.23.
•
P.
A. M. Dirac, The Principles of Quantum Mechanics Oxford University
Press, 1930.
Network
34
anal/sis
delta function d\t)
and synthesis
by the equations
=
<J(0 =
6\t) dt
f:
most important property
Its
is
(2.46)
1
for
t
*
(2.47)
the sifting properly, expressed by
f(t)d(t)dt=f(0)
f
J—
(2.48)
i
we will examine the unit impulse from a nonrigorous
Those who prefer a rigorous treatment should refer to
Appendix B for development of this discussion. The material in that
appendix is based on the theory of generalized functions originated by
G. Temple. 8 In Appendix B it is shown that the unit impulse is the
In this section
approach.
derivative of the unit step
r
At
first
..
-
o(0
glance this statement
is
=
„
if
/0 ao>l
(2.49)
,.
(0
doubtful. After
all,
the derivative of the
zero everywhere except at the jump discontinuity, and
it does
not even exist at that point! However, consider the function g t (i) in Fig.
2.25. It is clear that as e goes to zero, g ( (t) approaches a unit step, that is,
unit step
is
lim ge(t)
«(0
(2.50)
«-»o
Taking the derivative of g t(t), we obtain £',(0. which
is
defined by the
equations
g'«(0
as
shown in
««
>
e 4+1 .
-e
= 0;
;
t<0, t>€
Now let « take on a sequence of values e
such that
(
Consider the sequence of functions {g' c< (0} for decreasing
Fig. 2.26.
*V0
et (t)
FIG.
(2.51)
US.
Unit step when
«
— 0.
FIG.
1M.
Derivative of
g((t)
in
Fig. 2.25.
*
G. Temple, "The Theory of Generalized Functions," Proe. Royal Society, A, 228,
1955, 175-190.
Signals and
waveforms
35
h-\W>
£
8it
m
«s «s
n
i
i
5
n
*
FIG. 127. The sequence {gfi (t)}.
values of c { , as
shown
in Fig. 2.27.
The sequence has the following
property:
J'«t>0g' (t)dt
(i
*1<0
where
tx
and
/,
are arbitrary real numbers.
there corresponds a well-behaved function
g't t(t).
As
€f
=
l
(2.52)
For every nonzero value of e,
(i.e., it does not "blow up")
approaches zero,
g'«(o+)-
00
«,-»0
(2.53)
so that the limit of the sequence is not defined in the classical sense.
Another sequence of functions which obeys the property given in Eq.
2.52
is
m
We
the sequence {/«//)}
Fig- 2.28.
now define the unit impulse
of all sequences of functions which obey Eq. 2.52. In
6{t) as the class
particular,
we
define
r*t>o
J*«i>o
«t)A-lim
«i<0
g' t((t)dt
e<-»0«/«i<0
a
4hm
r*» >o
UQdt
(2.54)
Network
36
analysis
and synthesis
FIG. 2.28. The sequence {/£ ,(0>
It
should be stressed that this
stated previously,
From
is
is
not a rigorous definition (which, as
found in Appendix B) but merely a
the previous definition
we can
heuristic one.
think of the delta "function" as
having the additional properties,
<5(0)
<5(f)
=
=
oo
(2.55)
for
t^O
waveforms
Signals and
Continuing with
the impulse
is
this heuristic treatment,
say that the area "under"
unity, and, since the impulse is zero for
d(t)dt
J—
Thus
we
=
6
6{i)
dt
\
=
the entire area of the impulse
/*o-
S(t) dt
=
we have
(2.56)
1
is
"concentrated" at
r+oo
(5(f)
dt
=
t
t
=
=
is
0.
Con-
zero,
s(t)
J0+
J-oo
# 0,
t
Jo—
<x>
sequently, any integral that does not integrate through
as seen by
37
s(t)
= Au(t- a)
(2.57)
A,
The change of scale and time shift properties
discussed earlier also apply for the impulse
function.
The
s(t)
yields
derivative of a step function
= Au(t — a)
(2.58)
an impulse function
B'(t)
s\t)
= A6(t- a)
A
(2.59)
which is shown in Fig. 2.29. Graphically, we
represent an impulse function by an arrowhead pointing upward, with the constant
A written
Note that A is the
A d(t - a).
multiplier
s'(t)
= Ad(t-a)
next to the arrowhead.
FIG. 2.29
area under the impulse
Consider the implications of Eqs. 2.58 and 2.59. From these equations
see that the derivative of the step at the jump discontinuity of height A
we
an impulse of area A at that same point t = T. Generalizing on this
argument, consider any function /(0 with a jump discontinuity at / = T.
Then the derivative, /'(0 must have an impulse at t = T. As an example,
yields
consider /(0 in Fig. 2.30.
At
t
=
T,f(t) has a discontinuity of height A,
M
T
FIG. 2.30. Function with discontinuity at
t
T.
38
Network
which
is
analysis
and synthesis
given as
A =/(r+) -f(T-)
Let us define
same shape
fx(t)
(2.60)
as being equal to f(t) for
t
<
as/(/), but without the discontinuity for
t
T,
and having the
>
T, that
is,
Ut)=f(t)-Au{t-T)
The
derivative
A0
is
then
A0 -AC) + ^
The following example
the function /(f)
(2.61)
illustrates this
<K*
point
-
T)
(2.62)
more
clearly.
In Fig. 2.31a,
is
fit)
= Au(t-a)-A u(t - b)
(2.63)
Its derivative is
fit)
= Ad(t - a) - Ad(t - b)
(2.64)
and is shown in Fig.
2.316. Since/(/) has
two discontinuities, at t = a and
The coefficient of
— b, its derivative must have impulses at those points.
the impulse at = b is negative because
t
t
f(b+) -fQ>-)
- -A
(2.65)
m
A
8(0
4
a
b
t
2
(a)
r<*>
/
t
L
(a)
g'(t)
A'
2
2
6
a
t
~0]
'-A
t
1
(b)
FIG. 2.31. (a) Square pulse,
of squairepulse.
(b) Derivative
FIG .2.32.
(a) Signal,
(*)] derivative.
Signals and
waveforms
39
As a second example,
consider the function g(i) shown in Fig. 2.32a.
obtain #'(0 by inspection, and note that the discontinuity at /
1
produces the impulse in #'(0 of area
We
=
£0+)-£(l-) =
2,
(2.66)
as given in Fig. 2.32*.
Another interesting property of the impulse function is expressed by the
integral
+
7(0 S(t-T)dt= f(T)
f
This integral
1
5* T.
is easily
evaluated if
that
consider that d(t
—
T)
=
for all
Therefore, the product
f{t)d(t-T)
If/(r)
we
(2.67)
is
we
single-valued at
t
=
=
aXLtitT
T,f(T) can be factored from the
(2.68)
integral so
obtain
/(T)
d(t
|
J— O0
-T)dt= f(T)
Figure 2.33 shows /(r) and d(t - T), where /(/)
If/(0 has a discontinuity at / = T, the integral
is
(2.69)
continuous at i
=
T.
+00
Jf(t)Ht-T)dt
J
o
not denned because the value of f(T)
the following examples.
is
Example
is
not uniquely given. Consider
2.1
/(0
- 1*«*
T)dt
- e*» r
+0O
Example
Z2
J
e
~&*Hf
(2.70)
-00
/(/)
= sin t
£>''('- 5)*-^
(2.71)
Network
40
analysis
and synthesis
f(t)
FIG. 2.34. Impulse scanning.
Consider next the case where/ (0 is continuous for
us direct our attention to the integral
-
f(t)d(t-T)dt=f(T)
f
oo
< <
t
Let
oo.
(2.72)
-oo to +oo,
which holds for all / in this case. If T were varied from
of this sort
operation
An
entirety.
its
in
reproduced
then/(0 would be
paper with
of
sheet
a
moving
by
function/(f)
the
scanning
corresponds to
across a plot of the function, as shown in Fig. 2.34.
higher order derivatives of the unit step function.
Let us
we represent the unit impulse by the function ft(i) in Fig. 2.28,
a thin
slit
now examine
Here
-* 0, becomes the unit impulse. The derivative of/£ (0 is given
derivative of the
in Fig. 2.35. As e approaches zero,/' £ (0 approaches the
in Fig. 2.36.
seen
unit impulse d'(t), which consists of a pair of impulses as
to zero.
equal
is
doublet,
The area under d'(t), which is sometimes called a
which, as
e
Thus,
6'(t)
r
The
dt
=
other significant property of the doublet
I
1
/(») d'(t
(2.73)
is
-T)dt= -f\T)
rt (t)
f2
t
—€
t
1
(2
FIG. 2.35. Unit doubles as
<=
-* 0.
FIG. 2.36. The doublet
d'(t).
(2.74)
Signals and
waveforms
41
where f'(T) is the derivative of/(0 evaluated at t = T where, again, we
assume that/(/) is continuous. Equation 2.74 can be proved by integration
by parts. Thus,
*/
—
f(t)d'(t-T)dt=f(t)d(t-T)\
=
It
(t)d(t-T)dt
-00
(
«/— oo
(2.75)
-f'(T).
can be shown in general that
f°°
•r—
oo
/(O
-T)dt = (-l)y »'(r)
<B,
<5
(
(<
(2.76)
where (5< n and/ <"> denote nth derivatives. The higher order derivatives of
can be evaluated in similar fashion.
>
d(t)
Problems
2.1
Resolve the waveforms in the figure into odd and even components.
s(t)
s(t)
M
1
1
i
1
t
-v
+1
t
(a)
a>)
m
s(t)
1
?
2
\
1
1
1
2
t
-
PRO B. 2.1
I
\
1
t
Network
42
analysis
and synthesis
2.2 Write the equation for the
functions.
waveforms
in the figure using shifted step
fit)
2
1
/
A
2
t
(b)
f(t)
fit)
2
-
2
1.5
1
2
1
(d)
(c)
PROB.
2.2
in Prob. 2.2 and write the equations
impulse functions.
and/or
step
shifted
using
for the derivatives,
2.4 For the waveform /(r) given in the figure, plot carefully
23
Find the derivative of the waveforms
for a value of
t
>
i
f(r)dr
T.
f(t)
f(t)
2*Wl
EMt)
>*
-E6(t-T)
PROB. 2.4
PROB. 2.5
Signals and
2^ For
the waveform /(r),
must be so that
shown
waveforms
in the figure, determine
43
what value
K
P/(0<//-0
(«)
J— 00
j"7<f)<fr-o
<*)
J0+
2.6
For the waveforms shown
time functions, that
is,
/",«(/
in the figure, express in terms of elementary
- Q, d(t -
*
/,),
CU.1.1.
Sketch *u_
r
*
rL\
j * x
..
the ..
waveforms
for (6) and (c) neatly.
+
(„)/(/) (*)/'(/) (c) f
/(T) A«f—
J— 00
JW
m
20
N
"*
At
2
o
i
(0
(ii)
PROS.
2.7
2.4
Prove that
(a)
<*'(*)
(6)
--<}'(--X)
-<»(a;)
= x d'(x)
/•oo
(c)
;*)
«&
-o
J— 00
2.8
The waveform /(0
in the figure is defined as
/W-^(/-e)*,
= 0,
Show that
as e
;!
o<;/£«
elsewhere
— 0,/(/) becomes a unit impulse.
.
*
3
1
J" 5
t
Network
44
analysis
and synthesis
Plot
2.9
(a)
%°s
(ft)
t
2.10
sgn (cos 0;
£t £2n
'
Evaluate the following integrals
r
(fl )
- TO u(t - T2) dt;
<*t
Ti <T^
J— CO
/•go
a,)
<5(a>
I
—
a> )
cos
o>f rfo>
J— 00
OO
J—
2.11
,
[<5(0-/4<5(r-7'1 )+2/l<5(f-7 2)]e-^«"A
f
(c)
00
Evaluate the following integral
f " sin
2.12
(t
- l\ +
S'
('
"
f)
The response from an impulse sampler
=
r(t)
Plot
tU
KO for
P
<, t <.
sin
n s(t
l
+
*'<'
»>]
*
given by the equation
is
K = 0,
-Kj\dt;
~
1, 2, 3,
.
.
IT.
If the step response of a linear, time-invariant
determine the impulse response h(f), and plot.
2.13
system is r s (0
2.14 For the system in Prob. 2.13 determine the response
excitation
=
due to a
2e
'
u(t),
staircase
j(0-i«(/-*r)
*=o
Plot both excitation
and response
functions.
impulse response of a time-invariant system
determine the response due to an excitation
2.15
If the
e(t)
Plot
=
2S(t
-
1)
-
2d(t
e(t).
M
PROB. 2.16
-
2)
is
W) = «~* "(')»
Signals and
2.16
The
unit step response of a linear system
<x(r)
(a)
Find the response
(b)
Make
=
(2e-
2t
-
waveforms
45
is
1) «(/)
r(f) to the input /(f).
a reasonably accurate sketch of the response.
dimensions.
Show
all
pertinent
chapter
3
The frequency domain:
Fourier analysis
3.1
INTRODUCTION
One of
signals.
the most
If
common
classes of signals encountered are periodic
T is the period of the signal, then
s(t)
= s(t±nT)
7i
= 0,l,2,...
In addition to being periodic, if s(t) has only a
tinuities in any finite period and if the integral
"T
I
is finite
(where a
is
an arbitrary
\s(t)\
(3.1)
number of discon-
finite
dt
can be expanded
real number), then s(t)
into the infinite trigonometric series
s(t)
=^+
+
+ a t cos 2<ot +
b x sin cot + b t sin "hot +
a x cos
tot
•
•
•
•
•
„
2\
•
m=
2w/r. This trigonometric series is generally referred to as the
Fourier series. In compact form, the Fourier series is
where
s(t)
= ?* + 2 («» cos not +
2
b n sin
ntot)
(3.3)
«— i
apparent from Eqs. 3.2 and 3.3 that, when s(t) is expanded in a
Fourier series, we can describe s(t) completely in terms of the coefficients
It is
46
The frequency domain: Fourier
analysis
47
harmonic terms, a* alt a„ . , bx, b
These coefficients cona frequency domain description of the signal. Our task now is to
derive the equations for the coefficients a( , b t in terms of the given signal
function s(t). Let us first discuss the mathematical basis of Fourier series,
the theory of orthogonal sets.
of
its
.
.
stitute
ORTHOGONAL FUNCTIONS
3.2
Consider any two functions
Then
and/»(/) that are not identically zero.
x (f)
if
r,
f
JTx
/i(0/*(0dt
=O
(3.4)
we
say that/i(r) and/^/) are orthogonal over the interval [Tlt TJ. For
example, the functions sin t and cos t are orthogonal over the interval
nlit
^
r <;
<f>&)
(«
<j>
+
H(t)}.
Consider next a set of real functions {&(*),
l)2n.
If the functions obey the condition
(*.
4>i)
s
V<(0
I
4>M dt =
i*j
0,
(3.5)
then the set {<f>t } forms an orthogonal set over the interval [7\, TJ. In
Eq. 3.5 the integral is denoted by the inner product (^ 4 , <f> ). For convent
ience here, we use the inner product notation in our discussions.
The set {^J is orihonormal over [Tu 7",] if
=
The norm of an element
l&l
<f>
k in
the set
3.1. The Laguerre
time domain approximation,
1
W. H.
t-.j
{^J
is
defined as
- (4, &)* - (£%»\0 *)*
We can normalize any orthogonal
term <f>k by its norm ||^J|.
'
(3.6)
1
set
{<f>x, <f>t,
. . ,
fa} by dividing each
set,1
is
which has been shown to be very useful in
orthogonal over [0, <»]. The first four terms
Kautz, "Transient Synthesis in the Time Domain," Trans.
No. 3 (Sept. 1954), 29-39.
Theory, CT-1,
(3-7)
IRE on
Circuit
Network
48
analysis
and synthesis
of the Laguerre set are
- e~**
*i(0
Ut) -
To show
- 2(fl01
*,(/)
= e-o'H - MfO) + Hatf\
4>t(t)
= e-°'[l -
far
(3.8)
+ 6(flf)* - Ka/)8
]
that the set is orthogonal, let us consider the integral
f"
Letting t
e-*[l
Ut) hiO * = f V*"[l - MpQ +
2(«0*] dt
(3.9)
= at, we have
«i,«-;;J'V a-4T+2,*>rfr
,,
(3.10)
-ill-®"©]The norms of &(/) and &(/) are
M
,«_(J[-^.*)
-
11*1
(
f
-^
"^[l ~ 4(«0 + 4<a
a
') ]
Uo
(3,.,
*}
J
(3.12)
V2a
It is
not
difficult
to verify that the
norms of all the elements
equal to l/V2aT Therefore, to render the Laguerre
each element fa by
3.3
set
in the set are also
orthonormal,
we
divide
1/ V2a.
APPROXIMATION USING ORTHOGONAL FUNCTIONS
In this section
we
explore
some of the uses of orthogonal functions in
The principal problem is that of
the linear approximation of functions.
approximating a function/(0 by a sequence of functions/B(f) such that the
mean squared error
e
= lim
n-»oo
*W) - /»(0]
f
JTi
2
dt
-
(3.13)
The frequency domain: Fourier
When
Eq. 3.13
we
is satisfied,
49
analysis
say that {/„(f)} converges in the mean to
fit).
To examine
the concept of convergence in the
mean more
must first consider the following definitions:
Definition 3.1 Given a function/(f) and constant/;
>
we
closely,
for which
T,
\f(t)\*dt<ao,
(
JTi
we
say that /(f)
[T»
Tt
is
integrable
L*
in [7\,
TJ, and
we
write /(/)
e L»
in
].
Definition
12
tions integrable
If/(f)
eL'in [Tlt Tt], and {/„(f)> is a sequence of func-
L* in [Tlt TJ, we say
that if
Bm.P|/(0-/^)r*-0
then
p—
{/«.(*)}
The
mean of order/? to /(f). Specifically, when
mean to /(f).
converges in the
2 we say that
{/„(f)} converges in the
principle of least squares
Now let us consider the case when/„(f) consists of a linear combination
of orthonormal functions
4>i,<l>
/„(0
<f>
n-
= I *<&(<)
(3.14)
<-i
Our problem
is
to determine the constants at such that the integral
squared error
11/
- All* - f *W) - M9f dt
(3.15)
JTx
is
a minimu m. The principle of least squares states that in order to attain
( must have the values
m inimum squared error, the constants a
Ci
Proof.
must
set
= ]Kt)Mt)dt
We shall show that in order for
a{
=
ct for every
H/-/.II*
i
=
1, 2,
. .
,
.
\\f
(3.16)
— /„|| s to be minimum
,
we
n.
= </,/) - 2{f,fJ + (/„,/„)
=
11/11*
-22a,(/,&)+2a,W
(3.17)
Network
50
Since the set
(/,&).
analysis
{&}
and synthesis
orthonormal, |«\| a
is
=
1,
and by
definition, ct
=
We thus have
I/-/-I'
Adding and subtracting
i/-/.!"
=
-
l/l'
- 22«A + i
(318)
'*
2 c«* gives
8
ii/n
- i*"
- 2i«A
+ i«*'
+ i^
_1
<-l
<-l
<_1
»
(3.19)
n
We see that in order to
at
set
= c<.
coefficients
The
minimum
attain
integral squared error,
we must
defined in Eq. 3.16, are called the Fourier
coefficients c t ,
of/(*) with respect to the orthonormal
Parseval's equality
Consider fn(t) given in Eq. 3.14.
set
{+&)}.
We see that
ft
I L/j(or*-i«'
since
320>
This result is known as Parseval's
important in determining the energy of a periodic signal.
& are orthonormal functions.
equality,
3.4
<
and
is
FOURIER SERIES
Let us return to the Fourier series as denned earlier in this chapter,
aft)
_ £s + V (a , cos nmt +
2
b n sin nmt)
(3.21)
£Ti
From our discussion of approximation by orthonormal functions, we can
by a
see that the periodic function s(t) with period T can be approximated
is,
that
s(t),
to
mean
the
in
Fourier series sn(t) such that s n(t) converges
lim
im
-»oo
where a
is
any
real
mean squared error
r^MO-s-O)]*^ —
number.
||j(0
(
322>
Ja
We know, moreover, that if n
- *»(0II* is minimized when
is finite,
are the Fourier coefficients of s(t) with respect to the orthonormal set
/cos koat
sin k<ot \
.__«,«
the
the constants a{ , b t
The frequency domain: Fourier
51
analysis
m
fYVtW\
-3x
-2x
-x
FIG.
x
2x
3.1. Rectified sine
3x
t
wave.
in explicit form the Fourier coefficients, according to the definition given
obtained from the equations
earlier, are
*)A
fl0=s
rJ«
2 r
s(t)
-If
mean, when
s(t)
cos kmt dt
s(0 sin tor
We should note that because the
s(t) in the
(323)
«/f
(3.24)
(3.2S)
Fourier series sn(t) only converges to
jump discontinuity, for example,
contains a
= ^±I±J(£o=)
S .0,)
( 3. 26)
At any point
tt that s(t) is differentiable (thus naturally continuous)
converges to a(/J.*
As an example, let us determine the Fourier coefficients of the fully
rectified sine wave in Fig. 3.1. As we observe, the period is
n so
that the fundamental frequency is m
2. The signal is given as
*»(*i)
T=
=
s{t)
Let us take a
derived,
=
= A |sin
and evaluate between
we have
(3.27)
t\
and
ir.
Using the formula just
.
K = ~ Jo «(0 sin 2nt dt =
|
(3.28)
TT
«,
2A C*
&A
=—
sin tdt = —
\
IT
W
Jo
2[>
(3.29)
s(0 cos 2nt dt
IT
Jo
1
-An*
1
AA
(3.30)
• For a proof see H. F. Davis Fourier Series and Orthogonal Functions,
Allyn and
Bacon, Boston, 1963, pp. 92-95.
Network
52
analysis
Thus the Fourier
of the
series
5(/)
and synthesis
wave
rectified sine
—±—
- £i(l + f
is
cos 2nt)
(3.31)
EVALUATION OF FOURIER COEFFICIENTS
3.5
In this section we will consider two other useful forms of Fourier series.
In addition, we will discuss a number of methods to simplify the evaluation
of Fourier coefficients. First, let us examine how the evaluation of
symmetry considerations. From Eqs. 3.23-3.25
which give the general formulas for the Fourier coefficients, let us take
a = —T/2 and represent the integrals as the sum of two separate parts,
coefficients is simplified by
that
is,
rr/t
2
T
ss
ro
s(0 cos nmt dt
s(t)
cos mot dt
J
"
J-T/S
T/a
JO
2 rf
—
-i
+
T,t
nmt dt
s(t) sin
I
TUo
+
f
°
s(f) sin
I
(3.32)
i
nmt dt
J
J-T/i
Since the variable (0 in the above integrals is a dummy variable, let us
t in the
substitute x
t in the integrals with limits (0 ; r/2), and let x
=—
=
integrals with limits
an
bn
(—T/2;
= —2
C
T/ *
[s(x)
I
TJ °
=
T Jo
+ s(—x)] cos nmx dx
(3.33)
T/
—2 C
I
Then we have
0).
*
[s(a:)
— s(— *)] sin nmx dx
Suppose now the function is odd, that is,
a„ =
for all n,
s(x)
=
—s(—x), then we
see that
and
T/2
-if*
nmx dx
s{x) sin
tJo
This implies that,
sine terms.
s(x)
On
if a
function is odd,
its
Fourier series will contain only
the other hand, suppose the function
= s(—x), then b n =
an
(3.34)
is
even, that
is,
and
=—
I
TJo
s(x) cos
nmx dx
(3.35)
Consequently, the Fourier series of an even function will contain only
cosine terms.
The frequency domain: Fourier
anal/sis
53
FIG. 3.2
Suppose next, the function
s(t)
,(,
obeys the condition
± l)--<0
by the example in Fig. 3.2.
only odd harmonic terms, that is,
as given
«„
and
Then we can show
= bn = 0;
-if
TJo
T Jo
s(t)
(3.36)
that s(t) contains
neven
cos nmt dt
(3.37)
5(0 sin nmt
n odd
dt,
With this knowledge of symmetry conditions, let us examine how we
can approximate an arbitrary time function s(t) by a Fourier series within
an interval [0, 71. Outside this interval, the Fourier series sn(t) is not
required to fit s(t). Consider the signal s(t) in Fig. 3.3. We can approximate s(t) by any of the periodic functions shown in Fig. 3.4. Observe
that each periodic waveform exhibits some sort of symmetry.
Now let us consider two other useful forms of Fourier series. The first
is the Fourier cosine series, which is based upon the trigonometric identity,
C„ cos (na>i + 8J
=
C„ cos mot cos 0„
^|
—
C„ sin nmt sin 0„
T
FIG. 3J. Signal, to be approximated.
t
(3.38)
Network
54
anal/sis
and synthesis
m
A
^
^
^
^*«^T
-T--
^"'
t
(a)
m
A
"*•-«..
-r'- "~«^
f~"-~.
°
t
-A
(6)
m
A
**-
-T)
t
*
I**"
-A
(e)
FIG. 3.4. (a) Even function cosine terms only, (ft) Odd function
Odd harmonics only with both sine and cosine terms.
sine terms only,
(c)
We can derive the form of the Fourier cosine series by setting
= C„ cos 0„
bn = — C„ sin 0„
a„
and
We then obtain Cn and 0„ in terms
0,
(3.40)
of an and bn, as
c,-(fl.'
Ca =
(3.39)
+
M
a
(3.41)
—(-!:)
combine the cosine and sine terms of each harmonic in the original
we readily obtain from Eqs. 3.38-3.41 the Fourier cosine series
If we
series,
f(t)
= C + Ct cos (mi + 00 +
+ C.cos (3fl>f + 0,) +
•
C, cos (2o>f
•
•
+
0.)
+ Cn cos(nmt + J +
•
•
•
(3.42)
The frequency domain: Fourier
55
anal/sis
should be noted that the coefficients C„ are usually taken to be positive.
a term such as —3 cos 2cot carries a negative sign, then we can
use the equivalent form
It
If however,
— 3cos2a>f =
For example, the Fourier
was shown to be
of the fully
series
= ^(l + f
s(0
2ir\
Expressed as a Fourier cosine
s(0
3 cos (2<u/
n=i
+ tt)
(3.43)
rectified sine
wave in
Fig. 3.1
-^—
cos 2nt)
— An
(3.44)
,
I
1
series, s(t) is
= ^[l + 1 —£— cos (2nt + *)]
2wL
n-i4n — 1
(3.45)
J
Next we consider the complex form of a Fourier series. If we express
cos ncot and sin mot in terms of complex exponentials, then the Fourier
series
can be written as
(3.46)
= — + 2 ( "* ~ J&" **"* +
2
fl
" + ^* *-'"')
2
n-l\
n-l
2
/
If we define
P»
=
"~ n
'
2
2
'
2
then the complex form of the Fourier series
is
= A + i(/ff„e*"" + /*-„«-'"•")
5(0
"^
„
- 2
n—
(
34«)
PSoo
We can readily express the coefficient fi„ as a function of s(t), since
/?
Pn
a«
= —l
—
~ Jb »
C
T
I
T Jo
=i
T
f
T Jo
Equation 3.49
and Eq. 3.48
is
is
s(0(cos nott
s(t)e-
— j sin na»0 dt
(3.49)
mt dt
i'
sometimes called the discrete Fourier transform of
the inverse transform of P„(na))
=
/?„.
«(/)
56
Network
analysis
and synthesis
Cb
Ci
Ci
C2
Ch
<h
Ci\
c*.
w
-Au -3« -2w -w
FIG.
3.5.
2(i>
Amplitude spectrum.
-2T
¥ 2T
HF -T •4
-A
FIG.
3d)
3.6.
Square wave.
,C*
4a>
The frequency domain: Fourier
Observe that /?„
is
usually complex
=
/*n
The
real part
of fi„
Re /?„,
- -1
and the imaginary part of /?„
T Jo
+yIm£,
(3.50)
«(0 cos tuot dt
(3.51)
is
T Jo
It is clear that
and can be represented as
CT
JImfl,-^f
function in n.
57
obtained from Eq. 3.49 as
is
Re Pn
Re/8B
anal/sis
s(0sinn«>r<fc
(3.52)
Re /?„ is an even function in r, whereas Im /S„ is an odd
The amplitude spectrum of the Fourier series is defined as
ImP«M
|/JJ-(Re"fc.+
and the />/uue spectrum
is
(3.53)
denned as
6,
= arctan^
(3.54)
It is easily seen that the amplitude spectrum is an even function and the
phase spectrum is an odd function in n. The amplitude spectrum provides
us with valuable insight as to where to truncate the infinite series and still
maintain a good approximation to the original waveform. From a plot of
the amplitude spectrum, we can almost pick out by inspection the nontrivial terms in the series. For the amplitude spectrum in Fig. 3.5, we see
that a good approximation can be obtained if we disregard any harmonic
above the third.
As an example, let us obtain the complex Fourier coefficients for the
square wave in Fig. 3.6. Let us also find the amplitude and phase spectra
of the square wave. From Fig. 3.6, we note that s(t) is an odd function.
Moreover, since s(t
T/2) = — s(t), the series has only odd harmonics.
From Eq. 3.49 we obtain the coefficients of the complex Fourier series as
—
n
Ae-'Mi
--
TJo
dt--\
Ae' ,nat dt
TJt/»
(3.55)
(I
— 2e~ linmT/i)
-)-
jnmT
Since ncoT
e~
,tuaT
)
= nlir, /?„ can be simplified to
/».
- rr- a - 2*"*" +
«"*"*)
( 3 - 56 >
58
Network analysis and
synthesis
Amplitude
1
1
-3
-5
-
I
1
1
1
3
5
u
<a)
Phase
1
-5
-3
-1
2
(b)
FIG.
3.7. Discrete spectra
Simplifying
/5»
one step
of square wave, (a) Amplitude. (*) Phase.
we
further,
obtain
n odd
(3.57)
jnir
=
The amplitude and phase
n even
spectra of the square
wave are given in
Fig. 3.7.
OF FOURIER COEFFICIENTS USING
UNIT IMPULSES
34 EVALUATION
In this section
we make use of a basic property of impulse functions to
is
simplify the calculation of complex Fourier coefficients. This method
restricted to functions which are made up of straight-line components only.
Thus the method applies for the square wave in Fig. 3.6. The method is
based on the relation
f" /(*)«(«-
J— 00
TO* -/(TO
(3.58)
Let us use this equation to evaluate the complex Fourier coefficients for
,fmt
e~
, we have
the impulse train in Fig. 3.8. Using Eq. 3.58 with/(f)
-
'•
=
~ *~ 2V,,
e
d%
T fH' - 2 V'"" '~T
<
(359>
The frequency domain: Fourier
Al
A..
A
A,i
_L
r
1
2
FIG.
3.8.
59
analysis
.
_L
3r
T
2T
Impulse
sr
T
train.
We
see that the complex Fourier coefficients for impulse functions are
obtained by simply substituting the time at which the impulses occur into
the expression, e~,m" t .
In the evaluation of Fourier coefficients, we must remember that the
P„ integral are taken over one period only, i.e., we consider
only a single period of the signal in the analysis. Consider, as an example,
the square wave in Fig. 3.6. To evaluate /?„, we consider only a single
period of the square wave, say, from t
to
limits for the
=
t
=
T, as
wave
is
shown in Fig. 3.9a. Since the square
not made up of impulses, let us
differentiate the single period
wave to
can
give
s'(t),
now evaluate
as
shown
of the square
in Fig. 3.96.
We
the complex Fourier coeffi-
cients for the derivative s(t),
made up of impulses
s(t) is
>(t)
A
which clearly
is
alone. Analytically, if
-A
given as
s(0
T
f
(<»)
- 1
»——
inert
P«e
(3.60)
00
then the derivative of 5(0
•W
is
A
A|
(3.61)
Here,
we
define a
new complex
coefficient
(3.62)
or
jnm
(3.63)
-2A
a function which
consists of impulse components alone, then
If the. derivative s'(i)
we
is
simply evaluate y n first and then obtain
3.63. For example, the derivative
of the square wave yields the impulse train
/?„
from Eq.
(6)
FIG. 3.9. (a) Square wave
over period [0, 71. (b) Derivative of square wave over
period
[0, T\.
Network
60
and synthesis
analysis
in Fig. 3.96. In the interval
s'(t)
=
Then the complex
[0,
A6(t)-
T], the signal s'(f ) is given as
d(t
2^4
- |) + A d(t -
T)
(3.64)
coefficients are
Tvo
(3.65)
= —(\— 2e- linmT/* +
}
e~
inmT
)
T
The Fourier
of the square wave are
coefficients
_
a
y»
jnco
(3.66)
— ^
—
(i
2e~
linaTI * )
+
e~
inmT
)
jnmT
which checks with the solution obtained in the standard way in Eq. 3.55.
If the first derivative, /(f), does not contain impulses, then we must
differentiate again to yield
= f
n——
s »(f)
(3.67)
(3 - 68)
1
For the
[0,
triangular pulse in Fig. 3.10, the second derivative over the
T]
is
s»(0
The
inmt
K =jno>Yn = (P ™)* P«
where
period
Xne
oo
coefficients
= ^Uf) - 2d(f - |) +
A n are
Xn
now
=-
**
- d]
( 3 - 69 >
obtained as
T
( s"(i)e-
inat
dt
TJo
(3.70)
2A
which
,j
_ 2e-OfK»27*>
i
-fnmT\
e
simplifies to give
X =s
"
=
&A
—
T*
.,
n odd
(3.71)
n even
The frequency domain: Fourier
analysis
61
A&1
3T
T
2
(a)
24 »Y0
T
r
T
2"
t
2A
T
2A
(b)
[>"<'>
2A
(C)
FIG. 3.10. The triangular wave and
From
A_
we
its
derivatives.
obtain
P»-r.
(Jmnf
=
2A
(3.72)
nodd
nV
=
neven
A
slight difficulty arises if the expression for s'(t) contains an impulse
in addition to other straight-line terms. Because of these straight-line
terms
we must
differentiation,
differentiate
once more.
However, from
we
this additional
obtain the derivative of the impulse as well.
presents no difficulty, however, because we know that
°%(r) d'(t
-
T)
J— GO
-
-s\T)
(3.73)
I
so that
f " d'(t
J— 00
-
T)e- im"dt
This
= jna>e-"u T
'
(3.74)
—
Network
62
analysis
and synthesis
We can therefore tolerate doublets or even higher derivatives of impulses
la. Its derivative
in the analysis. Consider the signal s(f) given in Fig. 3.1
as
expressed
be
*'(<), shown in Fig. 3.116, can
S '(0
The second
=
M (0
-
|[
-
u(t
f )]
derivative s"(t) consists of
+ «0 - 2«(l - f)
0.75)
a pair of impulses and a pair of
doublets as given by
mas
shown
|[aco
in Fig. 3.11c.
-
-
*(t
f )]
+ *W -
2d'(t
1)
We therefore evaluate A„ as
,-inmt
s"(t)e
dt
tJo
(3.77)
— 2. (i _ e-(*~»r/») + 1™± (i _ 2e- UnaTin )
s'(t)
2
1
-
.,
r
'
t
:
(3.76)
t
J
(W
(o)
-*'('-!)
«"W
pit)
r
?«(*-i)
to
FIG. 3.11
t-M(l-f)
The frequency domain: Fourier
The complex
coefficients /S„ are
now
63
analysis
obtained as
Qtonf
±— (i _ e-o»»Z7«) + _!_ n _ 2e-(^«»r/a))
=
janT
(jconT)*
Simplifying,
(3-78)
'
we have
1
Pn=-—
mr +
t
—
3
"odd
j2irn
(3.79)
=
j
n even
j2irn
In conclusion,
it
must be pointed out that the method of using impulses
to evaluate Fourier coefficients does not give the d-c coefficient, a /2 or
/3
.
We obtain this coefficient through standard methods as given by Eq. 3.23.
3.7
THE FOURIER INTEGRAL
In this section
we
extend our analysis of signals to the aperiodic case.
We show through a plausibility argument that generally, aperiodic signals
have continuous amplitude and phase spectra. In our discussion of Fourier
series, the complex coefficient (i n for periodic signals was also called the
discrete Fourier transform
1
CT
s(i)e-
M"'*dt
TJ-i
-T/2
and the inverse
(discrete)
transform was
*(0
From
= 2
n=—
ln,'
Knfo)e
M
(3.81)
oo
the discrete Fourier transform
spectra which consist of discrete lines.
in the spectrum
(3.80)
we obtain amplitude and phase
The spacing between adjacent lines
is
A/ -
(n
+
l)/„
-
«/„
=i
(3.82)
As the period T becomes larger, the spacing between the harmonic lines in
the spectrum becomes smaller. For aperiodic signals, we let T approach
infinity so that, in the limit, the discrete spectrum becomes continuous.
64
Network
and synthesis
anal/sis
We now define the Fourier integral or transform as
s(/)
=
lim
T—x>
£W£ = r
Jo
A/-»0
The
inverse transform
-«"
s(()c
it
(3.83)
J— oo
is
<0-f"
J—
s
S(J)e
iU,t
(3.84)
df
00
Equations 3.83 and 3.84 are sometimes called the Fourier transform pair.
5~x denote
denote the operation of Fourier transformation and
If we let
inverse transformation, then
7
*</>-*-<0
s(t) = ?-i-S(f)
In general, the Fourier transform S(f)
S(f)
The
real part of
S(f)
is
is
(3 . 85)
complex and can be denoted as
= ReS(f)+ j Im S(f)
(3.86)
obtained through the formula
ReS(/)
=
J[S(/)
+
=
|"
cos 2-nft dt
S(-/)]
(3.87)
s(t)
J— CO
and
the imaginary part through
= £ [SCO -
Im S(f)
S(-/)]
2/
(3.88)
=_
°°
j
J— CO
The amplitude spectrum of S(f)
A(f)
and the phase spectrum
=
is
s (0 sin 2tt/(
dt
defined as
[Re SC/) 2
+ Im £(/)*]*
(3.89)
is
«"-"— Sin
**»
Using the amplitude and phase definition of the Fourier transform, the
inverse transform can be expressed as
s(t)
=
°°
f
A(f) cos
J— 00
Let us examine some examples.
[lirft
-
<Kf)] df
(3.91)
The frequency domain: Fourier
MS)
65
anal/sis
<Kf)
A
-f
+/
(a)
((,)
FIG. 3.12. Amplitude and phase spectrum of A d(t
Example
—
/,).
3.2.*
sit)
= A6\t-td
S(f)~ ("
A^t- t9)er^t* dt
J— 00
- Ae-i**">
Its
amplitude spectrum
(3.92)
is
A(f)=A
while
its
phase spectrum
(3.93)
is
<Kf)
-
(3.94)
-2^//o
as shown in Fig. 3.12.
Example 33.
Formally,
Next consider the rectangular function plotted in Fig.
we define the function as
HSyW
1
\x\
>
inverse transform of rect/is defined as sine
-1
3r
[rect/]
(3.95)
-W
recta;/
The
3.13.
the rect function.
t
(pronounced
sink),
» sine t
•W/2
e>*"*df
(3.96)
-I-Jf/2
sinirWr
Vt
* It should be noted here that the Fourier transform of a generalized function is also
a generalized function. In other words, if e C, (^ <f>) £ C. For example, J" <5(f ) =
1, where 1 is described by a generalized function 1„(/). We will not go into the formal
details of Fourier transforms of generalized functions here. For an excellent treatment
of the subject see M. J. Lighthill, Fourier Analysis and Generalized Functions, England,
Cambridge University Press, 19SS.
•
<f>
•
Network
66
analysis and synthesis
recti
1.0
-*
"
w
+
2
2
FIG. 3.13. Plot of rect function.
1.0
0.8
0.6
sine
t
0.4
v
0.2
\\
0.0
\
-0.2
-0.3
±
~w
£.
w
3.
~w
1.
~w
FIG. 3.14. The sine
e(t)
W
/
2.
W
3.
W
4.
W
5.
W
curve.
r(t)
Excitation
System
Narrow pulse
Wide band
*(0
Excitation
i.
-L
~w
Narrow pulse
r(t)
System
Response
10 Ti
10 Tx
Wide pulse
Narrower band
Wide pulse
FIG. 3.15. Illustration of the reciprocity relationships between time duration and
bandwidth.
The frequency domain: Fourier
67
anal/sis
From the plot of sine t in Fig. 3.14 we see that sine t falls as does |r| _1 ,
with zeros at t
n\W, n
We also note that most of the
1, 2, 3, .
energy of the signal is concentrated between the points
\\W t
\\W.
=
=
.
.
—
< <
Let us define the time duration of a signal as that point, t , beyond which
the amplitude is never greater than a specified value, for example, e
We
can effectively regard the time duration of the sine function as t
1/ W.
The value W, as we see from Fig. 3.13, is the spectral bandwidth of the
.
=±
rect function.
We
see that if
W increases,
decreases.
t
The preceding
example illustrates the reciprocal relationship between the time duration
of a signal and the spectral bandwidth of its Fourier transform. This
concept
narrow
is
quite fundamental.
pulses,
i.e.,
It illustrates
why
in pulse transmission,
those with small time durations, can only be trans-
mitted through filters with large bandwidths; whereas pulses with longer
time durations do not require such wide bandwidths, as illustrated in
Fig. 3.15.
3.8
PROPERTIES OF FOURIER TRANSFORMS
In this section
we
some important
consider
properties of Fourier
transforms.
Linearity
The
linearity property
transform of a
of Fourier transforms states that the Fourier
signals is the sum of their individual Fourier
sum of two
transforms, that
is,
Hcx *i(0 + c
= Cl S (/) + c
t sjflj]
x
2
St(f)
(3.97)
Differentiation
This property states that the Fourier transform of the derivative of a
signal isjlnf times the Fourier transform of the signal itself:
J-s\i)=j2-nfS{f)
(3.98)
?- s ™(t) = (j2*frS(f)
(3.99)
or more generally,
The proof is obtained by taking the
derivative of both sides of the inverse
transform definition,
00
s'(0
= 7;f
S(J)e»*<*df
at J -oo
(3.100)
-f
J—
CO
j2nfS(f)e»*"df
Network
68
Similarly,
it is
and synthesis
analysis
easily
that the transform of the integral of s(t)
shown
f
?
[
5(t)
'
dr\
= ^-
(3.101)
S(J)
j2nf
J
LJ-oo
is
Consider the following example
j(0
Its
Fourier transform
= «-°'«(0
( 3 - 102>
is
=
s(/)
r^vHtyr^dt
f"CO
J—
(3.103)
a
The
derivative of s(t)
s '(t)
Its
Fourier transform
+ j2irf
is
.
3(0
- aer** «(0
(3.104)
is
?[''«)]
-1-
j2nf
a+j2wf
a+j2irf
(3.105)
= j2nfS(f)
Symmetry
The symmetry property of Fourier transforms
states that if
= X(f)
^ X(t) = *(-/)
3"
then
•
(3.106)
x(t)
(3.107)
•
This property follows directly from the symmetrical nature of the Fourier
transform pair in Eqs. 3.83 and 3.84.
Example
3.4.
From the preceding
^
It is
•
section,
we know
that
= rect/
(3.108)
= sine ( -/) = sine/
(3.109)
sine /
then simple to show that
&
•
rect
t
which conforms to the statement of the symmetry property.
the Fourier transform of the unit impulse,
property
&
•
8(t)
=
1.
From
Consider next
the symmetry
we can show that
?•
1
=
8(f)
(3.110)
shown in Fig. 3.16. The foregoing example is also an extreme illustration
of the time-duration and bandwidth reciprocity relationship. It says that zero
time duration, 8(t), gives rise to infinite bandwidth in the frequency domain;
while zero bandwidth, 8(f) corresponds to infinite time duration.
as
The frequency domain: Fourier
m
analysis
69
HD
-/
-t
o
FIG. 3.1*. Fourier transform of /(f)
=
f
1.0.
Scale change
The
scale-change property describes the time-duration
-M*«0
[.(*)]
Proof.
We prove
and bandwidth
It states that
reciprocity relationship.
this
(3.111)
property most easily through the inverse trans-
form
J-\\a\ S{af)]
=
\a\
f" S(af)e™* df
(3.112)
J— oo
Let/'
= af;
then
*^l«l
S(f')]
=
\a\
f " S(/V*"'
J— co
( '/o,
—
a
(3.113)
As an example,
consider
T[e* "(01
(3.114)
J2vf+a
then
Sy
1
\a\
«(01
j2iraf
+
a
(3.115)
1
j2nf+\
ifo>0.
Folding
The
folding property states that
J-W-01
= s(-f)
(3.116)
Network
70
anal/sis
The proof follows
An
example
and synthesis
directly
from the
definition of the Fourier transform.
is
3=V u(-t)]
^—
=
(3.117)
l-j2vf
Delay
is delayed by an amount t in the time domain, the corresponding effect in the frequency domain is to multiply the transform of the
undelayed signal by er iz"ft that is,
If a signal
<>,
(3.118)
For example,
&[e- aU
-'° )
u(t-t
)]
-}2rftt>
=
a
(3.119)
+ j2nf
Modulation
The modulation or frequency
shift property of Fourier transforms
a Fourier transform is shifted in frequency by an amount f
the corresponding effect in time is described by multiplying the original
states that if
signal
,
by e iufot that
,
is,
3^[S(f - /<,)]
=
e
mM
(3.120)
s(t)
Example
3.5.
Given S(f) in Fig. 3.17a, let us find the inverse transform of
5"" 1
Si(f) in Fig. 3.176 in terms of s(t) =
S(f). We know that
Sxif)
=
S(f -f )
+ 5(/ + /„)
(3.121)
-/
+/
h
-/
*f
(b)
FIG.' 3.17.
Demonstration of amplitude modulation.
The frequency domain: Fourier
Then
analysis
3^ 5,(/) - *>*".« s{t) + er'*'U* g(t) = 2s(t) cos 2nfy
71
(3.122)
Thus we see that multiplying a signal by a cosine or sine wave in the time domain
corresponds to shifting its spectrum by an amount
±f In transmission terminology /„ is the carrier frequency, and the process of multiplying s(t) by cos
.
2itfy
is
called amplitude modulation.
Parseval's
theorem
An
important theorem which relates energy in the time and frequency
domains is Parseval's theorem, which states that
"
«i(0 St(0 dt
I
J—
00
= f"
The proof is obtained very simply
f"
si(0 »i(0 dt
=
J— no
Sx(/) S,(-/) df
as follows:
f " s„(t) dt f "
-f"
•*—
(3.123)
SiC/V""* d/
Si(/)4rf"
s&Vu"dt
(3.124)
J—ao
oo
sMSA-fldf
-f"
J—
00
when ^(r) = *j(f), we have a corollary of Parseval's theorem
known as PlancheraVs theorem.
In particular,
f " aty)
J— 00
* = J—
f"
1
ISC/)!
4T
(3.125)
00
If j(/) is equal to the current through, or the voltage across a
resistor, the total
energy
1-ohm
is
I
s
2
(0 dt
We see from Eq. 3.125 that the total energy is also equal to the area under
the curve of \S(f)\*.
or energy spectrum.
Thus
\S(f)\* is
sometimes called an energy density
Problems
3.1 Show that the set {1, sin nnt/T, cos nnt/T}, n = 1,2, 3,
, forms an
orthogonal set over an interval [a, a + 2T], where a is any real number. Find
the norms for the members of the set and normalize the set.
.
.
.
Network
72
3.2
analysis
and synthesis
Fourier
Given the functions/^) and/i(0 expressed in terms of complex
senes
/l(/)=2«»^
n— — oo
"
B,,
/K0-2A.'*-'
where both/i(0 and/j(0 have the same period T, and
«„
show
= Kl e**\
A.
-
e*»
that
-
«oA>
+ 22 l«g»»l cos(flB -
Note that
For the periodic
4>J
m * -n
\0,
33
lft»l
signals in the figure, determine the Fourier coefficients
f(f>
—»-
-«-
+T
-T
(h)
fit)
A
T
5
(e)
PROB.3.3
A
T
The frequency domain: Fourier
analysis
3.4 For the waveforms in Prob. 3.3, find the discrete amplitude
spectra and plot.
73
and phase
3.5 For the waveforms in Prob. 3.3 determine the complex Fourier coefficients
using the impulse function method.
3.6
Find the complex Fourier
coefficients for the function
shown
figure.
A AA
ST,
3T.
—r
~r
Tj
T,
T
-t
PROB.
3.7
L,
3T,
51
~T
~i
3.6
Find the Fourier transform for the functions shown in the
figure.
m
f(t)
1
/
/
-T
\
^C*
*T
t
t
(b)
(«>
f(t)
f<t)
V
.1
y
/
3V
t
(c)
Id)
PROB.
3.8
3.7
Find the Fourier transform for
(a)
/« - A <3(0
(*)
/(f)
— A sin atf
t
in the
74
Network
anal/sis
and synthesis
3.9 Prove that (a) if /(f) is even, its Fourier transform F(ja>) is also an even
function; (6) if /(/) is odd, its Fourier transform is odd and pure imaginary.
F(ju)
B
B
—
+«0
(i)
PROS.
3.10
Find the inverse transform of
F(jm)
shown
domain?
as
3.10
in the figure.
= P 6(fo -
What can you
Wo)
+ P 8{(o +
eo )
say about line spectra in the frequency
chapter
4
Differential equations
INTRODUCTION
4.1
This chapter
is
devoted to a brief study of ordinary linear differential
We will concentrate on the mathematical aspects of differential
equations.
equations and leave the physical applications for Chapter
equations considered herein have the general form
5.
The differen-
tial
/MO, *'('),
where
t.
•
•
•
,
*< B >(0,
t]
=
(4.1)
the independent variable and x(t) is a function dependent upon
superscripted terms ««>(/) indicate the rth derivative of x(t) with
t is
The
respect to
t,
namely,
«>/a
x(t)
=
d(i)
x(t)
~df~
(42)
=
F
in Eq. 4.1 is x(t) and must be obtained as an
of /. When we substitute the explicit solution x(t) into F,
the equation must equal zero. If fin Eq. 4.1 is an ordinary
linear differThe
solution of
explicit function
ential equation,
it is
(B>
«» * (0
given by the general equation
+ a^x^Xt) +
+ «, x\t) + oo x(t) =f(t) (4.3)
The order of the equation is n, the order of the highest derivative
term.
The term/(0 on the right-hand side of the equation is Xht forcing function
or driver, and is independent of x(i). When /(f) is identically
zero, the
•••
equation is said to be homogeneous; otherwise, the equation is
nonhomogeneous.
In this chapter we will restrict our study to ordinary, linear
differential
equations with constant coefficients. Let us now examine the
meanings of
these terms.
75
Network
76
Ordinary.
anal/sis
and synthesis
An ordinary differential equation is one in which there is only
one independent variable
our case,
(in
t).
As a result there is no need
for
partial derivatives.
conConstant coefficients. The coefficients a n , o^, ...,a„ alt a„ are
stant, independent of the variable t.
the
differential equation is linear if it contains only terms of
Linear.
For
Eq.
4.3.
by
given
as
derivatives,
first degree in x(t) and all its higher
A
example, the equation
3x'(t)
is
a linear
+ 2z(0 = sin
On the other hand,
differential equation.
3[*
W
+ 2*(f) x\t) + 4x(0 =
nonlinear, because the terms [x'(t)]*
is
(4.4)
t
and
5t
x(t) x'(t) are nonlinear
(4.5)
by the
definition just given.
important implication of the linearity property is the superposition
and x^t) are
x
property. According to the superposition property, if x (0
and/,(0,
functions/^)
forcing
for
equation
solutions of a given differential
of
combination
linear
were
any
function
then, if the forcing
An
respectively,
as
JUt) and/,(0 such
AO-afM + btff)
(4-6)
the solution would be
x(r)-a«i(0
(4 - 7 )
+ &*40
that the
where a and b are arbitrary constants. It should be emphasized
kept in
be
should
and
important
extremely
is
superposition property
mind
4.2
in
any discussion of linear
differential equations.
HOMOGENEOUS LINEAR
DIFFERENTIAL EQUATIONS
homogeneous,
This section deals with some methods for the solution of
let us find
First,
linear differential equations with constant coefficients.
the solution to the equation
x'(t)
- 2*(<) =
(4.8)
Now, with a little prestidigitation, we assume the solution to be of the form
*(0
where
is
C is any arbitrary constant.
truly a solution of Eq. 4.8.
we
=
Ce"
Let us check to see whether x(t)
(4.9)
=
Ce2 *
Substituting the assumed solution in Eq.
obtain
2Ce* f
- 2Ce»* =
tA
4.8,
tn
.
(4.10)
77
Differential equations
It
can be shown, in general, that the solutions of homogeneous, linear
equations consist of exponential terms of the form C<e p <*. To
differential
vt
obtain the solution of any differential equation, we substitute Ce for x(t)
for
which
the
equation
is
in the equation and determine those values ofp
zero. In other words, given the general equation
an *<">(*)
we
=
let x(t)
+••• +
Ce", so that Eq.
4.11
+ a, x(f) =
«! x'(t)
(4.1 1)
becomes
(4.12)
+ a# + a,) =
+ a^tf*- +
pt
Since e cannot be zero except at/? = — oo, the only nontrivial solutions
1
Ce'Kanp"
•
•
when the polynomial
for Eq. 4.12 occur
H(p)
•
- <W" + a-tf"-1 +
•
+
*
'
W+
«o
-
(4.13)
Equation 4.13 is often referred to as the characteristic equation, and is
denoted symbolically in this discussion as H(p). The characteristic equation
is zero only at its roots. Therefore, let us factor H(j>) to give
H(j>)
= an(p- PMp- Pi)---(p- p^
(4.14)
solutions
t .... Cw_ 1c*«-»' are all
of Eq. 4.11. By the superposition principle, the total solution is a linear
combination of all the individual solutions. Therefore, the total solution
of the differential equation is
From Eq. 4.14, we note that C^e"**, Cxe***
x(t)
Q**'
+
•
•
+
•
<:_!«»-»«
(4.15)
, C
,
x,
B_i are generally complex. The solution x(t) in Eq.
, Cn_x are uniquely
not unique unless the constants C„, Cls
specified. In order to determine the constants Cit we need n additional
pieces of information about the equation. These pieces of information are
usually specified in terms of values of x(t) and its derivatives at t
0+,
where
C C
= CV' +
.
.
.
4. IS is
.
.
.
—
and are therefore referred to as initial conditions. To obtain ft coefficients,
"-1
, a;
>(0+). In a number
we must be given the values <r(0+), x'(0+),
(
.
of special cases, the values at
/
= 0+.
x'(0— ),
.
to solve
t = 0—
.
.
are not equal to the values at
*(0— ),
=
values
at
In
order
determine
the
f
0+
we
must
^"-^(O— ),
for the constants C{
This problem arises when the forcing
If the initial specifications are given in terms of
.
.
,
.
an impulse function or any of its derivatives. We will
discuss this problem in detail in Section 4.4.
For example, in Eq. 4.9 if we are given that *(0+) = 4, then we obtain
the constant from the equation
function /(f)
is
x(0+)
= Ce*=C
(4.16)
p
Network
78
so that x(t)
is
analysis
and synthesis
uniquely determined to be
x(t)
Example
Find the solution for
4.1
+
*'(/)
given the
= 4e*
initial
5x\t)
+ 4a</) =
(4.17)
conditions
a<0+)-2
= -1
x'(0+)
From the given equation, we first obtain the characteristic equation
Solution.
= p* +
H(p)
which factors into
+ 4X/> +
(/>
+
5p
1)
=
4
(4.18)
=
(4.19)
The roots of the characteristic equation (referred to here as
step = -1 p = —4. Then x(t) takes the form
characteristic values)
;
= Qe-* +
x{t)
From
the initial condition
«(0+)
=
x(0+)
2,
we
(4.20)
C^e-**
obtain the equation
Q+C
=2 =
(4.21)
a
In order to solve for Cx and C, explicitly, we need the additional initial condition
x'(0+) = -1. Taking the derivative of x(i) in Eq. 4.20, we have
= -Qe-' - 4Q*-4
X '(f)
Atf =0+,a;'(0is
a:'(0+)
Thus the
final solution is
Next,
we examine
multiple roots.
root p
=p
(4.23)
a
we find
that
=i
a*/)
the case
(4.22)
-Q -4C
= -1 =
Solving Eqs. 4.21 and 4.23 simultaneously,
Ci
*
C% = —J
= %r* - \e~**
when
Specifically, let
(4.24)
the characteristic equation H(p) has
us consider the case where H(p) has a
of multiplicity k as given by
H(p)
= an(p -
k
)
(p
-pd---(p-pn)
(4-25)
be left to the reader to show that the solution must then contain k
terms involving e" * of the form
It will
x(i)
=
CV"»«
+ C01 te*>* +
+
+ C^*** +
Cnt'e***
•
+ C^t*-^"*
Cje "' +
+ Cne*'*
•
•
1
•
•
•
(4.26)
where the double-scripted terms in Eq. 4.26 denote the terms in the solution
due to the multiple root, (p — #,)*•
79
Differential equations
Example 4.2
Solve the equation
*»(/)
- Sx'(t) +
16*(r)
a<0+)«2 and
with
Solution.
The characteristic equation
-
H{p) ~p*
Since H(p) has a double root at/>
x(t)
In order to determine
Q
=4,
final solution is
"=4
is
+
16
= (p - 4)»
(4.28)
the solution must take the form
and C„ we evaluate
a<r)
x(f)
(4.29)
and
»'(') at r
= 0+
to give
= C1 =2
(4.30)
=4Q +C =4
*'(<>+)
Another
*'(<>+)
(4.27)
« C^* + Cy*"
«(0+)
Thus the
ip
=
2
= 2c*' - 4te*'
interesting case arises
(4.31)
when H(p) has complex conjugate
roots.
Consider the equation
H{p)
- a (p -pMp-pJ
(4.32)
2
where />! and />, are complex conjugate roots, that
=
Px,Pi
The
solution x(t) then takes the
x(t)
Expanding the term
x(t)
e*"
f
=
is,
o±jo>
(4.33)
form
Ci«<*W">*
+
<V—'»>'
by Euler's equation,
= C e*(cos cot +j sin cat) +
t
x(t)
(4.34)
can be expressed as
Cje^cos
cor
—y sin cat)
(4.35)
which reduces to
x(t)
= (Q +
C,y* cos
cor
+j(C1
Let us introduce two new constants,
more convenient form
M
x
—
CJe'*
sin cor
and Af„ so that
(4.36)
x(r)
may be
expressed in the
x(t)
where 3/x and
M
= Mje'* cos cor + Aftf"' sin cat
are related to the constants
t
M! =
Cx and C2 by the equations
d + C,
^. =y(C! - C2)
The
constants Afj
initial
conditions.
and
M
t
(4.37)
are determined in the usual
(4.38)
manner from
Network
80
and synthesis
analysis
Another convenient form for the solution x(t) can be obtained if we
and <f>, defined by the equations
introduce still another pair of constants,
M
Ml = MsiD *
= M cos
we
obtain another form of x(i), namely,
t
With the constants
M and
<f>
+ 2*'(/) + 5x(t) =
=
a<0+)
The
*'(<>+)=
l
H(p)
characteristic equation
+ 2/7 +
H(p) =/>*
5
= (p +
is
+j2Kp +
1
x\t) is
t
The
= 0+
/
=
a>
2.
(4.43)
(4.44)
= Mx( -«"' cos 2/ - 2e"' sin 20 + M£ -e~* sin 2/ + 2c"« cos 2t)
=
(4.45)
+, we obtain the equation
=
«'(<)+)
-
x(t)
we had used
=
+ 2MS
-Afj
Solving Eqs. 4.44 and 4.46 simultaneously,
Thus the final solution is
If
(4.42)
derivative of x(i) is
*'(*)
At
-j2)
= M&-* cos 2t + Mtf-* sin It
*(0+) - 1 - Mt
x(t)
At
1
we have a = -1 and
so that, assuming the form of solution in Eq. 4.37,
Then
(4.41)
conditions
initial
Solution.
(4.40)
<f>)
Solve the equation
x\t)
with the
<f>
- MC* sin (a>t +
x{i)
Example 4.3
(4.39)
M
<r*(cos 2/
we
find
(4.46)
M
x
=
1
and
+ £ sin 2t)
the form of x(t) given in Eq. 4.40,
M, =
+$.
(4.47)
we would have
obtained the
solution
x(t)
-
Vfe-* sin
[2t
+
tan"1 (2)]
Now let us consider a differential equation that
have discussed concerning characteristic values.
Example 4.4
The
*<«(/)
The
initial
differential
+ 9*
(4
>(f)
equation
(4.48)
illustrates everything
is
+ 32x<»(0 + 5Sx'(t) + 56<r'(0 + 24a</) =
conditions are
a!
U)(0+)=0
x'(0+)
= -1
5c
(0+)
=
x'(0+)
=•
(S,
we
l
a<0+)
-
1
(4.49)
Differential equations
The characteristic equation
Solution.
H(p)
81
is
= p* + 9/ + 32/>» + SSp* + 56p + 24 -
(4.50)
which factors into
H(p)
- (p +
1
+yi)0»
+
From H(p) we immediately write x(t)
a<0
- Mi*-
*
cos
/
-;1)0>
1
+ 2)ty +
=
(4.51)
as
+ M^r* sin f + Qc-** +
Since there are five coefficients,
3)
Qte"**
+ Qe-3
*
(4.52)
we need a corresponding number of equations
to evaluate the unknowns. These are
a<0+)
a;'(0+)
x'(0+)
a;«)(0+)
*<«((> +)
- Mx + C + C8 = 1
- -Mx + Mt - 2C - 3Ct + Q - -2M, + 4C - 4Q + 9Ct = -1
= 2Mj, + 2Mt - 8C + 12Q - 27C, =
= -4M1 + 16C + 81C, - 32CX =
Solving these five equations simultaneously,
Mi -
M
t
so that the final solution
C =
-1
(4.53)
1
we obtain
1
Q -i
=0
C.
is
a</)
- $e~' sin /+«-*+ Jte-**
(4.54)
We have seen that the solution of a homogeneous, differential equation
may
take different forms depending
upon
the roots of
its
characteristic
equation. Table 4.1 should be useful in determining the particular form of
solution.
TABLE
4.1
Forms of Solution
Roots of H(p)
1.
2.
3.
Single real root, p
=p
Root of multiplicity, k, (p -*)*
Complex roots at/>, , = a ±jm
e p,t
C„e*i*
+
Cje'i*
X
Me°* sin (mt
4.
Complex roots of multiplicity k
*tP*.6 = " ±j<>>
+
•
M C* cos mt + Mge"*
•
+ CfcV'-W
sin ait
+ +)
M^* cos mt + Mjte"* cos »/ + ••
+ Aft-if*-1**' cos mt + NtfP* sin mt
+ Nite?* sin mt H
+ Nk_1tk- sin mt
V
Network
82
4.3
analysis and synthesis
NONHOMOGENEOUS EQUATIONS
a nonhomogeneous
equation is one in which the forcing function f(t) is not
identically zero for all t. In this section, we will discuss methods for
obtaining the solution x(t) of an equation with constant coefficients
As we mentioned in the introduction to this chapter,
differential
an *<»>(*)
+ a^ x»-«(0 +
•
•
+a
•
=f(t)
x(t)
(4.55)
Let x p (f) be a particular solution for Eq. 4.55, and let x e (t) be the solution
in Eq. 4.55.
of the homogeneous equation obtained by letting f(t)
—
It is readily
seen that
<t)
is
(4-56)
e
also a solution of Eq. 4.55. According to the uniqueness theorem, the
solution x(t) in Eq. 4.56
is
= 0+. 1
mentary function; and x(t)
nonhomogeneous
the unique solution for the
equation if it
In Eq. 4.56, x p(t)
differential
t
= xjf) + x (0
satisfies
is
is
the specified
initial
the particular integral; x a(t)
conditions at
is
the comple-
the total solution.
Since we already know how to find the complementary function x e (t),
we now have to find the particular integral x p(t). In solving for xp(t), a
very reliable rule of thumb is that x p(t) usually takes the same form as the
forcing function if/(0 can be expressed as a sum of exponential functions.
Specifically,
example,
xp{t) assumes the form of /(/) plus all its
= a sin cot, then xjf) takes the form
For
derivatives.
if/(f)
x P(t)
= A sin cot + B cos cot
that must be determined are the coefficients A and B
of the terms in xjf). The method for obtaining x p(t) is appropriately
called the method of undetermined coefficients or unknown coefficients.
The only unknowns
In illustrating the method of unknown coefficients,
let
us take/(0 to be
/(0=<*e"
where a and
/S
A
is
the
then assume x p (t) to have a
is,
xjf)
and
We
are arbitrary constants.
similar form, that
(4.57)
unknown
coefficient.
= Ae"
To
(4.58)
determine A,
we simply
substitute
the assumed solution x p(t) into the differential equation. Thus,
AefiXaJ"
+
a,,.^"-
1
+
•
•
•
+ aj +
a
)
= oe"
(4.59)
•See, for example, C. R. Wylie, Advanced Engineering Mathematics (2nd
McGraw-Hill Book Company, New York, 1960, pp. 83-84.
ed.),
Differential equations
83
We sec that the polynomial within the parentheses is the characteristic
equation H(p) with /»
Consequently, the unknown coefficient is
/?.
obtained as
=
A = -^provided that H(fi) jt
Example 4.5
0.
Determine the solution of the equation
+ 3x'(t) + Ttff) - 4e*
**(r)
with the
initial conditions,
Solution.
The
=
a<0+)
1,
x'(0+)
= — 1.
+ 3/> + 2 = (p + 2Xp +
so that the complementary function
*JLi)
For the forcing function /(f)
1)
is
= cie-* + c*-*
— 4e«, the constants in Eq. 4.60 are
a
= 4,
=> 1.
^ = _i_ = ?
Then
H(l)
Thus we obtain
xp(t)
3
= |e*
total solution is
*(')
To
(4.61)
characteristic equation is
Hip) -/>»
The
(4.60)
- *c(0 + *„(0 = Cxe-« + Cir" + |e«
evaluate the constants
Q and C„ we substitute the given
(4.62)
initial conditions,
namely,
40+) =
*'(0+)
Solving Eq. 4.63,
we find
1
"Q+Cg+f
(4 ' 63)
= -1 - -Ci - 2C, + f
that
Q=
a</)
. _ e-' +
— 1,
C,
= $.
£«-»
Consequently,
+ fe*
(4.64)
C
C
we solve for the constants x and t from the
conditions for the total solution. This is because initial conditions are
not given for xe(t) or *„(/), but for the total solution.
It
should be pointed out that
initial
Next,
let
us consider an example of a constant forcing function/(f)
4.60 if we resort to the artifice
= a.
We may use Eq.
/(r)
=a=
ae*°
(4.65)
Network
84
that
we
is, j8
= 0.
analysis
For the
and synthesis
differential
equation in Example 4.5 with/(/)
*(0
and
= Cltr* + C*r« + 2
(4.67)
the forcing function is a sine or cosine function,
consider the forcing function to be of exponential form and
When
the
= 4,
see that
method of undetermined
fit)
coefficients
and Eq.
4.60.
we can still
make use of
Suppose
= oe* = a(cos <ot + j sin oat)
(4.68)
*
then the particular integral x vlit) can be written as
x Plit)
From the
-
Re
MO + J Im MO
we can show
superposition principle,
=
fit) =
fit)
if
if
o>t
then
a sin tot
then
a.
cos
(4 ' 69)
that
= Re MO
*,(0 = Im * rt(f)
x„(0
mt or a
Consequently, whether the excitation is a cosine function a cos
iat
<xe
driver
exponential
;
an
fit)
use
can
we
or,
sine function a sin
integral.
particular
resulting
the
of
part
imaginary
or
real
the
then we take
=
Find the particular integral for the equation
Example 4.6
x'it)
Solution.
First, let
+ 5*'(0 + 4a<<) = 2 sin 3*
(4.70)
us take the excitation to be
/i(0
= 2e»*
(4.71)
so that the particular integral *«.(/) takes the form
»rt(0
- Ae*
(4.72)
From the characteristic equation
H(p)
we determine
the coefficient
H(j2)
Then
MO
A
=•/>*
to be
_
-5 +jlS
=,
—L^tan-Hs)--]
5 VlO
1^ o^ ^*2
is
+5p +4
*«« -'5VI0
1
1
(4.73)
474>
<
:
Differential equations
and the
particular integral x„(t) for the original driver /(f)
85
= 2 sin 3/ is
- Im xpl(t) = —-= sin [3f + tan"1 (3) 2
xjf)
5V10
*]
(4.75)
There are certain limitations to the applicability of the method of
undetermined coefficients. If /(f) were, for example, a Bessel function
7 (f)> we could not assume xv(t) to be a Bessel function of the same form
(if it is a Bessel function at all). However, we may apply the method to
forcing functions of the following types
= A; constant.
= A(t n + Vi' +
+ V + b );
pt
3. /(f) = e
real
or
complex.
p
!•
/(0
M
2. /(f)
•
•
•
'
n, integer.
;
4.
Any function formed by multiplying terms of type
For the purposes of
linear
network
analysis, the
1, 2,
method
or
3.
more than
is
adequate.
Suppose the forcing function were
/(f)
The
particular integral
xp(t)
where the
4.4
AND
In this section
p = a+jm
can be written as
= (AJ* +
coefficients
STEP
= At*e pt
A& A±_lf
4^ +
.
.
. ,
...
A A
lt
+ A 1t + A )e*
f
(4.76)
are to be determined.
IMPULSE RESPONSE
we
will discuss solutions of differential equations with
step or impulse forcing functions. In physical applications these solutions
are called, respectively, step responses and impulse responses.
quantities, the step
and impulse responses of a
As
physical
linear system are highly
of system performance. In Chapter 7 it will be shown
is given by
its impulse response.
Moreover, a reliable measure of the transient
behavior of the system is given by its step and impulse response. In this
section, we will be concerned with the mathematical problem of solving
for the impulse and step response, given a linear differential equation with
significant measures
that a precise mathematical description of a linear system
initial
conditions at
From Chapter 2,
f
= 0—.
recall that the definition
«(f)
=
=
1
^
f <
f
of the unit step function was
»
Network
86
analysts
.
and synthesis
and the unit impulse was shown to have the properties:
d(t)
=
oo
=
and
r
=
t
tj*0
<3(0 df
=
1
JoJo-
in addition,
we have
the relationship
d(t)
= i*@
dt
As
the definitions of d(t) and u(t) indicate, both functions have dis-
=
continuities ait
0.
In dealing with initial conditions for step and impulse
we must then recognize that the
x"(f), etc., may not be continuous at t
drivers,
solution x{t)
x'(t),
= 0.
and
its
derivatives
In other words,
may be
it
that
*<«>(()-) ji z<«>(0+)
x (»-i)(0_)
x(0-)
In
many
^ a;<"-"(0+)
5*
*(0+)
physical problems, the initial conditions are given at
However, to evaluate the unknown
have the initial conditions at t = 0+. Our
constants of the total solution,
conditions at
t
t
is
= 0—
we must
task, then, is to determine the
= 0+, given the initial conditions at = 0—
discussed here
t
.
The method
borrowed from electromagnetic theory and
is
often
referred to as "integrating through a Green's function."*
Consider the
differential
a n *<»>(t)
equation with an impulse forcing function
+ a*.,. xt«-\t) +-- + a„x(f) = A d\t)
(4.77)
To
insure that the right-hand side of Eq. 4.77 will equal the left-hand
x0) must contain an impulse.
one of the terms z (B, (f), * (n_1) (0>
The question is, "Which term contains the impulse ?" A close examination
shows that the highest derivative term x ln) (t) must contain the impulse,
because if * <n-1) (f) contained the impulse, x (B) (f) would contain a doublet
C d'(t). This argument holds, similarly, for all lower derivative terms of
*-u
(n>
(f) contains the impulse, then s
x(i). If the term 3
(0 would contain a
{n~*
that,
We
conclude
therefore
step and x
for an impulse
(0> a ramp.
at i = 0.
discontinuous
derivative
terms
are
highest
the
two
forcing function,
side,
•
•
•
(
,
'The Green's function is another name for impulse response; see, for example,
Morse and Feshbach, Methods of Theoretical Physics, McGraw-Hill Book Company,
New York,
1952, Chapter 7.
87
Differential equations
For a step forcing function, only the highest derivative term is discontinuous
at
t
=
0.
=
0— , our task is to
t
determine the values a; (B) (0+) and * <n - 1, (0+) for an impulse forcing
function. Referring to Eq. 4.77, let us integrate the equation between
Since initial conditions are usually given at
t
= 0—
and
)xM
an
(t)
dt
Jo-
AP
==
t
0+, namely,
+ a^ f ^'-"(O dt +
Jo-
•
•
•
+
a ]x{t)dt
Jo-
= A f %*)<*'
Jo—
(4-78)
.
After integrating,
aB [*(»-»(0+)
we
obtain
- *<»-i>(0-)] + a„-i[* (B-
a,
- *<-»(0-)] +
(0+)
•
•
•
=A
(4.79)
We know that all derivative terms below (n —
are continuous at
1)
t
= 0.
Consequently, Eq. 4.79 simplifies to
an [x<»-1 >(0+)
xl
so that
- ^—"(O-)] = A
^v(0+) = — + * *-"(0-)
(
We must next determine * (n> (0+).
in Eq. 4.77
At
t
all
(4.81)
= 0+, the differential equation
is
+ a^ *<"-»(0+) + •• + Oo *(0+) =
derivative terms below (« — 1) are continuous, and
an *<">(0+)
Since
(4.80)
(4.82)
•
since
we
have already solved for z(B-1) (0+), we find that
*<">(<)+)
- - -±- [0^x^X0+) +
•
•
•
+
ai x'(0+)
+
*(0+)]
flo
<*n
(4.83)
For a step forcing function A u(t), all derivative terms except x<">(f), are
(B)
continuous at t
0. To determine a;
(0+), we derive in a manner
similar to Eq. 4.83, the expression
=
*<»>(0+)
= - - -±
The process of determining
is
-1>
[a^1 x(B
initial
(0+)
+
conditions
•
•
•
+
ao*(0+)]
when the
an impulse or one of its higher derivatives can be
(4.84)
forcing function
simplified
by the visual
—
Network
88
shown in Eqs.
process
its
analysis
and synthesis
and 4.86. Above each derivative term we draw
Note that we need only go as low
4.85
associated highest-order singularity.
as a step in this visual aid.
+
OnX^Kt)
a„x<«\t)
It
oB_1a: (B 1 »W
+
OaV- ^) + (Wf^O +
1
should be noted that
if a derivative
doublet—it
for example, a
+ a^^-'KO + ••+ ao^O =
<*'(')
+ flo^W = *0
•
(4.85)
(4.86)
term contains a certain singularity
also contains all lower derivative terms.
For
example, in the equation
+
*'(/)
we assume
+
3»'(0
=
2a</)
(4.87)
4<5'(/)
the following forms for the derivative terms at
t
= 0:
= A S'(t) + B 8{f) + C «(0
x'(f) - A 8(t) + B u(t)
x(t) = A u(t)
*"(0
Substituting Eq. 4.88 into Eq. 4.87,
A d'(t) + B (5(0 +
Or
in a
Cu(t)
we
(4.88)
obtain
+ 2A u(f) - 4d'(t)
(4.89)
+ (C + 3B + 2A) u(0 - 4d'(0
(4.90)
+ I A 8(t) +
3B «(r)
more convenient form, we have
A d'(t) +
Equating
(B
+
3A)
like coefficients
<J(0
on both
sides of the Eq. 3.90 gives
A=
=
C + 3B + 2A =
B+
3A
4
(4.91)
89
Differential equations
from which we obtain
B = — 12
C=
and
Therefore, at
28.
t
= 0,
it is
true that
= M(t) - 12(5(0 + 28«(0
x'(0 = 4d(0 - 12«(0
*(*) = 4u(t)
**(*)
The w(0 terms
conditions at
we
t
in Eq. 4.92 gives rise to the discontinuities in the initial
= 0. We are given the initial conditions at t = 0—
evaluate the A, B,
conditions at
example,
t
(4.92)
Once
.
C coefficients in Eq. 4.88, we can obtain the initial
= 0+ by referring to the coefficients of the step terms.
For
if
= -2
*'(0-) - -1
*»(<)-) = 7
*(<)-)
Then from Eq. 4.92 we obtain
x(0+)
=-2 + 4 = 2
= -1 - 12 ex -13
**(()+) = 7 + 28 = 35
x\0+)
(4.93)
The total solution of Eq. 4.87 is obtained as though it were a homogeneous
equation, since d'(t) =
for t j*s 0. The only influence the doublet driver
has
is
to produce discontinuities in the initial conditions at i
evaluated the
initial
conditions at
t
= 0+,
we can
= 0.
Having
obtain the total
solution with ease. Thus,
z(0
From
Eq. 4.93
we
= C <r* +
x
C*tr«
(4.94)
readily obtain
*(0
= (-9er* +
ll<r-«) u(t)
(4.95)
The total solution of a differential equation with a step or impulse forcing
For a step
is obtained in an equally straightforward manner.
function
forcing function, only the highest derivative term has a discontinuity at
t = 0. Since we do not need the initial condition of the highest derivative
term for our solution, we proceed as if we were solving a standard nonhomogeneous equation with a constant forcing function. For an impulse
driver, once we determine the initial conditions at t = 0+, the equation is
solved in the same manner as a homogeneous equation.
90
Network
Example 4.7
analysis
and synthesis
Find the step and impulse response for the equation
+ 4x'(t) +
2x"(f)
where /(f)
=
<5(f)
=
and/(f)
I0x(0 =/(f)
«(f), respectively.
The
initial
conditions at
f
= —
are
=
a<0-)
Solution.
Let us
contains an impulse
a ramp, and
we use Eq.
*'(0
Note that we
=
=
x"(0-)
find the impulse response.
and a step
;
f
=
note that the x" term
Thus x(0+)
0.
x term contains
= x(0— ) = 0. To
4.81
- - + *'(0 -) -
+)
=\
+
\
a2
actually need only x(0
+) and
+) to evaluate the constants
we proceed to the comple-
x'(0
Next,
mentary function x e(t). The characteristic equation
- 2(p* + lp +
- lip +
5)
is
+jlKp +
1
(4.96)
2.
2.
for the second-order differential equation.
H(p)
We
the x' term contains a step ; the
therefore continuous at
is
obtain a;'(0+),
first
*'(0-)
Since H(p) has a pair of complex conjugate roots,
1
-j2)
we use a standard form
= Me~* sin (2f + #
Substituting the initial conditions at f = 0+, we obtain
x c (i)
a<0+)
x'(0+)
from which we find ^
denote here as xt(t), is
=
(4.98)
= J = 2Mcos $ -
M=
**) =
\.
M
(4.99)
sin
<f>
Thus the impulse response, which we
1«~* sin 2f u{i)
Next we must solve for the step response xu{t). For convenience,
complementary function as
The
particular integral
constant /(f)
=
1
is
(4.100)
let
= e-KA^ sin 2f + A t cos 2f)
us write the
(4.101)
evaluated by considering the forcing function a
so that
a,
The
for
=0 = Afsin^
and
xe(t)
(4.97)
jt) « _!_ = J-
(4.102)
total solution is then
a<f)
- (A t sin 2f + At cos 20*?-* + ^
(4.103)
Since x'(t) and x(t) must be continuous for a step forcing function,
a
<0+)=a<0-)=
a;'(0+)-a; , (0-)
=0
(4.104)
Differential equations
Substituting these initial conditions into x(i)
and *'(/), we find that Ax
-0.05 A t
-0. 1 .
Therefore, the step response
*„(*)=
d/dt
=
=
C*
cos 2,)]*)
WO—
L-
*,«>—
(4.105)
>*,(/)
FIG.
Note
4.1.
that the impulse response
step response are related
and the
by the equation
x,
=
(0
i.
*„(<)
at
(4.106)
Wecan demonstrate Eq.4.106by thefollowingprocedure.
xu(t)
> *j(0
C*
L_
-«-*(0.5sin2/+
0.1[1
d/dt
> *»(0
*,(')
is
91
Let us substitute
into the original equation
2
^
Differentiating both sides,
_|
«.(*)]
from which Eq. 4.106
Generalizing,
equation,
we
+ 4 ^ x.(0 +
«.(*)
. h(0
10*„(0
(4.107)
we have
+ 4 1[| *„(o] +
=
io[^ *„(*)]
<H0
(4.108)
follows.
see that, if we have the step response for a differential
we can
obtain the impulse response by differentiating the step
can also obtain the response to a ramp function f(t) —
A p(t) (where A is the height of the step) by integrating the step response.
The relationships discussed here are summarized in Fig. 4.1.
response.
We
INTEGRODIFFERENTIAL EQUATIONS
4.5
In this section,
we
an
will consider
integrodifferential equation of the
form
a„ *"(t)
+
a,.!
*-\t)
+
+
a
x(t)
+
*
a_t f x(r) dr
= /(/)
(4.109)
Jo
where the coefficients {a„, a1K.1
, a_ t } are constants.
In solving an
equation of the form of Eq. 4.109 we use two very similar methods. The
first method is to differentiate both sides of Eq. 4.109 to give
,
a n x (n+1 \t)
+ a^ xM (t) +
The second method
consists
.
.
•
•
.
+
•
a x\t)
+
a_t
of a change of variables.
x(t)
-/'(»)
(4.U0)
We let y'(f) = x(t);
Eq. 4.109 then becomes
a, y(n+1)(t)
+
(n
*„_! y \t)
+
•
•
•
+
a„
y'(t)
+
a_x
y(t)
= f(t)
(4.111)
1
:
Network
92
analysis and synthesis
Note that from Eq. 4.110 we obtain x(t) directly. From Eq. 4.111, we
obtain y(t), which we must then differentiate to obtain x(t). An important
point to keep in mind is that we might have to derive some additional
the
initial conditions in order to have a sufficient number to evaluate
unknown
constants.
Solve the integrodifferential equation
Example 4.8
x'(t)
+
+2f
3a<0
*(t)
dr
=
(4.112)
5«(0
is x(0 — ) = 1.
Since the characteristic equation of Eq. 4.112 is of second degree,
obtain *'(<)+) from the
need an additional initial condition x'(0+).
The
initial
condition
Solution.
we
We
given equation at
t
= +
+ 3»<0+) + 2
Since x(t) is continuous at / = 0,
a< T) dr
*'(0+)
=
(4.113)
5
J
<H
a<T)</T=0
(4.114)
«(0+)-a<0-)-
(4.115)
I
and
=
x'(0 +)
Therefore,
-Method
1.
The complementary function
+ 3*'(0 + 2a<0 =
is
initial
conditions for
= Car* +
a<0+) and
x(t)
2.
