ANALYSIS\n
NETWORK\n
ANALYSIS\n
By\n
M.
E.
VAN
VALKENBURG\n
Professor of Electrical
Engineering
\nUniversity
PRENTICE-HALL,
Englewood
of
Illinois\n
INC.\n
Cliffs, NJ.\n
PRENTICE-HALL ELECTRICALENGINEERINGSERIES
L.
\nW.
Ph.D.,
Everitt,
Editor\n
by
1955,
\302\251
Copyright,
INC.
\nPRENTICE-HALL,
\nEnglewood Cliffs,
N.J.\n
this
rights
reserved.
No part of
book\n
BE REPRODUCED
IN ANY FORM,
BY MIMEO\302\254
\nGRAPH
OR ANY OTHER MEANS, WITHOUT PER\302\254
\nMISSION
IN WRITING
FROM THE PUBLISHERS.\n
All
MAY
First
Library
of
\nCatalog
Card
Congress
No.:
Fourth
Fifth
June,
July,
printing
IN
June,
THE
UNITED
61082\n
STATES OF
1958\n
1958\n
November,
printing
1956\n
1957\n
August,
printing
Sixth printing
PRINTED
1955\n
November,
printing
Second printing
Third
56-5609\n
1969\n
AMERICA\n
PREFACE\n
\nnetworks
\nmatter
\nconcerned
\nequations.
\nit
to the
is designed for use as an introduction
of electric
study
the so-called pole and zero approach.
from
be divided
into four parts. (1) Chapters 1
are
may
with
definitions
and with the formulation
of equilibrium
The
first
contains
a discussion of approximation as
chapter
the
to networks,
between the network abstrac\302\254
relationship
book
This
relates
The subject
through3
the
and
\ntion
Here the reader
physical
system.
feet firmly on
is giventhe opportunity
the ground before he becomesinvolved
in
The
details of analysis.
elements are introduced,theirlaws
their
combination
into
networks
discussed. Writing of
\nformulated,
for networks
is treated in Chapter 3.\n
\nequilibrium
equations
to
do
with the solutionofequilibrium
4 through
8 have
Chapters
(2)
in general,
equations
by both classical
integrodifferential
\nequations,
the
transform
method.
\nand
Chapter 8 amplifies the relation\302\254
Laplace
the
time
domain
and
the
between
frequency domain.\n
\nship
These
first
topics classified under the
encompass
eight
chapters
of electric
circuits. In the remaining
of
transient
\nheading
analysis
is
in
this
exploited
unifying concepts of transient
background
\nchapters,
and
sinusoidal
response
by the use of the poles
\nresponse
steady-state
\nto
his
plant
\nthe myriad
In Chapters
(3)
\nfrequency,
The
(4)
zero
\nand
remaining
approach
to
9 through 11, the readeris
complex
transfer
and
and
functions,
functions,
poles
are devoted to applications of the pole
chapters
to network
analysis.
Chapters 12 and 13 relate to
introduced
impedance
zeros.\n
and
networks
\nreactive
functions.\n
network
of
zeros
\nand
include
Foster\342\200\231s
reactance
theorem
and
filters
is an
14
\nstudied from the image parameter point of view.
Chapter
In
the
last
two
\nintroduction to stagger-tuned amplifier-networks.
block
the
of
and
the
representation
systems by
diagrams
\nchapters,
are studied.
References are given at the
of
feedback
systems
\nstability
\nend
each
in a more advancedor more
of
for those interested
chapter
\ndetailed
study.\n
the bookdates
to most of the contents of
\nback
to the
1920\342\200\231s. It
has
not been until recent years, however, that
\nthe
and
zero
has been widely taught in graduate
pole
approach
\nand
used
extensively
by electrical
engineers in industry in such
\nas
circuit
electronic
and automatic
control. This pole
circuits,
design,
literature
The
relating
schools
areas
\nand
\ncurriculum
zero
approach
in
many
is
now
different
its way into the
finding
areas of study and in many
iii\n
undergraduate
ways.\n
PREFACE\n
iv\n
the
\nat
\nthe
this
of
I
\nat
\tam
Stanford
\nis
also
A.