(4.1 16)
Letting y'(t)
becomes
We know that
obtain
(4.117)
5<K0
+
the
MO +
y'(P+)
C*r*>
*'(<>+),
- 4e~' -
= x(t),
y"(0
we
then
*.(0
Method
- 3s(0 +) = 2
Differentiating both sides of Eq. 4.112,
x"(t)
Using the
5
(4.118)
we
obtain the total solution
(4.119)
3e-*«
original
2y(t)
=
= «(0+) -
differential
equation then
5«(0
1
(4120)
,„,„,,
(4.121;
j,'(0+)= *'(<>+)= 2
From Eq.
4.120, at t
= 0+, we obtain
y(0 +)
Without going into
-
i[5
- y'(0 +) - 3y'(0 +)] -
details, the total solution
j<0
(4.122)
can be determined as
- -4e- + |e-*« + \
(4- 123 >
93
Differential equations
Differentiating y{t),
we
obtain
<i)
4.6
= y'(f) = 4e~' - 3<?-*'
(4.124)
SIMULTANEOUS DIFFERENTIAL EQUATIONS
Up to this point, we have considered only differential equations with a
single
with
dependent variable
In this section,
x{t).
more than one dependent
variable.
we
will discuss equations
We shall limit our discussion to
equations with two unknowns, x(t) and y(t).
The methods
described here,
however, are applicable to any number of unknowns. Consider
system of homogeneous equations
+ a, x(0 + & y'(t) + y(t) y'(0 + \ y(t) = o
yi *'(0 + Yo *(0 +
«x *'(0
first
the
fi,
(4.125)
<$!
where a„
f} it y it d i are arbitrary constants. The complementary function
obtained by assuming that
so that the characteristic equation
H(p)
=
is
given by the determinant
+
+ Yo>
(a^
oto)
(ftP
is
+ A)
(<V + to)
<fc/>
(4.126)
The
roots of H(j>) are found by setting the determinant equal to zero,
that
is,
- OV + PoXYiP + Vo) =
It is seen that a nontrivial solution of H(p) =
exists only if
(«tf>
+
(«*
««)(V
+
*•)
+ ««KV +
<>a)
* OV + &XftP + y<0
(4.127)
(4.128)
Assuming that the preceding condition holds, we see that H(p) is a seconddegree polynomial tap and can be expressed in factored form as
H(p)
where
= C(p- Po)(p - />0
C is a constant multiplier.
The complementary
= K^e** + K***
y(t) + Kf
x(t)
W
(4.129)
functions are
(4.130)
Network
94
anal/sis and synthesis
and the constants
K K K Kt are determined from initial conditions.
t,
x,
9,
As in the case of a single unknown,
tf Po
If
=
/>i»
if H(p)
has a pair of dduble roots;
i.e.,
then
x(t)
=
(#,
+
y(t)
=
(tf,
+ Kt t)e*<*
K^e***
(4.131)
H(p) has a pair of conjugate roots,
Pi
= a ±jto
Pi*)
then
x(t)
=
M^e"*
sin (cot
+
<^)
y(f)
=
Mte" sin (tor
+
<f>J
(4.132)
Consider the system of equations
Example 4.9.
2x'(t)
+ 4x(t) + y'(t) -y(f)=0
(4.133)
x\t)
with the
initial
+
2x(0
+ y'(0 + y(t) =
conditions
*'(0+)«2
y '(o+)=-3
a<0+)=0
Solution.
The
=
(4.134)
1
characteristic equation is
Evaluating the determinant,
#0)
so that
i/(0+)
2»
+4
»
-
1
p
+2
/»
+
1
we find
that
=/>*
+
5/>
(4.135)
+ 6 = (p + 2Xp + 3)
y{t)
=
a</)
- KtfT* + J>-»
Kjfi-"
+ A^-8
(4.136)
'
(4.137)
With the initial conditions, z'(0+) - 2,a<0+) = 0, we obtain K, =2,Kt = -2.
From the conditions y'(0+) - -3, y(0+) = 1, we obtain Kt = 0, JCg = 1.
Thus the final solutions are
(4.138)
y(.t)
»-s«
Differential equations
Next,
95
us determine the solutions for a set of nonhomogeneous
We use the method of undetermined coefficients
here. Consider first an exponential forcing function given by
the set of
equations
let
differential equations.
+ aye + piy + p = Ne°
Yi*' + y<P + b y' + d& =
We first assume that
xp(t) = Ae"
alX '
'
t
iff
(4.139)
x
yv(t)
(4.140)
= Be9
*
Then Eq. 4.139 becomes
+ xJA + (^0 +
(.YJ + YM + O*i0 +
Qtfl
The determinant
for the set of equations 4.141
H(8)
+
+ Y*
«i0
= A(0) =
where H(6)
is
(4.141)
is
PiO
«<,
YiO
.
=N
d )B =
o)B
W
the characteristic equation with p
=
+ Po
+\
0.
(4.142)
We now determine
#4he undetermined coefficients A and B from A(0) and its cofactors, namely,
A=
*A
1X (0)
A(0)
(4.143)
JVA ia(e)
B=
A(0)
where
A„ is
Example
the i/th cofactor of A(0).
4.10.
Solve the set of equations
2x'
+ Ax + y' - y = 3e**
(4.144)
x'+2x+y'+y=0
= 1, a<0+) = 0, y'(0+) = 0, y(0+) = -l.
The complementary functions *c(f) and ye(i), as well as the characequation H(p), were determined in Example 4.9. Now we must find
given the conditions x'(0+)
Solution.
teristic
A and B in the equations
The
*,(/)
- Ae**
y,(t)
- Be**
characteristic equation withy*
(4.145)
= 4 is
2(4)
H(A)~
(4)
+4
+2
(4)-l
(4)
+
l
42
(4.146)
Network
96
analysis
and synthesis
Then we obtain from Eq. 4.143 the constants
A
A The incomplete
B - -t
solutions are
*(f)
y(t)
= KiT* + Ktfr** + &e»
- K&-* + Kie-* - fe"
(4.147)
f
Substituting for the initial conditions,
we
obtain
finally
s
y(0=K-4«" '-3«4')
(4.148)
Example 4.11.
Solve the system of equations
+ 4x + y' + ly =
x' +x +y' +3y =
lx'
given the
initial
Solution.
First
a<0-)
= *'(<>-) = jKO-) = y'(0-) -
we find
the characteristic equation
=
A(/,)
-
2/>
/>
+4 / +
7
+
3
(4.150)
p +
1
simplifies to give
H(p)
=(pt +2p+5)=(p +
The complementary
and
-'
functions x c (t)
xe(t)
y e (t)
The
(4.149)
56(f)
conditions
J5T(Cp>
which
5«(/)
= ^e
-'
= jBiC
j/
(f)
l
+j2)(p
+
1
-j2)
(4.151)
are then
cos It
+ A&-* sin 2/
cos 2/
+ B^r* sin 2f
(4.152)
particular solutions are obtained for the set of equations with t
>
0,
namely,
+ 4x + y' + ly =
*' +
+ y' + 3y «-
2x'
5
(4.153)
a:
coefficients, we assume that x v and y,, are
Since the forcing function 5 can be regarded
as an exponential term with zero exponent, that is, 5
5e°', we can solve for
a with the use of the characteristic equation H(p) with p = 0. Thus,
x and
Using the method of undetermined
constants:
C
as,
= Q;
yv
= Ca
.
—
C
#(0)
=
A(0)
=
4
7
1
3
=
5
(4.154)
-
Differential equations
—
5AU(0)
n
c
Dd
= 5(3)
^--5(or
97
3
(4.155)
c,«
The
general solution
is
- ^e-' cos 2r + A^r* sin 2/ + 3
= fl^* cos 2/ + B&-* sin 2r - 1
y(t)
In order to find
The
5
1
then
x(t)
y'(0+).
-
A lt At Blt B* we
,
values for
need the values a<0+), x'(0+), y(0+),
first obtained by integrating the
a<0+) and y(0+) are
» 0— and f
original differential equations between t
/•0+
+ 4* + y' + ly)dt -
0+. Thus,
5«(f)<//
Jo-
Jo/•0+
(*'
We know
=»
/VH-
(2*'
impulses at
(4.156)
+ x + y' +
3y*<ft
(4.157)
-
f»H-
5d(t)dt
that only the highest derivative terms in both equations contain
f
- 0. Moreover, both x(t) and y(t)
« 0. Therefore, in the integration
contain, at most, step dis-
continuities at t
r
r
(4a;
After integrating,
+ ly) dt =
(4.158)
(x
+
3y)rff»0
we obtain
2a<0+)+y(0+)=0
(4.159)
*(0+)+y(0+)«5
Solving,
we find
a<0+)
To
find x'(0+)
and y'(0+), we
the original equations at
t
= -5
y(0+)
-
10
(4.160)
substitute the values for
= 0+.
a<0+) and y(0+) into
Thus,
- 20 + j/'(0+) + 70 = 5
x\0+) - 5 + y'(0+) + 30 =
x'(0+) = -20
y'(0+) = -5
2z'(0+)
so that
Substituting these values into Eq. 4.156,
(4.161)
(4.162)
we eventually obtain the final solutions
x(t)
- ( -%e~* cos It -
y(/)
- (1 \e~* cos 2/ + 3e~* sin It -
14e~* sin It
+
3) u(t)
(4.163)
1) «(f)
Network
98
analysis
and synthesis
Problems
Show
4.1
that
and
3^(0
= Mxe~* cos It
xjj)
m,
M&-* sin It
are solutions for the equation
+ 2x'(t) + 540 =
x'(t)
Show
that
x1
+
xt
is
also a solution.
Determine only the form of the solution for the equation
4.2
+ 4x'(0 + 340
+ 8x'(t) + 540
*"(/) - 540
x'(0 + 540
x"(t) + 6x'(0 + 2540
z'(0 + 6x'(.0 + 940
conditions 40+) = 1,
**(0
(a)
x"(t)
(6)
(c)
(d)
(e)
(/)
4.3 Given the initial
solutions for
=
=
=
=
=
=
*'(0+)
=
—1, determine the
+ 2540 =
&r'(0 + 16x(t) =
x\t)
x'(t) + 4.8lx'(t) + 5.7640 =
(a)
+
+
x'(t)
(b)
(c)
4.4
Find only the particular
fo'(0
integrals for the equations
x"(t)
(b)
(c)
(d)
(e)
(/)
4.5 Given the initial conditions
solutions for
(a)
x"(t)
(.b)
x"{t)
(c)
*"(/)
4.6
(a)
initial
(b)
'
determine the
+ 4*'(0 + 340 = 5e~* sin It
+ 6x'(t) + 254/) - 2 cos /
+ 8a;'(0 + 1640 = 2
A system is described by the differential equation
y'(0
The
= e-8
= 2 sin 3f
= e~* sin It
= <r"/2
=6
= 2e-* + 3e~at
40+) = 1, x'(0+) =0,
+ 7x'(t) + 1240
x'(t) + 3x'(t) + 240
*'(0 + 2x'(t) + 540
x"(t) + 2*'(0 + Sx(t)
x'(t) + 5.0x'(0 + 6.25x(t)
**(/) + fo'(0 + 540
(a)
+
3y'(/)
conditions are y(0-)
Given the
differential
=
1;
- a(0
y'(0-) = 2. Find
+
2y(t)
equation
as<*>(0
+
14*'(0
+ 840 -
6<K0
y(0+) and y'(0+).
Differential equations
=
with the intial conditions *(0-)
12, *'(<>-)
*<*>«)+), *'(0+), *'(0+), and a<0+).
4.7 Given the initial conditions x'(0—)
for the equations
= 6,
and x"(0-)
= **(0-) = 0,
+ 2x'(/) + 2a</) = 3<K0
+ 7s'(0 + 12a<r) - 5«(0
condition x(0— ) = —2, solve
(a)
=
99
-7, find
find the solutions
x'{t)
x"{t)
(*)
4.8
Given the
initial
the integrodifferential
equations
+ 4 f a<r) </t =
(«)
x'(t)
+
(b)
x'(t)
+ 2tft) + 2 f x{r) dr = | e-«
(c)
*'(')
+ &<0 + 9 f *(t) oV = 2u(t)
4.9
Given the
with s(0-)
y'(0+).
4.10
set
5a</)
*V) + *(/) + y{t) =
«(/)
=
<K0
*C)+y'(0 + 2y(0
- *'(0-) = y(o-) = y'(0-) = 0,
find *(0+), z'(0+), y(0+),
Solve the set of equations
+ 3*(0 + y'(t) +
*'« + *(') + y'W +
All initial conditions at f = 0— are zero.
6y(t)
6y(0
= */)
= «(')
Solve the set of equations
+ 2a<0 + y'(0 - y(i) =
+ a</) + y'(0 + 2y(r) =
2s'(0
x'(0
The
/
of equations
2*'(0
4.11
2 sin
initial
2<5(0
3e~* uSj)
conditions are
a#)-)
=
l
*'(0-)=-l
y(0-)=0
j,'(0-)=0
and
chapter
Network
analysis:
5
I
INTRODUCTION
5.1
In this chapter we will apply our knowledge of differential equations to
will assume
the analysis of linear, passive, time-invariant networks.
that the reader is already familiar with Kirchhoff's current and voltage
We
and with methods for writing mesh and node equations for a-c or d-c
1
We will, therefore, consider only briefly the problem of writing
circuits.
mesh and node equations when the independent variable is time t. The
problems in this chapter have the following format: Given an excitation
signal from an energy source and the network, a specified response that
is a current or voltage in the network is to be determined. When relating
laws,
these problems to the mathematics in Chapter 4, we shall see that,
physically, the forcing function corresponds to the excitation ; the network
is
described by the differential equation; and the
is
the response.
unknown
variable x(t)
will be twofold. First, we must write the
problems of the network using Kirchhoff's current and voltage
laws. Next, we must solve these equations for a specified current or
voltage in the network. Both problems are equally important. It is
useless, for example, to solve a differential equation which is set up in-
The problems encountered
differential
conditions are incorrectly specified.
The usual type of problem presented in this chapter might generally be
0, which connects an energy
switch is closed at t
described as follows.
correctly, or
whose
initial
=
A
(voltage or current) source to a network (Fig. 5.1).
closing at
t
=
is
the energy source whose output
The analog of a switch
is e{t) u(t).
Before the
1
For a comprehensive treatment, see H. H. Skilling, Electrical Engineering Circuits
(2nd Edition), John Wiley and Sons, New York, 1965.
100
Network
analysis:
I
101
*«0
/\
Energy
source'
FIG.
switch
Network
5.1.
Switching action.
closed, the currents
and voltages in the network have known
are the initial conditions. We must then
determine the values of the currents and voltages just after the switch
closes (at / = 0+) to solve the network equations. If the excitation is not
an impulse function or any of its derivatives, the current and voltage
variables are continuous at t = 0. For an impulse driver the values at
/ = 0+ can be determined from methods given in Chapter 4.
Having
obtained the initial conditions, we then go on to solve the network
is
values. These values at
differential equations.
only for
Since
/
t
= 0-
Unless otherwise stated,
all
the solutions are valid
^ 0+.
we are dealing only with linear circuits,
it is essential that we bear
the all-important principle of superposition. According to the
superposition principle, the current through any element in a linear circuit
in
mind
m
with n voltage and
current sources is equal to the algebraic sum of
currents through the same element resulting from the sources taken one
at a time, the other sources having been suppressed. Consider the linear
network depicted in Fig. 5.2a with n voltage and
FIG. 5 Jo
m
current sources.
102
Network
anal/sis and synthesis
FIG. 5.2b
Suppose we are interested in the current iT(f) through a given element Z,
as shown. Let us open-circuit all the current sources and short-circuit
n — 1 voltage sources, leaving only »/0 shown in Fig. 5.26. By iVj(t),
we denote the current through Z due to the voltage source »/f) alone. In
similar fashion, by iCk(t) we denote the current through Z due to the
current source ik(t) alone, as depicted in Fig. 5.2c.
FIG. 5.2c. Superposition in linear
By
circuits.
the superposition
Network
principle, the total current
algebraic
i
T(t) due to
of the sources
all
is
103
I
equal to the
sum
h = \ + '.,+
5.2
analysis:
+
••
+
I.,
t
H
+
»,,+ •••
+
(5.1)
NETWORK ELEMENTS
In this section,
we
will discuss the voltage-current (v-i) relationships
that exist for the basic network elements.
relationships,
it is
important to assign
Before
we examine
these
arbitrary reference polarities for
first
the voltage across an element, and a reference direction of flow for the
current through the element.
assume that the
For the purposes of our
positive polarity for voltage
arrow, as shown by the resistor in Fig.
5.3.
is
discussion,
at the tail
we
of the current
Now, let us review the voltage-
current relationships for the resistor, the inductor, and the capacitor,
which we
first
discussed in Chapter
1.
Resistor
The
resistor
shown
in Fig. 5.3 defines a linear proportionality relation-
ship between v(t) and
i(t),
namely,
K0 - r
i(0
where
m
(5.2)
= Gt*0
R
R is given in ohms and G in mhos.
Capacitor
For the capacitor shown
in Fig. 5.4a the v-i relationships are
dt
(5.3)
<t)
= -\
i(r)
CJo-
dr
+ »o(0-)
-o
'
m
+
Ut)\
tR
ZZC
**>
v(t)
i(*>\
v(t)
={=C
(a)
FIG.
5.3. Resistor.
FIG. 5.4. (a) Capacitor.
with initial voltage.
(b)
(A)
Capacitor
Network
104
anal/sis and synthesis
J(t)_
<
ml\
v(t)
(J).©-)
5.5. (a) Inductor.
C is given in farads.
v(t)
(b)
(a)
FIG.
°
(6)
Inductor with
initial current.
initial value t> c (0— ) is the voltage across
action. It can be regarded as an
switching
the capacitor just before the
Fig. 5.46. We should point out
shown
in
source,
as
voltage
independent
except impulses and
excitations
for
aU
t>
also that i>c(0— )
c(°+)
where
The
=
derivatives
of impulses.
Inductor
The inductor in Fig. 5.5a describes a dual relationship between voltage
and current when compared to a capacitor. The v-i relationships are
»-'5
(5.4)
m
i
f ir)dr+iL(0-)
LJo-
where L is given in henrys. The initial current iL(0— ) can be regarded as
an independent current source, as shown in Fig. 5.56. As is true for the
voltage across the capacitor, the current through the inductor is similarly
continuous for all t, except in the case of impulse excitations.
When the network elements are interconnected, the resulting i-v equations are integrodifferential equations relating the excitation (voltage or
current sources) to the response (the voltages and currents of the elements).
There are basically two ways to write these network equations. The first
way is to use mesh equations and, the second, node equations.
Mesh equations are based upon Kirchhoff's voltage law. On the mesh
basis, we establish a fictitious set of loop currents with a given reference
direction, and write the equations for the sum of the voltages around the
loops. As the reader might recall from his previous studies, if the number
of branches in the network is B, and if the number of nodes is N, the
+ 1.*
number of independent loop equations for the network is B —
N
1
See Stalling, op.
cit.
Network
anal/sis:
I
105
must, in addition, choose the mesh currents such that at least one mesh
current passes through every element in the network.
We
F.™mpl« 5.1.
In Fig.
a network is given with seven branches and five nodes.
1 = 3 independent mesh equations. The directions
i* /» are chosen as indicated. We also note that the
5.6,
We therefore need 7 - 5 +
of the mesh currents
ilt
Vi(t)
Li
</W-
I
»c,
»Ci"
'1
biwQ
\
fCi'
»»
»R 2
vatQ
0^9-r Qhfi-i
FIG. 5.6
capacitors in the circuit have associated initial voltages. These initial voltages
are assigned reference polarities, as shown in the figure. Now we proceed to
write the
mesh equations.
Meshix
:
«i(0
Mesh
-
c
I
if*
«» dr - — J^t) dr
it :
vCl(0-)
Mesh
- *1 h(0 + —
l
*C,(0-)
- vcfii-) = -
1
C*
^ J^r) dr+Li
—
di.
it :
-vJLO
+ vCt(0 -) =
^P
i/r) dr
+
^r
J\(t) dr+Ki tf
(5.5)
three unknowns, ilf it , and /», we can determine the branch
and the voltages across the elements. For example, if we were required
the branch currents i Cl and iCt through the capacitors, we would use the
After
we find the
currents
to find
following relationships
*<>*-'*-*•
(5.6)
»'c,
-
»i
- '»
Alternatively, if the voltage vj.t) in Fig. 5.6
v*t)
is
our objective, we see that
- tff)4
(5.7)
Network equations can also be written in terms of node equations,
which are based upon Kirchhoff's current law. If the number of nodes in
Network
106
anal/sis
and synthesis
^
Datum
FIG. 5.7
N, the number of independent node equations required
3
We can therefore
for the complete solution of the network is JV — l.
node
voltages will be
the
network.
All
the
datum
node
in
select one
the network
is
datum node.
Consider the network in Fig. 5.7. Let us write a set of node equations
for the network with the datum node shown. Since the number of nodes
1
2 independent node equations.
in the network is
3, we need JV
These are written for nodes vx and vs , as given below.
positive with respect to this
— =
N—
Node
v^.
i
Node
j
B (t)
-
i
L(0~)
= G1 Vl (t) + j(* Vl (r) dr-jl* v t(r) dr
L Jo-
L Jo-
ty.
i(0-)
. --
!
*
v t (r)
L Jo-
dr
+ 7 Jf
L
' V 2(r) dr
o-
+C
^& + G
t
t> 2
(»)
(5.8)
dt
Further examples are given in the following sections.
5.3
INITIAL
AND
In this section,
FINAL CONDITIONS
we consider some methods for obtaining initial conditions
for circuit differential equations.
We also examine ways to obtain particular
networks with constant (d-c) or sinusoidal (a-c) excitations.
In the solution of network differential equations, the complementary
function is called the transient solution or free response. The particular
integral is known as the forced response. In the case of constant or periodic
oo is the steady-state or final
excitations, the forced response at t
integrals for
=
solution.
*
See Stalling, op.
cit.
Network
analysis:
107
I
There are two ways to obtain initial conditions at / = 0+ for a network:
through the differential equations describing the network, (b) through
knowledge of the physical behavior of the R, L, and C elements in the
(a)
network.
Initial
conditions for a capacitor
For a capacitor, the voltage-current relationship
i(r)dr
»c(0+)
If
v
i(t)
d°—
at
t
= 0+
is
+ vc(0-)
(5.9)
Jo-
=
does not contain impulses or derivatives of impulses, »c(0+)
If? « the charge on the capacitor at t
0—, the initial voltage is
=
)•
»o(0+)
-
vc<0-)
=
(5.10)
clude that
is no initial charge on the capacitor,
»o(0+) = 0. We conwhen there is no stored energy on a capacitor, its equivalent
circuit at t
= 0+ is a short circuit.
When
there
This analogy is confirmed by examining
As a result of the conservation of
charge principle, an instantaneous change in voltage across a capacitor
implies instantaneous change in charge, which in turn means infinite
the physical behavior of the capacitor.
current through the capacitor.
Since
we
never encounter infinite current
in physical situations, the voltage across a capacitor cannot change
instantaneously. Therefore at
voltage source
initial
if
an
initial
= 0+, we can replace the capacitor by a
t
charge
exists,
or by a short
circuit if there is
no
charge.
Example 5.2. Consider the R-C network in Fig. 5.8a. The switch is closed at
t = 0, and we assume there is no initial charge in the capacitor. Let us find
the initial conditions i(0 +) and i'(0 +) for the differential equation of the circuit
Vu(.t)=Ri{t)
+
e£',(t)«/t
(5.11)
c
£_
I
i(t)
)
T
>B
V
I
R-C network
R-C network,
'
»«>+))
T
m
(b)
>R
T
Equivalent circuit at <
W
FIG. 5A. (a)
A-
Equivalent circuit at /
- 0+
— 0+.
Network
108
The
and synthesis
analysis
equivalent circuit at
- 0+
t
is
given in Fig. 5.86, from which
we
obtain
«>+>-5
To
obtain /'(u+),
we must
refer to the differential equation
V8(t) -Ri'(t)
At
t
(5.12)
+
i(0
(5.13)
C
» (0+)
-Jt/'(0+) +'
= 0+ we have
(5.14)
c
We then obtain
i'(0+)-
V
-i(0+)
RC
(5.15)
Fig. 5.8a is
final condition, or steady-state solution, for the current in
obtained from our knowledge of d-c circuits. We know that for a d-c excitation,
is
a capacitor is an open circuit for d-c current. Thus the steady-state current
The
i„(0
Initial
= »'(°°) -
conditions for an inductor
For an inductor, the voltage-current
1
f°
relationship at
= 0+
t
is
iKr)dT+iL(0-)
(5.16)
If there is no
iL(0+) — i£(0-).
which corresponds to an open circuit at
_ o+. This analogy can also be obtained from the fact that the current
t
through an inductor cannot change instantaneously due to the conserva-
If v(t) does not contain impulses, then
initial current,
L(0+)
i
= 0,
tion of flux linkages.
S3 In Fig. 5.9a, the switch closes at t = 0. Let us find the
conditions i(0+) and i'(0+) for the differential equation
Example
L^+JI*)
L
o.c.
-o
t
9
(a)
FIG.
5.9. (a)
R-L network.
s=:
o-
i(o+)J
W
(*) Equivalent circuit at / *=
0+.
initial
Network
From
the equivalent circuit at
= 0+, shown in Fig.
f
5.9b,
analysis:
we
i(0+)=0
+)
/'(0
109
see that
(5.18)
We then refer to the differential equation to obtain <"(0+).
V-L i"(0 +) + R i(0 +)
Thus
I
(5.19)
V R
- - - - /(0 +)
(5.20)
as
The
—V
L
steady-state solution for the circuit in Fig. 5.9a is obtained through the
knowledge that for a d-c source, an indicator
/„(')
=/(«>)
is
a short
circuit.
=•£
(5.21)
Example 5.4. In this example we consider the two-loop network of Fig. 5.10.
As in Examples 5.2 and 5.3, we use the equivalent circuit models at t — +
and t = oo to obtain the initial conditions and steady-state solutions. At
f =
the switch closes. The equivalent circuits at t = 0+ and / =• <x> are
shown in Fig. 5.1 la and b, respectively. The initial currents are
*i(0+)
ig(0+)
The
=
V
^~
*i
(5.22)
=
steady-state solutions are
«») - j^r^ = «")
<5 - 23 >
Final conditions for sinusoidal excitations
When
a pure sinusoid, the steady-state currents and
same frequency as the
excitation. If the unknown is a voltage, for example, 1^(0. the steady-state
the excitation
is
voltages in the circuit are also sinusoids of the
solution
would take the form
»i,(0
m
-
I
KOo)l
sin
K< - <K<*J\
(5.24)
and \V(jcoJ\ and ^(<u )
and phase of vl9(t). A similar expression would
hold if the unknown were a current.
To obtain the magnitude and phase, we follow standard procedures in
a-c circuit analysis. For example, consider the R-C circuit in Fig. 5.12.
The current generator is
( 5 -25)
W) - (Jo sJ n «*) "(0
where
is
the frequency of the excitation,
represent the magnitude
1
Network
10
analysis
and synthesis
*2
VWV
j
+
h(0
|
-VW\r
^=:C
)
3
kM
)
FIG. 5.10
«2
JVW\r
V!=
—
ww*
»2(0+)
»i(0+)J
Equivalent circuit at
t
I
O.C.
= 0+
WW-
-WWo
o.c.
g
Equivalent circuit at t
FIG.
5.1
1,
(a) Equivalent circuit at t
= 0+.
itfO
Y
FIG. 5.12
(6)
=
«>
Equivalent circuit at
/
=
oo.
'
Network
form shown
If the steady-state voltage takes the
analysis:
I
III
in Eq. 5.24,
IHM>)I
|rOo)l
(5.26)
h
(G a
#o>
and
so that
We
5.4
vjLO
refer to this
=
^w
tan"
co
1
2/~*\W
*C*j
^
J
(G>+
problem in Section
(5.27)
G
"
sin (°>o<
tan"
1
^)
(5.28)
5.5.
AND IMPULSE RESPONSE
STEP
As an
-
)
+
introduction to the topic of solution of network differential
equations, let us consider the important problem of obtaining the step and
impulse responses for any voltage or
As we shall see in
current in the network.
Chapter
the step and impulse responses
7,
are precise time-domain characterizations
of the network. The problem of obtaining
the step and impulse response
follows
energy,
:
is
stated as
Given a network with zero
we
initial
are required to solve for a
specified response (current or voltage)
F,e * 5 * 3
due
to a given excitation function u(t) or <3(0,
which either can be a current or a voltage source.
|
If the excitation
is
—
a step of voltage, the physical analogy is that of a switch closing at
time t
which connects a 1-v battery to a circuit. The physical
analogy of an impulse excitation is that of a very short (compared to the
time constants of the circuit) pulse with large amplitude.
The problems involved can best be illustrated by means of examples.
Consider the series R-C circuit in Fig. 5.13. The differential equation of
=
the circuit
—
is
KO =
We assume Vq(0—) =
x'(t) for j'(0 in
0.
*«)
= R 1(0 + —
l(r) dr
7
J»~
CJi>C
I
ii
Since Eq. 5.29 contains an integral,
(5.29)
we substitute
the equation, giving us
d(t)
= Rx'(t) + ±z(t)
(5.30)
1
Network
12
analysis and synthesis
Integrating both sides between
0—
and
0+
gives
*>+>-The
(5.31)
characteristic equation is
H(p)
and with
little
effort
- Rp +
(5.32)
£
we have
_ L ,-t/RC n(0
*(0-
m-
so that
R^'RC
6"'*
(5.33)
"®]
(5 ' 34)
We
thus arrive at the current impulse
in Fig. 5.14a.
as the result of an impulse voltage excitation. In the process
we have also obtained the step response x(i) in Fig. 5. 146 since, by definition,
the derivative of the step response is the impulse response. The reader
which
is
response
shown
i(t)
should check this result using a step excitation of voltage.
In the second example consider the parallel R-C circuit in Fig. 5.12
7 «(/), where /„ is a constant.
driven by a step current source i(f )
=
m
'
£aro
R*C
(*)
FIG. 5.14. (a) Impulse response of
R-C circuit.
(A)
Step response of R-C circuit.
Network
analysis:
I
113
»M
FIG. 5.15. (a) Step response of parallel
R-C circuit.
(A)
Impulse response of parallel
Jt-C circuit.
Assuming zero
initial
conditions, the differential equation
Jo „(0
from which we obtain the
=
Gt<0
steady-state value
C^p
(5.35)
characteristic equation
H(p)
The
+
is
of KO
=
Cp
+G
(5.36)
is
(5.37)
H(0)
G
m
/o
FIG. 5.1*. Pulse excitation.
1
Network
14
analysis
and synthesis
m
FIG.
5.17.
Thus the complete solution
r(0
From the
initial
to pulse excitation.
for the voltage step response
is
= (Ke-i/BC + 7oR) u(t)
(5.38)
= 0, we obtain K =
IRC u(t)
t>(0 = JoR(l - e-*
)
condition »(0+)
—I^R
so that
(5.39)
Differentiating Eq. 5.39 gives us the voltage impulse response
f(0
The step and impulse responses
= l* e -tlRC «(0
are plotted
(5.40)
on
Figs. 5.15a
-
T)]
and
b.
Suppose
the excitation in Fig. 5.12 were a pulse
KQ = IMt) shown
in Fig. 5.16.
u(t
Then, according to the superposition and time-
invariance postulates of linear systems, the response
K0 =
which
5.5
is
V*[(l
shown
(5.41)
-
e-" BO) u(t)
- (1 -
e-
(i
would be
- T)IBC
)u(t
-
T)]
(5.42)
in Fig. 5.17.
SOLUTION OF NETWORK EQUATIONS
we
our knowledge of differential equations
There are two important points in
network analysis: the writing of network equations and the solution of
these same equations. Network equations can be written on a mesh, node,
or mixed basis. The choice between mesh and node equations depends
largely upon the unknown quantities for which we must solve. For
In
this section
will apply
to the analysis of linear networks.
Network
instance, if the
mesh
write
unknown
quantity
On
equations.
across a certain element, then
the choice
quite arbitrary.
is
a branch
is
the other hand,
if
current,
for example,
it is
we wish
node equations are
If,
analysis:
I
115
preferable to
to find a voltage
In many cases
to find the voltage v
better.
we wish
across a resistor R, we can either find v directly by node equations or find
the branch current through the resistor and then multiply by R.
Example
network
source
in Fig.
is e(t)
when
5.18,
the voltage
= 2e-°-« ufjt) and »c(0-) = 0.
The
Solution.
*(/)
Find the current /(f) for the
5.5.
differential
= Rl(t) + ^ j
equation
i(r)dr
is
+ v^O-)
=fcC-if
(5.43)
Or, in terms of the numerical values,
we
have
2«~«" u{t)
= i(0 + 2
»(t)
dr
(5.44)
Differentiating both sides of Eq. 5.44,
2(5(0
To
obtain the
the limits
/
initial
= 0—
and
we
obtain
dm + 2i(t)
- e-° u u(t) =
dt
condition i(0+),
t
FIG. 5.18
= 0+
we must
to give «(0+)
(5.45)
integrate Eq. 5.45 between
= 2. From
the characteristic
equation
H(p)=p+2*°0
we obtain
/<#)
If
(5.46)
the complementary function as
we assume the
particular integral to be iv(t)
A = The incomplete
= JCr«
solution
_1
//(-0.5)
(5.47)
= Ae~°
bt
,
then
we
obtain
2
(5.48)
3
is
= Ke~* - §<r° •"
From the initial condition i(0+) = 2, we obtain the final solution,
Kt) = (f«-* - f*-°-") «(0
»(0
(5.49)
(5.50)
As we already noted in Section 5.3, in the solution of network differential
is called the free response, whereas
a particular integral, and in the case of constant or
oo is the steady-state
periodic excitation, the forced response at /
solution. Note that the free response is a function of the network elements
equations, the complementary function
the forced response
is
=
1
Network
16
alone and
is
analysis
and synthesis
On
independent of excitation.
the other hand, the forced
response depends on both the network and the excitation.
positive elements,
It is significant that, for networks which have only
the free response
is
made up of only damped exponential and/or
sinusoids
characwith constant peak amplitudes. In other words, the roots of the
example,
For
parts.
real
zero
or
negative
have
all
teristic equation H(p)
0.
a
a ±j(o, then Re (pj
is a root of H(p) written as
x
if
= ^
=
p
/»!
fact is intuitively reasonable because, if a
This
a response that
bounded excitation produces
exponentially increasing, then conservation of energy is
passive
not preserved. This is one of the most important properties of a
parts
real
whose
characteristic equation contains only roots
is
network. If a
network
are zero or negative, and if theyo> axis roots are simple, then the
unstable*
is
network
the
which it describes is said to be stable; otherwise,
will be discussed
Stability is an important property of passive networks and
in greater detail later.
Example
For the R-C network
5.6,
Eq. 5.25, find the voltage
p^O-) -
v(t)
in Fig. 5.19 with the excitation given
across the capacitor;
it is
given that v(0-)
by
-
0.
have already obtained the particular integral in Eq. 5.28.
on a
us find the complementary function. The differential equation
Solution.
Now
let
node
basis is
We
C -r + Gv = I
(5.51)
sin mtiKt)
at
from which we obtain the characteristic equation as
H(p)
-Cp+G
(5.52)
vJfl-KT* *®
so that
and the incomplete
solution
(5-53)
is
- to-™*) + . °o)>c w sin {<* - tan"* J «(/)
(G +
-g
^
obtain
—
—
we
t<0
)
0,
condition
From the initial
*/)
r
/
a>C\
.(o^-KO^-Jf-^r^fe^^l^'V)From the argand diagram in
Fig. 5.20
we see that,
Consequently,
This is not a formal definition of stability,
but
it
(5.54)
suffices for the
moment.
(555)
Network
analysis:
117
I
lm
A
A
«—o—
«C
1
Be
MG.SM
From the complementary function
the circuit is T - C\G - RC.
in Eq. 5.53,
we see that the time constant of
kinds of free responses
Next, let us examine an example of the different
relative values of the
on
depend
that
equation
of a second-order network
in Fig. 5.21 ; let us
network
the
given
are
we
network elements. Suppose
equation
differential
the
for
v^t)
find the free response
(5.58)
'
Differentiating
both sides of Eq.
VJf)
The
LJ*-
dt
S.S8,
we have
- C p"(t) + G »'(0 + £ <0.
(5.59)
characteristic equation is then
H(p)
- Cp» + Gp + 1 =
In factored form, H(p)
c(j>»
+ 1 P + j^)
(5.60)
is
#(p)
= G(p-J>i)0>-/>i)
(5.61)
»£©-)
nasai
118
Network
and synthesis
analysis
v(t)
K
1
+ K2
t
FIG. 5.22. Overdamped response.
A±B
Pi
where
2C
P*
LCI
2l\c)
(5.62)
There are three different kinds of responses depending upon whether
real, zero,
case
B is
or imaginary.
Bis
1.
real, that is,
a,>±
lc
(§1
then the free response
is
vdt)
= Kie- U-B)t + K e- u+BU
(5.63)
%
which is a sum of damped exponentials. In this case, the response is said
to be overdamped. An example of an overdamped response is shown in
Fig. 5.22.
case
2.
5=
0, that is,
(-T-\Cl
LC
= p = —A
At
vc(t) = (^ + K t)e-
then
p1
so that
When B
g
(5.64)
t
=
0,
the response
is critically
damped, as shown in Fig.
v(t)
o
FIG. 5.23. Critically damped response.
t
5.23.
Network
analysis:
I
119
FIG. 5.24. Underdamped response.
case
3.
Letting
B is imaginary,
this case, the
damped
is,
®
B = jfl, we have
»c(0
In
that
=
response
e
is
LC
(*i sin
said to
fit
+ Kt cos
(5.65)
fit)
be underdamped, and
is
shown by the
oscillatory curve in Fig. 5.24.
Example
5.7.
In this example we discuss the solution of a set of simultaneous
network equations. As in the previous examples, we rely upon physical reasoning
rather than formal mathematical operations to obtain the initial currents and
voltages as well as the steady-state solutions. In the network of Fig. S.25, the
switch S is thrown from position 1 to position 2 at t = 0. It is known that
prior to t = 0, the circuit had been in steady state. We make the idealized
assumption that the switch closes instantaneously at / = 0. Our task is to
find «i(0
and
/,(/)
The values of the batteries
and the element values are given as
after the switch position changes.
Vl and V% are Vx = 2 v, Vt =
3 v;
L=
C=
The mesh equations
for
ix{t)
Rt = 0.5 n
= 2.o n
lh,
jig
if,
and
it(i)
after
t
=
are
Vt =Li\{t)+Rl i1{i)-R1 i£)
-vc(0-)= -/Mito +
(5.66)
l
cL ^T)dr+{R
k
l
+R^i^t)
(5.67)
Network
120
analysis
and synthesis
t.o^o—nnnr^
?2
-
v2 ^:
hit))
FIG. 5.25
Since Eq. 5.67 contains an integral,
we differentiate it to
give
= -Rt i\(0 + ^ i#) + (*i + **) i'#)
(5.68)
Using Eqs. 5.66 and 5.68 as our system of equations, we obtain the characteristic
equation
Lp +'*!
-Ri
H(p)-
-+(R1+ RJp
-RiP
<569>
- URx + *lp* + (| + *i*l)p + §
Substituting the
dement values
H(p)
The
into H(p),
- 2.5p* + 4p +
1.5
we have
- 2.5(p + lXp+ 0.6)
(5.70)
free responses are then
(5.71)
The
steady-state solutions for the
mesh
currents are obtained at
considering the circuit from a d-c viewpoint. The inductor
circuit and the capacitor is an open circuit; thus we have
is
/
»
oo
by
then a short
(5.72)
Now let us determine the initial currents and voltages, which, incidentally, have
sources are not
the same values at / - 0- and t - 0+ because the voltage
Network
impulses. Before the switch is thrown at
analysis:
I
121
= 0, the circuit with Kx as the voltage
f
source was at steady state. Consequently,
(0-)-K1 «2v
ro
«0-)-^-4amp
(5.73)
«0-)-0
We next find i\(0+) from Eq. 5.66 at / - 0+.
Vt ~Li\{0+) + 4*i<P+) - RiW+)
Substituting numerical values into Eq. 5.74,
«'
From Eq.
5.68 at / <-
1
amp/sec
*!+*. /'i(0+)
values of
initial
-
we find
0+, we obtain similarly
*'«>+)
With these
1(0+)
(5.74)
it(t)
and
i/t),
- 0.2 amp/sec
we can
(5.75)
quickly arrive at the final
solutions
/i(r)
i/f)
which are plotted in Fig.
- (0.5*-* - 2.5e-° « + 6) i<0
- ( -0.5tf-« + 0.5*-°«) n(f)
5.26.
6.0
5.5
h(t)
—
5j0
45 4.0,
o.io'
0.08
-
\«w
0.06
0.04
O02
r
(>
i
i
1
1
1
0.5
1.0
IS
2.0
1
2.5
FIG. 5.26
3.0
33
1
1
40 43
1
1
M
Network
122
and synthesis
analysis
ANALYSIS OF TRANSFORMERS
5.6
According to Faraday's law of induction, a current ix flowing in a coil L x
induce a current it in a closed loop containing a second coil Lt The
may
.
for
inducing the
sufficient
conditions
current
are: (a) part of the flux
it
Ox
in
Lx must be coupled magnetically
coil L a (b) the flux O x must be
the coil
to the
;
changing with time.
In
this section
we
containing a device
will analyze circuits
made up of two
known as a
magnetically coupled coils
transformer. In Fig. 5.27, the schematic
of a transformer is given. The Lt side
of the transformer is usually referred to
as the primary coil and the L t side as the secondary coil. The only distinction
between primary and secondary is that the energy source is generally at
the primary side.
The transformer in Fig. 5.27 is described mathematically by the
FIG. 5.27. Transformer.
equations
VO-L^ + M^
at
at
(5.77)
, x
,
,
du
di t
.
,
at
at
M
is the mutual inductance associated with the
where
and is related to Lx and L2 by the relationship
flux linking
Lt to Lt
M = k4IT
x
The constant
bounded by the
K in
limits
Eq. 5.78
(5.78)
z
called the coefficient of coupling.
1. If \K\
1, then all of the flux
is
=
<, \K\ <,
,
It is
Ox in
Lj is linked magnetically to L % In this case, the transformer is a unitycoupled transformer. If
0, the coils L x and L s may be regarded as
two separate coils having no effect upon one another.
coil
.
K=
For
circuits
with transformers,
for the mutually induced voltages
we must
establish reference polarities
dijdt.
Usually, the references are
M
on the input and output leads of a transformer,
shown by the dots on the schematic in Fig. 5.28. The reference dots are
given by small dots painted
as
placed at the time of manufacture according to the procedure outlined
here.
voltage source v is connected to the primary Lx side of the
voltmeter is attached on the secondtransformer, as shown in Fig. 5.28.
terminal
is assigned the dot reference to
the
side,
ary. At the primary
A
A
Network
analysis:
I
123
Voltmeter
ttransformeri
An experiment
FIG. 5.28.
to determine dot references.
which we connect the positive lead of the voltage source. The dot reference
is placed on the secondary terminal at which the
voltmeter indicates a
positive voltage. In terms of the primary current i the positive
voltage at
lt
the secondary dot is due to the current i flowing into the dot
on the
t
primary side. Since the positive voltage at the secondary dot corresponds
to the current
we can think of the dot references
both currents are flowing into the dots or away
from the dots, then the sign of the mutual voltage term Mdijdt is positive.
When one current flows into a dot and the other away from the second dot,
the sign of Mdijdt is negative.
/,
flowing into that dot,
in the following way.
If
N
t
and
N
linkages of Ly
and
it
t
If
number of turns of coils L^ and L% then the flux
and iV.O,, respectively. If both ix
dots, the sum of flux linkages of the transformer is
are the
and L, are given by
flow into the
,
N&
2 $ linkages = N&i + N^
however, one of the currents, for example,
other ig flows out of the other dot, then
If,
ilt
(5.79)
flows into a dot, and the
2 $ linkages = N& - N Q
t
t
(5.8O)
An
important rule governing the behavior of a transformer is that the
sum of flux linkages is continuous with time.
The differential equations for the transformer in Fig. 5.29 are
V u(t) = Lt i\(t) + Rt i&) +
Q
=
M
i\(t)
+
R* i8(0
FIG. 5.29
M
i'tf)
+ L, I'M
(581)
Network
124
analysis
and synthesis
Integrating this set of equations between
/
= 0—
and
t
= 0+
results in
the determinant
- /i(o-)] MM>+) - «o-)] _
"Q
M\ii(f>+) - ix(0-)] LJMP+) - '.(0-)]
Za&(0+)
By
(I^L,
If
t
-
-
M»)[i1(0+)
K<
obtain
*1 (0-)][i,(0+)
-
/,(0-)]
>
M*, that
is,
fl
(5.83)
1,
(0+)
»*(0+)
and
K<\
*<1
- /,((>-),
-
ig(0-),
For the transformer circuit in Fig. 5.29,
Rt -= 3Q, R, 80, and Jlf - 1A. The excitation is V - //0-) - 0.
f (/) and i#), assuming that /x(0-)
ir««ph»
5.8.
-
x
Solution.
The
differential
6«(0
The
-
then the currents must be continuous at
equal to zero. Thus
in order for the determinant in Eq. 5.82 to be
L^Lx
-
we
evaluating this determinant,
(5.82)
(5.84)
(5.85)
Lt 6ii(0-
1A,
L»
-
2h,
Let us find
equations for the circuit are
- i'i(0 + 3i\(0 + /',(<)
- iVO + 2i'&) + 8i,(0
characteristic equation is given
(5.86)
by the determinant
Wj
+8
= 0> + 3)(2p + 8)-/>»
= (^+14/» + 24)
-
(/>
2^
+ 2)(p +
(5.87)
12)
Thus the complementary functions are
= Kx<r« + A^e-"'
= K*r« + JV"m
Ie (0
ile{t)
(5.88)
i
obtain the particular integral, or steady-state solution,
physical reasoning to arrive at
To
V_
hJ®—t
2
Oamp
we
rely
upon
amp
(5.89)
Network
analysis:
I
125
—
Since the excitation does not contain an impulse (and since I^L,
jt 0, as we shall see later), we can assume that
d(0+) and ^(0+) are
Af1
also zero.
Then using Eq.
5.86
we can find i\(0+) and i'J0+).
+ f,(0+)
o -r»<p+) + 2i',(o+)
1(0+) - 12 and i'£0+) - 6.
6«'"i(0+)
Solving,
ditions,
(5.90)
'
we find i'
With
we obtain the final solutions of ix(t) and %(/)
these initial con-
(5 ' 91)
«0 - (-Ir* + *e-«0 «(0
Suppose, now, I^L, - M», that
K - then i,(0 and £(/) need not be
continuous at — 0. In fact, we will* show that the currents are discontinuous at >
for a unity-coupled transformer. Assuming that K = 1,
is,
1,
f
r
consider the
mesh equation
*t 'i(0+)
for the secondary at i
= -M i'^O-H) - L, .',(0+)
(5.92)
--?[!* i'!(0+) + M i',(0+)]
(5.93)
The mesh equation of the primary
V - Rt
it
(0+)
+
[L,.
= 0+
i\(0+)
side then
becomes
+ M iV0+)) = Rr h(Q+) -^R, iJ0+)
M
(5.94)
We
need an additional equation to solve for ^(0+) and /g(0+). This
provided by the equation
£iPi(0+)
-
it
+ MM>+) - i^O-)] =
5.82. Since x (0-) = i^0~) = 0,
(0-)}
which we obtained from Eq.
i
is
(5.95)
we
solve
Eqs. 5.94 and 5.95 directly to give
^3
1^(0+)
R\JL%
wo+)
=™RL
1
Consider the following example.
the element values are
£i«=4h,
At-8 0,
M = 2h,
—
+ R^Li
t
(5.96)
+ R^
For the transformer
L,= lh
R9 =*3Q
K=10v
in Fig. 5.29
Network
126
anal/sis
and synthesis
Assuming that the circuit is
t
= 0, let us find h(t) and
at steady state before the switch
The
/2 (0-
is
differential equations written
closed at
on mesh
basis are
10-4^
+ 8^ + 2^
dt
dt
(5.97)
dt
The
characteristic equation
dt
is
4(p
+
2p
2)
H{p)
2p-
- 20p +
H(p)
which yields
(p
24
=
+
(5.98)
3)
= 20(p + |)
(5.99)
so that the complementary functions are
(5.100)
The
we
particular integrals that
t
(r)
obtain by inspection are
= ±1 = 12 =
5
8
(5.101)
hit)
The
initial
conditions are
*i(0+)
i*(0+)
We then find Kx =
VL,
=
RtLi
+ R^
-0.75 and
20
(5.102)
-VM
-
-!2-a5
=
K = - 1.0 so that
t
= (-O.lSe-11 +
W) = -er^u{t)
*
ijff)
We
see that as
state value
of
t
approaches
1.25.
-1.0
infinity
it(t)
1.25) u(t)
-» 0, while
(5.103)
it (t)
goes
its
steady-
Network
analysis:
127
I
Problems
5.1
Write the mesh equations for the network shown.
—nnnp—
I
— —nnnp—
»Cl <0->
Pi
*^\{=
»'l(0-)
»2<0-)
i
*~-,
Ci
*1«
Q«Wb(«
»!
«te1
(0-)=J;C2
PROB.
5.2
Write a
set
D
J8 2
5.1
of node equations to solve for the voltage vjt) shown in the
figure.
W
ii
wm{(T)
SJlo
Ci
*2>
= = G>
»2(0
»c <0-)
PROB.
5.2
53 The network shown has reached steady state before the switch S is
opened at t = 0. Determine the initial conditions for the currents
1-fy) and
4(0 and their derivatives.
R
o^f o
vW-
—
^WV
»2WJ
±C
1
PROB. 5.3
5.4
The network shown has reached steady
from a to
b.
derivatives.
state before the switch
moves
Determine the initial conditions for iL(t) and vdt) and their
Determine also the final values for iL(t) and vc(t).
first
—
128
Network
analysis and synthesis
1
10 v
T
S
lh_
vc (t)
o—
^
*t
5v-
==
If
:io
T
PROB.
5.4
5.5 The network shown has reached steady state before the switch moves
from a to b. Determine the initial conditions for the voltages vx(,t) and vt(t)
and their first derivatives. Determine also the final values for v^t) and vjj).
b
t«0
vi(t)
-
J:-
-
s
T
a
+
*i
=C
r
R2 *
v 2 (t)
i
PROB.
5.6
For the network shown ii{0— )
0- < / < ».
»(/);
(a)
Find
(A)
Show
(c)
5.5
= 0.
that v(t) approaches an impulse as
Find the strength (area) of the impulse.
G
-» 0.
ItfOQ)
PROB. 5.6
5.7 For the network shown,
and sketch Jbto; 0- < r < «.
»(r)
-
d(t)
- tr*u(t)
and t<0-)
= 4.
Find
—
Network
10
I
v/W-
analysis:
I
129
o—
—m
\k*t)
•1Q
W(J)
PROB.
5.8
5.7
For the network shown, before the switch moves from a to
Find the current /(f).
b, steady-
state conditions prevailed.
6<?
PROB.
5.9
For the
been in steady
5.8
shown, switch S is opened
Find /(f); 0— <f < ».
circuit
state.
at
f
=
after the circuit
had
>Tnnp
/^
PROB.
U\
f
5.9
5.10 Find the current /(f) in the network shown when the voltage source is a
unit impulse. Discuss the three different kinds of impulse response waveforms
possible depending upon the relative values of R, L, and C. All initial conditions
are zero at/ —0—.
1
130
Network
anal/sis
and synthesis
—
i
L
nnRp
Qm
*>J
PROB.
5.10
5.11 An R-C differentiator circuit is shown in the figure. Find the requirements for the R-C time constant, such that the output voltage v^t) is approxi-
mately the derivative of the input voltage.
uoW
PROB. 5.
1
5.12 An R-C integrator circuit is shown in the figure. Find the requirements
for the time constant such that the output t> (f) is approximately the integral of
the input voltage.
S
A/V
O"
ci
"«)
PROB.
5.13 Find the free response for
source; (b) a voltage source.
i(f)
vo(t)
5.12
in the figure
shown
for (a)
a current
Network
analysis:
131
I
Source
PROB. 5.13
At t = 0, the switch goes from position 1 to 2. Find /(/), given
= e~* sin It. Assume the circuit had been in steady state for t < 0.
5.14
e(t)
PROB.
5.15
initial
that
5.14
For the circuit shown, switch S closes at t = 0. Solve for l(t) when the
conditions arc zero and the voltage source is e(t) = sin (cat + 0). What
should be in terms of R, C, and
term is zero?
m
so that the coefficient of the free response
R
-VNAr—
I"
^)
PROB.
5.15
5.16 For the circuit shown, the switch S moves from a to b
and sketch v^t) for - < / < «. The circuit is in a steady
at
f
= 0.
state at
Find
f
= 0.
Network
132
analysis
l
and synthesis
MA—
+
°
->
t
Xs
1
PROB.
5.17
For the
circuit
shown,
i(r)
=
5.16
- 4«r« «(0-
Find
i<f);
- </<
°°.
g
»w
JQ^
lf=t=
PROB.
ih
5.17
in Prob. 5.4.
5.18 Find the complete solution for i L(t) and vdO
Find
t - 0.
5.19 For the circuit shown, the switch is closed at
energy.
initial
zero
oo.
Assume
<t <
ijj) for
10Q
irff)
ana
lh
10Q
PROB. 5.19
For the transformer circuit shown, the switch
and i2(f). Assume zero initial energy.
S
5.20
ijft)
5i>.
Af=lh
PROB. 5.20
closes at
t •= 0.
Find
Network
\
Q>
30
PROB.
lh<3 1hg>lh
analysis:
133
60^02(0
5.21
switch S opens at
5.21 For the transformer circuit shown, the
°°. Assume zero initial energy.
/
the voltage far) for
- < <
I
t
= 0.
Find
chapter
The
6
Laplace transform
THE PHILOSOPHY OF TRANSFORM METHODS
6.1
In Chapters 4 and
ential equations.
discussed classical methods for solving differsolutions were obtained directly in the time domain
5,
The
we
of solving the differential equation, we deal with
time
at
every step. In this chapter, we will use Laplace
functions of
transforms to transform the differential equation to the frequency domain,
where the independent variable is complex frequency s. It will be shown
that differentiation and integration in the time domain are transformed
into algebraic operations. Thus, the solution is obtained by simple
since, in the process
algebraic operations in the frequency domain.
•
a striking analogy between the use of transform methods to
solve differential equations and the use of logarithms for arithmetic
operations. Suppose we are given two real numbers a and b. Let us find
There
is
the product
c=axb
(6.1)
by means of logarithms. Since the logarithm of a product is the sum of the
logarithms of the individual terms,
log
so that
If a
and b were two
we have
C = log a X b = log a + log b
C = logr1 (log a + log b)
six-digit
(6.2)
(6.3)
numbers, the use of logarithms would probably
facilitate the calculations, because logarithms transform multiplication
into addition.
is the use of transform methods to solve integroConsider the linear differential equation
An analogous process
differential equations.
yW))
-/(»)
134
(6.4)
The Laplace transform
135
Algebraic
Differential
equation^
equation
Transform
FIG.
6.1.
X(s)
^
x(t)
Solve
Invert
Philosophy of transform methods.
the forcing function, x(t) is the unknown, and y(a<0) is the
differential equation. Let us denote the transformation process by 7X-),
where /(0
and
Eq.
is
be the frequency variable.
we have
When we
let s
6.4,
= W)]
(6.5)
domain functions are given by
capital letters, let us write
7W*(0)1
Since frequency
Eq-6.5as
Y(X(s),s)
where X(s)
=
transform both sides of
TftOL
*W -
= F(s)
T[f(t)\,
and
(6.6)
Y(X(s), s)
is
an algebraic
equation in s. The essence of the transformation process is that differential
equations in time are changed into algebraic equations in frequency.
We
can then solve Eq. 6.6 algebraically to
obtain X(s). As a final step, we perform
an inverse transformation to obtain
*(0
-
1
r-
(6.7)
[*(*)]
Differential
m
equation
EM
System
m
RM
function
In effecting the transition between the
FIG. 6.2. Linear system.
time and frequency domains, a table of
very
transform pairs MO, X(s)} can be
diagram outlining the use of transform methods is given
helpful.
in Fig. 6.1. Figure 6.2 shows the relationship between excitation and
response in both the time and frequency domains.
A
THE LAPLACE TRANSFORM
6.2
The Laplace transform of a function of time/(/)
CLf (01
= Hs) =
{"/(&-" dt
is
defined as
(6.8)
no-
where s
is
the complex frequency variable
s
=
a +ja>
(6.9)
This definition of the Laplace transform is different from the definition
given in most standard texts, 1 in that the lower limit of integration is
1
M.
New
E.
Van Valkenburg, Network Analysis, 2nd Ed. Prentice-Hall, Englewood Cliffs,
Jersey, 1964.
Network
136
/
= 0—
instead of
that/(t)
t
and synthesis
= 0+. We
thus take into account the possibility
may be an impulse or one of its higher derivatives.
=
C[d(0]
analysis
for the
It is clear that
0+ definition, whereas for the 0- definition, C[($(0] =
*
•
In the case when no impulses or higher derivatives of impulses are involved,
Therefore, all of the
it was shown in Chapter 4 that/(0-) =/(0+).
treatment
of Laplace trans"strong results" resulting from a rigorous
lower
limit
also
apply for the 0—
a
0+ as
forms* obtained by using t
=
definition.
In order for a function to possess a Laplace transform,
it
must obey the
condition
f
\f{t)\e-"dt<
oo
(6.10)
Jo-
Note that, for a function to have a Fourier
must obey the condition
for a real, positive a.
form,
it
f—
•/
|/(0I
dt<
oo
trans-
(6.11)
i
As a result, a ramp function or a step function will not possess a Fourier
transform, 3 but will have a Laplace transform because of the added
t%
convergence factor er a *. However, the function e will not even have a
Laplace transform. In transient problems, the Laplace transform is
preferred to the Fourier transform, not only because a larger class of
waveforms have Laplace transforms but also because the Laplace transform takes directly into account initial conditions at t = 0— because of
the lower limit of integration in the Laplace transform. In contrast, the
Fourier transform has limits of integration (— oo, oo), and, in order to take
into account initial conditions
function must take a form
due to a switch closing
as/(0 u(t)
+ /(0-)
(3(0,
at
t
= 0, the forcing
where /(0-) repre-
sents the initial condition.
The
inverse transform C-lrjFT(.0]
/(0
is
= -M
FWds
(6.12)
Alt} Jn—iaa
a real positive quantity that is greater than the a convergence
factor in Eq. 6.10. Note that the inverse transform, as defined, involves a
where at
is
D. V. Widder, The Laplace Transform, Princeton University Press, Princeton, 1941.
These functions do not possess Fourier transforms in the strict sense, but many
possess a generalized Fourier transform containing impulses in frequency; see M. J.
Lighthill, Fourier Analysis and Generalized Functions, Cambridge University Press, New
*
*
York, 1959.
The Laplace transform
137
complex integration known as a contour integration.1 Since it is beyond
the intended scope of this book to cover contour integration, we will use a
partial fraction expansion procedure to obtain the inverse transform.
by
find inverse transforms
recognition,
we must remember
To
certain basic
transform pairs and also use a table of Laplace transforms. Two of the
most basic transform pairs are discussed here. Consider first the transform
of a unit step function
u(t).
Example 6.1. /(/)-«(/).
(6,13)
Next,
let
us find the transform of an exponential function of time.
Example 6.2. /(/)
~ e°<«(01
F(j)-J
With
these
e#e-«dt=s
(6.14)
—a
two transform pairs, and with the use of the properties of Laplace
we discuss in Section 6.3, we can build up an extensive table
transforms, which
of transform
6.3
pairs.
PROPERTIES OF LAPLACE TRANSFORMS
In this section we will discuss a number of important properties of
Laplace transforms. Using these properties we will build up a table of
transforms. To facilitate this task, each property is illustrated by considering the transforms of important signal waveforms. First let us
discuss the linearity property.
Linearity
finite sum of time functions
forms of the individual functions, that is
The transform of a
t[| /,«]=
is
the
sum of the trans-
2 OK')]
(615)
This property follows readily from the definition of the Laplace transform.
Example 63. /(/)
= sin <ot.
Expanding
fit)
4
For a
by Euler's
identity,
we have
= ^.(e** - r+*)
Goldman, Transformation Calculus and
Englewood Cliffs, New Jersey, 1949, Chapter 7.
lucid treatment, see S.
Transients, Prentice-Hall,
sin cat
(6.16)
Electrical
1
38
Network
anal/sis
and synthesis
The Laplace transform of /(f)
cisoidal e ±iat terms. Thus
is
the
sum of
1/1
=-
C[sin cat]
—— \
1
)
— yto
:
2/\*
the transforms of the individual
+./«/
*
m
=
**
+
(6.17)
«>*
Real differentiation
Given that £[/(f)] = *(*), then
Eflwhere /(0-)
is
By
Proof.
sF(s)-/(0-)
the value of/(f) at
(6.18)
= 0-.
f
definition,
£[A0]=J*V'7'(0<*f
Integrating Eq. 6.19
by
parts,
£[/'(')]
(6.19)
we have
= «-7(0
+
/(Oe-'df
s
(6.20)
JoSince e~" -»
£[/"(')]
as
f
-*
oo,
and because the
= ^)» we have
t[f'(t)]
Similarly,
c
we can show
|"£7(0"l
=
n
s
f(s)
integral
on the right-hand
= sF(s)-f(0-)
side
is
(6.21)
for the nth derivative
_
«-i
s
/(0 _)
_ S "-V(0-)
/
(
- u (0-)
(6.22)
where f
1
"-"
is
the («
that differentiating
by
—
f
l)st derivative
in the time
of /(f) at
domain
is
f
= 0-. We
thus see
equivalent to multiplying
complex frequency domain. In addition, the initial conditions
by the terms / (i) (0— ). It is this property that
transforms differential equations in the time domain to algebraic equations
in the frequency domain.
by
s in the
are taken into account
Example
6.4a.
/(f)
= sin cot.
Let us find
By
=
C
— j sin
cot)
(6.23)
cgsin<of]=,(^)
(6.24)
C[cos
cot]
the real differentiation property,
we have
The
i(_^ = _I 5
c[cosa,,]=
sothat
=
In this example, note that sin <o(0—)
Example
6.4b.
unit impulse.
/(/) = «(/). Let
We know that
Laplace transform
139
(6.25)
0.
us find £[/'(')], which
the transform of the
is
=-
Ct«(0]
(6.26)
s
C[*0]
Then
since
= *(j) =
1
(6.27)
= 0.
«(0— )
Real integration
= F(s)> then the Laplace transform of the integral of /(f)
If £[/(*)]
flfa)
divided by
that
s,
c[£/(r)dr]=^
By
Proof.
c|"
by
P
LX
parts,
/(t)
f
-»
-.C
[£-
/(T) dT
]
d'
(6,29)
obtain
=-
—f
/(t) dr
s Jo-
J
as
e_rt
/(t) dr]
we
rfrl
LJoSince e~" -*
(6.28)
definition,
c
Integrating
is
is,
oo,
and
"
+-
o-
°°
f
e-'/(f) dt
(6.30)
s Jo-
since
f
=0
I
/(t)<*V
Jo-
(6.31)
It-o-
we then have
c[£/(r)dr]=^
Example 6.5.
We know that
Since
Let us find the transform of the unit ramp function, p(t)
j
h(t)
dr
C[«(0]
=
(6.32)
" t u(t).
(6.33)
p(f)
=-
(6.34)
s
then
C[p</)]
-
-^ . _
£[«(')]
1
(6.35)
Network
140
and synthesis
analysis
Differentiation by •
Differentiation
by
multiplication
by
s in the
in the
t
complex frequency domain corresponds to
domain, that
W)] =
From the
Proof.
is,
-^
(6-36)
as
of the Laplace transform, we see that
definition
±1® = f°7(0 ± e-* dt - - f " tf(t)e-* dt Example 6.6.
Given /(O
-Z[t f(t)]
= g-*\ whose transform is
m—rrz
let
us find
C[te~"']-
the preceding theorem,
By
638>
(
we find that
•*»
«"-i--s(rh)-<rb
Similarly,
we can show
that
(640)
"a-^-crrb* 5
where n
is
a positive
Complex
By
integer.
translation
the complex translation property,
/=X5
where a
is
Proof.
(6.37)
Jo-
ds
Jo-
ds
if
F(s) <= £[/(/)], then
- a) = C[e°7(0]
< 6 - 41 )
a complex number.
By
definition,
£1*7(0]
Example 6.7.
-J"
^/(Oe"" dt
Given /(t)
=
J"
= sin <»/, find
.
e"
(M)
7(0 dt
£[«-»' sin <»/].
qsinorfl-^^
= F(s -
a)
(6.42)
Since
643>
<
by the preceding theorem, we find that
Similarly,
we can show that
c^coso,.]-
^;^;^
(«-45)
The Laplace transform
141
Real translation (shifting theorem)
Here we consider the very important concept of the transform of a
P(s), then the transform
shifted or delayed function of time. If C[/(/)]
of the function delayed by time a is
=
£[/('
Proof.
By
(6.46)
= f V*7(r - a) dt
(6.47)
a)]
definition,
C[/(r
- a) u{t -
new dummy
Introducing a
C[/(t) m(t)]
= e-** F(s)
- a) u(t -
-
°°
f
Jo-
a)]
variable t
= — a, we have
t
= g— ['°f(r)e-
T
e-lT^f(r) dr
dr
= <f~ F(s)
(6.48)
Jo-
in Eq. 6.48/(t)u(t) is the shifted or delayed time function; therefore
the theorem is proved.
It is important to recognize that the term e~"' is a time-delay operator.
If
we
are given the function
e~"
G(s),
and are required to
C-i[r~G(,)]-
we can
discard er*' for the
moment,
find
ft (0,
find the inverse transform
E-HG(*)]-«('X
and then take
into account the time delay
by
setting
g(t-a)u(t-a)=g1(t)
(6.49)
Example 6.8a. Given the square pulse /(f) in Fig. 6.3, let us first find its transform F(s). Then let us determine the inverse transform of the square of F(s),
i.e., let
us find
Mt) =
Solution.
The square pulse
is
/(/)
Its
Laplace transform
is
C-i[F«(5)]
(6.50)
given in terms of step functions as
- «(/) -uit-a)
(6.51)
then
fit)
F(s)=\(l-e~«)
Squaring F(s),
F\s)
we
(6.52)
obtain
- -5 (1 - 2e-» + e"*")
(6.53)
To find the inverse transform of F\s), we
need only to determine the inverse transform
FIG.
4.3.
Square pulse.
Network
142
analysis
and synthesis
of the term with zero delay, which
is
&]-
C-4-J -fN(<)
Then
C-»[F*(*)]
=
/«(/)-
2(*
-
a)u(t
(6.54)
- a) +(t - 2a) «(/ - 2a)
(6.55)
so that the resulting waveform is shown in Fig. 6.4. From this example, we see
that the square of a transformed function does not correspond to the square
of its inverse transform.
AW
f(t)
K6
K2
Ki
Xb
it
Kq,,
1
I
a
FIG.
FIG.
Triangular pulse.
6.4.
Example
6.8b. In Fig.
of a train of impulses
fit)
6.5, the
6.5.
Impulse
output of an ideal sampler
= *o <K0 + *i <K' ~ Tt) +
The Laplace transform of this impulse
C[/(/)l
3Ti 4Ti 5Ti 6Ti
2T!
Ti
2a
•
•
"
is
7Ti
train.
shown.
It
consists
+ *» *' - « ri)
(
656>
train is
- *o + *!*-•*» + ^-2,Tl +
•
•
•
In dealing with sampled signals, the substitution z =
represent the transform of the impulse train as
+ *„e-*» r
e ,r »
is
i
often used.
(6.57)
Then
we can
Ai1
W)]-*. + T
A.
+^ +
A«
••+?
(6.58)
in Eq. 6.58 is called the z-transform of /(/). This transform
widely used in connection with sampled-data control systems.
The transform
PERIODIC
is
WAVEFORMS
The Laplace transform of a periodic waveform can be obtained in two
ways (a) through summation of an infinite series as illustrated in Example
6.8d, and (*) through the formula derived below.
:
£[/(*)]
-}
f(tyr"dt
+Jt
f{t)e-'*dt
+
(6.59)
flufUT
f(t)e-*dt
InT
+
The Laplace transform
Since /(0
is
periodic, Eq. 6.59 reduces to
£[/«]
=
T
dt
\ f{i)e-'*dt
Jo-
+
e- T [
+ e-'
e-nT f*f(t)e-* dt
= (1 +
e-T
—
l-e-
+ e-aT +
+
+
)
(6.60)
T
\ f{t)e-*dt
Jo-
T
~
J
T Jo- f(f)e-'*dt
\
Given the periodic pulse
6.8c.
determine
f(t)e-*dt
\
Joo-
Jo-
Example
143
train in Fig. 6.6 let us use Eq. 6.60 to
Laplace transform.
its
f°
1
dt
-1
(6.61)
1 1
s
1
- €~"
- e-' T
m
1
T
a
FIG.
Example
6.8d.
Fig. 6.6 using
T+a
2T+a
2T
6.6. Periodic pulse train.
In this example we calculate the Laplace transform of /(f) in
summation of an infinite series. The periodic pulse train can be
represented as
/(/)
Its
=
iKt)
- iKt - a) + u(t - T) - u[t - (T + a)]
+ u(t - IT) - u[t - (2T + a)] +
Laplace transform
F(s)
=
-
(1
•
•
{pM)
is
- *-" + e~tT - e-<T+«)» +
.
•
.)
(6.63)
=s
[1
- e~" +
e~,T(l
- e-») +
*-2 « T(l
- «-") +
•
•
•]
Network
144
which
analysis
and synthesis
simplifies further to give
F(s)
=
-
(1
- «-^"Xl + e- T + e~UT +••)
(6.64)
s
We then see that F(s) can be given in the closed form
1 1 - e-™
Other periodic pulse trains also can be given in closed form. The reader
end of this chapter.
is
referred to the problems at the
At
the end of the chapter
is
a table of Laplace transforms. Most of
the entries are obtained through simple applications of the properties just
discussed. It is important to keep those properties in mind, because many
transform pairs that are not given in the table can be obtained by using
For example, let us find the inverse transform of
these properties.
F(s)
=,
™
+
j_
(s
ay
+
(6-66)
,
car
Since the s in the numerator implies differentiation in the time domain,
we
can write
"
F(s)-a
(s
Using the differentiation property,
+
we
ay
+
.
(6.67)
co'
obtain
y
dt
/
Note that e~"
6.4
sin
mt at
t
(6.68)
= (—«r sin oat +
= 0— is zero.
to
cos oot)e~*
USES OF LAPLACE TRANSFORMS
Evaluation of definite integrals
The Laplace transform
integrals.
is
often useful in the evaluation of definite
occurs in the evaluation of
An obvious example
-/:•c-**sin5rdr
If we replace tr %%
of sin
5/,
which
(6.69)
by er**, the integral /then becomes the Laplace transform
is
Z[S ux5t]
= -rj-s + 25
(6.70)
The Laplace transform
Replacing J by
2,
we have
I
=
¥TTs
Perhaps a more subtle example
=
/
First,
we note
that
t
eriW
%
is
is
(6 - 71)
i
the evaluation of
W
+
J
=
,W dt
(6.72)
an even function; therefore
= 2jVC-2'df
/
From
145
the table of Laplace transforms
we
(6.73)
see that
and the transform of
= f tV**Jo-
(6.75)
vm -^tv
(<76)
/(0
Later
we
consists
will see that the partial-fraction expansion of Z[f(t)] in Eq. 6.76
of three terms for the multiple root (s + If, as given by
pr//A1
ELKO]
—
= 0.25 "
—
0.25
Taking the inverse transform of Eq.
f{t)
Now
t
=
1,
^i - j^
0.5
~
6.77,
= 0.25(1 - e-« -
we
2te-**
.-_
(6.77)
obtain
- 2rtr-«)
observe that the definite integral in Eq. 6.73
that
1
11(f)
(6.78)
is
equal to 2/(f) at
+ 0.5 = 0.162
(6.79)
is,
t
= 2/(1) =
-2.5<r*
Solution of integrodifferential equations
In Section 6.3 we said that the real differentiation and real integration
properties of the Laplace transform change differential equations in time
to algebraic equations in frequency. Let us consider some examples
using Laplace transforms in solving differential equations.
Example
6.9.
Let us solve the
differential
equation
+ 3x'(t) + Irtf) - 4e*
given the initial conditions a<0-) = 1, x'(Q—) = — 1.
*"(/)
(6.80)
Network
146
Solution.
analysis
and synthesis
We first proceed by taking the Laplace transform of the differential
equation, which then becomes
[i
2
X(s)
-sx(0-)- x'(0 -)] +
3[s X(s)
- x(0 -)] + 2X(s) =
Substituting the initial conditions into Eq. 6.81
(**
(s*
We then obtain
+
3s
+ 2) X(s)
and
—
:
s
simplifying,
—
4
-
we have
+5+2
(6.82)
1
X(s) explicitly as
™-jr=mrm+%
To
(6.81)
find the inverse transform x(t)
= f'^W],
(6 - 83)
we expand
X(s) into partial
fractions
Solving for X_i,
The
A^i,
and
Kt algebraically, we obtain
K_i = § JTi = -1 Kt = |
final solution is thf inverse
a<0
transform of X(s) or
=
fe*
- e-' + fe-»
(6.85)
In order to compare the Laplace transform method to the classical method of
is referred to the example in Chapter 4,
solving differential equations, the reader
Eq. 4.64, where the differential equation in Eq. 6.80
Example
6.10.
Given the
2x'(f)
set of
+
is
solved classically.
simultaneous differential equations
4a<r)
+ y'(t) + 7y(0 = 5u(t)
+ y'(t) + 3jK0 = 56X0
(6.86)
+ x(t)
with the initial conditions x(0 -) = y(0 -) = 0, let us find x(j) and y(t).
x'(t)
Solution.
Transforming the
2(s
(s
set
of equations,
+ 2) X(s) + (* + 7)
+
1)
X(s)
+ (* +
Solving for X(s) and Y(s) simultaneously,
X(s)
we
obtain
Y(s)
--
3) Y(s)
s
=
(6.87)
5
we have
+ 5^
(5/*)A u
A
(6.88)
Y(s)
(5/5)Au
+ 5A*
The Laplace transform
where
zyth
A
is
the determinant of the set of equations in Eq. 6.87, and
More explicitly,
cofactor of A.
X(s)
-5i»
in partial fractions,
- 30s +
the
15
(689)
we have
Kx
+ Kz
K-s
^> = 7 + ?T27T5
Multiplying both sides of Eq. 6.90 by s and letting *
K1
is
is
™~-*FT2r+sr
Expanding X(s)
A tf
147
<690)
= 0, we find
=sX(s)\,_ =3.
Kt and Ks are then obtained from the equation
-8s
3
- 36
*« "-7+5 + 5
(691)
A further simplification occurs by completing the square of the denominator
of X(s), that
is,
s*
As a
result of
+ 2s +
5
=
(s
1)*
+4
(6.92)
Eq. 6.92, we can rewrite X(s) as
3
*<'>-"
8(i
fr
so that the inverse transform
«(/)
In similar fashion,
Wrt
y(s)
The
+
we
_
=
=
(3
+ 1) + 14(2)
+ iy+gy
(693)
is
-
8c-' cos 2/
-
14e-' sin It) «(/)
(6.94)
obtain Y(s) as
"* + 17
1
+
~s 7T27T5 - -; +
*
+ l)+3(2)
+ i)* + (2)*
llfr
|
(,
(6,95)
inverse transform is then seen to be
y (j)
= ( _ i +1 i e-* cos 2f +
3<r* sin It) u(t)
(6.96)
is also solved by classical methods in Chapter 4, Eq. 4.163.
note one sharp point of contrast. While we had to find the initial
conditions at t
0+ in order to solve the differential equations directly
in the time domain, the Laplace transform method works directly with
This example
We
=
= 0—
In addition, we obtain both the complementary function and the particular integral in a single operation when we
use Laplace transforms. These are the reasons why the Laplace transform
method is so effective in the solution of differential equations.
the
initial
conditions at
t
.
Network
148
analysis
and synthesis
PARTIAL-FRACTION EXPANSIONS
&5
As we have seen, the ease with which we use transform methods depends
upon how quickly we are able to obtain the partial-fraction expansion of a
given transform function. In this section we will elaborate on some simple
and effective methods for partial-fraction expansions, and we discuss
procedures for (a) simple roots, (b) complex conjugate roots, and
(c)
multiple roots.
should be recalled that if the degree of the numerator is greater or
equal to the degree of the denominator, we can divide the numerator by
the denominator such that the remainder can be expanded more easily
It
into partial fractions.
Consider the following example:
=
m=
s»
+
D(s)
Since the degree of N(s)
is
s'
3s*
+ 3s + 2
+
2s
+
2
greater than the degree of D(s),
we
divide D(s)
into N(s) to give
F(s)
'
= 5 + 1s
+2s +
(6.98)
2
Here we see the remainder term can be easily expanded into partial
However, there is no real need at this point because the
denominator s2 +^2s + 2 can be written as
fractions.
s*
+ 2s +
2
= (5 +
1)*
+
1
(6.99)
We can then write F(s) as
so that the inverse transform can be obtained directly from the transform
tables,
namely,
C-W)] =
From this
form
example,
<5'(0
+
3(0
+ er*(sm t - cos i)
we see that intuition and a knowledge of the trans-
table can often save considerable work.
examples in which intuition plays a dominant
Example
6.11.
(6.101)
Find the
Consider some further
role.
partial-fraction expansion of
™
2*
=
+3
(JTWTT)
(6102)
The
If we see that F(s)
(s
Find the
+ 1) + (s + 2)
+ Dfr+g
(6103)
fr
then the partial-fraction expansion
6-12.
149
can also be written as
*W-
Exanpfe
Laplace transform
is trivially
partial-fraction expansion
of
s +5
j
+5
F(s) = , T -,
(6-105)
We see that F(s) can be rewritten as
= _i_ T _i_^
mW = fe±a±i
+
+
+
2)»
(s
(s
(6 106)
2)»
(s
2)
Real roots
Now let us discuss some formal methods for partial-fraction expansions.
First
we examine a method
for simple real roots.
m=
- s tt^t:
)(s — s0(s — St)
(6-107)
:
,
(s
where s<» st, and st are distinct,
Expanding F{s) we have
F(s)
Consider the function
real roots,
and the degree of N(a)
= -£*- + -^- + -^S — Si
S — Si
S — S
Let us first obtain the constant A,.
j ) to give
of the equation by (s
< 3.
(6.108)
We proceed by multiplying both sides
—
(s
— s«)F(s) = K
H
s
If we let 5
=*
in Eq. 6.109,
we
—
1
s
«!
—
.
obtain
Ji-(#-J»)J^)|^
Similarly,
we
(6.109)
s,
see that the other constants can
(6110)
be evaluated through the
general relation
A,
1
6.13.
= (*-*) ft*) |_.,
(6.111)
Let us find the partial-fraction expansion for
J +2j - 2
.g» + J^- + J^_
FV- j(*+2X*-3)
FW
*
s+2+3-3
'
° Z)
(6.112)
K
l
c
150
Network
anal/sis
and synthesis
Usin gEq. 6.111, we find
=
Ao
sF(s) |_>
+ 2s -2
+ 2X* - 3)
s*
~
(*
~
«-o
+2s -2
-3) »— a
s*
Kl
1
3
+ 2s - 2
s2
*"
(6.113)
1
sis
5
13
~15
3
s(s+2)
Complex
roots
Equation 6.111 is also applicable to a function with complex roots in
its denominator. Suppose F(s) is given by
F(s)
=
JVC*)
D&Xs -
*
Ki
s-x-jfi
where A^/Di
is
- JPK' - * + JP)
+
the remainder term.
K1 =
JVi(s)
s-x+jp
(6.114)
Djis)
Using Eq. 6.111, we have
ypDM+m
(6.115)
-irfD^-m
where we assume that j = a ± jP are not zeros of Z>i(s).
It can be shown that the constants Kx and K% associated with conjugate
roots are themselves conjugate. Therefore, if we denote K± as
(6.116)
K2 = A-jB = K *
then
If
1
we denote the
we see that
inverse transform of the
=
complex conjugate terms as
——— + ———
fx(t),
A(0
i(o
(6.117)
_i
r
=
e"'(K ie "»
=
2c**(A cos Bt
+
Kft-"*)
- B sin pi)
(6.118)
The Laplace transform
151
A more convenient way to express the inverse transform f^t) is to introduce the variables
M and
denned by the equations
<X>
M
= 2A
McosO = -25
where
A
terms of
and
B
M and
obtain
the inverse transform
<I>,
When
we note
lt
=
6.116.
In
(6. 120)
that
M cos O + jM sin O =
Mf =
6.14.
Kr in Eq.
is
-2B + ;2,4 = 2;XX
related to the original function F(s),
Example
(6.119)
= Me?* sin (0t + O)
M and $ from K
Me'*
<1>
and imaginary parts of
are the real
/i(0
To
sin
we
see
(6.121)
from Eq. 6.115 that
*<«+Jfl
(6 . 122)
Let us find the inverse transform of
w-vr^rkr+ii
For the simple root s
=
—2, the constant
(6123)
.
K is
K = (*+2)F(*)|,__8 =
i
(6.124)
For the complex conjugate roots
**+2j+5=(* +
We see that
a
= — 1,
Me**
The inverse transform
C-i[F(*)]
1
+j2Xs
+
1
-y'2)
(6.125)
= 2, thus
«*
=
+3
=
2(*+2) i—1+*2
is
—
V5
c-i(tan-»j+W»)
(6.126)
then
=
- e-**
+ —= <r* sin \^
~\~ tan-1 2j
(6.127)
7
= - e~u
5
2
-p e~*
V5
Multiple roots
Next let us consider the case
—
-1
tan 2)
which the partial fraction involves
examine two methods. The first
the second does not.
repeated or multiple roots.
requires differentiation;
cos (2/
We
in
will
Network
152
analysis
and synthesis
A
Method
Suppose we are given the function
W=
^
F(S)
(6.128)
- s )" ^(s)
The partial fraction expansion
with multiple roots of degree n at s = j
(s
.
of F(s)
is
Kt
K
Kx
(s-s ) n
(s-so)""
..... *«-i
.
s-s
(s-s )»-8
1
*i(?)
J>i(«)
(6.129)
where N^jDjUs) represents the remaining terms of the expansion. The
problem is to obtain K^K^..., K^. For the K^ term, we use the method
cited earlier for simple roots, that
Ko
However,
if
we were
is,
= (s-s rF(s)\„
(6.130)
t
to use the
same formula
to obtain the factors
K ,..., #„_!, we would invariably
t
dition.