\nto
of
\nniques.
\nengineering
Dale
I
at the University
chapters\342\200\224whose
questions
and
class\302\254
on the book.
for
Philip
Weinberg.\n
M. E. Van
Urbana,
Utah\342\200\224and
Indebtedness
of Stanford University and
Baker of the University of. Utah
and offering helpful suggestions, and
many imprints
Dr. Glen Wade
Doran
of
editor of this series.\n
to study network synthesis under Professor
my good fortune
Tuttle,
Jr., at Stanford University. In preparing this book,I
of approach and his teachingtech\302\254
influenced
by his method
the friendly cooperation of the electrical
also acknowledge
the
of Utah, particularly Professors
at
staff
University
and
been
servomechanisms.\n
and
students
some
manuscript
Harris
F.
\nhave
the
L. Everitt,
was
\nDavid
and
W.
Dean
It
\nL.
to
Baker
parts
\nreading
left
have
acknowledged
Don
\nDr.
my
for
University
discussions
\nroom
to
students
The
analysis,
indebted
deeply
in
for junior and senior
which has been offered since 1949.
been to provide background material for
as communications
engineering, pulse
subjects
system
power
\ntechniques,
by using it
developed
for a course
notes
such
of
study
been
has
Analysis
of Utah,
course
has
University
\nobjective
Network
classroom
of
form
\nthe
of
material
The
Illinois\n
Valkenburg\n
CONTENTS\n
1.
1- \t1.
1-4.
2.
current,
The
relationship
\ntance
parameter,
resistance
The
a
as
circuit,
1-11.
20.
\nsentation,
for
Conventions
2- \t1.
element
Active
3.
Network
3- \t1.
3-3.
41.
node
53.
5- \t1.
Initial
conditions
initial
of
differential
6-2.
72\n
The principle of
individual
in
of derivatives,
6- \t1.
\ntion,
61.\n
72.
equations,
The integrating
4-3.
conditions,
Equations/
Solution
42. 3-4.
Equations
4-5.
Differential
97.
analysis,
differential
interpretation
evaluating
and loop
4-2. General and
factor, 74. 4-4.\n
superposition,
79.\n
in Networks
Conditions
\nmetrical
6.
77.
constants,
Initial
currents
for currents,
the
on
72.
solutions,
\nparticular
5.
for
Definitions
Time
33.\n
Branch
directions
Differential
First-Order
3-2.
40.
Positive
network
Resistive
3-10.
4- \t1.
circuits,
loop basis, 43. 3-5. Loop analysis
44.
3-6. Formulating equations on
coupled
coils,
47.
3-7. Duality,
51. 3-8. General network
3-9.
The solution
of equations by determinants,\n
basis,
\nequations,
4.
for coupled
40\n
equations
with
circuits
58.
2-2. Current and voltage
(or geometry), 29.\n
topology
27.
conventions,
Network
Kirchhoff\342\200\231s-equations,
\nFormulating
\nthe
27\n
Equations
\ncurrents.
\nof
22.\n
Networks
Describes
convention
dot
The
and power,
Energy
2-3.
28.
\nconventions,
2- \t4.
1\n
2. 1-3. Electric\n
1.
1-2. Electric
charge,
Sources
of energy; electric potential, 3. 1-5.\n
of field and circuit concepts, 5. 1-6.
capaci\302\254
5.
1-7. The inductance parameter, 10. 1-8.\n
1-9. Approximation of a system\n
17.
parameter,
1-10. Other approximations in circuit repre\302\254
18.
Introduction,
The
2.
of the CircuitConcept
Development
elements,
86. 5-3.
84.
A
5-2.
procedure
Geo\302\254
for\n
87.\n
Continued
a second-order
The standard
equations,
84\n
102.
homogeneous
97\n
differential
equa\302\254
form of the solutionof second-order\n
6-3. Higher-order
homogeneous\n
v\n
CONTENTS\n
VI\n
differential
differentialequations,113. 6-5.
The
of
method
7.
an
in
\ncharge
The
undetermined
the\n
by
integral
particular
6-6. Capacitor
113.
coefficients,
series network,
BLC
of nonhomogeneous\n
6-4. Solution
111.
equations,
118.\n
125\n
Transformation
Laplace
7-1. Introduction, 125.
7-2.The Laplace
127.\n
transformation,
for the Laplace transformation, 129. 7-4.\n
theorems
7- \t3. Basic
of problems with the Laplace transfor\302\254
of the
solution
Examples
fraction expansion, 134. 7-6. Heavi\302\254
Partial
7-5.
131.
\nmation,
\nside\342\200\231s
expansion
the
\nby
8.
7-8. The initial
and
final
145.\n
theorems,
Domain and the
in the Time
Topics
of total solutions
Examples
143.
transformation,
Laplace
\nvalue
7-7.
138.
theorem,
153\n
Domain
Frequency
8-2. Other unit functions:the\n
157.
8-3. The Laplace transform
and
doublet,
ramp,
impulse,
161.
and
8-4. The convolution
\nfor
shifted
functions,
singular
Fourier
8-6. Complex exponen\302\254
8-5.
167.
series, 170.
\nintegral,
179.
8-7.
The frequency
Fourier
\ntial
form
of the
spectra
series,
and
The
Fourier
con\302\254\n
8-8.
181.
\nof periodic
waveforms,
integral
8-9. Fourier transforms and
183.
spectra,
tinuous-frequency
8-
\ntheir
9.
unit
The
\t1.
to
relationship
and
Impedance
9-
\t1.
The
and
admittance,
9-4.
200.
186.\n
....
Functions
Admittance
of
\nbinations,
transforms,
Laplace
concept
\nimpedance
153.
function,
step
194.
frequency,
complex
197.
Th&venin\342\200\231s
Series
9-3.
and
theorem
194\n
9-2. Transform
and parallel
Norton\342\200\231s
com\302\254
theorem,\n
205.\n
10.
10-
\t1.
214\n
Functions
Network
and
Terminals
terminal
pairs,
214.
10-2. Driving-point\n
216.
10-4.
Poles\n
functions,
immittances, 215. 10-3.
and zeros, 219. 10-5. Restrictions on poleandzerolocations
10-6. Time-domain
behavior from the pole and
222.
s-plane,
Procedure
for finding network functions for
225.
10-7.
plot,
Transfer
in\n
zero\n
\ngeneral
11.
networks,
two-terminal-pair
Sinusoidal
Steady-State
230.\n
Analysis
from Pole-Zero\n
Configurations
240\n
sinusoid, 241. 11-2. Magnitude\n
and
of network
phase
functions, 245. 11-3. Sinusoidalnetwork
\nfunctions
in terms
cir\302\254
of poles and zeros, 252. 11-4. Resonance,
\ncuit
255.
11-5. Asymptotic change of magni-\n
Q, and
bandwidth,
11- \t1. Radian
frequency
and the
vii\n
CONTENTS\n
with
tude
the
Reactive
12-
networks,
Use of
12-8.
302.
317.
and
T
cal
13-
w
half
m-derived
The
13-10.
346.
\t1. Shunt
tuned
Block
for
diagrams
block
closed-loop
16.
transformations,
gram
representation
16\t1.
the
Routh
Feedback
rule,
criterion,
Nyquist
tem,
Index\n
415.
for
block
of physical
systems,
Systems
\t\n
402.\n
410\n
System stability in terms\n
16-3.
The Routh criterionor\n
equation,
16-4. The Hurwitz criterion, 418. 16-5.The\n
419.
16-6. Application
to a closed-loopsys\302\254\n
systems,
characteristic
394\n
diagrams, 394. 15-2. Block\n
396. 15-3. Open-loop and\n
elements,
399. 15-4. Block dia\302\254
diagram
equivalents,
400.
15-5. Limitations
in the block dia\302\254\n
in Feedback
Stability
of
\t\n
operations
electrical
\ngram
363\n
372.\n
polynomials),
Basic
\t1.