Ku
arrive at the indeterminate 0/0 conn
s ) and
Instead, let us multiply both sides of Eq. 6.129 by (s
—
define
F1(s)s(s-s rF(s)
(6.131)
Thus
Fx(s) -
A,
+ Kt(s - s ) +
•
+ K^s - So)""1 + *(•*)(' - s*Y
•
(6.132)
where R(s) indicates the remaining terms. If we
s,
we
differentiate
Eq. 6.132 by
obtain
Fi(s) - *i + 2K^s 4
as
s )
+
•
•
•
+
*„_!(/!
-
D(s
- so)""* +
(6.133)
Kx = — Ft(s)
It is evident that
(6.134)
ds
On the same basis
and
Kt =
;
F^s)
(6.135)
2 ds
in general
K = -tWs)!
!«-«•>
.
t
j\ ds'
J
= 0,1, 2
n
-1
(6.136)
The
Example
6.15.
Laplace transform
153
Consider the function
(6 - ,37)
'W-STTlJi
which we represent-in expanded form as
™-(7Ti? + (JTI? + 7TT + 7
The constant A
=
for the simple root at s
<«'»>
is
A=sF(s)\^=-2
To obtain
the constants for the multiple roots
Fjis)
= (s +
l)
(6.139)
we first find Ft(s).
=
8 F(s)
S
-^
(6.140)
s
Using the general formula for the multiple root expansion, we obtain
d
1
^i=T7T
1 ds
!
fs
-2\
\
s
/
2
I
!_!
„,_„,.
=2„
-3
« |__!
I
<6142)
—
*-i^)L-(-5)L-;
IW __2 gl +
sothat
^
i+
-i
1
-5
(6.144)
Method B
for arriving at the partial-fraction expansion for
The second method
no differentiation. It involves a modified power
Let us consider F(s) and F^s), defined in Eqs. 6.128
s
Joand 6.131 in Method A. We define a new variable/; such that/>
multiple roots requires
series expansion.*
= —
Then we can write F^s) as
*(p
Dividing N(p
+s)
ascending powers ofp,
fiG*
+ tf-£?T^
by
D%(p + s
we
obtain
),
(6145)
with both polynomials written in
+ s ) = K + K p + Ktpt +-- +
1
X^j,""1
+
n
f"f
+ *o)
0l(l>
(6.146)
* I
am
Institute
indebted to the late Professor Leonard O. Goldstone of the Polytechnic
of Brooklyn for showing
me this method.
Network
154
The
anal/sis
and synthesis
original function F(s)
F(P
+
°o)-
is
F (p + s ) by the equation
related to
x
-,„-%»-!+
pn
+
+
,
Dl(p
+ So)
(6.147)
Substituting 5
F(s)
—
=
s
= p in Eq. 6.147, we obtain
—£*— +
g
/
\ Bl +
•
•
•
-^ + -£\
+
(6.148)
We
have thus found the partial-fraction expansion for the multiple-root
terms. The remaining terms KjlD^s)] still must be expanded into partial
fractions. Consider Example 6.16.
Example 6.16
*wUsing the method just given, F^s)
F^s)
Setting p
=s+
1,
we
-
(s
fr
+
(6149)
ijUa
is
+ If F(s) =
-^
(6.150)
then have
- 1) into a series as given in Eq. 6.146 requires dividing
the numerator 2 by the denominator 1 + p, with both numerator and denominator arranged in ascending power of p. The division here is
The expansion of Ft(p
2
- lp + 2/
l+/0"2
2 +2/>
-2^
-2p-2f
Since the multiplicity of the root
we have three terms in the quotient.
AT = 3, we terminate
We then have
is
the division after
FX{P -l)=2-2p+2p*- j£-x
The original function F(p — 1) is
F(P
~
1)
(d -1)
=-FJx7T- -
2
-
2
? ?
2
(6.152)
2
+~ -pTl
p
(6153)
The Laplace transform
Substituting s
+
1
= />, we have
m
Example
- (7TT? " (7TI? + (7TT) " JT2
As a second example,
6.17.
F(,)
Since
we have two
sets
= 0;
(6155)
=^T1?
= 0, and s =
—1) we have a
Let us arbitrarily choose to
s here, we do not have to make any substitutions.
=
since /j
(6154)
consider the function
of multiple roots here
F&) is then
(at s
— Ooi s = — 1.
choice of expanding F^s) about s
expand about *
155
+
J
2+s
2
™=(7TT? = rT2FT7
Ft(j), we obtain
Expanding
-
F^s)
is
then
F(s)
+ 4)
^ + iy
3* + 4
+
+^
- 3j +
2
2
F(s)
3
-^ -
j
We must now repeat this process for the
Fortunately
we
3s
answer
POLES
(6.158)
term
+
1)*
is
+4 = 3Qr +
+
1)»
(*
1)
+
+
1
iy
~
s
1
3_ +
+ I (s +
(6159)
1)»
then
"W-l-i+TTT + frTi?
6.6
(6.157)
.
see that the term can be written as
(s
final
s\3s
+4
3*
(s
The
<6156>
(6160)
AND ZEROS
In this section
we
will discuss the
many
implications of a pole-zero
description of a given rational function with real coefficients F(s).
We
be the roots of the denominator of F(s). The
zeros of F(s) are defined as the roots of the numerator. In the complex s
plane, a pole is denoted by a small cross, and a zero by a small circle. Thus,
define the poles of F(s) to
for the function
_ ^-i+;D(s-i-Jl)
+ l)\s+j2)(s-j2)
—
s =
1
(double)
s = —y*2
(s
the poles are at
s
=
+y*2
K
156
Network
and synthesis
analysis
jo
|
« plane
j2 i
.t
-<r
o
,1
—i_W_
Double
pole.*.
i
i
I
I
12
i
-3 -2 -1
o
->2 x
-J<0
FIG.
and the zeros are
diagram of F(s).
6.7. Pole-zero
at
s
=
s
=
1
S
=
00
-j\
zeros of F(s) are shown in Fig. 6.7.
Now let us consider some pole-zero diagrams corresponding to standard
example, the unit step function is given in the complex
The poles and
signals.
For
frequency domain as
C[u(0]
=-
(
6162>
s
and has a pole at the origin, as shown in
signal e—*, where <x > 0, has a transform
S
function cos
single pole at s
<a
i,
=
+
C[coso,
(6.163)
Oq
-o-* as indicated in
whose transform
The exponential
—
C[e—']
which has a
Fig. 6.8a.
Fig. 6.86.
The
cosine
is
t]
= -r-1-~t
(6-164)
=
±7<u »
poles at s
has a zero at the origin and a pair of conjugate
diagram
pole-zero
the
shows
6.Sd
Figure
as depicted in Fig. 6.8c.
is
corresponding to a damped cosine wave, whose transform
Z[e-"* cos
co t ]
g°
= — s +r=-*
(s +
r+
<r
(6.165)
o>
i
The
Laplace transform
157
a
I
(a)
(b)
]W
m»)'
""
M,
«+«,
*******
-a
x
— -jao
id)
FIG.
6A
Poles and zeros of various functions.
From these four pole-zero diagrams, we note that the poles corresponding to decaying exponential waves are on the — a axis and have zero
imaginary parts. The poles and zeros corresponding to undamped
on the/o> axis, and have zero real parts. Consequently, the
and zeros for damped sinusoids must have real and imaginary
sinusoids are
poles
parts that are both nonzero.
Now let us consider two exponential waves/x(0 = er'i* and/i(f) = er***,
where at
6.9a.
>
<r
x
> 0, so that/,0) decays faster than/iO), as shown in Fig.
The transforms of the two
functions are
*i(»)
1
=
s
+
ax
(6.166)
1
F&) =
s
+
a,
by the pole-zero diagram in Fig. 6.96. Note that the further
the pole is from the origin on the —a axis, the more rapid the exponential
decay. Now consider two sine waves, sin atjt and sin co t t, where <o t >
«>! > 0. Their corresponding poles are shown in Fig. 6.10. We note here
as depicted
m
JU>
l
s plane
*2(«K
Fi(»K
— 0"1
—V2
(a)
FIG.
*.*. Effect
(b)
of pole location upon exponential decay.
a
-
Network
158
analysis
1
and synthesis
JU
1
X./C02
JO)
X
,
«2
>«i
X
«1
a
X
si*
(
-jwi
X
«2*
<
-j(i>2
FIG.
FIG. 6.10. Pole locations corresponding to sin to,/ and sin mj.
6.1
represents frequency of
that the distance from the origin on theyco axis
frequency.
oscUlation; the greater the distance, the higher the
responses fx {f) and
time
the
compare
let
us
thumb,
of
rules
these
Using
shown in Fig.
corresponding to the pole pairs {* lf s 1 *} and fo, s a *}
a (0
sinusoids.
damped
see that both pairs of poles corresponds to
/
6.11.
We
than/a(f)
sinusoid/i(0 has a smaller frequency of oscillation
s t Also,
of
part
because the imaginary part of * x is less than the imaginary
time
The
Res
because
2
1
(t) decays more rapidly than fx (t)
The damped
.
fz
responses /x(0
>Kcs
and/a(0
are
shown
in Fig. 6.12.
f2 (t)
h
- lme«»*
h») =
\^" ~~-
—
——
•C-
lme«»'
___ __
-
=*-
S ^-^
(b)
in Fig. 6.11.
FIG. 6.12. Time responses for poles
.
The Laplace transform
159
ju
« plane
-
m
at
a
-m
M
(b)
FIG. 6.13. Effect of right-half plane poles upon time response.
Let us examine more closely the effect of the positions of the poles in
upon transient response. We denote a
pole pt as a complex number (
at +jco i For a given function F(s)
p
with only first-order poles, consider the partial-fraction expansion
the complex frequency plane
=
F(s)
The
=
_*!L_
s-
+ _*!_ +
(6.167)
s-P«
-Pi
Po
inverse transform of F(s)
/(*)
.
is
= K e»« + Kie*« +
= K^"*e im,t + K1e'
•+K
lt
e
iait
ne"*
+• + K^e*""*
In/0), we see that if the real part of a pole is positive, that
the corresponding term in the partial-fraction expansion
K e"
i
i
*e
is,
at
(6.168)
> 0, then
iatt
an exponentially increasing sinusoid, as shown in Fig. 6.13a.
thus
see that poles in the right half of the s plane (Fig.
6.13*) give rise to
exponentially increasing transient responses.
system function that has
poles in the right-half plane is, therefore, unstable. Another
is
We
A
unstable
situation arises if there
is
a pair of double poles on theyw
axis,
such as for
the function
F(
"-(7f^'
whose pole-zero diagram
F(s)
is
shown
in Fig. 6.14a.
(6.169)
The inverse transform of
is
/(0
= -— sin co
2io
t
(6.170)
160
Network
analysis
and synthesis
Vjao
*-.7«o
f(t)
m ^-
sin (dot
(b)
(a)
FIG. 6.14. Effect of double zeros on
they'd) axis
upon time
response.
which is shown in Fig. 6.146. It is apparent that a stable system function
cannot also have multiple poles on theytu axis.
=
N(s)lD(s).
Consider the system function H(s)
numerator and denominator polynomials, we obtain
H(s)
=
- Zq)(s - Zi)
- P )(S - Pi)
Hg(s
(S
•
•
•
•
•
If
- g„)
(S
Pm)
•
(s
we
factor the
(6.171)
H(s) is completely specified in terms of its poles and zeros
and an additional constant multiplier #„. From the pole-zero plot of H(s),
obtain a substantial amount of information concerning the
It is clear that
we can
whether the
behavior of the system. As we have seen, we can determine
the jm axis
and
poles
for
plane
right-half
the
checking
by
system is stable
its transient
concerning
information
obtain
We
poles.
for multiple
discuss
behavior from the positions of its poles and zeros; and, as we will
diagram also gives us significant.information
We thus
concerning its steady-state (Jm) amplitude and phase response.
see the importance of a pole-zero description.
Suppose we are given the poles and zeros of an excitation E(s) and the
response
system function H(s). It is clear that the pole-zero diagram of the
and
H(s)
of
diagrams
pole-zero
the
of
R(s) is the superposition
in Section 8.1, the pole-zero
function
£(j).
Consider the system function
W"
and the
(6.172)
(i
- ftK» " A)
excitation
s
—
(6.173)
pt
The Laplace transform
Then
the response function
H(s)
161
is
=
- Zq)(s - «i)
- p )(s - pj(s - p»)
gpEofr
(s
(6.174)
R(s) contains the poles and zeros of both H(s) and £(*),
except in the case where a pole-zero cancellation occurs. As an example,
let us take the system function
It is clear that
H(s)
=
2(5
(s
+
+
2+j4)(s
1)
(6.175)
+ 2-j4)
whose pole-zero diagram is shown in Fig. 6.15c. It is readily seen from the
pole-zero diagrams of the excitation signals in Fig. 6.18a that the step
response of the system has the pole-zero diagram indicated in Fig. 6.156.
The response to an
tion of Fig.
we
zero plot,
plot
6. 1 5c.
excitation signal 3 cos It has the pole-zero representa-
To specify completely the given function F(s) on a pole-
indicate the constant multiplier of the
numerator on the
itself.
Let us examine the significance of a pole or zero at the origin. We
that dividing a given function H(s) by s corresponds to integrating
the inverse transform h(t)
C- l [//(*)]. Since the division by s corresponds
know
=
to a pole at the origin,
we
see that a pole at the origin implies
an
inte-
gration in the time domain.
Because the inverse transform of H(s) in
Fig. 6.15a is the impulse response, placing a pole at the origin must
correspond to the step response of the system. In similar manner, we
deduce that a zero at the origin corresponds to a differentiation in the
time domain. Suppose we consider the pole-zero diagram in Fig. 6.15c
with the zero at the origin removed; then the resulting pole-zero plot is
the response of the system to an excitation E sin It. Placing the zero at
the origin must then give the response to an excitation
Et cos It, which, of
ju
J(0
X
ju
J*
K-2
JT-2
JT-6
j2::
^r
1
-2 -1
a
-2 -1
-j2x
-;4-
X
(a)
FIG. 6.15. (a) System function,
excitation e(j)
= 3 cos 2i.
-74
-74-
(b)
(jb)
Response to unit step
(0
excitation,
(c)
Response to
162
Network
course,
is
true.
differentiator
analysis
We
and synthesis
therefore conclude that the system function of a
must have a zero
have a pole at the
an integrator must
at the origin; that of
origin.
EVALUATION OF RESIDUES
6.7
Poles and zeros give a powerful graphical description of the behavior
of a system. We have seen that the poles are the complex frequencies of
the associated time responses.
What
role
this question, consider the partial fraction
i
where we
s
=
shall
s
—
do the zeros play? To answer
expansion
Si
assume that F(s) has only simple poles and no poles
at
oo.
The
inverse transform
is
/(0
=
fV
(6-177)
i
time response /(r) not only depends on the complex
on the constant multipliers t These constants t
are called residues when they are associated with first-order poles. We will
show that the zeros as well as the poles play an important part in the
It is clear that the
K
frequencies s( but also
determination of the residues
Earlier
we
discussed a
K
{
K
.
.
number of
methods for obtaining
different
the residues by partial-fraction expansion. Now we consider a graphical
method whereby the residues are obtained directly from a pole-zero
diagram. Suppose
we
f(s)
are given
= 4>(8 ~ goX5 ~ ^
(s - p )(s - Pi)
•
where
m>
n and
all
F(s)
Our
task
is
•
•
•
•
(s
( 6 .i78)
the poles are simple. Let us expand
-^ +
= -&- +
S — Po
* —
Pi
to determine the residues A,.
^ = (s-ft)^(s)
~ 2 »)
- pj
(s
•
•
•
F(s)
as
+ -£=* — Pm
(6-179)
We know that
— Zq)(p< - zi) (Pi - z«)
(Pi - Pm)
(Pi ~ Po)-" (Pi - Pi-^iPi - ft+l)
A
•
(Pi
'
'
-
*
*
(6.180)
The Laplace transform
When we interpret the Eq. 6.180 from a
complex-plane viewpoint, we see that each one
z f) represents a vector drawn
of the terms (p (
JU
ozo
Pix
—
from a zero
z t to the pole in question,
—
Similarly, the terms (p f
where
/>„),
i
p
163
t.
^ k,
from the other poles to the
pole p^ In other words, the residue t of any
represent vectors
a
P0
K
pole
p
t
is
equal to the ratio of the product of
the vectors from the zeros to
p
t,
to the prod-
uct of the vectors from the other poles to
To
illustrate this idea, let
p
{.
us consider the pole-
o
A*
*1
FIG. 6.16. Poles and zeros
of F(s).
zero plot of
F(s)
=
(S
given in Fig. 6.16.
The
F(s)
Ms - zp)(s - Zl)
- p )(s - p )(s - ft*)
partial-fraction expansion of F(s)
- K
s —
(6.181)
x
Kl
Pt,
s
—
pt
I
is
^1
(6.182)
-Pi-
where the asterisk denotes complex conjugate. Let us find the residues A"
and Kx by means of the graphical method described. First we evaluate x
by drawing vectors from the poles and zeros to/^, as shown in Fig. 6.17a.
K
The
residue
K
x is
then
K1 _
XqAB
(6.183)
CD
where symbols in boldface represent vectors. We know that the residue
of the conjugate pole^* is simply the conjugate of x in Eq. 6.183. Next,
K
FIG. 6.17. Determining residues by vector method.
Network
164
and synthesis
analysis
to evaluate
ju
C-3
K& we draw vectors from
poles and zeros to
/>
,
We see that
6.176.
n
K - A °RL
(6.184)
MN
-,—
the
as indicated in Fig.
With the use of a ruler and a protractor,
determine the lengths and the angles of
i
-1
we
the vectors so that the residues can be determined quickly and easily. Consider the fol-
-jl
lowing example. The pole-zero plot of
_
55
F(s)
FIG. «.I8. Pole-zero diagram
+
(«
of-
2)(s
+ 1 - jl)(« + 1 + jl)
F(s).
is
shown
(6.185)
The
in Fig. 6.18.
F(s)
at
—1 +jl
are
is
Kt*
=
s
First let us evaluate
partial-fraction expansion of F(s)
K
"*"
s
+
1
+ jl
(6.186)
s
+2
The phasors from the poles and zeros to
x.
shown
+ - jl
1
in Fig. 6.19a.
Kx =
3
X -p
We see that Kt is
^2/135°
J2 1*5°_ X
2/90°
_3
2
JU
C-3
n
AIVr
-2/\«&
-<r
180*
nJ
'
V90*
r—
—
(a)
— -a
(b)
FIG. 6.19. Evaluation of residues of F(s).
the pole
The
From
Fig. 6.19*
we find the value of the
3
^_
K,=!
Laplace transform
x 2_
= _3
V2 /-13S° x 72 7+135°
=
-
+ l-jl
s
THE
6.8
In
INITIAL
this section
transforms.
The
value of /(/) at
infinity,
that
t
AND
we
+
is
1—
s
+ l+jl
s
(6.187)
+2
FINAL VALUE THEOREMS
will discuss
two very
useful theorems of Laplace
value theorem.
It relates the initial
first is
the
= 0+
to the limiting value of sF(s) as s approaches
initial
is,
lim f(t)
The only
Kt to be
residue
Therefore, the partial-fraction expansion of F(s)
F(s)
165
restriction is that /(f)
= lim s F(s)
(6.188)
must be continuous or contain,
at most,
a
=
step discontinuity at / == 0.
In terms of the transform, P(s)
t[f(t)];
this restriction implies that F(s) must be a. proper fraction, i.e., the degree
of the denominator polynomial of F(s) must be greater than the degree of
the numerator ofF(s). Now consider the proof of the initial value theorem,
which we give in two
(a)
From
parts.
The
function /(f)
the relationship
Cf/'(0I
we
obtain
Therefore
continuous at
is
t
= 0,
that is,/(0— )
=/(0+).
=/ V(0«r" dt = s F(s) - /(0-)
hm C[/'(01 -
lim s F(s)
»-»oo
•-»»
- /(0-) =
(6.189)
(6.190)
Um s F(s) = /(0-) = /(0+)
(6.191)
«-»oo
(b)
The function/(0 has a step discontinuity
at
/
= 0.
Let us represent
f{t) in terms of a continuous part/x(0 and a step discontinuity
as shown in Fig. 6.20.
can then write /(/) as
D «(/),
We
/(0-/i(0
where
D =/(0+) — /(0-).
The
+ J>«W
derivative of/(f)
f'{t)=f\(f)
+
Do{t)
(6^2)
is
(6.193)
1
Network
66
anal/sis
and synthesis
Du(t)
fi(t)
fit)
D
FIG. 6.20. Decomposition of a discontinuous function into a continuous function
plus a step function.
Since fx (t)
is
continuous at
t
=
we know from
0,
lim s F^s)
= A(0-) = /(0-)
Taking the Laplace transform of both
s F(s)
which
simplifies to
part (a) that
of Eq. 6.193,
sides
(6.194)
we have
- /(0-) = s F^s) - M0-) + D
s F(s) <= s F (s) + D
t
(6.195)
(6.196)
Now, if we take the limit ofEq. 6.196 as s-+ oo and let/(0+) -/(0-) =
D we have
(6.197)
lim s F(s) = lim s F x (s) + /(0+) - /(0-)
By Eq.
6.194,
we then
obtain
limsF(s)=/(0+)
Example
6.18.
(6.198)
Given the function
m=
2(s
s*
+
1)
+2s +
(6.199)
5
us find/(0 -f ). Since the degree of the denominator is greater than the degree
of the numerator of F(s), the initial value theorem applies. Thus
let
lim s F(s)
Since
—
2(s
+
\)s
lim
C-H^Cj)]
=2
=2c-*cos2r
(6.200)
(6.201)
we see that /(0+) =2.
Example
6.19.
Now let us consider a case where the initial value
theorem does
not apply. Given the function
f(t)
we see that/(0+)
=
3.
=
(5(0
+ 3r*
The transform of/(f)
F(s)
=
1
(6.202)
is
3
+
s
+
(6.203)
1
,
The Laplace transform
lim sF(s)
so that
= lim si 1 +
.-co
*-*«,
\
——
+
s
-
=
oo
*
Next we consider the final value theorem, which
(6.204)
I
/
states that
= lim s F(s)
lim /(0
167
(6.205)
provided the poles of the denominator of F(s) have negative or zero real
parts, i.e., the poles of F(s) must not be in the right half of the complexfrequency plane. The proof is quite simple. First
=
f'(t)e-' dt
- /(0-)
s F(s)
i:
Taking the
limit as s
->
in the Eq. 6.206,
/'(«) dt
=
lim
j;
Evaluating the integral,
/(oo)
(6.206)
we have
s F(s)
- /(0-)
(6.207)
«-»o
we
obtain
- /(0-) - hm 5 F(s) - /(0-)
Consequently,
/(oo)
= lim s F(s)
(6 .208)
(6.209)
»-»o
Example
6.20.
Given the function
=
/(/)
3«(/)
+ 2e~*
(6.210)
which has the transform
F(s)
s
let
= ii±i
*(s + l)
»
=l +
+
s
1
us find the final value/( oo). Since the poles of F(s) are at s
the final value theorem applies. We find that
(6 . 211)
=0 and s = — 1
we see
]imsF(s)=3
which
is
the final value of /(/) as seen
Example
6.21.
we see that
Given
from Eq.
f(t)
lim/(f)
But from
sF(s)
= le*
= oo
lim * F(s)
=
6.210.
(6.213)
(6.214)
7s
=
s
we have
(6.212)
—
;
(6.215)
1
(6.216)
«-»o
We see that the final value theorem does not apply in this case, because the pole
= 1 is in the right half of the complex-frequency plane.
s
Network
168
analysis and synthesis
TABLE
6.1
Laplace Transforms
F(s)
fit)
1.
FW-JT/Oy**
fit)
2.
«i/i(0
3.
|/W
4-
^/<»
+ «./««)
CxFxC^+fljF^)
»fW
-/(0-)
dn
5.
a" Fis)
f/W
!*w
rfT
6.
f
f fir)drda
Jo- Jo-
7.
(-0"/W
8.
9.
10.
fit -a)u(t
- a)
>(,)
e-°«F(*)
**/X0
F(s
8(t)
1
- a)
dn
1
12.
«(/)
13.
/
14.
-r
s
1
1
n\
1
15.
«-««
16.
/?-« (
17.
sin cof
*
1
e-««
- J f^'f^O-)
+
a
1
_ e-ft)
a>
The Laplace transform
TABLE
6.1 (cont.)
s
18.
cos
19.
sinh at
20.
cosh a/
21.
£ "'sin or
22.
*""' cos
cor
5»
+ <0»
a
5
i*-fl»
0>
(*
+
(5
(* + «)
+ «)* +
mt
+
CO
1
co»
1
e-**t*
23.
(j
24.
—
25.
—/„(«/); n -0,1,2,3,...
t
«*)
+
a)**"1
*
.
sin co/
2co
0*
+
<*»»)»
(5*
+
a«)W[(i»
1
(Bessel function of first kind,
nth order)
26.
J-«
(t/)-*
T(*
27.
r*
(A:
+
1)
need not be an integer)
Problems
6.1
Find the Laplace transforms of
(a)
/(0-sin(orf+|8)
(6)
/(/)
(*)
= *-<*«> cos (at + P)
f(t) .- (r» + 1>-"
A" is a real constant
/(/) - JT*;
/« - (** + «- 0^**
(/)
f«)-td'(t)
(g)
/W - aW* - 4)]
(c)
(</)
6 J,
>1.
Find the Laplace transforms of
- cos (t -
*/4) u(f
(a)
/(/)
(6)
/(0-cos(/-W4)«(0
-
w/4)
+
a»)H
- 5]-»
169
170
6.3
Network
analysis
and synthesis
Find the Laplace transforms for the waveforms shown.
Kt)
fit)
4
2
t
3
1
t
-1
(a)
6.4
(b)
Find the Laplace transforms for the derivatives of the waveforms in
Prob. 6.3.
6.5
Find the inverse transforms for
(a)
F(s)
(6)
{S)
2s
=
(s
"
+
+9
3X*
+ 4)
5s - 12
+ 4s + 13
4* + 13
s*
GO
all the inverse transforms may be obtained without resorting to
normal procedures for partial-fraction expansions.
Note that
.
The
Laplace transform
171
Find the Laplace transform of the waveforms shown. Use (a) the infinite
(6) the Laplace transform formula for periodic waveforms.
Both answers should be in closed form and agree.
6.6
series
method and
fit)
fit)
A
1
uv\\A
3T
2T
T
2T
3T
t
-A
lb)
la)
*t)
+1
r/
2
T
>
\
/
2T
t
-1
PROB. 6.6
(e)
6.7
Id)
Evaluate the definite integrals
w
t
(«)
sin It dt
%
t
6.8
Solve the following differential equations using Laplace transforms
(a)
x'{t)
(b)
x'(t)
(c)
x"{t)
id)
x"{f)
+ fceXO + 9x(t) = cos 2t
+ 3x-(/) + 2a<0 = (5(0
+ 2^(0 + 5a<0 = u(t)
+ Sx'(t) + MO = {e~* + <r**) «(0
(e)
It is
sj-CO
given that
6.9
cos 3f dt
a<0— )
= as'fl)— )
+ 2as'(0 = «(0 + *~* «(0
= for all the equations.
Using Laplace transforms, solve the following
sets
equations:
(«)
2*'(0
x'(0
(b)
2x'{t)
+ 4a<0 + y'(f) + 2y(t) = <J(0
+ 3a</) + y'(f) + 2y(r) =
+ «(/) + y'{t) + 4jK/) = 2e-«
+ jr*0) +
The initial conditions are all zero at t = 0—
*'(0
8y(0
=
(5(0
of simultaneous
172
6.10
Network
anal/sis
and synthesis
Find the inverse transforms for
H0-
(«)
+ 9X' +
+a
(s*
j
(»)
3)
s+P
(*+!)»
(c)
S*
+
s
l
s*+2s +2
tf>
+ 3* + 1
s* +s
s + 5
~(s + W* + 2)»
*»
to
F(s)-
(/)
ns)
(g)
F(s)
(A)
F(s)~
.
s*
6.11
+ lX*+2)*
3j» - J* - 3s + 2
(*
sKs-W
Find the inverse transforms for
+«-*'
1
(«)
sH.s+2)
_ «r-s«
+ 3s + 2
->*
(ft)
F(')
=
(c)
*"(*)
=
se
**
re-"
s
+0
write
6.12 Given the pole-zero diagrams shown in (a) and (b) of the figure,
What
can
polynomials.
of
quotients
Fjs)
as
and
(s)
the rational functions t
you say about the relationship between the poles in the right-half plane and the
signs of the coefficients of the denominator polynomials?
F
FiM
ju
-A
A
>ij0.6
1
Fzb)
*
>>0.6
1
-1
+1
H-/D.6
1
|->D.6
-A
-A
(«
to
PROS. 6.12
*
*
—
The Laplace transform
173
6.13 For the pole-zero plots in Prob. 6.12, find the residues of the poles by
the vector method.
6.14
For the pole-zero
plots
shown
in the figure, find the residues of the
poles.
-2
j2
:
jl<
<
-1
-jln
-J2
(a)
(b)
ja
f-H n
-<r-3
-2
-x-
-1
-1
+1
M
W
PROB. «. 14
6.15 Two response transforms, R^s) and R/s), have the pole-zero plots
shown in (a) and (b) of the figures, respectively. In addition, it is known that
»i(0 +) - 4 and /-,(/) - 4 for very large t. Find r^t) and rjf).
Sd»)
ja
—
n
X
I
1
-V -2
-l
a
6
-1?
i:
'
o
i
x
(a)
-n
(b)
PROB. 4.15
Network
174
6.16
It is
analysis
and synthesis
given that
_.
F(s)
as*
=
+ bs + c
+ 2j* +s +
s*
1
and c such that/(0 +) =/'(0 +) = /'(0 +) = 1
Using the initial and final value theorems where they apply, find/(0 +)
and/(oo) for
Find
a, b,
6.17
(6)
F(*)
c»
+
3X*
(*
+
IX*
s
j
(c)
-
1)
s* + 3s + 2
FW - «* + 3* + 3j +
2
5<s
<«
+ 4)
+ 4X* +
8)
^-"cr+ixr+Q
1
7
chapter
Transform methods
network
7.1
in
analysis
THE TRANSFORMED CIRCUIT
In Chapter 5 we discussed the voltage-current relationships of network
elements in the time domain. These basic relationships may also be
represented in the complex-frequency domain. Ideal energy sources, for
example, which were given in time domain as v(t) and i(f)> may now be
represented by their transforms V(s)
resistor, defined
by the
=
»(0
is
= £[t<01
and
I(s)
= Cfl(f)].
*«(')
(
denned in the frequency domain by the transform of Eq.
=
V(s)
For an inductor, the
The
v-i relationship
RI(s)
7.1,
71 )
or
(7.2)
defining v-i relationships are
w- !
1
(7.3)
»(0
= 7p
Transforming both equations,
F(s)
/(5)
we
t<T)dT-M(0-)
obtain
= sLI(s) - Li(0-)
=
1
J-F(S)
sL
+ ^—;
i(O-)
s
175
CM)
176
Network
analysis and synthesis
IM
IM
1
g
f\HO-)
VM
)i»(0-)
FIG.
7.1. Inductor.
The transformed
Fig. 7.1.
circuit representation for an inductor
For a capacitor, the defining equations are
v(t)
=-
'
i(r)
I
C Jo-
dr
is
depicted in
+ v(0- )
(7.5)
*>- c
5
The frequency domain counterparts of these equations
are then
F« -J; /« + *&=>
sC
I(s)
=
sC
s
V(s)
(7. 6)
- C o(0-)
as depicted in Fig. 7.2.
From this
analysis
we
see that in the complex-frequency representation
and admittances in
For
example,
series or parallel with energy sources.
from Eq. 7.4, we see
that the complex-frequency impedance representation of an inductor is sL,
and its associated admittance is XjsL. Similarly, the impedance of a
capacitor is 1/sC and its admittance is sC. This fact is very useful in circuit
analysis. Working from a transformed circuit diagram, we can write
mesh and node equations on an impedance or admittance basis directly.
the network elements can be represented as impedances
-%
IM
IM
•c
VM
+
=PC (l)
•K0-)
FIG. 7.2. Capacitor.
V<»
Transform methods
in
network
analysis
177
vs(t)
vi(t)
FIG. 7.3
The process of solving network differential equations with the use of
transform methods has been given in Chapter 6. To analyze the circuit
on a transform basis, the only additional step required is to represent all
the network elements in terms of complex impedances or admittances with
associated initial energy sources.
Consider the example of the transformer in Fig. 7.3. If we write the
defining equations of the transformer directly in the time domain, we have
(7.7)
Transforming
this set
of equations, we obtain
V&) = sLy I^s) V&) = sMIM -
+ sM I£s) - M ijp-)
M ^(0-) + sL, Us) - U i,(0-)
Ly i^O-)
(7.8)
This set of transform equations could also have been obtained by representing the circuit in Fig. 7.3 by its transformed equivalent given in Fig. 7.4.
In general, the use of transformed equivalent circuits is considered an
easier
way
to solve the problem.
v2W
FIG. 7.4
1
178
Network
analysis
and synthesis
—
£2
-nnnp
fi,(0-)
+__
+
12
BC (0-)
FIG. 7.5
Example 7.1. In Fig. 7.5, the switch is thrown from position 1 to 2 at t = 0.
Assuming there is no coupling between L^ and Z*, let us write the mesh equations
from the transformed equivalent circuit in Fig. 7.6. The mesh equations are
—
— £-.+.)«
*->•
-l
+ A'i,(0-) = -^/
1(5)
—'m^Qi
—^
«I2
«Li
(7.9)
»
+
—^>—
vuo'
1
1
L1M0-)
1*1140-)
QviW
FIG. 7.6
Example
f
= 0.
7.2.
In Fig. 7.7, the switch
Just before the switch
amp, »o(0— )
= 2 v.
is
is
thrown from position
thrown, the
Let us find the current
initial
i(t)
S
Qv(t)
>'
1
FIG. 7.7
to 2 at time
after the switching action.
lh
-—x
1
conditions are /^(O— )
=2
Transform methods
3
i—vw
^
—
network
179
anal/sis
2
=0
»
nnnp-
f
in
/i«\
FIG. 7.8
Since the switch
is
closed at
formed
= 0, we can regard the S-v battery as an equivacircuit is
The mesh equation
circuit.
Solving for
/
The
lent transformed source 5/s.
I(s),
5
2
s
s
+2
now redrawn
in Fig. 7.8 as
for the circuit in Fig. 7.8
(3
a
trans-
is
+s+ ?) 70)
C7.10)
we have
+3
1
(7.11)
+ IX* + 2) s + 1 s + 2
i(0 = C-HAO] = e~* + «-»
(7.12)
Consider the network in Fig. 7.9. At t = 0, the switch is opened.
2s
(j
Therefore,
Examine 73.
Let us find the node voltages
vjCf)
and
v%(t) for the circuit. It
Z,=Jh
G = 1 mho
C=
V=
is
given that
lf
1
v
L
i<0-)
+
C=fcoc (0-)
FIG. 7.9
Before
we
substitute element values, let us write the
node equations for the
transformed circuit in Fig. 7.10. These are
NodeV1
:
*_
_ (l(0-0 +
Node
Vt
:
CPc(o-)
/
\
= (sc
+ -)ki(o
1
-- vM
^-4™ (£«)•»
1
(7.13)
Network
180
anal/sis and synthesis
V2
•G
FIG. 7.10
If
values into the set
=
(t(0— ) =
of node equations, we have
1
had been in steady
amp. Substituting numerical
the switch opening the circuit
we assume that prior to
then we have v^O-)
state,
1 v,
1
(*+^K (i)-?K/f)
-
1
(7.14)
Simplifying these equations,
s
-
we obtain
1
1
= (s* + 2)Vt(s) - 2V&)
. -2Vx{s) + (s + 2)V£s)
Solving these equations simultaneously,
we have
J+l
s +1
+ 2s + 2 " (s + l)1 + 1
* +2
s +2
""
(s + 1)» + 1
s* + 25 + 2
KlW ."
Krf,)
(7.15)
s»
(7.16)
so that the inverse transforms are
Vl(t)
7.2
THEVENIN'S
= e-« cos
/,
»*(/)
- <r*(cos t + sin /)
(7.17)
AND NORTON'S THEOREMS
In network analysis, the objective of a problem is often to determine a
branch current through a given element or & single voltage across
an element. In problems of this kind, it is generally not practicable to
write a complete set of mesh or node equations and to solve a system of
equations for this one current or voltage. It is then convenient to use two
single
very important theorems on equivalent circuits,
and Norton's theorems.
known
as Thevenin's
1
Transform methods
in
network
181
anal/sis
-3X
Networks
n voltage
<S)
sources
Zi(s)
in current sources
FIG.
7.1
Thevenin's theorem
From the standpoint of determining the current I(s) through an element
can be
of impedance Z^s), shown in Fig. 7.1 1, the rest of the network
replaced by an equivalent impedance Ze(s) in series with an equivalent
voltage source V£s), as depicted in Fig. 7.12. The equivalent impedance
Zt(s) is the impedance "looking into" from the terminals of Zt(s) when
all voltage sources in JV are short circuited and all current sources are
open circuited. The equivalent voltage source V,(s) is the voltage which
appears between the terminals 1 and 2 in Fig. 7.1 1, when the element Z^s)
N
N
is
removed or open
theorem is
circuited.
that the elements in
The only requirement
for Thevenin's
Zx must not be magnetically coupled to any
element in N.
The proof follows. The network
current sources.
in Fig. 7.11 contains n voltage
and
m
We are to find the current I(s) through an element that
not magnetically coupled to the rest of the circuit, and whose impedance
1
According to the compensation theorem, we can replace Z^s)
x (s).
by a voltage source V(s), as shown in Fig. 7.13. Then by the superposition
principle, we can think of the current /(*) as the sum of two separate parts
is
is
Z
Let the current It(s) be the current due to the n voltage and
<¥-
m
current
fl
VM
\2M
FIG.
1
7. IX
^
Thevenin's equivalent
Zi(e)
circuit.
Sons,
See H. H. Skilling, Electrical Engineering Circuits, 2nd Ed. John Wiley and
New York,
1965.
182
Network
analysis
and synthesis
we short circuit the source V(s), as shown in Fig. 7.14a.
equal to the short-circuit current 7gc Let IJs) be the
current due to the voltage source V(s) alone, with the rest of the voltage
sources short circuited and current sources open circuited (Fig. 7.146).
current and n voltage
With the
sources alone;
Therefore I^s)
i.e.,
is
.
m
sources removed in Fig. 7.146,
Network
N
it
m
voltage sources
current sources
FIG.
7.
^
that the network
)V(s)
7aC0
is
N
is
we
see
passive so that
related to the source V(s)
by the
relation
««)
13
=-
V(s)
(7.19)
Zm(s)
the input impedance of the circuit at the terminals of the
where Zm(s)
is
source V(s).
We can now write /(*) in Eq.
/(s)
= Ms) -
Since Eq. 7.20 must be satisfied in
7.18 as
(7.20)
Zm(s)
all cases,
consider the particular case
when we open circuit the branch containing Zx(j). Then /($)
is the open-circuit voltage Kocfa). From Eq. 7.20 we have
he(s)
= Foc(s)
Z ln(s)
=
and
V(s)
(7.21)
Transform methods
in
network
analysis
183
FIG. 7.14b
so that
we can
rewrite Eq. 7.20 as
Zla(s)I(s) -
KocCs)
=
V(s)
(7.22)
In order to obtain the current I(s) through Zx (s), the rest of the network
can be replaced by an equivalent source Ve(s)
Voc(s) in series,
with an equivalent impedance Ze (s)
Zm(s), as shown in Fig. 7.12.
N
=
=
Example 7.4.
Let us determine by Thevenin's theorem the current I^s) flowing
through the capacitor in the network shown in Fig. 7.1S. First, let us obtain
LH0-)
'c<°-)
FIG. 7.15
by opening all current sources and short circuiting
Then, we have in the network in Fig. 7.16, where
Z,(s)
Z,(s)
Next we find
Ve(s)
=R+sL
all
voltage sources.
(7.23)
by removing the capacitor so that the open-circuit voltage
1 and 2 is V,(s), as shown in Fig. 7.17.
We readily
between the terminals
184
Network
analysis
and synthesis
-rnnp-
>i
u
—o2
FIG. 7.16
determine from Fig. 7.17 that
F,Cs)
-/(*)*+ Z/(0-)-
M0-)
(7.24)
By Thevenin's theorem we then have
/l(j)
VJto
~ Z.{s) + Z(s) =
7(j)Jt
+ Lt(0
-) -Pq(0-)/s
(7.25)
A+sL+ 1/sC
FIG. 7.17
7.5. For the network in Fig. 7.18, let us determine the voltage vjt)
0, and we
across the resistor by Thevenin's theorem. The switch closes at t
0.
assume that all initial conditions are zero at t
Fvampk
-
=
8
Lj
Li
=£c
Qv(t)
itftf
FIG. 7.IS
terms of its transformed representation,
almost determine by inspection that the
can
which is given in Fig. 7.19. We
network to the left of AT in Fig. 7.19 is
the
of
source
voltage
Thevenin equivalent
First let
us redraw the
circuit in
VM
LiS
+ \jsC
(7.26)
I
Transform methods
Lis
La»
in
network
analysis
185
*—
* Vo«
FIG. 7.19
and the input impedance to the
left
of AT is
slM/sC)
(7.27)
V&)
We know that
Therefore
V&)
"
ZJis)
RVisYMsC)
(R + sL^sLi + lIsC)
Finally
=
»o(0
(7.28)
+R
+ LJC
(7.29)
(7-30)
e-HKtCi)]
Norton's theorem
When
tance
is
required to find the voltage across an element whose admitYifs), the rest of the network can be represented as an equivalent
it is
admittance Y£s) in parallel with an equivalent current source I,(s), as
shown in Fig. 7.20. The admittance Yt(s) is the reciprocal of the Thevenin
impedance. The current I,(s) is that current which flows through a short
circuit across Y^s).
From
Fig. 7.20,
Vx{s) =
The element whose admittance
is
/.(*)
(7.31)
Yt(s) + YM
Yt must not be magnetically coupled to
any element in the rest of the network.
Consider the network in Fig. 7.21. Let us find the voltage across the
capacitor by Norton's theorem. First, the short-circuit current source
Ie (s) is found by placing a short circuit across the terminals 1 and 2 of the
Vi
—91
(£)«•>
Yds)
FIG. 7.20
Yito
Network
186
anal/sis
and synthesis
-nppp>
Pi
.1
•«c
IM(
<P
'
FIG. 7.22
FIG. 7.21
capacitor, as
shown
From
in Fig. 7.22.
e
The admittance
Fig. 7.22 l£s)
is
(7.32)
sL+R
Y,(s) is the reciprocal
'
Then
'««)
w
of the Thevenin impedance, or
(7.33)
sL+R
the voltage across the capacitor can be given as
V(s)
Example 7.6.
that v(t)
=
Yt(s) + Yds)
(sL
+ R)sC +
(7.34)
1
In the network in Fig. 7.23, the switch closes at / = 0. It is given
' and all initial currents and voltages are zero.
Let us find the
" O.le-6
/g(/) by Norton's theorem.
The transformed circuit is given in Fig. 7.24. To find the Norton equivalent
current source, we short circuit points 1 and 2 in the network shown. Then
current
Ie(s)
is
the current flowing in the short circuit, or
Us) =
0.1
V(s)
Rt +
The equivalent admittance of the
Y,(s)
J?!
circuit as
+sL
(s
5)
+
S)(s
+
viewed from points
+ sRxC +
R± +sL
s*LC
1
- sC +
1
Us + RILXs +
sL
o-iTJoTP-wv
c=J=t
FIG. 7.23
1
0.5s*
1
(7.35)
10)
and 2
+ 5s +
* + 10
is
10
(7.36)
Transform methods
in
network
analysis
187
FIG. 7.24
I^s)
is
IM
then
+ GJ
(7.37)
+ 6)
(7.38)
/<(')
BJLYJto
1
(S
+
sy(s
By inspection, we see that /«(«) can be written
+ 6) - (s + 5)
(s + 5)V +6)
1
(*
/•(*)
as
(s
+ 5)*
(*
+ 6X* +
(7.39)
5)
Repeating this procedure, we then obtain
1
1
//*)
(*+5)»
Taking the inverse transform of IJs), we
j+5 + j+6
finally
(7.40)
obtain
«0 = ('«"" - «"" + e"*')«(0
7.3
(7.41)
THE SYSTEM FUNCTION
As we
discussed earlier, a linear system
e(t) is related to
the response
tit)
by a
is
one in which the excitation
linear differential equation.
When
the Laplace transform is used in describing the system, the relation between
the excitation E(s) and the response R(s) is an algebraic one. In particular,
when we
related
discuss initially inert systems, the excitation
by the system function H(s) as given the
R(s)
We
how
= E(s)H(s)
and response are
relation
(7.42)
a system function is obtained for a given network,
can be used in determining the system response.
As mentioned in Chapter 1, the system function may assume many forms
and may have special names such as driving-point admittance, transfer
will discuss
and how
this function
188
Network
analysis
and synthesis
T
1
'R
'«w(D
^
a?-r
FIG. 7.25
This is because the
impedance, voltage or current-ratio transfer function.
voltage
form of the system function depends on whether the excitation is a
or
current
specified
or current source, and whether the response is a
functions.
We now discuss some specific forms of system
voltage.
Impedance
a current source and the response is a voltage,
excitation and
then the system function is an impedance. When both
we have a
terminals,
pair
of
response are measured between the same
is
impedance
driving-point
impedance. An example of a
When
the excitation
is
driving-point
given in Fig. 7.25, where
m = m = R+
HKS)
I,(s)
«I?22L
(7.43)
sL+l/sC
Admittance
a voltage source and the response is a current,
admittance IJV, is
H(s) is an admittance. In Fig. 7.26, the transfer
obtained from the network as
When
the excitation
is
H(s)=
m=
——
l_
(7.44)
Voltage-ratio transfer function
When the excitation is a voltage source and the response is also a voltage,
the voltagethen H(s) is a voltage-ratio transfer function. In Fig. 7.27,
ratio transfer function
V (s)lVt(s) is obtained as follows. We first find the
sL
i—nnnp
*°t
— —wvfit
1(
T
Si-
J2W
^=r
FIG. 7.26
Transform methods
in
ZiM
)yt
m
M
Z*8)
network
anal/sis
189
T
1
FIG. 7.27
current
Va(s)
I{s)sa
Z {s)+Zt(s)
VJs) = Z&)I(s)
(7.45)
x
Since
V (s)
Vg(s)
then
(7.46)
ZJs)
ZM + Z
(7.47)
t(s)
Current-ratio transfer function
the excitation is a current source and the response is another
current in the network, then H(s) is called a current-ratio transfer function.
As an example, let us find the ratio IJIa for the network given in Fig. 7.28.
Referring to the depicted network, we know that
When
I„(s)
= IM + /„(s),
Eliminating the variable Ilt
(7.48, 7.49)
sC
we find
(7.50)
so that the current-ratio transfer function
Jofr)
_
I„(s)
1/sC
R + sL+1/sC
h(s)
7'
W Q)
is
4=ic
FIG. 7.28
(7.51)
Network
190
anal/sis
and synthesis
FIG. 7.29
From the preceding examples, we have seen that the system function is a
the
function of the elements of the network alone, and is obtained from
let
us
network by a straightforward application of Kirchhoff's laws. Now
system
the
obtain the response transform R(s), given the excitation and
the
function. Consider the network in Fig. 7.29, where the excitation is
that
assume
We
current source i„(t) and the response is the voltage v(t).
0. Let us find
the network is initially inert when the switch is closed at t
=
the response V(s) for the excitations:
1-
i„(t)
= (sin a>ot)u(i).
2.
i„(0
is
3.
i,(0 has the
First,
we
the square pulse in Fig. 7.30.
waveform
in Fig. 7.31.
obtain the system function as
s
1
H(S)
sC
1.
it (t)
- (sin ou
+
1/sL
f) u(t).
+G
C[s
8
+
(7.52)
s(G/C)
The transform of i„(t)
+
1/LC]
is
(7.53)
so that
V(s)
<Oo
= /„(s)H(s) =
s*
+
a> »
C[s*
+
s(GIC)
+
(7.54)
1/LC]
ig(t)
ig(t)
2
1
A
a
<i
FIG. 7.30
t
t
FIG. 7.31
)
Transform methods
For the square pulse
2.
in Fig. 7.30,
i„(0
transform
Its
is
in
network
anal/sis
can be written as
ig {t)
= u(t) - u(t - a)
= i (i _
j,(s)
191
(7.55)
e-«)
(7.56)
s
The response
V(s) is therefore given as
=
V(s)
—+
C[s*
—
1
p-0*
-
s(G/C)
+
(7
57);
l
1/LC]
Note that in obtaining the inverse transform C-1 [F(j)], the factor e-°*
must be regarded as only a delay factor in the time domain. Suppose we
rewrite V(s) in Eq. 7.57 as
K(s)-^(l-0
(7.58)
s
Then
if
we denote
the inverse transform by v^t),
we
r{^]
=
*<0
(7.59)
obtain the time response
-
v(t) <= Vl (t)
Observe that »x(0 in Eq. 7.59
is
Vl (t
- a) u{t - a)
(7.60)
the response of the system to a unit step
excitation.
The waveform
3.
tf
and
V(s)
its
is
transform
then V(s)
in Fig. 7.31 can
=
we denote by
v,(t)
+
It (s)
(l
as
\
If
^^
=
+———
+ 1 _ Cl) —
+
«(0
- «(0
-
u(t
a
is
=
be represented as
-
a)
(7.61)
a
-
1 1
s \
as
as
3
as J C[s
(7.62)
/
1
s(GIQ
+
the response of the system to a
(7
K
63)
excitation,
we
1/LC]
ramp
'
see that
C-1 [F(s)]
= Vl (t) + - Vi(t) - - v (t a
where vt(t)
is
t
a
the step response in Eq. 7.59.
a) u(t
-
a)
(7.64)
192
Network
Let us
now
response R(s)
analysis and synthesis
discuss
some
further ramifications of the equation for the
= H(s) E(s).
Consider the partial-fraction expansion of
R(s):
=
R(s)
T
S —
i
, v
!s — s
+
By
A<
Si
i
(7.65)
t
where s( represent poles of H(s), and s, represent poles of E(s). Taking
the inverse transform of R(s), we obtain
rW^ZA^ + ZB^"*
(7.66)
The terms Af?** are associated with the system H(s) and are called free
response terms. The terms B^'* are due to the excitation and are known
as forced response terms. The frequencies st are the natural frequencies of
the system and j, arc the forced frequencies. It is seen from our discussion
of system stability in Chapter 5 that the natural frequencies of a passive
network have real parts which are zero or negative. In other words, if we
jo> { , then <x< <, 0.
<rt
denote st as st
=
+
initially inert network in Fig. 7.32, the excitation is
Let
us find the response v (t) and determine the free and
i cos t df).
forced response parts of y (r). The system function is
Example
For the
7.7.
va(t) =•
Since V,{s)
K (*)
1/sC
VJis)
R +MsC
V,(s)
is
the response
is
2
+
2\s*
s
+
(7.67)
2
(7.68)
1/
then
W- VM™ =
0.4
(st
+ Ws + 2)
J+2 +
We next obtain the inverse transform
vjt) - -0.4* " + 0.4 cos t + 0.2 sin t
JK>
«-2Q
-WW
C-Jf=i= uoW
FIG. 7.32
0.4*
*»
+ 0.2
+l
(7.69)
(7.70)
1
Transform methods
It is
apparent that the free response
0.4 cos
As a
r
in
network
analysis
193
-0.4e~*, and the forced response
is
is
+0.2 sin/.
us consider the basis of operation
for the R-C differentiator and integrator shown in Figs. 7.33a and 7.336.
We use the Fourier transform in our analysis here, so that the system
function is given as H(ja>), where o> is the ordinary radian frequency
final topic in
variable.
Consider
our discussion,
first
let
the system function of the differentiator in Fig.
7.33c.
VJjw)
_ jaRC
jmRC +
R
_
R+
V,(J(o)
1/jcoC
(7.71)
1
Let us impose the condition that
coRC
«
(7.72)
We have, approximately,
*W» =! jcoRC
Then
the response
(7.73)
VJjm) can be expressed as
(7.74)
Taking the inverse transform of
V^ijco),
we
obtain
nMsiRC-vJtl)
(7.75)
at
Note that the derivation of Eq. 7.75 depends upon the assumption that the
R-C time constant is much less than unity. This is a necessary condition.
Next, for the
function
R-C
integrator in Fig. 7.33ft, the voltage-ratio transfer
is
V/J*)
VJijm)
=
l/ja)C
1/jfoC
+R
1
=
jwCR
+
(7.76)
1
C
hgW
VO<t)
(a)
FIG. 7.33. (a)
(b)
R-C differentiator.
(6)
R-C integrator.
1
Network
94
If
we assume
analysis
that
and synthesis
o>RC )J>
1,
then
VM^T^zVjUa)
jcoRC
Under
these conditions, the inverse transform
v
so that the
7.4
(7.77)
is
(t)~-±-r Va(r)dT
(7.78)
RC Jo
R-C circuit in Fig.
7.336
is
approximately an integrating
circuit.
THE STEP AND IMPULSE RESPONSES
In this section we will show that the impulse response h(t) and the system
we can obtain step and
impulse responses directly from the system function.
We know, first of all, that the transform of a unit impulse (5(f) is unity,
i.e., C[<5(0]
1. Suppose the system excitation were a unit impulse, then
the response R(s) would be
function H(s) constitute a transform pair, so that
=
R(s)
We
= E(s) H(s) = H(s)
(7.79)
thus see that the impulse response h(t) and the system function H(s)
constitute a transform pair, that
is,
= H(S)
Z-^[H(s)] = h(t)
Z[h(t)]
(7.80)
Since the system function is usually easy to obtain, it is apparent that we
can find the impulse response of a system by taking the inverse transform
of H{s).
Example
7.8.
Let us find the impulse response of the current
The system function is
/(/) in
the
R-C
circuit in Fig. 7.34.
~
Simplifying H(s) further,
VM ~ R + 1/*C ~ R(s + 1/RO
we have
^^O-JTw)
The impulse response
Ht)
which
is
shown
is
=
(7 ' 81)
(7 - 82>
then
C-i[/f(5)]
in Fig. 7.35.
= i ^r) - -L e-wj
„(,)
(7.83)
Transform methods
network
in
analysis
195
i(t)
m
)9(t)
/Z-t/RC
_
1
R*C
FIG. 7.34
FIG. 7.35. Current impulse
R-C network in
response of
Fig. 7.34.
Since the step response is the integral of the impulse response, we can
use the integral property of Laplace transforms to obtain the step response
as
•W-rf?]
where
oc(f)
denotes the step response.
Similarly,
(7.84)
we
obtain the unit
ramp
response from the equation
«^-m
(7.85)
where y(t) denotes the ramp response. From this discussion, it is clear
that a knowledge of the system function provides sufficient information to
obtain
all
the transient response data that are needed to characterize
the system.
Example
7.36.
7.9.
Let us find the current step response of the R-L circuit in Fig.
is the response and V (s) is the excitation, the system function
g
Since I(s)
is
V„{s)
Therefore H(s)/s
is
h{s)
s
The
(7.86)
R+sL
step response </) is
=
n_ i \
=i
s(R+sL) R\s
s + RjLJ
now
i
(7.87)
obtained as
1
a(/)=-(l -e-W*X)ii(/)
To
check
this result, let us consider the impulse response of the
(7.88)
R-L
circuit,
Network
196
which we found
analysis
in
and synthesis
Chapter
5.
1
(7.89)
Kt)
The
step response
the integral of the impulse response, or
is
a(0
»</t=-(i -*-w*>0«(i)
(7.90)
-J>
if we know the impulse response of an initially inert
the response of the system due to any other
determine
can
linear system, we
excitation. In other words, the impulse response alone is sufficient to characterize the system from the standpoint of excitation and response.
It is readily
seen that
R
-VSMAK»)
V,M{
FIG. 7.37
FIG. 7.36
system
7.10. In Fig. 7.37, the only information we possess about the
in the black box is: (1) it is an initially inert linear system; (2) when v&) = 6(t),
Example
then
With this information, let us determine what the excitation vjf) must be in order
to produce a response v£f)
- te~" u(f).
First,
we determine the system function
to be
H(s)
=
V^s)
1
Vjs)
7T2
2*4-5
t j+3
We next find the transform of te~* «(*)•
Vjs) - titer"] excitation is then
VAs)
Simplifying
2(a
(7.92)
+3)
(7.93)
+2)*
found from the equation
JW "
H(s)
and expanding V&s)
K<W =
The system
°
+ 2X«
1
<»
The unknown
(j
(s
+ 2Ks + 3)
2.5X* + 2)»
(7.94)
20 +
into partial fractions,
1
(s + 3)
+ 2.5X« + 2) " s + 2
we have
0.5
*
+ 2.5
(7.95)
excitation is then
vtf)-{e-*t -0.5e-*
M)u(t)
(7.96)
Transform methods
7.5
in
network
analysis
197
THE CONVOLUTION INTEGRAL
In this section we will explore some further ramifications of the use of
the impulse response Mj) to determine the system response r(t). Our
discussion is based upon the important convolution theorem of Laplace
(or Fourier) transforms. Given two functions /,(*) and//*), which are
zero for t
0, the convolution theorem states that if the transform of
/
x (f) is
<
Ft(s),
and
if
convolution of/i(f)
WTO,
that
the transform of f£i)
and/t(r)
is
F£s), the transform of the
is,
- t)/,(t) dr] = F1(s) F&)
t[ j*Mt
where the
is
the product of the individual transforms,
is
integral
I
ft(t
(7.97)
— t)/,(t) dr
the convolution integral or folding integral,
and
is
denoted operationally
as
fm -
Let us prove that C[/, */J
Proof.
C|/i(0V^0]
From
t)/^t) dr
J
dt
(7.99)
the definition of the shifted step function
- t) =
1
=
we have the
Then Eq.
- t)/,(t) dr =
r>t
(7.100)
f 7i(«
" t) «(* - t)A(t) dr
(7.101)
Jo
7.99 can be written as
C[/i(0Vi(0]
= f "e- f 7i(* - t) «(t - r)Mr) dr dt
x
(7.102)
Jo
Jo
let
r£t
identity
f/i('
Jo
we
(798)
= F&. We begin by writing
= V"[ J/iO ~ r)Mr) dr]
u(t
If
- fflVMt)
= — r so that
t
'
-.t
e
=
e-^»f,>
(7 103)
Network
198
analysis
and synthesis
then Eq. 7.102 becomes
£[/i(0* /*(')]
=
I*" f"'f1 (x)u(x)fi (T)e-°'e->*dTdz
Jo Jo
=
f 7i(«)
Jo
u(x)e-" dx f 7»(r)e- w dr
Jo
= F (s)F2(s)
(7.104)
1
The separation of the double integral in Eq. 7.104 into a product of two
integrals is based upon a property of integrals known as the separability
property. 2
-2
Example 7.11. Let us evaluate the convolution of the functions/i(f) = e ' u(t)
and fift) = t u(t), and then compare the result with the inverse transform of
F^s) FJs), where
(7.105)
W
The convolution
variable
/
—
of_/i(f)
t for
=
and/2(f)
p
obtained by
is
first
substituting the
dummy
in/i(/), so that
/
fiff
-
T)
=
c-* (
'-T)
u(t
-
(7.106)
t)
Then/iWVaOis
~
f *fi(f
T)/*( T)
=
rfT
Integrating
let
^T = «~a*
T« 2r
dr
(7.107)
|
Jo
by parts, we obtain
fi(0* fiff)
Next
_2
r« " _T)
I
Jo
Jo
=
({-\+\ *-*) «W
us evaluate the inverse transform of F^s) F£s).
(7-108)
From Eq.
7.105,
we
have
C-Vito *a(*)l = (^ - ^ + \ e~*)
so that
An
important property of the convolution integral
equation
„
.
t
flit
•
See, for example, P. Franklin,
New
York, 1940.
Jo
A
Treatise
is
(
71 10)
expressed by the
t
- r)/a(r) dr =
Jo
Sons,
"(')
MMA -
r)
dr
(7.111)
on Advanced Calculus, John Wiley and
/
Transform methods
This
is
readily seen
from the relationships
network
in
W
= *i(')
CWOViC)] - Wi(«)
CC/i(0V*«]
and
199
analysis
(7-112)
(7113)
To give the convolution integral a more intuitive meaning, let us examine
the convolution or folding process
from a graphical standpoint. Suppose
we
JA(t)
=
Mr)
= t «(t)
take the functions
shown
as
- ,
and
in Figs. 7.38a
obtaining the integral
*
7.39a.
,
.
m(t)
(7.114)
In Fig. 7.38, the various steps for
.,
A('-t)/2(t)«V
Jo
= «(— t).
are depicted. Part (b) of the figure shows /i(— t)
ft(t
in part (c) merely advances
product
A(|
is
shown in part (d).
_
-
- u(t - t)
t)
/x (—t)
.
t)/
(t)
The
_
(7.115)
by a variable amount
t.
_ t)t u(t)
<f
function
Next, the
(?
n6)
We see that the convolution integral is the area under
the curve, as indicated by the cross-hatched area in part
convolution integral has a variable upper limit,
under the curve of/i(f
T)ft(r) for all t. With
Fig. 7.38a", the area under the curve is
—
AV.-/(0 =
we must
t
(</).
Since the
obtain the area
considered a variable in
(7.117)
!J«(0
as plotted in part (e) of the figure.
In Fig. 7.39,
same
we
result as in
taking the inverse
by folding g(r) about a point t, we obtain the
Fig. 7.38. The result in Eq. 7.117 can be checked by
transform of F^s) Fjfa), which is seen to be
see that
c- 1 [F1 (s)f»W]
=
=
£-1
l3j
m(o
(7118)
2
Let us next examine the role of the convolution integral in system
From the familiar equation
analysis.
R(s)
we
= E(s) H(s)
(7.119)
obtain the time response as
r(t)
where e(r)
is
=
£"*[£(«) H(s)]
the excitation
and
=
('e(r) h(t
- t) dr
(7.120)
k(r) is the impulse response of the system.
T
200
Network
analysis
and synthesis
Mr)
Mr)
1
T
/
/
T
(a)
(a)
\M-t)
M-r)
1
—
r
T
(b)
M-t)
Mt-r)
1
1
—T
t
T
t
T
(e)
hM M-t}
M-r) Mr)
—
t
T
t
T
id)
ffh
«
M
FIG. 7.39
FIG. 7.38
Using Eq. 7.120, we obtain the response of a system directly in the time
domain. The only information we need about the system is its impulse
response.
EVro-pW. 7.12a.
Let us find the response /(f) of the R-L network in Fig. 7.40 due
to the excitation
c(t)
- 2<r* y(f)
The impulse response for the current
is
.AW-|*-«ra«BW
Therefore, for the
R-L
circuit
(7.121)
(7.122)
under discussion
Kt) «*-"«(*)
(7-123)
Transform methods
in
network
Using the convolution integral, we obtain the response
Kt)
-
J
V~
t)
Kr) dr
i(i)
201
analysis
as
- 2j V"«-rtr* </r
(7.124)
- 2«-< f V"*</t - 2(«-* - *-*)«(0
VvWvV-
W
0*"
FIG. 7.41
PIG. 7.40
The ideal amplifier in Fig. 7.41 has a system function H(s) — K,
s 7.12b.
where Kis a constant. The impulse response of the ideal amplifier is then
Kt)-Kd(t)
(7.125)
Let us show by means of the convolution integral that the response r{t)
to the excitation e(t) by the equation
K0-JT4)
Using the convolution
K0
integral,
is
related
(7.126)
we have
«(t)*('-t)«/t-a:
r)dr-Ke(f)
(7.127)
r-
a variable in the expression for r(t), we see that the ideal amplifier in
is an impulse-scanning device which scans the input «(/) from
domain
the time
/ —
to f — oo. Thus, the response of an ideal amplifier is a replica of the
input e(f) multiplied by ibcgain IT of the amplifier.
Since
7.«
/ is
THE DUHAMEL SUPERPOSITION INTEGRAL
In the Section 7.5, we discussed the role that the impulse response plays
in determining the response of a system to an arbitrary excitation. In
this section we will study the Duhamel superposition integral, which also
describes
an input-output
relationship for a system.
The superposition
integral requires the step response a(f) to characterize the system behavior.
plan to derive the superposition integral in two different ways.
We
Network
202
The
simplest
is
analysis
and s/nthesis
examined
We
first.
relationship
R(s)
begin with the excitation-response
= E(s) H(s)
(7.128)
Multiplying and dividing by s gives
R(s)
=
^
•
E(s)
s
(7.129)
5
Taking inverse transforms of both sides gives
C- 1 [«(5)]
=
C-1
)
[^- -s£(s)]
(7.130)
= £-ip£)l* t-i [s£(s)]
which then
yields
K0 =
a(0*[e'(0
+ *(0-)d(0]
= e(0-) a(f) +
where
and
e'(t) is
<x.(t)
is
the derivative of
e'(r) a(f
e(0— )
7.13.
Duhamel superposition
Let us find the current
the voltage source
- t) dr
(7.131)
the value of e(i) at
is
Equation 7.131
the step response of the system.
referred to as the
Example
e(t),
'
J
/
is
=
0—,
usually
integral.
i(t) in
the
R-C circuit
in Fig. 7.42
when
is
(7.132)
as
shown
The system function of the R-C circuit
in Fig. 7.43,
H(s)
Us)
=
s
V„(s)
R(s
Therefore the transform of the step response
M£)
s
is
+
(7.133)
IjRC)
is
1
R(s
+
(7.134)
l/RC)
vg (t)
»*WI
M
m
Slope
a A2
zizc
FIG. 7.42
FIG. 7.43
Transform methods
in
network
analysis
203
Taking the inverse transform of H(s)/s, we obtain the step response
1
~-tlRC
<t)
From the excitation in Eq.
is
see that e(0
(7.135)
— ) =0 and
= A 1 d(f) + A t u{t)
v'^i)
The response
we
7.132,
«(0
(7.136)
then
i(0
-jTj'V,(r)a(t -r)dr
-
A x Hr)«(t
<5(t) o(/ -
j
t) dr
r)dr
+
\
At <x(f -
r)dr
Jo-
Jo-
e-Ur-r)IRC fc
ic
/{
(7.137)
Jo_
- [J e-"* + /<* C(l - e-**
)] u(t)
from the Example 7.13, e(0— ) = 0, which is the case in many
Another method of deriving the Duhamel integral
avoids the problem of discontinuities at the origin by assuming the lower
limit of integration to be t = 0+. Consider the excitation e(t) shown by
the dotted curve in Fig. 7.44. Let us approximate e(t) by a series of step
As we
see
transient problems.
functions, as indicated in the figure.
We can write the
staircase approxi-
mation of e(0 as
e(t)
= e(0+) h(0 + &E1 u(t- At)
+ AE, - 2At) +
i/(f
•W
•
•
•
+ A£n u(t - /iAt)
'11
Signal
}a£3
s
/
-A£2
hABi
«<P+>
I
0h-AT-»|
2At
_L
3At
4At
FIG. 7.44. Staircase approximation to a
t
signal.
(7.138)
Network
204
analysis
and synthesis
where AE^ is the height of the step increment at
t
«=
*At. Since we assume
the system to be linear and time invariant, we know that if the response to a
t). Therer) is X< a(f
unit step is <x(f), the response to a step K+ u(t
fore we can write the response to the step approximation in Fig. 7.44 as
—
K0 =
If
At is
K0 -
e(0+) a(0
small, r(t)
+ AEX a(/ - At)
+ A£^a(r — 2At) +
•
•
—
+ A£„a(f - nAf)
•
(7.139)
can be given as
e(0+) o(0
+
—At
AE
* oc(*
- At) At
(7.140)
+ *h «(, - 2At)At +
•
•
+ ^aO -
•
«At)At
At
At
which, in the limit, becomes
AT
n
r(t)
-
«(0+) «(0
lim 2
+ Ar-»0<-©+
~
tV
AT
'
At) At
1»-»CO
(7.141)
= e(0+) a(0 + f e'(r) «(« - t) dr
Problems
In the circuit shown
sketch^;)7.1
v(t)
.- 2«(f)
and /£(0-)
-2
amps.
Find and
100
r
VWfe«|
I
vwQ
1000
5h
PROB.7.1
after having been at
7.2 The switch is thrown from position 1 to 2 at t
t
1e~' sin fit.
for a long time. The source voltage is vjif)
1
»V
(a)
Find the transform of the output voltage
(b)
Find the
initial
vtf).
and final values of vjt)-
plane
Sketch one possible set of locations for the critical frequencies in the *
transform.)
and write tint form of the response v&). (Do not take the inverse
(c)
Transform methods
in
network
anal/sis
205
MtOB. 7.2
13
/(f)
for
In the circuit shown, all initial currents and voltages are zero.
using Thevenin's theorem.
t >
Find
51(0
PROS. 7.3
7.4 In the circuit shown in (a), the excitation is the voltage source e{t)
described in (6). Determine the response i(t) assuming zero initial conditions.
e(t)
10
*t)
o
'S*
A
4=i*
1/2
t
-1
W
M
PROS. 7.4
13
Determine the expression for vji) when /(/)
Use transform methods.
conditions.
— Hi), assuming zero initial
Network
206
anal/sis
and synthesis
PROB. 7.5
7.6
The
opened.
0.5 sin
circuit
shown has zero
Find the value of the
Vlt
n(0.
The excitation
initial
resistor
is i(t)
=
energy. At t =
the switch S
that the response is v(t)
X such
te~ **
is
=
u(t).
m
i
v(t)
=r*
PROB. 7.6
7.7 Use transform methods to determine the expressions for i^t) and /j(/)
-10 '.
Assume zero initial
lOOe
in the circuit shown. The excitation is v(f)
energy.
=
1000
K=\
PROB. 7.7
circuit shown, the switch S is opened at / = 0. Use Thevenin's
to determine the output voltage t^f). Assume zero initial
theorem
or Norton's
7.8
energy.
For the
Transform methods
in
network
207
analysis
PROB. 7.8
7.9
6u(0>
For the transformer shown, find i^t) and «,(/). It is given that e(t) =
and that prior to the switching action all initial energy was zero; also
M«=1A.
PROB. 7.9
7.10
For the circuit shown, find /,(/), given that the circuit had been in steady
state prior to the switch closing at
10Q
WWW
/
=
0.
Kml
I
100
—WW
400
1
vl=
2hojlo8h
«2
)
j=z=_40v
PROB. 7.10
7.11
cos 20
Find
u(t).
/,(/)
using Thevenin's theorem.
Assume zero
initial
energy.
The
excitation is e(t)
=
100
Network
208
anal/sis and synthesis
l
—TVTP-
y/W
2h
10O
•300
•7.50
|6h
PROB. 7.11
7.12
«(/),
Determine the transfer function H(s)
find v t{i).
Assume zero
initial
= V^IV^s). When
v^t)
conditions.
10
r-nAAA-
Y
1
*
if
h
=j=
T
v
*
G) 5V
i»
=r
3f
fOS v3
PROB. 7.12
7.13
find v(t)
Using (a) standard transform methods and
when
/(/)
(b) the
convolution integral,
» 2e~* u(t). Assume zero initial energy.
—mo—
m
lf=t
io:
ih<
PROB. 7.13
7.14
linear system is shown in the figure. If the
determine the response values r(l) and HA,)
The impulse response of a
excitation were
<*/) * 3e"*'«(0,
using graphical convolution.
m
PROB. 7.14
Transform methods
in
network
analysts
209
is given as H(s) = l/(s* + 9)*. If the excitation
determine the response #•(')• (Hint: Use the convolution
integral to break up the response transform R(s).)
7.15
were
The system function
c(/)
7.16
= 3S'(t),
Solve the following integral equations for
(a)
«(0
+
(*)
sinr
=
f(f
-
t)x(t)«/t
x(t).
=
1
a<T)e-<«-f >«/T
J
«(/)
7.17
+j
Using the convolution
x(jt
—
r) e-*
dr
=* t
integral find the inverse transform of
K
(«)
(s
m
+ aXs + b)
2fr+2)
(s*
+ 4)»
(s*
+
s
F(s)
<<0
7.18
"
By
1)«
graphical means, determine the convolution of /(/)
(i.e., determine/(/)*/(/)).
shown
in the
figure with itself
M
T
PROB.7.18
7.19
Using
(a) the convolution integral
integral, find v(t) for e(t)
and
(b) the
Duhamel
superposition
= 4e-**u(t). Assume zero initial conditions.
PROB.7.19
210
7.20
Network
Using
integral, find
analysis
(a) the
«<0 for
and synthesis
convolution integral and (A) the Duhamel superposition
= 2e-s * «(/). Assume zero initial conditions.
e(t)
PRO B. 7JO
=
K(i)//(*) has the
7.21 For the circuit in (a) f the system function H(s)
and C. If the excitation »'(/)
poles shown in (c). Find the element values for
has the form shown in (b), use the convolution integral to find v(t).
R
m
(b)
C-0.5 ju
«W
-«r
-2
-1
to
PROS. 7.21
7.22 The excitation of a linear system is x(r), shown in (a). The system
impulse response is Hf), shown in (ft). Sketch the system response to x(t).
neat, carefully dimensioned sketch will
(No equations need be written.
A
suffice.)
1
Transform methods
in
network
anal/sis
21
(a)
h(t)
t
l
(b)
PROS.
7.23
is /(/)
7.22
A unit step of voltage is applied to the network and the resulting current
= 0.01*^' + 0.02 amps.
(a)
Determine the admittance Y(s) for
this network.
(b)
Find a network that
admittance function.
will yield this
PROS. 7.23
PROB.
/
7.24
7.24 The current generator delivers a constant current of 4 amps. At
the switch 5 is opened and the resulting voltage across the terminals 1, 2 is
=
c(r)
(a)
(b)
=6e-«'
+
12v.
Find Z(s) looking into terminals 1, 2.
Find a network realization for Z(s).
chapter
8
Amplitude, phase, and delay
8.1
AMPLITUDE AND PHASE RESPONSE
In this section we will study the relationship between the poles and
zeros of a system function and its steady-state sinusoidal response. In
other words, we will investigate the effect of pole and zero positions upon
the behavior of H(s) along the jw axis. The steady-state response of a
system function
is
given by the equation
H(j<o)
where
=M {my*™
(8.1)
an
or magnitude response function, and
odd
and
is
an
^(o>) represents the phase response,
is
Af(a>) is the amplitude
even function in o>.
function of to.
The amplitude and phase response of a system provides valuable
information in the analysis and design of transmission circuits. Consider
the amplitude and phase characteristics of a low-pass filter shown in
Figs. 8.1a
and
8.1 A.
The
cutoff frequency of the filter is indicated on the
c. It is generally taken to be the "half-
amplitude response curves as
m
power" frequency at which the system function \H(jt»a)\
is
equal to 0.707
amplitude \H(j(Omu)\. In terms of decibels, the halfpower point is that frequency at which 20 log \H(j(o c)\ is down 3 db from
20 log |#(/a)max)|. The system described by the amplitude and phase
characteristics in Fig. 8.1 shows that the system will not "pass" frequencies
of the
maximum
Suppose we consider a pulse train whose
We know
amplitude spectrum contains significant harmonics above co
block all
will
but
m
below
harmonics
that the system will pass the
c,
distorted
will
be
train
pulse
output
the
harmonics above (oc Therefore
harmonic
higher
many
because
train,
pulse
original
when compared to the
terms will be missing. It will be shown in Chapter 13 that if the phase
that are greater than <o a .
.
.
212
Amplitude, phase, and delay
213
Amplitude
response
0707
-«b
«fc
(a)
Phase response
FIG.
response
8.1.
Amplitude and phase response of low-pass
then
<f>{w) is linear,
from the phase response
filter.
minimum pulse distortion will result.
in Fig. 8.1 that the phase
<f>(a>)
is
We see
approxi-
mately linear over the range —(o c <,w<, +co c If all the significant
harmonic terms are less than coc then the system will produce
minimum phase distortion. With this
example, we see the importance of an
.
,
-VWVAr
amplitude-phase description of a system.
In the remaining part of this chapter,
we
concentrate
will
on methods to
ViW
obtain amplitude and phase response
curves,
£=r
V2 (s)
both analytically and graph-
ically.
To
curves,
obtain
we
let
amplitude
s
=ju>
and
phase
FIG.
84
in the system
and express H(ja>) in polar form. For example, for the amplitude
and phase response of the voltage ratio VJVX of the R-C network shown
function,
in Fig. 8.2, the system function
H(s)
=
is
V^s)
Letting «
ljRC
F^s)
= jot, we see that H(Jco)
H(Jm)
s
+
l/RC
(8.2)
is
l/RC
jm
+
l/RC
(8.3)
214
Network
analysis
and synthesis
M(w)
FIG.
8.3.
Amplitude and phase response of R-C network.
In polar form HQco) becomes
HO) = (to
l/RC
2
+
1/K
--/tan-i
2
C2)*
a>BC_
=
M(ft>)e'*
(0,)
(8.4)
The amplitude and phase curves are plotted in Fig. 8.3. At the point
(o = 0, the amplitude is unity and the phase is zero degrees. As a> increases,
When co *= l/RC, the
—45°. This point is the half-power
Finally as <o -» oo, Af(o>) approaches
the amplitude and phase decrease monotonically.
amplitude
is
0.707 and the phase
is
point of the amplitude response.
zero and ^(g>) approaches —90°.
Now let us turn to a method to obtain the amplitude and phase response
from the pole-zero diagram of a system function. Suppose we have the
system function
H(s)
Ms - Zq)(s -
=
(s
p
)(s
-
Pi)(«
Zi)
- Ps)
^
H(jco) can be written as
HUco)
-
— *,)0«
— *P
Zq)0'<»>gp
0«>
Pi)(J°> - P*)
Po)0«>
jco — z or jco — p corresponds
(8.5)
ApO'cu
=
(g6)
to a vector
Each one of the factors
t
}
from the zero zi or pole/*, directed to any point jco on the imaginary axis.
Therefore,
if
we
express the factors in polar form,
jco
- z, =
rV»",
jco- p,=
M,e'*'
(8.7)
i(y><>+v>i-a»-»i-ti)
(g 8 )
then H(j(o) can be given as
u
jj(jca)
as
shown
in Fig. 8.4,
=
j4 °
N°Nl
e
MoMiMj
where we note that
0! is negative.
Amplitude, phase, and delay
FIG.
8.4.
In general,
215
Evaluation of amplitude and phase from pole-zero diagram.
we can express the amplitude response
M(co) in terms of the
following equation.
n
H
M(.) = ^-
vector magnitudes from the zeros to the point
XI vector magnitudes from
Similarly, the phase response
is
on
thejco axis
the poles to the point on the/co axis
given as
n
^(eu)
=J
angles of the vectors from the zeros to the/co axis
m
—2
It is
angles of the vectors from the poles to thejco axis
important to note that these relationships for amplitude and phase
are point-by-point relationships only. In other words, we must draw
vectors from the poles and zeros to every point on thejco axis for which we
wish to determine amplitude and phase. Consider the following example.
F(s)
=5*
4s
4s
+ 2s + 2
(s
+ H-jl)(s+l-jl)
(8.9)
Let us find the amplitude and phase for FQ2). From the poles and zeros
2, as shown in Fig. 8.5. From
of F(s), we draw vectors to the point to
=
216
Network
anal/sis
and synthesis
m
J2
V5/
——
/\45*
x
>
n
/
i
Vio7
>•
/
i
-l
a
/
——
/\71.8*
x
FIG.
8.5.
-n
1
.
Evaluation of amplitude and phase from pole-zero diagram,
the pole-zero diagram,
it is
MQ2)
and
2
= 4(
,_
,- )
=
= 90° - 45° - 71.8° =
<f>(j2)
With the values M(j2) and
four additional points,
clear that
<f>Q2)
1.78
-26.8°
and the amplitude and phase
we have enough
at three or
information for a rough estimate
JU
X
r
\M5'
-
yi
XX90*
-s.90*
2
1.0
a
'
x
/S
—1A5 —
-yi
(a)
FIG.
8.6.
—
J
I
Br.
-j»
(6)
Determining amplitude and phase at zero and very high frequencies.
Amplitude, phase, and delay
217
=
=
of the amplitude and phase response. At w
0, we see that the vector
magnitude from the zero at the origin to to
0, is of course, zero.
0. From Eq. 8.9 for F(s), F(j0) is
Consequently, M(j0)
=
lim F(ja>)
4Q0)
=
(8.10)
(l+Jlxl-Jl)
o>>0
this equation, we see that the zero at the origin still contributes a
90° phase shift even though the vector magnitude is zero. From Fig.
45°
45°
90°.
90°
is #0)
8.6a we see that the net phase at <w
From
approximately equal to
coJfii,0°,
,
where
Extending
y>
=
+
all
1,
the vectors are
Then
~ 4eoh _
ca k
#a>»)
toh
as seen in Fig. 8.66.
M(w»)
and
-
=
=
mh
Next, at a very high frequency
<»»
~ 90° - 90° - 90° =
-90°
this analysis for the frequencies listed in
Table
8.1,
we
obtain
and phase as given in the table. From this table we
sketch
the
amplitude
and phase curves shown in Fig. 8.7.
can
Next let us examine the effect of poles and zeros on the jto axis upon
values for amplitude
frequency response. Consider the function
F(s)
= s* +
s* +
1.03
1.23
_ (s + jl.015)(s - jlMS)
~ (s+ ;1.109Xs - jl.109)
(8.11)
^
/
Z
Amplitude
/
/
10
+90
0123456789
—
-90
Frequency,
w
»-
FIG. C7. Amplitude and phase response for F(s)
in
Eq.
8.9.
10
218
Network
analysis and synthesis
TABLE
Frequency,
8.1
Amplitude
CO
Phase, degrees
1.0
1.8
25.8
1.5
2.0
-5.3
-50.0
-66.0
-78.5
3.0
1.3
5.0
0.8
10.0
0.4
whose pole-zero diagram is shown in Fig. 8.8. At co <= 1.015, the vector
from zero to that frequency is of zero magnitude. Therefore at a zero
on the /o> axis, the amplitude response is zero. At co = 1.109 the vector
from the pole to that frequency is of zero magnitude. The amplitude
response is therefore infinite at a pole as seen from Eq. 8.11. Next,
consider
i
<
the
phase response. When
apparent from the pole-
JU
co
y'1.109
zero plot that the phase
co
<J./1.015
>
1.015,
1.015
it is
and o><
from the zero
at
co
is
When
zero.
1.109, the vector
=
1.015
is
now
pointing upward, while the vectors from
the other poles and zeros are oriented
O-./1.015
:
in the
same
direction as for co
We see that at a zero on the yw
-y'1.109
<
1.015.
axis, the
phase response has a step discontinuity
of +180° for increasing frequency.
Similarly, at a pole on the jco axis, the
phase response is discontinuous by —180°. These observations are
illustrated by the amplitude and phase plot for F(s) in Eq. 8.11 for the
frequency range 0.9 <; co ^ 1.3, shown in Fig. 8.9.
With a simple extension of these ideas, we see that if we have a zero at
z — — a ± jco t , where a is very small as compared to co t , then we will have
a dip in the amplitude characteristic and a rapid change of phase near
co = co t as seen in Fig. 8.10. Similarly, if there is a pole atp = — a ± jco t
with a very small, then the amplitude will be peaked and the phase will
decrease rapidly near co
co t as seen in Fig. 8.11. A contrasting situation
occurs when we have poles and zeros far away from they'd) axis, i.e., a is
large when compared to the frequency range of interest. Then we see that
these poles and zeros contribute little to the shaping of the amplitude and
phase response curves. Their only effect is to scale up or down the overall
amplitude response.
From stability considerations we know that there must be no poles in
the right half of the s plane. However, transfer functions may have zeros
FIG. 8.8
,
,
=
,
Amplitude, phase, and delay
1.015
FIG.
8.9.
in the right-half plane.
and
8.126.
219
1.109
Amplitude and phase far F(s) in Fig.
8.8.
Consider the pole-zero diagrams in Figs. 8.12a
Both pole-zero configurations have the same poles; the only
difference is that the zeros in (a) are in the left-half plane at s
=—
while the zeros in (b) are the mirror images of the zeros in
(a),
1
±]1,
and are
= +1 ±jl. Observe that the amplitude responses of the
two configurations are the same because the lengths of the vectors correspond for both situations. We see that the absolute magnitude of the
located at s
5.
<E
NX"
E
£
FIG. 8.10. Effect of zero very near
FIG.
the/co axis.
the/o> axis.
8.1 1. Effect
of pole very near
220
Network
analysis
and synthesis
JW
X
J2
o
-
>2
yi
1
-n
X
-y'2
o
-J2
(a)
(a)
<r
1
o-;i -
FIG. 8.12.
o
1
-2
-l
X
-
fl
a
l
1
-2
X
(b)
Minimum phase function,
(b)
Nonminimum phase
function.
phase of (b) is greater than, the phase of (a) for all frequencies. This is
because the zeros in the right-half plane contribute more phase shift
(on an absolute magnitude basis) than their counterparts in the left-half
system
plane. From this reasoning, we have the following definitions.
function with zeros in the left-half plane, or on the jco axis only, is called a
minimum phase function. If the function has one or more zeros in the
A
a nonminimum phase function. In Fig. 8.13, we see
the phase responses of the minimum and nonminimum phase functions
right-half plane,
in Figs. 8.12a
it is
and
^
8.126.
Let us next consider the pole-zero diagram in Fig. 8.14. Observe that
the zeros in the right-half plane are mirror images of the poles in the
100
y«- Minim urn phase
l-ioo
1„
^^/" Nonminimum phase
— YY)
-360
10
12
FIG. 8.13. Comparison of minimum and nonminimum phase functions.
Amplitude, phase, and delay
left-half plane.
Consequently, the vector
drawn from a pole to any point
co x
JO>
on
>2
they a> axis is identical in magnitude with
the vector drawn from its mirror image
apparent that the amplitude
response must be constant for all frequencies. The phase response, however, is
anything but constant, as seen from the
amplitude and phase response curves
to
<»!.
221
*
Kml
~
n
O
It is
given in Fig. 8. 1 5 for the pole-zero config-
1
l
-2 -i
1
x-yi
o
FIG. 8.14. All-pass function.
uration in Fig. 8.14.
A system function whose poles are only in the left-half plane and whose
zeros are mirror images of the poles about the jm axis
function.
The networks which have
is
called
an
all-pass
all-pass response characteristics are
often used to correct for phase distortion in a transmission system.
8.2
BODE PLOTS
In this section we will turn our attention to semilogarithmic plots of
amplitude and phase versus frequency. These plots are commonly known
as
Bode plots. Consider the system function
H(s)
=
Ms)
(8.12)
D(s)
1.0
1
1
1
1
2
4
6
8
u
—
1
10
*-
+ 180
l -I*
-360
10
FIG. 8.15. Amplitude and phase of all-pass function in Fig.
12
8.14.
Network
222
analysis
and synthesis
W
Ka
>l
3
S
togu-
K2
l*KI
Magnitude
(a)
K positive
Phase
K negative
e-
—IT
logw(b)
FIG.
We know that the
Magnitude and phase of constant.
8. 16.
amplitude response
M(«)
-
is
|HC/»)I
- j^}
(8.13)
\D(J°>)\
If
we
express the amplitude in terms of decibels,
20 log M(w)
= 20 log
\N(jco)\
In factored form both N(s) and D(s) are
- 20 log
|
D(jco)\
(8.14)
made up of four kinds of factors:
K
(a)
a constant,
(b)
a root at the origin, s
(c)
a simple real root, s
(d) a
we have
+
a
complex pair of roots, s2
+ 2<w +
a*
+
/?*
To understand the nature of log-amplitude plots, we need only examine the
amplitude response of the four kinds of factors just cited. If these factors
are in the numerator, their magnitudes in decibels carry positive signs.
If these factors belong to the denominator, their magnitudes in decibels
carry negative signs. Let us begin with case (a).
(a)
The factor K. For the constant K, the db loss (or gain)
20 log
K = Kt
is
(8.15)
Amplitude, phase, and delay
223
K
t is either negative if \K\ < 1, or positive if \K\ > 1. The
phase response is either zero or 180° depending on whether is positive or
negative. The Bode plots showing the magnitude and phase of a constant
The constant
K
are given in Fig. 8.16.
The factor
(b)
The
s.
loss (gain) in decibels associated with
±20 log
(zero) at the origin is
Thus the plot of magnitude
to.
versus frequency in semilog coordinates
±20
db/decade or
is
constant for
+
For convenience,
a.
±2Qlog
shown
^
shown
+^=
The phase
in Fig. 8.18a.
Arg
as
is
Bode
plots in Fig. 8.17,
=
at to
1,
we
and the phase
is
all co.
(c) The factor s
magnitude is
as
a straight line with slope of
±6 db/octave. From the
see that the- zero loss point (in decibels)
a pole
in decibels
O+
let
us set a
±2Qlog(o)S
+
=
1.
Then the
, )M
(g
lfi)
is
l)*1
=
±tan~x
a>
(8.17)
in Fig. 8.186.
A straight-line approximation of the actual magnitude versus frequency
curve can be obtained from examining the asymptotic behavior of the
factor y<o
1. For to
1, the low-frequency asymptote is
«
+
20 log
For
to
)»
1,
\j(o
+
\
a<1 s* 20 log
the high-frequency asymptote
201og|7«o
0.1
1|
02
03
0.5
+
1.0
Frequency,
FIG.
8. 17.
=
db
(8.18)
is
20 log
l|
0.7
1
m db
2.0
3.0
(8.19)
5.0
a
Magnitude and phase of pole or zero at s
= 0,
7.0
10.0
224
Network
anal/sis
and synthesis
12
(«+!)->.
2
8"
/
1
2
025
0.5
(ii
—
1
fc-
(b)
FIG. «.!«. Magnitude and phase of simple real pole or zero.
Amplitude, phase, and delay
TABLE
a> =-
Approximation,
Error,
db
db
db
2 octaves below
octave below
break frequency
octave above
2 octaves above
J
=i
o) = 1
m =2
o)
o-4
8.2
Actual
Magnitude,
Frequency
225
Straight-Line
±0.3
±0.3
±1
±3
±1
±3
±7
±6
±12.3
±12
±1
±0.3
we saw in (b), has a slope of 20 db/decade or 6 db/octave. The
1, which we designate
low- and high-frequency asymptotes meet at m
as the breakfrequency or cutofffrequency. The straight-line approximation
is shown by the dashed lines in Fig. 8. 1 8a. Table 8.2 shows the comparison
between the actual magnitude versus the straight-line approximation. We
which, as
=
see that the
maximum
error
is
at the break frequency
co
=
1,
or in un-
=
a.
normalized form: m
For quick estimates of magnitude response, the straight-line approximation is an invaluable visual aid. An important example of the use of
these straight-line approximations is in the design of linear control
systems.
roots. For complex conjugate roots, it is consymbols so that we can use the universal curves
adopt
standard
venient to
(d)
Complex conjugate
We describe the conjugate pole (zero) pair in terms
that result therefrom.
of a magnitude
as
shown
a>
measured from the negative
and an angle
in Fig. 8.19.
Explicitly, the
real axis,
parameters that describe the pole
which we call the undampedfrequency of oscillation,
known as the damping
(zero) positions are a>„
and £
= cos 0,
If
factor.
the
pole (zero) pair
/»
is
given in terms of its real and imaginary
parts,
/>i,.
a and
ft
- -« ±P
are related to £
(8-20)
juo
following:
=
-J"«o
g>
cos
=
to«£
P = <u
sin
=
o),
a
Vw*
and co by the
(8.21)
-/woVW 2
Returning to the definition of the
cos d, we see that
FIG. 8.19. Pole location in terms of
damping factor, £
=
VI
— £*
£ and
<ot .
Network
226
the closer the angle
the angle 6
is
=
1
is
to
7r/2,
Bode
the
plots for the conjugate pole (zero) pair, let us set
The magnitude
for convenience.
±20 log
|1
and the phase
is
—
a)
2
is the damping factor. When
damping factor is nearly unity.
the smaller
nearly zero degrees, the
To examine
«>
and synthesis
analysis
+]2t,a>\
<f>(w)
then
is
= ±20 log [(1 =
a> 8) 8
+ 4£ <u 8]*
8
(8.22)
2gn>
-i
tan"
(8.23)
l-a>8
we examine the low- and high-frequency asymptotes of the magnitude,
we see that the low-frequency asymptote is decibels; the high-frequency
If
asymptote (for <o y> 1) is ±40 log co, which is a straight line of 40
db/decade or 12 db/octave slope. The damping factor £ plays a significant
part in the closeness of the straight-line approximation, however. In
Fig. 8.20 the asymptotic approximation for a pair of conjugate poles
(a> = 1) is indicated by the dashed line. Curves showing the magnitude
=
=
=
We
1.0 are given by the solid lines.
0.6, and £
0.1, £
see
0.6 is the straight-line approximation a close one.
that only for £
for £
~
Universal curves for magnitude and phase are plotted in Figs. 8.21 and
8.22 for the frequency normalized function
G(*)
=
(sK)8
+ 2£(s/a) ) +
(8.24)
1
We see that the phase response, as viewed from a semilog scale, is an odd
function about
eo/a>
1
=
1
.
The phase at co
1
=
<w
is
1
—90° or —n\2 radians.
1
1
+ 10
—
+5
^x^^-r-o.6y
—
-b
~~
Asymptote
-10
-15
0.1
0.5
1.0
Frequency,
^\
vft.
A.
,
1
0.2
—
^\
r-io-^N.
o>
—*
2.0
1
5.0
FIG. 8.20. Magnitude versus frequency for second-order pole.
10.0
Amplitude, phase, and delay
qp'IfflOl-
227
228
Network
anal/sis
and synthesis
woaiv
s
8
CM
1
1
Amplitude, phase, and delay
229
For a conjugate pair of zeros, we need only reverse the signs on the scales
of the magnitude and phase curves.
KVraMpig 8.1. Using Bode plot asymptotes, let us construct the magnitude
versus frequency curve for
0.1s
GW
(8.25)
'(Ht
+
16X10*
10»
+
1
two first-order break frequencies at o> «- and a> — 50. In
addition, there is a second-order break frequency at m — 400. With a quick
calculation we find that £ — 0.2 for the second-order factor. The asymptotes
are shown in Fig. 8.23. The magnitude and phase plots are given in Fig. 8.24
through a microfilm plot computer program.
We see
there are
-12db/oetave
FIG. 8.23. Asymptotes for G(s) in Eq. 8.25.
8.3
SINGLE-TUNED CIRCUITS
We will now study a class of circuits whose system functions can be
described by a pair of conjugate poles. These circuits are called singletuned circuits because they only need two reactive elements an inductor
and a capacitor. The undamped frequency of oscillation of the circuit is
—
then
«w
= {LC)-
l/i
.
circuit in Fig. 8.25,
W=
H(S)
An
example of a single-tuned
whose
!o(i>
V£s)
=
circuit is the
R-L-C
voltage-ratio transfer function is
1/LC
1/sC
R+
sL
+
ljsC
s*
+ (K/L)s +
(8.26)
1/LC
230
Network
anal/sis
and synthesis
FIG. 8.24a. Magnitude of G(s) in Eq. 8.2S.
FIG. 8.246. Phase of G(s) in Eq. 8.2S.
FIG.
8.25. Single-tuned circuit.
Amplitude, phase, and delay
231
FIG. 8.26
The poles of H(s)
are
'
(8.27)
2\LC
2L
where we assume that (R*IL*)
J} J
< (4/LC).
In terms of a and
/J
in Eq. 8.27,
//(5) is
ff(s)
=
+^
(s + a+j/f)(s + a-7/8)
»'
(8.28)
From the pole-zero diagram of H(s) shown in Fig. 8.26, we will determine
the amplitude response \H(Jm)\. Let us denote the vectors from the poles
to they'w axis as |MX and |M,| as seen in Fig. 8.26.
can then write
We
|
|HO)l
where
K = a* +
j8*
(8.29)
|MJ
|M,|
and
iMii
|M,|
- [«• + (» + m*
- [a» + (a, - W\"
In characterizing the amplitude response, the point
\H(jco)\ is
aspects.
(8.30)
eu
=
a>msx, at
which
maximum, is highly significant from both the analysis and design
Since \H(ja>)\
is
always positive, the point at which \H(Jco)\* is
point at which \H(ja>)\ is maximum.
maximum corresponds exactly to the
Since \H(jm)\* can be written as
«
Network
232
analysis
|H0«>)|
8
and synthesis
=
+ F?
+ (a, + /?)*][«* +(«,-j8)8
(«' +
8
4
«> + 2tt>\%* - f) + (a +
(«'
[a*
]
(8.31)
/?*)'
/?*)•
we can find a>max by taking the derivative of H(ja>)
setting the result equal to zero. Thus we have
|
dJH(U»>)\*
__
<fo>«
eoSuz
£,
^
(8.32)
/**)
=
8
jS
-
(8.33)
—
(8.34)
Expressed in terms of the natural frequency of oscillation
damping factor
and
*
)l
°f
dm
we determine
a>*
<8*)]
V
* |H
From the equation
with respect to
+ /8*)W + 2(q' + («* + /W
+ 2a» "
(a'
K
*
\
«o
and the
wj^ is
- (o) Vl
-£*)*-
1
(C^o)
= <(\ ~ 2£*)
(8-35)
Since comax must always be real, the condition for a>max to exist, i.e., the
condition for \H(jw)\ to possess a maximum, is given by the equation
21* <,
(8.36)
1
<
=
45°.
cos 0, o)max does not exist for
so that t <, 0.707. Since £
In
this
When 8 45°, we have the limiting case for which ©max exists.
the
same
poles
have
0.707 and the real and imaginary parts of the
case, £
=
—
magnitude,
i.e.,
a
=
/?,
or
£«>.
=
«>
Vl
-
£*
=
(8-37)
= 0. This is the lowest
> 0.707, or
We see from Eq. 8.34 that, when a /J, then o>max
frequency at which o)max may be located. For £
£«>,
comax
is
imaginary;
it
> <Vl - P
<8 - 38 >
therefore does not exist To summarize, the key
that the imaginary part of the pole must be greater
point in this analysis is
or equal to the real part of the pole in order for a>max to exist. Interpreted
graphically, if we draw a circle in the s plane with the center at —a and the
then the circle must intersect theyVo axis in order for comax
radius equal to
0,
as seen in Fig. 8.27. Moreover, the point at which the circle interfrom the triangle
sects the positive jm axis is comax- This is readily seen
to
exist,
Amplitude, phase, and delay
233
JU
«max
>
i
/
i
<r
1
H-«-»>
i
1
1
-"max
-70 x
FIG. 8.27. Peaking
with sides a,
/S,
By
a>max in Fig. 8.27.
circle.
the Pythagorean theorem,
we find
that
(oJLx
The circle described in
Fig. 8.27
is
- j8* - a*
called the peaking circle.
the peaking circle intersects the jot axis at
When a
>
/J,
(8.39)
o>
= 0,
the circle does not intersect the /to axis at
therefore, a>max cannot exist.
When
of the
a>
=w
,
=
/?,
all (Fig. 8.286);
the imaginary part of the pole
much greater than the real part, i.e., when /J J>
the/a> axis at approximately
When a
as seen in Fig. 8.28a.
is
then the circle intersects
the natural frequency of oscillation
a,
circuit (Fig. 8.28c).
A figure
of merit often used in describing the "peaking" of a tuned
Q, which is defined in pole-zero notation as
circuit is the circuit
(8.40)
2£
2 cos
From this definition, we see that poles near the ja>
axis (£ small) represent
high-g systems, as given in Fig. 8.28c, and poles far removed from the jo)
axis represent low-Q circuits (Fig. 8.28a). Although the Q of the circuit
given by the pole-zero plot of Fig. 8.286 is theoretically defined, it has no
practical significance because the circuit does not possess a maximum
point in
its
amplitude.
By means of
point,
which
\H(jw c)\
is
the peaking circle,
the frequency
- 0.707 |#(./aw)|.
co c
we can
also determine the half-power
at which the amplitude response
is
234
Network
and synthesis
analysis
ju
jco
JP
J0
Umax
j- *Jmm
—a\
y
J
— «max
-jftk-
(a)
(6)
FIG. 8.28. Examples of peaking
undetermined,
(c)
p
circles,
» a, eomax ~ p.
(a)
= P, ma
a
0.
(6)
a
>
fi,
»„
We will now describe a method to obtain co c by geometrical construction.
Consider the triangle in Fig. 8.29, whose vertices are the poles {pi,pi*}
and a point to, on they'w axis. The area of the triangle is
= fa
Area (£piPi*aid
In terms of the vectors
be expressed as
|MJ and
(8.41)
|M,| from the poles to
\MX
Area (&PiPi*<ot) =-
co it
the area can also
|M,| sin y
>
\
(8.42)
2
where
y> is
the angle at
m
it
as seen in the figure.
we
Ju
From Eqs.
8.41
see that the product |Mi|
|M 2
is
equal
to
|MX
|
|M,|
= -^
sin
\H(ja>i)\
where
(8.43)
y>
Since the amplitude response
=
is
(8.44)
K is a constant, then
.Ksin
FIG. 129
and 8.42,
|
|H(M)I
y>
(8.45)
20*
235
Amplitude, phase, and delay
For a given pole pair {puPi*} the parameters /J, a, and ^Tare prespecified.
we have derived \H(Jm)\ in terms of a single variable parameter,
Therefore,
the angle
y>.
When the angle y>
= tt\2 rad, then sin =
rp
|H(/«>max)|
When v
=
w/4 rad, then sin
y>
\H(ja>J\
= coc
so that a>t
o) a
Let us
.
= Wmax, and,
—
(8.46)
= 0.707 and
= 0.707 I^C/o^nu)!
Let us consider
.
first
=
1, to,
now a geometric construction to
draw the peaking
circle as
shown
in Fig. 8.30.
obtain
We will
A the point at which the peaking circle intersects the positive
Now we draw a second circle with its center at A, and its radius
equal to AB, the distance from A to either one of the poles, as seen in
Fig. 8.30. The point where this second circle intersects the/a) axis is w c
denote by
real axis.
.
The reason is that,
at this point, the inscribed angle is
y>x
= w/4 because it
equal to one-half the intercepted arc, which, by construction, is w/2.
o)max
0, the half-power point m c is also called the half-power
bandwidth of the tuned circuit. In Fig. 8.31a the half-power point 's given
is
=
When
when
(Umax
= 0.
For a high-g
highly peaked at
\HQ0)\
m=
—
where comw
<*><» the amplitude is
shown in Fig. 8.316. In this case, if
are two half-power points m Ci and co
0i
circuit,
<Um»x, as
< 0.707 |//(/ft>max)|, there
Peaking
'circle
FIG.
8.30.
Geometric construction to obtain half-power point.
Network
236
anal/sis
and synthesis
0.707
Iff,
(a)
FIG.
8.31. (a)
(b)
Low-G
circuit response,
(b)
High-Q
circuit response.
a>max, as seen in Fig. 8.316. By the construction process
we obtain the upper half-power point o> Ci It can be shown1
a>max is the geometric mean of m Ci and a> Ct that is,
about the point
just described,
that the point
.
,
(8.47)
As a
result, the
lower half-power point
°>Oi
is
= a>m»x
(8.48)
0>Ct
The bandwidth of the system for a high-g circuit of this type is described by
BW = m Ct —
(o
(8.49)
0i
In design applications these high-g circuits are used as narrow bandpass
filters.
Finally, there are certain aspects
of the phase response of high-g circuits
that are readily apparent. In Fig. 8.32
we
see several steps in the process
=
is 0, as seen
of obtaining the phase response. The phase shift at eo
it rad, as shown in part (c)
oo, the phase shift is
from Fig. 8.32a. At <a
a> =i «>max, the phase
of the figure. Finally, in the neighborhood of a
plane
is approximately
lower
half
in
the
shift resulting from the pole
is controlled
region
in
this
in
phase
change
The
(Fig.
8.32&).
tt/2
0,
in the
response
the
phase
seen
that
readily
It
is
the
pole
in large by
px
region of a> has the greatest negative slope, as seen from a typical phase
—
=
=
=—
—
.
response of a high-g circuit shown in Fig. 8.33.
1
See for example, F. E. Terman, Electronic and Radio Engineering, McGraw-Hill
Book Company, New York,
1953.
Amplitude, phase, and delay
?
237
J<*
Hi
b)0
k
(*)
(b)
(e)
PIG. S.32. Several steps in obtaining phase response for high-Q
>
circuit,
(«) o><
a.
an example to illustrate our discussion of single-tuned
us find the amplitude response for the system function
Finally, as
circuits, let
#(s)
34
=
s*
Now we
determine the
and also the amplitudes
maximum and
|/f(/*comax)|
(8.50)
+ 6s + 34
and
half-power points
awx
and
<o c ,
\H(jo> c)\. In factored form, H(s) is
34
H{s)
+
(8.51)
+ 3-;5)
and the poles of H(s) are shown in Fig. 8.34. We next draw the peaking
circle with the center at j = — 3 and the radius equal to 5. At the point
where the circle intersects thejfa> axis, we see that eom»x — 4. To check this
— a* gives
result, the equation <»*„, =
cBm»x = (5* - 3*)* = 4
(8.52)
(s
3+;5)(s
/J*
-180*-
FIG. 8.33. Phase response of high-Q
circuit.
Network
238
analysis
and synthesis
The amplitude
6.78
= <or
|H(/4)|
|£f(/a>max)| is
then
34
-
(3+j9)(3-jl)
=—=
1.133
(8.53)
30
The
A
point
at
which the peaking circle
intersects the positive real axis is located
at s
= 2.0.
draw a
With the center
of radius
circle
AB
at A,
we
(equal to
At the point C where
new circle intersects theyeo axis, we
have o) c By measurement, we find
5-\/2 in this case).
this
.
to c
FIG. 8.34. Peaking
tion example.
~ 6.78
(8.54)
Let us check this result. Referring to
Fig. 8.34, we know_that the line segment
circle construc-
AB is of length 5>/2 it follows that AC
The line segment AO is of length
(8.55)
AO = 5 — 3 = 2 units
;
is
also 5-v/2 units long.
_.
Then
co c is
given as
co c
Finally,
we
=
sl(AC)1
U
|H(j'6.782)|
n
8.4
is
V46
=
6.782
(8.56)
=
(8.57)
obtain \H(j<o c)\ as
1
which
- (AO)* =
=
34
/
V(34
-
.
46)*
+
—
,
(6V46)
.
S
0.802
precisely 0.707 \H(j(Ow**l)\-
DOUBLE-TUNED CIRCUITS
In Section
poles.
8.3,
we studied the frequency response for a pair of conjugate
Now we turn our
attention to the amplitude response of
circuit.
pairs
We will consider
V2 (s)
FIG. 8.35. Double-tuned
two
we analyze here is the
of conjugate poles in a high-g situation. The
double-tuned or stagger-tuned circuit given in Fig. 8.35.
circuit
*
Amplitude, phase, and delay
239
the special case when the R, L, and C elements in the primary circuit are
equal in value to their counterparts in the secondary. Since the primary
and secondary inductances are equal, the mutual inductance is
M=KL
(8.58)
K
In this analysis we assume the coefficient of coupling
to be a variable
parameter. Let us determine the amplitude response for the voltage-ratio
transfer function V^V^s). From the mesh equations
V^s)
=^R +
S
L+
/.(s)
J^j
- sM /2(s)
(8.59)
= -sM h(s) +
{r
+
sL
+ -M
/,(*)
we readily determine
& + WQs + 1/LC]* -
Vito
Using tuned-circuit notation, we
s
4
K*
}
l
'
set
=-
2£<o
(8.61)
H(s) can then be written as
H(s)
^RM'L
-
\
(1
-
K*)(s*
\
+ -^2_ , + -ȣ-)( f + -fe_ s + _22jL\
1 + K
l + K/\
l-K 1-KI
(8.62)
If
we
*M
A=
set
,
I?(l
then
we can
^
K*)
= ^2*
1 - K*
(8.63)
K
'
write
H(s)
=
(s
where
-
-
{ Sl , Sl *}
11
l
SiXs
-
—-
sfXs
s»)(s
^S_ ± j„J—L+ K J "ll + K
(8.64)
-
s,*)
(1
+
£-_]
K)*J
(8.65)
1
l
-k
°Li
-x
ci
—
«
jc)*J
Let us restrict our analysis to a high- Q circuit so that £*
1 . Furthermore,
let us assume that the circuit is loosely coupled so that AT« 1. Under
Network
240
analysis
these assumptions,
and synthesis
we can approximate
the pole locations by discarding
Then the poles
the terms involving £* under the radicals in Eq. 8.65.
approximately as
fai» *i*} can be given
(8.66)
Similarly, {s„ *,*}
can be given as
{s„ s,*}
The
~ -£o> ± jo) ^l +
—J
pole-zero diagram of H(s)
poles
— £<u
is
(8.67)
given in Fig. 8.36. The real part of the
comparison to the imaginary parts for
is greatly enlarged in
FIG. 8.M. Poles and zeros of a dottble-tuned
circuit.
Amplitude, phase, and delay
Note that we have a triple zero at the
of the vectors in Fig. 8.36, the amplitude response is
clarity purposes.
origin.
^4|Mo|'
241
In terms
(8.68)
]H(J(0}1
JM.I |M,| |M.| IMtl
Since the circuit
is
high- 2 in the vicinity of
a>
= a>
,
we have
|M,|-|M4 |=i2|M |-2o,
so that in the neighborhood of
(8.69)
<u
Aco
|HO»)|S'
4IMJIM.I
It is evident that
(8.70)
the amplitude response
of \H(jm)\ in the neighborhood of a>
depends only upon the pair of vectors
IMJ and |M,|. The double-tuned problem has thus been reduced to a singletuned problem in the neighborhood of
Consequently, we can use all the
on the peaking circle that were
derived in Section 8.3. Let us draw a
o)
.
results
peaking
circle
s
=
with the center at
-£«„ +/«>,
(8.71)
FIG. 8.37. Peaking circle for doubletuned circuit.
and with a radius equal to tw^T/2, as
shown in Fig. 8.37. The inscribed
y> then determines the location of the maxima and half-power points
of the response. Without going into the derivation, the amplitude response
can be expressed as a function of y according to the equation
angle
|HO)l
-
A<o sin
4a)
sin
y
K(ta>
)
2(1
Referring to the peaking circle in Fig. 8.37,
let
-
y
(8.72)
K*)
us consider the following
situations:
> WiKfl:
In this case, the peaking circle never intersects the jot
axis; y> is always less than ir/2 (Fig. 8.38a), and the amplitude response
never attains the theoretical maximum
1.
a>
£
iW =
*-=
2(1
-K*)
as seen by the curve labeled (a) in Fig. 8.39. In this case,
circuit is said to
be undercoupled.
(8.73)
K < 2£, and the
Network
242
anal/sis
and synthesis
jo>
«8lM
2
\
i
-f«o
T
7
WO
y
A
/
\
ly\ w
2
x^^
(a)
FIG. 8.38.
2. o>,£
=
single point
(b)
(a)
Here the peaking
m=
(Fig. 8.38ft).
At
lmax
(c)
Undercoupling. (6) Critical coupling,
co^Kjl:
I
(c)
Overcoupling.
circle intersects
the amplitude
the/eo axis at a
equal to #max
in Eq. 8.73. In this case £
Kj2, and we have critical coupling.
3. <u £
cogA/2 The peaking circle intersects theytu axis at two points.
a>
a>
,
is
=
<
o> x
and
to,,
:
as seen in Fig. 8.38c.
The
intersecting points are given
by the
equation
<*>!,*
max
=
tt>
± «>
(8.74)
.[(!H*
(
Consequently, the amplitude response attains the theoretical maximum
Hum* at two points, as shown by curve (c) in Fig. 8.39. In this situation,
the circuit is said to be overcoupled.
Note that
have their
and critically coupled curves
The overcoupled curve, however, is
in Fig. 8.39 the undercoupled
maximum
points at
me
.
"a mix
FIG. 8.39.
(a)
Undercoupled
case, (A) Critically
coupled case,
(c)
Overcoupled case.
Amplitude, phase, and delay
uCl
uCt
<<«
FIG. 8.40. Half-power points of overcoupled
maximum at o>! and u>t
we can determine
struction
critical
coupling,
the half-power points by using the geometrical con-
method given in Section 8.3. Observe that there are two
co c and co c
as shown in the overcoupled curve in
The bandwidth of the circuit is then
half-power points
Fig. 8.40.
circuit.
In the case of overcoupling and
.
243
,
BW =
The
Example &2.
a>
Ct
— m Ci
voltage-ratio transfer function of
(8.75)
a double-tuned
circuit is
given as
H™
As?
=
(*
+2
-HylOOX*
+2
+2
-ylOOX*
+yl06X>
+2
-/106)
(8/76)
From H(s), let us determine the following: (a) the maximum points <ot ma and
<o t „„; (b) the 3 db bandwidth BW; (c) the damping factor £; (d) the coefficient
of coupling*; (e) the gain constant A; and (/) the maximum of the amplitude
response //max.
Solution, (a) The natural frequency of oscillation o> is taken to be approximately halfway between the two poles, that is, <o = 103 radians. In the neighborhood of »„, we draw the poles s — —2 +/100 and s = —2 +/106, as
shown in Fig. 8.41 . From the peaking circle centered at the point s = — 2 + /<».,
shown
in Fig. 8.41,
we
obtain
mt max
so that
(b)
— mo =
<°i
max
=
°*o
«»i
m*r
=
«»o
Next we draw a
— 2* = 2.236 radians
+ 2.236 =
— 2.236 =
circle centered at
s
<o
V(3^2)»
-
(8.77)
105.236 radians
(8.78)
100.764 radians
=
we have
this circle intersects theyVu axis,
">c,
"^3*
1
+ja> with radius 3V2. Where
Ct so that
1
- 4.123 radians
(8.79)
Network
244
and synthesis
anal/sis
FIG. 8.41. Peaking
The
db bandwidth
3
is
(c) The damping factor
from which we obtain
The
coefficient
which
o»j)
{ is obtained
8.2.
103
of coupling
-
8.246 radians
from the real part of the poles
«
t
4
(d)
Example
then
BW - 20»Ol -
circle,
circle for
— u0.0194
'" 15^
(8.80)
£a>
=
2,
(8.81)
K is obtained from the radius of the peaking
is
o>oA:
(8.82)
K = — = 0.0582
We thus have
(e)
The gain constant
A
is
equal to
2{«.„A:
*
(f) Finally, the
1
- K*
maximum
(8.83)
2(2X0.0582)
1
- (0.0582)*
amplitude
flmiz
Hmmx
= "•"**
(8.84)
is
1
=
2(1
-**)
-0.5009
(8.85)
Amplitude, phase, and delay
ON
83
POLES
AND ZEROS AND
245
TIME DELAY
time delay? How do we relate it to frequency response? We
attempt to answer these questions in this section. First consider the
transfer function of pure delay
(8.86)
<r tT
H(s)
What is
will
=
any excitation e(f) produces an
which is delayed by time T with respect
shown by the Laplace transform relationship,
For a system described by Eq.
identical response signal e(t
to the excitation. This
Bis)
is
—
8.86,
T),
- C[e(/ -T)] = er' T C[e(0]
(8-87)
Let us examine the amplitude and phase response of the pure delay.
From the equation
*-***
< 8 - 88)
HO) -
we
obtain the amplitude response
\H(Jw)\
and the phase response
We
see that the delay
response, that
<f>(a))
T is
=
-
(8.89)
1
—coT
(8.90)
equal to minus the derivative of the phase
is,
T= -
d<f>(w)
(8.91)
dco
The magnitude, phase and delay characteristics of H(ja>) = erlaT are
given in Fig. 8.42a, b, and c.
If we define delay as in Eq. 8.91, we can readily deduce that for the
response to be nearly identical to the excitation, the system amplitude
response should be constant, and its phase response should be linear over
the frequency range of interest. If the phase is not linear, we have what is
known as delay distortion. To visualize delay distortion more clearly,
M<»)
A(«)
*te)
(a)
(h)
\
FIG. a.42. Amplitude, phase, and delay of ideal delay function,
(ft)
Phase, (c) Delay.
(e)
(a)
Amplitude,
Network
246
anal/sis
and synthesis
we recall from Fourier analysis that any signal is made up of different
frequency components. An ideal transmission system should delay each
frequency component equally. If the frequency components are delayed
by different amounts, the reconstruction of the output signal from its
Fourier components would produce a signal of different shape as the input.
For pulse
applications, delay distortion is
an
essential design considera-
tion.
Let us next examine how we relate delay, or envelope delay (as it is
sometimes called) to the poles and zeros of a transfer function. For any
transfer function
n (»-««)
H(s)=*tf
(8.92)
IT(»-Pi)
= —a
with zeros at z €
t
±jcot and poles
=—
at/>,
<r
y
±jco i , the phase for
real frequencies is
lr
\
<K<»)
? tan-1 W ±
=
=2
.
ot
*-i
Envelope delay
J-l
1 to
.
tan"
+ CO,
x
'
(8.93)
a,
h o' + (c ± (of + A o* + (« ±
~
(894)
co,)*
see that the shapes of the delay versus frequency characteristic are
the same for
all
poles and zeros.
the poles, positive delay.
The zeros
However,
transfer functions with zeros alone.
Now
let
pole atp
contribute "negative" delay;
do not have
Ls is the only
linear physical systems
The inductor H(s)
=
=
exception. Its phase
is #a>)
ir/2; thus the delay is zero.
us consider the delay due to one singularity, for example, a
—a +jcoQ The delay due to the one pole is
=
.
The following points are
1.
.IL
-J
is
dm
We
fflj
*
*
pertinent:
The maximum delay due
to this pole
is
Am = and occurs at
about
co
=
<w
to
.
=
co
.
The delay
versus frequency curve
(8.96)
is
symmetric
247
Amplitude, phase, and delay
FIG.
2.
8.43.
The frequency
Graphic construction to obtain delay bandwidth.
at
which the delay
A
o>x
3.
The
"effective delay
simply 2a
.
graphically
The
=
co
bandwidth"
is
±
is
half the
maximum, or
l/2or
,
is
(8.97)
ff
then
eo
—
<r
<m<
to
+
cr
or
The upper and lower half-bandwidth points can be obtained
by drawing a circle with center at a> = co, and radius a
.
intersections of the circle with the jco axis are the half-bandwidth
shown in Fig. 8.43.
The product of the maximum delay and delay bandwidth
points, as
4.
2.
Thus,
if
we wish
to obtain large delay
theyeo axis, the effective delay bandwidth
5.
The delay of an
is
always
by placing zeros or poles near
is
then very narrow.
all-pass function is twice the delay
due to the poles
alone.
The delay versus frequency curve
We
for the pole
is
shown
in Fig. 8.44.
see that the delay-bandwidth concept is only useful for a
rough
approximation, since the delay versus frequency characteristic only falls
as 1/to. To calculate the delay versus frequency characteristic for a
a number of poles and zeros, it is convenient to
obtain the delay curves for the individual singularities and then to add the
separate delays. It is not as desirable to obtain the total phase response
transfer function with
and then
differentiate numerically.
Finally,
when
it
should be pointed out that envelope delay only has meaning
the phase response goes through the origin. If it does not, there
a frequency-shift component in addition to the delay that
account for analytically.
is
is
hard to
Network
248
wo -
4<?o
">o
analysis
and synthesis
— 3»o wo - 2<ro wo — ^o
FIG.
«o +
«o
8.44. Plot
ffo
«o +
2<ro
wo +
3<ro
wo + 4»o
<*
of delay versus frequency.
Problems
8.1
and
Find the poles and zeros of the impedances of the following networks
on a scaled s plane.
plot
o
—nnnr*
|f-
4h
io
Z(s>-
la)
o
M/*
1(-
20
iM
ZM-
m
20
4h
ZW-
PROB. 8.1
&2 The
circuit
shown
in the figure
is
a shunt peaking
video amplifiers,
(a)
Show that the admittance
m
Y(s) is of the
K(s
form
- s^s — s»>
(s
-*,)
Express slt sa , and ss in terms of R, L, and C.
circuit often
used in
Amplitude, phase, and delay
-
-
-
249
10"* mhos, find
-10 -ylO», and Y(jB)
-10 +yi0», st
(6) When st
the values of R, L, and C and determine the numerical value of *,.
=£c
Y(,>
:r
PROS.
U
Find the amplitude and phase response for the following functions and
83
sketch.
(«) F(s)
(*) F(s)
+K
s
s
+K
s
(c) F(s)
Note
8.4
id) F(s)
.
K
and o> are positive
Given the function
that
7T °v
quantities.
<**">
C(») +71*.)
determine the amplitude and phase of G(jm) in terms of ^, B, C, D.
the amplitude function is even and the phase function is odd.
Show that
85 By means of the vector method, sketch the amplitude and phase response
for
s-l
+0.5
j
(a) F(s)
-
sis
+
(C) F(*)
s
10)
s
(b)
F(s)
's*
s
(e)
F(s)
-
'
5
(«/)
+2s +2
Plot
F(5)
s*
+1
-1
,
^)F(j) - (,/2xV+ 9)
8.6
+
on semilog paper
F(*)
the
=
x
W
F(5)
=
-2s +
5*
x
F(,)
2
- 2j + 5
-(TTW+T)
=
s*
(j
+2s +5
+
2X*
+
Bode
plots of magnitude
100(1
+ 0.5*)
s(s
F(s)
«
<*>
1
50(1
(1
+2)
+ 0.025jX1 + O-IJ)
+ 0.05*X1 + 0.01s)
1)
and phase for
250
8.7
Network
Plot
analysis
on semilog paper
(«)
W
8.8
and synthesis
F(*)
the
Bode
plots of magnitude
10005
5 10~ 5*
and phase for
(1
+ O.OOZsXl +
(1
200(1 + 0.05s)
+ 0.02jX1 + 4 10-** + 10-V)
=
•
+ 10~V)
•
For the function
*W-3 + 2s + 5
s*
determine o> mu , \F(jco m *x)\, the half-power point
amplitude and phase response.
a>
and
,
\F(ja>
)\.
Sketch the
8.9 For the circuit shown, determine the current ratio IJIg and find: (a)
the point €<>,„„, where its amplitude is maximum; (b) the half-power point <oc ;
1. Use geometric construction.
(c) the point to„ where |/i(a) M)//B(<o u)|
=
2.5 h
PROS.
8.10
A network function consists of two poles at p
ltt
jco t , as given in the figure.
Af\to)
8.9
is
maximum
at
o>
Show
m* =
/•<*
=
r^*^' -9
'
= — at ±
that the square of the amplitude response
|cos 20|.
PROB.
8.10
8.11 In connection with Prob. 8.2 plot the poles and zeros of the impedance
Find, approximately, the maximum point of the
l/l%s).
function Z(s)
amplitude response. In addition, find the bandwidth at the half-power points
and the circuit Q.
=
Amplitude, phase, and delay
251
8.12 The pole configuration for a system function H(s) is given in the figure.
From the plot, calculate:
(a)
(6)
The undamped frequency of oscillation
The bandwidth and Q.
—
f
a>
yioo
-5
<T
h
-y'100
PROB.
8.12
In connection with Prob. 8.10 determine the ratio Af\m aM)lM*(0).
8.14 Determine the amplitude and phase response for the admittance Y(s)
of the circuit shown. Is the peaking circle applicable here? What can you say
about the shape of the amplitude response curve in a high-g situation? Determine the bandwidth of the circuit and the circuit Q.
8.13
20
o——WW
lh
^SW"
YM-
1
=rif
PROB.
8.14
8.15 For the overcoupled case of a double-tuned circuit, derive an expression
for the peak-to-valley ratio that is, M(o> m »x)/M(a> ), where A#(-) denotes amplitude. Use the notation in Section 8.4. {Hint: see Prob. 8.13.)
8.16
For the voltage
ratio of a double-tuned circuit
™S
fri v\
(s
+ 4 +y50X* + 4 -y50X* + 4
+j6Q)(.s
+4
-y60)
252
Network
analysis
and synthesis
and
circle to determine the maximum and half-power points
coupling K.
the circuit Q. Find the gain constant A and the coefficient of
and halfa x i mu
8.17 For the double-tuned circuit shown, determine the
of
power points and the circuit Q. Find the gain constant A and the coefficient
Use the peaking
m
coupling K.
10
"9
f
M=5xl<r 6 h
PROB.
8.18
Determine the delay at
a =
8.17
0, 1,
and 2 for
1
(«)
F(s)
(b)
F(s)
(c)
F(s)
(<*)
F(s)
(e)
*
s+2
= s-3
s
W
+3
3*
=
(s
+
s*
s + 1
+ Is + 5
=
IX*
+ 2)
(s+2Ks*+2s+ 2)
m
chapter
Network
analysis
9
II
NETWORK FUNCTIONS
9.1
network theory, the word port has a special meaning. A
regarded as a pair of terminals in which the current into
one terminal equals the current out of the other. For the one-port network
shown in Fig. 9.1, 1=1'. A one-port network is completely specified
when the voltage-current relationship at the terminals of the port is given.
For example, if V = 10 v and 1 = 2 amp, then we know that the input
or driving-point impedance of the one-port is
In
electric
port
may be
Zi„
= — = sa
(9.1)
Whether the one-port is actually a single 5-Q resistor, two 2.5-Q resistors
in series, or two 10-Q resistors in parallel, is of little importance because
the primary concern
is
the current-voltage relationship at the port.
Consider the example in which /
admittance of the one-port is
Yto =
=
2s
+
V=
1
;
then the input
^ = 2s + 3
which corresponds to a 2-f capacitor in
its
3 and
(9.2)
parallel with
a $-Q
resistor in
simplest case (Fig. 9.2).
I
+
=!=2f
One-port
network
V
I'
FIG.
PIG. 9.2
9.1
253
•io
254
Network
analysis and synthesis
h
h
1
+
Two-port
network
Vi
2
v2
2'
r
FIG. 9.3
Two-port parameters
A general two-port network, shown in Fig. 9.3, has two pairs of voltagecurrent relationships. The variables are Vx , V2 , Ju lz Two of these are
dependent variables ; the other two are independent variables. The number
of possible combinations generated by four variables taken two at a time
describing a two-port
is six. Thus there are six possible sets of equations
.
network.
We will discuss the four most useful descriptions here.
The z parameters
A particular set of equations that describe
z-parameter equations
Vx =
a two-port network are the
z^ + zj
t
(9 3)
V2 =
Z
7
21 l
+
S
wJ%
In these equations the variables Vx and Vt are dependent, and Iv I2 are
independent. The individual z parameters are denned by
V,
z ll
h
h
J,=0
7,=0
(9.4)
v,
«21
h
'M
/!=0
7,-0
observed that all the z parameters have the dimensions of impedance.
Moreover, the individual parameters are specified only when the current in
one of the ports is zero. This corresponds to one of ports being open
circuited, from which the z parameters also derive the name open-circuit
It is
Vi
zb
FIG. 9.4
Vi
Network
anal/sis
255
II
parameters. Note that z u relates the current and voltage in the 1-1' port
only; whereas z M gives the current-voltage relationship for the 2-2' port.
Such parameters are called open-circuit driving-point impedances. On
and z2l relate the voltage in one port to
the current in the other. These are known as (open-circuit) transfer
the other hand, the parameters z liS
impedances.
As an example,
in Fig. 9.4.
let
T circuit
us find the open-circuit parameters for the
We obtain the z parameters by inspection
Zll
h
=
+z
6
= zb + z
e
z«
7,=0
z ti
7,-0
(9.5)
*1*
=Z
h
=Z
6
7,-0
_^2
«*l
=
h
7»=0
Observe that z ia
z 21
When the open-circuit transfer impedances of a
two-port network are equal, the network is reciprocal. It will be shown
later that most passive time-invariant networks are reciprocal. 1
Most two-port networks, whether passive or active, can be characterized
by a set of open-circuited parameters. Usually, the network is sufficiently
complicated so that we cannot obtain the z parameters by inspection, as
.
did for the T circuit in Fig. 9.4. The question is now, "How do we
obtain the z parameters for any circuit in general?" The procedure is as
we
We write
a set of node equations with the voltages at the ports
and other node voltages within the two-port Va V4
Vk
as the dependent variables. The independent variables are the currents It
and I& which we will take to be current sources. We then proceed to
write a set of node equations.
follows.
Vx
and
Vt
,
,
h = nn V + n^Vi H
- n.1^1 +
+
1
- "*iFi +
1
One important exception
is
r*
+ nu
lk V„
rk
+ nSk V
rt
+ ns»
Sk V
,
.
.
.
,
t
l
+ "**n
the gyrator discussed later in this chapter.
(9.6)
Network
256
analysis
and synthesis
where n it represents the admittance between the
»«
If the circuit is
n it
=n
ft
.
equations,
As a
AM
,
= Gu +
made up of R-L-C
result, the ijth
and/th nodes, that
is,
1
+
sCu
ith
(9.7)
sL,•il
elements only, then
it is
clear that
cofactor of the determinant of the node
must be equal to
they'ith cofactor,
A
ti ,
that
This result leads directly to the reciprocity condition z21
is,
A w = AH
= z^,
as
.
we
shall see.
Returning to the
and Vt We obtain
set
of node equations in Eq.
9.6, let
us solve for
Vx
.
Vl
~ A 1+ A
2
(9.8)
In relating this
parameters,
it is
last set
of equations to the defining equations for the z
clear that
z
u=
»ii
^
An
A
ztt
= A8a
(9.9)
Since for a passive network
A^ = A w
T
,
it
follows that za
network is then reciprocal.
As an example, let us find the z parameters of the Pi
First, the node equations are
= z12
,
the
circuit in Fig. 9.5.
h"{TA + Yc)V -Yc V
1
Ia
CD
=-Yc V
1
l
(9.10)
+ (YB + Yc)Vt
Vi
v2
K
YB
FIG.
9S
d>
Network
The determinant
for this set of equations
anal/sis
II
257
is
Ay = YA YB + YA YC + YB YC
(9.11)
In terms of Ay, the open-circuit parameters for the Pi circuit are
+
^~ YBAy
Yc
Zai
~Ay
(9.12)
*
Now
~
Zu
Ay
YA + YC
Ay
us perform a delta-wye transformation for the circuits in Figs.
9.4 and 9.5. In other words, let us find relationships between the immittances of the two circuits so that they both have the same z parameters.
let
We readily obtain
Zl«
-z
«M
- z„ + z - ^
*11
- za + z -
-±s-
e
r,
.
-T
YB + Yc
b
z
Zo
We then find
(9.13)
Ay
Ay
-&
Ay
(9.14)
Zc
Ay
The y parameters
Suppose we were to write a
Fig. 9.3.
Then the voltages
and the currents Ix and
set
of mesh equations for the two port in
Vx and V2 would become independent sources,
/»
would be just two of the dependent mesh
mesh equations
currents. Consider the general set of
- «uA + «iA +
V% — »»mA + "hJ* H
= msi/i+
Vi
= m uJ +
l
"
+ «i
r-
m^/j
+ W.A
(9.15)
+ m kkTk
sum of the impedances in the /th mesh and m ti is
common impedance between mesh i and meshy. We note here again
where
the
•
•
m
it
represents the
258
that for
Network
anal/sis
and synthesis
m — mH
an R-L-C network,
it
for all
i
and / Thus
reciprocity
holds.
Solving the set of mesh equations for /x and 7a ,
we
obtain the following
equations.
The equations of 9.16
h~ A
x
A
*
+
A
*
(9.16)
A
define the short-circuit admittance parameters as
h = VuVi + yit v*
h = y i^i + ynVt
(9
17 )
S
= A„/A for all i andy.
Let us find the y parameters for the bridged-T circuit given in Fig. 9.6.
The mesh equations for the circuit are
where ytj
s
'.
=A+
(-
+
we
A=
A
\s
s
(9.18)
I
obtain
2(2s
-A
+ ~h
s
s
In straightforward fashion
l)h
+
1)
ls%
+ 4s +
1
(9.19)
S
A
-A - - ls + 2s +
*
FIG. 9.6
1
Network
The
analysis
259
II
short-circuit parameters are then
+ 4s + 1
2(2s + 1)
8
2s + 25+1
y«i = Vi* — 2(2s + 1)
When ylx = yn or zu = za2 the network is symmetrical. 2
yu
=
sfa
=
2s
8
(9.20)
,
Returning to Eq. 9.17, which defines the y parameters,
parameters are expressed explicitly as
911
we
see that the
y
_h
Vy Fi-0
/,
3/m
v* Fi-0
(9.21)
h
K
h
Vn
Vm
Ft-0
v* Fi-0
y parameters are also called short-circuit admittance
apparent. In obtaining yu and y a , the 2-2' port must
be short circuited, and when we find yn and ylt , the 1-1' port must be
short circuited, as shown in Figs. 9.7a and 9.76.
As a second example, let us obtain the y parameters of the Pi circuit in
The reason
parameters
Fig. 9.5.
that the
is
To
now
obtain y a and y M ,
we
short circuit terminals 2-2'.
We then
have
A
Vu
y»i
yn, *2i
°
(9.22)
= -Yc
,Jz
m yn Z3.
h>i
(b)
(a)
FIG. 9.7
*A
and
symmetrical network
is
easily recognized because
by interchanging the
2-2' port designations, the network remains unchanged.
1-1'
2
Network
260
analysis
and synthesis
We next short-circuit terminals
1-1' to obtain
= YB + Yc
^12 = -Yc
^22
,g
23)
The h parameters
A
set
of parameters that are extremely useful in describing transistor
h parameters given by the equations
circuits are the
Vi
I»
The
= *uA + hi*V*
— "tl-M. "22'
individual parameters are defined
h
by the
relationships
-h
Fj=0
h
"21
7l-0
1*
=~
h
"89
see that h lt
and h 91 are
Jl-0
and hx% and h^
The parameter hn can be interpreted
short-circuit type parameters,
are open-circuit type parameters.
as the input impedance at port
seen that h u
is
(9.25)
=h
F,-0
We
(9.24)
"I"
1 with port 2 short circuited.
merely the reciprocal of ylt
It is easily
.
hu
=
(9.26)
yu
The parameter h 22
is
an open-circuit admittance parameter and
to z M by
h«a
=—
is
related
(9.27)
Both the remaining h parameters are transfer functions; h 21 is a shortand h 12 is an open-circuit voltage ratio. Their
relationships to the z and y parameters is discussed later in this chapter.
For the Pi circuit in Fig. 9.5, the h parameters are
circuit current ratio,
fcll
1
=
Ya
h 12
h ai
+ YC
Yc
YA + YC
= — Yc
YA + YC
YA YC
v„
YA +YC
=
.
fc_
(9.28)
Network
analysis
II
261
h
h
+
1
t
NIC
^
FIG.
9.8.
1
1
Negative impedance converter with load impedance.
=
—hlt . This is the reciprocity
Observe that for the Pi circuit, h^
condition for the h parameters and can be derived from their relationships
to either the z or y parameters.
Next
let
us consider the h parameters of an ideal device called the
negative impedance converter (NIC), which converts a positive load
impedance into a negative impedance at its input port.* Consider the
NIC with a load impedance ZL shown in Fig. 9.8. Its input impedance is
Z.„
= -Z,
(9.29)
= ¥*
(9.30)
which can be rewritten as
¥i
The following voltage-current
relationships hold for the
NIC.
Vx - kVt
(9.31)
A = kh
If
we
interpret Eq. 9.31 using
h parameters, we arrive at the following
conditions.
h vt
=
—=
(9.32)
k
«21
We see that since h u
j&
—h tl
,
the
NIC is nonreciprocal.
NIC is
In matrix notation, the h matrix of the
k
»ii
(9.33)
i
.k
The NIC
is
is
.
a convenient device in the modeling of active
circuits.
not, however, a device that exists only in the imagination.
*
For a
Chapter
NIC, see L. P. Huelsman, Circuits,
and Linear Vector Spaces, McGraw-Hill Company, New York, 1963,
lucid discussion of the properties of the
Matrices,
4.
It
Practical
Network
262
realizations of
and synthesis
anal/sis
NIC's have been achieved using
an article by Larky. 4
transistors.
Some of
these are described in
The ABCD parameters
Let us take as the dependent variables the voltage and current at the
port
1,
and define the following equation.
A
C
LA.
(9.34)
ara
This matrix equation defines the A, B, C,
D parameters, whose matrix is
known as the transmission matrix because it relates the voltage and current
at the input port to their corresponding quantities at the output. The
reason the current It carries a negative sign
is
that
most transmission
engineers like to regard their output current as coming out of the output
port instead of going into the port, as per standard usage.
In explicit form, the ABCD parameters can be expressed as
*
A=
B= - Yi
h
lt=0
Fa=0
(9.35)
±
C=
r«=o
From
these relations
we
see that
A
represents
an open-circuit voltage
a short-circuit transfer impedance; C is an opencircuit transfer admittance; and D is a short-circuit current ratio. Note
that all four parameters are transfer functions so that the term transmission matrix is a very appropriate one. Let us describe the short-circuit
transfer functions B and D in terms of y parameters, and the open-circuit
transfer functions A and C in terms of z parameters. Using straightforward algebraic operations, we obtain
transfer function; J?
is
A=^
B = _J_
«n
J/21
i-
D = _2u
z 21
Vtl
(9.36)
c=
For the
ABCD
parameters, the reciprocity condition
is
expressed by the
equation
YA
det
B~]
C D
= AD-BC=l
* A. I. Larky, "Negative-Impedance Converters,'
CT-4, No. 3 (September 1957), 124-131.
Trans.
(9.37)
IRE on
Circuit Theory,
Network
263
II
h
h
Let us find, as an example, the
parameter for the ideal trans-
ABCD
former in Fig.
analysis
+
+
whose denning
9.9,
Vi
equations are
Vl - nVt
/1
<*SD
= i(_ j8)
FIG.
If we express Eq. 9.38 in matrix form,
9.9. Ideal transformer.
we have
n
(9.39)
1
h
n
so that the transmission matrix of the ideal transformer
A
B"
is
~n
^s
C D
(9.40)
I
L
«J
Note, incidentally, that the ideal transformer does not possess an impedance or admittance matrix because the self- and mutual inductances
are infinite. 5
For the
ideal transformer terminated in a load
impedance shown in
Fig. 9.10a, the following set of equations apply.
(9.41)
nZ,
Taking the
ratio
of
Vx to I we find the input impedance at port
l3
Zl
Thus we
see that
an
=£=
h
n*ZL
ideal transformer is
the load element were an inductor
L (Fig.
1
to be
(9.42)
an impedance transformer. If
1 we would see
9.106), at port
an equivalent inductor of value n 2L. Similarly, a capacitor C at the load
would appear as a capacitor of value C\n* at port 1 (Fig. 9.10c).
As a second example
*
For a
indicating the use of the transmission matrix in
detailed discussion concerning ideal transformers, see
Introduction ta
Modem
Network
Synthesis,
M.
E.
John Wiley and Sons,
Van Valkenburg,
New
York, 1960.
Network
264
anal/sis and synthesis
h
m:li
+
v2
(»)
c
n?i,
c=£
FIG. 9.10. Ideal transformer as an impedance transformer.
network
ABCD
analysis, consider the
Fig. 9.5.
A
— Ys
parameters of the Pi circuit in
+Yc
Yc
B=
1
Yc
YA YB + YB YC + YA YC
C=
Yc
Y
+Y
A
C
—
(9.43)
n
If
we check for
reciprocity
AD-BC = (YA +
from Eq.
YC)(YB
we
9.43,
+ Yc) - (YA YB + YB YC + YA YC
)
a
Y
In
-&*!
9.2
see that
(9.44)
RELATIONSHIPS BETWEEN TWO-PORT PARAMETERS
The relationships between two-port parameters are quite easily obtained
because of the simple algebraic nature of the two-port equations. For
example, we have seen that /t u = l/y u and h sa = l/z 2 ». To derive h lt in
terms of open-circuit parameters, consider the z parameter equations
when port
1 is
open
circuited:
= 0.
*\ = z
V = z22/2
/x
i2-'a
t
(9.45)
Network
Therefore
we have
= -y1
h lt
analysis
II
= 2H
265
(9.46)
V.
Similarly, since h
n
defined as a short-circuit type parameter,
is
we can
derive h 21 in terms of y parameters as
h a ==
**
(9.47)
We can express all the A parameters as functions of the z parameters or y
An easy way to accomplish this task is by finding out
parameters alone.
what the
and y parameters themselves.
and y parameters are not simply
reciprocals of each other (as the novice might guess), since one set of
parameters is defined for open-circuit conditions and the other for shortCertainly,
relationships are between the z
by
their very nature, the z
circuit.
The
z
and y
relationships
can be obtained very
easily
by using matrix
notation. If we define the z matrix as
[Z]
=
(9.48)
L*si
and the y matrix as
[11
yu yn\
=
(9.49)
J/ti
In simplified notation
we can write
the two sets of equations as
=
[/] =
[V]
and
Replacing
[7]
in Eq. 9.50
Vt»]
[Z][7]
(9.50)
[Y][V]
(9.51)
by [Y][V], we obtain
[V]
=
[Z][Y)IV]
(9.52)
so that the product [Z][Y] must yield the unit matrix [U]. The matrices
[Y] and [Z] must therefore be inverses of each other, that is,
1
[ZTr
From
z
the relationship,
=
and
[Y]
we can
[IT-^IZ]
(9.53)
find the relations between the individual
and y parameters.
Zjg
Zji
A«
A*
A.
(9.54)
A.
Network
266
anal/sis and synthesis
TABLE
9.1
Matrix Conversion Table
M
[*]
n
z 12
^
z 22
*22
z ia
z
[A]
[71
s/«
3/12
A
A12
A
AT
a,
A,
Am
Agg
C
C
3/a
3/u
Agi
1
1
£>
\
A,
A*
Agg
c
C
1
A„
D
Ay
3/n
J/12
Au
Au
B
£
1
/<
3/22
An
An
A*
2/21
An
J?
J?
1
3/12
B
An
AT
A12
D
Ah.
Agg
D
D
C
D
^
B
C
D
ft
w.
M
A,
A,
«M
A,
«ii
A,
z 12
A,
Z 22
«*2
3/ii
3/u
«a
1
2/21
A,
[A]
Z 22
3/ll
«ii
A.
3/22
1
A»
An
«21
Z21
3/21
3/21
An
Agi
1
z22
\
3/u
AM
1
3/21
A21
Agi
*m
m
z
where A„
1
n
««1
Vii
3/21
= zuz g — ZigZax
g
and
;
«ii
= T~
Zia
=—
J/22
=—
(9.55)
where
Av =
^n2/ 22
—
Vi^/n-
2/12
Zg!=
Using these
-
?21
identities
we can
derive the h
or ABCD parameters in terms of either the z or y parameters. Table 9.1
provides a conversion table to facilitate the process. Note that in the
tMe
9.3
b. T
= AD- BC
(9.56)
TRANSFER FUNCTIONS USING TWO-PORT PARAMETERS
will examine how to determine driving-point and
of
a two-port by use of two-port parameters. These
transfer functions
broad categories. The first applies to two-ports
two
into
functions fall
In this section
we
Network
analysis
II
267
without load and source impedances.
These transfer functions can be
For example, let us derive
the expressions for the open-circuit voltage ratio VJiVx by using paramz
eters first and y parameters next. Consider the z parameter equations
for the two-port when port 2 is open circuited.
described by
means of z or y parameters
Vt =
alone.
z ai/j
(9.57)
Vx - z^
If we take the ratio of V, to
Vx we obtain
,
Yl
=
Vt
By
letting
%L
(9.58)
*ii
/2 of the second y parameter equation go to zero,
we
derive
the open-circuit voltage ratio as
(9.59)
V\
In similar manner
two-port as
we can
Via
derive the short-circuit current ratio of a
(9.60)
Is
and
_ _«M
(9.61)
The open- and short-circuit transfer functions are not those we usually
deal with in practice, since there are frequently source and load impedances to account for. The second category of two-port transfer
functions are those including source or load impedances. These transfer
functions are functions of the two-port parameters z, h, or y and the
source and/or load impedance. For example, let us derive the transfer
IJVX of a two-port network that is terminated in a resistor
of/? ohms, as given in Fig. 9.1 1. For this two-port network, the following
equate apply.
admittance
_
+
h
*?
+
+
Vi
Two-port
network
FIG. 9.11
V2
268
Network
analysis
and synthesis
1
"~l
l°-
Zll
^-^
+1*12*2
1+
s~~-
v2
Vi
l'o-
2'
FIG.
By
Two-port equivalent.
9.12.
eliminating the variable
Vt we obtain
,
yJR
V1 yu + 1/R
the same. YS1 is
rsl
and ytl are not
of the two-port network terminated in a
Note that
resistor
(9.63)
the transfer admittance
R, and y n
We must be
is
the transfer
make
admittance when port 2 is short circuited.
nature.
similar
this distinction in other cases of a
In order to solve for transfer functions of two-ports terminated at
either port by an impedance ZL , it is convenient to use the equivalent
circuit of the two-port network given in terms of its z parameters (Fig.
parameters (Fig. 9.13). The equivalent voltage sources z lg/a
9.12) or
careful to
y
and z„Ix in Fig. 9.12 are called controlled sources because they depend
6
upon a current or voltage somewhere in the network. Similarly, the
current sources y lt Vt and y^Yx are controlled sources. For the circuit
in Fig. 9.12, let us find the transfer impedance Zm = VJIU with port 2
"1
a.
1
1
+
1
y2iVil
Y2
y22
yi
'
1
I
FIG.
v2
9.13.
2'
Two-port equivalent.
« For a lucid treatment of controlled
2nd Ed., McGraw-Hill Book Company,
sources, see E. J. Angelo, Electronic Circuits,
New York,
1964.
Network
terminated in a load impedance
the /* mesh,
ZL
If
.
write the
269
II
mesh equation
for
we have
-«tA-(*ii
Since
we
anal/sis
+ Zx )li
(9.64)
Vt = —ItZL we readily obtain
,
z« -
r = ^rr
It also is clear that the current-ratio transfer
two-port network
< 9 - 65 >
function for the terminated
is
-*
h
In similar fashion,
we
"*»
+ zz,
=
(9.66)
Z22
obtain the voltage-ratio transfer function for the
circuit represented in Fig. 9.13 as
Via.
(9.67)
^8 + yM
Next, suppose
we
are required to find the transfer function
the two-port network terminated at both ends, as
shown
VJV„
for
in Fig. 9.14.
We first write the two mesh equations
(9.68)
Next,
we
solve for It to give
/t
=
+ 2ll)(^2 + «m) - *1**21
Vt = —RtI%, we may now arrive
(9.69)
(«1
From
the equation
at the following
solution.
Y*=_Ml =
V,
V„
?«£t
{R1
+
z^)(R t
+ z,,) - znZlt
*u
r^aro £>*
FIG. 9.14
(9.70)
Network
270
anal/sis and synthesis
WW—^O1-^h
—MAA"
-£
Z22-Z12
211-Z12
(Z21-«12Ml
v2
»Z12
Vl
FIG. 9.15. Two-port equivalent
Note that the equivalent
circuit
with one controlled-voltage source.
of the two-ports in Figs. 9.12 and
circuits
9.13 are not unique. Two other examples are given in Figs. 9.15 and 9.16.
Observe that the controlled sources are nonzero in these equivalent
circuits
only
the circuit
if
nonreciprocal.
is
Finally, let us consider the hybrid equivalent circuit
shown in
Fig. 9.17.
Observe the voltage-controlled source A gl F2 at port 1 and the currentcontrolled source Ag^ at port 2. Let us find the input impedance Z ln
.
The
pertinent equations are
Vi
=
+ hi*V*
h lxh
(9.71)
Va = -ZLh = ~(h 2lh + h^V^ ZL
and
Solving Eq. 9.72 for
(9.72)
Vt we find
,
V = —
htiZiJi
1
+
Substituting V, in Eq. 9.73 into Eq. 9.71,
t^
(9.73)
h&ZL
we have
hiah« z &
( i.
\
1
+
\
j
(9.74)
h^LjJ
(9.75)
so that
h
1
fc
MZ£
h
yi2
o-ii-
Vi
-WW.yu + yn
-<
o
,73
y22
+ yi2<
Q)
f
Vi
I
FIG. 9.1*. Two-port equivalent
circuit
with one controlled-current source.
Network
FIG. 9.17. Hybrid equivalent
anal/sis
271
II
circuit.
We can easily check to see that Eq. 9.75 is dimensionally correct since Au
has the dimensions of impedance, h 22 is an admittance, and h 12 h tl are
,
dimensionless since they represent voltage and current ratios, respectively.
9.4
INTERCONNECTION OF TWO-PORTS
In this section
we
will consider various interconnections of two-ports.
We will see that when a pair of two-ports are cascaded, the overall transmission matrix is equal to the product of the individual transmission
matrices of the two-ports. When two two-ports are connected in series,
their z matrices add ; when they are connected in parallel, their y matrices
add.
First let us consider the case in
ports
Na and N
h
which we connect a pair of two-
in cascade or in tandem, as
shown
in Fig. 9.18.
We see
that
ra-ra
k
The transmission matrix equation
z
12a
i
+
Vi
for
N
tt
N
a is
ara
(9.77)
hb
+'
\
Vsa
Vn
—
(9.76)
*i
+
Nb
—
•
FIG. 9.18. Cascade connection of two ports.
v2
-
Network
272
analysis and synthesis
h r"
h
h'
9-*-
DC
Vi
L.
<
Za
V2
V2
'
*b
Vi'
9
Gyrator__
FIG. 9.19. Gyrator in tandem with
Correspondingly, for N„
T circuit.
we have
(9.78)
Substituting the second matrix into the
first,
we
obtain
Ma-eara
We
see that the transmission matrix of the overall two-port
<network
is
simply the product of the transmission matrices of the individual twoports.
As an example,
let
us calculate the overall transmission matrix of a
T network shown in Fig. 9.19. The ideal
an impedance inversion device whose input impedance Zta is
its load impedance ZL by
gyrator 7 in tandem with a
gyrator
is
related to
Zm =
=
a*YL
21
(9.80)
Zi
The constant a
in Eq. 9.80
is
defined as the gyration resistance.
regard the gyrator as a two-port,
its
Vx =
If
we
denning equations are
a(-/«)
h - -a vt
(9.81)
' B. D. H. Tellegen, "The Gyrator, a New Electric Network Element," Phillips
Research Repts., 3 (April 1948), 81-101, see also Huelsman, op. eit., pp. 140-148.
Network
so that the transmission matrix of the gyrator
"0
analysis
in
II
is
a
1
.a
We see that for the gyrator
J
AD-BC<=-1
(9.82)
a nonreciprocal device, although it is passive. 8
The overall transmission matrix of the configuration in Fig. 9.19 is
obtained by the product of the individual transmission matrices
Therefore the gyrator
A
is
za
B
C D
+z
»
V* + V. + V.
i
La
(9.83)
a
V* + Vc + V.
+ *»
z«
ah
az h
we check
If
the configuration in Fig. 9.19 for reciprocity,
we
see that for
transmission matrix in Eq. 9.83
AD-BC=-\
We
thus see that any reciprocal network connected in tandem with a
gyrator yields a configuration that
Next consider the
N
b
(9.84)
situation in
are connected in parallel, as
is
nonreciprocal.
which a pair of two-ports
shown
in Fig. 9.20.
Na
and
Let us find the
FIG. 9.20. Parallel connection of two ports.
*
Most gyrators are microwave
devices that depend
upon the Hall effect in
ferrites.
274
Network
anal/sis
and synthesis
lc
FIG. 9.21
y parameters for the overall two-port network. The matrix equations
for the individual two-ports are
V
V
L/J
and
From
UJ
Fig. 9.20,
we
J/llo
Vila'
-Vila
2/*2a-
~2/ll6
2/l2&
-2/216
2/226-
>1«"
W
(9.85)
>«"
(9.86)
UJ
must hold.
see that the following equations
~ V\a = V\h
h = Ao + Aft
V\
V*
=
^2o
=
'26
(9.87)
A = Ao + I»
In connecting two-ports in series or in parallel,
we must be
careful
not altered when
connected in series or parallel with another two-port. For example, when
we connect the two-ports in Fig. 9.21 in parallel, the impedances Z, and
Zg will be short circuited. Therefore, to insure that a two-port network
that the individual character of a two-port network
is
does not interfere with the internal affairs of the other, ideal transformers
are used to provide the necessary isolation. In matrix notation, the sum
of the individual [/J matrices of the two-ports in parallel must equal the
[/] matrix of the overall two-port network. Thus we have
(9.88)
4
Vila
2/««
2/21a
Vtia
J/116
2/126
2/226-
y**
Network
*
II
275
/&,
/la
+
anal/sis
+
+
3
JV«
e>
V2a
Ideal
Vi
Vz
In
lib
C
+
+
Nb
Vu
Vu,
^
'-'
FIG. 9.22. Series connection of two ports.
so that the y parameters of the overall two-port network can be expressed
in terms of the y parameters of the individual two-ports as
If
we connect
|yii
y«1
Ly«
y»J
two-ports in
_
+
lytia +
[vna
series,
as
yita
Vtu
Vin
(9.89)
ytu>-
in Fig. 9.22, we can express
network in terms of the z param-
shown
the z parameters of the overall two-port
eters
+
y*ta +
Vub
of the individual two-ports as
pil
^t]
LZ»i
2tJ
+ 2m
z Mo + z*i J
Z
*11&
_ pUa + «U»
Zflb
Lzaia + Z
Zl*.
"l
(9.90)
222&-1
21ft
We may summarize by the following three points.
When
two-ports are connected in parallel, find the y parameters
and, from the y parameters, derive the other two-port parameters.
2. When two-ports are connected in series, it is usually easiest to find
the z parameters.
3. When two-ports are connected in tandem, the transmission matrix
1.
first,
generally easier to obtain.
is
As a
final
in Fig. 9.6.
example,
let
us find the y parameters of the bridged-T circuit
We see that the bridged-T circuit would be decomposed into
a parallel connection of two-ports, as shown in Fig. 9.23. Our task is to
The y parameters
find the y parameters of the two-ports
a and
b
of b are obtained by inspection and are
N
first
N
.
N
= y*u, = — i
= y*n = s
Viiii
ym>
N
a is
a
T circuit so
that the z parameters can be obtained
(9.91)
by
inspection.
276
Network
analysis
and synthesis
These are
zllo
— 2Mo — S +
l
(9.92)
— Z«]„
ZlSn
*12o
*Mo
1
We then find the y parameters from the equations
Az„
+
2s
1
(9.93)
yi2a = y«io = t
2 12a
Since both JV and
N
b
now
1
we know that ylt — yat
The y parameters for the bridged-T
obtained as
3/ii
tflt
=
+ Viu
= s(s + 1)
2s + 1
—
Vila
1
=
2s*
+ 4s + 1
+ 1)
2(2s
(9.94)
Vl2a
+
Vltb
sf_
2s
9.5
+
are symmetrical two-ports,
for the overall bridged-T circuit.
circuit are
2s
+
1
_
1
2
+
2s
+
2(2s
+
l)
2s*
1
INCIDENTAL DISSIPATION
As we have
seen, the system function H(s) of an
R-L-C network consists
of a ratio of polynomials whose coefficients are functions of the resistances,
inductances, and capacitances of the network. We have considered, up
to this point, that the inductors
and capacitors are
dissipationless;
i.e.,
Network
analysis
II
277
no parasitic resistances associated with the L and C. Since, at
high frequencies, parasitic losses do play an important role in governing
system performance, we must account for this incidental dissipation
there are
somehow.
An
effective way of accounting for parasitic resistances is to
the pure inductances and capacitances with incidental
dissipation by associating a resistance r, in series with every inductor L { ,
"load
down"
and, for every capacitor Ct , we associate a resistor whose admittance is t ,
g
as depicted in Fig. 9.24. Suppose we call the system function of the
network without parasitic dissipation (Fig. 9.24a) H(s), and the system
function of the "loaded" network H^s) (Fig. 9.246). Let us consider the
the relationship between H(s) and H^s) when the dissipation is uniform,
i.e.,
in a
manner such
that
Lt
(9.95)
Ct
where the constant a is real and positive.
When a network has uniform dissipation or is uniformly loaded, the
sum of impedances in any mesh of the unloaded network
m
if
<=
R +
4
sL{
+—
(9.96)
becomes, after loading,
m'a
= JR, + sL +
t
= R + L&i +
{
r,
+
a)
+
(9.97)
1
C (s +
4
a)
Si
c,
L-vw-J
HM
Hi(s)
(a)
FIG. 9.24.
* H(s + a)
(b)
(a) Original
network.
(A)
Loaded network.
Network
278
analysis and synthesis
Jf
-It-
=*=**
2Q
1Q
<2h
FIG. 9.25
Similarly,
on node
basis, if the
original (unloaded)
network
n if
admittance between any two nodes of the
is
=G +
sCt
t
1
+
(9.98)
then the same node admittance after loading
»'„
=G +
sCt
=G +
Q(s
t
+
g(
is
1
+
(9.99)
t
1
+ a) +
+ a)
L,(s
Since any system function can be obtained through
tions,
H(s
H(s
+
+
it is
mesh or node equabecomes
readily seen that the original system function H(s)
«) after the
network has been uniformly loaded, that
is
/^(s)
=
a).
Consider the following example. Let us
the unloaded network in Fig. 9.25.
1
2
Vu
,
s
,
3
By
s
1
4
2
find the
,
y parameters for
we have
7s
12
= 1 + — + - - 1 + —7—
4
2s
Via
first
inspection
(9.100)
4s
= »»i=-4
Then, for a loading constant a = J, the loaded network is shown in Fig.
9.26. In parallel with the capacitor Cx = J f, we have an admittance
gl
In parallel with the capacitor
gt
= *C1 = $mho
(9.101)
Ct =
is
J
f,
the associated admittance
— *Ca = imho
(9.102)
Network
anal/sis
II
279
80
-vw
L-\H
if
J
20-
*
6Q<
< 1Q
=*f
10
1 2h 1
3
—
o—
= J.
FIG. 9.26. Loaded network a
In series with the inductor
L=
r,
2
=
h,
we have a
olL
resistor
=1Q
(9.103)
Now we determine the y parameters for the loaded network to be
1
>
.
s
,
1
,
s
= i + Ks + i) +
- 1 + A(* + t)
y'«
.
1
K*
+
i)
=i+
+1 4
+
D* + 2
=i+
4(s + |)
2a-
8
(a
'-- (H~*H)
(9.104)
We
see that the y parameters of the loaded network could have been
obtained from the y parameters of the unloaded network by the relation-
ship
H
t (s)
=
H(s
+
a).
We will make use of the uniform loading concept to prove an important
theorem concerning network
9.6
realizability in
Chapter
10.
ANALYSIS OF LADDER NETWORKS
In this section we will consider a simple method of obtaining the
network functions of a ladder network in a single operation.* This
method depends only upon relationships that exist between the branch
* F. F. Kuo and G. H. Leichner, "An Iterative Method for Determining Ladder
Network Functions," Proc. IRE, 47, No. 10 (Oct. 1959), 1782-1783.
280
Network
analysis
and synthesis
z,
>
p
'i
*2
*
I
*
^
^*4
*J*i
FIG. 9.27. Ladder network.
currents
and node voltages of the
in Fig. 9.27, where
all
all
ladder.
Consider the network shown
the series branches are given as impedances and
If the v denote node
denote branch currents, then the following relationships
the parallel branches are given as admittances.
voltages
and
i
apply.
vi+1
=
il+i
Zi+s +
vi+>
These equations form the basis of the method we discuss here.
To illustrate this method, consider the network in Fig. 9.28, for which
the following node voltage and branch current relationships apply.
- yJA v^s)
VJL») - '*(*) U') + v&) = [1 + Z^*) Yt (s)] v&)
h(s) - Y2 (s) VJLs) + Us)
= {Y&)[1 + ZJs) Us)] + Y,{s)} VAs)
V (s) = I (s)Z (s)+Va(s)
= {zjlyai + z,yj + rj + (i+ z,n)} v&)
/•(*)
1
1
(9.106)
1
Upon examining the set of equations given in Eq. 9.106, we see that
each equation depends upon the two previous equations only. The first
equation, Vt
V% is, of course, unnecessary. But, as we shall see later,
it is helpful as a starting point. In writing these equations, we begin at
=
l
+
Vi
h
Zi
\-^H
z3
Y2
Y*
T
v2
j.
FIG. 9.28
Network
analysis
II
281
the 2-2' port of the ladder
and work towards the 1-1' port. Each succeeding equation takes into account one new immittance. We see,
further, that with the exception of the first two equations, each subsequent
is obtained by multiplying the equation just preceding it by the
immittance that is next down the line, and then adding, to this product
the equation twice preceding it. For example, we see that /,(» is obtained
by multiplying the preceding equation V&) by the admittance Yt(s).
The next immittance is ZJis). We obtain KB(i) by multiplying the previous
equation IJs) by ZJs) to obtain /gZjj then we add to this product the
equation twice preceding it, V^s), to obtain Va
It Z,
Vt The process
equation
=
+
.
then easily mechanized according to the following rules (1) alternate
writing node voltage and branch current equations; (2) the next equation
is obtained by multiplying the present equation by the next immittance
is
:
(as
we work from one port
to the other),
and adding to
this
product the
results of the previous equation.
Using
obtain the input impedance Z{n(s) by
by the equation for I^s). We obtain the
function V£s)IVi(s) by dividing the first equation
of equations,
this set
we
dividing the equation for V^s)
voltage-ratio transfer
P*C0 by the
last
equation
as transfer immittances
if
we
let
VJs) term is
K2(j) =
We obtain other network functions such
manner. Note that
In taking ratios of these equacanceled. Therefore, our analysis can be simplified
every equation contains
tions, the
V-fe).
and current
ratios in similar
Vt(s) as a factor.
1.
equation of the set were a current variable fj(s) instead of the
voltage Vi(s), the subsequent equations would contain the current variable
I£s) as a factor, which we could also normalize to 7/s)
1. An example
If the
first
=
in which the
equation
a current rather than a voltage equation
may be seen by determining the y parameters of a two-port network.
Before we embark upon some numerical examples, it is important to
note that we must represent the series branches as impedances and the
shunt branches as admittances. Suppose the series branch consisted of a
resistor
R=
first
1 £2
of the branch
is
in parallel with a capacitor
z(s) =
J f Then the impedance
.
-^L
= _i+R
+
llsC
and must be considered as a
ladder.
C=
is
Similarly, if a shunt
an inductor L^
=
1
s
2
(9 . 107'
v
)
single entity in writing the equations for the
branch consists of a resistor R1
branch is
= 2Q and
h, then the admittance of the
y(s)
=
7T~
2 + s
( 9 - 108 >
,
282
Network
analysis and synthesis
in this discussion is that we must use the total impedance
or admittance of a branch in writing the equations for the ladder.
The key point
Example 9.1. Let us find the voltage ratio VJVlt the current ratio Ijlu the
input impedance Zx = VJIlt and the transfer impedance Zn = VJI-i for the
network in Fig. 9.29. First, we must represent the series branches as impedances
and the shunt branches as admittances, as shown in Fig. 9.30. The branch
current and node voltage equations for the network are
h
h
—h*-
-/fflnp-*—
f
3s
3h
Vi
2«i
Vi
V2
2f=t=
*h
FIG. 9.30
FIG. 9.29
V^s)
=
Us)
= 2s
Va(s) =
V,
1
=
3s(2s)
+
1
-(6s*
s
+
1)
6s*
+
1
(9.109)
It(s)
=
VM
+-
14s
s
+
14g+
+ l=6s* + 57+-l
6s2
\)
{
The various network
+ 2s =
S*
functions are then obtained.
7 —
(a)
—
— 6^+57^+8
3
14s
s*
Yi
<*)
+2s
+
6s»
57s8
+
8
(9.110)
2s*
(c)
li
** ~
id)
Example
network
9.2.
Find the
14s*
/x
+2
"
14s2
+
2
short-circuit admittance functions
y n and
j/2 i
for the
in Fig. 9.31. To obtain the short-circuit functions, we must short
circuit the 2-2' port. Then we represent the series branches as impedances and
the shunt branches as admittances so that the resulting network
is
given as
is
Network
rVW*-i
h
l
V.
analysis
II
283
I.
2a
Vi
2f
»lh =j=2f
FIG. 9.31
The
illustrated in Fig. 9.32.
/,
Ja
pertinent equations are
=1
Va =2
= K2) + 1 = * +
1
(9.111)
+
2$*
h
l/5\
5/2
We now obtain our short-circuit functions as
2
12
7J + 5 +
h
1
yn
-,'
la
V°
1
«» +
+
h
~^~~5
»
(9.112)
2
's.aa
Vi
2»
2
+l
«
o—
1
FIG. 9.32
A
number of other
on ladder networks have appeared
but a few, there are the works of
Bubnicki," Walker," and Dutta Roy. 14
contributions
in the recent literature.