15-
356.\n
363. 14-2. Stagger-\n
network,
14-3. Overstaggered amplifiers\n
networks,
Diagrams
theorem,
\t\n
amplifier
365.
peaked
amplifier
(Chebyshev
15.
bisection
Bartlett\342\200\231s
Networks
Amplifier
14-
328. 13-6. Constant-A filters, 331.\n
338. 13-8. Image impedance of\n
filter,
344. 13-9. Composite filters,\n
(or L) sections,
of termination, 351. 13-11. Lattice\n
problem
networks,
13-12.
353.
filters,
310\n
m-derived
The
\t7.
(Filters)\n
transfer
Image
networks,
Networks
310. 13-2. Image impedance, 312.\n
network,
314. 13-4. Application to LC\n
function,
13-5. Attenuation
and phase shift in symmetri\302\254\n
ladder
The
\t1.
13-3.
normalized frequency,304.\n
I-Pa ir Reactive
Two-Termina
13-
14.
12-4.
reactive
of
274\n
12-2. Separation
for\n
property
The
four
reactance
function\n
279.
for reactance functions, 287.\n
Specifications
of reactive
networks, 288. 12-6. Cauer form\n
296.
12-7. Choice of network
realizations,\n
form
Foster
....\n
12-3.
networks,
\t5.
An
274.
networks,
284.
forms,
11-6.
263.\n
Networks
Reactance
12-1.
reactive
poles and zeros,260.
lattice,
symmetrical
One-Terminal-Pair
12.
13.
in terms of
frequency
\napplication:
410.
16-2.
411.
423.\n
435\n
CHAPTER
1-1.
THE
OF
DEVELOPMENT
1\n
CONCEPT\n
CIRCUIT
Introduction\n
of science is the continual bringingtogether
wide
of facts to fit into a simple, understandabletheory
\nof a
variety
as possible. The name
account
for
as many
\nthat
will
observations
scheme
has
American
chemist and univer\302\254
been
used
the
\nconceptual
by
James
for the theory or picture that results.*
Conant
\nsity
president
the
most
familiar
scheme to students of science
\nPerhaps
conceptual
we
our
\nand
is
that
of
atomic
take
the
theory from which
engineering
con\302\254
\npicture of the electron and of electric charge. Otherimportant
methods
the
of
One
are
schemes
\nceptual
man\342\200\224the charging
of
\nhistory
and
electricity
Although
in
\nlodestone
navigation\342\200\224it
means
chemical
by
\nproduced
in developing
a
Volta
about 1800
and
Galvani
by
\ndiscovery
the
made
was
progress
\nsignificant
of energy and conservation of charge.\n
were recognized early in the
magnetism
of amber
by friction, the use of
was
not until the nineteenth century that
conservation
\nafter
Volta.
\nrent,
and
Oersted
identified the
1820,
measured
the force caused
Ampere
\nFaraday,
\nful
In
and
from
\nresulting
\ntual
Maxwell\342\200\231s
\nfrom
magnetic
with
discovered
electric
cur\302\254
1831
induction.
The success of Maxwell\342\200\231s concep\302\254
the
by
persistent agreement of results deduced
for a period of over 100
with
observation
and
charge
is evidenced
scheme
magnetic field
by the current. In
together to form a success\302\254
by the English physicist James Clerk Maxwell
has come to be known,
as the scheme
equations,
are explained in terms of fields
phenomena
Maxwell\342\200\231s
electric
\nall
of time
interval
were brought
experiments
scheme
conceptual
1873.
\nin
Henry,
independently
other
and
\nThese
short
relatively
In
and
experimentation.
simplified
greatly
\nImportant discoveries were made in a
conceptual scheme. The
that electricity could be
equations
current.
\nyears.\n
another
\ncuit?
\nrelated?
\nof
the
*
\nHaven,
do we now embark
upon a study
same
the
electric cir\302\254
phenomena,
conceptual
as a question,
how are the two concepts
important
Equally
The
answer
to the first of our questions is the practicalutility
circuit
As a practical matter, we are not ofteninterested\n
concept.
view
In
\nof
James
of
Maxwell\342\200\231s
B.