Bashkow,10 O'Meara,11
To name
10
T. R. Bashkow, "A Note on Ladder Network Analysis," IRE Trans, on Circuit
Theory, CT-8, (June 1961), 168.
11
T. R. O'Meara, "Generating Arrays for Ladder Network Transfer Functions,"
IEEE Trans, on Circuit Theory, CT-10, (June, 1963), 285.
11
Z. Bubnicki, "Input Impedance and Transfer Function of a Ladder Network,"
IEEE Trans, on Circuit Theory, CT-10, (June,
"F. Walker, "The Topological Analysis
1963), 286.
of Non-Recurrent Ladder Networks,"
No. 7, (July, 1964), 860.
14
S. C. Dutta Roy, "Formulas for the Terminal Impedances and Transfer Functions
of General Multimesh Ladder Networks," Proc. IEEE, 52, No. 6, (June, 1964), 738.
Proc. IEEE, 52,
—
284
Network
analysis
>
and synthesis
Because of the recursive nature of the equations involved, the digital
ideal tool for the analysis of ladder networks. In Chapter
15 a detailed description is given of a digital computer program based
upon the algorithm described in this section for evaluating ladder network
computer is an
functions.
Problems
9.1
Find the
z,
y, h,
and
T parameters of the networks shown in the figures.
Some of the parameters may not be defined for particular
—
c
circuit configurations.
-OQ—
PS
Y
'
o
PROB.
9.2
Find the
(e)
(b)
(a)
z,
y, h,
lo
and
9.1
T parameters for the networks shown in the figures,
—iA/W
1
\AAA/
<
i
o2
Network
93 Find the z parameters for the lattice and bridge
shown. (The results should be identical.)
9.4
For the
lattice
and bridge
terms of the admittances
Ya =
analysis
and
Yb
=•
MZb
285
circuits in the figures
circuits in Prob. 9.3 find the
1/Z„
II
y parameters
in
.
93 For the circuit shown, find the voltage-ratio transfer function VJVX and
the input impedance x = V^lx in terms of the 2 parameters of the two-port
network
and the load resistor RL .
N
Z
1
h
I2
2
+
Vi
+
N
Rl<
-
v2
_
1
2'
1'
PROB. 9S
Network
286
anal/sis
and synthesis
9.6 For the cascade connection of two-ports depicted in the figure, show that
the transfer impedance Zj 8 of the overall circuit is given in terms of the a parameters of the individual two-ports by the equation
In addition, show that the short-circuit admittance yri
V-a
Vu.b
+
Vita
h
+
+
Na
Vi
Nb
PROB.
9.7
figure.
of
given by
VnJti»
= -
h
is
Lx
,
v2
PROB.
9.6
9.7
Find the z and y parameters of the transformer (nonideal) shown in the
Determine the T- and w-equivalent circuits for the transformer in terms
L2 , and M. {Hint: Use the z parameters for the T-equivalent circuit,
and the y parameters for the w-equivalent circuit.)
9.8 The inverse hybrid parameters of a two-port network are defined by the
equation
eh
*ii
gl*
<f22j
g parameters in terms of either the z or y parameters and give a
physical interpretation of the meaning of these parameters; i.e., say whether a
parameter is an open- or short-circuit parameter, and whether it is a driving
point or transfer function. Finally, derive the conditions of reciprocity for the
g parameters.
Express the
9.9
Prove that for a passive reciprocal network
AD - BC =
1
where A, B, C, Dare the elements of the transmission matrix.
9.10 Find the z and h parameters of the
sented by its T-circuit model.
-VA
common
emitter transistor repre-
— —Wv-^O
-o
i
r
mh
.
+
J2
V2
Vi
PROB. 9.10
Network
9.11
anal/sis
II
287
The circuit in part (a) of the figure is to be described by an equivalent
shown in part (b). Determine Zeq in (b) as a function of the elements
input circuit
and voltages
in (a).
R,
9.12
of the
Find the
T parameters
of the configurations shown in parts (a) and (b)
figure.
9.13 Find the y parameters of the twin-T circuit in Prob. 9.2c by considering
the circuit to be made up of two T circuits in parallel.
9.14
Find the z parameters of the
circuits
shown.
Network
288
1
o
anal/sis
and synthesis
h
>—
Y„
h
2
-o
+
V2
Vi
V
1
2*
Ideal
transformer
(a)
h
h<
Z„
2
o
V2
Vi
Ideal
<b)
transformer
2'
PROB.
9.14
PROB.
9.15
Ideal
transformer
PROB. 9.16
i
Network
analysis
II
289
9.15
Find the
9.16
Find the z parameters of the
9.17
For the circuit in Prob. 9.26 determine the y parameters of the uniformly
z
parameters of the circuit shown.
circuit
shown.
loaded circuit derived from the original circuit with the dissipation a =0.1.
Plot the poles and zeros of both cases.
9.18 Find the transfer impedance VJIX for the circuit in part (a) and the
voltage ratio VJV1 for the circuit in part (b). Plot the poles and zeros for the
transfer functions obtained.
—
h
o—>- /Tnnn^-r-^voTP
}h
.
+
in
:£4f
Vi
10
V2
(a)
—WW-
2
Vi
+
2h
lf=:
if;*:
:ia
v2
PROB.
(6)
9.19
method
Find the
short-circuit parameters for the ladder
network
9.18
utilizing the
in Section 9.6.
10
HWv—
20<
20
_lh
Hwv—i-'Tnnp-^
=t=if
if
PROB.
9.1*
30
|—VvV-|
Yf-rifippJfL£-
20-
if
Vi
if==
1Q<
V2
4f=r
PROB. 9.20
9.20 Determine the voltage ratio VJVlt the current ratio IJIlt the transfer
impedance VJIlt and the driving point impedance Fi//a for the network shown.
—
chapter 10
Elements
<|>f
realizability
theory
10.1
CAUSALITY
AND
STABILITY
In the preceding chapters we have beep primarily concerned with the
problem of determining the response, giver^ the excitation and the network;
this problem lies in the domain of network analysis. In the next five
chapters we will be dealing with the problem of synthesizing a network
given the excitation E(s) and the response R(s).
any synthesis problem is the system function
The
starting point for
—3
(10.1)
from a given system function.
procedur^
is to determine whether H(s)
step in a synthesis
There are two important
network.
physical
passive
can be realized as a
we mean that a
causality
stability.
By
and
causality
considerations
Our task is
The first
to synthesize a network
voltage cannot appear between any pair of terminals in the network
before a current is impressed, or vice ver^a. In other words, the impulse
response of the network must be zero for
h(t)
As an example,
is
causal, whereas
=
t
<
0, that is,
<0
(10.2)
e- «(0
(10.3a)
the impulse response
h(t)
=
h(t)
= e- "
290
(10.36)
Elements of
realizability
theory
291
W-T)
(a)
FIG.
not causal.
is
(b)
(b) Realizable
Nonrealizable impulse response,
10.1. (a)
impulse response.
In certain cases, the impulse response could be made
by delaying it appropriately. For example, the impulse
realizable (causal)
response in Fig. 10.1a
seconds,
we
not realizable.
is
If
we
find that the delayed response h{t
delay the response by
—
T)
is
T
realizable (Fig.
10.16).
In the frequency domain, causality is implied when the Paley-Wiener
criterion 1 is satisfied for the amplitude function \H(jw)\. The Paley-Wiener
criterion states that a necessary and sufficient condition for an amplitude
function \H(ja>)\ to be realizable (causal)
is
that
dco
<
|log \H(ja>)\
£
1
+o>a
oo
(10.4)
The following conditions must be satisfied before the Paley-Wiener
criterion is valid: h(t) must possess a Fourier transform H(ja>); the
square magnitude function \H(jw)\* must be integrable, that is,
\H(jco)\ *dco
r.
<
oo
(10.5)
The physical implication of the Paley-Wiener criterion is that the amplitude
\H(jco)\ of a realizable network must not be zero over a finite band of
frequencies. Another way of looking at the Paley-Wiener criterion is that
the amplitude function cannot
fall off
to zero faster than exponential
For example, the ideal low-pass filter in Fig. 10.2 is not realizable
because beyond m c the amplitude is zero. The Gaussian shaped curve
order.
|HO)l
shown
in Fig. 10.3
is
=
(10.6)
not realizable because
|log|tf0a>)||
1
e~°
=
o*
(10.7)
R. E. A. C. Paley and N. Wiener, "Fourier Transforms in the Complex Domain,"
Soc. Colloq. Pub., 19 (1934), 16-17.
Am. Math.
Network
292
analysis
and synthesis
\H(ja>)\
\H(ju)\
FIG.
FIG.
10.2. Ideal filter char-
f"
J-*
so that the integral
is
not
finite.
10.3.
Gaussian
filter
character-
istic.
acteristic
On the other hand,
«>
1
+
,
(10.8)
dea
i
eof
the amplitude function
1
(10.9)
\H(jo>)\
vrr
co-
does represent a realizable network. Ih fact, the voltage-ratio transfer
function of the R-C network in Fig. 10.4 has an amplitude characteristic
given by \H(Jm)\ in Eq. 10.9.
For the ideal filter in Fig. 10.2, the inverse transform A(f) has the form
h{t)
=A
sin <o c t
(10.10)
nt
A is a constant. From the sin xfx curve in Fig. 3J4, we see that
nonzero for t less than zero. In fact, in order to make h(t) causal,
it must be delayed by an infinite amount. In practice, however, if we
delay h(i) by a large but finite amount t d such that for t < the magnitude
of h(t — t^ is less than a very small quantity e, that is,
where
h(t) is
\Kt
~
Q\
<
e
*<0
—wwv
1Q
Vi(s)
lr.^z
FIG,. 10.4
v
w
Elements of
we then can approximate
zero for
t
< 0.
—
h(t
(For a more
realizability
t£ by a causal response h^t) which
an
excellent treatment
then for a bounded excitation
If a network
words, if
other
In
bounded.
is
also
(0
is stable,
where
stable,
Cx and
C, are
then from
<
the convolution integral
<
the response
0<.t<co
C*
real, positive, finite quantities.
IKOI
by Wallman.*)
e(t)
0^r<oo
KOKCi
KOI
is
detailed discussion of the Paley-Wiener
criterion, the reader is referred to
then
293
theory
we
CiJ"|*(t)I dr
If a linear system
is
obtain
<
C,
(10.11)
Equation 10.1 1 requires that the impulse response be absolutely integrable,
or
""|/i(t)|
dr
<
(10.12)
oo
/;
One important requirement for h(t) to be absolutely integrable
impulse response approach zero as
t
approaches
is
that the
infinity, that is,
lim/i(r)-"0
<-»00
can be said that with the exception of isolated impulses, the
impulse response must be bounded for all t, that is,
Generally,
it
|A(0I<C
all*
(10.13)
where C is a real, positive, finite number.
Observe that our definition of stability precludes such terms as sin ay
from the impulse response because sin a> f is not absolutely integrable.
These undamped sinusoidal terms are associated with simple poles on
the yw axis. Since pure L-C networks have system functions with simple
poles on the jm axis, and since we do not wish to call these networks
unstable, we say that a system is marginally stable if its impulse response
approach zero as t
is bounded according to Eq. 10.13, but does not
approaches infinity.
In the frequency domain, the
stability criterion requires that the
system
G. E. Valley Jr. and H. Wallman, Vacuum Tube Amplifiers, McGraw-Hill Book
Company, New York, 1948, Appendix A, pp. 721-727.
•
Network
294
anal/sis
and synthesis
function possess poles in the left-half piane or on theyw axis only. More-
on theyVo axis must be sini]pie. 3 As a result of the requirement of simple poles on they'w axis, if H{,s) is given as
over, the poles
H(s)
=
a„s
bm
n
m
s
+
+
a n_ 1 s
n~ 1
+
m~ l
b m _ lS
+
+! +
•
•
•
+a
bis + b
fl x s
(10.14)
then the order of the numerator n cannot exceed the order of the denominator m by more than unity, that is, n — m <£ 1. If n exceeded m
by more than unity, this would imply that at s =jo> = oo, and there
would be a multiple pole. To summarise, in order for a network to be
stable, the following three conditions oii its system function H(s) must
be satisfied:
1.
H(s) cannot have poles in the right-fealf plane.
2.
H(s) cannot have multiple poles in
3.
The degree of
the denominator by
Finally,
it
tjhey'w axis.
the numerator of H(s) cannot exceed the degree of
more than
unity.
should be pointed out that a rational function H(s) with poles
in the left-half plane only has an inverse transform h(t), which
t
< 0.*
In
this respect, stability implies causality.
is
zero for
Since system funct'ons
of passive linear networks with lumped elements are rational functions
with poles in the left-half plane or jco axis only, causality ceases to be a
problem when we deal with system functions of this type. We are only
concerned with the problem of causality when we have to design a filter
for a given amplitude characteristic such as the ideal filter in Fig. 10.2.
We know we could never hope to realize exactly a filter of this type
because the impulse response would not^ be causal. To this extent the
Paley- Wiener criterion is helpful in denning the limits of our capability.
10.2
HURWITZ POLYNOMIALS
In Section 10.1
its
we saw
that in order fot a system function to be stable,
poles must be restricted to the left-half plane or theyeo axis. Moreover,
the poles
on
theyco axis
must be simple. The denominator polynomial of
the system function belongs to a class
*
i tt
In Chapter 6
it
of]
polynomials
was shown that multiple poles on
known
they'co axis
gave
as Hurwitz
rise to
terms as
sin wot.
4
G. Raisbeck, "A Definition of Passive Linear Networks in Terms of Time and
Energy," /. Appl. Phys., 25 (Dec., 1954), 15ip-1514. The proof follows straightforwardly from the properties of the Laplace transform.
Elements of
polynomials.
A
polynomial P(s)
is
theory
realizability
295
said to be Hurwitz if the following
conditions are satisfied:
1.
2.
is real when s is real.
The roots of P(s) have real
P(s)
As a
result
of these conditions,
P(s)
then
all
then
ac,
parts which are zero or negative.
must be
= (s+
Hurwitz because
hand,
is
all
real
if s {
;
=
<x
(10.15)
+ j@ is a root of P(s),
4
The polynomial
negative.
P(s)
a Hurwitz polynomial given by
= ansn + a^s"-1 +-- + a1s + a
the coefficients a {
must be
if P(s) is
1)(j
+
1
+js/2)(s
+
-js/2)
1
of its roots have negative real parts.
=
G(s)
-
(s
l)(.y
+
2)(j
+
(10.16)
On the
other
(10.17)
3)
not Hurwitz because of the root s = 1, which has a positive real part.
Hurwitz polynomials have the following properties:
is
1. All the coefficients a { are nonnegative.
This is readily seen by
examining the three types of roots that a Hurwitz polynomial might
have. These are
= — y<
s = ±.jm
s — —cti±.jPi
s
yt
a-i
The polynomial P(s) which contains
Since P(s)
is
= (s +
y< Xji
and
positive
and
positive
(Of real
t
P(s)
real
+
real
these roots can be written as
+
<»*)[(*
oO*
+
m
•
•
the product of terms with only positive coefficients,
(10.18)
it
follows
must be positive. A corollary is that between
the highest order term in s and the lowest order term, none of the coeffithat the coefficients of P(s)
cients
may be
zero unless the polynomial
must not be zero
is
even of odd. In other words,
the polynomial
is neither even
nor odd. This is readily seen because the absence of a term a{ implies
cancellation brought about by a root s — y< with a positive real part.
2. Both the odd and even parts of a Hurwitz polynomial P(s) have
roots on the ja> axis only. If we denote the odd part of P(s) as n(s) and
the even part as m(s), so that
fln-i. a«-2>
•
•
»
fl»» <*i
P(s)
if
= n(s) + m(j)
(10.19)
Network
296
analysis and synthesis
then m(s) and n(s) both have roots on tne jw axis only. The reader is
8
referred to a proof of this property by Guillemin.
3. As a result of property 2, if P(s) is either even or odd, all its roots are
on
they'co axis.
The continued
4.
parts or the even to
quotient terms.
y(j)
= m(s)ln(s),
fraction expansion of the ratio of the
odd
odd
to even
parts of a Hurwitz polynomial yields all positive
—
n(s)lm(s) or
Suppose we denote the ratios as y>(s)
then the continued fraction expansion of y>(s) can be
written as
y(s)
^
= q xs +
«««
(10.20)
+
«ss
+
where the quotients qlt q%,..-,qn must be positive
6
P(s) = «(j) + m(s) is Hurwitz.
sion,
we must perform a
series
To
if
the polynomial
obtain the continued fraction expan-
of long divisions. Suppose
V<s)
= 2®
y>(s) is
(10.21)
n(s)
is of one higher degree than n(s). Then
obtain a single quotient and a remainder
where m(s)
m(s),
we
if
we
divide n(s) into
n(s)
The degree of the term R^s) is one lower than the degree of «(*). Therefore
if we invert the remainder term and divide, we have
»(°L
Inverting
=q . + M2l
(io.23)
and dividing again, we obtain
RM^qj + M)
*,(*)
(10.24)
*.(«)
E. Guillemin, The Mathematics of Circuit Analysis, John Wiley and Sons, New York,
An excellent treatment of Hurwitz polynomials is given here.
*
proof can be undertaken in connection with L-C driving-point functions; see
*
1949.
A
E. Van Valkenburg, Introduction to Modern Network Synthesis, John Wiley and
Sons, New York, 1960.
M.
Elements of
We
realteability
theory
297
see that the process of obtaining the continued fraction expansion
simply involves division and inversion. At each step we obtain a
then invert the
quotient term q<s and a remainder term, R {+1 (s)/RJ[s).
into RJs) to obtain a new quotient.
remainder term and divide
There is a theorem in the theory of continued fraotions which states that
of
ip(s)
We
R^s)
the continued fraction expansion of the even to
odd or odd
to even parts
7
finite in length.
Another theorem states that, if
of a polynomial must be
odd
to even or even to odd parts
the
expansion
of
fraction
the continued
terms,
then the polynomial must
quotient
yields
positive
of a polynomial
JV(s).*
That is, if we write
factor
multiplicative
within
a
Hurwitz
to
be
fTO-irO^M
(10.25)
then F(s) is Hurwitz, if W(s) and F^s) are Hurwitz. For example,
test whether the polynomial
F(s)
is
- i* + s* +
5s*
+
3s
+4
let
us
(10.26)
Hurwitz. The even and odd parts of F(s) are
n(s)
We now
= s* +
3s
= "»W/«C0 by
and then inverting and dividing again, as given by
perform a continued fraction expansion of yfa)
dividing n(s)
by
m(s),
the operation
s*
+
+ 5s* + 4(s
s* + 3s*
2s* + 4)j» + 3j(*/2
j» + 2*
3s)s*
s)2s*
+ 4(2s
2s*
4)j(j/4
so that the continued fraction expansion of y(s)
*,)
^
= 2& - 5 +
»(«)
is
1
+
(10.28)
l
s/4
'
1
See Van Valkenburg, be. eit.
W(s) is a common factor in mfc) and
n(*).
Network
298
Since
all
analysis
and synthesis
the quotient terms of the continued fraction expansion are
positive, F(s) is Hurwitz.
Example
10.1.
Let us
whether the polynomial
test
= 5s + 2s9 +
GO)
is
+6
3s
(10.29)
Hurwitz. The continued fraction expansion of n(s)lm(s)
is
obtained from the
division
+
2s2
6)5*
s*
+ 3* (s/2
+ 3s
We see that the division has been terminated abruptly
+ 3s. The polynomial can then be written as
by a
common
factor
5s
G(s)
We know
+ 3s is
s*
that the term
+
(s*
2/s
10.2.
is
-
+
>H)
3s)
1
which we referred to
= 57 + 2s* +
2s5
The term
«(*)
•*
wiO)
2
s*
+
3s
is
the mul-
earlier.
now
non-Hurwitz.
is
+ s* + 4s3 +
fraction expansion of F(s) is
(10.30)
Since the multiplicative factor
Hurwitz.
Next consider a case where W(s)
F(s)
The continued
+
Hurwitz.
is
also Hurwitz, then G(s)
tiplicative factor fV(s),
Example
1
=
85*
+
8*
+4
(10.31)
obtained.
1
'
1
fr
+
j^t)
(1 °- 32)
(S4-+-4)
We thus see that
fV(s)
=
s*
W(s)
It is clear
that F(s)
Example 103.
is
+ 4, which can be factored into
= (s* + 2s + 2X*8 -2^+2)
(10.33)
not Hurwitz.
Let us consider a
F(s)
more obvious non-Hurwitz polynomial
= 5* + 5s +
2t*
+
35
+2
(10.34)
Elements of
The continued
realizability
theory
299
fraction expansion is
3
j
+
3s)S*
s*
+ 2s* + 2 (s
+ 3s*
-j*+2)j» +
3*(-,j
5s)-5*+2(-,s/5
-4*
2}5iXfs
We see that F(j) is not Hurwitz because of the negative quotients.
Example
10.4. Consider the case where F(s) is an odd or even function. It is
impossible to perform a continued fraction expansion on the function as it
stands. However, we can test the ratio of F(s) to its derivative, F'(s)* If the
ratio F(s)IF'(s) gives a continued fraction expansion with all positive coefficients,
then F(s)
is
Hurwitz. For example,
F(s)
then F'(s)
F'(s)
is
if
F(s)
is
given as
= s7 + 34s + 2s3 + *
= 7s* + 15** + 6s» +
(10.35)
1
(10.36)
Without going into the details, it can be shown that the continued fraction
expansion of F(s)IF'(s) does not yield all positive quotients. Therefore F(s) is
not Hurwitz.
POSITIVE REAL
10.3
In
this section
known
we
FUNCTIONS
of a class of functions
These functions are important because
will study the properties
as positive real functions.
A
they represent physically realizable passive driving-point immittances.
function F(s) is positive real (p.r.) if the following conditions are satisfied:
is, F(a) is real.
of F(s) is greater than or equal to zero when the real
greater than or equal to zero, that is,
1.
F(s)
is
2.
The
real part
part of s
is
real for real s; that
Re
[F(s)]
^
for
Re s
^
Let us consider a complex plane interpretation of a p.r. function.
Consider the s plane and the F(s) plane in Fig. 10.5. If F(s) is p.r., then a
point
map
<r
on
F(s) plane.
*
the positive real axis of the s plane
onto, a point
F(p^ which must be on the
would correspond
to,
or
positive real axis of the
In addition, a point st in the right half of the s plane would
See Guillemin, he.
cit.
Network
300
analysis
and synthesis
j\mF
JU
F(t) plane
• plane
«i
o
F(sO
(TO
F(ao)
o
FIG.
10.5.
Ref
Mapping of s plane onto F(s) plane.
map onto a point Ffo) in the right half of the F(s) plane.
In other words,
maps onto the
right half of the F(s) plane. The real axis of the s plane maps onto the
real axis of the F(s) plane.
A further restriction we will impose is that F(s) be rational. Consider
the following examples of p.r. functions:
for a positive real function, the right half of the s plane
= Ls (where L is
a real, positive number) is p.r. by definition.
function,
then L is an inductance.
impedance
If F(s)
an
2. F(s) = R (where R is real and positive) is p.r. by definition. If F(s)
is an impedance function, R is a resistance.
3. F(s) = K/s (K real and positive) is p.r. because, when s is real, F(s)
is real. In addition, when the real part of s is greater than zero, Re (s) =
1.
F(s)
is
a>0.
Re
Then
Therefore, F(s)
is p.r.
Ka
(f)"
If F(s) is
o*
+
>0
(10.37)
a>
an impedance
function, then the corre-
a capacitor of I IK farads.
We thus see that the basic passive impedances are
Similarly, it is clear that the admittances
sponding element
is
Y(s)
=K
7(5)
=
Y(s)
= K
p.r.
functions.
Ks
(10.38)
s
are positive real if
K
is
real
and
positive.
We now show that all driving-
point immittances of passive networks must be
p.r.
The proof depends
the following assertion: for a sinusoidal input, the average power
dissipated by a passive network is nonnegative. For the passive network
upon
Elements of
in Fig. 10.6, the average
power
Average power
by the network
dissipated
=
I
Re
realizability
[ZJtjco)]
|/|»
theory
301
is
^
(10.39)
We then conclude that, for any passive network
Re
^
[Zjytt))]
We can now prove that for Re s =
a
(10.40)
^ 0, Re Z^o- + jm) ;> 0.
Consider
the network in Fig. 10.6, whose driving-
point impedance
/
is ZJ^s).
Let us load
the network with incidental dissipation
such that if the driving-point impedance
of the uniformly loaded network is
Passive
network
Zi»W
ZxCO, then
Zfc)
= ZJis +
a)
where a, the dissipation constant,
impedance of a passive network,
is
real
Re Z^jco)
Re ZJa
so that
no.
(10.41)
and
^
positive.
10.6
Since Z^s)
is
the
(10.42)
+ jm) ^
(10.43)
Since a is an arbitrary real positive quantity, it can be taken to be a.
Thus the theorem is proved.
Next let us consider some useful properties of p.r. functions. The
proofs of these properties are given in Appendix D.
1.
If F(s) is p.r., then l/F(s)
driving-point impedance
admittance,
is
is
is
p.r.,
also p.r. This property implies that if a
then
its
reciprocal, the driving-point
also p.r.
functions is p.r. From an impedance standpoint,
two impedances are connected in series, the sum of the
impedances is p.r. An analogous situation holds for two admittances in
parallel. Note that the difference of two p.r. functions is not necessarily
The sum of p.r.
2.
we
see that if
p.r.
;
3.
for example, F(s)
The poles and
=s—
1/s is
not
p.r.
zeros of a p.r. function cannot have positive real
i.e., they cannot be in the right half of the s plane.
Only simple poles with real positive residues can exist on the yco axis.
5. The poles and zeros of a p.r. function are real or occur in conjugate
pairs. We know that the poles and zeros of a network function are
parts,
4.
functions of the elements in the network. Since the elements themselves
are real, there cannot be complex poles or zeros without conjugates
because this would imply imaginary elements.
Network
302
The
6.
may
analysis and synthesis
highest powers of the numerator
differ at
=
zeros at 5
most by
and denominator polynomials
and
unity. This condition prohibits multiple poles
oo.
The lowest powers of the denominator and numerator polynomials
may differ by at most unity. This condition prevents the possibility of
7.
=
multiple poles or zeros at s
8.
The necessary and
real coefficients F(s) to
0.
sufficient conditions for
be
p.r.
a rational function with
are
(a) F(s)
must have no poles in the
(b) F(s)
may have only simple poles on theyco axis with real and positive
right-half plane.
residues.
(c)
Re F(jm)
^
for
Let us compare this
a.
all
new
definition with the original
one which requires
the two conditions.
1.
F(s)
2.
Re F(s)
is
real
^
when s is real.
0, when Re s ^
0.
In order to test condition 2 of the original definition, we must test
every single point in the right-half plane. In the alternate definition,
we test the behavior of F(s) along the
apparent that testing a function for the three conditions
condition (c) merely requires that
jm
axis.
It is
given by the alternate definition represents a considerable saving of
effort, except in simple cases as F(s)
1/j.
=
Let us examine the implications of each criterion of the second definition.
Condition
we test the denominator of F(s) for roots in
we must determine whether the denominator of
(a) requires that
the right-half plane,
i.e.,
is Hurwitz. This is readily accomplished through a continued fraction
expansion of the odd to even or even to odd parts of the denominator.
The second requirement condition (b) is tested by making a partial
F(s)
—
—
and checking whether the residues of the poles
on theya) axis are positive and real. Thus, if F(s) has a pair of poles at
s = ±ja> lt a partial fraction expansion gives terms of the form shown.
fraction expansion of F(s)
S
The
— jCOi
s
+ jco^
residues of complex conjugate poles are themselves conjugates. If
as they must be in order for F(s) to be p.r.— then
the residues are real
K = Ki* so that
—
t
S
— JO)
1
S
+ JCOx
s*
+
(O*
Elements of
K
If
x is
found to be
reaiizability
positive, then F(s) "satisfies the
theory
303
second of the three
conditions.
In order to
test for the third
first find the real part of F(ja>)
let
we must
To do this,
condition for positive realness,
from the
original function F(s).
us consider a function F(s) given as a quotient of two polynomials
F(s)
=
^
(10.45)
6(s)
We
can separate the even parts from the odd parts of P(s) and Q(s) so
that F(s)
is
M
8( S )
+
JV 2(s)
where Mj(s) is an even function and N£s) is an odd function. F(s) is now
decomposed into its even and odd parts by multiplying both P(s) and
Q(s)
by A/g
- Ns so that
r(s)
_ M + NiM*- N*
Mt + JVa Ma - Nt
= M M8t - N * Nt
M 8 - Nt
i
t
We
x
MM
see that the products
MN
t
and
MN
s
x
are
odd
If
we
let s
= joy, we
=
M
Nx - Mj Nt
M* - Nt
(10.47)
t
*
JV,
Nt
are even functions, while
Therefore, the even part of F(s)
^-^
is
(10.48)
is
Odd
while the
and
z
x
functions.
Ev[F(s) ]
and the odd part of F(s)
M
t
[F(s)]
=
^VY'
see that the even part of
odd part of the polynomial
is
(10.49)
any polynomial
is real,
imaginary, so that if F(jw)
is
written as
F(ja>)
it is
clear that
and
=
Re
[F(jo>)]
Re
[F(ja>)]
j Im
[F(ja>)]
+ j Im [F(jo>)]
(10.50)
= Ev [F(s)] \_ja
= Odd [F(s)] \^, a
(10.51)
(10.52)
Therefore, to test for the third condition for positive realness,
we
determine the real part of F(ja>) by finding the even part of F(s) and then
joy. We then check to see whether Re F{joy)
for all <w.
letting s
=
^
Network
304
anal/sis and synthesis
A(u)
Single
root
FIG. 10.7
FIG. 10.8
The denominator of Re F(j(o)
M
a(ja>)*
That
is
always a positive quantity because
- JVaO)2 =
M^to)*
+ N ((o)* ^
(10.53)
2
an extra j or imaginary term in N^jco), which, when
—1, so that the denominator of ReF(ja>) is the sum of
two squared numbers and is always positive. Therefore, our task resolves
into the problem of determining whether
is,
there
is
squared, gives
A(a>)
4 M O) MJjai) -
Ntito) NJJm)
x
£
(10.54)
If we call the preceding function A{m),
we see that A(a>) must not have
of the type shown in Fig. 10.7; i.e., A(co) must never
have single, real roots of to. However, A(a>) may have double roots
(Fig. 10.8), because A(t») need not become negative in this case.
As an example, consider the requirements for
positive, real roots
F(s)
s
s*
to be p.r.
First,
we know
the left-half plane or
that, in
on they'd)
+
+
bs
a
+
(10.55)
c
order for the poles and zeros to be in
axis, the coefficients a, b, c
must be greater
=
or equal to zero. Second, if *
0, then F(s) will possess poles on the
jco axis. We can then write F(s) as
F(s)
3
-
+
°
•
S*
+C
'
5*
+
(10.56)
C
We will show later that the coefficient a must also be zero when b = 0.
Let us proceed with the third requirement, namely,
the equation
Re F(ja>)
^ 0. From
- JVxO) NAjo>) £
a(-(o* + c) + 1x0*^0
A(co) = (b — a)afi + ac ;>
M^o)) Mfim)
we have
which
It is
of
simplifies to
(10.57)
(10.58a)
(10.586)
evident that in order to prevent A(m) from having positive real roots
eu,
b must be greater than or equal to
a, that
is,
b^a. As
a
result,
Elements of readability theory
when b
= 0,
fulfilled in
1. a, b,
2.
b
^
=
then a
0. To summarize, the conditions that must be
order for F(s) to be positive real are
^ 0.
c
a.
Fx(s) =
We see that
,
sr
is p.r.,
305
s
+
+
3s
\+
(10.59)
,
2
while the functions
F«<S) =
(ia60)
TT1
+2
s
As a second example,
are not p.r.
us determine the conditions for the
let
biquadratic function
F(s)
=
TT^TT
+ b^ + b
<
ia62>
s"
We
will assume that the coefficients au Aq, b u b are all real,
to be p.r.
positive constants. Let us test whether F(s) is p.r. by testing each require-
ment of the second
definition.
of the denominator * x and b are positive, the
denominator must be Hurwitz. Second, if bx is positive, we have no poles
on the yeo axis. Therefore we can ignore the second condition.
The third condition can be checked by first finding the even part of
First, if the coefficients
F(s),
which
is
(s
_
The
s*
+
+
O
+
(«* +
[(a„
—
)
ft„)
t»i
-
bof
s
fliftjs'
-
ft x
+
a &.
(10.63)
V
real part of F(j(o) is then
Re
[F o»1
0)
~ °:y 18+
- W< - ««"
t
!
—
)" +
to
CO
O
+
(
flA
dO.64)
t»i
We
see that the denominator of
Re
[F(jca)] is truly
always positive
so it remains for us to determine whether the numerator of
ever goes negative. Factoring the numerator, we obtain
m, t =
(a.
+
b
)
-
a 1b1
±I^
[(flo
+
feo)
Re
_ aAf _ 4flA
[F(jco)]
(1Q65)
Network
306
analysis and synthesis
There are two situations in which
Re
[F(jco)]
does not have a simple real
root.
When the quantity under the radical sign of Eq.
1
10.65
is
zero (double,
(complex roots). In other words,
real root) or negative
+ *o) - «AP - 4aA <.
+ b ) - aAP <, 4a<A>
(a + K) - cA ^
(a + b ) - «A <, 2y/a^
(10.66)
K«o
or
( 10 - 67)
[(fl
If
then
*A ^ (V^ - yfb
or
(10.68)
(10.69)
)*
(10.70)
+ ^-aA<0
aA - (a + b ) <, 2VaA
< aA - (a +
fa, + * ) - aA <
If
(a,
then
but
aA ^ (\Za — V*o)
so again
The second
when
2.
real root is
situation in
eo*
g
b
)
2
[F(jco)]
and
does not have a simple
when
(10.75)
(10.76)
we have
«A - 0*o + *o) > 2VoA > («o + b - aA
aA > (V^ - VV)
)
Thus
We
(10.73)
(10.74)
+ *o) - aAl 8 - 4aA >
(a + b ) - dA <
[0*o
10.75
(10.72)
in Eq. 10.65 is negative so that the roots are imagi-
nary. This situation occurs
From Eq.
which Re
(10.71)
8
thus see that Eq. 10.70
is
a necessary and
(10.77)
(10.78)
sufficient condition for
a
biquadratic function to be positive real. If we have
aA = (V^o~-V6o>
we will have double zeros for
Consider the following example:
then
F(s) =
s
s
We see that
aA = 2
so that F(s)
is p.r.
X
5
^
(s/a
Re
g
a
+
T
+
00.79)
[F(jeo)]
2s
!
5s
+
T
+
~
25
(10-80)
16
- V^ )2 = (V25 - Vl6 )*
(10.81)
Elements of
realizability
theory
307
The examples
illustrate the
just given are, of course, special cases. But they do
procedure by which functions are tested for the p.r. property.
Let us consider a number of other helpful points by which a function
might be tested quickly. First, if F(s) has poles on they«w axis, a partial
fraction expansion will show if the residues of these poles are positive
and
real.
For example,
F(S)
+5
TTTT,
+ 1)
3s 8
=
<
ia82>
s(s*
has a pair of poles at s
=
±j\. The partial fraction expansion of F(s),
-2s
5
+& = -rr7
+
F
s
2
l
= ±j is negative.
shows that the residue of the poles at s
is
not
(io.83)
s
Therefore F(s)
p.r.
and admittances of passive time-invariant networks
use of our knowledge of impedances
connected in series or parallel in our testing for the p.r. property. For
example, if Z^s) and Z^s) are passive impedances, then z\ connected in
parallel with Z, gives an overall impedance
Since impedances
are p.r. functions,
we can make
Z(s)
Since the connecting of the
the passivity of the network,
that if Fjis)
and FJs) are
i77^
= Zi(s) +
two impedances in
we know
that Z(s)
p.r. functions,
<
Z^s)
parallel has not affected
must
also be p.r.
p.r.
(1085)
Consequently, the functions
and
where a and
We see
then
m = JMM.
must also be
1084>
K are
real
and
F(s)
= -£*s + a
(10.86)
F(s)
=—
s + a
(10.87)
—
positive quantities,
must be
p.r.
We
then
observe that functions of the type
s
+
s
+a
a
fi
must be
p.r. also.
s
+a
(10.88)
Network
308
analysis
and synthesis
Finally, let us determine
whether
= _*L_
s + a
jr( s)
=
F(s)
V
two terms must
we conclude
^—
(10.90)
+ a/Ks
Therefore, the
ctfKs are p.r.
be p.r. Since the reciprocal of a p.r. function
that F(s)
sum of
is
the
also p.r.,
is p.r.
ELEMENTARY SYNTHESIS PROCEDURES
The
is
s/K
s/K and
see that the terms
10.4
(10.89)
If we write F(s) as
is p.r.
we
o,K£0
basic philosophy behind the synthesis of driving-point functions
up a p.r. function Z(j) into a sum of simpler p.r. functions
to break
Ziis), Z4s), .... ZJs), and then to synthesize these individual Z,(j) as
elements of the overall network whose driving-point impedance is Z(j).
Zis)
= Us) + Z£s) +
---
+ Zn(s)
(10.91)
"breaking-up" process of the function Z(s) into the
One important restriction is that all Zj(s) must be
sum
p.r. Certainly, if all Z 4(s) were given to us, we could synthesize a network
whose driving-point impedance is Z(s) by simply connecting all the ZJs)
First, consider the
of functions Z£s).
in series. However, if we were to start with Z(s) alone, how would we
decompose Z(s) to give us the individual Z/j)? Suppose Z(«) is given in
general as
""
+ ""-i* 1 + ••• + «»' + « . Jfi>
1
+ hs + ba Q(s)
ms" + t^s"- +
= (that b = 0).
pole
at
s
Consider the case where Z(s) has a
W=
7( *
fl
1
-s "
•
fe
•
is,
divide P(s)
by
we can denote
Q(s) to give a quotient D/s
as
Z1(^)
and
Za(s).
Z(s)
= - + R(s)
(1(>
92)
•
and a remainder
R(s),
Let us
which
D£
s
=
Z^s)
+ Z&)
previous discussions, we know that Z^
Are Z! and Z, p.r. ?
the p.r. criteria given previously.
Consider
p.r.
IsZa(5)p.r.?
is
From
(10.93)
=
D/s
ZgCO must have no poles in the right-half plane.
2. Poles of ZJLs) on the imaginary axis must be simple, and their
residues must be real and positive.
1.
3.
Re
[ZJJa>)]
^
for all m.
Elements of
realizabiiity
Let us examine these cases one by one. Criterion
1 is satisfied
the poles of ZJ&) are also poles of Us). Criterion 2
same argument.
theory
is satisfied
309
because
by
this
A
simple partial fraction expansion does not affect the
residues of the other poles. When s —jo>, Re [Z(y«w)
D/jco]
0.
Therefore we have
=
=
Re ZAjco)
From the
a- partial
foregoing discussion,
= Re Zjja>) ^
it is
fraction expansion can be
seen that
if
made such
the form Kfs and the other terms combined
(10.94)
Us) has a pole
at s
that one of the terms
still
remain
= 0,
is
of
p.r.
A similar argument shows that if Us) has a pole at s = oo (that
— m = 1), we can divide the numerator by the denominator to give a
is,
n
quotient Ls and a remainder term R(s), again denoted as Z^s) and Zt(s).
Us)
Here ZJs)
is
= Ls + R(s) = Z^s) + Z^s).
also p.r. If
ZXjs)
has a pair of conjugate imaginary poles on
±jto1 , then Z{s) can be
the imaginary axis, for example, poles at s
expanded into
(10.95)
=
partial fractions so that
(10.96)
Re
Here
so that
=Re (_i^L_\ =0
(_2*M
Zj(.y) is p.r.
Re [Z(/<u)] is minimum at some point m^ and if the value of
Re ZijcDf) = Kf as shown in Fig. 10.9, we can remove a constant K<, Kt
from Re [Z(ya>)] so that the remainder is still p.r. This is because Re [Z(/o>)]
Finally, if
still be greater than or equal to zero for all values of w.
Suppose we have a p.r. function Z(s), which is a driving-point impedance function. Let Z(s) be decomposed, as before, so that
will
Z(s)
ReZfjwi
=
74s)
+
ZJs)
(10.98)
Network
310
anal/sis
and synthesis
Zi(s)
ZzM
Z(s)
^
>
FIG. 10.10
where both 2^ and Zj are p.r. Now let us "remove" ZjCs) from Z(,s) to
give us a remainder Z^s). This removal process is illustrated in Fig.
10.10 and shows that removal corresponds to synthesis of ZjO).
Example
10.5.
Consider the following
Z(s)
function
p.r.
= s* +2s +6
s(s + 3)
(10.99)
We see that Z(s) has a pole at s = 0. A partial fraction expansion of Z(s) yields
*(*)
=
2
*
+
j
+3
j
-2&)+2fr)
If
(10.100)
we remove Zt(s) from Z(s), we obtain Z2(s), which can be shown by a resistor
in parallel with an inductor, as illustrated in Fig. 10.11.
*
o
r—
f
1(
1Q
Z<s)
i"
Za
FIG.
10.11
Example 10.6
7s
+2
n»>- 2j +4
(10.101)
where ]%?) is a p.r. function.
Let us synthesize the network by first removing min [Re
part of Y(jco) can be easily obtained as
8
+
Y(jco)].
The
real
14o>*
(10.102)
We
see that the
minimum of Re [Y(ja>)]
min [Re YQw)] =\. Let us then remove
occurs at
to
Yx =\ mho
= 0,
and
is
from
Y{s)
and denote
equal to
Elements of
the remainder as YJs), as
p.r.
is
because
shown
we have removed
theory
3
1
The remainder function Y^s) is
in Fig. 10.12.
only the
realizability
minimum
real part of YJjm).
Y^s)
obtained as
3s
1
(10.103)
It is readily
seen that YJp)
Thus the
capacitor.
final
is
made up of a J-ii resistor in series with a
is that shown in Fig. 10.13.
f-farad
network
:jo
^20
>20
Y*'}
"
Y(s)
>-
YM
,
]
"
=§f
>
J
FIG. 10.12
Example
10.7.
FIG. 10.13
Consider the
p.r.
impedance
fo8
+
3**
Z{s)
+
3s
+
(10.104)
The real part of the function is a constant, equal to
of 1 il, we obtain (Fig. 10.14)
Z1(s)=Z(s)-l =
The
reciprocal of
1
+3s
6s*
unity.
3^ +
6s*
Removing a constant
1
+ 3s
(10.105)
Zt(s) is an admittance
no) -
which has a pole at s
expansion of Yjfs);
=
This pole
<x>.
fit
is
8
+3s
3s*
+
removed by finding the
r1(i)=25+
5?4n
—WWW
10
o
Z(s)
„
(10.106)
1
Zi(s)
FIG. 10.14
partial fraction
(10107)
Network
312
analysis and synthesis
—WWW
in
o
Z(3)
FIG. 10.15
and then by removing the term with the pole
2 farads in parallel with Y£s) below (Fig.
Y&) The
reciprocal of YJs)
Tito
-
25
=
at *
10.15).
-
oo
YJs)
to give a capacitor of
is
now
obtained as
(10.108)
^--j-j
is
Zjfc)
=
which is, clearly, an inductor of 3 h in
network is shown in Fig. 10.16.
3*
+
1
(10.109)
•
series with
a capacitor of
1
farad.
The
final
o
www10
ZM
3h
Z\z!=.
FIG. 10.16
These examples are, of course, special cases of the driving-point synproblem. However, they do illustrate the basic techniques involved.
In the next chapter, we will discuss the problem of synthesizing a network
with two kinds of elements, either L-C, R-C, or R-L networks. The synthesis techniques involved, however, will be the same.
thesis
Problems
10.1
Test the following polynomials for the Hurwitz property.
(«)
s8
(ft)
5*
to
S1
+5* +2s + 2
+5* +5 + 1
w
5»
+ 5s + 5* + 5
+ 45* + 55 + 2
to
5s
+25»
V)
7
5
+5
+ 25* + 2/ + 5
Elements of readability theory
31
10.2 Determine whether the following functions are p.r. For the functions
with the denominator already factored, perform a partial fraction expansion
first.
(a)
F(»)
+
**
=
F(s)
(&)
+
IX*8
(*
+
+
2X«
1X5
«*
+
3«*
(s
F(s)
(c)
W
F(s)
(«)
F(s)
+2s +4
2s*
-
(s
=
!
+4*
5»
.*»
=
5**
-
+ 4)
+ 3)
+4
+
3*
1
l
Suppose Fj(5> and Ft(s) are both p.r.
= FjKs) - FJ.3) is also p.r.
103
F(s)
+
+s
+
**
+ 2)
Discuss the conditions such that
10.4 Show that the product of two p.r. functions need not be p.r. Also show
that the ratio of one p.r. function to another may not be p.r. (Give one example
of each.)
10.5
GivenZfr)-
*+
*\
X for Z(s) to be a p.r. function?
(a)
What
(A)
Find A* for
(c)
Choose a numerical value for
10.6
are the restrictions
Re [ZQot)]
Prove that
if
on
-
to have a second-order zero at
10.7
Z(»)
Z(s) by
10.8
WW®
positive real.
———
— -j—
first
= 0.
X and synthesize Z(s).
Zt(s) and Zt(s) are both p.r.,
ZUi -
must also be
co
is
p.r.
Determine min [Re Z(/o>)] and synthesize
removing min [ReZ(/co)].
Perform a continued fraction expansion on the ratio
W=
5»
S8
+ 2j* + 3j + 1
+ 5s + 2* + 1
What does the continued expansion imply if Y(s) is the driving-point admittance
of a passive network ? Draw the network from the continued fraction.
3 4
1
Network
analysis
and synthesis
the
10.9 The following functions are impedance functions. Synthesize
impedances by successive removals of/« axis poles or by removing min [Re (/»)].
(«)
+ 4s
7*T2
j + i
(*)
s(*
s8
2?
<c>
W
s*
+
2)
+4
27+1
+ 3s +
s» + l
1
chapter
1
Synthesis of one-port networks
with two kinds of elements
In this chapter we will study methods for synthesizing one-port
networks with two kinds of elements. Since we have three elements to choose
from, the networks to be synthesized are either R-C, R-L, or L-C
networks.
We will proceed according to the following plan. First we will discuss the
properties of a particular type of one-port network,
synthesize it. Let us first examine some properties of
functions.
I
PROPERTIES OF
I.I
L-C
and then we
L-C
will
driving-point
IMMITTANCE FUNCTIONS
Consider the impedance Z(j) of a passive one-port network.
Let us
represent Z(s) as
„,
.
Z(S)
where
N N
x,
s
M M
lt
are
t
odd
=
M^s)
i^
M
a (s)
+ JVj(s)
±^i
+ N&)
are even parts of the numerator
parts.
The average power
/J J j-j
(U1)
and denominator, and
by the one-port is
dissipated
= J Re [Z(jd)] \I\*
Average power
(1 \ .2)
where J is the input current. For a pure reactive network, it is known
that
the power dissipated is zero. We therefore conclude that
the real part of
Z(/to)iszero; that
is
Re Z(»
where
Ev Z(s)
=
Ev 2ija>)
=
= tf'('W)-^')^)
Mt\s)-Nt\s)
315
(11.3)
,,
n
(1L4)
,
Network
316
In order for
anal/sis
Ev Kjio)
and synthesis
= 0, that
MJlJm)
is,
M
either of the following cases
80'o>)
~ W») fUja>) =
must hold:
M
M
= - Nt
«N
t =
(a)
(11.5)
x
(b)
In case
(a), Z(i) is
and in case
We
(11.6)
t
see
(A)
from
this
Z(s)
=
^
Z(s)
-
^
(H.7)
(H-8)
development the following two properties of L-C
functions:
'"
Zjj^s) or YxcC*) is tne ratio of even to odd or odd to even P013
nomials.
have only imaginary
2. Since both Mis) and N£s) are Hurwitz, they
or Y^s) are on
Z^s)
of
zeros
and
poles
the
that
follows
roots, and it
1.
the imaginary axis.
Consider the example of an
L-C immittance
a,*'
+
a*1
+
function given by
a,
(u
9)
know,
Let us examine the constraints on the coefficients a { and *,. We
coeffireal,
the
positive
to
be
first of all, that in order for the impedance
impedance
an
that
know
also
We
cients must be real and positive.
qo is
function cannot have multiple poles or zeros on theyw axis. Since
the
and
numerator
the
of
powers
highest
the
axis,
defined to be on theyw
if
example,
For
unity.
most,
at
differ
by,
can
polynomials
denominator
of
the
order
highest
the
then
2n,
is
numerator
the
of
order
highest
the
denominator can either be 2n - 1 so that there is a simple pole at s = oo,
s = oo.
or the order can be 2n + 1 so that there is a simple zero at
differ
by
can
Similarly, the lowest orders of numerator and denominator
Z(s)
at
of
zeros
or
at most unity, or else there would be multiple poles
s
= 0.
Another property of the numerator and denominator polynomials is
for example, the next
that if the highest power of the polynomial is In,
powers must
succeeding
the
and
In
2,
highest order term must be
any missing
be
cannot
There
through.
way
the
differ by two orders all
Synthesis of one-port networks
317
no two adjacent terms of either polynomial may differ
by more
than two powers. For example, for Z(s) given in Eq.
11.9, if ft, = 0,
then Z(s) will have poles when
terms,
i.e.,
V+
so that the poles will be at *
=
(^V
*-
*i*
and
=
(11.10)
at
"*"
*
= 0,1,2,3
dm)
It is clearly seen that
none of the poles sk are even on theyto axis, thus
one of the basic properties of an L-C immittance function.
From the properties given in Eq. 11.11, we can write a general L-C
impedance or admittance as
violating
Z(S)
=
Expanding Z(j) into
2(.)
+ 'Ofr' + « *')•••(»' + «**)•••
+
tf coW + <»*) .•(,» + «,,«)...
Jft'
we
partial fractions,
< 1112>
obtain
= S + -2p-j + -*&- +
.
.
.
+
KkS
(1L13)
K
where the t are the residues of the poles. Since these poles are
all on
theyco axis, the residues must be real and positive in order for
Z*» to be
positive real. Letting s =jm, we see that Z(;o>) has zero
real part, and
can thus be written as a pure reactance jX(o>). Thus we have
\
—
ft),
ft)
1
/
ft)
= J x(<°)
(11.14)
Differentiating X(a>) with respect to
+K +
dco
Since
all
<o*
the residues
K
t
<°
are positive,
dX(io)
a>,
(«,,»-cV
it is
.
IT*
A
is
similar development
also positive, that
is,
we have
+
seen that for an
(1U5)
L-C function,
.
shows that the derivative of Im [Y{jw)]
<
1116>
= B(m)
Network
318
and synthesis
anal/sis
Consider the following example. Z(s)
is
given as
+
_
™^'(s'
+ co^ +
**»*
Letting s
= jco, we obtain
(11.18)
to*)
X{a>) as
= ; x(co) =
zo«»)
<o a*)
^—^—^ <}
Kco(-coa
+j
+
2
cog )
(11.19)
=
0,
Let us draw a curve of X(co) versus o>. Beginning with the zero at co
<o
as
encountered
let us examine the sequence of critical frequencies
next
the
positive,
always
increases. Since the slope of the X(io) curve is
large or
critical frequency we encounter is when X(co) becomes infinitely
from
goes
and
sign
changes
X(co)
a>
As we pass 2 ,
the pole is at a> 2
+
.
pass through any critical frequency,
to -. In general,
seen from the wayy X(to) is written in
as
of
sign,
there is always a change
through a> 2 , with the slope of X(a>)
pass
After
we
the last equation.
the next critical frequency is the zero
that
see
easy
to
is
positive,
it
always
Thus, if an impedance function is an L-C immittance, the poles
whenever we
at «,.
and zeros of the function must alternate. The particular X(a>) under
powers
discussion takes the form shown in Fig. 11.1. Since the highest
the
of the numerator and the denominator always differ by unity, and
oo,
s
=
and at
lowest powers also differ by one, we observe that at s =
frequency, whether a zero or a pole.
and a zero at
is a zero at s
external
called
oo
are
and s
s
are
frequencies
critical
critical frequencies, whereas the remaining finite
to t
and
a>
cog,
example,
previous
3,
referred to as internal. Thus, in the
there
is
always a
critical
For the example just discussed, there
= oo. The critical frequencies at s =
=
=
are internal critical frequencies.
admittance
Finally, let us summarize the properties of L-C impedance or
functions.
1.
ZiC(s)
or
YLC&s)
is
the ratio of
odd
nomials.
IM
FIG.
I
I.I
to even or even to
odd poly-
9
Synthesis of one-port networks
3
1
X(w)
+2
+1
+3
FIG.
I
I.Z
The poles and zeros are simple and lie on theyeo axis.
The poles and zeros interlace on they'eo axis.
The highest powers of numerator and denominator must
2.
3.
4.
unity; the lowest powers also differ
by
There must be either a zero or a pole at the origin and
5.
The following functions are not L-C
Ks(s
Z(5)«
by
infinity.
for the reasons listed at the
+
2
2
left.
4)
+ l)(s + 3)
3
s + 4s + 5s
Z(s) =
3s* + 6s
2
K(s + l)(s* + 9)
z( =
2
(s + 2)(s + 10)
(s
differ
unity.
a
5
2.
2
1.
(11.20)
2
On
the other hand, the function Z(j) in Eq. 11.21,
diagram
is
shown
in Fig. 11.2,
is
_ 2(s +
»
1)(s
s(s*
11.2
SYNTHESIS OF
L-C
whose pole-zero
an L-C immittance.
»
+
9)
(11.21)
+ 4)
DRIVING-POINT IMMITTANCES
We saw in Section 11.1 that an L-C immittance is a positive real function
with poles and zeros on the jco axis only. The partial fraction expansion
of an
L-C function
is
expressed in general terms as
s
s
2
+
+ Kxs
(11.22)
cog
The synthesis is accomplished directly from the partial fraction expansion
by associating the individual terms in the expansion with network elements.
If F(s)
is
an impedance
l/Ko farads; the term
term I^/s represents a capacitor of
an inductance of K„ henrys, and the term
Z(s), then the
Kx s
is
Network
320
and synthesis
anal/sis
1
,
2X,
tf.h
Wi
1
f
o—^HTHf—
—l0Q0>—
'-vOOOv-'
2X1
'
2«! h
7T5" n
.
Z(s).
FIG.
2^5/(5
2
+
o) 42) is
11.3
a parallel tank circuit that consists of a capacitor of
\f2Kf farads in parallel with an inductance of IKjcof. Thus a partial
fraction expansion of a general L-C impedance would yield the network
shown in Fig. 11.3. For example, consider the following L-C function.
2(s
_,.
8
Z(S)
+ l)(s* +
*- + 4)
9)
(11.23)
A partial fraction expansion of TUjs) gives
Z(s)
=
¥s
2s
+ * + -72-
We then obtain the synthesized network in Fig.
(11.24)
11.4.
method is based upon the elementary
synthesis procedure of removing poles on the jco axis. The advantage
with L-C functions is that all the poles of the function lie on the jco axis
so that we can remove all the poles simultaneously. Suppose F(s) in
Eq. 11.22 is an admittance Y(s). Then the partial fraction expansion of
Y(s) gives us a circuit consisting of parallel branches shown in Fig. 11.5.
For example,
The
partial fraction expansion
W
s( S «
(s
2
+ 2)(s» + 4)
+ l)(s* + 3)
HI—
2h
if
Z<8)
FIG. 11.4
Synthesis of one-port networks
321
K„lziz
Y(s),
FIG. 11.5
The
partial fraction expansion of Y(s) is
r(s)
is
is
=s+
s
+
8
3
s*
+
(11.26)
1
from which we synthesize the network shown in Fig. 11.6. The L-C
networks synthesized by partial fraction expansions are sometimes called
Foster-type networks. 1 The impedance form is sometimes called a Foster
series network and the admittance form is a Foster parallel network.
A useful property of L-C immittances is that the numerator and the
denominator always differ in degree by unity. Therefore, there is always
a zero or a pole at $ = oo. Suppose we consider the case of an L-C
impedance Z{s), whose numerator is of degree In and denominator
is of degree 2n — I, giving Z(j) a pole at s = oo. We can remove this
pole by removing an impedance L^ so that the remainder function Z^s)
is still L-C:
ZjCO
=
The degree of the denominator of Z 2(j)
2/i
—
degree by
1.
degree
-
ZCO
is
LiS
2n
—
(11.27)
1,
but the numerator
is
of
because the numerator and denominator must differ in
oo. If we
Therefore, we see that Z 8 (s) has a zero at s
2,
=
have a pole at s = oo,
which we can again remove to give a capacitor C*s and a remainder
Ya(s), which is
invert
Zt(s)
to give
Yt(s) =
l/Z^s),
Ys(s)
will
Y&) = Y2(s) - C2s.
(11.28)
2h<
lf=t
YM.
if:
FIG.
1
a. M. Foster,
11.6
"A Reactance Theorem," Bell System Tech. J., No. 3 (1924), 259-267.
—
322
Network
i
analysis and synthesis
Ll
£3
c*3=
Cizk
FIG.
1
FIG. 11.8
1.7
=
We readily see that F3(s) has a zero at s oo, which we can invert and
remove. This process continues until the remainder is zero. Each time
we remove a pole, we remove an inductor or a capacitor depending upon
whether the function is an impedance or an admittance. Note that the
final structure
of the network synthesized is a ladder whose series arms
whose shunt arms are capacitors, as shown in Fig. 1 1 .7.
are inductors and
Consider the following example.
s
We
see that Z(s) has a pole at s
+ 4s1 + 3
= oo, which
we can remove by
dividing the denominator into the numerator to give a quotient 2s
remainder
Z,(j), as
shown
Then we have
in Fig. 11.8.
Zt(s) -
Z(s)
-
s
= -44s +
2s
s
=
Observe that Z^s) has a zero at s
the pole at infinity.
y3(s),
as
may
first
and a
10s
+ 4s +
a
(11.30)
3
Inverting Z^s),
oo.
we
again remove
Then we realize a capacitor of J farad and a remainder
be seen in Fig. 11.9.
+
Y3(s)=Y£s)-±s= *f
s
4
4s + 10s
3
Removing
of | h and
the pole at s
=
oo of
Z^j)
=
llYs(s),
Zt(s) - Z3(s) -~s =
3
2h
o
r
—
Y3 (s)
if-
FIG.
1
1.9
gives a series inductor
FTT1
fs + 3
Ginr*
(11.31)
(11.32)
i
Synthesis of one-port networks
323
/toot^
2h
Remainder
Z4>
FIG. 11.10
in Fig. 11.10. The admittance y4(s) = 1/Z4(j) has a pole at
which we remove to give a capacitor of f farad and a remainder
y6(j) = 3/2y, which represents an inductor of f h. Removing this inductor gives us zero remainder. Our synthesis is therefore complete and
the final network is shown in Fig. 11.11.
Since we always remove a pole at s = oo by inverting the remainder
and dividing, we conclude that we can synthesize an L-C ladder network
by a continued fraction expansion. The quotients represent the poles at
s = oo, which we remove, and we invert the remainder successively until
the remainder is zero. For the previous example, the continued fraction
as
s
shown
=
oo,
expansion
s*
+ 4s2 +
is
3)25°
2s
5
+
+
+
fo +
4s3 +
12s3
3
16s(2s<-->
Z
6s
I0s)s*
s*
+
+
4s 2
+
3(ts <->
+
3)4s3
Y
5.2
2->
is'
4s3
+
+
I0s(is<->Z
is
3(|$ <->
Y
3.2
2s
3)2y(fj*->
Z
2s
We see that the quotients of the continued fraction expansion give the
elements of the ladder network. Because the continued fraction expansion
-'innp—
2h
fh
**4=
FIG.
f"
ffi
11.11
324
Network
anal/sis an
synthesis
always inverts each remainder and divides, the successive quotients alternate between Z and Y and then Z again, as shown in the preceding
expansion. If the
initial function is an impedance, the first quotient must
an impedance. When the first function is an admittance,
the first quotient is an admittance.
Since the lowest degrees of numerator and denominator of an L-C
admittance must differ by unity, it follows that there must be a zero or a
pole at s = 0. If we follow the same procedure we have just outlined,
and remove successively poles at s = 0, we will have an alternate realization in a ladder structure. To do this by continued fractions, we
arrange both numerator and denominator in ascending order and divide
the lowest power of the denominator into the lowest power of the numerator; then we invert the remainder and divide again. For example, in
the case of the impedance we have
necessarily be
(s*
i*
z( S)
The continued
+ l)(s* + »
3)
j
z
;r
s(s* + 2)
(H.33)
fraction expansion to give the alternate realization
is
+ 58)3 + As* + s\3I2s <-> Z
3 + fs
fs* + s*)2s + ^(4/5$ *-> Y
2s + $s*
isP^s* + 5*(25/2s <-> Z
2s
2
|j
2
i£
The final synthesized network is shown in Fig.
11.12. The ladder networks
Cauer ladder networks because W. Cauer2 discovered
the continued fraction method for synthesis of a passive network.
Note that for both the Foster and the
Cauer-form
realizations, the number of
o
|(
1(
L
L
elements is one greater than the number of
^f
§f
internal critical frequencies, which we defined
5 h oj
oj
realized are called
1
5h
*p
*p
[
[
FIG.
1
1.12
previously as being
all
the poles
and zeros
of the function, excluding those at s — and
00. Without going into the proof of the
s
=
•Wilhelm Cauer, "The Realization of Impedances with Prescribed Frequency
Dependence," Arch. Electrotech., 15 (1926), 355-388.
Synthesis of one-port networks
325
can be said that both the Foster and the Cauer forms give
elements for a specified L-C driving-point
function. These realizations are sometimes known as canonical forms.
statement,
PROPERTIES OF R-C DRIVING-POINT IMPEDANCES
11.3
The
properties of
known
properties of
onto the
we
it
minimum number of
the
will
R-C driving-point impedances can be derived from
L-C functions by a process of mapping the ja> axis
— a axis. 8 We
assume that
all
will
not resort to this formalism here. Instead,
driving-point functions that can be realized with
two kinds of elements can be realized in a Foster form. Based upon this
assumption we can derive all the pertinent properties of R-C or R-L
driving-point functions.
Let us consider
first
the properties of
R-C
driving-point impedance functions.
L-C impedance given in
obtain a Foster realization of an R-C impedance by
simply replacing all the inductances by resistances so that a general R-C
impedance could be represented as in Fig. 11.13. The R-C impedance,
as seen from Fig. 11.13, is
Referring to the series Foster form for an
Fig. 11.3,
we can
Z(s)
- ^ + K„ + -&_ + -&- +
s
s + a
s +
g
R a = K„, C = \\K Rx = Kja
x
=
(11.34)
cr
where C
l/A^,
x
X,
lf and so on. In
order for Eq. 11.34 to represent an R-C driving-point impedance, the
constants K\ and at must be positive and real. From this development,
two major properties of R-C impedances are obtained, and are listed in
the following.
Ci
Hf-
—IH
Co
-AMr-
R2
*i
FIG. 11.13
*
M
Sons,
.
E.
Van Valkenburg,
New York,
Introduction to
1960, pp. 140-145.
Modern Network
Synthesis,
John Wiley and
Network
326
and synthesis
anal/sis
R-C driving-point impedance are on the negative
can be shown from a parallel Foster form that the
poles of an R-C admittance function are also on the axis. We can thus
conclude that the zeros of an R-C impedance are also on the —a axis.
2. The residues of the poles, Kit are real and positive. We shall see later
that this property does not apply to R-C admittances.
1.
The
poles of an
real (—or) axis.
It
R-C impedances are on the — a axis, let
along the — a axis." To find the slope,
Since the poles and zeros of
us examine the slope of
dZ{a)lda,
we
first let
Z(c) with respect to
Z(<r)
s
a.
Z(<r)
=
a in Z(s), and then we take the
Thus we have
= £ + K. + -&- + _&_ +
a
a+
a + a
o-j
—da— =
and
It is clear
Let us
? H
*
a*
1
(a
+
.
•
.
now look
(11.35)
%
+
?
1
(a
atf
+
+
1-
o%?
•
•
•
^^^0
da
that
derivative of
(n
^^
36}
(11-37)
at the behavior of Z(s) at the
two points where the
a = <a — and at
a = (o = oo. This is readily done by examining the general R-C network
in Fig. 11.13 at these two frequencies. At a = 0, (d-c), if the capacitor C
is in the circuit, it is an open circuit and there is a pole of Z(s) at a = 0.
real axis
If
C
is
and the imaginary
not in the
circuit,
axis intersect, namely, at
then Z(0)
is
simply the
sum of all
the resistances
in the circuit.
= R + R2 +
+ R„
(11.38)
because all of the capacitors are open circuits at a = 0.
At a = oo, all the capacitors are short circuits. Thus, if R^ is in the
circuit, Z(oo) = R^. If R x is missing, then Z(oo) = 0. To summarize
Z(0)
these last
t
two statements, we have
f
Z(0)
C
oo,
present
=
C„ missing
Z(oo)
If
we examine
the
=
o,
Rn
missing
Rm
present
two cases for Z(0) and Z(oo), we
Z(0)
£
Z(oo)
see that
(11.39)
Synthesis of one-port networks
Next,
let
327
us see whether the poles and zeros of an R-C impedance
We have already established that the critical frequency
function alternate.
nearest the origin
must be a
zero.
must be a pole and the
Therefore, if Z(s)
=
Z( S)
(5!
(s
critical
frequency nearest a
=
oo
given as
is
+
+
ga)(s
ffOO
+
+
g4)
(11.40)
or,)
Then, if Z(y) is R-C, the singularity nearest the origin must be a pole
which we will assume to be at s = — gx the singularity furthest from
the origin must be a zero, which we will take to be s = — g
Let us plot
4
;
.
z , ff) =
(g
(g
versus
— g,
Z(0)
equal to a positive constant
is
beginning at a
=
+
+
+
a^ia +
ga)(ff
g4)
(11.41)
a3)
and extending to a
Z(0)
= ^*
= — oo.
At a
= 0,
(11.42)
—
Since the slope of Z(g) is always positive as
a increases, Z(g) must
increase until the pole s
is
reached
(Fig.
11.14). At a
x
= -a
Z(g) changes sign, and
reached.
= — <V
We
= —au
negative until the next critical frequency is
see that this next critical frequency must be the zero,
is
—
Since Z(g) increases for increasing
a, the third critical
frequency must be the pole s
g8 Because Z(g) changes sign at g„
the final critical frequency must be the zero, s
g4 . Beyond a
g4 ,
the curve becomes asymptotic to Z(oo)
1.
From this analysis we see
that the poles and zeros of an R-C impedance must alternate so that for
*
= —
.
=—
=
Z(»)
—a
a*
FIG. 11.14
—
=—
Network
328
anal/sis
and synthesis
the case being considered
oo
In addition,
we see
>
<r4
>
>
<r8
or
g
>
Oi
^
(11.43)
0.
>1
that
(11.44)
a, ex.
which shows that Z(0) > Z(oo).
To summarize, the three properties we need to recognize an
impedance are:
1.
Poles and zeros
2.
The
lie
on the negative
The
An
residues of the poles
example of an
= — oo must be a zero.
must be
R-C impedance
Z(s)
= (s +
real
and
positive.
is:
+ 4)(s +
+ 2)(s + 6)
l)(s
s(s
The following impedances
and they alternate.
must be a pole whereas
real axis,
singularity nearest to (or at) the origin
the singularity nearest to (or at) a
3.
R-C
8)
(11.45)
are not R-C.
= (s + l)(s + 8)
(s + 2)(s + 4)
(s + 2)(s + 4)
Z(s) =
(s + D
(s + l)(s + 2)
Z(s) =
s(s + 3)
Z(s)
(11.46)
Let us reexamine the partial fraction expansion of a general
impedance.
F(s)
= ^ + K + -^i- +
S + Ot
s
a
•
•
•
R-C
(11.47)
Instead of letting F(s) represent an impedance, consider the case where
F(s) is an admittance Y(s). If we associate the individual terms in the
expansion to network elements,
we
then obtain the network shown in
FIG. 11.15
Synthesis of one-port networks
329
We sec that an R-C impedance, ZBC(s), also can be realized
R-L admittance Y^is). All the properties of R-L admittances are
Fig. 11.15.
as an
the same as the properties of
to specify whether a function
R-L
R-C impedances.
is
to be realized as
It is therefore
important
an R-C impedance or an
admittance.
SYNTHESIS OF R-C IMPEDANCES
11.4
OR
R-L
ADMITTANCES
We
postulated in Section 11.3 that the Foster form realization exists
an R-C impedance or an R-L admittance. Since Foster networks are
synthesized by partial fraction expansions, the synthesis is accomplished
with ease. An important point to remember is that we must remote the
minimum real part of Z(/a>) in the partial fraction expansion. It can be
shown* that min [Re Z(/co)] = Z(oo), so that we have to remove Z(oo)
as a resistor in the partial fraction expansion. In cases where the numer-
for
ator is of lower degree than the denominator, Z(oo) = 0. When the
numerator and the denominator are of the same degree, then Z(oo) can
be obtained by dividing the denominator into the numerator. The
quotient is then Z(oo). Consider the following example.
F(s)
The
=
3(a
+
2)(s
s(s
+
partial fraction expansion of the
s
s
+ 4)
remainder function
+
(11.48)
3)
+
3
is
obtained as
(11.49)
3
where F(oo) = 3. If F(s) is an impedance Z(s), it must be an R-C impedance and it is realized in the series Foster form in Fig. 11.16. On the
other hand, if F(s) represents an admittance, we realize Y(s) as an R-L
network in the parallel Foster form (Fig. 11.17).
-\[
8
f
f
AV
30
10
If
Z(s)
FIG. 11.16
4
Van Valkenburg,
loc
cil.
i
330
Network
analysis
and synthesis
30.
*k:
Jo:
lh.
Y(s)
FIG. 11.17
An alternate method of synthesis is based on the following fact. If
we remove min Re [Z(yto)] = Z(oo) from Z(y), we create a zero at s = oo
for the remainder ZjO). If we invert Z x (j), we then have a pole at $ = oo,
which we can remove to give Za(s). Since min Re [ Y^jco)] = Yt(co), if
we remove Y2(co), we would have a zero at s = oo again, which we
again invert and remove. The process of extracting Z(oo) or r(oo) and
the removal of a pole of the reciprocal of the remainder involve dividing
we see that the whole
by a continued fraction expansion.
the numerator by the denominator. Consequently,
synthesis process can be resolved
The quotients represent the elements of a ladder network.
For example, the continued fraction expansion of F(s) in Eq.
j2
+
+
l&y
35*+
9s
3s)3s*
9s
+
24(3
+
24>*
11.48
is
+ 3s($s
s* + f*
is)9s
+
24(27
9s
24)is(s/72
4*
If F(s) is
an impedance Z(j), the resulting network is shown in Fig.
an admittance Y(s), we have the R-L network of Fig. 11.19.
11.18. If F(s) is
-Wv
30
Z(s)
270
—
jinpr^-
±o:
If:
Y(s)_
FIG. 11.18
FIG. 11.19
1
1
.
Synthesis of one-port networks
33
J PROPERTIES OF R-L IMPEDANCES AND
R-C ADMITTANCES
II
The immittance
parallel Foster
that represents a series Foster
R-C admittance
F(s)
is
R-L impedance or a
given as
= KK s + K + -&- +
s + a
•
•
•
(11.50)
t
The
significant difference
pedance
R-C impedance and an R-L imR-C "tank"
between an
that the partial fraction expansion term for the
is
circuit is Kj(s + at); whereas, for the R-L impedance, the corresponding
term must be multiplied by an s in order to give an R-L tank circuit
consisting of a resistor in parallel with an inductor.
The properties of R-L impedance or R-C admittance functions can be
much
functions.
the same manner as the properties of R-C impedance
Without going into the derivation of the properties, the more
significant
ones are given in the following:
derived in
Poles and zeros of an
R-L impedance or R-C admittance are located
and they alternate.
2. The singularity nearest to (or at) the origin is a zero. The singularity
nearest to (or at) s = oo must be a pole.
3. The residues of the poles must be real and negative.
1
on
the negative real axis,
Because of the third property, a partial fraction expansion of an R-L
impedance function would yield terms as
Ki
s
+
(11.51)
o
t
This does not present any trouble, however, because the term above does
not represent an R-L impedance at all. To obtain the Foster form of an
R-L impedance, we will resort to the following artifice. Let us first expand
into partial fractions. If Z(s) is an R-L impedance, we will state
without proof here that the partial fraction expansion of 7^s)js yields
positive residues. 6 Thus, we have
2i_s)js
Z°>=*2 + K +
a>
s
*
he.
Actually,
cit.
s
^-+-s
+
(11,52)
a{
ZBL{s)js has the properties of an R-C impedance;
see
Van Valkenburg
Network
332
and synthesis
analysis
°-VW
ifi
40.
Z(s)
&»={=
to:
T
FIG.
FIG. 11.20
where *o,
Z(s) in
Ku
.
.
.
,
K„ ^
the desired form
multiply both sides by s, we obtain
Consider the following function:
we
If
0.
11.21
for synthesis.
m-
2(s
(s
+ l)(s + 3)
+ 2)(s + 6)
(11.53)
an R-L impedance or an R-C admittance because it
of
the first two criteria cited. The partial fraction expansion
F(s) represents
satisfies
F{s)
is
F(s)
=
2
s
so
we see
F(s)js,
+
that the residues are negative.
on the other hand,
F(s)
s
2
2
+
s
The
(11.54)
6
partial fraction expansion of
is
_ 2(5 +
s(s +
If we multiply both sides
14
i
-
- 3)
2)(s + 6)
by
s,
we
4
t
s
s
+
4
+
2
s
+
(11.55)
6
obtain
is
1
i.
t
IX*
Is
+
s
+
(11.56)
6
an impedance Z(s), it is synthesized in series Foster
Y(s),
form, giving the R-L network in Fig. 11.20. If F(s) is an admittance
11.21.
Fig.
shown
in
then the resulting network is the R-C network
To synthesize an R-L impedance in ladder form, we make use of the
If we remove Z(0) from Z(s), the
fact that min Re [Z(Jm)] = Z(0).
If F(s) represents
remainder function
Zx(s) will have
a zero at s
FIG. 11.22
= 0.
After inverting Z^s),
Synthesis of one-port networks
333
YM
FIG. 11.23
we can then remove
the pole at s
=
0.
Since the value Z(0)
by dividing the lowest power of the denominator
is
obtained
power
into the lowest
term of the numerator, the synthesis could be carried out by a continued
fraction expansion by arranging the numerator and denominator polynomials in ascending order and then dividing. For example, the following
function is either an R-C impedance or an R-C admittance.
= 2(5 + l)(s + 3) = 6 + 8s + 2s*
12 + 8s + s*
(s + 2)(s + 6)
The continued
12
+
8*
fraction expansion of F(s)
is
+ s*)6 + is + 2s*(l
6 + 4s + js*
4s + fs )12 + Ss + s\3ls
12 + §J
ls + s*)4s +
4s +
2
f**(f
£r»
iV)& +
**(49/5s
aS
OAAA
S*
As*
Y(s),
an impedance function, the resulting network is the R-L network
If, on the other hand, F(s) is an R-C admittance
the network is synthesized as in Fig. 11.23.
11.6
SYNTHESIS OF CERTAIN
If F(s) is
shown
in Fig. 11.22.
Under
certain conditions,
R-L-C
R-L-C
FUNCTIONS
driving-point functions
may be
syn-
thesized with the use of either partial fractions or continued fractions.
For example, the function
z(')~
s
a
4- 2s
+
2
VL+ it
+
S'
s
1
( 1L58)
334
is
Network
analysis
and synthesis
neither L-C, R-C, nor R-L. Nevertheless, the function can be synthesized
by continued
s*
fractions as shown.
+ s+
+ 2s + 2(l«-Z
+ s+1
s + iy + s +
S2 + s
l>*
s*
i(s
+ l(s + \*-Z
s+l
1>
The network derived from
this expansion is given in Fig. 11.24.
In another case, the poles and zeros of the following admittance are
on the negative real axis, but they do not alternate.
Y(s)
The
An
Y(s)
is
(11.59)
partial fraction expansion for Y(s)
FIG. 11.24
Since one of the residues
= (s + 2)(s + 3)
(s + l)(s + 4)
negative,
=1+
s+1
we cannot
s
+4
all
is
(11.60)
use this expansion for
Y(s)ls and then
method would be to expand
multiply the whole expansion by s.
synthesis.
When we
alternate
multiply by
y(a)
I
s
s
s,
we
i
6
S
s
+
s
1
(11.61)
+4
obtain,
h
ito-*—
2
+
V
s
1
s
(11.62)
+4
Y(s) also has a negative term. If we divide the denominator of
this negative term into the numerator, we can rid ourselves of any terms
Note that
with negative signs.
y(j)
=
3_(l__U + J!
2
\3
s
+
1/
h
6
s
+
1
s
+4
s
+
-
(11.63)
Synthesis of one-port networks
335
60-
lofa:
Y(s)
in
At 4=
FIG. 11.25
The network
that
realized
is
from the expanded function
is
given in Fig.
11.25.
expand Y(s) by continued fractions, we see that negative
However, we can expand Us) = l/l%s) by continued
fractions, although the expansion is not as simple or straightforward as
in the case of an R-C function, because we sometimes have to reverse the
If
we
try to
quotients result.
order of division to
make
fraction expansion of Z(s)
6
+
5s
the quotients
all positive.
The continued
is
+ s*)4 + 5s + s%i
4 + ¥* + !**
|s + is*)6 +
6 +
5s
+ s%n/5s
is
¥*
+
Hi
i**
+ Hs
Asy^s
+ **(¥
¥*
s*)&s(6I15s
As we
see,
the division process giving the quotient of 1/3 involves a
reversal of the order of the polynomials involved.
network
is
given in Fig. 11.26.
—WV
o
AAAr
£h«
-K-
:&°
Z(t)
FIG. 11.26
The
resulting ladder
W
Network
336
J
and synthesis
anal/sis
In the beginning of this section,
conditions can an
R-L-C
it
was
stated that only
under special
driving-point function be synthesized with the
use of a ladder form or the Foster forms. These conditions are not
given here because they are rather involved. Instead, when a positive
and it is found that the function is not synthesizable
by using two kinds of elements only, it is suggested that a continued
fraction expansion or a partial fraction expansion be tried first.
real function is given,
Problems
11.1
(a)
Which of the following functions are L-C driving point impedances?
Why?
4X* +
A_,, " jQ» ++ 9X*»
+ 25)
8
(6) Synthesize the realizable
_„
16)
(**
w=
There
is
113
no need to
.
fr*
+
+
8)
*fc»+4)
L-C impedance.
+ 9Xi* + 25)
+ 4X*« + 16)
1X5*
jCi"
calculate the element values of the four networks.
Synthesize the
L-C driving-point impedance
=
Z(s)
in the
1X«*
two Foster and the two Cauer networks
that could be used to synthesize the following
7(
+
impedances in a Foster and a Cauer form.
Indicate the general /orm of the
11.2
(j»
Z*S)
'
form shown in the figure,
and farads.
i.e.,
6s*
+ 425* + 48
18j» + 48s
J+
determine the element values of the network
in henrys
=FC8
PROB.
II
11.4 There exists an L-C network with the same driving-point impedance as
the network shown in the figure. This alternate network should contain only
two elements. Find this network.
i
Synthesis of one-port networks
-nnnp
2h
337
—
lh«
4=if
2W
If:
PROB.
11.5
The input impedance
11.4
for the network
2*»
Zta=
shown
+2
"i»+2s*+25 +
Z
2
is an L-C network: (a) Find the expression for
Foster series form.
If
is
Z
.
(6) Synthesize
Z
in
a
2b
PROB.
II
&
11.6 Indicate which of the following functions are either R-C, R-L, or
impedance functions.
s*
+2s
(a)
z(,)
" FT4?~+1
(b)
ZW
~
„
.
(c)
Z(5)=
W
Z(4)
(e)
Z(5)
„,
s*
+ 4* + 3
2
+ 45 + 3
s
?T6F+8
5*
.
=
+ 5s + 6
«+,
j«
+ Ss* + 6
** +3
L-C
338
Network
analysis
and synthesis
An impedance function has the pole-zero pattern shown in the figure.
Z(— 2) = 3, synthesize the impedance in a Foster form and a Cauer form.
11.7
If
ju
-5
-3
-1
PROB.
11.7
11.8 From the following functions, pick out the ones which are
tances and synthesize in one Foster and one Cauer form.
+ l)Qr + 3)
+ 2X* + 4)
sjs + 4X* + 8)
is + lXs + 6)
= 4js +
2js
Y(s)
(s
sis
= (s +
Y(s)
l)fr
3)
+2)
l)fr
sis
+
R-C admit-
+ 4)
+2)
11.9 Find the networks for the following functions. Both Foster and ladder
forms are required.
Z(s)
(«).
-
is
+
(b)
3is
Z(s)
+
For the network shown,
find
1
V„
2+Y
2s»
Synthesize
+ 4)
3
sjs* +3)
+s* +6s +
1
Y as an L-C admittance.
PROB.
11.11
+
Y when
K9
=
4)
+2)
1)(*
s
11.10
+
1)(*
s(s
11.10
Synthesize by continued fractions the function
s
*w- 5
s*
+ 2s» + 3j + 1
+ s* +2s + 1
Synthesis of one-port networks
339
11.12 Find the networks for the following functions in one Foster and one
Cauer form.
(s
Z(s)
11.13
+
2X*
20 +
+ 4)
+ 4)
0.5)(*
Synthesize the following functions in Cauer form.
+s + 1
+ J8 +s
s* + s*+2s + l
Z(s)
V+4» + 3*«+a + l
4^ + 3^+4^+2
Z(s)
2s» +5
(s + 2X* + 4)
Synthesize Z(s) = i
i\/ v ^ mto tne form
(s + IX* + 5)
s*
Z(s)
+2s*
5»
11.14
.
r—*Tnnp
—
—
in the figure.
6
If
i
i
shown
Zfr).
PROB.
11.15
sents
Of the
11.14
three pole-zero diagrams shown, pick the diagram that repre-
an R-L impedance function and synthesize in a
series Foster
JO)
-4
-3 -2
-l
ju
-4
-3
-2
-1
JO)
-4 -3 -2 -1
PROB.
11.15
10
«•
form.
Network
340
analysis
and synthesis
11.16 Synthesize a driving-point impedance with the pole-zero pattern shown
in the figure in any form you choose. (Hint: Use uniform loading concepts.)
JU
+
*i
LU
PROB.
11.17
Following are four successive approximations of tanh
3s
(a)
s*+3
+ 105*
+ 455* + 105
IPs*
(c)
s*
11.16
s.
+ 15s
+ 15
s
s* + 1055 + 945*
W) 15** + 4205s + 945
s*
(*)
6s*
Synthesize networks for the functions above whose input impedances approxi-
mate tanh*.
chapter 12
Elements of transfer
function synthesis
12.1
PROPERTIES OF TRANSFER FUNCTIONS
A transfer function is a function which relates the current or voltage
another port. In Chapter ? we
at one port to the current or voltage at
terms of the opendiscussed various descriptions of two-port networks in
parameters y u . Recall that for
circuit parameters z„ and the short-circuit
open-circuit transfer imthe two-port network given in Fig. 12.1, the
pedances «!, and Zu were defined as
h l/i-o
(12.1)
*1 lli-0
voltage-ratio transfer
In terms of the open-circuit transfer impedances, the
function
is
given as
T,
« £«
Yl
(12.2)
ratio
In terms of the short-circuit parameters, the voltage
Yl
is
= _ 2»
z"
of the overall
£ = _?S*_
h
341
*«
to be
(12.3)
When the network is terminated at port two by a resistor R,
Fig. 12.2, the transfer impedance
shown
+*
network
as
shown
in
is
(12.4)
Network
342
h
h
+
and synthesis
analysis
+
Two-port
network
Vi
V2
—
—
FIG.
The
FIG. 12.2
12.1
transfer admittance of the overall structure in Fig. 12.2
Y„
= -' =
Vi
where
G=
l/R.
y»2
is
(12.5)
+G
When both ports are terminated in resistors,
Fig. 12.3, the voltage-ratio transfer function
shown in
z ai°2
*a
V.
as
V^Vg is
u
(z
+
+
RMtn.
*«)
(12.6)
- 2gi«i8
Other transfer functions such as current-ratio transfer functions can
and short-circuit parameters. In
Chapter 10 we discussed the various properties of driving-point impedances such as z u and z 28 This chapter deals with the properties of the
transfer immittances z M and y tl for a
also be described in terms of the open-
.
passive reciprocal network.
us
discuss
certain
First, let
properties
which
apply to all transfer functions of passive
linear networks with lumped elements.
We denote a transfer function as T(s).
1.
FIG. 12.3
T(s) is real for real s. This property
satisfied
is
when
T(s)
is
a rational
function with real coefficients.
2. T(s)
has no poles in the right-half plane and no multiple poles on
=
If T(s) is given as T(s)
P(s)lQ(s), the degree of P(s) cannot
exceed the degree of Q(s) by more than unity. In addition, Q(s) must be
they'to axis.
a Hurwitz polynomial.
3.
that
Suppose P(s) and Q(s) are given in terms of even and odd
is,
ns)
where M£s)
is
=
m = MM±IM
m +
parts,
(12 7)
.
w,(5)
e(S)
8( S )
even and N£s) is odd. Then T(ja>) is
MiC/oO
MJja>)
The amplitude response of
\T(j<o)\
+
+
A^
NMa>)
T(ja>) is
=
UfA/co)
+ Nt*(ja>)]
(12.9)
Elements of transfer function synthesis
and
is
an even function in m. The phase response
Arg TQm)
is
m—m
= arctan
=
343
'«—
0, we see that the phase response is an odd function in <o.
us discuss some specific properties of the open-circuit and
short-circuit parameters.
If arg T(jO)
Now
let
1. The poles of zgl (,s) are also the poles of z (j) and z {s). However,
u
it
not all the poles of z u (s) and z M (y) are the poles ofzn (s). Recall that in
Chapter 9 we defined the z parameters in terms of a set of node equations
as
A
Z^s)
=
^
— Z«l — ^12
2.o
A
is no cancellation between each numerator and denominator of z
u,
and z lg then the poles are the roots of the determinant A, and all three
functions have the same poles. Consider the two-port network described
by the black box in Fig. 12.4a. Let z' u z' a2 and z' lt be the z parameters
of the network. Let us examine the case when we attach the impedances
Zx and Zg to ports one and two, as shown in Fig. 12.46. The z parameters
If there
zlt ,
,
,
,
for the two-port network in Fig. 12.46 are
= z ii + Zj
Z 22 — Z 22 + Zj
Z 12 = Z 12
z ii
poles of z u include the poles of Z^ the poles of z
M
include the poles of Zj. However, the poles of z
lt include neither the
poles of Zi nor Z,. Consequently, we see that all the poles of z„ are also
It is clear that the
poles of z u
and z„. The
h
+
l_
*\l> 2*22
z'tt
reverse
h
is
not necessarily
true.
£
I,
+
v2
r
Vi
*
—<*'ll. *'22
«'l2
w
(a)
FIG. 12.4
3j
-o
v2
Network
344
anal/sis
and synthesis
*
*
A-i
r^
Vi
Y!
-
4"
y'n, y'22
y'12
"
3f
h
h
h
o
/2
>
*
Y2
V2
^T
7
*h
U»
Vi
V*
FIG. 12.6
FIG. 1X5
poles of yis(s) are also the poles of y u(s) and y M(s). However,
not all of the poles of yu(s) and y M(s) are the poles of yM(s). This property
is readily seen when we examine the two-port network in Fig. 12.5.
2.
The
The y parameters are
.
v
,
= y'»* +
yit = v'm
y**
y%
y12(s) do not include the poles of either
Consider the network in Fig. 12.6. The y parameters are
Clearly, the poles of
2
ViM = -s +
y»(s)
Yt
and
Yt
.
3s
= - + 3s
s
yis(s)
=
-3s
and s = 00, whereas
at s =
Observe that yu(s)
=
00.
s
pole
at
has
a
ylt(s) only
Let us
3. Suppose yu (s), y M(s), and y 18(s) all have poles at s = st
denote by u the residue of the pole at st of the function y u (s). The
residue of the pole s = Sj, of y w(s) will be denoted as km , and the residue
of the same pole of yia (s) will be denoted as k lt Without going into the
proof, 1 a general property of L-C, R-C, or R-L two-port networks is that
and yM(s) have poles
.
Jfc
.
k ak M
s
=
(12.11)
*
as the residue condition. For example, for the
network in Fig. 12.6, the residue condition applied to the pole at
3*
0, we have
0; whereas for the pole at s
00 gives 3 x 3
is fulfilled
condition
0*
residue
the
see
that
we
Thus
0.
8
This equation
L-C
- klt ^
2x4 —
is
known
—
=
=
= >
for both poles.
l
For a general discussion, see M. E. Van Valkenburg, Introduction
Network Synthesis, John Wiley and Sons, New York, 1960, pp. 305-313.
to
Modem
Elements of transfer function synthesis
C
345
/2
FIG. 12.7
+
-T7>
Vi
c
h
<
o
+
Vj
—
o
FIG. 12.8
111
A
ZEROS OF TRANSMISSION
zero of transmission
transmission, there
is
is a zero of a transfer function. At a zero of
zero output for an input of the same frequency.
For the network in Fig. 12.7, the capacitor is an open circuit at s = 0,
so there is a zero of transmission at s = 0. For the networks in Figs.
=
12.8 and 12.9, the zero of transmission occurs at s
±//VZc. For
the network in Fig. 12.10, the zero of transmission occurs at s
—l/RC.
In general, all the transfer functions of a given network have the same
=
zeros of transmission, except in certain special cases. For example if
««(*) has a zero of transmission at s
su than y«(j), V^lV^s), etc.,
=
o
+
h
k
+
v2
FIG. IX*
FIG. 12.1*
346
Network
analysis and synthesis
*1
+
z3
Zx
-
y2
Vi
ZB
-
Y<1
FIG.
^6
12.11
=
s^ This fact is clearly seen when we examine
have a zero at s
the relationships between the transfer functions. For example, we have
will also
Z«i
yn
—
yuVn.
and
-
(12.12)
Vi^ti
Vn-
(12.13)
In addition, the voltage- and current-ratio transfer functions can be
expressed in terms of the z and y parameters as
Yl
— 5ll
h — Usk
(12.14)
In Chapter 8 we saw that transfer functions that have zeros of transmission only on the jco axis or in the left-half plane are called minimum
phase functions. If the function has one or more zeros in the right-half
plane, then the function is nonmimimum phase. It will be shown now that
any transfer function of a passive reciprocal ladder network must be
minimum phase. Consider the ladder network in Fig. 12.11. The zeros of
transmission of the ladder occur at the poles of the series branch impedances or at the zeros of the shunt branch impedances. Since these
branch impedances are themselves positive real, the poles and zeros of
these impedances cannot be in the right-half plane. Consequently, the
transfer functions of ladder networks must be minimum phase. For the
network in Fig. 12.12, a transfer function would have two zeros of
FIG. 12.12
Elements of transfer function synthesis
347
FIG. 12.13
=
due to the elements Lx and C8 It would also have a
= due to Clt a zero at s = —l/Ra Ct due to
the parallel R-C branch, and a zero of transmission at s = jULtCtf*
due to the L-C tank circuit. It is seen that none of the transmission
zeros are in the right-half plane. We also see that a transfer function may
transmission at s
oo
.
zero of transmission at s
possess multiple zeros
on
the ja> axis.
In Section 12.4 it may be seen that lattice and bridge circuits can easily
be nonminimum phase. It can also be demonstrated that when two ladder
networks are connected in parallel, the resulting structure may have
right-half-plane zeros. 2
12.3
SYNTHESIS OF
Y„
AND Zn WITH A
I-S2
TERMINATION
In this section we consider the synthesis of an L-C ladder network
with a 1-Q resistive termination to meet a specified transfer impedance Zji
or transfer admittance Ytl . In terms of the open- and short-circuit
parameters of the
L-C circuit, Z^C?) can be
Zm
and
Ytl(s) is
—
'21
+
I"*-'
Before
and
(12.15)
1
Vn
y«
as depicted in Figs. 12.13
expressed as
+
(12.16)
i
12.14.
we proceed with the actual details of the synthesis,
it is
necessary
two important points. The first deals with the ratio of the odd
to even or even to odd parts of a Hurwitz polynomial Q(s). Suppose
to discuss
FIG. 12.14
«
Van Valkenburg,
op.
cit.,
Chapter
11.
Network
348
Q(s)
analysis and synthesis
given as
is
Q(s)
where M(s)
is
=
M{s)
the even part of Q(s),
+ N(s)
and N{s)
(12.17)
is
the
odd
part.
We know
that the continued fraction expansion of M(s)lN(s) or N(s)IM(s) should
yield all positive quotients. These quotients can, in turn, be associated
with reactances. Therefore it is clear that the ratio of the even to odd or
the
odd
to even parts of a Hurwitz polynomial
is
an L-C driving-point
function.
The second point to be discussed is the fact that the open-circuit transfer
impedance ztl or the short-circuit transfer admittance yn of an L-C circuit
that in an L-C
is an odd function. To show this, we must remember
90° out of phase with the
circuit with steady-state input, the currents are
voltages.
Thus the phase
voltages or input voltages
shifts between the input currents and output
and output currents must be 90° out of phase,
or
and
=
Arctan^=±Jrad
(12.18)
Arctan^=±?rad
(12.19)
=
Re j/ 21(;a)) for an L-C network. In order for the
so that Re z n (jw)
of an L-C two-port
real parts to be equal to zero, the functions z« and y ai
network must be odd.
Suppose, now, that the transfer admittance
of two polynomials*
Y11 =
where P(s)
circuit
is
either
^ = ^+
M{s)
Q(s)
even or odd.
Y21 is given as the quotient
(12.20)
N(s)
Now, how do we
determine the short-
parameters y n and y M from the Eq. 12.20 to get
ya
Y" =
i+y»
it
into the
form
(12.21)
We divide both the numerator P(s) and the
N(s), the even or the odd part of Q{s).
or
denominator Q(s) by M(s)
even, we divide by N(s) so that
is
P(s)
must be odd, if
Since
The answer
is
quite simple.
ytl
P(s)/N(s)
In
1
+
[M(s)IN(s)]
(12>22)
Elements of transfer function synthesis
From this we
—
yn
obtain
349
(12.23)
_M(s)
y"
On the
other hand,
if P(s) is
JV(s)
odd,
we
divide
by M(s) so that
Z(!VMW_
+ [tf(s)/M(s)]
y„
1
(12 24)
.
M(s)
(12.25)
yM
M(5)
We assume that P(s), Mis), and N(s) do not possess common roots. For
our purposes, we will consider only the synthesis of Yn or Ztl with zeros
=
oo. In a ladder network, a zero of
of transmission either at s = or s
= corresponds to a single capacitor in a series branch
or a single inductor in a shunt branch. On the other hand, a zero of
oo corresponds to an inductor in a series branch or
transmission at s
a capacitor in a shunt branch. In terms of the transfer impedance
transmission at s
=
Zn(s)
= fW =
+
*('"
1
««-i«"-
+
•
•
•
+
"i«
+
«»>
(12 26)
.
=
implies that the coefficients an_ x ,
ax , a are all zero. The number of zeros of Z,^) at s = oo
is given by the difference between the highest powers of the denominator
and the numerator, m — n. We know that n can exceed m by at most
unity, while m can be greater than n by more than one. For example,
, a lt a„ = 0, we know that the
if m — n = 2, and n = 3 with «„_i,
and two zeros
transfer function has three zeros of transmission at s =
the presence of n zeros of Z, x (.y) at s
a„_,
.
.
.
=
oo.
of transmission at s
We can now proceed with the matter of synthesis.
following example.
sr
We
+
3s
+ 4s + 2
see that all three zeros of transmission are at j
numerator P(s)
is
a constant,
it
Consider the
=
oo.
Since the
must be even, so we divide by the odd
350
Network
analysis
and synthesis
part of the denominator s*
+ 4s. We then obtain
a
+ 4s
g
— 3ss + 2
+ 4s
s
z as
(12.28)
i
We see that both zsl and z2g have the
same poles. Our task is thus simpliz 22 so that the resulting network
synthesize
must
where
we
fied to the point
requires
that we first examine the
z
This
of
zeros
transmission
has the
21
possible structures of the networks which have the required zeros of
transmission and see if we can synthesize z 22 in one of those forms. For
the example that we are considering, a network which gives us three zeros
of transmission at s = oo is shown in Fig. 12.15. We can synthesize z22
to give us this structure by the following continued fraction expansion
.
of
l/z 22 .
3j«
+
2)i»
+
4s(ls+-Y
s»+
|s
^)3j*
+
2(&s +-Z
3£*
2Y£s(is^Y
from the l-£2 termination toward the input end,
the final network takes the form shown in Fig. 12.16. Examining the
network more closely, we see that it takes the form of a low-pass filter.
Since z 22
is
synthesized
Thus the specification of all zeros at s
=
oo
is
equivalent to the specification
of a low-pass
As a second example, consider the transfer impedance
filter.
*«>Since the numerator of
Zgl
(,s)
is
,+
»+
<12S)
4.
an odd function, we have to divide both
-nRPP-
-ffflHT1
<D T
FIG. 12.15
+2
}
FIG. 12.16
Elements of transfer function synthesis
351
FIG. 12.17
numerator and denominator by the even part of the denominator so that
*n
-«*
=
3s*
*«
""
+2
s*
=
+ 4s
T
+2
(12.30)
3s*
at s =
is a high-pass
by a continued fraction expansion of «M .
The network that gives three zeros of transmission
structure
The
which
is realized
final realization is
shown
in Fig. 12.17.
Finally, consider the transfer admittance
Yn(s)
+ 4s + 2
which has two zeros of transmission at s = 0, and one zero
Since the numerator is even, we divide by s* + 4s so that
3s' + 2
y« = -rrr
s + 4s
s*
+
(12.31)
3s*
at s
=
oo.
„~„„x
(12.32)
The question remains
as to how we synthesize y a to give a zero of transmission at s
oo and two zeros at s
0. First, remember that a parallel
inductor gives us a zero of transmission at s
0. We can remove this
parallel inductor by removing the pole at s
of y n to give
=
=
—
=
Vi
=
1
i/—
Vu
2s
5s/2
—
= ~—
s* + 4
(12.33)
If we invert ylt we see that we have a series L-C combination, which gives
us another transmission zero at s
0, as represented by the f-farad
capacitor, and we have the zero of transmission at s
oo also when we
remove the inductor of f h. The final realization is shown in Fig. 12.18.
=
=
H(2h<
)Vt
FIG. 12.18
352
Network
analysis
and synthesis
1
1Q<
Network
Zn
\)
I
(a)
An important point to note in this synthesis procedure is that we must
place the last element in series with a voltage source or in parallel with a
current source in order for the element to have any effect upon the transfer
function. If the last element
is
denoted as
Zn
,
then the proper connection
of Z n should be as shown in Fig. 12.19.
12.4
SYNTHESIS OF CONSTANT-RESISTANCE NETWORKS
In this section we will consider the synthesis of constant-resistance twoport networks. They derive their name from the fact that the impedance
looking in at either port is a constantresistance R when the other port is terminated in the same resistance R, as depicted
Two-port
network
in
Fig.
works
12.20.
Constant-resistance
net-
are particularly useful in transfer
because when two
networks with the
network.
same R are connected in tandem, as shown
loads down the other. As a result, if the
network
neither
in Fig. 12.21,
voltage-ratio transfer function of a is V^V-l and that of b is VJVS the
voltage-ratio transfer function of the total network is
function
FIG.
12.20. Constant-resistance
synthesis,
constant-resistance
N
N
Y*Y»
,
(12.34)
Equation 12.34 implies that, if a voltage-ratio transfer function is to be
ratio could
realized in terms of constant-resistance networks, the voltage
Elements of transfer function synthesis
*.
t.
FIG.
Na
V2
353
+
«+
R<
Nb
12.21. Constant-resistance
networks in tandem.
could be
be decomposed into a product of simpler voltage ratios, which
tandem.
in
connected
then
realized as constant-resistance networks, and
For example, suppose our
objective is to realize
- z«)(s - zi)(s - z,)
- p )(s - Pi)(« - P«)
K_
K(s
Va
(s
We
in terms of constant-resistance networks.
individual voltage ratios
can
(12.35)
first
synthesize the
_ K (s - 2 )
s — Po
V.
\\
v*
= K^s
s
Y*
s
VhjVa
p,
in
.
Although there are many
works,
-
and then connect the three networks
as constant-resistance networks
realize
(12.36)
Pi
-zj
= Xa(s
Vt
tandem to
-*i)
—
different types of constant-resistance net-
we will restrict ourselves to networks of the bridge- and lattice-type
structures as
shown
in Figs. 12.22o
and 12.226. These networks are
the input and output ports do not possess comclose examination, we see that the bridge and
a
Upon
terminals.
mon
are identical circuits. The bridge circuit is
12.22
Figs.
in
circuits
lattice
merely the unfolded version of the lattice. Consider the open-circuit
balanced structures;
i.e.,
parameters of the bridge circuit in Fig. 12.23.
impedance
z
u
First
we determine
the
as
«ii
Next we determine the
transfer
=
(12.37)
impedance zn , which can be expressed as
= V* ~ V
h
*'
z ai
(12.38)
Network
354
analysis and synthesis
+
'.
(a)
+"
(b)
FIG.
and
is
12.22. (a) Bridge circuit,
obtained as follows.
We first obtain the current /' as
r_
Next we
(6) Constant-resistance lattice.
Za + Z b
Vt - Vt =
find that
_h
Vi
-
(Zb
(12.39)
2
- Z )I'
tt
= (Z -Za)|
6
FIG.
12.23. Analysis
of bridge
circuit.
(12.40)
Elements of transfer function synthesis
^ = Z> ~ Z
*
so that
From
Now
(12.41)
the lattice equivalent of the bridge circuit
us consider the
355
we
note that z M
= zu
.
terminated in a resistance
R, as shown in Fig. 12.226. What are the conditions on the open-circuit
parameters such that the lattice is a constant-resistance network? In
other words, what are the conditions upon z u and z n such that the input
impedance of the lattice terminated in the resistor R is also equal to R7
let
In Chapter 9
we found
lattice circuit that is
that the input impedance could be expressed as
Zu
Since z u
= *ii - -SjJ-r
Zn + R
= zu for a symmetrical network, we have
Z" "
In order for
Zu
z 11
lattice
network,
- z«ia =
*
(12.44)
we then have
iKZ,
which
(1243)
+*
= R, the following condition must hold.
*u*
For the
(12.42)
+ Z,)* - (Za - Z»)»] = **
(12.45)
— R*
(12.46)
Z^Lb
simplifies to give
Therefore, in order for a lattice to be a constant-resistance network,
Eq. 12.46 must hold.
Next, let us examine the voltage ratio VjVg of a constant-resistance
lattice whose source and load impedances are equal to R (Fig. 12.24).
From Chapter 9 we can write
z^R
V»
V9
(zu
+ K)(zM + R) - ztfr*
FIG. I2J4. Double-terminated
lattice.
(12.47)
Network
356
which
anal/sis
simplifies to
and synthesis
=
IS-
5*&ii
(zn
V.
+ Kf - **i
a
= __Zt£__2
2Rz11 + 2R
In terms of the element values of the
V,
_
V„
From
(12.48)
lattice,
we have
- Za)R
R(Z b + Za) + 2R*
l(Z b
(12.49)
the constant-resistance condition in Eq. 12.46,
V,
UZ b - (RVZh)]R
_
Vt
we obtain
+
R[Z b
(R*/Z»)]
+
- R*)
+ R8) + 2RZ
a
a
_ j(Zt - R )
(Z + R?
2R a
KZ»»
(Z 6a
h
b
=
* (Z >
~
R)
(12.50)
Zb + R
In Eq. 12.50, the constant multiplier J comes about from the fact that
the source resistance
R acts as a voltage divider.
G(s)
we can
express
Z6 in Eq.
=
^
12.50 in terms of
HP +
If
we
let
(12.51)
G as
G(s)]
1-G(5)
"
In terms of Z„, the voltage ratio can be given as
V2
Va
1R-Z
a
In the following examples, we will usually
Example
12.1.
The voltage
(12.53)
2R + Za
let
R
be normalized to unity.
ratio is given as
2l=l£JlI
2s + 1
V„
(12.54)
Elements of transfer function synthesis
we
which, as
recall, is
with Eq. 12.50,
an
all-pass transfer function.
By
associating Eq. 12.54
we have
(12.55)
Z„ =>s,
Since
Z^Za =
1,
we then
Za --s
obtain
(12.56)
We see that Z„ is a 1-h inductor and Za is a 1-farad capacitor.
is
shown
357
The final network
in Fig. 12.25.
If
FIG. 12.25
Example
12.2.
Let us synthesize the all-pass function
V ~2s +
'
whose pole-zero diagram
is
shown
l s
_
2s
V„
let
First,
we
~
+
Since the portion
l
1
us concentrate on synthesizing the function
Vt
j*
-2s +2
a
s*
+2s+2
V~
separate the numerator
even parts. Thus
(12.57)
+ 2s+2
in Fig. 12.26.
Yl
has already been synthesized,
s*
1
B
(12.58)
and denominator function
into
odd and
we have
V*
Va
2
=
(f
(5*
+ 2) - Is
+ 2) + 2s
(12.59)
jut
x
-<T
yi
-1
X
+1
-fl
-
FIG. 123*
«
Network
358
If
we
analysis
and synthesis
divide both numerator
and denominator by the odd part
Vt
[(*»
[(5*
We then see that
+ 2)/2t] + 2)/2*] +
a* + 2
s
2s,
we
1
(12.60)
1
1
=2 +I
which consists of a $-h inductor in
pedance Z„ is then
series with
*-JT5
and
is
obtain
(12.61)
a
1 -farad
capacitor.
The im-
(1262)
recognized as a J-farad capacitor in parallel with a 1-h inductor. The
VjVa is thus realized as shown in Fig. 12.27. The structure that
voltage ratio
FIG. 12.27
VJVg in Eq. 12.S7 is formed by connecting the
networks in Figs. 12.25 and 12.27 in tandem, as shown in Fig. 12.28. Finally,
it should be pointed out that constant-resistance lattices can be used to realize
other than all-pass networks.
realizes the transfer function
FIG. I2JS
Next,
12.29.
let
us consider the constant-resistance bridged-T network in Fig.
network are all equal to R ohms, the
If the resistances in the
network has constant-resistance
if
ZA - *
(12.63)
359
Elements of transfer function synthesis
ED—
AMr
R
-vwR
Vi
.*.
Zb
Vt
FIG. VIM. Constant-resistance bridged-T
Under
circuit.
the constant-resistance assumption, the voltage-ratio transfer
function can be given as
V*
*
_
(12.64)
Zt + R
R + Za
Vx
Exanple 123.
z>
_
Let us synthesize the voltage ratio
V.
s*
+
l
j*+25 +
(12.65)
l
as a constant-resistance bridged-T network terminated in a 1-G resistor. First
let
us write
VJV1 as
V1
Vt
1
+
1
+
[2*/(*»
(12.66)
1)]
Is
so that
Zb
and
We
recognize
circuit.
(12.67)
+l
~ s* + l
*»
Za
as a parallel
The final network
is
L-C tank
shown
circuit
(12.68)
and Z» as a
in Fig. 12.30.
2h
r-CH
-a/v\
10
—i-vwk
10
10!
:2f
FIG. I2J0
series
L-C
tank
Network
360
Example
analysis
and synthesis
Let us synthesize the voltage ratio
12.4.
(s
Vs
Fx "
in terms of
At
first,
(5
+ 2X* + 4)
+ 3X3* + 4)
two constant-resistance bridged-T
we break up
circuits
(12.69)
connected in tandem.
the voltage ratio in Eq. 12.69 into two separate voltage
ratios
S+1
5+3
Va
Vi
+4
5
and
va
For the voltage
ratio
~
(12.70)
(12.71)
35+4
VJVi, we have
s
+
2
5+3
=
zbl +
(12.72)
1
1
so that
Zn
=»
s
For the voltage
+ 2 and
ratio
zal ~5+2
vjva we have
5+4
35+4
AnH
(12.73)
1
1
+zot
(12.74)
25
jr_.
FIG.
5+4
12.31
(12.75)
i
I
Elements of transfer function synthesis
and
Zha
The final
synthesized network
is
—
shown
s
361
+4
(12.76)
'
2*
in Fig. 12.31.
Problems
12.1 Give an example of a network where: (a) a transfer function has
multiple zeros on they'to axis; (b) the residue of a pole of a transfer function on
thsjo> axis is negative.
12.2
Show
that the residue condition holds for the networks
shown
in the
figure.
—
I
lo-
lh
'TflHP
—
3h
10
lo-^VVW-
-o2
af
-o2
Hf:io
=fc*f
iq:
-o2'
l'o-
-o2'
l'o(b)
(a)
PROB.
123 For the network shown,
and plot on a complex plane.
find
i
12.2
by inspection the zeros of transmission
—nnnp—
ff
lf^=
^=2f
For the networks in the figure
are equal to
when Z„Zt
R*.
12.4
ln
R
.20
£f:
lh
S10
PROB.
Z
in.
-
12.3
show
that the driving point impedances
362
Network
anal/sis
and synthesis
s
R
JC
Vi Zfci-*
zb
*£ *
(a)
%
Vi
Zta -ii
*^
PROS.
123 For the networks in Prob.
v2
12.4
12.4, find the voltage-ratio transfer functions
yjvv
12.6
For the network shown
in the figure (a)
Yl.
V
(b) Synthesize
1
2+Y
.
+ 2)
+ 2s + 2
O.S(*»
K ™
s*
10
PROB.
(a)
that
Ywhen
l\
12.7
show
<
V2
12.4
For the constant-resistance bridged-T circuit, show that if Z
Zh — 1,
tt
then
L_
Vt
n
i
+za
Elements of transfer function synthesis
(b) Synthesize
Za and Zh if
_
£_
V ~
l
12.8
363
s*
s*
+3s +
+45* +5j
2
+2
Synthesize the following voltage ratios in one of the forms of the
networks in Prob. 12.4
<«)
V,
s+2
Vl
*
Vt
3
Vx " 2** + 2* + 6
3(jr + 0.5)
Vt
Vx " 4j + 1.5
(e)
12£
+
2(s*+3)
Synthesize iV with termination resistors /tt
=4fi,i?1
~into
give
12j*
Yl
"
V,
15i»
+ 7* + 2
PROB. IL9
12.10
For the network
in Prob. 12.9, realize network
X
Yl
(a) Synthesize AT. as
_L_
+3
2 2s
F„
Na to give
a constant-resistance
lattice. (R =- 1 n.)
a constant-resistance ladder as in Prob. 12.4.
as a constant-resistance bridged-T circuit. (R -
(6) Synthesize AT. as
(c) Synthesize A',
12.11
(«)
Synthesize the following functions into the form
Z
1
s*
(*)
(c)
id)
Z„- *»
+3s* +3s
+2
s
+
3s*
+
3s
+
2
3s
+
2
s*
'a
s*
+
s*
+ 3s* + 3s +2
(*
+ 2)«
3s*
+
s*
Y„-
s*
(e)
'
shown
(Jt
1
-
1
Q.)
Q.)
in the figure
364
Network
anal/sis
and synthesis
PROB.I2.II
12.12
(«)
(b)
12.13
circuits.
Synthesize as a constant-resistance lattice terminated in a 1-Q resistor.
-s+1
+s + 1
h ~ s* - 3s + 2
vt s* + 3s +2
via s* - 20s* + 5s Vx(s) ^ s* + 20s* + 5s +
V,
s*
Vx
s*
20
20
Synthesize the functions in Prob. 12.8 as constant-resistance bridged-T
chapter 13
Topics
13.1
THE
FILTER DESIGN
design
in filter
PROBLEM
In the preceding chapters we examined different methods for synthesizing
a driving point or transfer function H(s). Most problems have as their
initial specification an amplitude or phase characteristic, or an impulse
response characteristic instead of the system function H(s). Our problem
is to obtain a realizable system function from the given amplitude or
phase characteristic. For example, a typical design problem might be to
synthesize a network to meet a given low-pass filter characteristic. The
specifications might consist of the cutoff frequency m c the maximum
allowed deviation from a prescribed amplitude within the pass band, and
the rate of fall off in the stop band. We must then construct the system
function from the amplitude specification. After we obtain H(s), we
proceed with the actual synthesis as described in the Chapter 12. Another
problem might consist of designing a low-pass filter with a linear phase
characteristic within the pass band. Here, both amplitude and phase are
specified. We must construct H(s) to meet both specifications. Problems
of this nature fall within the domain of approximation theory. In this
chapter we will consider selected topics in approximation theory and then
present examples of filter design where both the approximation and the
synthesis problems must be solved.
,
13.2
THE APPROXIMATION PROBLEM
NETWORK THEORY
The
IN
essence of the problem
is the approximation of a given funcby another function/^t; «!,..., a„) in an interval xx ^ x ^ xt
The parameters ax
in the approximating function are fixed
, <x„
tion/(rr)
.
,
.
.
.
365
Network
366
by the
/at*;
analysis and s/nthesis
•
<*i»
•
•
»
*n)» the following error criteria are
Least squares. The value of /(alf
1.
When we
particular error criterion chosen.
and w(x)
.
.
.
,
e
let
= f{x) —
most common:
aj is minimized where
a weighting function which stresses the error in certain sub-
is
intervals.
Maximally flat. The
2.
at
first
n
—
\
derivatives of e are
made
to vanish
= *o.
x
3.
Chebyshev. The value of p
where
/*
=
is
minimized in the interval *i<,x <,xt
|«ln«x-
The value of
<, xt
4. Interpolation.
in the interval x^£x
e is
made
to vanish at a set of n points
.
is chosen, we must determine the particular
function. This depends upon whether we
approximating
form of the
time or frequency domain. Suppose /(a;)
the
in
approximate
choose to
represents a magnitude function in the frequency domain and the approximating function is to be rational in co*; then
After an error criterion
/„(*;
«!,..•,*»)=
„
.
—
(13.1)
where * = to*. In addition, the values of xk must be restricted to insure
^ 0. In the time domain, /(&) might represent an
that/ (x; alf
,
system to be synthesized. In the case of an R-C
of
a
impulse response
we
have
function,
transfer
.
.
.
O
fa(x; 04,
where x
=
t.
Since an
...
,
oc„)
=
«**'
+
oe,
e"«*
+
•
•
•
(13.2)
R-C transfer function must have its poles on the
<x t , k even, are restricted to negative real
negative real axis, the values of
numbers.
The keystone of any approximation problem
lies in
the choice of a
suitable error criterion subject to realizability restrictions.
can be simplified
when some of the
oc's
The problem
are assigned before applying the
error criteria. All the error criteria cited, except the Chebyshev, can then
be reduced to a set of linear algebraic equations for the unknowns
«ii
• • • >
*»•
Time-domain approximation
The principal problem of time-domain approximation
consists of
approximating an impulse response h(0 by an approximating function
Topics
in filter
367
design
h*(t) such that the squared error
e
= ["[fc(0-**(')f *
minimum.
is
A generally effective procedure in time-domain approximation utilizes
orthonormal functions ^(r). 1 The approximating function h*{t) takes
the form
**(0-Z«.«0
k=l
(13.3)
so that the error
Jo
«*&(<)] dt
L
*-i
minimized when
is
«.=r
k
h(t)<f>k (t)dt
=
l,2,...,n
(13.4)
we saw in Chapter 3. If the orthonormal set is made up of a
of exponentials e**', then the approximate impulse response
as
,»ri
sum
(13.4)
*=1
#*(*)=!
has a transform
*=1 s
Realizability is insured if in the
k
=
1, 2,
.
.
.
,
n.
—
(13.5)
sk
orthonormal
Synthesis then proceeds
set {a te***},
Re ^
^ 0;
from the system function H*(s)
obtained in Eq. 13.S.
Frequency-domain approximation
In frequency-domain approximation
the
principal
problem is to find a
whose magnitude
rational function H(s)
|/f(/Yu)|
approximates the ideal low-pass
-1
+1
characteristic in Fig. 13.1 according to
a predetermined error criterion. In the
next few sections we examine several
W. H. Kautz, "Transient Synthesis in the
Theory, CT-1, No. 3, September 1954, 29-39.
1
FIG . I3X
Ideal
,
fihcrchar_
acteristic.
Time Domain," IRE
Trans, on Circuit
Network
368
different
analysis
and synthesis
ways to approximate the
the maximally flat or
ideal low-pass:
Butterworth approximation, the equal-ripple or Chebyshev approximation,
and the optimal or Legendre approximation. Another major problem is
that of obtaining a transfer function H^s),
whose phase
is
approximately
approximately flat over a given range of frequencies.
Here again, there are two different methods the maximally flat or the
equal-ripple methods. Our discussion will center around the maximally
flat method. The joint problem of approximating both magnitude and
linear or whose delay
is
:
phase over a given frequency range
is
possible, but will not be discussed
here.
13.3
THE MAXIMALLY FLAT LOW-PASS
APPROXIMATION
In Chapter 10
we saw
FILTER
that the ideal low-pass
filter
in Fig. 13.1
is
not
not zero for t < 0.
However, if we use a rational function approximation to this low-pass
filter characteristic, the Paley- Wiener criterion will be automatically
satisfied. We will therefore restrict ourselves to rational function approxirealizable because
its
associated impulse response
is
mations.
In low-p.ass filter design, if we assume that all the zeros of the system
function are at infinity, the magnitude function takes the general form
where
K
is
the d-c gain constant and/(<u 2)
to give the desired amplitude response.
/(«)
a
)
=
is
the polynomial to be selected
For example,
a)
if
2"
(13.7)
then the amplitude function can be written as
M(»)
=
,.
(1
We see that M(0) =
eo.
K,,,
+ o>\2nn)^
and that M(a>)
is
monotonically decreasing with
In addition, the 0.707 or 3-decibel point
M(l)
= -^
< 13 - 8 )
is
alln
at to
=
1
for
all n,
that
is,
(13.9)
is thus seen to be <w = 1. The parameter n conof approximation in both the pass band and the stop
band. Curves of M(m) for different n are shown in Fig. 13.2. Observe
The
cutoff frequency
trols the closeness
Topics
Radian frequency,
0.3
0.2
0.1
0.6
0.4
369
design
In filter
a
0.8
5^ ,v
-2
-4
•\
-6
n=3
-8
n=5
n
\VV
=7
\
-10
I*
-12
\\\
A
-14
^ \
-16
\
\\
-18
-20
\\
-22
\
\
-24
\
\
\
\
-26
FIG.
13.2.
Amplitude response of Butterworth low-pass
filters.
that the higher n is, the better the approximation. The amplitude approximation of the type in Eq. 13.8 is called a Butterworth or maximally flat
response. The reason for the term "maximally fiat" is that when we
expand M(to) in a power series about <o = 0, we have
Jlf(eo)
We
(0
- Ao(l -
l<o
see that the first 2n
= 0.
For
«>
»
1,
tn
—
+ |co«» 1
A<o«"
+
i^eo""
+
•
•
•)
(13.10)
derivatives of M{o>) are equal to zero at
the amplitude response of a Butterworth function
can be written as (with Kg normalized to be unity)
M(w) =!
—
<w»l
(13.11)
to"
We observe that asymptotically,
Af(to) falls off as
wrn
response. In terms of decibels, the asymptotic slope
20 log M(co)
= — 20/i log <w
is
for a Butterworth
obtained as
(13.12)
Network
370
analysis
and synthesis
Consequently, the amplitude response
6m db/octave or 20n db/decade.
One
How
question remains.
from only the amplitude
follows.
falls
asymptotically at a rate of
do we obtain a
transfer function H(s)
characteristics M(co)l
The procedure
is
as
We first note that the amplitude response M{m) and the complex
system function H(jco) are related by
M\w) =
If
we
new
define a
(13.13)
function A(s 2) such that
A(s 2)
we
H(j<o) H(-jco)
M
see that
2
-
H(-s)
H(s)
(<o)
=
(13.14)
^(-ot) 2)
(13.15)
From A(— eo 2) all we need do is to substitute s* = —to2
Then we factor A(s2) into the product H(s) H(—s). Since
to give «(s 2).
the poles
and
zeros of H(s) are the mirror images of the poles and zeros of H(—s), we
simply choose the Hurwitz factors of A(js) as H(s). An example will serve
Consider the third-order (n
to clarify this discussion.
=
3)
Butterworth
response given by
JW
l
—
\OJ)
1
+ to
1
-
(13.16)
6
1
"We see that AC?2)
a
«(s )
is
1
=
1
Factoring
«(**),
(13.17)
(13.18)
--(s8)8
we obtain
1
1
«(**)
=
1
+
2s
+
2s
2
+
»
s'
1
-
2s
+
2s
2
-
s*
= H(s)H(-s)
(13.19)
We then have
if(s)
1
=
s
8
+
2s
s
+
2s
+
1
1
(»
+
l)(s
+ i + A/3/2X* + i - A/3/2)
(13-20)
The poles of #(y) and fT(-*) are shown in Fig. 13.3. Observe that the
poles of H(—s) are mirror images of the poles of H(s), as given by the
theorem on Hurwitz polynomials in the Chapter 2.
Topics
13.3.
Poles of H(s)
design
371
-^ «f-»>
HM^.-
FIG.
in filter
H(-s)
for
an n
= 3 Butterworth filter.
For a Butterworth response, the poles of H(s) H(—s) are the roots of
(-1)V B = -1
=
The poles
«*«*-"»
or simply by
sk
=
e
Expressing st as s k
=
=
=
e*<»-u/ft*
e
-
(13.21)
1
(13.22)
n odd
fc
ak
In
2
„ even
i(i/n) '
«<»+'-i>/an] I
=
0, 1,
by
s k are then given
sk
k
+ ja> k
,
= o, 1, 2,
,2n
(13.23)
the real and imaginary parts are given
by
ow
= cos 2fc + nn-
1
ir
=
.
sin
2n
/2fc
-
I
\
1\tt
/2
1
n
(13.24)
ct)v
=
2k
sin
+ n-
l
IT
=
cos
In
/2fc-
1
I
\
1\tt
—
n
12
It is seen from Eqs. 13.22 and 13.23 that all the poles of H(s) H(—s) are
located on the unit circle in the s plane, and are symmetrical about both
the a and the jco axes. To satisfy readability conditions, we associate the
poles in the right-half plane with H(—s), and the poles in the left-half
plane with H(s).
As an example, consider the construction of an H(s) that gives an n
4
Butterworth response. From Eq. 13.23, it is seen that the poles are
given by
=
st
=
^(«fc+*)/8]»
e"
(13.25)
Network
372
H(s)
anal/sis and synthesis
then given as
is
H($)
=
(S
If
we
+
i{
e ™')(s
+
13.2 for n
(,
e>
*>*)
_
2
+
0.76536s
+
+
l)(s*
=
1
to n
=
8,
+
1.84776s
simplify the use of Butterworth functions, H(s)
and
(13 26)
"
+
and expand, we obtain
—
(s
To
«*««»«Xa
express sk in complex form
H( s\
13.1
+
««»«.)(,
in factored
(12 27)
1)
given in Tables
is
form as in Eq.
13.27, or
multiplied out as
H(s)
=
-^
a ns
n
+
fln-xs"-
TABLE
1
+
•
•
•
+
a ts
+
(13.28)
1
13.1
Butterworth Polynomials (Factored Form)
n
+1
+ V2s + 1
(**+* + 1)(* + 1)
(s2 + 0.76536* + IX** + 1.84776* + 1)
8
2
(* + 1)(* + 0.6180* + IX* + 1.6180* + 1)
2
(** + 0.5176* + IX* + ^2* + IX*2 + 1.9318* + 1)
2
2
2
(* + IX* + 0.4450* + IX* + 1.2456* + IX* + 1.8022* + 1)
(s2 + 0.3986s + IX*2 + 1.1110* + IX*2 + 1.6630* + IX*2 + 1.9622* +
1
s
2
5s
3
4
5
6
7
8
TABLE
1)
13.2
Butterworth Polynomials *
n
a*
«i
«8
«4
«s
««
1
1
2
V2
1
3
2
2
4
2.613
3.414
2.613
5
3.236
5.236
5.236
3.236
6
7
3.864
7.464
9.141
7.464
3.864
4.494
10.103
14.606
14.606
10.103
4.494
8
5.126
13.138
21.848
25.691
21.848
13.138
a
=
!•
«7
«8
1
1
1
1
1
5.126
1
Topics
13.4
OTHER LOW-PASS
In Section 13.3,
low-pass
filter
we examined
characteristic.
approximants in
FILTER
design
373
APPROXIMATIONS
the maximally
We
in filter
approximation to a
flat
will consider other low-pass filter
this section.
equal-ripple approximation
have seen that the maximally flat approximation to the ideal lowpass filter is best at co
0, whereas, as we approach the cutoff frequency
We now
<w = l, the approximation becomes progressively poorer.
consider an approximation which "ripples" about unity in the pass band
and falls off rapidly beyond the cutoff co = 1. The approximation is
The Chebyshev or
We
=
equally
good
at
co
=
ripple approximation.
and co = 1, and, as a result is called an equalThe equal-ripple property is brought about by the
use of Chebyshev cosine polynomials defined as
= cos (n cos-1 co)
= cosh (n cosh-1 co)
C^co) = 1
For n = we see that
and for n = 1, we have
Ci(<*>) = <°
C„(a>)
|co|
<, 1
|co|
>
(13.29)
1
(13.30)
(13.31)
Higher order Chebyshev polynomials are obtained through the recursive
formula
_
(B 32)
CJfu) =
^ ^^ ^ ^
Thus
for n
In Table
= 2, we obtain Cg(co) as
Cg(co) = 2<o((o) —
= 2eo 8 - 1
13.3,
1
Chebyshev polynomials of orders up to n
TABLE
Chebyshev polynomials
n
(13.33)
=
10 are given.
13.3
Cn(m) = cos (n cos-1 <o)
1
1
CO
2
2co*
4e»3
3
4
5
6
7
8
9
10
-1
— 3o>
8a>* - 8a>* + 1
16a> 6 - 20a>8 + 5o>
32o>* - 48<o« + 18a>* - 1
64o> 7 - 112a>« + 56eo8 - lea
128a>8 - 256w« + 160w4 - 32o>* + 1
256o>» - 576o»7 + 432o)« - 120a.8 + 9a>
512a» 10 - 1280w8 + 1120a>* - 400eo* + 50«* -
1
Network
374
FIG.
The
and synthesis
analysis
13.4. C,(co)
and
Chebyshev polynomials.
C,(u>)
pertinent properties of Chebyshev polynomials used in the low-pass
filter
1.
approximation are:
The
zeros of the polynomials are located in the interval
as seen by the plots of
2.
Within the interval
exceeds unity; that
3.
C3(o))
Beyond
values of
is,
and
the absolute value of
\w\ <, 1,
|C„(eo)| <,
the interval
<
\a>\
|eo|
<, 1,
CJio)) in Fig. 13.4.
for
1
1,
\co\
CB (co)
never
<, 1.
|C„(w)| increases rapidly for increasing
\a>\.
Now, how do we apply
the Chebyshev polynomials to the low-pass
approximation? Consider the function e 2 C„%w), where e is real and
and e2
small compared to 1. It is clear that e 2 Cn 2 ((o) will vary between
filter
in the interval
e2
^
\u>\
Now we
1.
C„ 2(g>). This new function
slightly greater
than unity, for
we
the function which
add
<,
|<w|
will associate
IHO"))!*
=
1
<
1
to this function
varies between
1.
1
and
1
+
making
e2,
it 1
+
a quantity
Inverting this function,
we obtain
with \H(jco)\ 2 ; thus
+ *C n\a>)
(13.34)
\H(ja>)\ 2 oscillates
about unity such that the
Outside this interval,
Cn2(w) becomes very large so that, as to increases, a point will be reached
2
approaches zero very rapidly with
where e2 CB2(e>)
1 and \H(jcu)\
further increase in <o. Thus, we see that \H(jw)\ 2 in Eq. 13.34 is indeed a
Within the interval
maximum value is
1
|eo|
1,
and the minimum
is 1/(1
+
e 2).
»
approximant for the ideal low-pass filter characteristic.
Figure 13.5 shows a Chebyshev approximation to the ideal low-pass
<u <, 1, \H{jw)\ ripples
filter.
We see that within the pass band
suitable
^
between the value
1
and (1
+
1
e 2 )- -*.
The ripple height or distance between
Topics
(l
in filter
-^—^-—
+ e2) *
«
l
FIG.
1
3.5. Chebyshev approximation to low-pass
maximum and minimum
in the pass
<o
=
because
\H(jl)\
1, |H(;«>)| is
Cn\\) =
where
e*
is
given as
1
(13.35)
(!+«>"
=
+ ^a
Cn*{(o) »
1
(13.36)
is,
for
\<o\
^
1,
as a> increases,
we
reach a point
so that
]ff(Mlas-
>
O)
eC n(a>)
The
filter.
1.
In the stop band, that
io k ,
band
=1—
Ripple
At
375
design
(13.37)
0) k
loss in decibels is given as
Loss
But for large
<w,
CB(co)
= -20 log10 H(jm
^ 20 log e + 20 log CB(o)
\
)\
can be approximated by
its
(13.38)
_1
leading term 2" <w",
so that
Loss
= 20 log6 e + 20 log5 2*- g>"
= 20 log e + 6(n - 1) + 20/t log a>
1
(13.39)
Network
376
We
anal/sis and synthesis
Chebyshev response also falls off at the rate of 20/i
initial drop of 20 log e + 6(n — 1) decibels. However,
in most applications, e is a very small number so that the 20 log e term is
see that the
db/decade after an
actually negative. It is necessary, therefore,
to compensate for this decrease in loss in
band by choosing a
the stop
cosh 0k
sufficiently
large n.
From
the preceding discussion,
we
see
a Chebyshev approximation depends
upon two variables, e and n, which can be
determined from the specifications directly.
The maximum permissible ripple puts a
bound on e. Once e is determined, any
that
sinh
0*
desired value of attenuation in the stop
band fixes n.
The derivation of the system function
H(s) from a Chebyshev amplitude approximation \H(jco)\ is somewhat involved and
not be given here. 2 Instead, we will
simply give the results of such a derivation.
will
First
we
introduce a design parameter.
FIG. 13.6. Locus of poles of
Chebshev filter.
fih
—
- sinh
*
n
where n
is
-
(13.40)
e
and e is the factor
The poles, sk = ak + jcok of the equal-ripple
located on an ellipse in the s plane, given by
the degree of the Chebyshev polynomial
controlling ripple width.
approximant H(s) are
sinh
,
cok
+
a
/S fc
cosh*
=
(13.41)
1
/?*
The major semiaxis of the ellipse is on the jco axis and has a value co =
±cosh pk The minor semiaxis has a value a = ±sirih(ik and the foci
are at co = ±1 (Fig. 13.6). The half-power point of the equal-ripple
.
,
amplitude response occurs at the point where the ellipse intersects the^eo
cosh fik . Recall that for the Butterworth response, the
axis, i.e., at co
half-power point occurs at co = 1. Let us normalize the Chebyshev poles
1 instead of at co
sk such that the half-power point also falls at co
cosh fi k ; i.e., let us choose a normalizing factor, cosh P k , such that the
=
=
=
*
Interested parties are referred to
Network
Synthesis,
M.
John Wiley and Sons,
E.
Van Valkenburg,
New
Introduction to
York, 1960, Chapter
13.
Modern
Topics
normalized pole locations
s' t
in filter
377
design
are given by
cosh
(th
cosh
/?*
cosh
t
= <**+./«>*
locations can be derived as
The normalized pole
„
= tanhu &
sin
•
<r't
/2fc
- lW
^
^j
(13.43)
/2fc-lW
a>t
= cos ^__j_
Comparing the normalized Chebyshev pole locations with the Butterworth
we see that the imaginary parts are the same,
while the real part a' k of the Chebyshev pole location is equal to the real
part of the Butterworth poles times the factor tanh /3». For example, with
0.444, the Butterworth poles are
n
3 and tanh /?,.
pole locations in Eq. 13.24,
=
=
*i- -1 +J0
j23
=
_0.5 ±/0.866
so that the normalized Chebyshev poles are given by
4 --1(0.444)+ JO
szs
=
-0.444
=
-0.5(0.444) ±;0.866
=
-0.222 ±y0.866
Finally, to obtain the denormalized
s' k
by cosh
/3 t that
,
+;0
Chebyshev
poles,
we simply
multiply
is,
Sk
=
(a\
+ /»',) cosh &
(1 3.44)
There is an easier geometrical method to obtain the Chebyshev poles,
given only the semiaxis information and the degree n. First we draw two
circles, the smaller of radius sinh /?* and the larger of radius cosh /?„ as
Network
378
and synthesis
anal/sis
Butterworth
pole locus
Chebyshev
pole locus
FIG.
shown in
13.7.
»
=
3
Chebyshev
poles.
radial lines according to the angles of
we draw
Fig. 13.7. Next,
filter
the Butterworth poles (Eq. 13.22) as shown.
Finally,
we draw
vertical
dashed lines from the intersections of the smaller circle and the radial lines,
and horizontal dashed lines from the intersections of the large circle and
the radial lines. The Chebyshev poles are located at the intersection of
the vertical and horizontal dashed lines, as shown in Fig. 13.7.
Consider the following example. We would like to obtain a system
function H(s) that exhibits a Chebyshev characteristic with not more than
ripple in the pass
1 -decibel
When we design for
down
1
band and is down
1-decibel ripple,
\H(jl)\
=
20 log —-t-zu
'(!+€»)*
1
We then obtain
(1
and
+
a
* )
e
Our next
task
is
to
=
1,
Solving for «,
1A
=
=
= -1
w=
2.
\H(j\)\ is
0.509
^ 20 log 0.509 + 6(n =
134 5)
(13.46)
(13.47)
from the 20 decibels at m = 2
can be given as approximately
we obtain «
(
0.891
to find n
Eq. 13.39 the loss
20
1)
+
specification.
20« log 2
(13.48)
n must be an integer, we let n = 3.
the pole locations are completely speci-
2.65. Since
With the specification of n and e,
Our next task is to determine
fied.
20 decibels at
that at
decibel so that
20 log
From
at least
we know
these pole locations.
First
we must
Topics
find
Pk
.
From
Eq. 13.40
in filter
379
design
we have
n
e
= I sinh"
1
(13.49)
1.965
=
0.476
In order to find the normalized Chebyshev poles from the Butterworth
poles, we must first determine tanh fik Here we have
.
tanh
From Table
13.1, the
*!
n
=
=
pk
=
tanh 0.476
= 0.443
(13.50)
3 Butterworth poles are
-1.0,
s 23
=
-0.5 ±y0.866
(13.51)
Multiplying the real parts of these poles by 0.443, we obtain the normalized
Chebyshev
poles.
s\
=
-0.443,
s' 2S
=
-0.222 ±/0.866
Finally, the denormalized Chebyshev poles are obtained by multiplying
the normalized ones by cosh (ik
1.1155 so that the denormalized poles
are sx
-0.494 and $2S
-0.249 ±/0.972.
=
=
H(s)
is
H(s)
=
then
=
°^02
(s
+ 0.494)(s + 0.249 - ;0.972)(s + 0.249 +
(13.52)'
v
0.502
=
s*
+
0.992s*
+
/0.972)
1.255s
+
0.502
In Fig. 13.8, the amplitude responses of the Chebyshev and an n
Butterworth filter are shown.
Monotonic
filters
with
=
3
optimum
cutoff
with Chebyshev niters, the following
can be said. The Butterworth response is a maximally flat, monotonic
response, whereas the Chebyshev response is equal ripple in the pass band.
In the stop band, the Chebyshev response falls off more rapidly than the
Butterworth (except when e is very, very small). In this respect, the Chebyshev filter is a better filter than the Butterworth. However, as we shall see
in Section 13.5, the transient response of the Chebyshev filter is very poor.
If we require sharp cutoff characteristics for a given degree n, however,
the Butterworth filter is quite unsatisfactory. In 1958, Papoulis 8 proposed
In comparing Butterworth
*
A. Papoulis, "Optimum
March
1958, pp. 606-409.
filters
Filters with
Monotonic Response," Proc. IRE, 46, No.
3,
380
Network
analysis
and synthesis
Radian frequency,
02
0.1
0.6
0.4
0.3
o>
0.8
1
iI
3
'-;>< \
-1
i
—2
\
—4
-\\
—5
\\
—6
I
Amplitude response of n
-/
-8
Chebyshev
—
1.0-db ripple
in
=
\
3
with
filter
pass band
Buttterworth respons>e (n
=
\
3)
\\
—9
\\
-10
\\
— 11
\\
I
— 12
\
\
— 13
\
-14
FIG.
13.8.
= 3 Chebyshev filter with 1.0-decibel ripple in pass
= 3).
Amplitude response of n
band and Butterworth response
(n
a class of filters called Optimum or
"L"
filters,
which have the following
properties:
1.
2.
The amplitude response is monotonic.
The fall-off rate at m cutoff is the greatest possible,
if monotonicity is
assumed.
3.
The
zeros of the system function of the
L filter
Recall that the magnitude response of a low-pass
infinity can be expressed as
M(o>)
filter
*•
=
[1
=
with
all
zeros at
(13.53)
+JV)l*
Let us denote the polynomial generating the
/(>*)
are all at infinity.
L„(ft> 8)
L filter by
(13.54)
Topics
in filter
design
381
Ln(mx) has the following properties:
The polynomial
=
L„(l) =
LJO)
(a)
(b)
1
^^0
(0
dco
dL
«0
>
2
>
=M
(M maximum)
dco
and c are the same as for the Butterworth generating polynomial/^*) = to2 ". Property c insures that the response M(a>) is monotonic and property d requires that the slope of L n(caP) at tu = 1 be the
Properties a, b
steepest to insure sharpest cutoff.
Papoulis originally derived the generating equation for the polynomials
Ln (for n
odd) to be
Ln((ot)=
where n = 2k
4
first kind
+
1
L
dx
a<p<(a;)
Llo
(13,55)
J
and the PJx) are the Legendre polynomials of the
=1
P^x) = x
P2(x) = K3* - 1)
P9(x) = i(5x* - 3x)
/>,>(*)
2
and the constants
a«
a, are given
by
=—=—=
3
(1356)
•
=
•
2k
5
+
= —1=
1
V2(fc
+
(13.57)
1)
Later Papoulis 6 and, independently, Fukada,' showed that the evenordered L n polynomials can be given by
W
n
•
•
r
rim*-!
(a>8)
(*
-J
=
2k
+
+
*
1)[2>«
-|
*>«(*)]
8
dx
(13.58)
2
E. Jahnke and F. Emde, Tables of Functions, Dover Publications, New York, 1945.
A. Papoulis, "On Monotonic Response Filters," Proc. IRE, 47, February 1959,
332-333.
•
M. Fukada, "Optimum
Trans.
IRE, CT-6, No.
3,
Filters of Even Orders with Monotonic Response,"
September 1959, 277-281.
:
382
Network
anal/sis and synthesis
where the constants at are given by:
Case 1 (k even)
a
«i
-
7
=
«»
a*
-
+
=
a
(13.59)
V(fc
+
l)(fc
+
2)
V(fc
+
l)(k
+
2)
polynomials up to n
=
7 together with
2fc
=
1
1
M=
Case 2 (k odd):
a*
3
flo
+l
=•=«.*
= «2
=°
7
2fc
Fukada tabulated the
dLjai^ldco evaluated at
the cutoff. This
To
is
LB(w 2)
to
shown
=
(13.60)
to give an indication of the steepness of
1
in Table 13.4.
L filter, we must
obtain the system function H(s) for the
factor the
equation for his2) and choose the Hurwitz factors as H(s).
(13.61)
For example, for n
=
3,
the magnitude response squared
M\a>)
is
1
=
1
+
L^co*)
(13.62)
1
1
Substituting
+
a>*
-
— <w = s we obtain
2
=
) = H(s) H(-s)
2
3«>
4
+
3eo
6
2
,
1
fc(s
1
TABLE
L„(a> 2
)
-
s*
-
3s
4
-
(13.63)
35*
13.4
Polynomials
dLn(l)
dm
Ln(a>2)
4
CO*
- 3o>4 + <u2
4
6eo 8 - 6o>* + 3to
10 - 40a>8
20o
+ 28a>8 - 8a>4 + to2
12 - 120a> 10 + 105e»8 - 40w* + 6o>4
50to
10 - 355o>8 + 105a>«
175*o M - 525a>» + 615«>
8
3o>8
12
18
24
-15eo4
+
2
a)
32
Topics
in filter
design
383
Frequency, rad/sec
02
0.1
0.3
0.4
0.6
0.8
^
-2
-4
1.0
—
Amplitude i esponsecrf
OpB mum vers
Butbsrworth fit ers
-6
\\
\\
-8
\\
\\
tf-10
0.4
|-12
—
-14
-16
0.5
Frequency
0.6
0.8
i
1.0
—
1.2
\
\ \
\\
y
-1.0
\\
\\
\
\
»
\
-18
\
-20
\
ft-
y
\
\
A
-22
Explanded
scale
-24
\
\
\\
-26
FIG.
After
we
13.9.
Amplitude response of Optimum versus Butterworth
factor A(s*),
we
if(s)
filters
obtain
0.577
=
s*
+
1.31s
2
+
1.359s
+ 0.577
(13.64)
where the numerator factor, 0.577, is chosen to let the d-c gain be unity.
poles of H(s) are sx = -0.62; s2>3 = -0.345 ±;0.901. The amplitude response of third-order Optimum (L) and Butterworth niters are
compared in Fig. 13.9 Note that the amplitude response of the Optimum
filter is not maximally flat, although still monotonic. However,
the cutoff
characteristic of the Optimum filter is sharper than the cutoff of the
The
Butterworth
filter.
Linear phase filters
Suppose a system function
is
given by
H(s)
where
= Ke~'T
K is a positive real constant.
Then
(13.65)
the frequency response of the
system can be expressed as
H{jm)
=
Ke-"" T
(13.66)
Network
384
analysis and synthesis
so that the amplitude response M(<co) is a constant K,
sponse
#w)
is linear
in
= -mT
{e(t),
re-
(13.67)
The response of such a system
co.
the transform pair
and the phase
to
an excitation denoted by
E(s)} is
R(s)
so that the inverse transform
= K E(s)e-' T
r(t)
(13.68)
can be written as
KO-c-W)]
= Ke(t -
T)u(t
-
(1369)
T)
We see that the response r(t) is simply the excitation delayed by a time
and multiplied by a constant. Thus no
signal distortion results
T,
from
transmission through a system described by H(s) in Eq. 13.65. We note
further that the delay T can be obtained by differentiating the phase
response
<f>((o),
by
<u;
that
is,
Delay
_ - f^2> - r
(13.70)
dco
Consequently, in a system with linear phase, the delay of the system is
obtained by differentiating the phase response #co).
system with linear phase and constant amplitude is obviously desirable
from a pulse transmission viewpoint. However, the system function H(s)
A
in Eq. 13.65
a delay
only realizable in terms of a lossless transmission line called
If we require that the transmission network be made up of
is
line.
lumped elements, then we must approximate H(s)
function in s. The approximation method we
due to Thomson. 7
= Ke~ ,T by
a rational
shall describe here is
We can write H(s) as
(13.71)
£o
sinh
sT + cosh sT
Let the delay T be normalized to
where K is chosen such that H(0)
denominator of H(s) by sinh s
and
unity and let us divide both numerator
=1.
to obtain
Kjsmhs
coth
'
W.
E.
Thomson, "Network with Maximally
1952, 256-263.
s
+
(13?2)
1
Flat Delay," Wireless Engrg., 29, Oct.
If sinh s
and cosh * are expanded
coshs
in
power
Topics
in filter
series,
we have
s*
s*
s*
= l++-+-+
From these series expansions, we
385
---
(1373)
,'
.'
6
»
s
s
sinh5- S + - + - + - +
s
design
'--
then obtain a continued fraction expan-
sion of coth s as
coth
s
=-+
3
S
+
s
(13.74)
5
1
~*
z+...
5
If the continued fraction is terminated in
n terms, then H(s) can be
written as
where
Bn(s) are f?es.«>/ polynomials defined by the formulas
2*0=1
Bt = s+1
From these
formulas,
(13.76)
2?„
- (2/i - 1)3^ + s*Bn_t
we
obtain
= s* + 3s + 3
2?, = j» + 6s* + 15s +
if,
(13.77)
15
Higher order Bessel polynomials are given in Table 13.5, and the roots of
Bessel polynomials are given in Table 13.6. Note that the roots are all in
the left-half plane. A more extensive table of roots of Bessel polynomi8
als is given by Orchard.
The amplitude and phase response of a system function employing an
unnormalized third-order Bessel polynomial
W=
Ms)
s*
+
^+
6s*
15s
+
(13.78)
15
• H. J. Orchard, "The Roots of Maximally Flat Delay Polynomials"
on Circuit Theory, CT-12 No. 3, September 1965, 452-454.
IEEE Trans,
Network
386
analysis
and synthesis
TABLE
13.5
Coefficients of Bessel Polynomials
n
h
bo
h
bt
b*
*»
h
b7
1
1
1
1
2
3
3
1
3
15
15
4
105
105
10
1
5
945
945
6
45
420
105
15
1
6
10,395
10,395
4,725
1,260
210
21
1
7
135,135
135,135
62,370
17,325
3,150
378
28
1
1
are given by the solid lines in Figs. 13.10 and 13.11. These are compared
with the amplitude and phase of an unnormalized third-order Butterworth
function given by the dotted lines. Note that the phase response of the
constant-delay function is more linear than the phase of the Butterworth
function. Also, the amplitude cutoff of the constant-delay curve is more
gradual than that of the Butterworth.
TABLE
13.6
Roots of Bessel Polynomials
Roots of Bessel Polynomials
1
2
3
4
-1.0 +y0
-1.5 ±y0.866025
[-2.32219 +y0
— 1.83891 ±yl.75438
i
/ -2.89621 ±y0.86723
\ -2.10379 ±y2.65742
-3.64674 +j0
-3.35196 ±yl.74266
[-2.32467 ±/3.57102
C
C
[
-4.24836 ±y0.86751
-3.73571 ±/2.62627
-2.51593 ±y4.49267
-4.97179 +j0
-4.75829 ±yl .73929
-4.07014 ±y3.51717
1-2.68568 ±y5.42069
r
i
A
Topics
0.1
0.4
0.2
in filter
Frequency, rad/sec
0.6
0.8 1
-2
387
design
4
3
5
6 7
\
>^
V
-4
\
-Bessel
\
-6
\
-8
\
\
,-10
B it* srv fO rth
h
!
2
—
\
\
\
-14
\
\
-16
\
-18
V
<
\
-20
\
\
\
\
-22
-24
\
\
\
\
-26
FIG. 13.10. Amplitude response of n
= 3 Bessel and Butterworth filters.
Frequency, rad/sec
0.2
0.4
FIG.
0.6
13.1 1.
0.8
1.0
1.2
Phase responses of low-pass
1.4
filters.
1.6
1.8
2.0
388
13.5
Network
analysis
and synthesis
TRANSIENT RESPONSE OF LOW-PASS FILTERS
we
compare the
transient response of the filters
discussed in Section 13.4. In particular,
we will compare the step response
In this section
will
of the niters according to the following figures of merit:
1.
Rise time
t
R
The
.
rise
time of the step response
the time required for the step response to rise
is
defined here as
from 10%
to
90%
of
its
final value as depicted in Fig. 13.12.
Ringing is an oscillatory transient occurring in the response
of a filter as a result of a sudden change in input (such as a step).
quantitative measure of the ringing in a step response is given by its
2. Ringing.
A
settling time.
3. Settling time.
The
settling
step response does not differ
time
is
from the
that time
final
t,
value by
beyond which the
more than ±2%,
as depicted in Fig. 13.12.
Delay time is the time which the step response
of its final value as shown in Fig. 13.12.
5. Overshoot.
The overshoot of the step response is defined as the
difference between the peak value and the final value of the step response
4.
Delay time,
requires to reach
t
D
.
50%
(see Fig. 13.12) expressed as
a percentage of the
8
FIC.
tB
=
12
10
Time
t,
sec
13.12. Figures of merit for step response.
delay time.
final value.
ta
= rise
time;
/.
=
setting time;
Topics
in filter
389
design
Most of the foregoing figures of merit are related to frequency response,
particularly bandwidth and phase linearity. Some of the quantities, such
related to each other but have
Let us examine qualitatively the
relationships between the transient response criteria just cited and
and delay time are intimately
as rise time
rather tenuous ties with overshoot.
frequency response.
Rise time and bandwidth have an inverse relationship in a filter. The
wider the bandwidth, the smaller the rise time; the narrower the bandwidth, the longer the rise time. Physically, the inverse relationship could
be explained by noting that the limited performance of the filter at high
down the abrupt rise in voltage of the step and prolongs
the rise time. Thus we have
frequencies slows
TB X
BW = Constant
(13.79)
Rise time is a particularly important criterion in pulse transmission. In
an article on data transmission,* it was shown that in transmitting a pulse
of width 7\ through a system with adjustable bandwidths, the following
results were obtained:
Rise Time
Bandwidth
(/.
- UTJ
(milliseconds)
0.5
f.
The
table
shows a
2/«
0.25
3/c
0.16
4/«
0.12
y.
0.10
definite inverse relationship
between
rise
time and
bandwidth.
A
definition of time delay
is
given by Elmore as the
first
moment
or
centroid of the impulse response
Jo
-r
t
h{t) dt
(13.80)
provided the step response has little or no overshoot. Elmore's definitio
of rise time is given as the second moment
TB
= \2nj\t - TDf h(t) a]
(13.81)
•R. T. James, "Data Transmission—The Art of Moving Information,"
Spectrum, January 1965, 65-83.
IEEE
Network
390
anal/sis and synthesis
These definitions are useful because we
can obtain rise time and delay time directly
from the coefficients of the system function
H(s). Without going into the proof, which
is in Elmore and Sands, 10 if H(s) is given as
=
His)
FIG.
13.13.
1
2
1
R-C network.
+ ° lS + a *s2 +
+ b s + 6js +
x
'
+ a*s n
+ b ns m
(13.82)
the time delay
and the
rise
time
bl
—
ax
af
+
2(a 2
TD =
TD i%
TR =
For the R-C network
-
{2tt[V
in Fig. 13.13, H(s)
H(s)
=
- b )]}*
(13.84)
2
is
R
V(s)
I(s)
1
+
(13.85)
sRC
TD = RC_
TB = yJlnRC
so that
It
(13.83)
is
(13.86)
should be emphasized that Elmore's definitions are restricted to step
responses without overshoot because of the moment definition. The more
general definition of rise time is the 10-90% one cited earlier, which has
no formal mathematical
Overshoot
excess gain
is
definition.
generally caused
by "excess" gain
we normally mean a magnitude
at high frequencies.
characteristic with a
By
peak
such as the shunt peaked response shown by the dashed curve in Fig.
13.14. A magnitude characteristic with no overshoot is the magnitude
characteristic of
an R-C interstage shown by the
solid curve in Fig. 13.14.
Log frequency
FIG.
10
V,
13.14.
Comparison of shunt-peaked and simple R-C magnitudes.
W. C. Elmore and M. Sands, Electronics, National Nuclear Energy
McGraw-Hill Book Company, New York, 1949, pp. 137-138.
1,
Series,
Div.
Topics
in filter design
391
1.2
1.1
1.0
0.9
?0.8
N
=5 0.7
E
5
c
I
0.6
0.5
In
f 6.4
I
J 3 ~A> = 7/il=lo"
0.3
0.2
0.1
2
4
6
8
10
Time
FIG.
13.15.
t,
12
16
14
18
20
sec
Step response of normalized Butterworth low-pass
filters.
The step responses of the n = 3, n = 7, and n = 10 Butterworth filters
are shown in Fig. 13.15. Note that as n increases, the overshoot increases.
This
because the higher order Butterworth niters have flatter magnitude
(i.e., there is more gain at frequencies just below the cutoff).
Ringing is due to sharp cutoff in the filter magnitude response, and is
accentuated by a rising gain characteristic preceding the discontinuity.
is
characteristics
an n = 3 Bessel (linear phase) filter is compared
an n = 3 Chebyshev filter with 1-decibel ripple in
Fig. 13.16. We cannot compare their rise times since the bandwidths of
the two filters have not been adjusted to be equal. However, we can
compare their ringing and settling times. The Chebyshev filter has a
sharper cutoff, and therefore has more ringing and longer settling time
than the Bessel filter. Note also the negligible overshoot of the Bessel
The
step response of
to the response of
filter
that
The
is characteristic
of the entire class of Bessel
decision as to which
filter is
best depends
filters.
upon
the particular
In certain applications, such as for transmission of music,
not important. In these cases, the sharpness of cutoff may be the
situation.
phase
is
dominant factor so that the Chebyshev or the Optimum filter is better
than the others. Suppose we were dealing with a pulse transmission
system with the requirement that the output sequence have approximately
the same shape as the input sequence, except for a time delay of T
Tt Tx , as shown in Fig. 13.17a. It is clear that a filter with a long rise
=
—
time
is
not suitable, because the pulses would "smear" over each other
—
Network
392
1.1
1.0
/
t
t
0.9
1
0.8
and synthesis
analysis
'
i
N
/
^ —^ s
/
—
!
1
0.7
1
•S 0.6
a
3 Bessel fitter
-Step response of n
-Step response of n « 3 Chebyshev
filter with 1-db ripple
nr.
a 0.5
i
<0.4
1
|
0.3
/
0.2
~r
0.1
ly
1
4
2
8
6
12
10
14
16
18
20
Time, sec
FIG. 13.16. Comparison of filter transient responses.
System
with short
rise
and
settling
Ti
times
Input pulse train
ALA
T2
Output pulse
train
(a)
System
with long
rise
and
settling
Ti
times
Output pulse train
Input pulse train
(b)
FIG. 13.17. Smearing of pulses in systems with long
rise
and
setting times.
as seen in Fig. 13.176. The same can be said for long settling times.
Since a pulse transmission system must have linear phase to insure undistorted harmonic reconstruction at the receiver, the best filter for the
system
13.6
is
a linear phase
filter
with small
rise
and
settling times.
A METHOD TO REDUCE OVERSHOOT
IN FILTERS
method to reduce the overshoot and ringing of a
filter
filter step response. The step response of a tenth-order Butterworth
We
is
about
overshoot
18%.
the
seen
that
is shown in Fig. 13.18. It is
We
present here a
J
Topics
393
in filter design
\2
1.0
J
0.8
tep re sponsc
I
0.6
I
0.4
/
/
02
1
«.'
J
Second
derivative
\
"^ ^>
r
/
\
\
-1.2
8
10
12
Time t, sec
16
14
18
20
FIG. 13.18
note that after the
first
peak, the ringing of the step response has an
approximate sinusoidal waveshape. Let us now consider the second
derivative of the step response shown by the dashed curve in Fig. 13.18.
Beyond the first peak of the step response, the second derivative is also
(approximately) sinusoidal, and
is
negative
when
when
the step response
is
than unity.
If we add the second derivative to the step response, we reduce the overshoot and ringing. 11
Suppose a(0 is the step response and H(s) is the system function of the
greater than unity,
filter.
The
and
positive
the step response
is less
corrected step response can be written as
a1(0
= «(0 +
«^
L>
(13.87)
where K is a real, positive constant. Taking the Laplace transform of
Eq. 13.87, we have
tW*)]
-
Xs
L—
(13.88)
H(s)
s
11
F. F.
Kuo, "A Method to Reduce Overshoot
No. 4, December 1962, 413-414.
Theory, CI-9,
in Filters," Trans.
IRE on
Circuit
Network
394
analysis and synthesis
1.2
/%
1.0
^
'
--».
^-.^
»
k
i.
0.8
lo.6
—-
<E
—
Tenth order Butterworth
Zeros at
Zeros at
0.4
= i
1.0
a>= *
1.5
a)
7
Jtt
0.2
y/f
Jl
8
12
10
Time t, sec
16
14
18
20
FIG. 13.19
From
s
=
Eq. 13.88
we
see that
by adding a pair of zeros on the jm
axis at
±//v K, the overshoot and ringing are reduced.
l/\K must in general be greater than
For normalized Butterworth filters the
bandwidth is co = 1 so that K <, 1. The factor AT also controls the amount
of overshoot reduction. If K is too small, adding the zeros on they'co axis
For low-pass
niters, the factor
the bandwidth of the system.
-o
-2
-4
-6
-8
-10
.o
-o
-12
c -14
<3
_i 6
-18
-20
—«
"
——
—
.—
Tenth order Butterworth
Zeros at co = t 1.0
Zeros at w = * 1.5
Zeros at co = ± 1.8: with
original bandwidth also
equal to 1.8
-22
-24
I
-26
-28
0.1
0.2
0.3
0.4
0.6
0.8 1.0
Radian frequency,
FIG. 13.20
u
1.5
2.0
3.0
Topics
in filter
395
design
will have negligible effect. Therefore the zeros should be added somewhere near the band edge. Figure 13.19 shows the effects of adding
zeros at <w = ±1 (i.e., right at the band edge), and at m = 1.5. We see
that the further away the zeros are placed from the band edge, the less
they will have. The addition of the zeros will decrease the 3-decibel
bandwidth of the filter, however, as seen in Fig. 13.20. Therefore, a
effect
compromise must be reached between reduction
in
bandwidth and
reduction in overshoot.
An effective way to overcome this difficulty is to scale up the bandwidth
by a factor of, for example, 1.8. Then the zeros are placed at co = ±1.8,
which will reduce the 3-decibel bandwidth to approximately its original
figure <w = 1.0, as shown in Fig. 13.20. The overshoot, however, will be
reduced as though the zeros were at the band edge.
A MAXIMALLY FLAT DELAY AND CONTROLLABLE
MAGNITUDE APPROXIMATION
13.7
we
examine an interesting result, which is due to
phase approximation with controllable magnitude. In Section 13.4 we discussed approximation of a
flat delay using Bessel polynomials. The resulting rational approximant
was an all-pole function (all transmission zeros at s = oo), whose denominator was a Bessel polynomial. There was no control of the magnitude using the all-pole approximant. In Budak' s method the magnitude
is controllable, while the phase is as linear as the standard Bessel approxiIn this section
Budak. 1 * The
will
result deals with linear
mation.
Budak's approximation is obtained by introducing the parameter k to
two parts such that
split e~~* into
*-
= pSr.
<
k
<. 1
(13.89)
and then approximate independently c-** and e~lk~lU with all-pole Bessel
polynomial approximations. Thus the resulting approximation for e~*
will have Bessel polynomials for both numerator and denominator. The
poles of the e~ * -1) ' approximant will be the zeros in the final approximant,
(
while the poles of the e~** approximant remain as poles in the final
approximant. For realizability, the degree of the c- <t-1) * approximant
should be less than the degree of the e ** approximant.
11
A. Budak,
Trans,
"A Maximally Flat Phase and Controllable Magnitude Approximation,"
of IEEE on
Circuit Theory,
CT-12, No.
2,
June 1965, 279.
Network
396
anal/sis
and synthesis
A0
M(o>)
1.0
1.0
-0.6
«£T*
0.8
0.8
0.7^
-1.0
ft
0.6
0.6
(Bessel)"^
A=1.0_
0.4
0.4
k-
0,8-
=
02
km 0.7-j^
*-0.6tC
0.2
2
4
3
3
(a)
(b)
FIG.
As an example,
13.21
consider the approximation with three zeros and four
poles.
105
(ks)*
+
10(fcs)
8
+ 45(fcs)* +
[(fe
(13.90)
105
-
l)sf
+
6[(fc
-
l)s]
a
+
15[(ik
-
l)s]
+
(13.91)
15
then perform the operations as indicated by Eq. 13.89 to obtain
-._ 7{[(fc -
6
£
+
15
g-tt-i).^
We
105(jfcs)
—
(ks)*
1)5]'
+
+
- l)s]« + 15[(fe - l)s] + 15}
+ 45(fcs)» + 105(Jb) + 105
6[(k
10(fcs)
3
In Fig. 13.21a the magnitude characteristic of Eq. 13.92 is plotted with
The phase characteristic is given in terms of deviation
as a parameter.
of phase from linearity A<f> = eo — <£(co) and is shown in Fig. 13.216.
The improvement in phase linearity over the all-pole Bessel approximation
k = 1 is shown by these curves. Note as the bandwidth is increased
(A:
decreasing), phase linearity is improved.
response of Eq. 13.92 also with
decreasing
k
A:
Figure 13.22 shows the step
as a parameter.
Since the effect of
increases bandwidth, the corresponding effect in the time
domain is to decrease rise time.
Budak also observes that as k decreases from unity, the poles and zeros
migrate to keep the phase linear. The zeros move inward from infinity
along radial
lines,
while the poles
move outward along
radial lines.
Topics
in filter
design
397
l.l
1.0
k> 0.6-
0.9
0.8
*>«0.7-
-km
).8
0.7
0.6
_*-1.0
(Bessel)
0.5
%
0.4
0.3
02
0.1
0.0
§
-0.1
-0,2
02
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
t
FIG. 13.22
13.8
SYNTHESIS OF LOW-PASS FILTERS
Given the system function of the low-pass filter as derived by the
methods described in Section 13.4, we can proceed with the synthesis of
the filter network. If we consider the class of filters terminated in a 1-Q
load, and if we let the system function be a transfer impedance,
Z«(s)
or a transfer admittance
= -^a-
Yn(s) =
1
+
(13.93)
(13.94)
y»
we can synthesize the low-pass filter according to methods given in
Chapter 12. For example, consider the n = 3 Optimum (L) filter function
Network
398
analysis
and synthesis
given as a transfer impedance
0.577
=
Z21W
s
+
8
1.31s
+
a
1.359s
+
(13.95)
0.577
We see that the zeros of transmission are all at infinity.
ator of
Z21
is
we
even,
divide both numerator
odd part of the denominator
+
s3
Since the numerand denominator by the
Thus we have
1.359*.
0.577
Za
s
Z S2
—
+
8
s
The
structure of the low-pass
infinity is given in
Chapter
reactance structure. This
we
fraction expansion of l/« 22
1.31s 2
8
+
s
+
(13.96)
0.577
1.359s
with three zeros of transmission at
filter
12.
1.359s
1.31s
We
must synthesize
z
M
to give the
it
accomplish through the following continued
:
+ 0.577 )ss + 1.359s (0.763s
s8 + 0.440s
0.919s ) 1.31s8
1.31s
+ 0.577 ( 1.415s
8
0.577) 0.919s ( 1 .593s
0.919s
The optimum filter is shown in Fig.
For the n
13.23.
= 3 Butterworth filter
given by the transfer impedance
Z2l(s)
we have
z n (s)
8
2s
s
+
2s
(13.97)
+ 2s-rl
,.
=
s
We
~ss +
2s
8
+
1
-t» 7T7s
(13.98)
then synthesize « M (s) by a continued fraction expansion to give the
filter
shown
in Fig. 13.23.
(a)
Optimum
(b) Butterworth
filter
FIG. 13.23
filter
Topics in
filter
>ia v2
c7 :i
FIG. 13.24. Canonical form for
niters described in
Tables 13.7, 13.8, and 13.9.
In Tables 13.7, 13.8, 13.9 are listed element values (up to n
single-terminated Butterworth, Chebyshev (1-decibel ripple),
filters,
respectively. 13
transfer
impedance
desired,
we
vice versa.
399
design
=
7) for
and Bessel
These apply to the canonical realization for a
Z21(s)
shown
simply replace
all
The element values
in Fig. 13.24.
If a
shunt capacitors by
all
Y21(s)
realization is
series inductors
and
carry over.
we will consider some examples of synthesis of double
To stimulate the curiosity of the reader, note that the
transfer function V2 jV of the network in Fig. 13.25 is
In Chapter 14,
terminated
filters.
voltage-ratio
precisely the n
=
3 Butterworth function.
Recall that in Chapter 12,
when we cascaded two
networks, the overall system function
H
H
H
(s)
constant-resistance
was the product of the
in-
We
can apply this property to
networks which are not constant-resistance if we place an isolation amplifier between the networks, as shown in Fig. 13.26. Since pentodes provide
the necessary isolation, our task is simplified to the design of the individual
structures H^s),
2 (s),
n (s), which we call interstage networks.
dividual system functions
H
Some common
.
.
.
t (s)
,
2 (s).
H
shown in Fig. 13.27. In Fig.
a structure known as the shunt-peaked network is shown. The transfer impedance of the shunt-peaked network is
interstage structures are
1 3.27a
Za(s)
FIG. 13.25. n
=
= -1
13,
+
RjL
3 double-terminated Butterworth
u More extensive tables are given in L.
and Synthesis, Chapter
s
(13.99)
Cs 2 + sR/L+(llLC)
filter.
Weinberg's excellent book, Network Analysis
McGraw-Hill Book Company, 1962.