Conant,
why
success,
scheme
for the
Science and
Common
1951).\n
1\n
Sense
(Yale
University
Press,
New
\nfavors
analysis
The
desired.
and
\nanswer
\nbasic
as charge,
answer
to
such
\nquantities
\nif
as we are in voltagesand currents.The
circuit
in terms of voltage and current from which
so much
fields
in
justification.
involves
\neral
concept
we
\nbefore
basic
blocks:
the
\nnomena,
primitive
of
scheme
\nconceptual
interconnected
of
\ninclude
sources
\ncircuit
functions
of
a
of supply
point
the
called
\nversion
that are not
It is important
charge
and
included in the
that we understand the
more
of circuit
limitations
of
function
the
However,
equations.
energy.
gen\302\254
theory\342\200\224
in terms
of
We regard charge
and
the
circuit
in describing electrical phe\302\254
denominators
in terms of which we can build our
quantities
the electric
circuit. A physical circuit is a sys\302\254
Here we use the word apparatus to
apparatus.
components, loads, etc. A
energy. Energy transfer is
transfer.
In the circuit, energy is transferred
(the source) to a point of transformationor con\302\254
be
(or sink). In the process, the energy may
wires,
energy,
connecting
to transfer
and transform
charge
by
\naccomplished
\nfrom
other
common
least
the
as
Maxwell\342\200\231s
helpful to definethe
building
\nenergy
\ntem
theory.
do
approximations\342\200\224the
our subject.\n
develop
It will be
\ntwo
field
these
of
\nnature
as
approximations
of
concept
fields, energy, power, etc. can be computed
our second question will require a longer
circuit
concepts arise from the same
Briefly,
facts
experimental
\ncircuit
Chap. 1\n
CIRCUIT CONCEPT\n
THE
OF
DEVELOPMENT
2\n
load
\nstored.\n
Electric
1-2.
charge\n
when
\namber
\ntrified\342\200\235
is capable
for
\ntechnique
(and
inverse
the
\nestablishing
Our
producing
France
in
\nlomb
briskly
and
with the discovery about 600 b.c.that
rubbed
with a piece of silk or fur becomes
\342\200\234elec\302\254
of attracting
small pieces of thread. This same
was used
centuries
later by Cou\302\254
electricity
is credited
Greece
of
Thales
by Cavendish in England) in
of charged bodies.\n
law of attraction
of the nature of charge is basedon
independently
square
understanding
present-day
We picture the atom as
nucleus
surrounded
of
a positively
charged
by negatively
In the neutral
electrons.
atom, the total charge of the nucleus
to
the
total
charge of the electrons. When electronsare
equal
becomes positively charged.
from
a substance,
that substance
substance
with
an excess of electrons is negatively
this
The
basic
unit
of charge is the charge of the electron.
is so small, the practical unit of the coulomb
is used.
\ncharge
\ntron
has
a charge
of 1.601 X 10~19coulomb.\n
\nthe
scheme
conceptual
of the
atomic
theory.
\ncomposed
\ncharged
\nis
\nremoved
\nA
charged.\n
Because
The
elec\302\254
1-3. Electric
current\n
The
\nto
another
phenomenon
is described
of transferring
by the
term
in a circuit
electric current. An electric
charge
from one point
current\n
\na
cross-sectional
motion
random
A
boundary.
3\n
electric
across
charge
of electrons in
unless there is a net transferof
a current
constitute
not
\ndoes
CIRCUITCONCEPT\n
time rate of net motionof
as the
defined
be
may
OF THE
DEVELOPMENT
1-4\n
Art.
a metal
charge
with
\ntime.\n
In
current*
the
form,
equation
is\n
(1-1)\n
the
If
charge
q
current
the
\nonds,
\nAndr6
Ampere).
it follows
\nlomb,
in sec\302\254
is given in coulombs and the timet is measured
is measured in amperes (after the Frenchphysicist
cou\302\254
Since
the electron
has a charge of 1.601X 10-19
that a current of 1 amperecorresponds
to the motion of
of a
cross
section
X 10-19) = 6.25 X 1018electrons
any
\nconducting path in 1 sec.\n
In terms
of the atomic theory conceptual scheme, all
as
made
In a solid, some
relatively
\npictured
up of atoms.