400
Network
analysis
and synthesis
TABLE
13.7
Normalized Element Values for a Single Terminated
Butterworth Filter
Cx
n
Lt
c,
L*
Q
Lt
1
1.000
2
0.707
1.414
3
0.500
1.333
1.500
4
0.383
1.082
1.577
5
0.309
0.894
1.382
1.694
1.545
6
0.259
0.758
1.202
1.553
1.759
1.553
7
0.222
0.656
1.055
1.397
1.659
1.799
C7
1.531
TABLE
1.558
13.8
Normalized Element Values for a Single Terminated
Chebyshev Filter with l-db Ripple
Cx
L,
C,
L4
Cs
4
1
0.509
2
0.911
0.996
3
1.012
1.333
1.509
4
1.050
1.413
1.909
1.282
5
1.067
1.444
1.994
1.591
1.665
6
1.077
1.460
2.027
1.651
2.049
1.346
7
1.083
1.496
2.044
1.674
2.119
1.649
TABLE
C
1.712
13.9
Normalized Element Values for a Single Terminated
Bessel Filter
n
Q
Lt
c,
Z*
c.
1
1.000
2
0.333
1.000
3
0.167
0.480
0.833
4
0.100
0.290
0.463
0.710
5
0.067
0.195
0.310
0.422
0.623
6
7
0.048
0.140
0.106
0.225
0.301
0.382
0.170
0.229
0.283
0.036
A.
Q
0.560
0.349
0.511
1
Topics
in filter design
401
Pentode
Interstage
Interstage
Interstage
network
network
network
FIG.
Pentodes used as isolation amplifiers.
13.26.
We sec that Ztl(s) has a real zero and a pair of poles which may be complex depending
upon
the values of R, L,
R-C interstage is shown, whose
transfer
Zn(s) Observe that
all
the
filter
and C. In Fig. 13.27A, a simple
impedance is
1
Cs +
(13.100)
1/RC
transfer functions considered
up
to this point
made up of pairs of conjugate poles and simple poles on the —a axis.
is clear that if we cascade shunt-peaked stages and R-C stages, we can
are
It
and C elements to give the desired response characteristic.
The only problem is to cancel the finite zero of the shunt-peaked stage.
For example, if we wish to design an amplifier with an n = 3 low-pass
Butterworth characteristic, we first break up the system function into
adjust the R, L,
complex pole pairs and
real pole terms, as given
Z«(s)
=
by
1
+ s + l)(s + 1)
1
s+
1
s* + s+ls + ls +
(s*
(13.101)
•
l
We
then associate the individual factors with shunt-peaked or simple
stages and solve for the element values. The n *= 3 Butterworth
amplifier is given in Fig. 13.28.
R-C
ci
ci ^*
.R
(o>
FIG.
13.27. (a)
Shunt-peaked interstage.
(b)
(f>)
R-C interstage.
Network
402
analysis
and synthesis
FIG.
Butterworth amplifier.
13.28.
MAGNITUDE AND FREQUENCY NORMALIZATION
13.9
In Section 13.8,
cutoff frequency of
we
discussed the synthesis of low-pass niters with a
rad/sec and a load impedance of
1 Q. Filters designed
with these restrictions are considered to be normalized in both cutoff
frequency and impedance level. We will now discuss methods whereby
the normalized
1
which meet arbitrary
Let us denote by a
subscript n the normalized frequency variable s n and the normalized
element values L„, jR„, and C„. The normalized frequency variable s n is
related to the actual frequency s by the relation
filters
cutoff frequency
can be converted into
and impedance
filters
level specifications.
s„
=
(13.102)
where <u the normalizing constant, is dimensionless and is often taken to
be the actual cutoff frequency.
Since the impedance of an element remains invariant under frequency
normalization, we obtain the actual element values from the normalized
values by setting the impedances in the two cases equal to each other. For
example, for an inductor, we have
,
s„L„
From
this
equation
we then
= sL =
a>o5 n
L
(13.103)
obtain the denormalized value of inductance
as
L-^
(13.104)
from the impedance l/sn C„ of a frequency normalized capacitor
we obtain the denormalized value of capacitance through the equation
Similarly,
C„,
1
sn
Cn
_
1
(13.105)
sC
"
Topics
so that the actual value of the capacitance
C=
in filter
design
403
is
—
(13.106)
«>o
Since resistances, ideally, are independent of frequency, they are unaffected by frequency normalization.
Consider,
impedance
impedance
next,
level
Z is
impedance denormalization.
=
Z
where R
Rn
,
is
Suppose the actual
should be Rq ohms instead of 1 Q. Then a denormalized
related to a normalized impedance Z n by
2?oZ„
(13.107)
taken to be dimensionless here. Thus, for a normalized resistor
the denormalized (actual) resistance
R=
RoR„
For an inductance, the corresponding
sL
is
(13.108)
relationship
= R^sL n)
so that the actual inductance value
(13.109)
is
L = R Ln
Similarly, for a capacitor
is
(13.110)
we have
so that the actual capacitance
_L
= Jk
sC
sC„
(13.111)
is
C=
^
(13.112)
R
For combined frequency and magnitude denormalization, we simply
combine the two sets of equations to give
R = RoR n
C=
R^o
(13.113)
L = ^O^n
co a
Let us consider an actual example in design. In Section 13.8, we
Zn with an n 3 Butterworth amplitude
characteristic with a cutoff frequency of 1 rad/sec and a load impedance of
1 ii. Let us redesign this filter for a cutoff frequency of 10* rad/sec to work
synthesized a transfer impedance
=
Network
404
and synthesis
anal/sis
mh
66.7
Q>
^=0.3 nf
=4=0.1 nf
500 a
FIG. 13.29. Denormalized low-pass
into a load of 500 Q.
From the
original
V2
filter.
network in Fig. 13.23, we take the
element values and denormalize with the normalizing factors,
=
and Rt
Then
<o
=
10*
500.
the denormalized element values are
R=
C,=
- 500
500*£
i
500004)
=
0.1 /*f
(13.114)
L = 1(500) = 0.0067 h
10,000
C,
I
=
ST
4
500(10
The
final design is
13.10
shown
=
0.3^f
)
in Fig. 13.29.
FREQUENCY TRANSFORMATIONS
Up to this point, we have discussed only the design of low-pass filters,
while neglecting the equally important designs of high-pass, band-pass,
We
not by
as a
introducing new design procedures but through a technique
low-pass
normalized
frequency transformation, whereby, beginning from a
Using frequency transfilter, we can generate any other form of filter.
and band-elimination
filters.
will
remedy
this situation here,
known
formations, the elements of the normalized low-pass filter are changed
into elements of a high-pass, band-pass, or band-elimination filter.
Analytically,
a frequency transformation simply changes one L-C
I^C driving-point function. Therefore,
driving-point function into another
the transformation equations
must be LrC functions themselves. Also,
low-pass filters, the transformation
equations include built-in frequency denormalization factors so that the
the
resulting networks need only be scaled for impedance level. Consider
since
we proceed from normalized
Topics
in filter design
405
simplest transformation equation, that of low-pass to high-pass, which
s
= a>
is
(13.115)
where sn represents the normalized low-pass frequency variable, s is the
regular frequency variable, and to is the cutoff frequency of the high-pass
filter. In terms of real and imaginary parts, we have
a
+ jm =
«>o
ff» +./'«>«
(13.116)
we are interested principally
we let <r„ = so that
Since
axis,
in
how
the j<o n axis
maps
into theycu
co
(13.117)
a>.
which is the equation that transforms normalized low-pass filters to
denormalized high-pass filters. From Eq. 13.117 we see that the point
wn = ±1 corresponds to the point m = ±a> It is also clear that the
transformation maps the segment \a>„\ ^ 1 on to the segments denned by
.
M
mo ^
^ °°> as shown in Fig. 13.30.
Now let us see how the frequency transformations change the network
For convenience, let us denote the normalized low-pass network
elements with a subscript n, the high-pass elements with a subscript h, the
band-pass elements with a subscript b, and the band-elimination elements
with a subscript e. For the low-pass to high-pass case, let us first consider
the changes for the capacitor C„. The transformation is given by the
elements.
ju
JVn
+j
„ plane
/<*>
* plane
<r*
-J
FIG.
13.30.
-Jox>
Low-pass to high-pass transformation.
Network
406
anal/sis and synthesis
equation
= LhS
C„s n
For the inductor
L n we
,
fo
(13.118)
C,
have
(13.119)
Chs
s
We
observe that a capacitor changes into an inductor and an inductor
changes into a capacitor in a low-pass to high-pass transformation (Fig.
13.3 1).
The element values of the high-pass filter
normalized low-pass
filter
Lh
1
=
(13.120)
<o
Cn
(o
Ln
Ch =
and
(13.121)
From
Consider the following example.
Butterworth
filter
with
pass
filter
level
of 500 Q.
given in Fig. 13.23,
cutoff frequency
its
From
the low-pass
High-pass
Low-pass
are given in terms of the
elements as
cu
let
the normalized third-order
us design a corresponding high-
=
filter,
and the impedance
we can draw by inspection the
10« rad/sec
Band-pass
Band-elimination
1*1
o-^fiflp-o
1{—
o
Ca
=
1_
«<>£»
o
Cbl
npnr- -H(BW
BW
2L»
<">0
Cd = LnBW
r„. BW
Lh woC
n
o-'lHRP-o
-
La
o-j^nnp-
Cc2
CnBW
Cb2=
FIG.
13.31.
-gyp
Element changes resulting from frequency transformations.
Topics
Ch =
1.5
x 10~ 9
in filter
407
design
f
Hf-
FIG.
13.32. Transformation of low-pass filter in Fig. 13.23 into high-pass
high-pass-filter circuit
shown
^
in Fig. 13.32.
Its
filter,
element values are:
6
10 (i)
1
(13.122)
.
son
4
10 (f)
Next, let us examine the low-pass to band-pass transformation (also
an L-C function):
S
where,
if
-
+ -°)
= BW
S(\a>
s I
<
13123 >
m ct and wp, denote the upper and lower cutoff frequencies of the
filter, BWis the bandwidth
band-pass
BW^cocz-cOd
and
<o
is
the geometric
mean of <o C2 and
m =
co cl
Vo)c2«>ci
The low-pass
to band-pass transformation
the segments
m^ ^
|a»|
^ eu^,
(13.124)
shown
maps
(13.125)
the segment
in Fig. 13.33.
\a> n \
<,
1
to
The normalized
low-pass elements are then modified according to the following equations
**.
-
°^
bs + BWs
BW
(13.126)
408
Network
anal/sis and synthesis
\jo>
«0
«C1
+j
«» Plane
s plane
<Tn
-J
-<oo
FIG.
13.33.
Low-pass to band-pass transformation.
We
note that the inductor L„ is transformed into a series-tuned tank,
shown in Fig. 13.31, whose elements are given as
=
LM
BW
BW
CM =
«>0
The capacitor C„ is transformed
whose elements are
(13.127)
Ln
into a parallel-tuned tank (Fig. 13.31),
LfiS
— BW
(13.128)
C/h9
Let us transform the third-order Butterworth low-pass
BW =6x10*
rad/sec.
We
=
draw the band-pass
(T\ isg
13.34.
in Fig. 13.23
filter
Li
FIG.
filter
with a 1-ii impedance level, whose bandwidth is
4 x 10*
rad/sec, and its band-pass is "centered" at w
into a band-pass
Band-pass
filter
shown
filter
in Fig. 13.34
by the
rules
Ct
—
nroip
1(-
d=c3
"=r-Ci
yCi
Sjij
i*
transformed from low-pass
filter
v2
in Fig. 13.23.
Topics
given above.
The element
in filter
values of the band-pass
filter
design
409
are given in the
following equations:
10
= 0.75
U « 6x
x io*m)
*
**
x KT'h
(4
Ci-
—X
X
6
10*
10-*f
12
-^— = ^xl(T
6
x
10*
6
X
4
h
9
(13.129)
-
X 10»At)
(4
^
10*
6X104
=
(4
x
10-*f
32
= o.25
x l(T*h
X lO^f)
Q = 6x
= 0.25
X KT*f
10*
through the transfor-
Finally, the band-elimination filter is obtained
mation
s.
=
BW
Is
\O)
BW and
(13.130)
eu
\
S I
manner similar to that for the band-pass
segment of the _/«>,, axis in Fig. 13.35a
the
maps
filter. The transformation
in Fig. 13.356. For the low-pass
axis
ja>
on
the
shown
onto the segments
where
a>
are defined in a
je>
Jf>n
wo
-J
-«o
-Wei
(a)
(b)
FIG. 13.35. Low-pass to band-elimination transformation.
410
Network
analysis
and synthesis
to band-elimination transformation we, therefore, have the following
element changes:
Ls
n
"
1
(s/L n BW)
A
+
ML BWs)
n
1
Ctls +
(1/L a s)
(13.131)
1
Cn s n
C nBW\a>
= Lsis
-\
°
C<gS
Observe that the normalized low-pass inductor goes into a paralleltuned circuit and the capacitor C„ goes into a series-tuned circuit, as
shown in Fig. 13.31. In Table 13.10, we have a composite summary of the
various transformations.
TABLE
13.10
Table of Various Frequency Transformations
Transformation
Equation
Low-Pass to
High-pass
Band-pass
Band-elimination
sn
=
BW
fe + ?)
Problems
13.1
figure.
Z =
VJIX for the filter shown in the
n
in order for \Z21 (ja>)\ to be maximally flat?
Find the transfer impedance
What should L be
PROB.
13.1
;
1
Topics
13.2 Find the poles of system functions with n
Butterworth characteristics. (Do not use the tables.)
Show
13.3
response
is
at
that the half-power point of a
m =
cosh
«
for e
/?*
in filter
=
»
3,
design
= 4,
41
and n
=
5
Chebyshev low-pass amplitude
1.
Determine the system function for the following
Ripple of \ db in band |o>| <> 1
(b) at co = 3, amplitude is down 30 db.
13.4
filter specifications:
(a)
Compare the slopes
13.5
(a)
/(a>*)
(6)
/(a,*)
(c)
/(«>»)
at
to
=
of the following polynomials (for n
= a>*»
- iC„(2a»* = Ln(«»8).
Determine the polynomials
13.6
1
1)
+J =
L4(co 2) and
—
3):
Q«(«.)
!»(«>*).
13.7 Expand cosh j and sinh s into power series and find the first four terms
of the continued fraction expansion of cosh */sinh s. Truncate the expansion at
« = 4 and show that H(s)
KjB^s).
13.8 Synthesize the n
ated in a l-O load.
Synthesize the low-pass
13.9
will
=
= 3 linear phase filter as a transfer impedance terminfilter,
which,
when terminated
have a transfer admittance whose poles are shown in the
PROB.
in
a
1-il resistor,
figure.
13.9.
Determine the asymptotic rate of falloff in the stop band of: (a)
filters; (b) linear phase filters.
13.11 Synthesize the n = 3 and n = 4 Butterworth responses as transfer
impedances terminated in a load of 600 SI with a cutoff frequency of 10* rad/sec.
13.12 Synthesize a Chebyshev low-pass filter to meet the following speci13.10
optimum
fications:
RL =
600
Q
(a)
load
(b)
&-db ripple within pass band
resistor,
(c) cutoff
frequency
(d) at 1.5
x
=
5
x 105
rad/sec
10* rad/sec, the magnitude
must be down 30 db.
Network
412
13.13
analysis
Synthesize n
and synthesis
= 3 Optimum and linear phase filters to meet the following
specifications:
(a) load resistor
(b) cutoff
13.14
=
10*
frequency
-
Design an n
Q
10* rad/sec.
=4
Butterworth amplifier with the following
specifi-
cations:
(a)
impedance
(ft)
cutoff frequency
13.15
level
= 500 ft
= 10* rad/sec.
Synthesize a high-pass
filter
in a 10*-O load,
whose amplitude
frequency of
=
a>
for a given transfer admittance terminated
Optimum (L), with a cutoff
characteristic is
10* rad/sec.
Synthesize: (a) a band-pass filter; (b) a band-elimination filter, with
8 x 10* and o>ci
maximally flat (n
4) amplitude response with ce
13.16
=
2 x
10*.
w-
=
chapter 14
The
14.1
INCIDENT
AND
scattering matrix
REFLECTED
POWER FLOW
we will devote our attention to certain power relationand two-port networks. The characterization of a network in
terms of power instead of the conventional voltage-current description is a
helpful analytical tool used by transmission engineers. It is especially
important in microwave transmission problems where circuits can no
longer be given in terms of lumped R, L, and C elements. In the powerflow description, we are concerned with the power into the network, which
we call the incident power and the power reflected back from the load,
which is the reflected power. A convenient description of the network in
terms of incident and reflected power is given by the scattering matrix,
which is the main topic of discussion in this chapter.
It is convenient to think of incident and reflected power when dealing
with transmission lines. Therefore, we will briefly review some concepts in
transmission line theory. For a more comprehensive treatment of transmission lines, the reader is referred to any standard text on wave propagation. 1 Consider the transmission line shown in Fig. 14.1. The voltage
at any point down the line is a function of x, the distance from the source.
The parameters which describe the transmission line are given in the
In this chapter,
ships in one-
following:
R=
G=
L=
C=
1
resistance per unit length
conductance per unit length
inductance per unit length
capacitance per unit length
See, for example, E. C. Jordan, Electromagnetic
Prentice-Hall,
EngUwood Cliffs, New
Jersey, 1950.
413
Waves and Radiating Systems,
Network
414
anal/sis
i
and synthesis
l-
h:
FIG.
14.1.
Transmission
line.
Given these parameters, we can now define the impedance per unit
length as
Z = R+joL
(14.1)
and the admittance per unit length as
Y=G+jcoC
The
characteristic
impedance
Z
of the line
is
(14.2)
given in terms of
z = Vz/r
and the propagation constant
Z and
Fas
(14.3)
is
y
=
sfZY
(14.4)
equations for
these definitions in mind, let us turn to the general
line
the
x
down
point
any
the current and voltage at
With
. Vfi->* + Vsyx
I(x) = !£*" - iy
V{x)
(14.5)
Z
Z
the
/ refer to the incident wave at point * and
14.5
Eqs.
Solving
x.
wave
at
reflected
the
to
terms with subscript r refer
and reflected
simultaneously, we obtain explicit expressions for the incident
The terms with the
subscript
waves
^«*[K(*) + Z,/(*)]
Vre*x = h[V(x)-Z
Consider the case when a transmission line of length
its characteristic impedance, that is,
V(L)
_
ZtfQ
(146)
l(x)]
L
is
terminated in
(14.7)
The
Then we
see that the reflected
wave
scattering matrix
415
is zero.
f>»L
=
(14.8)
yL
cannot be zero, we see that the coefficient VT is identically zero
As a result, the reflected wave at any point x is zero. Also,
the impedance at any point x down the line is equal to Z as seen from
Eq. 14.5 with Vr = 0. With these brief thoughts of transmission lines in
mind, let us turn our attention to the main topic of this chapter, namely,
Since e
for this case.
the scattering parameters.
I4J
THE SCATTERING PARAMETERS FOR A
ONE-PORT NETWORK
For the one-port network shown in Fig. 14.2a, consider the following
The incident parameter a is defined as
definitions.
(14.9)
and the
reflected parameter b, is defined as
(14.10)
where R^
is
an
arbitrary, positive,
dimensionless constant called the
For the transmission line described in
Section 14.1, if the characteristic impedance Z = /{<,, then we can describe
reference impedance factor.
the incident parameter in terms of the incident voltage as
a
Similarly,
Ve~T"
- ^=-
(14.11)
b can be expressed in terms of the
b
reflected
wave as
= ^=r
(14.12)
VZ„
/
+
V
•
One-port
network
(a)
a *
,
b
One-port
network
m
FIG. 142. Scattering parameters of a one-port network.
Network
416
Thus we
reflected
To
analysis
and synthesis
a and b do indeed describe an incident-
see that the parameters
wave
relationship in a one-port network, as depicted in Fig. 14.26.
give further
meaning to a and
b,
consider the power dissipated by the
one-port network
P = iRe
VI*
(14.13)
where /* denotes the complex conjugate of /. From Eqs. 14.9 and 14.10
we solve for V and / in terms of the incident and reflected parameters to
_
give
K=(a + b)V«
-
a
Then the power
dissipated
(1414)
b
by the one-port network
P = Haa* -
is
bb*)
(14.15)
-KM* -1*1*)
where, again, the asterisk denotes complex conjugate. The term Jaa*
can be interpreted as the power incident, while $bb* can be regarded as the
power reflected. The difference yields the power dissipated by the one-port
network.
The incident parameter a and the reflected parameter b are related by the
equation
b
where
S is
coefficient.
= Sa
(14.16)
more commonly, the reflection
and b we can make the following
called the scattering element or,
From
the definitions of a
substitution:
Solving for S,
where
we obtain
S
Z is the impedance of the
= Z ~ **°
Z+R
one-port network
Z=
excited
(14.19)
|
A further useful result is that when the impedance R<, is
impedance Z, the reflected parameter b — 0.
For the one-port network
(14.18)
by a voltage source
set
equal to the
V9 with a source
The
we
when we choose the
impedance Rq to be equal to
resistor R,, as depicted in Fig. 14.3,
will
show
reference
i
AAA
"VW
that
+
One-port
network
V
!>
-&
The proof
By
follows.
incident parameter a
is
(,4 20>
-
definition, the
FIG. 14.3
given as
^)
-K£+^)
From
Fig. 14.3,
we have
V,
1
and
-IRB
=
we
(14.21)
=V
(14.22)
(14.23)
Ra
Substituting these equations for
14.21,
417
scattering matrix
+Z
V and / into the expression for a in Eq.
obtain
_
_
l
VBR,
17.,
2V^
RqV,_~\
\
,
r r,+z
}
^+ zJ
(14.24)
R' + zJ
2VV
in a oneConsider once more the expression for the power dissipated
network:
port
V
(14.25)
P = l(aa* - bb*)
Factoring the term aa*
=
\a\*
from within the parentheses,
2 \
P
becomes
|a|V
(14.26)
a
=
When we
ance,
i.e.,
l^- (i-isi»)
choose the reference impedance to be equal to the source
and
R„, then S =
when R*
resist-
=
n _ nil!
~Z~ ~ on
2
ePA. ~
iR„
(14.27)
Network
418
analysis
and synthesis
**
PA
where
~»vVA
°
Av
>
^
*"
^*^
represents the available gain or
power of a voltage source Va with
a source resistance R„. For the case of a
available
I
one-port network, the available gain is
defined as the power dissipated in the oneport network when the impedance of the
network
FIG. 14.4
Z is
equal to the resistance of the
As a result of this definition, we
source/?,.
see that for the one-port
Fig. 14.4, the
power
dissipated in
Pa
Z
with
Z = Rt
network shown in
is
= ^-
The
available gain thus represents the
terminals of the voltage source.
(14.28)
maximum
available
power
at the
From this discussion, it is apparent that the value of the reference impedance Rq should be chosen equal to the source impedance R„. A
standard procedure is to assume a 1-Q source impedance and denormalize
when
necessary, that
is, let
S
where
Next,
z
us
let
=
(14.29)
—
(14.30)
some of the important properties of the
a one-port network.
briefly consider
scattering parameter
1.
- 2^-i
z + 1
S for
The magnitude of S along theyco
for a passive network, that
axis
is
always
less
or equal to unity
is,
|S(/«>)| <. 1
(14.31)
This property follows from the fact that the power dissipated in a passive
network is always greater or equal to zero. Since the power can be
expressed as
P-^fl-ISft^O
we
see that
2.
For a
\S(ja>)\*
1
(14.33)
network |S(jo>)| = 1. This property follows from the
power dissipated in a purely reactive network is zero.
reactive
fact that the
^
(14.32)
The
3. For an open circuit 5=1, and for a short
shown to be true from the equation
S
For an open
therefore
Before
5=
=
circuit z
oo,
circuit
S=
—
1.
= ^|
z + 1
so that
419
scattering matrix
This
is
(14.34)
5=1.
For a short
circuit z
= 0;
—1.
we proceed
to the next property, let us consider the following
definition.
A
DEFINITION.
bounded real function P(s)
defined by the con-
is
ditions
|^)|
(«)
(6) F(s)
In
^ AT
is real
Rea^O
for
when
s is real.
K denotes any positive real constant.
(a),
4. If z
— Z/Rq
a positive
is
real function, then
5
is
a bounded
real
function.
The proof follows from
S
From
the equation,
= (z-
l)/(z
the positive real condition (a)
+
Re z(s)
1).
^ 0, when Re s ^ 0,
we
see
that
s
so that
when Re j
|S(s)|
;> 0. (b)
Im z(s)-[l- Re z(s)]
jlmz(s)+ [I + Rez(s)]
j
+
= ( Im'^
llm , z(s) +
[l
+
Re2(S)]«/
y
'
3° f
Where s is real, z(s) is real. Then
S = (zmust be real. Thus the
bounded real function.
14.3
U-Re^ *
(14.35)
l)/(2
+
1)
scattering parameter for a passive
network
is
a
THE SCATTERING MATRIX FOR A
TWO-PORT NETWORK
In this section we will extend the concepts developed for one-port
networks discussed in Section 14.2 to two-port networks. In the two-port
network shown in Fig. 14.5, we are concerned with two sets of incident and
reflected parameters {at , *x } at the 1-1' port, and {a,, b } at the 2-2' port.
t
These parameters are defined in similar manner as for the one-port
420
Network
and synthesis
anal/sis
-o
lo—>-
-*—o2
Ol
o-
02
v2
Two-port
network
Vi
*i«
62
-o
FIG.
network, that
D2'
l'o-
14.5. Scattering
o-
parameters for a two-port network,
is,
(14.37)
"•
=
l(vfe
where
R01
and
w
i?
+vr"'')
and output
are the reference impedances at the input
ports respectively.
The
scattering parameters
Stj for
the two-port network are given by the
equations
bt
=
(14.38)
iSjitfi
+
Si2a2
In matrix form the set of equations of Eqs. 14.38 becomes
r
6i
i
=
Si1
\
^im
(14.39)
(14.40)
where the matrix
is
called the scattering matrix of the two-port network.
From
Eqs. 14.38,
we see that the scattering parameters of the two-port network can be
expressed in terms of the incident and reflected parameters as
Sn
S«
=
Si.
=h
a* 01—0
=
SM
=
a* ai—0
(14.41)
The
scattering matrix
I
I
In Eqs. 14.41, the parameter
5M
is
Su
is
called the input reflection coefficient;
the forward transmission coefficient;
and SM
coefficient;
is
421
Slt
is
the reverse transmission
the output reflection coefficient.
Observe that
all
four scattering parameters are expressed as ratios of reflected to incident
parameters.
Now let us examine the physical meaning of these scattering parameters.
First, consider the implications
of setting the incident parameters at and a t
to zero in the defining relations in Eqs. 14.41. Let us see what the conimplies in the definition for the forward reflection coefficient
dition a4
=
«i
tlt^O
Figure 14.6 shows the terminating section of the two-port network of Fig.
14.5 with the parameters a, and b a of the 2-2' port shown. If we treat
the load resistor
Rt as a one-port network with scattering parameter
S8
where Rgx
is
=
(14.42)
R»
+
Ro»
the reference impedance of port 2, then oi
and b% are
related
by*
aa
= SA
When the reference impedance R^
St becomes
is set
(14.43)
equal to the load impedance Rt,
then
S,=
Rot
Rat
— Ri*8 =
+ Rto*
(14.44)
«
under this condition. Similarly, we can show that, when
so that at
ax
0, the reference impedance of port 1 is equal to the terminating
—
*
From
reflected
the viewpoint of the load resistor /?„ the incident parameter
parameter is a x .
is
b, and the
Network
422
impedance
and synthesis
anal/sis
R01 =
We see
as a result of this discussion that the
merely imply that the reference impedances
are chosen to be equal to the terminating resistors Rt and R t ,
(i.e.,
conditions a x
R01 and /?M
=
/?i).
and aa
=
respectively.
Next,
let
us consider the relationship between the driving-point imped-
ances at ports
1
and 2 and the
reflection coefficients
denote the driving-point impedances at ports
.
Let us
(14.45)
-M
a
S
the equation
5U and Su
and 2 as
h
h
From
1
(14.46)
l lai-0
We can write
S„
=
*K W*oi)
which reduces
Similarly,
easily to
Sn
=
we have
These expressions
tell
us
if
— Rqi
Z + Rqx
z» — Rq»
Z + R oi
+
(14.47)
Rnli)
2q
(14.48)
x
Rt-Rot
t
h,_h„
we choose
(14.49)
the reference impedance at a given
port to equal the driving-point impedance at that port, the reflection
coefficient
of that port will be zero, provided the other port
is
terminated
in its reference impedance.
Next, let us derive some physically meaningful expressions for the
forward and reverse transmission coefficients Sn and Slt Consider the
.
definition for <S81
S
•321
-*l
——
(14.50)
a«—
implies that the reference imseen, the condition a, =
pedance Roi is set equal to the load impedance Rt , as seen in Fig. 14.7.
If we connect a voltage source Vgl with source impedance Rn = Rlt
As we have just
JtogaJfa
FIG. 14.7
)
The
then
we can
express
ax as
< 14- 5I >
""vfe
Since a,
423
scattering matrix
= 0, we have the equation
y
—^
=
from which we obtain
bt
Consequently,
—>/*»'»
(14.53)
= "(-/= - V*»z«
(14.54)
v*«
Finally,
we can
express the forward transmission coefficient as
_
s
vjj**_
(14.55)
Krl^ ^* Ri-fi.i,Ri-Boi
In similar fashion, we find that when port 1 is terminated in Rqi = R t and
when a voltage source V,t with source impedance Rt is connected to port 2,
then
S
"-
?MF
(1456)
We see that both Slt and SM have the dimensions of a voltage-ratio transfer
function. Indeed, if
It is
R^ =
R^, then
Slt and Stl are simple voltage ratios.
Sn = Slt
seen that for a passive reciprocal network,
.
Now let us consider as an example the scattering matrix of the n: 1
ratio
Recall that for an ideal transformer
ideal transformer in Fig. 14.8a.
'Vi-jtV*
A---/,
n
(14.57)
»
Assuming first that Jin'— Rn
1, let us find 5U by terminating the
2-2' port in a l-Cl resistor, as shown in Fig. 14.86. Then
-^-
-
1
(14.58)
Network
424
anal/sis
and synthesis
1
h
_Jf_
1»:1|
+
—2o
+
V2
Ideal transformer
(a)
FIG.
14.8.
Determination of scattering parameters for an ideal transformer,
From Eq.
14.48,
we have
Next we terminate the
as
we
^ = n*
Z1 =
so that
did for
Su
Sn
(14.59)
= n*-l
n* + l
1-1' port in
a 1-G
(14.60)
resistor (Fi£. 14.8c).
We obtain,
,
SM
= (1/w*)
(!/«*)
-1_1+1 1+
n*
n
1
(14.61)
The
425
scattering matrix
We obtain Sn by connecting a voltage source Vgl with a source impedance
= Q at the 1-1' port and terminating the 2-2' port with a resistance
- «*» the equivalent circuit
jRo, = 1 ii, as seen in Fig. 14.8rf. Since V^h
jRo!
1
of the ideal transformer as seen from the voltage source as a 1-Ii impedance in series with an «*-ohm resistance (Fig. 14.8c). Then Vx can be
expressed in terms of Vtl , as
K,=
Since
Va =
(14.62)
+
n8
l
Vjn, we have
Kin
Since
R^ = R^ =
1
Q,
(14.63)
+
n*
l
SM is
2F,
2it
n*
r »i
+
(14.64)
l
We can show in similar fashion that
2n
+
n
(14.65)
1
Therefore the scattering matrix for the ideal transformer
v-1
n*
+
l
As a second example,
let
transmission line of length
shown in
Fig. 14.9. If we
L
+
given as
In
n*
+
l
(14.66)
l-n»
In
M*
is
l
fl*
+l
us find the scattering matrix for a lossless
its characteristic impedance, as
terminated in
assume that
R^ =
!*«
= Z* then the reflection
coefficients are
c
= ?oZlZo _ Q
z, + z
(14.67)
x^-°
Zo
1'
FIG,
14.9. Lossless
transmission line.
V»
Network
426
This result
anal/sis and synthesis
not implausible, because a transmission line terminated in
impedance has zero reflected energy. To determine Su ,
we terminate the line in Z at both ends and connect at the 1-1' port a
voltage source Vtl , as depicted in Fig. 14.9. Since the transmission line
has zero reflected energy, that is,
its
is
characteristic
then
From
the equation
Z> x
=
v,
= *>- ri
a2
=
(14.68)
~ 2 \*J vgl
Sn - 2e-*
(14.69)
Stl
we obtain
we
In similar fashion,
(14.70)
find that
Sit
-
2e- rL
(14.71)
Therefore the scattering matrix for the lossless transmission line
r
~~
14.4
is
2«-" z, i
o
(14.72)
VW-
J
PROPERTIES OF THE SCATTERING MATRIX
Having defined the
Section 14.3,
matrix.
let
From
scattering
matrix of a two-port network
in
us consider some important properties of the scattering
the general restriction for a passive network that the net
power delivered to
P=
all
we
ports must be positive,
H«i«i*
+
a*fh*
-
W
-
obtain the condition
W) £
(14.73)
Equation 14.73 follows from the fact that the power delivered to the 1-1'
port
is
Pi
-
KW -
and the power delivered to the 2-2' port
P,
The
total
power
-
i(^»*
-
delivered to the network
(14.74)
bib,*)
is
W)
is
(14.75)
then
P = P1 + Pt
which is exactly the expression in Eq.
delivered to the network is
(14.76)
14.73. In matrix notation, the
P - Hla'fW ~
lb*flb]}
£
power
(14.77)
The
where
T denotes the transpose operation,
lb]
=
[S][a],
then
[b*]
Equation 14.77 can
now
IP
=
-
T
=
[a*]
427
and
-a
-a
[a]
Since [b]
scattering matrix
(14.78)
T [S*] T
(14.79)
be rewritten as
{[a*]
T [a]
T
la*] llu]
- [a*] T [S*] T [S][a]}
- [S*] T [Smal >
This then implies that the determinant of the matrix
must be greater or equal to zero, that is,
Dct
-
[[«]
T
(14.80)
[[«]
—
[5*] T [S]]
£
[S*] [S]]
(14.81)
Consider the special but, nevertheless, important case of a lossless
P 0, so that
network. In this case
=
[S*] T [S]
=
[u]
(14.82)
A matrix satisfying the condition in Eq. (14.82) is unitary.
For a
lossless
two-port network
From
this equation,
we have
the following conditions for the scattering
matrix
Sj»*Su
+ Stl *Stl =
+ Sn'Su =
Sn *Si,
+ Sn *S„ =
Su*Su
+
Su*S11
Note that Eqs.
network
14.85
is reciprocal,
and
then
S„*S„
-
1
(14.84)
(14.85)
(14.86)
(14.87)
1
14.86 are conjugates of each other.
If the
Stl = Slt and
+
IVHI +
|Sii(/«>)|»
1
=
|5„(/«))|« -
|S„(/«>)|»
1
(14.88)
1
428
Network
analysis and synthesis
Two-port
network
A
<*2
FIG. 14.10
from which
it
\SJ(ja>)\ <. 1
(i.e.,
=
=
follows that for a lossless reciprocal network |Su (ya>)|
1. Also it is clear that when \Sjj<o)\
and \Su(ja>)\
^
=
there is a zero of transmission), then \Su (ja>)\
all the power that has been delivered to the
when
condition states that
1.
This
network
back to port 1-1'.
At this point, it might be profitable to discuss why we use scattering
matrices. What are the advantages of the scattering description over
conventional descriptions? Let us discuss three major reasons for the
from port
1-1'
is
reflected
scattering formalism.
networks do not possess an impedance or admittance matrix.
ideal transformer has no Z or Y matrix because its
elements are not finite. However, as we have seen, the ideal transformer
8
can be described by a scattering matrix. Carlin states that all passive
1.
Many
For example, an
networks possess scattering matrices.
2. At high frequencies, incident and reflected parameters play dominant
roles in problems of transmission, while voltage-current descriptions are
relegated to the background. Then the scattering matrix is necessarily the
more powerful description of the system. Note that the voltage standingwave ratio (VSWR) is given in terms of a reflection coefficient S as
VSWR =
i±M
(14.89)
In networks where power flow is a prime consideration (e.g., filters),
the scattering matrix is very useful. For example, in the network given in
Fig. 14.10, ifPA represents the available power from the generator and Pt
is the power dissipated in the load R t , then we can.show that the magnitude3.
squared forward transmission coefficient
is
is«o)i*-£*
pA
We will discuss this point in more detail in
< 14 - 90>
Section 14.5.
* H. J. Carlin, "The Scattering Matrix in Network Theory," Trans. IRE, CT-3,
No. 2, June 1956, 88-96; see his extensive bibliography.
The
429
scattering matrix
INSERTION LOSS
14.5
we described the forward and reverse transmission
terms of voltage ratios. Perhaps a more appropriate
In Section 14.4,
coefficients in
description of a transmission coefficient
is in terms of a power ratio rather
than a voltage ratio. In this section we will show that |SM(/«o)|* and ]SU
(J(o)\* can be expressed in terms of power ratios. We will then introduce
the very important concept of insertion loss and finally relate |£ti(/<»)l* to
the insertion power ratio.
Consider the equation for
Stl
in the two-port
network shown in Fig.
14.5,
From
we
the equation
obtain
l-S^OOl*
= Sn(/w)Sa *(/a>)
|M-)|-
=
(14.92)
l^>»
I^iO«»)IV8*x
(14.93)
Pal
where Pt is the power dissipated by the load Rt, and
gain of the generator Vgl Similarly, we have
PAl is
the available
.
|Si.O)l*
We
see that both \Su (j(o)\*
relate the
power dissipated
and
- £-
ISutyco)!* are
(14.94)
power
transfer ratios
at a given port to the available
power
which
in the
other port.
Now
us examine the idea of insertion
loss. Consider the network
Between the terminals 1-1' and 2-2' we will insert a
two-port network, as shown in Fig. 14.116. Let us denote by VM the
voltage across the load resistor R, before inserting the two-port network,
and by Vt the voltage across Rt after inserting the two-port network. A
measure of the effect of inserting the two-port network is given by the
insertion voltage ratio IVR, which is denned as
let
shown in
Fig. 14.1 la.
IVR
4
^V
(14.95)
t
is
Another method of gaging the effect of inserting the two-port network
power dissipated at the load before and after inserting the
to measure the
430
Network
analysis
and synthesis
t
-J
vW^-
r
(a)
Hi
2
-i_
+
r
Inserted
)y»i
$
*2* >V2
two-port
network
?
r
lb)
FIG. 14.11
two-port network. If PM is the power dissipated at the load before the
two-port network is inserted, and if Pt is the power dissipated after insertion, then the insertion power ratio of the two-port network is denned as
e
If
we
u
_ Ew
take the logarithm of both sides,
a
where a
is
(14.96)
we
=10 log
obtain
^
(14.97)
the insertion loss of the two-port network. In terms of the
we can calculate Px from the relation
circuit given in Fig. 14.11,
Then
Pm is
V„
=
P»
=
V«
|K„|
dissipated
(14.98)
8
2R t
*.
2(R t
The power
R>
by the load
ir.il'
+
(14.99)
RJ*
after inserting the two-port
network
is
given by
P.= Hi!!
2R t
(14.100)
The
The
insertion
power
ratio
2a
_
when
431
can then be expressed as
£_80
v
_ w„r
P.-VXOk +
In the special case
scattering matrix
V
<14101)
the source and load impedances are equal, that
is,
Rx = R2 = Rn = R^
(14.102)
the reciprocal of the squared magnitude of the forward transmission
coefficient in Eq. 14.93 is equal to the insertion power ratio
Pgo
When Rt
j&
Rt
|S*iO)|
,
then
PjQ
4R1 R 2
(*i
In any event,
i\Sa (jai)\ t
power
we
(14.103)
a
+
1
2
R*) |s*iO)l
(14.104)
a
see that the magnitude-squared transmission coefficients
and {S^jai)]* can be regarded physically as equivalent
insertion
In Section 14.6, we will use this relationship in the synthesis
of double-terminated filter networks.
14.6
ratios.
DARLINGTON'S INSERTION LOSS FILTER SYNTHESIS
In this section,
we will consider a filter synthesis procedure first proposed
classic paper in 1939. 4 We will use scattering matrix
notation to describe the essence of Darlington's original work. Our
coverage will be restricted to the class of low-pass filters which are termi-
by Darlington in a
=
=
nated in equal source and load impedances, Rn
Roi R , as shown in
For normalizing purposes we will let /^ be equal to 1 12.
Fig. 14.12.
FIG. 14.12
4
S. Darlington, "Synthesis
Loss Characteristics,"
/.
of Reactance 4-Poles which Produce Prescribed Insertion
Math. Phys., 18, 1939, 257-353.
-
Network
432
analysis and synthesis
Recall that when the source and load
impedances are equal, then the insertion
power ratio is equal to the reciprocal of
|SM (yG>)| 2 , that
is,
1
20
(14.105)
|S21 (>>)| a
P*
Expressed as a loss function, the insertion
power
ratio is
A =10 log %*
P
=
-101og|S 21(»| a
db
(14.106)
In circuit design, the specification of an
A (Fig. 14.13a) is equivalent
insertion loss
to the specification
squared transmission coefficient shown in Fig.
of the
14.13ft.
amplitude-
One of the most
ingenious techniques given in Darlington's synthesis procedure is the reduction of insertion loss synthesis to an equivalent L-C driving-point
synthesis problem. This technique can be developed in terms of scattering
parameters. Our initial specification is in terms of |Su (/<u)|. For an L-C
two-port network
\Sn(jcoW
Next,
Sn(s)
is
l
the equation
Su
=1-
8
\S ia (Jo>)\
= ————
+
^i
we
|S*iO)l
(14.107)
obtained from the magnitude-squared function
Su (s)Sii(- s)
Then from
=1-
2
\
im ^,
(14.108)
(14.109)
«o
obtain the driving-point impedance
Z^s)
1
=R
1
+ Su(s)
- Su (s)
(14.110)
We then synthesize the network from Z^s).
discussion here to low-pass filters given by the
our
We
lossless ladder structure terminated at both ports by 1-Q resistors in
Butterworth or
Fig. 14.14. These low-pass filters can take the form of a
shown
in Fig. 14.12.
will restrict
Chebyshev
specification for |Ssl (/a))|
|S21
2
,
that
=
0)f _
1 +
is,
2
(14.111)
,2n
CO
-
The
10
scattering matrix
433
Li
/-
~*np
|
—
T-nnnp
+
c2 z
T-
C-1-—
FIG.
14.14.
Canonical form for double-terminated low-pass
+
1
where
10.
Zi(s)
niters.
C B8(<u)
c
v2
(14.112)
Cn(co) represents an nth-order Chebyshev polynomial.
Example
14.1.
Let us synthesize a low-pass
filter
for the specification
1
._.
"*W-TT^
(14.113)
which represents a third-order Butterworth amplitude characteristic. The load
and source impedances are R^ = R^ = 1 Q. First we find |5u(y'o>)| s as
IWI* = !-—-*
1
Lettingy'eo
= s in
|Su
(/a>)| a
,
we
(14.114)
+0)6
obtain
<Sii(»>S_(-*)
*•
= 1
-s*
(14.115)
which factors into
Sn(s) Sil(- S) sothat5u(5)is
Next,
8
A-*
)
"
g +2> +
^ + ^ 1^ +2j2 _ ^
_______
(14.116)
a*
sll(f) =
(14 117)
.
Zx(j) is obtained from the equation
Zfy)
=
1
1
+ Su(s)
- SuM
2s*
+2s* +25
2s?
+
2s
+
+
1
(14.118)
1
Network
434
analysis and synthesis
FIG. 14.15
We next perform a Cauer ladder expansion for Z^s).
2s*
+ 2* + l)^ + 2s» + 2s +
l(s
2s*+2s*+s
s
+
1)2**
2s*
+2s +
+
l{2s
Zs
1)5
The low-pass
filter is
An equivalent
we
realization for the double-terminated filter is obtained if
Su(5)
= YM - G
G = 1 mho
1
1
14.2,
and
14.3
(14.119)
Yt(s) + G
Y1(s) =
The canonical
1(5
thus synthesized in the structure shown in Fig. 14.15.
use the equation
Then, assuming
+
+ Sn (s)
- Su(s)
realization for Yx(s) is
list
shown
element values (up to n
(14.120)
in Fig. 14.16.
=
Tables 14.1,
7) for double-terminated
Butterworth, Chebyshev (1-db ripple) and Bessel filters, respectively.
These apply to the canonical realization for Yt(s) given in Fig. 14.16. If
a ZjCs) realization in Fig. 14.14 is desired, we simply replace all shunt
capacitors by series inductors
FIG. 14.16.
and
vice versa.
Canonical form for
filters
in Tables 14.1, 14.2,
and
14.3.
The
TABLE
scattering matrix
435
14.1
Normalized Element Values for a Double-Terminated
Butterworth Filter (Equal Terminations)
n
Ci
1
2.000
Lt
C3
Lf
Lg
Cj
C?
2
1.414
1.414
3
1.000
2.000
4
0.765
1.848
1.848
5
0.618
1.618
2.000
1.618
0.618
6
0.S18
1.414
1.932
1.932
1.414
0.518
7
0.44S
1.248
1.802
2.000
1.802
1.248
1.000
0.765
TABLE
0.445
14.2
Normalized Element Values for a Double-Terminated
Cheb/shev Filter with -decibel Ripple (Equal Terminations)
I
71
Cx
L*
C8
2.024
c5
L*
1
1.018
3
2.024
0.994
5
2.135
1.091
3.001
1.091
2.135
7
2.167
1.112
3.094
1.174
3.094
TABLE
J-6
c7
1.112
2.167
14.3
Normalized Element Values for a Double-Terminated
Bessel Filter (Equal Terminations)
C3
La
Cj
C3
Lt
Le
Cj
1
2.000
2
1.577
0.423
3
1.255
0.553
0.192
4
1.060
0.512
0.318
0.110
5
0.930
0.458
0.331
0.209
6
7
0.838
0.412
0.316
0.236
0.148
0.051
0.768
0.374
0.294
0.238
0.178
0.110
Note that the even orders
are not given. This
is
0.072
for the double-terminated
0.038
Chebyshev niters
filters do not
because the even-ordered Chebyshev
meet realizability conditions for minimum insertion loss at s = 0. 5 We
have only given tables for equal source and load terminations. For other
possible realizations, the reader should consult L. Weinberg's excellent
book. 8
£
L. Weinberg, Network Analysis
New York,
'Ibid.,
1962, p. 589.
Chapter
13.
and
Synthesis,
McGraw-Hill Book Company,
Network
436
anal/sis
and synthesis
Problems
14.1 Determine the reflection coefficient
in the figure.
S for
the one-port networks
shown
?o
°-VW—r—vW—
4=tt'
w
For the one-port_network in Fig. 14.3, let iJ, . R
If the incident
B
is a = VJl^R^ find the reflected parameter
A.
14.3 For the network in Prob. 14.1, determine
\S(jio)\.
Show that the
scattering elements S for the networks in Prob. 14.1 are
bounded real functions.
14.4 For each of the networks shown, find the
scattering matrix for ik, =
14.2
.
parameter
-o2
(a)
(b)
X
+
Vi
—/2-*-
^-J-?-
Gyrator
/i
A
V
L
Negative
impedance
converter
\
*!,
+
^ W*
*1
2
r
Vi
o—
w
PROB.
14.4
1.
z
T
w
+
v2
—
The
scattering matrix
437
14J5 Find the insertion voltage ratio and insertion power ratios for each of
the networks shown. These networks are to be inserted between a source
impedance R„ = 2 ft and a load impedance RL = 1 ft. From the insertion
power
ratio, find |S»i(j«>)|*.
2h_
i
2_h
o-'TRRTH-^WnP-02
Jsjf
io
—
_lh_
.
nptpP 1
=f=if
=r=i'
l'o-
(6)
2h
40
:*f
PROB.
14.6
Synthesize low-pass niters for the specifications
(a)
|SaQ'«>)l*
(«
l-W)!
=
1
+0)*
1
+afl
1
2
Synthesize an equal-ripple low-pass filter such that 20 log |5n(/a>)| has
most i-db ripple in the pass band and an asymptotic falloff of 12 db/octave in
14.7
at
14.5
the stop band.
chapter 15
Computer techniques
in
circuit analysis
15.1
THE USES OF DIGITAL COMPUTERS
IN
CIRCUIT ANALYSIS
The advent of the high-speed computer has made
routine many of the
computational aspects of circuit theory.
Digital computers have become widely used in circuit analysis, time and
frequency-domain analysis, circuit (filter) design, and optimization or
formerly tedious and
We
iterative design.
difficult
will discuss these aspects in general in this section.
In succeeding sections,
we
will
discuss
some
specific
circuit-analysis
computer programs.
Circuit analysis
The primary objective of a linear circuit-analysis program is to obtain
responses to prescribed excitation signals. These programs are based on
many different methods: nodal analysis, mesh analysis, topological
formulas, and state variables. Most of them can handle active elements
such as transistors and diodes by means of equivalent circuit models.
The
state-variable
programs based upon Bashkow's
tion 1 perform their calculations directly in the time
A
matrix formula-
domain
via numerical
and matrix inversion. The outputs of these programs provide
impulse and step response in tabular form. If the excitation signal were
given in data form, the state-variable programs would calculate the
response directly in the time domain.
integration
1
T. R. Bashkow,
Theory, CT-4, No.
"The A Matrix—New Network Description," IRE Trans, on Circuit
3,
September 1957, 117-119.
438
Computer techniques
The majority of
circuit-analysis
in circuit anal/sis
439
programs, however, perform their
calculations in the frequency domain. The program user is only required
to specify the topology of the network, the element values, and what
transfer functions he wishes to obtain. The computer does the rest. It
calculates the specified transfer functions in polynomial form, calculates
the poles and zeros of these functions, and can also provide transient
response and steady-state response, if desired. With versatile inputoutput equipment, the output can also provide a schematic of the original
network, as well as plots of time- and frequency-response characteristics.
Time- and frequency-domain analysis
The time- and frequency-domain analysis programs can be used in
conjunction with the circuit-analysis programs or independently. The
time-domain programs depend upon solving the convolution integral
FIG. 15.1. Frequency response of fifth-order Butterworth filter.
FIG. 15.2. Phase response of fifth-order Butterworth filter.
440
FIG.
Network
15.3.
analysis
and synthesis
Impulse response of fifth-order Butterworth
filter
evaluation by Laplace
transform.
This approach obviates the necessity of finding roots of
It has the advantage that the excitation signal
need not be specified analytically, but merely in numerical form.
The frequency-domain programs usually consist of finding transient
and steady-state responses, given the transfer function in factored or
numerically.
high-order polynomials.
The program user must specify the numerator and
denominator polynomials of the transfer functions, the types of transient
response he wishes (i.e., impulse or step response), and the types of
steady-state responses he wishes (amplitude, amplitude in decibels, phase,
delay, etc.). In addition, he must specify the frequency and time data
points at which the calculations are to be performed. This may be done
in two ways. If he requires evenly spaced data, he need only specify the
minimum point, the increment, and the number of points. If he wishes
unfactored form.
to obtain data at certain points, he
must supply the
list
of data points at
the input.
Examples of outputs of a steady-state and transient analysis computer
program are shown in Figs. 15.1, 15.2, and 15.3. In Fig. 15.1, the magnitude of a fifth-order Butterworth filter is plotted via a microfilm plotting
subroutine. Figure 15.2 shows the phase of the filter, while Fig. 15.3
shows its impulse response.
Computer techniques
In Section 15.2
we
examine further
will
in circuit analysis
details
441
of a typical steady-state
analysis program.
Circuit
The
(filter)
filter
design
design programs are probably the most convincing argument
for the use of computers in circuit design. Designing insertion loss filters"
to meet certain amplitude requirements requires considerable numerical
calculation even in the simplest cases.
insertion loss
filter
design
is clearly
a
The use of
digital
logical alternative.
computers in
The amount of
programming time for a general filter synthesis program is considerable.
However, the ends certainly justify the means when large numbers of
filters must be designed to meet different specifications.
An outstanding example of a digital computer program for filter design
The
is the one written by Dr. George Szentirmai and his associates.*
complete in that it handles the approximation as well as the
synthesis problem. It is capable of dealing with low-, high-, and band-pass
poles
filters with prescribed zeros of transmission (also called attenuation
maximally
or
equal-ripple
either
for
or loss peaks). There are provisions
flat-type pass-band behavior, for arbitrary ratios of load to source im-
program
is
pedances, and for predistortion and incidental dissipation.
In addition, Dr. Szentirmai has a modified program that synthesizes
low- and band-pass filters with maximally flat or equal-ripple-type delay
in their pass band, and monotonic or equal-ripple-type loss in the stop
bands.
In the specifications, the designer could specify both the zeros of
4
If
transmission (loss peaks) and the network configuration desired.
his specifications include neither, the computer is free to pick both configuration and zeros of transmission. The computer's choice is one in
which the inductance values are kept at a minimum. The program was
written so that the same network could be synthesized from both ends.
Finally, the computer prints out the network configuration, its dual,
the normalized element values, and the denormalized ones. It also
provides information such as amplitude and phase response, as well as
plots of these responses obtained from a microfilm printer.
Figures 15.4, 15.5, 15.6, 15.7, 15.8, and 15.9 show the results of a
band-pass
»
filter
synthesis using Dr. Szentirmai's program.
R. Saal and E. Ulbrkh,
Circuit Theory, CT-5,
"On
December
the Design of Filters
by
Figure 15.4
Synthesis," Trans.
IRE on
1958, 287-327.
Digital Computer Program Package for
of the First Allerton Conference on Circuit and System
Theory, November 1963, University of Illinois.
•
G. Szentirmai, "Theoretical Basis of a
Filter Synthesis," Proceedings
4
Saal and Ulbrich, op.
cit.
Network
442
analysis
and synthesis
BAND PASS FILTER SYNTHESIS
CASE NUMBER
3.1
OEGREE OF FILTER
MULTIPLICITY OF PEAK AT ZERO
MULTIPLICITY 0F PEAK AT INFINITY
NUMBER OF FINITE PEAKS BELBM THE BANO
NUMBER #F FINITE PEAKS ABBVE THE BANO
EQUAL RIPPLE PASS BANO REQUESTED
PASS BANO RIPPLE MAGNITUDE
L0HER PASS BANO E06E FREQUENCY
MID-BAND FREQUENCY
UPPER PASS BANO EDGE FREQUENCY
13
3
2
1
3
0.09000
DB.
1.0000000E+04 CPS.
1.3416408E*04 CPS.
i.80oooooe*04 cps.
"ARBITRARY" STOP BANO REQUESTED
NUMBER 0F SPECIAL POLES «
l.OOOOOOOE+00
MA
1
TERMINATIONS
INPUT TERMINATION
OUTPUT TERMINATION
6.0000000E«02 OHMS
0.
OHMS
LOWER FINITE STOP BANO PEAKS
FREQUENCY CPS
NORMALIZED
6.2000000E+03
4.621 2071E-01
VALUE OF M
0.6229266
UPPER FINITE STOP BANO PEAKS
FREQUENCY CPS
NORMALIZED
2.0700000E+O4
1.5428869E+00
2.31000006*04
l.7217723E*00
3.5200000E+04
2.6236531E+00
VALUE OF H
2.3788111
1.9296560
1.4968796
COMPUTER HILL SPECIFY CONFIGURATION
LAST INDUCTOR IS A SERIES BRANCH
FIG. 15.4
gives the specification of the problem.
The pass-band magnitude
is
equal ripple with ripple magnitude of 0.05 db, and the degree of the
is
As we
to be 13.
to be
filter
indicated in Chapter 14, odd-degree, equal-ripple
are nonrealizable. In this example the designer utilized an ingenious
filters
—
—
an extra pole to accomplish the synthesis. The program logic
then provided two extra zeros: one to cancel the extra pole and the other
to provide the odd degree. The extra pole is called a special pole in Fig.
—1.0.
15.4, and is located at s
Further specifications call for the lower band-edge frequency to be 10*
device
=
cycles;
the upper, 1.8
Vl.8 X
10*
=
X
1.3416408
zeros of transmission
10* cycles;
X
at/=
10* cycles.
(three),
and the midband frequency to be
In the stop band, there are to be
/=
oo (two),
and four
finite
zeros
of transmission: one below the pass band and three above the pass band.
The positions of these finite zeros of transmission are chosen by the
designer as indicated by the notation "arbitrary" stop band requested.
Computer techniques
CASE NUMBER
FMUtD
1.1
OtMEE OF FILTER
• 13
MM
EQUAL RIPPLE PASS SAND
LONER -PASS-OAM EOSE FREQUENCY
UPPER PASS-BANO EOCE FREQUENCY
MID-BAND FREQUENCY
1
2
0.0500
•ARBITRARY' STEP
DB
cps
• 1.3416408E*04 CPS
• i.aooooooE»04
100
300
NORMAL IZEO
UNNORHALIZEO
1.0000000E*00
6.0000000E*02
3.0T04337E-01
6.0706102E-09
S.9892B16E-03
3.85TA228E-09
2.9723019E»00
C
9. 089744 OE-08
1.0083380E»00
4.44298O6E»00
..L.e..:
7.1783955E-03
9.1797212E-08
TERMINATION
803
PEAK FREQUENCY
3.5200000E+04
901
PEAK FREQUENCY
4.2000000E»03
801
2.6982703E-08
1.364T491E«00
••••
L C
9.3749272E-01
7.S1321S3E-01
7.9423894E-01
C
1.44M444E*00
i""c
2.T112T46E-01
t.2441S6*E»00
3.S2A810TE-03
1.9447656E-0B
PEAK FREQUENCY
2.0700000E»04
1.4931737E-08
100
2.8S46S14E-08
802
1.429T93TE-03
2.499S449E-08
PEAK FREQUENCY
2. 310000 0E»04
1.0127M7E»01
2.0022669E-07
400
9.97702ME-02
•.0165648E-04
200
100
l.A3787S9E»Q0
c
3.2382 800E-08
4.2943194E-01
L
4.4B148I6E-03
300
TERMINATION
SHORT
SNORT
(AND)
- i.ooooooos»o* cps
CONFIGURATION SPECIFIED QV COMPUTER
•03 901 *01 100 002 *00 200
L*"e
443
REALIZATION FROM A SDMI CIRCUIT ADMITTANCE
•
MULTIPLICITY OF PEAK AT
MULTIPLICITY OF MAX AT INFINITY -
T.8921337E-01
1.8499769E-0I
in circuit analysis
FIG. 15.5
The terminations
filter
are:
input
=
60 Q, output
= 0Q
terminates in a short circuit. In this example, the
which means the
filter
configuration
inductance configuration. 8
chosen by the computer, and is a minimum
In Fig. 15.4 there are, in addition, listings of the finite zeros of transmission (loss peaks), which the designer specified.
Figure 15.5 is a printout of the configuration of the filter as shown by
the dotted lines flanked by the associated element values, both normalized
Since there are four
(left column) and unnormalized (right column).
finite zeros of transmission, there must be associated four L-C tank circuits.
is
S
W.
Saraga,
"Minimum
30, July 1953, 163-175.
Inductance or Capacitance Filters," Wireless Engineer,
Network
444
FREQUENCY
IN CVCt.ES
16000
16250
16300
16730
17000
17250
17500
17750
10000
10250
16500
16750
19000
19250
19500
19750
20000
20250
20500
20750
21000
21250
21500
21750
22000
22250
22500
22750
25000
25250
anal/sis
and synthesis
VOLTAGE RA
TIO IN 08
IN OEGREES
5.716532
5.704379
5.707644
5.729113
5.750516
5.743140
5.708979
5.720983
5.702947
4.131556
-1.119811
-7.753694
-14.148566
-20.210264
-26.118735
-32.100833
-38.473164
-45.657895
-56.325030
-71.701988
-59.444189
-57.483949
-57.567501
-58.703992
-60.546362
-63.101161
-66.602572
-71.867460
-63.403713
-80.597817
145.549717
128.958686
111.534102
92.934806
72.763652
50.659800
26.219204
356.194819
322.320795
272.913465
225.689589
195.922988
177.085862
163.748654
153.507803
145.227029
138.296917
132.354567
127.166241
302.572268
298.458582
294.740796
291.334706
288.250341
285.388013
282.735723
280.267304
277.961066
275.798866
93.765362
PHASE
NAG
REAL I IN
IN 8HNS
2 IN
IN 8HNS
134.691
130.664
125.883
119.931
112.687
104.489
95.717
REAL V IN
IN
56.981
61.193
64.762
67.928
71.154
74.696
77.919
79.138
78.767
87.140
111.450
139.624
164.004
184.583
202.605
218.952
234.147
248.501
262.211
275.407
286.178
300.592
312.696
324.536
336.138
347.527
358.726
369.753
380.621
391.345
65. 165
67.626
40.032
16.386
3.462
1.725
.541
.167
.049
.013
.002
.000
.000
.000
.000
.000
.000
.000
.000
.000
.000
.000
.000
NILNMS
6.2973
6.2766
6.2814
6.3129
6.3445
6.3337
6.2835
6.3011
6.2748
4.3533
1.2913
-2797
.0641
.0158
.0040
.0010
.0002
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
INA6 V IN
IN NILNHOS
-2.6641
-2.9395
-3.2315
-3.5756
-4.0061
-4.5277
-5.1151
-5.8552
-7.3083
-9.4759
-6.7826
-7.1517
-6.0967
-5.4176
-4.9357
-4.5672
-4.2708
-4.0241
-3.8137
-3.6310
-3.4701
-3.3268
-3.1980
-3.0813
-2.9750
-2.8775
-2.7876
-2.7045
-2.6273
-2.5553
FIG. 15.6
20
(
1
-20
/
-40
1
I
1
J
-60
-80
f\
1
/
/
'
-100
-120
0.000
0.400
Add 1.00000£
0.800
+
1.200
1.600
2.000
03 to abscissa
Frequency, cycles
2.400
x
FIG. 15.7
10*
2.800
3.200
3.600
4.000
Computer techniques
445
in circuit analysis
7
--
6
5
7
4
X
JE
;
J
3
2
1
I
-1
0.000
0.400
Add 1.00000S
0.800
+
1.200
2.000
1.600
03 to abscissa
Fl6queflcy
^
2.800
2.400
3.200
3.600
4.000
x „,
FIG. 15.8
0.800
0.600
0.400
r
s
0.200
J
0.000
(*"
-0.200
|
/
-0.400
i
--
/
i
f
\
-0.600
'*
\
1,
-0.800
u¥
-1.000
0.000
0.400
Add 1.00000E
0.800
1.200
+ 03 to abscissa
1.600
2.000
}nmm
^
FIG. 15.9
2.400
x
10
4
2.800
3.200
3.600
4.000
Network
446
The peak
anal/sis
and synthesis
the
column entitled "termination,"
on the same line two columns to
frequencies are listed under the
and the associated L-C tank
circuits are
left.
is a listing of a small portion of the frequency response,
which was calculated after the filter had been synthesized. Figure 15.7
Figure 15.6
a plot of voltage ratio in decibels versus frequency.
Note that 1000
must be added to every frequency value in the abscissa. This
merely reflects an idiosyncrasy of the plot routine.
Figures 15.8 and 15.9 show the real and imaginary parts of the input
admittance of the filter. Note particularly the shapes of these characteristics. If a number of these band-pass filters were connected in parallel
and if the pass bands of the filters were adjacent but nonoverlapping,
then the input conductance of the filter system would be essentially
constant over the entire pass band, while the input susceptance would
cancel out, as shown in Figs. 15.10a and b. This then means that the
input admittance of the filter system would be real and could then be
driven by any arbitrary source impedance without fear of reflections.
Filters obtained using computer programs of the type we have just
described are rapidly supplanting conventional hand- and handbookdesigned filters. The filter synthesis programs provide filters that are
orders of magnitude more sophisticated than niters designed by the
is
cycles
Frequency
(a)
FIG. 15.10. Effects of paralleling several band-pass
filters
with adjacent pass bands.
Computer techniques
447
in circuit anal/sis
Moreover, the designs are completed in minutes
a typical cost of twenty dollars rather than two
thousand (not including initial programming costs, of course).
conventional manner.
rather than days
and
at
Optimization
Network design cannot always be accomplished by way of
analytical
Quite often networks are designed through a trial-and-error
process. The network designer begins with a set of specifications. He
then selects a network configuration, and makes an initial guess about
the element values. Next he calculates or measures the desired responses
and compares them with the specifications. If the measured responses
differ by a wide margin from the specified responses, the designer changes
the values of the elements and compares again. He does this a number
of times until (hopefully) the measured responses agree with the specified
means.
responses to within a preset tolerance.
made to converge, sometimes quite
one uses a method of steepest descent.* To get a rough idea
of the steepest descent method, suppose there are n parameters in the
network. Let us regard each parameter x{ as a dimension in an n,xn)
dimensional euclidean space. We define a function f(x u xa xs
that assigns a functional value to each point of the euclidean space. We
then ask, at point p t what direction of motion decreases the value of
f(xlt xit *»,...,*„) most rapidly? The function f(xt) may be defined as
a least squared error, or as an absolute error between calculated response
and specified response. The direction of steepest descent is the direction
denned by the gradient
This process of cut and try can be
rapidly, if
,
,
.
.
.
,
+ T-*» +
mdf=T-*i
OX
OXi
---
2
+ ^*n
OX
Therefore, incremental value of change for each parameter
6xt
(15.D
n
is
=-C—
(15.2)
OXi
where
C
is
a constant. After
all
incremental value in Eq. 15.2, a
are obtained.
An
parameters have been changed by the
gradient and new incremental values
new
The process continues until
by C. L. Semmelman7
excellent paper
the
optimum
is
obtained.
describes a steepest descent
* Charles B. Tompkins, "Methods of Steep Descent," in Modern Mathematics for the
Engineer (Edwin F. Beckenbach, ed.), Chapter 18, McGraw-Hill Book Company, New
York, 1956.
7
C. L. Semmelman, "Experience with a Steepest Descent Computer Program for
Designing Delay Networks," IRE International Convention Hec, Part 2, 1962, 206-210.
448
Network
analysis
and synthesis
Fortran program used for designing delay networks. The specifications
Rt given at the frequency data points/,. The program
are the delay values
successively changes the parameters
xi so
that the squared error
<*,)=i[K,-n*<,/,)]
a
(i5.3)
i-X
minimized. In Eq. 15.3, T{x it ft ) represents the delay, at frequency ff
of the network with parameters xt
In order to use the program, the network designer must first select the
initial values for the parameters x (
He must also provide the specified
delays R, and the frequency data points/^. The program provides for 128
match points and 64 parameter values. It is capable of meeting requirements simultaneously in the time and frequency domains. The designer
is not restricted to equal-ripple approximations or infinite Q requirements.
He is free to impose requirements such as nonuniform dissipation and
range of available element values on the design. For related methods of
optimization, the reader should refer to a tutorial paper by M. R. Aaron. 8
is
,
.
.
Machine-aided design
The concept of real-time
man and computer holds
network design. Multiple-access computing
systems, such as the project
of Massachusetts Institute of Technology, make cut and try design procedures practicable. The initials
could describe either the term multiple-access computer or the term
machine-aided cognition. In an article describing the
computer
system, 9 Professor R. M. Fano states "The notion of machine-aided cognition implies an intimate collaboration between a human user and a
computer in a real-time dialogue on the solution of a problem, in which
the two parties contribute their best capabilities."
A simple multiple-access computer system is shown in Fig. 15.11.
There are n data links and n terminals connected to the central processor.
Located at each terminal is input-output equipment, such as teletypewriters, teletypes, and oscillographic displays. The sequence in which
the central processor accepts programs from the terminals is controlled by
a built-in queueing logic.
The main reason a multiple-access computer is effective is that the
central processor computes thousands of times faster than the user's
reaction time. When a user feeds in a program, it seems only a "moment"
much promise
interaction between
in the field of
MAC
MAC
MAC
8
M. R. Aaron, "The Use of Least Squares in System Design," IRE Trans, on
December 1956, 224-231.
•R. M. Fano, "The MAC System: The Computer
Spectrum, 2, No. 1, January 1965, 56-64.
Circuit
Theory, CT-3, No. 4,
Utility
Approach,"
IEEE
Computer techniques
in circuit analysis
Remote
Data
Memory
449
terminal
bank
link
Memory
Central
bank
processor
Buffer
Data
memory
link
.
1
Remote
terminal
Queueing
Data
logic
link
2
Remote
terminal
3
1
Memory
bank
Memory
bank
Data
Remote
terminal
link
FIG.
15.1 1.
n
Block diagram showing typical multiple-access computing system.
before he gets the results through the teletypewriter or oscilloscope
display. He examines the results, changes some parameters, and feeds the
program into the central processor again. The queueing time and processing time on the multiple-access computer may amount to only a
minute or two, which, for the user, is probably not significantly long.
Dr. H. C. So has written a paper on a hybrid description of a linear
10
in which he has shown that the hybrid matrix is ideally
ii-port network
suited for such problems such as multielement variation studies and
iterative design. Suppose there are n variable elements in the network.
These can be "extracted" and the rest the bulk of the network can be
—
—
described by the n-port hybrid matrix.
10 H. C. So, "On the Hybrid Description of a Linear i»-Port Resulting from the
Extraction of Arbitrarily Specified Elements," Trans. IRE on Circuit Theory, CT-12,
No. 3, September 1965, pp. 381-387.
'
Network
450
analysis
and synthesis
Command
,
Initialize
i
1
i
Frequency
range
Engineer
—
Control
Computer
*j
,
Parameter
i
adjust
'
•
Display
p
Output
FIG. 15.12. Iterative design of network using machine-aided cognition.
Dr. So has written a computer program to formulate the hybrid matrix
for the H-port network automatically. 11
The inputs to this program are
node connections specifying the network topology; (2) the impedance functions of the elements; and (3) the specifications of the
special elements to be extracted. This program was written with manmachine interaction in mind. The process is described in Fig. 15.12.
First the computer reads in the n-port program with initial specifications.
It performs the calculations and feeds information, such as transient or
steady-state responses, back to the engineer via a visual display console.
The engineer assesses the data and then changes the parameter values of
(1) the
the extracted elements, and, in
some
instances, the frequency range over
which the calculations are made. The process
is
repeated a
number of
times until the engineer has obtained the desired results. Such a program
could also be controlled by a steepest descent steering program
if
the
engineer wished to obtain his results with less "eyeballing."
Now let us examine some details of specific network analysis programs.
15.2
AMPLITUDE AND PHASE SUBROUTINE
Purpose
Our purpose
is
to
compute the amplitude and phase of a
rational
function
"(s) - ?Hr?
1
H. C. So, unpublished memorandum.
(154)
Computer techniques
in circuit anal/sis
451
over a set of frequencies (o^ In Eq. 15.4, C„ and C are real constants,
d
and E(s) and F(s) are polynomials in s = a +ja>. The program described
here computes the amplitudes M(co,) and phase #a> ) over a set of
t
frequencies cok where
,
JU
«W =
\CnE(jlOh)\
(15.5)
\Ca FOcok)\
and
<f>((ok)
= arctan
arctan
ft
Fj(ja> k)
(15.6)
where in Eq. 15.6 the
indicate
real
respectively.
<Ko>k)
<f>
*s
r
and
i
subscripts
and imaginary
Note that in Eq.
limited to the
<,it radians.
range
parts,
15.6,
—
it
<>
In order to circumvent
this restriction on the phase, we use the
following method to compute amplitude
-JW
FIG. 15.13. Calculation of magnitude and phase at a> ( due to zero z.
and phase.
Method used
Let H(s) be factored into poles and zeros such that
H(s)
= C n (s - z )(s - zt)
Cd(s ~ Pi)(s - P»)
t
(S-PJ
(15.7)
Consider the amplitude and phase due to any pole or zero, for example,
«
* +jP, shown in Fig. 15.13. The magnitude is
=—
M
s
and the phase
is
{a>,)
4>.{a>d
=
[««
=
arctan
The amplitude and phase of the
+
(<»<
- ft*}*
ft,
'~
(15.8)
"
l
|
overall function
(15.9)
is
M(wJ =
(15.10)
QIT^M)
and
<K°>i)
= 2 tJfOi) - j £,,(«>,)
(15.11)
452
Network
analysis and synthesis
where the subscripts p and
due to a pole and
z indicate the contribution
zero, respectively.
Input
1. The numerator E(s) and denominator F(s) can be read in either as
polynomials (high degree to low) or in terms of their roots. If numerator
and/or denominator are read in as polynomials, their roots will be printed.
If they are given in
terms of their roots, the program could generate
polynomials from these roots.
2.
The data points can be read in if unequally spaced;
only the
minimum
point and the spacing need be read
Read Nl, ND,
if equally
spaced,
in.
NN
KK, LI, L2, L3
Read CN,
KK =
Test
Read
W
L1
CD
KK
KK=1
Read WMIN,
(I)
=
Test LI
Read A (I), B (I)
Ll
=
DW
l
Read E(l)
J
L2
=
TestL2
Read C (I), D (I)
L2=l
Read
F(l)
*~r*
To main program
FIG.
15.14.
Flow chart of input
instructions for magnitude
and phase program.
Computer techniques
A flow chart listing the input instructions is given in
initions
453
in circuit analysis
Fig. 15.14.
Def-
of the symbols in the flow chart follow.
= number of frequency points
ND = degree of denominator
NN = degree of numerator
KK = indicates equally spaced data points. We
Nl
=
1
Wjqn, Aa>, and number of points.
indicates unequally spaced data points.
then read in only
We
then read in
all
to,.
LI
=0
Read
read
L2
=
=
1
Read
Read
read
in zeros (if
complex conjugate, both zeros must be
in).
in numerator polynomial, high degree to low.
in poles (if
complex conjugate, both poles must be
in).
= Read in denominator polynomial, high degree to low.
CN = numerator multiplier
CD = denominator multiplier
1
Output
The main output of
the program is the frequency response consisting
of the following columns: frequency in radians W(I); amplitude,
M(W(I)); magnitude in decibels, 20 log10 Af; and phase in degrees,
^(W(I)).
typical printout of an amplitude and phase program is given
in Fig. 1S.6, which shows the magnitude and phase of the band-pass filter
A
designed using Szentirmai's program.
15.3
A FORTRAN PROGRAM FOR THE ANALYSIS OF
LADDER NETWORKS
In this section we will discuss the methods and organization for a
Fortran computer program for the analysis of linear ladder networks.
The analysis proceeds by computing the voltage and current at the input
of successive
L
sections, beginning
with the terminating section.
branch impedances of the ladder by comimpedance
arms according to the instructions of the
C
individual programmer.
The poles and zeros and the frequency responses of the input impedance and voltage transfer ratio at the input of each L section are
found. In addition, the program provides for the analysis of short- and
open-circuited networks as well as for those with normal terminations.
Separate problems may be run consecutively if so desired.
The program
bining R, L,
calculates the
series
—
Network
454
vm
analysis and synthesis
Zm
v„ -l
Zn
°
la"
Vo
Vl
l
Zi
1
1
1
Zn-l
|
|
1
1
Ym
Yi
r»-i
Y„
l
1
o
FIG. 15.15
initially decomposed into separate L sections as in
L sections are then added successively to the terminating
form the complete network. With the addition of an L
The network
Fig. 15.15.
L
is
These
section to
and current
computed by the equations
section, the voltage
^n =
at the input of the resulting
network are
(15.12)
^n^n
+ Vn-1
which were originally discussed in Chapter 9 of this book. Thus the
program proceeds toward the front of the ladder requiring only the branch
immittances of the present L section and the voltage and current of the
previous
L
section to
make
its
calculations.
Suitable assumptions for the initial voltage
and current
V and
/ allow
for analysis of short- and open-circuited networks as well as for those
with normal terminations. These initial values of voltage and current are
determined by control instructions.
For the calculation of the voltage and current at an L section, the
branch impedances of the section are required. To simplify the preparation of the input data specifying these impedances, the following procedure
is used. Each branch impedance is formed by the addition in series or
parallel of basic R, L, C impedance arms. A basic R, L, C impedance
arm is a series combination of a resistance, an inductance, and a capacany or all of which may be absent. The impedance arms are speciitor
fied by the element values R, L, C and an instruction indicating how the
arm is to be combined.
The elements of the impedance arms are frequency and impedance
normalized to aid computation, and then the arms are combined as
specified to form the appropriate branch impedance.
The roots of the voltage and .current polynomials for each L section
—
are
computed by a root-finding
routine.
;
Computer techniques
455
in circuit analysis
For the normal load and open circuit termination, the frequency
responses of the input impedance, and voltage-transfer ratio are computed
for short-circuit termination, the input impedance and current gain are
computed. Provision is made for either a linear or logarithmic increment
The frequency boundaries and increments are read by the
program along with the input data.
According to instructions given by the programmer, the various calin frequency.
culations are printed for each
L section or only for the network as a whole.
Program operation
The ladder network-analysis program requires two sets of input data.
The first set consists of the control instructions. These tell the computer
which of the various options, such as information concerning polynomial
roots or frequency responses, are to be exercised.
In addition, the control instructions provide the computer with necessary
parameters such as normalizing factors, and the number of L sections in
the ladder. The second set of data determines the branch impedances of
the network by specifying the R, L, C impedance arms and their combining
instructions.
Impedance data
The data specifying
the branch impedances consist of an ordered series
an impedance arm and/or a comarm of any branch impedance requires no
of instruction cards, each determining
bining instruction.
The
first
combining instruction. The branch impedance is set equal to the impedance of this arm. Subsequent impedance arms are added in parallel
or series with the existing branch impedance, according to the combining
instructions of the arms. A blank card tells the computer that a complete
branch impedance has been formed and that the next arm begins a new
branch impedance. If blank cards are not properly inserted, the computer
will calculate
a single giant branch impedance.
the data determining the branch impedances isZx ljYlt
The order of
^a»
1/
Jg,
•
•
•
,
•
The values of R, L, C and the combining instruction XIN are written
on one card in the order XIN, R, L, C.
The values of XIN and the associated operation are shown in the
table at the bottom of page 456.
Although the instructions XIN = 1.0, 2.0 will suffice for the conmany branch impedances, they are inadequate for other
struction of
impedances.
For example, suppose the series combination of two tank circuits (Fig.
15.16) is desired as a branch impedance, this impedance may only be
Network
456
anal/sis
and synthesis
50m*f
lMf
2mh
.05 h
-^OOOy-
1
Tank
circuit
A
Tank
circuit
B
XIN
5.00000£
1.0
2.00000E
2.0
3.0
-
11
- 03
.05
.000001
2.0
4.0
Card
Blank
FIG. 15.16. Construction of branch impedances using instructions for XIN.
constructed with use of three
and an additional
Tank
circuit
more
instructions
(XIN
=
3.0, 4.0,
and
5.0)
of computer storage locations.
formed in the usual manner, but must then be tem-
set
A is
porarily stored during the calculation of tank circuit B.
The two tank
then added to form the final branch impedance. Examples of
impedances that cannot be formed are shown in Fig. 15.17. Observe that
circuits are
a series branch impedance of zero ohms often simplifies the formation of
branch impedances. Two blank cards will indicate a short-circuited
impedance.
XIN
Blank card
Operation
A complete
the next
branch impedance has been calculated, and
begins a new branch impedance.
arm
1.0
This arm is added in series with the existing branch
impedance.
2.0
This
3.0
This arm begins a new internal branch impedance.
Subsequent XIN = 1.0 and 2.0 refer to this new internal branch.
4.0
Add
arm is paralleled with the existing branch impedance.
the two internal branches in series to form the final
branch impedance.
5.0
Parallel the internal branches.
Computer techniques
HWV-i
in circuit analysis
457
r-VW-i
-Wrr-VArn
[-VS/Sr-j
L-vs/Sr-l
FIG.
15.17.
Branch impedances that cannot be formed by program.
Output
coefficients of the voltage and current polynomials are printed in
frequency normalized form. To calculate the denormalized coefficients,
divide by the normalizing frequency raised to the power of the corre-
The
sponding exponent.
and zeros of the voltage and current polynomials are in
frequency normalized form to facilitate zero-pole plots.
The
poles
Frequency responses appear in denormalized form. The frequency
variable is in radians per second. The impedance level is in decibels,
the phase in degrees. The gain and phase of the transfer ratios are
expressed similarly.
15.4
PROGRAMS THAT AID IN DARLINGTON
FILTER SYNTHESIS
Two double precision Fortran programs have been written that help
in the synthesis of double-terminated filters using the Darlington procedure. The mathematics of the programs is described here.
Given a forward transmission
coefficient
S8i(s)
-
KN(s)
D(s)
(15.13)
Network
458
we
analysis
and synthesis
generate an input reflection coefficient
=
Su(s) Sn(-s)
1
-
D(s)
Su(s) from
the equation
S»(s) S gl (-5)
D(-s)
-
K* N(s) N(-s)
(15.14)
D(-s)
D(s)
The denominator of Su(j) is D(s) because all the poles of Su(s) must be
in the left-half plane. The zeros of Sn(s) are not so restricted. If Sn(s)
Su(—s) has zeros &t—a±jb and a ± jb, the zeros of Sn(s) can be chosen
as either — a ±jb or a ±jb. The only difference is that certain choices
lead to niters with unity-coupled coils and others without. 12
Once the zeros of Sn(s) are chosen, the input impedance of the filter
is
then
Zi(s)
=
1
1
ReadCN
~
+
Sii(*)
(15.15)
Su(s)
Numerator
multiplier
I
Read
CD
Denominator
Number
ReadLN
I
Read
A (I),
B(I),N(I)
Read LD
multiplier
of zeros
(complex conjugate zeros
count as one zero)
Read
real
and imaginary
parts of zeros and
multiplicity,
N (I)
Number of poles
(conjugate poles
count as one pole)
Read
H(I),G(I), M(l)
Read
real
and imaginary
parts of poles and
multiplicity, P(l)
To program
FIG.
11
15.18. Input to
T. Fujisawa, "Realizability
Sn(5)
Theorem
Ladders without Mutual Induction," Trans.
1955, 320-325.
5u (—j) program.
Low Pass
on Circuit Theory, CT-2, December
for Mid-Series or Mid-Shunt
IRE
Computer techniques
459
in circuit anal/sis
Program descriptions
We
next proceed with brief descriptions of the two programs. The
program finds the roots of Su (s) S^—s) given the roots of N(s),
N(—s), D(s), and D(— s). A flow chart for the input of the program is
given in Fig. 15.18. Note that we must read in both the zeros of N(s) and
N(—s), although only half of a complex conjugate pair need be read in.
The same applies for the roots of Z>($) and /)(—$). The constant CN is
first
K* in Eq. 15.14.