\nfree
of the
the
attractive
nucleus;
6.24 x 1018electrons/sec
of current\n
on these
electrons
are exceed\302\254
\nAmpere
\nforces
/\n
are dis\302\254
small.
Such
electrons
\ningly
past
\n1/(1.601
substancesare
electronsare
flow
of
\nrate
time
in
\nrents
are
\nMost
metals
few
\nrelatively
\nelectrons
electrons
free
many
so
electrons,
that
large
cur\302\254
than
etc. There is
no sharp
is pos\302\254
insulators.
Conduction
In vacuum tubes, for example,
from one metallic plate to
vacuum
and
conductors
a
plastics,
mica,
glass,
through
pass
con\302\254
materials
are known as conductors.
are good conductors. Materials with
liquids
are known as insulators. Common insulat\302\254
materials
other
in
\nsible
some
free
between
line
\ndividing
a
Such
and
include
materials
\ning
of charge in
\nductor.\n
attained.
easily
Motion
1-1.\n
Fig.
materials, there are
In some
1-1,\n
\nPig.
area
sectional
Cross
\npassingfrom one atom to the next
\nas pictured
-\n
Charge
elec\302\254
free electrons,
these
of
free
is the
current
electric
An
\ntrons.
name
the
by
\ntinguished
solids.
partial
\nanother.\n
at
\npropagate
travel
\nductor
electron
\nfree
numerical
a
\nfor
Sources
1-4.
Another
The
example.)\n
of energy/
of
symbol
energy.
i for
electric
potential\n
upon which
our training
in
scheme
conceptual
\nconservation
\n*
that since some electric waves
misconception
the speed of light the electronsin a con\302\254
approximately
with
this same velocity.
The actual mean velocity of
is but
a few millimeters per second! (SeeProb. 1-2
drift
a common
is
There
By
current is
taken
from
the
our thinking is based is the
the methods of science,we
French
word
intensity.\n
become
immediately
\nnor
can be
this
\nthat
the
are
The familiarrotating
thermal
\nfrom
other
\nand
mechanical
an
\nent,
x
\nduce
Other
(4)
\njunction
\nenergy
by
mechanical
is converted
energy
and in turn, the thermal
energy
by burning coal.\n
produce electric energy by
energy.\n
The friction machines
energy.
exception
de Graaff generator
the Van
being
used by Coulomb
electric energy
produce
This
is little used
method
experimenters
early
\nverting
batteries
Electric
methods.
Electrostatic
(3)
chemical
from
chemical
\nconverting
a turbine,
by
energy
mechanical
the
Often
methods.
Voltaic
electric energy from
produces
converted
is
\nenergy
invented
generator
rotation.
of
\nenergy
(2)
1831
in
some
following:\n
(1) Magnetic induction.
\nFaraday
few
In order of economicimportance,
accomplished.
methods
these
cannot be created,
converted in form. Electricenergyis
form. There are relatively
ways
other
some
from
converted
\nenergy
\nof
that
but
destroyed,
states that energy
of energy
it can be
of conservation
law
\nThe
that creates energy.
scheme
of any
suspicious
Chap. 1\n
CIRCUIT CONCEPT\n
THE
OF
DEVELOPMENT
4\n
by
at
con\302\254
pres\302\254
used to
pro\302\254
in nuclear
Thermal
methods.
is produced by heat at a
electricity
of
dissimilar
metals
such as bismuth
and copper. Light
can
be
converted
into electric
energy by photoelectric
and
rays
used
in research
physics.\n
\ndevices.\n
two
\nexample,
in
\nimmersed
of energy
terms
in
same
\nthe
sources of electricenergyis
and charge. In one form
for
battery,
of these different
of each
function
The
metallic
acid.
causes
circuit
\\\n
+Terminal\n
-Terminal
at
\nmulate
\nCharge flow
in
\nence
Si\n
copper
the
\nOnce
Device
\nto
for supplying
bodies
charged
\nconversion of
energy
chemical
\nan
external
by
energy\n
1-2,
\nFig.
as
\npended
1-2.
Fig.
showing
\nbattery
\nper
\nused
Representation
unit
term
charge,\"
voltage).
electron
the
of
a
flow.\n
is given the
In the form
\nin
transporting
copper\342\200\224are
the
copper\n
to
accu\302\254
charge
electrodes.
is
Energy
negative
\nsuppliedto the
\302\261j\302\261\n
\nand
of
of zinc and
The formation
ions
External
one
and
zinc
of
electrodes\342\200\224one
sulfuric
dilute
of
charge
by
the
differ\302\254
of ionization of zinc
in the
chemical
reaction.
is closed by
circuit
battery
in
as shown
connection,
is ex\302\254
the
chemical
energy
work
for each unit of charge
the
around
the
charge
energy
\nexternal circuit. The quantity
or identically,
charge\"
\nper unit
\342\200\234
energy
\342\200\234work
name potential (or the more
of an equation,\n
commonly
W\n
V\n
(1-2)\n
Q\n
CIRCUITCONCEPT\n
OF THE
DEVELOPMENT
Art.1-5\n
5\n
in coulombs,
(or energy) in joules and q is the charge
v is in volts (after Alessandro Volta). The potentialof an
\nthe
potential
described by the term electromotive
source
is sometimes
force,
\nenergy
literature. We will avoiddesignating
\nabbreviated
emf, in the electrical
it
is
a
because
as
force
misleading and instead usethe terms
\npotential
If
or
\nvoltage
\342\200\230potential.\n
dw,
energy
the
dq is given a
of charge
amount
differential
a
If
\nin
work
is the
to
potential
differentialincrease
is increasedby the
of the charge
amount\n
(1-3)\n
this
If
by the current,
is multiplied
potential
dw
result
Thus
\np.
to be a
is seen
is the
power
time rate
The
In
=
(3)
each
physical
The capacitance
vi
(1-5)\n
j p
dt =
j
vidt\n
(1-6)\n
and circuit concepts\n
circuit conceptual
of three parameters.
scheme, we will followthe same
These steps are the following:\n
phenomenon.
parameter\n
substances\342\200\224for
separated
in
shown
those
\342\200\234action
The
phenomenon.
\tPhysical
(1)
a
at
Fig.
This
stances.
as
distance\342\200\235
in
experimental
fact.
+ +
of nature,
Coulomb
the
+ +
spatially\n
+
+
+
^
re-
+\n
i\n
Fig. 1-3. Chargedbodies,\n
a basic\n
found
two
\t\n
sub-
we
of charge on
\t\n
1-3\342\200\224causes
phenomenon
a property
presence
example,
form of a force between the two
gard
current,\n
We will discuss in a quantitative
\nmanner
an
electrical
which is observed by experi\302\254
phenomenon
this
in
We
will do
terms of charge and energy.\n
\nment.
Field
We will next discuss the interpretation of
interpretation.
in terms
\nthe
of a field quantity.\n
phenomenon
we will introduce a circuit param\302\254
Circuit
interpretation.
Finally,
\neter
to relate
relationship.\n
voltage and current in place of the field
(2)
an
the
for
The
(1)
1 -6.
of field
relationship
steps
is power
equation\n
developing
\nthree
which
of energy,
change
=
(1-4)\n
given by the integral
w
1-5.
of
as\n
P\n
of potential and
product
p
and energy is
_
~
~dt
Xdt
\ndq
the
dw
_
~
dq
dq/dt
that
this force was
of such
nature\n
that
\342\200\234like charges
\nforce varied
and
repel\342\200\235
\342\200\234unlike
to the
according
THE CIRCUITCONCEPT
OF
DEVELOPMENT
6
Chap.1\n
the
that
and
attract\342\200\235
charges
equation\n
9*?2
=
F
(1-7)\n
\n4irer2\n
In
this
\nto
point
charge,
the
permittivity,
in the
\ne is
meter
\nper
to
\nstrictly
\nto
mks
the
are
<72
measured
charges
that this equation applies
the equation may be applied
However,
charges
only.
of known
charge distribution
by vectorially adding all
point
geometry
any
and 91 and
system,
understood
be
should
It
coulombs.