The second program performs
the computations in Eq. 15.15 for
combinations of zeros of Sxl(s), given the fact that the zeros of
Su(s) need not be in the left-hand plane. The previous program finds all
the zeros of Sn (s) Su (— s). We must choose only half the number of
zero pairs for 5u(s). Certain combinations of zeros of Sn (s) Su (—s)
lead to filters with coupled coils; others do not. If one wishes to try a
number of combinations of zeros for 5u(f), the program has the facility
to enable him to do so. He need only supply those zeros of Sn(s) Su (— s)
in the right-hand plane, C(I) +y'B(I), and a list of constants S(I), which
different
are either 1.0 or
S(I)
•
—1.0 so
that the zeros of
— A(I) +y'B(I).
C(I) +y'B(I)
denominator polynomials of
The
Sxl(s) may
be represented as
routine forms numerator
Su (s)
= E(s)
(15.16)
F(s)
ReadLD
Read
ReadLN
Read
of pole pairs of
Su
count as one pair)
of
Sn
and
multiplicity,
M (I)
Number
of right-hand plane
zeros of
Sn (a) Sn(-s)
Real and imaginary parts of right-hand
C(l), G(l), N(l)
c
Number
(real poles
Real and imaginary parts of poles
H(I).G(I),M(I)
r
plane zeros and multiplicity,
N 0)
I
Read SO)
Multiplicative constants
(i 1.0) (each
combination of zeros has LN cards)
To program
FIG.
and
Su(s)
15.19.
Input to
Zta(j) program.
460
Network
analysis
and synthesis
and then computes the numerator and denominator of Z^s) as
ZM= i^ = m^m
The input
summarized on the flow chart
instructions are
(15 17)
.
in Fig. 15.19.
Problems
15.1
Organize a flow-chart for computing magnitude and phase given only
the unfactored numerator and denominator polynomials. Do not use a rootfinding subroutine.
15.2
Write a program to calculate the delay of
_,.
at the points o>
= 0,
1, 2,
.
.
.
,
3(5
+
1)
10.
Repeat Prob. 15.2 for a general transfer function given in terms of its
unfactored numerator and denominator polynomials. The frequency points co t
are to be read in by the computer.
15.4 Suppose you have calculated the phase of a transfer function at points
m = 0, 1, 2, .... 50. Devise an algorithm to test for phase linearity. The
deviation from phase linearity is to be called phase runoff.
15.5 Write a program to analyze a two-mesh network made up of only
15.3
R, L, and
15.6
C elements.
Repeat Prob. 15.5 for nodal analysis.
15.7 Write a program to calculate the step and impulse response of a linear
system whose system function contains only simple, real poles and zeros.
Repeat Prob. 15.7 if simple complex poles and zeros are allowed.
15.9 Write a program to calculate the residues of a transfer function with
multiple as well as simple poles. The poles and zeros are to be real.
15.8
appendix
A
Introduction to matrix algebra
A.I
FUNDAMENTAL OPERATIONS
Matrix notation
is
merely a shorthand method of algebraic symbolism
that enables one to carry out the algebraic operations
more
quickly.
The theory of
matrices originated primarily from the need (1) to solve
simultaneous linear equations and (2) to have a compact notation for
linear transformations
from one
As an example of (2),
of variables to another.
set
consider the set of simultaneous linear equations
«u*i
^l^X
+
+
ami^i
+
aM";3
+
+
amixa
+
«i**a
•
•
•
•
•
•
•
•
•
+
+
alnxn
"
a* »*»
= yx
—
= y*
**
(A.1)
+ amnxn = ym
These may express a general linear transformation from the xt to the yt
In general, mj&n. An example where the numbers of variables in the
two sets are unequal is that of representing a three-dimensional object in
two dimensions (in a perspective drawing). Here, m = 2 and n = 3.
.
Definition
A
an ordered rectangular array of numbers, generally the
The matrix is denned by giving
all its elements, and the location of each.
The matrix of the equations in Eq. A.1 written as
matrix
coefficients
is
of order
is
of a linear transformation.
mx
n.
(The
Oil
«12
fl gl
att
first
«ln
number here
461
is
the
number of rows; the
462
Network
second
is
the
analysis
and synthesis
number of columns.) The matrix may be denoted by a
A or by [aw ].
single capital letter
A
matrix
is
a single complete
entity, like
a position in chess.
Two
A
and B are equal only if all corresponding elements are the
same: atj = btj for all i andy. A matrix may consist of a single row or
single column. The complete matrix notation applied to Eq. A.1 is
matrices
<hi
olt
Oi„
r*r
On.
on
oin
xa
Xn
i
'mi
'ml
"yr
=
.
Vt
.
(A.2)
Vm .
says, if put crudely, that A operates on x t to yield y,. This
emphasizes the similarity between a matrix and a transformation. The
which
and y-matrices are column matrices.
and column matrices are called vectors (specifically, row vectors
and column vectors) and their similarity to the more usual type of vector
x-
Row
is
Here, vectors will be written as small
discussed later.
letters,
such as
that the elements of vectors need only one subscript, while
x
or y. Note
elements of matrices need two.
A.2
ELEMENTARY CONCEPTS
Square matrix
A
square matrix has the same number of rows as columns (i.e., a
matrix of order rat). The Y matrix is an example of a square matrix
(A.3)
Diagonal matrix
A
diagonal matrix
diagonal are zero
matrix
is
is
(i.e.,
a square matrix whose elements off the main
for 1 ?*/)• The following
one in which <xw =
diagonal.
'1
01
0-2
(A.4)
3
_0
Unit matrix
A unit matrix
a diagonal matrix for which a u
is
denoted as U. For example,
is
1
for
1
=j, and
is
U
0"
"1
U=
(A.5)
1
,0
1_
Appendix
A
463
Equality
Two matrices are equivalent if they
columns and
Suppose yx =
2,
in matrix form,
have the same number of rows and
the elements of corresponding orientation are equal
if
=
yt
—3, and ya
=
—6.
If we write this set
of equations
we have
"
2"
~Vi
= -3
y*
(A.6)
-6.
.y*.
Transpose
The transpose of a matrix A denoted as A r is the matrix formed by
interchanging the rows and columns of A. Thus, if we have
,
1
A=
-6
(A.7)
3
-CIS
A*
then
(A.8)
Determinant of a matrix
The determinant of a matrix
is defined only for square matrices and is
formed by taking the determinant of the elements of the matrix. For
example, we have
1
2
12
det
l
4j
f
L-5
5
=
14
(A.9)
4
Note that the determinant of a matrix has a particular
itself is merely an array of quantities.
value, whereas the
matrix
Cofactor
The
cofactor
deleting the fth
A i}
of a square matrix
is
the determinant formed by
row andy'th column, and multiplying by (—l)i+i
example, the cofactor
A tl
-n
An =
IS
.
For
of the matrix
(-iy*+i
x
6= -6
(A.10)
(A.11)
Adjoint matrix
The
A is formed by replacing each
and transposing. For example, for the
adjoint matrix of a square matrix
element of
A
by
its
cofactor
Network
464
and synthesis
analysis
matrix in Eq. A.10,
we have
=
adjA
h
i
l:
-l;
(A.12)
Singular and nonsingular matrices
A singular matrix is a square matrix A for which det
singular matrix is one for which det A ?* 0.
OPERATIONS
A.3
ON
A=
0.
A
non-
MATRICES
Addition
may be added
both matrices are of the same order.
added to the element of the second
Each element of the first
orientation
is the same. An example of
column
and
matrix, whose row
A.
13.
shown
in
Eq.
is
addition
matrix
Two
matrices
if
matrix
is
(A.13)
[-in-'-H-<-2
Thus
if
matrices A, B,
A+B+
The
associative
K are all
C
•
•
•
+K=
of the same order, then
+b +
[a u
•
•
it
•
+k
tj ]
(A.14)
and commutative laws apply.
Associative:
Commutative:
A + (B + C) = (A +
A+B=B+A
B)
+C
(A.15)
Multiplication by a scalar
We define multiplication by a scalar as
XA
=
Thus to multiply a matrix by a
A[a w ]
=
[AaJ
scalar, multiply
(A.16)
each of
its
elements by
the scalar.
Example
Al
"
2
-1
"
0"
1
-3 2
-
6
0"
-3
3
-9
6
(A.17)
Appendix
A
465
Linear combination of matrices
of addition and multiplication by a scalar are combined,
for two matrices of the same order we have
If the rules
oA
where « and
+
/SB
=
[*ait
+ pb
(A.18)
it ]
are scalars.
/9
Multiplication
In order for matrix multiplication AB to be possible, the number of
columns of the first matrix A must equal the number of rows of the second
matrix B. The product C will have the number of rows of the first and
A has
the number of columns of the second matrix. In other words, if
product
the
then
columns,
rows and n columns, and B has n rows and/>
m
C
will
have
m
rows and
p
columns.
The
individual elements of
C
are
given by
(A.19)
Example A.2
4
(A.20)
2
[-:-:;]-[-,:;]
-3
Example A3.
The system of equations
«u/i
ZtJi
+ Ws = Vx
+ z«/s = Vt
(A.21)
can be written in matrix notation as
(A.22)
caca-ra
We
see that systems of equations
can be very conveniently written in
matrix notation.
Matrix multiplication
is
not generally commutative, that
AmnBB „ * B^A,™
Observe that the product BA is not defined unless p =
of square matrices
is
generally not commutative, as
is,
(A.23)
m. Even a product
may be
seen in the
following example.
r
i
L-i
onr-i
2JL
o
<n
_ r-i
2j~L
i
oi
(A.24)
f]
Network
466
If
we
anal/sis
and synthesis
interchange the order of multiplication,
1
-1
-1
-2
-1
2
we
obtain
(A.25)
3
Because of the noncommutative nature of matrix multiplication, we
must distinguish between premultiplication and postmultiplication. In
BA, A is premultiplied by B; B is postmultiplied by A. For matrix
multiplication, the associative and distributive laws apply.
=
= ABC
+ C) = AB + AC
Associative:
A(BC)
Distributive:
A(B
(AB)C
(A.26)
Transpose of a product
The transpose of a product AB is equal to the product, in reverse
of the transpose of the individual matrices A and B, that is,
(AB)
(ABC)r
and
- B rA T
order,
(A.27)
= C T(AB) T = C^B^A 2
"
(A.28)
The product x Tx, if x is a column vector, is a scalar number equal to
the sum of the squares of the elements of x. Thus we have
xT x
The product xx r
Common
(a)
=
[ Xl
=
*i
xt
2
+ x*
T
is a square matrix C such that C = C
+
*,*
It is also
xx r
if
x
is
expression auxf
where
A
•
t
a and b are column vectors.
(b) The sum of squares x^
The
•••
aA + ajb +
typical element of a product matrix,
(c)
+
.
expressions in simple notation
The sum of products
vector).
(A.29)
x„]
may be
+ xt* +
•
•
•
•
•
+ anb„,
written as
+
x n*
+
a nnxn *
is
which are the
a^b or b ra where
x Tx
(x,
column
x rAx
(x,
column
thus
a row vector.
+ atixta +
•
•
•
is
a diagonal matrix with elements ait Similarly, the
T
T
expression aaxjfx
a^e^yt H
h a nnx„yn is x Ay or y Ax.
(d) An expression such as
vector),
is
.
+
2 Z aux x
i
t>
=
a,
Appendix
is
a quadratic form. This
«ii*i*
467
is
+
1a x%xx xt
+
««*!*
+
•
•
A
•
+
+
+
+ 2a ln x1 x„
2a u xt xa H
<J»»*«
2
h 2a, B a;2 x B
(A.30)
x TAx
=
Inverse
Division
is
not defined in matrix algebra. The analogous operation
that of obtaining the inverse of a square matrix.
matrix
obtain
= AA- = U
1
A-1 we first obtain the adjoint
,
the determinant of A.
that
inverse
A-1
is
of a
A is defined by the relation
A-*A
To
The
The
inverse
A-1
is
(A.31)
of A, adj A. Then we obtain
equal to adj A divided by |A|,
is,
A" 1
=
—
adj
A
(A.32)
|A|
Example A.4.
Let
A be given as
(A.33)
Its
determinant
is
|A|
=3
(A.34)
and the cofactors are
An = 1
Afn = —1
The
adjoint matrix
Au =
1
An =
2
(A.35)
is
adJA =
[-l
2j
(A.36)
-C
so that
A-1 is
-»C 1
-EH
(A.37)
Network
46.8
As a check we
and synthesis
anal/sis
see that
p
-r
}
K
"
r
2
i
'
V
2
-1
-l
i
(A.38)
P
_i
*
§
1
l
If the determinant
of the matrix is zero, then the inverse is not defined.
other words, only nonsingular square matrices have inverses.
In
SOLUTIONS OF LINEAR EQUATIONS
A.4
Consider a
set
of linear algebraic equations, to be solved simultaneously.
011*1
asi^i
«m*i
The h t are
we have
constants.
H
+
+ a^xt + •• +
r-
<*iss*»
+ a n **i +
•••
A
x
gives
In expanded form, this
.
In matrix notation
=
h
(A.40)
=A
x
(A.41)
Ax
Premultiplying by
(A.39)
+ a nnxn = h n
desired to solve for the xf
It is
-1
= hx
aSnx„ = hs
alnxn
h
is
x.
^si
J_
™ia
|A|
^32
-"m
(A.42)
Thus
|A|
*1
Oia
*is
ht
a«
On
ha
a S2
a»
|A|
and
similarly for
xt and *3
.
This
is
the familiar Cramer's rule for solving
such equations.
Example A.5.
Solve for x, y,
z.
x
2x
+2 =2
+y =
3y
+ « = -1
—
-* +
1
(A.43)
Appendix
In matrix form, these equations are written as
1
-1
2
1
-1
Also, |A| =» 10. Thus,
3
r
-&
Therefore
A.5
x
The following
is
r
y
h
,
=
i
A_1h
-r
r
r
2
2
i
7-2
3
-i
-0.4, z
(A.44)
-l
z
4
i
= 0.7, y =
REFERENCES
=
1
-2
z
X
»
469
Ax = h where
~x~
we have x =
*1
y
r
A
r
=
X
io
-4
(A.45)
9
= 0.9
ON MATRIX ALGEBRA
a short
list
of books on matrices that the reader might wish to
examine.
A. C. Aitken, Determinants and Matrices, 9th Ed., Interscience Publishers,
New
York,
1956.
R. Bellman, Introduction to Matrix Analysis, McGraw-Hill Book Company, New
York, 1960.
York,
R. L. Eisenman, Matrix Vector Analysis, McGraw-Hill Book Company, New
1963.
D. K. Faddeev and V. N. Faddeeva, Computational Methods of Linear Algebra, W. H.
Freeman and Company, San Francisco, 1963.
Publishers,
F. R. Gantmacher, Applications of the Theory of Matrices, Interscience
New
York, 1959.
Hohn, Elementary Matrix Algebra, The Macmillan Company, New York, 1964.
McGraw-Hill Book
L. P. Huelsman, Circuits, Matrices, and Linear Vector Spaces,
Company, New York, 1963.
and Sons, New York,
P. LeCorbeiller, Matrix Analysis of Electric Networks, John Wiley
F. E.
1950.
Survey of Matrix Theory and Matrix Inequalities, Allyn and
Bacon, Boston, 1964.
Sons, New York,
E. D. Nering, Linear Algebra and Matrix Theory, John Wiley and
M. Marcus and H. Mine,
1964.
1952.
of Matrices, Addison-Westey, Reading, Massachusetts,
Englewood Cliffs, NJ.,
L. A. Pipes, Matrix Methods for Engineering, Prentice-Hall,
S. Perlis, Theory
1963.
A. M. Trapper, Matrix Theory for
Electrical Engineering Students, Harrop,
London,
1962.
A. von Weiss, Matrix Analysis for
NJ., 1964.
Electrical Engineers,
D. Van Nostrand, Princeton,
i
appendix
B
Generalized functions and
the unit impulse
GENERALIZED FUNCTIONS
B.I
The
unit impulse, or delta function,
is
a mathematical anomaly. P. A.
M. Dirac, the physicist, first used it in his writings on quantum mechanics. 1
He denned
the delta function d(x)
£
d(x)
Its
most important property
f—
by the equations
d(x)
=
dx
=
1
(B.I)
for
x
jt
is
f(x)6(z)dx=f(0)
(B.2)
V
—
is continuous at x
0. Dirac called the delta function an
improper function, because there existed no rigorous mathematical justi-
where f(x)
it at the time. In 1950 Laurent Schwartz* published a treatise
The Theory oj Distributions, which provided, among other things,
a fully rigorous and satisfactory basis for the delta function. Distribution
theory, however, proved too abstract for applied mathematics and
fication for
entitled
1
P.
A. M. Dirac, The Principles of Quantum Mechanics, Oxford University Press,
1930.
1
and
L. Schwartz, Theorie des Distributions, Vols.
1951.
470
I
and
II,
Hermann
et Cie, Paris, 1950
Appendix B
physicists.
It
was not
until 1953,
when George Temple produced a more
elementary (although no
less rigorous)
ized functions,* that this
new branch of
Our treatment of
deserved.
theory through the use of generalanalysis received the attention
it
generalized functions will be limited to the
definition of the generalized step function
The treatment of
471
and
its
derivative, the unit
work of
Temple4 and Lighthill. 8
To get an idea of what a generalized function is, it is convenient to use
as an analogy the notion of an irrational number a beng a sequence {oc„}
of rational numbers <x n such that
impulse.
these functions follows closely the
a
= lim a„
n-»ao
where the
limit indicates that the points
the point representing a.
<x„
on
the real line converge to
on the
on the sequence {a„} defining
a.' We can also think of a generalized function as being a sequence of
functions, which when multiplied by a test function and integrated over
number a
irrational
(—00,
are actually performed
a finite limit. Before we formally define a generalized
important to consider the definition of (1) a testing function
oo) yields
function,
and
All arithmetic operations performed
(2)
it is
a regular sequence.
DEFINITION
A function #f) of class
B.I
everywhere, any
is differentiable
any of its derivatives are multiplied by
lim
[t
m
<f>
C[#f) G C]
number of times and
M(t)]-*0
t
raised to
for all
is
one that
that (2)
when
any power, the
m&fe^O
it
(1)
or
limit is
(B.3)
*->±oo
Any
testing function is
a function of class C.
The Gaussian function e-*'ln% is a function of class C. It is
obvious that if a function is of class C then all of its derivatives belong to class C.
Example B.l.
*G. Temple, "Theories and Applications of Generalized Functions," J. London
Math. Soc., 28, 1953, 134-148.
4
G. Temple, "The Theory of Generalized Functions," Proc. Royal Society, A, 228,
1955, 175-190.
6
M.
and Generalized Functions, Cambridge University
book to "Paul Dirac, who saw it must be
Laurent Schwartz, who proved it, and George Temple, who showed how simple
J. Lighthill,
Press, 1935.
true,
it
Fourier Analysis
Lighthill dedicated his excellent
could be made."
* This defines an irrational
number according to the Cantor definition. For a more
on real variables such as E. W. Hobson, The Theory of
Functions of a Real Variable, Vol. I, third edition, Chapter 1, Cambridge University
detailed account see
Press,
any
text
Cambridge, England, 1927.
Network
472
and synthesis
anal/sis
DEFINITION
A sequence {/„(*)}
B.2
to be regular if for any function
of functions of class
C
is
said
belonging to C, the limit
<f>(t)
W
lim(/„,#=lim
n-*ao
(
J— oo
n-*oo
fn(t)<Kt)dt
(B.4)
Note that it is not necessary that the sequence converge pointwise.
For example, the sequence {e~ nt\nlir)^} approaches infinity as n -» oo at
the point t = 0. However, the limit lim (/"„, <f>) exists.
exists.
n-»oo
DEFINITION
alent if for all
B.3
Two
C
6£
regular sequences {/„} and {g„} are equiv-
lim(/B ,#=lim(g B ,#
n-*oo
Example B.2.
regular sequences {e-n**(«/f) w }
The
(B.5)
n-*ao
{<r' ,/aB*(l/V2^i)} are
and
equivalent.
A
DEFINITION B.4
generalized Junction g is denned as a total, or
complete, class of equivalent regular sequences. The term total implies
here that there exists
no other equivalent
Any member
both g and the
to this class.
represent
denning g.
We denote this
Example B.3.
total class
example, {g n },
is sufficient
to
of equivalent regular sequences
symbolically by the form
1 '"4
Or*
},
g
~ {gn
}.
function
~ {e~**
The inner product
B.5
<f>(t)
€
{«--«»/»•*}]
{e-"'-}
same generalized function g
DEFINITION
g and a
regular sequence not belonging
class, for
All of the equivalent, regular sequences
[{*-'*/»•},
represent the
of the
C is defined
(g,
lnt
}.
<f>)
of a generalized function,
as
(g,#=lim f"gn(t)<Kt)dt
n-*oo
The inner product
is
(B.6)
J— co
often given the following symbolic representation.
°
(g,
Note that the
+)
=
f
(B.7)
g(t)<Kt)dt
integral here is used symbolically
and does not imply
actual integration.
DEFINITION
g and h are two
g
{g n } and h
representation g + h
{g „ +
B.6
If
resented by the sequences
defined by the
Note
~
that the set of sequences {g n
equivalent regular sequences
and
~
h; therefore
g
+ h n}
generalized functions rep-
~ {h
n },
the
sum g
+h
is
A»}.
represents a total class of
made up of the sum of sequences
defining
+ h is a properly defined generalized function.
g
D
DEFINITION
and a constant a
B.7
is
The product a.g of a
is
473
generalized function;
~ {#„}
defined by the representation ag <~ {«gn }.
DEFINITION
{#„}
Appendix B
g~
B.8 The derivative g' of a generalized function
{g' n }.
defined by the representation g'
~
Example B.4. For the generalized function
by
gx ~ {*-*
1/b *}
the derivative
is
represented
and
(g\, *)
= lim
f
"
(
- ^ e"**'"1
*(/)
(B.8)
«//
J
we have defined the operations of addition,
and differentiation. It must be pointed out that
the operation of multiplication between two generalized functions is not
In Definitions B.6, B.7, and B.8
multiplication
by a
scalar
defined in general.
We
next consider an important theorem, whose proof
Lighthill, 7
which
as a step function
Theorem B.l.
by a generalized function
for
^
€ C.
lent in terms
write
satisfying the condition
r
dt
<
co
(B.9)
N ^ 0, there exists a generalized function8 / ~ {/,(/)} such that
some
for all
1/(01
given in
equivalent.
Given any ordinary function /(f)
1'
is
any ordinary function, such
will enable us to represent
/ =/.
In other words, an ordinary function satisfying Eq. B.9 is equivaof inner products to a generalized function. Symbolically, we
If, in addition, / is continuous in an interval, then lim fn
=/
""*"
pointwise in that interval.
Furthermore,
multiplication,
it
and
can be shown that all the operations of addition, scalar
differentiation performed on both /and /yield equivalent
results, that is,
(«a
when
+
differentiation is permitted
m'
= («/i +
6ti'
on the ordinary
<M
function.
'
Lighthill, op. cit., Section 2.3.
*
Note that when we represent an ordinary function by generalized function equivalent,
we use a bold face
italic letter
to denote the generalized function.
Network
474
DEFINITION
total class
analysis
B.9
and synthesis
The
generalized step function u
of equivalent regular sequences
(«,#
= lim
n-*oo
{«„(*)}
is
defined as the
such that
f°uJit)4(t)dt
J— ao
-f"ii(t)#0A
(B.12)
J— 00
where
u(i) is the unit step defined in
Chapter
That
2.
{«„(/)} exists is
guaranteed by the previous theorem allowing representations of ordinary
u.
functions by generalized functions. Hence, we write u
=
Example B.5.
The sequence
/>o
t£0
which
is
plotted in Fig. B.l,
is
one member of the
class
(B.13)
of equivalent regular
sequences which represents the generalized step function.
2
3
4
5
6
FIG. B.l. The generalized step sequence,
DEFINITION
is
7
8
t
«,(/)•
B.I0 The unit impulse, or Dirac delta function <$(r),
{«'„(*)}.
denned as the derivative of the generalized step function d(t)
~
Appendix B
It
475
should be stressed that <J(f) is merely the symbolic representation for a
of equivalent regular sequences represented by {u' (t)}. Thus
n
total class
when we
we
write the integral
actually
£ mmdt
mean
f " <5(0 <Kt) dt
= (d,
J— oo
Example B.6.
<j>)
= lim
n-»oo
f " u' n(0 #t) dt
The sequence
^-(5-2r)«p[-i(J
,>o
+ |.)]
(B.15)
=
in Fig. B.2
is
(B.14)
J— oo
t
£0
one member of the class of equivalent regular sequences which
Other members of the class are the sequences
represents the unit impulse.
{e-»«*(«MK} and {^1*^(1/ VJ^n)}.
1.0
1.2
1.4
1.6
I
_1_
_L_
FIG. B.2. The generalized sequence,
«'„(/).
476
B.2
Network
anal/sis
and synthesis
PROPERTIES OF THE UNIT IMPULSE
Sifting
The most important property of the
unit impulse
is
the sifting property
represented symbolically by
J«<o
W(0*-/(0);
differentiable over [a,
where /is any function
Eq. B.16
is
(B.16)
(M,lj8|<«>)
The
/J].
left
hand
side of
defined formally by
f>
\ \t)f(t)dt
Js<0
slim
the sifting property
The proof of
n-»ao
is
(B.17)
\"*u'JLt)f{t)dt
Ja<0
obtained by simply integrating by
parts, as follows.
>O
lim
n—<x>
r u'„(0/(0 dt = lim u n(t)f(t)\- lim
J«<0
n-»oo
l«
n-»oo
f'ii,(0/'(0 dt
J*
-/tf)-f'limu.(0/'(OA
_,„
= /C8)-[/(/5)-/(0)]=/(0)
Pictorially
impulse
is
we
represent <$(0
centered at
t
by a spike as shown in
Fig. B.3.
If the
= a, then the sifting property is given symbolically
as
r
d(.t
-
fl
)/(f) dt
- /(a)
(|a|,
KO
FIG. B.3. The unit impulse.
HI
<
ao)
(B.19)
Appendix B
where/'(0 most exist over
are
infinite,
we
actually
[a,
Note that when the limits of integration
/?].
mean
(°°S(t)f(t)dts lira \\t)f{t)dt
J— oo
a-*— oo Ja
In the
477
sifting property, if
both
a,
(I
>
or a,
dt
<K0/(0
/?
(B.20)
< 0, then
=
(B.21)
/:
The proof of this property is similar to the original proof of
property, and will be left as an exercise for the reader.
the sifting
Integration
The
defining equations of the delta function according to Dirac are
f>0
d(x)
=
dx
1
/.«<o
(B.22)
d(x) = 0;
x
i*
These are actually properties of the delta function as viewed from the
generalized function standpoint. The proof can be obtained directly from
the sifting property. Suppose we have the integral
f>
1
and we
let f{t)
=
1.
/?
(B.23)
Then we have
Ia<0
If both a,
\t)f(t)dt=f(0)
d(t)dt=f(p)
=
(B.24)
l
are greater than zero or both are less than zero, then
1
This property
is
d(t) dt
stated symbolically
=
(B.25)
by the conditions
d(t)
=
for
/
ft 0.
Differentiation across a discontinuity
Consider the function /(0 in Fig. B.4. We see that /(f) has a discontinuity of A at t
T. If we let/i(0 =f(f) for t
T, and/i(f) =/(*)
A
for t
T, then we have
=
<
-
^
At)=MQ + Au(t-T)
(B.26)
Network
478
analysis
and synthesis
*M
•h(t)
FIG. B.4. Function with discontinuity.
Since /(0,/i(0» and u(t) satisfy the condition
lg(OI
dt <
&\N
I « (1 + O
for
some N; we can
co
represent these ordinary functions by generalized
functions
(ft)
Taking derivatives on both
=Mt) +
n(0
(B.27)
sides of Eq. B.27 yields
fV)=f\(.t)
+
(B.28)
Au'(.t)
which symbolically can be written as
no =A(o + A m
We
thus see that whenever
we
(B.29)
differentiate across
a discontinuity, we
obtain a delta function times the height of the discontinuity.
Example B.7.
The
step response of
Hf)
shown
in Fig. B.Sa.
an R-C network
is
= Ae-" T u(t)
The impulse response
l
is
shown
in Fig. B.Sb.
(B.30)
is
h'(t)=Ad(t)-j,e- l T u(t)
and
given as
(B.31)
Appendix B
479
FIC. B.5. Differentiation across a discontinuity.
Differentiation
The
derivative of
a delta function, which we
symbolically as d\t)
f"(i) exists over [a,
~
{u" n(t)}.
It
call
a doublet,
is
defined
has the following property, where
fit].
/V>o
<J'(0/(0 dt
J
J«<0 '
=
The proof is obtained through
-/'(0);
(|«|,
\fi\
successive integration
<
oo)
by
parts.
(B.32)
>0
lim
T" V(0/(0 dt = n-»oo
f «"B(0/(0 dt
Ja<0
Ja<0
fi>0
- Km
We see that since
«
'.(0/(0
lim u'Jfi)
—
lim
-lim
«<0
n-»oo
u' n (a)
>0
limH' B (0/0)'
n-*oo
«<p
(B.33)
Ct>9
u'n (t)f'(t)dt
J«<0
= 0,
=0
(B.34)
480
Network
analysis and synthesis
f(t)
FIG. B.6. The doublet
<$'(')
We then integrate by parts again so that
>0
-lim
f
u' B
(0/'(0^=-lim« n/'| +lim
['u n (t)f\t)dt
(B.35)
\yu{t)f{t)dt
-/'(/?)+
-fW +fW -/'(0) =
=
-/'(0)
In general, the derivative-sifting property can be stated symbolically as
*
«-(*-«)/(»)
J
J«<0
where/ <B+1) (0 exists over [a,
The generalized function
A -(-DT"'(«)
(B.36)
/3].»
d'(t)
is
pictorial representation of a doublet
sometimes called a doublet.
is
The
given in Fig. B.6.
Other properties of the
unit impulse
Dirac and others have obtained a host of identities concerning the unit
impulse. We will merely give these here without proof.
- ,5(0
d'(-t) - -<5'(0
t <3(0 = o
td'(t) = -d{t)
<5(-o
-a*) = i \a\-1 {d(t f(t)6(t-a)=f(a)d(t-a)
d(t*
(3)
(4)
(5)
a)
+
d(t
+
a)}
(6)
(7)
these properties are obtained through the inner product
with a testing function
•
(2)
m
diat)
The proofs of
(1)
The condition on/'" +1
<f>(t)
'
€ C.
is sufficient,
but not necessary.
appendix
C
Elements of complex variables
AND OPERATIONS
CI ELEMENTARY DEFINITIONS
A complex variable z is a pair of real variables (x, y) written as
z = x+jy
(CI)
where j can be thought of as v — 1.
The variable x is called the real part of z, and y is the imaginary part of z.
Written in simpler notation, we have
z
The
=
Re(z),
variable z can be plotted
y
= Im(z)
(C.2)
on a pair of rectangular coordinates. The
abscissa represents the x or real axis, and the ordinate represents the y or
imaginary axis. The plane upon which x and y are plotted is called the
complex plane. Any point on the complex plane, such as z = 3 +j2, can
be represented in terms of its real and imaginary parts, as shown in Fig. C. 1
From the origin of the complex plane, let us draw a vector to any point z.
The distance from the origin to z is given by
|x|
and
is
=
known
(x*
+
y»)*
(C.3)
as the modulus of
angle which the vector subtends
as the argument of z or
arg z
=
tan
=
-i*
V
x
is
The
z.
known
3 *
(C.4)
=
\z\,
we can
arg z and r
Letting 6
represent z in polar coordinates as
z
=
re»
(C.5)
481
FIG.
CI
Network
482
Expanding
anal/sis
this last
and synthesis
obtain
+jr sin 6,
z == r cos 6
(C.6)
= r cos 8
y = r sin
x
so that
The
we
equation by Euler's formula,
two complex numbers
rule for addition for
(a
where j* = — 1.
given as
(C.8)
+ jb)(c +jd) = (ac - bd) + j(ad + be)
(C.9)
If
we
+jd)
(c
When two complex numbers
(a
=
is
+ d)
+jb)
+
(C.7)
(a
+
are multiplied,
+j(b
c)
we have
express the complex numbers in polar form,
we
obtain
m
(c+jd) = r^ m
+ jb) =
(a
and
When we
we
(C.ll)
multiply the two numbers in polar form, then
r^rtf"*
If
(CIO)
ri e
divide these
two numbers
=
jv^**-
-*'
1
(C.12)
in polar form, then
7^ = 7^'^
(C13)
In rectangular coordinates, the operation of division can be expressed as
a+jb _ (a+ jb)(c - jd)
c+jd (c+jd)(c-jd)
ac + bd
.be — ad
J
c + d*
c* + d*
In connection with the modulus of a complex number,
(C.14)
it is
useful to note
the following rules:
l«i«2l
1*1*1*1
z-z*
where z*
is
= Uil
=
= |z|»
|«il
•
tal
•
l*i*l
=
the complex conjugate of z and
z*
is
2
l*il
(CIS)
defined as
= x +jy = x —jy
(C.16)
Appendix
C
483
The following rules deal with operations involving the conjugate definition
Zi
+ zt = zx * + z,*
^=
Finally, if z has
*!*•*.*
(C.17)
a modulus of unity, then
z
-1
(C.18)
z*
The operations of raising a complex number to the nth power, or taking
the nth root of a complex number, can be dealt with most readily by using
n
the polar form of the number. Thus, we have z B
(re1")"
r eSnB ,
=
and
z
= jMne sn»+tk,)M
=
k0l
n
_
(C
l
19)
ANALYSIS
C.2
If to each z
is
1/n
= x +jy, we assign a complex number w = u +jv, then w
a function of z or
w =/(*)
The following are examples of complex
(C.20)
functions,
i.e.,
functions of a
complex variable:
= 2z
w = log, z
w = 1/z
w = za + 4
w=
w
(C.21)
|z|
We
see that
w may
be complex, pure
real,
or pure imaginary, depending
upon the particular relationship with z. In general, the real and imaginary
parts of w are both functions of x and y. That is, if we let w = j + jv,
then
u-A*,V)
As an example,
let
we
v=f(x,y)
us find u and v for the function
w=
Simplifying,
and
z*
+4=
(x
+jy)*
+4
w
(C.22)
= z* + 4.
(C.23)
obtain
w
= (*« - y* +/'2ry) + 4
(C^4)
Network
484
and synthesis
analysis
jy
jy
Ax
J&
z
+ Az
Pathl
+ A?
Path 2
-jy
-jy
FIG. C.2
u
so that
The
«
derivative of a
=
FIG. C.3
a:*
— yl +
4
and
complex function /(z)
= 2«y
i>
is
(C.25)
defined as
(C.26)
If
one
restricts the direction
we have what
function
is
is
known
or path along which
Az approaches
zero, then
as a directional derivative. However, if a
to possess a derivative at
all,
complex
the derivative must be the
same
any point regardless of the direction in which Az approaches zero. In
other words, in order for/(z) to be differentiable at z = z we must have
at
,
= constant
(C.27)
dz
for all directions of approach of Az.
Consider the two directions
and C.3. For path 1, we have
/'(z)
in
=lim
which Az approaches zero in
,/(z
lim'
A*->0 Av->0
If
we
+ Az)-/(z)
Figs.
C.2
(C.28)
Az
substitute
Az
into Eq. C.28,
f'(z)
we
= Ax + j Ay
(C.29)
obtain
+ A* +j(y + Ay)] -f(x+jy)
As + j Ay
/(z) = u +jv
f(z + Az) = u + Aw +j(v + A»)
= lim
lim
f[x
(C.30)
Aac-»0 Aff-*0
Since
and
(C.31)
(C.32)
Appendix
we
C
485
finally arrive at
/'(«)
Am +jAi>
= lim
lim
a»-.o Av->o Aa;
+ j'Ay
—=—+
-*/Ao
= hm Au +
Ax
a*-»o
,.
For path
2,
9u
,
(C.33)
.
j
dx
9»
—
dx
we have
Au + Ad
Km " T J A
/
/'(z)=lim
a»-»oa«s-oAx
==lim
+ J Ay
AiL±M£
(C.34)
/'Ay
a»-o
dt>
.
dy
du
dy
we assume that the function /(z) is differentiable, the
must be independent of path. Thus, we have
Since
— _•— = —
dy
dy
From this
last equation,
we
4-
/
—
derivatives
(C 35)
9x
dx
obtain the Cauchy-Riemann equations, which
are
dv
_
du
dx
dy
du__dv
dx
dy
We have just seen that in order for a function to have a derivative, the
Cauchy-Riemann equations must hold. A function which is single valued
and possesses a unique derivative is called an analytic function. A set of
Cauchy-Riemann equations
For example, consider the function
sufficient conditions for analyticity is that the
are obeyed.
/(z)
/(z)
is
= z* + 4
(C.37)
analytic because
^ = 2x = ^
dy
9
dx
— = —2
= ——
dy
(C.38)
dx
.
Network
486
On the
anal/sis and synthesis
= z* is not analytic because
other hand,/(z)
=
u
x
— = +1
du
j.,
and
v
a
and
*>
SINGULARITIES
AND
(C39)
= -1
—
ay
ox
C.3
= —y
1
RESIDUES
If/(z) is analytic within a region or domain in the
complex plane except
then/(z) has an isolated singularity at z
Suppose /(z) has a
singularity at z , then we can expand /(z) about z in a Laurent series
at a point z
,
.
- z )B
(z
z
-z
+
Om(2
- So)"* +
*
•
*
(C.40)
In the expansion,
term a_i
is
if
m is finite,
then z
is
called a pole of order
m.
1
The
called the residue of the singularity.
Example C.l.
Consider the Laurent series for the function /(z)
We can expand e* in a power series to give
= e*\z
about
the pole at the origin.
—
*=
z
I/i
z\
+* +
!«•
)
+ -*»+
2!
3!
/
(C.41)
= l+l + Lz+l.z* + -z
According to the
2!
3!
Example C.2. Expand the function /(z)
and find the residue of the pole at z = 1
1
1
z(z-l)«
(z
-
1
Note
then *o
is
|z
—
1|
that if
an
<
,
(z
—
1.
l/z(z
—
l)2
equal to
1.
about the pole at
z
=
1,
[1
1)
- (z -
1)
+ (z -
1)»
- (z -
I)*
+
•
•
•]
-
1)»
+
•
•
•
is
equal to —1.
(C.42)
1)»
1
1
<
=
is
1
(«
- i)M + (z 1
for
=
of the pole at z
definition, the residue
„,
1)*
z
—
,
+
1
- (z -
1)
+
(z
1
Here, the residue of the pole at z
we have an
infinite
essential singularity.
=
number of nonzero terms with
1
negative exponents,
Appendix
Example C3.
=
Find the residues of the poles at s
and s
=
C
-1
487
of the
function
s
To find
we simply perform a
the residues,
«x
2
3
(C.43)
partial fraction
expansion
3
1
(C.44)
Thus the residue of the pole at
s = -lis +3.
C.4
+2
s =»
is
—3, and the residue of the pole at
CONTOUR INTEGRATION
In complex integration the integral is taken over a piecewise smooth path
C and is defined as the limit of an infinite summation
IC
f(z) dz
= lim 2 f(z
n-»oo j— 1
i)
A*/
(C.45)
where zt lies on C. Unlike the process of differentiation, the path along
which we take the integral makes a difference as to the ultimate value of
the integral.
Thus the
integral
r f(z)dz
(C.46)
upon whether we choose to inteshown in Fig. C.4. If we integrate along
in general, has different values depending
Q
or path C8 as
a closed path, say from a to b and then to a again, we are integrating along
a closed contour. The path shown in Fig. C.5 is an example of a closed
contour. The following theorem, known as Couch/ s residue theorem gives
grate along path
,
a method for rapid evaluation of integrals on closed paths.
^
Jy
jy
Closed
I
£.
'contour
X
FIG. C.4
Theorem C.l.
If
FIG. C.5
C is a simple closed curve in a domain
analytic except for isolated singularities at
z^, z,,
...
,
z„,
D, within which /(z) is
then the integral along
Network
488
the closed path
anal/sis
C is
I
where
K
t
and synthesis
f(z)dz
= 2n/(JKi +K. + ---+KJ
(C.47)
represents the residue of the singularity z ( .
Example C.4.
Consider the integral
s+2 ds
§ sH.s + 1?
along the
within the
circle
\s\
(C.48)
= 2, as given in Fig. C.6. Since there are two singularities
= and at s = —1, whose residues are respectively —3
circle, at s
and +3, then the integral along the circle
S
i
s\s
is
+ 2 ds - 2*j(-3 +
+ iy
3)
=
(C.49)
C=|.|-2
Example C.5.
C.7.
The
Find the
function /(«)
integral of f(s) along the closed contour
is
35+5
IX* +
'*"-(* +
in Fig.
(C.50)
2)
Jy
'
'
L
—
shown
given as
-2 -1
X
-Jy
FIG .
C7
Appendix
C
489
A partial fraction expansion of f(s) shows that
m —ri
+
+ + Txi
1
s
so that the residues are
1
and 2. The value of the
path within which both the
singularities lie is
j> /(s) ds
If f(z) is analytic in
along any closed path
=
iTTJil
a domain with no
is
zero, that
is
known
2)
integral along the closed
then
= 6tt/
singularities,
(C.52)
then the integral
is,
f(z)dz
if
This result
+
(C51)
2
s
=
as Cauchy's integral theorem.
(C.53)
D
Proofs of some theorems on
appendix
positive real functions
Theorem D.l.
If Z(s)
and W(s) are both
positive real, then Z(fV(s))
is
also
positive real.
When Re s £ 0, both Re Z(s) and Re W(s) ^ 0, then Re Z( W(s)) £ 0,
When s is real, both Z(s) and ff%s) are real, hence Z(W(s)) is real. Since
Proof.
also.
Z(W(s))
satisfies
Theorem D.2.
If Z{s) is positive real,
W(s)
Proof.
Theorem D.3.
Z(s)
Proof.
both conditions of positive realness,
=
1/* is positive real,
it is
positive real.
then Z(\/s) is positive real.
hence Z(tV(s)} = Z(l/s) is positive
real.
If lV(s) is positive real, then l/fV(s) is also positive real.
=
l/s is positive real,
=
hence Z(W(s))
l/fV(s) is positive real
by
Theorem D.l.
Theorem D.4. The sum of positive real functions is positive real.
Proof. Suppose Zx(s) and Z2(s) are both positive real. When Re s
ReZx k
then
ReZ2 ^
ReZj + ReZa = ReZ ^
so that
is
and
^ 0,
0.
Also, when s is real, both Zx and Z2 are real. The sum of two real numbers
a real number. Therefore, Zx + Zt is positive real.
Theorem D.5.
(i.e., lie
poles
and zeros of Z(s) cannot have
positive real parts
m
the right-half plane.
is a pole s
expansion about s so that
Suppose there
Proof
Laurent
The
in the right half of the s plane).
series
(*
- 5o)n
(S
- * )"
*
490
Let us
make a
Appendix
D
491
ReZ<»)
FIG. D.I
where n is real and
approximated by
finite.
In the neighborhood of the pole S& Z(s) can be
Z(s)
»,-^V(s - s n
)
We can represent Z(s) in polar form by substituting each term by its polar form;
(j - «o)+" = rV»» and k_n = Kt** so that
i.e., let
z<*) = ^ **~**.
Re Z{s) =
* cos (* -
nfl)
which is represented in Fig. D. 1 When 9 varies from to lit, the sign of Re Z(j)
change 2n times. Since ReZ(s) t.
when Re s 2. 0, it is seen that any
change of sign of Re Z(s) in the right-half plane will show that the function is not
positive real. Therefore, we cannot have a pole in the right-half plane. Since the
.
will
function 1/Z(s)
is
positive real if Z(s) is positive real,
it is
obvious that there
cannot be any zeros in the right-half plane also.
Theorem D.6.
Only simple poles with
positive real residues
can
exist
on the
ya> axis.
As a consequence of the
may exist on they'd) axis if n =
Proof.
poles
that the pole
and real.
is
Proof.
real
when
If
s
1,
and ^
simple and the condition
It is readily
seen that zeros
The poles and
Theorem D.7.
derivation of
on
=
seen that
it is
1
implies
implies that the residue is positive
they' to axis
must also be simple.
zeros of Z(s) are real or occur in conjugate pairs.
a complex pole or zero
is real.
<j>
Theorem D.5,
= 0. The condition n —
exists without its conjugate, Z(s)
As a result of this theorem and Theorem D.S,
cannot be
it is
seen that
both the numerator and denominator polynomials of Z(s) must be Hurwitz.
The highest powers of the numerator polynomial and the
denominator polynomial of Z(s) may differ by at most unity.
Proof. Let Z(s) be written as
Theorem D.8.
+ fln-iJ"-1 +
m + bn.lS"
*>m*
««?"
Z{s)
•
•
•
+ OyS + a.
+
+b
V
Pis)
Q(s)
Network
492
anal/sis
and synthesis
FIG. D.2
m — n ^ 2, when s = », Z(«) will have a zero of order 2 or greater at s = oo,
which is on the/w axis. Similarly, if n - m £ 2, then, at s = », Z(s) will have a
pole of order 2 or more at * = oo. Since Z(s) cannot have multiple poles or
zeros on the/a> axis, these situations cannot exist; therefore \n — m\ <.\.
If
Theorem D.9. The lowest powers of P(s) and Q(s) may differ by at most unity,
Proof. The proof is obtained as in Theorem D.8 by simply substituting 1/s
for s and proceeding as described.
Theorem D.10.
A rational function F(s) with real coefficients is positive real if:
(a) F(s) is analytic in the right-half plane.
(b) If F(s)
has poles on
they'a> axis,
they must be simple and have
real, positive
residues.
(c)
Re F(ja>) ^
for all
o>.
We need only show that these three conditions fulfill the same requirements as Re Z(s) ^ for Re s ^ 0. We will make use of the minimum modulus
Proof.
states that if a function is analytic within a given region, the
min imum value of the real part of the function lies on the boundary of that
region. The region with which we are concerned is the right-half plane which is
bounded by a semicircle of infinite radius and the imaginary axis with small
indentations for theyVu axis poles. If the minimum value on theya) axis is greater
than zero, then Re Z(s) must be positive over the entire right-half plane (Fig. D.2).
theorem which
appendix
An
aid to
of
E.I
E
the improvement
filter
approximation
INTRODUCTION
The introduction of an additional pole and zero in the second quadrant
of the complex frequency plane, and at their conjugate locations, can give
amplitude or phase corrections to a filter approximant over some desired
band of frequencies without significantly changing the approximant at
other frequencies. However, a cut-and-try procedure for finding the best
positions for such a pole-zero pair can be tedious.
visual aid is presented
herein which reduces the amount of labor required to make modest
corrections of this type.
Constant phase and constant logarithmic gain contours for the correction by a pole-zero pair1 are plotted on transparent overlays. One of these
may be placed over a suitably scaled sheet of graph paper representing the
complex frequency plane. Then the pair-shaped phase and gain corrections along the jm axis are indicated by the intersections of the overlay
A
contours with this axis. Corrections which best reduce the errors in the
original approximant are then sought by variation of the overlay position
and orientation. Either phase or amplitude may be corrected. However,
it is not always possible to simultaneously improve both the phase and the
amplitude characteristics of an approximant by a single pair-shaped
correction.
F. F.
Kuo and M.
CT-9, No.
1
4,
Karnaugh, reprinted from the
December
IRE Transactions on Circuit Theory,
1962, pp. 400-404.
This will be called, hereafter, a pair-shaped correction.
•493
Network
494
E.2
analysis
and synthesis
CONSTANT LOGARITHMIC GAIN CONTOURS
Suppose we begin with a transfer function G(s) with certain deficiencies
its amplitude or phase response. Let us consider the transfer function
of a corrective network G^s) such that the product
in
GAs)-GM<K»)
will
CE.1)
have better gain or phase characteristics. For the purposes of
we will restrict G^s) to have the form
this
paper,
O^-C <'-«)<'-«>
p)(s
(s
(E.2)
p)
where C is a constant, and q and p are a zero and a pole, respectively, in
the second quadrant of the complex frequency plane. If the correction
—p
is to be applied at sufficiently high frequencies such that s — qo*s
then
GMcsZtzA
(E.3)
s-p
Let us consider the effect when the pole-zero pair in Eq. E.3 is used to
augment any given rational function in the complex frequency plane.
The added
gain, in decibels,
D=
where we have neglected the
then
For
due to
20 log10
effect
fc
=
k
=
-q
s- p
s
of the constant
(lO)*'
20
E.3. If
we let
(E.5)
a
circle
with inverse points* at q and
is
kjp^q\
II
its
C in Eq.
(E.6)
P
and
(E.4)
s-p
fixed k, this is the equation of
p. Its radius
this pole-zero pair is
center
is
- *"l
(E7)
at
Press, Oxford,
*
E. C. Titchmarsh, The Theory of Functions, Oxford University
England, second edition 1939, pp. 191-192.
Appendix E
495
Let
(E.9)
Then
P
—
-
— £5©
II
(E.10)
k*\
(E.ll)
Here, s
is e. It is
is the midpoint between the pole and zero, and their separation
easy to see from Eq. E.l 1 that the center of each circle of constant
gain
is externally collinear with the pole and zero.
For the purpose of
drawing the family of constant gain contours, we may let s be the origin
of coordinates and the scale factor e may be set equal to unity.
Furthermore, only half the pattern need be drawn because the function
D has negative symmetry with respect to a reflection about the perpendicular bisector of the pole-zero pair.
E.3
CONSTANT PHASE CONTOURS
Figure E.la represents a pole at/>, a zero at q and an arbitrary point
complex frequency plane. When the pole and zero are used to
correct a given phase characteristic, the added phase at s is
s in the
4>
-
a
-
9,
(E.12)
where BQ and 6 9 are measured with respect to a single arbitrary reference.
In Fig. E.l A, a circle has been drawn through p, q, and s. Angle ^ is
FIG. E.l. Derivation of constant phase contours.
Network
496
and synthesis
analysis
equal to one half the subtended arc ps'q. Therefore, the arc qsp is a
constant phase contour. The angle at the center of the circle between cp
and the perpendicular bisector of chord pq is also equal to
All of the
<f>.
circular contours of constant phase have their centers on this perpendicular
bisector.
Note that minor arc ps'q
is
also a contour of constant phase, but the
phase angle
«
is
negative, as indicated
=
-
<f>
*
(E.13)
by the clockwise rotation from s'p to s'q. The
is herein taken to be counterclockwise.
convention for positive rotation
E.4
CONTOUR DRAWINGS
Sets of constant phase
drawn
in Figs. E.2
and
and constant logarithmic gain contours are
The curves are symmetric about the zero-
E.3.
decibel line, except that the gain curves are of opposite sign.
Therefore,
only one half of each figure has been drawn.
V
7.5*
«=^5*
-K s
~~yf
1
/i?\
9'
v>
I
10*
^.s\
lb"
30°
i
Ills
2\
1.71 1
5|
]l
4
|] .2
\
1
••
°-
0.7
0.6
,
'
1
/
TO*
/
20°
'
lb"
10"
9*
7.5"
^v
*Sf
v^jy
-J'l
J
/
rV.
FIG. E.2. Constant amplitude and phase contours.
/
/
db
0.5
Appendix E
497
FIG. E.3. Constant amplitude and phase contours.
The
zero-decibel line
is
joining the pole and zero.
the perpendicular bisector of the line segment
It is a gain contour of infinite radius.
through the pole and zero is a zero phase contour. Together,
form a useful reference system. The signs of the
phase and gain in the four quadrants formed by these axes are shown in
The
line
these perpendicular axes
Fig. E.4.
The corrections in
Figs. E.2
and E.3 do not carry algebraic signs because
Network
498
analysis
and synthesis
Zero gain
.1
G-
G+
<t>-
-x^-*-Zero phase
P
G-
G+
<t>+
FIG. E.4. Signs of the gain
G and phase
<f>
corrections.
only half of the symmetrical pattern has been drawn. The signs may be
obtained from Fig. E.4, and are important in selecting the orientation of
the pole-zero pair.
E.5
CORRECTION PROCEDURE
In correcting a given approximant, it is first necessary to plot the desired
magnitude and/or phase corrections versus frequency. When the contours
of Fig. E.2 or E.3 are overlaid on a second sheet representing the complex
frequency plane, the contour intersections with the ja> axis indicate the
corrections that will actually be realized.
There are several variables at the designer's disposal. The first is the
distance between the correcting pole and zero. Since the scale of the
contours in Figs. E.2 and E.3 is not specified, the frequency scale of
the underlying complex plane determines the distance between the pole
and zero. The second design variable is the location of the center of the
pole-zero pair. The distance of the center from a given band on theyeo
axis determines the
magnitude of the correction.
is the orientation of the pole-zero pair. Figure E.4
shows how the orientation affects the gain and phase corrections. As a
simple example, suppose one wishes to have zero phase correction at
co = 1.0, negative phase correction above and positive correction below
that frequency. The attack would be to point the zero phase axis at
co = 1.0 with the pole nearest to they'to axis. If it is desired to have equal
phase correction above and below co = 1.0, the zero phase axis should be
oriented parallel to the real axis of the complex frequency plane. If one
wishes to have more phase correction above co = 1.0 and less below, the
The
third variable
Appendix E
499
zero phase axis should be rotated clockwise with respect to the a (real)
axis.
We thus see that by varying the frequency scale of the complex frequency
plane, the position of the center of the pole-zero pair,
and the orientation
of the pole-zero pair, the pair-shaped correction can be made to approxi-
mate the desired correction.
It must be emphasized that the method suggested herein is an aid to
cut-and-try correction. As such, it is easier to use the method than to
precisely set down rules for applying it. However, a few rather general
may be helpful.
Unless the pole and zero both lie on the real axis, one must remember
that another pole and zero are located at conjugate positions. The contributions from both pole-zero pairs may be added algebraically. In most
practical cases, the desired correction will have a band-pass character.
statements
Therefore, only one pole-zero pair will normally contribute significantly
at
any frequency.
The shape and magnitude of the
way
The broadness
desired correction will dictate the
intersect the correction contours.
which the/co axis must
of the desired correction will dictate the proper scaling of the jco axis.
Usually, only a few trials are needed to fix the pole and zero locations for
in
the best
fit.
be found that a worthwhile correction can be made in either the
phase or the gain characteristic. Only fortuitously can they be improved
simultaneously by a single pair-shaped correction.
It will
Example E.1. The amplitude response of a third-order Butterworth filter is
given by the solid curve in Fig. E.5. It is desired to steepen the gain roll-off
near the cutoff frequency «o c = 1 This is done by increasing the gain just below
o) = m e and decreasing it above that frequency. Figure E.6 illustrates the type
of correction desired. Figure E.7 shows a pole-zero pair that achieves this type
of correction. The gain at to = 1 remains unchanged by this particular choice.
Other pole-zero pairs that "aim" the zero-decibel line at to = 1 but give asymmetrical corrections about that point might also be used. The dashed curve in
Fig. E.S shows the corrected gain.
.
A
different pole-zero pair, also shown in Fig. E.7, has been chosen to minimize
the deviation of the slope of the phase response from its slope at to = 0. This
pair-shaped correction decreases the phase for m < 0.7 and increases it for
co
> 0.7.
Figure E.8 shows the deviation of the phase responses from linear phase.
It is clear from Figs. E.5 and E.8 that the gain corrected approximant has a
poorer phase response than the original, while the phase corrected approximant
has a gentler gain roll-off than the original.
500
Network
anal/sis
and synthesis
2
0.2
0.5
0.3
Frequency,
0.7
1.0
1.5
2.0
u
FIG. E.5. Amplitude response of corrected and uncorrected Butterworth
This will not surprise the experienced
filter
designer.
It is possible,
filters.
however,
both gain and phase by using two pairshaped corrections. A useful approach to this objective lies in localizing the
gain correction further out of band, and the phase correction further in-band
than in the separate corrections just discussed. This can be done by shifting
the pole-zero centers to higher or lower frequencies and also by experimenting
with nonsymmetrical corrections.
to achieve moderate corrections in
FIG. E.6. Amplitude correction to steepen
fall-off
of third-degree Butterworth
filter.
Appendix E
—
Loss
1.2
1
correction
o
1.0
Butterworth
_
poles
0.8
X o
Phase
0.6
correction
0.4
0.2
_L
J_
j«
-1.0 -0.8 -0.6 -0.4 -0.2
\
-0.2
\
\
\
\
-0.4
\
\
\
-0.6
\
x o
\
-0.8
o
-1.0
X
-1.2
FIG. E.7. Poles and zeros of the original
filter
and correction equalizers.
20
>
10
With >hase :orrect on
^
s -10
^'
/
Tl ird-orc er
^^v
/
//
/
//
/
/
/
zL
Biitterwo rvn'/
\\ 1
\
-20
>
/w th amp litude
\
\
correct on
\
-30
\
/
/
-40
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
Frequency, rad/sec
FIG. E.8. Phase deviation from slope at zero frequency.
501
Network
502
E.6
anal/sis
and synthesis
CORRECTION NETWORK DESIGN
A
few words are in order concerning the synthesis of the equalizer from
the pole-zero pair obtained by the method described.
In order to provide optimum power transfer and to facilitate cascading
the correction network with the original
should be of the constant-resistance type.
to the bridged-T network in Fig. E.9,
given as
the correction network
filter,
We will restrict
whose voltage
our discussion
transfer function
G(s)
provided the network
equation
is
is
(E.14)
of the constant resistance type as given by the
Z (j)Z2(i) =
1
For normalization purposes, we
will let
R
»
(E.15)
J
R =
1
Q.
Let the pole-zero correction be written in general as
G(s)
Since the correction has
=
2
+as+a
fc(s* + b lS + b
s
minimum
t
phase,
we know
addition, since the d-c gain cannot exceed unity,
We
(E.16)
)
kb
that {a ( , b t
^a
^ 0}.
In
.
can express the impedance Z^s) in terms of G(s) in Eq. E.14 as
Zl(s)
=
_L_ 1= (±
+ (kb, - ajs + (fcb„ ~,
T~,
s + a ts + a
l)s*
G(s)
Since Z^s) must be positive real, the coefficients must
so that
all
a„)
(Ai')
be nonnegative
k- ^0
1
kbi
kb
—
—
<*i
a
^
^
(E.18)
Zi
So
Vi
So
So
Z2
So
FIG. E.9. Bridged-T network.
v2
Appendix E
503
-ju
FIG. E.I0. Poles and zeros of Zt(s).
Moreover, in order for a biquadratic driving-point immittance to be
positive real, the following condition
Details concerning the synthesis of
Let us plot the pole-zero pair of
must apply. 8
Z^j) are
also given in Seshu's paper.
Zt{s) as in Fig. E.10. We can represent
<£„ and 0,
Rack* has shown that in
the locations of the poles and zeros in terms of the polar angles
and
their distances
from the
order for in-band loss to be
origin,
n p and n t
.
finite
^<2
(E.20)
In addition he has shown that for an unbalanced bridged-T circuit, if
a = max [0M , <f>9] then the larger the angle a, the larger the in-band loss.
In particular, « should be less than 70° to restrict the in-band loss to
reasonable proportions.
If
one considers other network configurations
*S. Seshu, Minimal Realizations of the Biquadratic
December 1959, 345-350.
Trans, on Circuit Theory, CT-6,
4
A.
J.
Rack, private communication.
Minimum
Function,
IRE
Network
504
analysis and synthesis
such as lattice networks, it is conceivable that Rack's restrictions might
5
be relaxed. For lattice network design, Weinberg's method is applicable,
although here again, the in-band loss restrictions are severe.
E.7
CONCLUSION
We
have presented here a simple visual aid to the correction of the
amplitude or phase response of filters. The method has the advantage of
facilitating the commonly used cut-and-try approach to this problem. It
has the disadvantage, in many cases, of failing to provide simultaneous
amplitude and phase corrections in a single step.
1
L. Weinberg,
"RLC lattice
networks," Proc. IRE, 41, September 1953, 1139-1144.
Bibliography
SIGNAL ANALYSIS
Goldman,
S.,
McGraw-Hill,
Frequency Analysis, Modulation and Noise,
'
New
Javid,
York, 1948.
M. and
.
E. Brenner,
McGraw-Hill,
^w/y^w, Transmission and Filtering of Signals,
New York,
1963.
Lathi, B. P., Signals, Systems
New
and Communications, John Wiley and
5>ons,
York, 1965.
Lighthill,
Functions,
M. J., Introduction to Fourier Analysis and Generalized
Press, New
York, 1958.
Cambridge University
Circuits, Signals and
Mason, S. J. and H. J. Zimmerman, Electronic
1960.
Systems, John Wiley and Sons, New York,
McGraw-Hill, New
Applications,
Its
and
Integral
Papoulis, A., The Fourier
York, 1962.
Rowe, H.
E., Signals
and Noise
in
Communications Systems, D. van
Nostrand, Princeton, N.J. 1965.
Modulation and Noise,
Schwartz, M., Information Transmission,
Hill, New York, 1959.
McGraw-
NETWORK ANALYSIS
Allyn and Bacon, Boston,
Balabanian, N., Fundamentals of Circuit Theory,
Systems, Addison-Wesley,
Bohn, E. V., The Transform Analysis of Linear
Reading, Massachusetts, 1963.
Brenner, E. and
M. Javid, Analysis of Electric Circuits, McGraw-Hill,
York, 1959.
505
New
Network
506
analysis and synthesis
Brown, R. G. and J. W. Nilsson, Introduction to Linear Systems Analysis,
John Wiley and Sons, New York, 1962.
Brown, W. M., Analysis of Linear Time-Invariant Systems, McGraw-Hill,
New York, 1963.
Carlin, H. J. and A. Giordano, Network Theory, Prentice-Hall, Englewood
Cliffs, N.J., 1964.
W.
Cassell,
L.,
Linear Electric Circuits, John Wiley and Sons,
New York,
1964.
Chen,
W.
H., The Analysis of Linear Systems, McGraw-Hill,
New
York,
1963.
dePian, L., Linear Active Network Theory, Prentice-Hall, Englewood
Cliffs, N.J. 1962.
Friedland, B., O. Wing,
and R. B. Ash, Principles of Linear Networks,
McGraw-Hill, New York, 1961.
Gardner, M. and J. L. Barnes, Transients in Linear Systems, Vol. 1, John
Wiley and Sons, New York, 1942.
Guillemin, E. A., Introductory Circuit Theory, John Wiley and Sons,
New
York, 1953.
Guillemin, E. A., Theory ofLinear Physical Systems, John Wiley and Sons,
New York,
1963.
Harman, W. W. and D. W.
McGraw-Hill,
New York,
Lytle, Electrical
and Mechanical Networks,
1962.
W. H., Jr. and J. E. Kemmerly, Engineering Circuit Analysis,
McGraw-Hill, New York, 1962.
Huelsman, L. P., Circuits, Matrices and Linear Vector Spaces, McGrawHill, New York, 1963.
Kim, W, H. and R. T. Chien, Topological Analysis and Synthesis of Communication Networks, Columbia University Press, New York, 1962.
Ku, Y. H., Transient Circuit Analysis, D. Van Nostrand, Princeton, N.J.,
Hayt,
1961.
Legros, R. and A. V.
New
W.
R. and
York, 1952.
LePage,
Martin, Transform Calculusfor Electrical Engineers,
J.
Prentice-Hall, Inc.,
Englewood
S. Seely,
Cliffs, N.J., 1961.
General Network Analysis, McGraw-Hill,
Ley, B. J., S.G. Lutz, and C. F. Rehberg, Linear Circuit Analysis,
Hill,
New York,
Lynch,
W. A. and
Hill,
New York,
McGraw-
1959.
J.
G. Truxal, Introductory System Analysis, McGraw-
1961.
Paskusz, G. F. and B. Bussell, Linear Circuit Analysis, Prenctice-Hall, Inc.
Englewood
Pearson, S.
I.
and Sons,
Cliffs, N.J., 1964.
and G.
J.
New York,
Maler, Introductory Circuit Analysis, John Wiley
1965.
Bibliography
Pfeiffer, P. E.,
Reza, F.
507
New York, 1961.
McGraw-Hill, New
Linear System Analysis, McGraw-Hill,
M. and
S. Seely,
Modem Network Analysis,
York, 1959.
Sanford, R.
S.,
Physical Networks, Prentice-Hall, Englewood
Cliffs, N.J.,
1965.
Schwarz, R. J. and B. Friedland, Linear Systems, McGraw-Hill, New York,
1965.
Scott,
R.
E., Elements
of Linear
Circuits,
Addison-Wesley, Reading,
Massachusetts, 1965.
Seely, S.,
Seshu, S.
Sons,
Dynamic Systems Analysis, Reinhold, New York, 1964.
and N. Balabanian, Linear Network Analysis, John Wiley and
New York,
1959.
H. H., Electrical Engineering Circuits, Second Edition, John
Wiley and Sons, New York, 1965.
Van Valkenburg, M. E., Network Analysis, Second Edition, Prentice-Hall,
Englewood Cliffs, N.J., 1964.
Weber, E., Linear Transient Analysis, John Wiley and Sons, New York,
Skilling,
1954.
Zadeh, L. A. and C. A. Desoer, Linear Systems Theory, McGraw-Hill, New
York, 1963.
NETWORK SYNTHESIS
Balabanian, N., Network Synthesis, Prentice-Hall, Englewood
Cliffs, N.J.,
1958.
Bode, H. W., Network Analysis and Feedback Amplifier Design, D. Van
Nostrand, Princeton, N.J., 1945.
Calahan, D. A., Modern Network Synthesis, Hayden, New York, 1964.
Chen, W. H., Linear Network Design and Synthesis, McGraw-Hill, New
York, 1964.
Geffe, P. R., Simplified
Modern
Filter Design,
John F. Rider,
New York,
1963.
Guillemin, E. A., Synthesis of Passive Networks, John Wiley and Sons,
New York,
1957.
Guillemin, E. A., The Mathematics of Circuit Analysis, John Wiley and
Sons, New York, 1949.
Hazony, D., Elements of Network Synthesis, Reinhold, New York, 1963.
Kuh, E. S. and D. O. Pederson, Principles of Circuit Synthesis, McGrawHill, New York, 1959.
Matthaei, G. L., L. Young, and E. M. T. Jones, Microwave Filters,
Impedance-Matching Networks and Coupling Structures, McGraw-Hill,
New York, 1964.
Network
508
Saal, R.,
analysis
and synthesis
Der Entwurf von
Tiefpdsse, Telefunken
Skwirzynski,
J.
Filtern mit Hilfe des Kataloges Normierter
GMBH,
1961.
K., Design Theory
and Data for
Electrical Filters,
D. Van
Nostrand, Princeton, 1965.
Storer, J. E., Passive
Truxal,
Tuttle,
J.
D.
New York, 1957.
New York, 1955.
1, John Wiley and Sons, New York,
Network Synthesis, McGraw-Hill,
G., Control System Synthesis, McGraw-Hill,
F.,
Network Synthesis, Vol.
1958.
Van Valkenburg, M.
E., Introduction to Modern Network Syntheses,
John Wiley and Sons, New York, 1960.
Weinberg, L., Network Analysis and Synthesis, McGraw-Hill, New York,
1962.
Yengst,
W. C, Procedures of Modern Network Synthesis, Macmillan, New
York, 1964.
Name
Eisenman, R. L., 469
Elmore, W. C, 390
Ende, F., 381
Aaron, M. R., 448
Aitken, A. C, 469
Angelo, E. J., 268
Ash, R. B., S06
Faddeeva, V. N., 469
Faddev, D. K., 469
Fano, R. M., 448
Feshbach, H., 86
Foster, R. M., 321
Franklin, P., 198
Friedland, B. O., 506
Fujisawa, T., 458
Fukada, M., 381
Balabanian, N., 505, 507
J. L., 506
Bashkow, T. R., 283, 438
Bellman, R., 469
Bode, H. W., 221, 507
Bohn, E. V., 505
Brenner, E., 505
Brown, R. G., 506
Brown, W. M., 506
Bubnicki, Z., 283
Budak, A., 395
Bussell, B., 506
Barnes,
Gantmacher, F. R., 469
W. H., Jr., 506
Geffe, P. R., 507
Giordano, A., 506
Goldman, S., 137, 505
Hayt,
Calahan, D. A., 507
Goldstone, L. O., 153
Guillemin, E. A., 296, 299, 506, 507.
428
Cassell, W. L., 506
Cauer, W., 324
Chen, W. H., 506
Chien, R. T., 506
Carlin,
H.
I.,
Darlington,
S.,
Harman, W. W., 506
W. H., Jr., 506
Hazony, D., 507
Hobson, E. W., 471
Hohn, F. E., 469
Huelsman, L. P., 261, 469, 506
Hayt,
431, 457
Davis, H. F., 51
de Pian,
L.,
506
Jahnke, E., 381
James, R. T., 389
Javid, M., 505
Desoer, C. A., 507
Dirac, P. A. M., 33, 470, 471
Dutta Roy,
S.
Index
C, 283
509
Name
510
Jones, E.
Jordan, E.
Index
M. T., 507
C, 413
Reza, F. M., 507
Rowe, H.
E.,
505
27
Justice, G.,
Saal, R., 441
Karnaugh, M., 493
W.
Kautz,
Sands, M., 390
H., 47, 367
Sanford, R.
Kemmerly, J. E., 506
Kim, W. H., 506
Ku, Y. H., 506
Kuh, E. S., 507
Kuo, F. F., 279, 393, 493
Larky, A.
Semmelman, C.
Lathi, B. P., 505
Le
Seshu,
Corbeiller, P.,
Lighthill,
R.,
M.
.,
506
65, 136, 471, 473,
Lutz, S. G., 506
Lynch,
Lytle,
W.
447
Skwirzynski,
506
J.,
L.,
503, 507
Skilling,
Leichner, G. H., 279
Ley, B.
S.,
H. H., 100, 104, 106, 181, 507
J. K., 508
So, H. C, 449, 450
Storer, J. E., 508
Szentirmai, G., 441, 453
469
Legros, R., 506
Le Page, W.
507
Schwartz, M., 505
Schwarz, R. J., 507
Scott, R. E., 507
Seely, S., 506, 507
262
I.,
S.,
Saraga, W., 443
Schwartz, L., 470, 471
A., 506
D. W., 506
Maler, G. J., 506
Marcus, M., 469
Martin, A. V. J., 506
Mason, S. J., 505
Matthaei, G. L., 507
Mine, H., 469
Morse, P. M., 86
Nering, E. D., 469
Nilsson, J. W., 506
505
Tellegen, B. D. H., 272
Temple, G., 34, 471
Terman, F. E., 236
Thomson, W. E., 384
Titchmarsh, E. C, 494
Tompkins, C. B., 447
Tropper, A. M., 469
Truxal, J. G., 506, 508
Tuttle, D. F., 508
Ulbrich, E., 441
Jr., 293
Valkenburg, M. E., 135, 263, 296,
Valley, G. E.,
Van
297, 325, 329, 331, 344, 347, 376,
O'Meara, T. R., 283
Orchard, H. J., 385
A. C, 291
507, 508
Von
Weiss, H., 469
Papoulis, A., 379, 381, 505
Walker, F., 283
Wallman, H., 293
Paskusz, G. F., 506
Weber,
Paley, R. E.
506
Pederson, D. O., 507
Perlis, S., 469
Pfeiffer, P. E., 507
Pipes, L. A., 469
Pearson, S.
Rack, A.
J.,
I.
E., 507
Weinberg, L., 400, 435, 504
Widder, D. V., 136
Wiener, N., 291
Yengst,
W. C, 508
Young,
L.,
507
503
Raisbeck, G., 8, 294
Rehberg, C. F., 506
Zadeh, L. A., 507
Zimmerman, H., 505
Subject Index
ABCD
parameters, 262
Admittance, driving-point,
All-pass network, 221, 357
15,
187
Cauchy-Riemann equations, 485
Cauer ladder expansion, 324
Causality, 9, 290
Characteristic equation, 77
Amplifier, 11
impulse response of, 201
Amplitude response, 212, 342
Characteristic impedance,
computer program for, 450
evaluation by vector method, 215
Amplitude spectrum, 3
Analytic function, 485
Approximation problem, 17, 365
Available gain, 418
Available power, 418
Cbebyshev
414
Characteristic value, 78
filter,
373
approximation, 366
locus of poles, 376, 378
tables of, 373, 400, 435
transient response, 392
Chebyshev polynomial, 373
Circuit, bridged-T 258, 275,
502
double-tuned, 238
Band-elimination
filter
255
229
symmetrical, 259
reciprocal, 8,
transformation,
single-tuned,
409
Band-pass filter transformation, 407
Bandwidth, half-power, 235
spectral, 67, 389
Bessel
filter,
383, 435
phase response of, 387
tables of,
Coefficient of coupling, 122
Compensation theorem, 181
Complementary function, 82
variables, 481
Complex
analysis,
400, 435
483
484
487
differentiation,
392
Bessel polynomial, 386, 395
transient response of,
integration,
programs,
438,
439,
Biquadratic immittance, 305, 503
Computer
Black box, 15
441, 447 448, 450 453, 457
Constant-resistance network, 352, 502
Bode plots, 221
Bounded real function, 419
Break frequency, 225
Bridge circuit, 285, 354
Butterworth
filter,
368, 434, 500
amplitude response of, 369, 387, 394,
500
pole locus of, 371
step response of, 391, 394
tables of, 372, 400,
Contour
18,
487
integration,
Controlled source, 268
Convergence in the mean, 49, 51
Convolution integral, 197
Crest factoT, 27
Critical coupling, 242
Current source, 12
Cutoff frequency, 17, 225
435
Damping
factor,
225
d-c value, 25
Canonical form, 325, 400, 433
Capacitor, 13, 103, 176
Cauchy integral theorem, 489
Cauchy residue theorem, 487
Delay, 32, 212, 245, 384
distortion,
time, 388
511
245
Subject Index
512
Fourier transform, phase spectrum, 64,
Delta function, see Unit impulse
Delta-wye transformation, 257
Differential equation, 75, 145
65
properties of, 67
forcing function of, 75
homogeneous, 75, 76
integrodifferential equation, 91,
145
76
nonhomogeneous, 75
ordinary, 76
linear,
simultaneous, 93, 146
Differentiator,
193
18,
11,
Digital computer, 18,
438
Distributions, theory of,
Dot reference
470
for transformer,
123
Doublet, 40, 479
Duhamel superposition integral, 201
Duty
cycle,
symmetry conditions, 68
Free response, 106
Frequency, angular, 2
complex, 4
domain, 14, 134, 367
forced, 192
natural, 192
Frequency normalization, 18, 402
Frequency transformation, 18, 404
26
Energy density, 71
Energy spectrum, 71
Equal ripple approximation, see Chebyshev approximation
Essential singularity, 486
Gain contours, logarithmic, 494, 496
Gaussian filter, 291
Generalized functions, 34, 64, 470
g parameters, 286
Green's function, 86
Gyrator, 272, 287
Hall effect, 273
High-pass filter transformation, 405
Hurwitz polynomial, 294, 316 347
Hybrid (h) parameters, 260
Hybrid matrix, 261, 449
Excitation, 1
Faraday's law of induction, 122
Filter approximation, 368, 373,
379,
Ideal low-pass
filter,
457
441, 457
Filter design, 17, 365, 397, 433,
computer programs
for,
driving-point,
Flux linkage, 122
Forced response, 106
transfer,
1,
46,
15, 253,
50
amplitude spectrum, 57
complex form, 55
cosine series, 53
evaluation of coefficients, 52, 58
188
Impulse function, see Unit impulse
Impulse response, 44, 85, 111, 194
Incident power, 416
Incidental dissipation,
13, 77,
Initial conditions,
phase spectrum, 57
Insertion loss, 429
Fourier transform, 1, 63
amplitude spectrum, 64, 65
inverse,
63
64
of unit impulse, 65
of unity, 68
86 107, 176
value theorem, 165
Initial
conditions, 53
276
Inductor, 13, 104, 176
orthogonality conditions, 49
discrete, 56,
315
matrix, 254
transformer, 263
Foster network, 321, 325
Fourier series,
424
Immittance, 15, 315
Impedance, 15, 176
Final conditions, 106, 109
Final value theorem, 165
symmetry
292, 367
Ideal transformer, 263, 273, 275,
383, 493
filter synthesis,
Insertion
power
431
ratio,
430
voltage ratio, 429
Integrator, 11, 18, 193
Integrodifferential equation, 91, 145
Kirchhoffs laws, 100, 104
513
Subject Index
Ladder network, 279, 346, 347, 453
Laguerre polynomial, 48
Laplace transform, 1, 134
definition of, 135
inverse,
136
Node equations,
Norm, 47
255
106,
Normalization, frequency, 402
magnitude, 402
Norton's theorem, 180, 185
properties of, 137
Optimization techniques, 447
Optimum (L)
486
L-C immittance, 315
properties of, 315
synthesis of, 319
series,
379
filter,
amplitude response of, 380
polynomials, 382
Least squares, principle of, 49, 366
Legendre polynomial, 381
Linear phase filter, see Bessel
Linear system, 8
derivative property,
parameters,
254, 343
uses of, 144
Lattice network, 285, 354, 503
Laurent
impedance
Open-circuit
table of, 168
filter
Orthogonal set, 47
Orthonormal set, 47
Overcoupling, 242
Overshoot, 388, 392
Paley-Wiener criterion, 291
9
Parseval's equality,
ideal elements, 12
50
theorem, 71
ideal models, 10
Partial fraction expansion, 148
MAC,
Project,
conjugate poles, 150
multiple poles, 151
448
Machine-aided design, 448
Magnitude normalization, 18, 402
Matrix algebra, 461
definitions, 462
operations, 464
references, 469
Maximally flat response, see Butterworth filter
Mean squared error, 48
Minimum
inductance
transformation,
443
Minimum modulus theorem, 492
Modulation, amplitude, 70
Monotonic
filter,
see
Optimum
filter
Multiple-access computer, 448
Mutual inductance, 122
real poles, 149
Particular integral, 82
Passive network, 8
Peaking
233, 241
circle,
Peak-to-valley ratio, 251
Phase contours, logarithmic, 495
Phase response, 212, 343
computer program for, 450
evaluation by vector method, 215
linearity of, 213, 383
Phase
shift, 1
distortion,
213
minimum, 220, 346
spectrum,
3,
57
Phasor, 5
Plancheral's theorem, 71
Negative impedance converter (NIC),
261
Network, analysis, 7, 100, 175, 253
linear, 8,
n-port,
449
100
reciprocal, 8, 100,
255
symmetrical, 259
synthesis, 290, 315, 341, 397, 431
time-invariant, 9,
Pole-zero diagram, 156
Port, 12, 253
Positive real
100
passive, 8,
Pole, definition of, 155
100
(p.r.)
function,
16, 299,
490
Power, average, 301, 315
available, 418
Propagation constant, 414
Proportionality, principle of, 8
Pulse transmission, 389, 392
514
Q,
Subject Index
Spectra, continuous, 3, 63
236
circuit, 233,
57
line (discrete), 3,
Ramp
Stability,
function, 31
R-C admittance,
R-C impedance,
properties of, 331
properties of, 325
290
marginal, 293
Steady-state solution, 106, 109
Steepest descent method, 447
synthesis of, 329
Realizability conditions,
342,
116,
16, 290,
301,
Step function, see Unit step
Step response, 44, 45, 85, 111, 194
490
for driving-point functions, 301
filter
for transfer functions, 342
Reciprocal network,
8,
255
Rect function, 65
Reference impedance factor, 415
Reflected parameter, 415
Reflected power, 416
Reflection coefficient, 416, 421, 458
Remainder function, 308
Residue, definition of, 162
evaluation by vector method, 162
Residue condition for two-ports, 344
Resistor, 13, 103, 175
Response, critically damped, 118
overdamped, 118
underdamped, 119
Ringing, 388
Rise time, 388
R-L admittance, properties of, 325
synthesis of, 329
R-L impedance, properties of, 331
synthesis of, 331
Sampled-data system, 142
Sampler, 18, 142
element, 416
R-L-C
functions, 333
transfer functions, 347, 352
System function,
14, 187,
admittance
circuit,
Signal, continuous,
decomposition
parts, 21
parameters,
248, 400
23
into
even
20
46
symmetrical, 21, 54
Signum (sgn) function, 30
Thevenin's theorem, 180
Time constant, 24
Time delay, 32, 212, 245, 384, 388
Time delayer, 11
Time domain, 14, 134, 366
Time-invariant system, 8
Transfer function,
14,
486
16,
188,
266,
341, 352
admittance, 188, 347
impedance,
Transformed
16,
188, 347
175
Transformer. 122. 177
ideal, 263
circuit,
filters,
388
and odd
common
emitter,
Transmission
coefficient, 421,
Transmission
line,
286
457
413, 425
Transmission matrix, 262
Transmittance, 16
Two-port, 12, 16, 254, 264
cascade connection, 271
equivalent circuits of, 268, 271
matrix representation, 264
parallel connection of,
series
273
connection of, 275
Sine function, 65
Singularity,
194
Terminals, 12
Transistor,
257, 344
periodic, 20,
functions, 325, 329, 331
functions, 325, 329, 331
Transient solution, 106
Settling time, 388
deterministic,
397
synthesis,
functions, 315, 319
unity coupled, 125
matrix, 420
Shunt peaked
L-C
R-C
R-L
Transient response of low-pass
Scattering parameters, 415, 419
Short-circuit
Synthesis procedures elementary, 308
Undercoupling, 241
515
Subject Index
Undetermined coefficients, method
g2
Uniform loading, 278
Unit coupled transformer, 121
of,
Unit impulse (delta) functions, 33, 43,
47(j
derivative of,
Voltage ratio, 16, 188
Voltage source, 12
Voltage standing wave
ratio,
428
y-parameters, 257, 344
see also Short-circuit parameters
40
properties of, 34, 36, 476,
function, 31
Unit ramp
Unit step function, 28, 474
derivative of, 34, 37, 475
480
z -parameters, 254 343
see also Open-circuit parameters
z-transform, 142
Zero, definition of , 155
Zero of transmission, 345
C
o
This Wiley international Edition
gram of paperbound
text
is
part of a continuing pro-
books especially designed
dents and professional people overseas.
reprinting of the original
hardbound
It
is
m
o
O
z
D
m
g
o
for stu-
an unabridged
edition,
which
is
also
available from your bookseller.
Wiley International Editions
include
titles in
z
the fields of:
Chemical Engineering
Civil
Engineering
Electrical Engineering
Psychology
Agricultural Engineering
Biological Science
Metallurgy
Chemistry
Mechanical Engineering
Earth Sciences
Business
Management
(A
and Agriculture
Mechanics
Statistics
3
o
in
>
2
O
Mathematics
Physics
c/>
C/>
W
JOHN WILEY & SONS, INC., U.S.A.
TOPPAN COMPANY LTD., JAPAN
WILEY
TOPPAN
[P 10)