\nin
F is the force in newtonsdirectedfrom
point
charge
the
in
of
r is the separation
point charges meters,
value
of 8.854 X 10-12farad
the
having
free-space
equation,
\nforces.\n
\tField
(2)
force
a
\nof
force
\nThis
electric
an
called
of
field
the two chargedbodies.
since force is a vector
value\n
F\n
-
=
E
in terms
can be described
phenomenon
charge placed between
a vector quantity
charge,
unit
per
is
\nquantity,
This
interpretation.
on a unit
(1-8)\n
\nQ\n
a
As
+
+
+
be represented by
force that would be exerted on
this field may
aid,
conceptual
\nthe direction
of the
+
+ +
+
+
+
Electric
\n\342\200\234lines
of
field lines
charge
exploring
+\n
or
in
\nlines
are
illustrated
\nlines
are
conceptual
be
\nent.
1-4.
unit
Eqs.
Using
\nfield
be
may
Fig.
aids:
of as
thought
each
at
evaluated
positive\n
point. Such
1-4. These
should
they
actually
1-7-and
in
drawn
the
\nnot
Fig.
lines
being
pres\302\254
1-8, the electric
for a particular
force.\342\200\235\n
\nproblem.\n
\tCircuit
(3)
\nfrom
one
The
interpretation.
to the other
plate
work
necessary
of Fig. 1-4 may
W\n
=
J F
cos 6
to move a
be foundfrom the
dr\n
charge
equation\n
(1-9)\n
increment
of distance between the plates and 6 is the
An
the force and the direction of movement
\nangle between
expres\302\254
\nsion
for
the force
has been given by Coulomb\342\200\231s
law,
Eq.
1-7, which
dr
where
is an
dr.
\nmay
be
substituted
into
the
w
equation
=
j
We
are
\nthe
voltage
more
interested
between
w2
above to
cos 0 dr
give\n
(1\"10)\n
in the quantity work per unit charge,which
is
\342\200\224 =
=
the plates.
When <71
but
<7 (equal
<71
oppo-\n
Art.
site
the
on
charges
w
=
CIRCUIT CONCEPT\n
the expression for
plates),
V
THE
OF
DEVELOPMENT
1-6\n
f cos
(
=
potential
7\n
becomes\n
6 , \\
11
\\\n
<x-u>\n
\\-J
The
of this equation
can be evaluated for simplegeometry,or
integral
case
can be measured
fixed
For
any
by measuring q and
any
is given
the name
\ngeometry, the integral is a constant
elastance,
letter
this
the
S.
With
\nsymbolized by
l
wdr)\302\253
\nin
v.
which
definition,\n
=
v
C.
\nletter
the capacitance,
of S is
The reciprocal
be
may
=
and
\nlombs
Cv\n
(1-13)\n
In these equations, if q is
the unit of C is thefarad
(in
of capacitance.
v in volts, then
terms
the
by
represented
written\n
q
in
is
which
1-12
Equation
(1-12)\n
Sq
measuredin
cou\302\254
of Michael
honor
colorful
name
\nFaraday), and the unit for S bearsthe
daraf
(farad
The quantity C (or the quantityS) which
char\302\254
\nspelled backwards).
under
and
\nacterizes
the
the
study
permits
system
simplerelationship
v and
is
to
be
written
known
as
\nbetween
a
circuit
q
parameter,the
a
of
there
system,
capacitive
\nship
system.*\n
our objective, a
To reach
\nin
a
of
\ncapacitance
and
charge
current
relationship betweenvoltageand current
remains the task of studying the
relation\302\254
given by the
\n1
~
equation\n
dq
(1-14)\n
dt\n
with
\nlinearly
charge on a
initial
is an
there
If
the
time,
charge
q
is found
current
The
the
\ntime, giving
*
\nonly
that
on
\ncharge
should
It
for
the
any
\nTheory
may be
charge
increases
written\n
kt\n
by differentiating
(1-15)\n
the charge
with
to
respect
-
af
-k
<M6>\n
the current in the system is independent
initial
of
In going the other direction,computing
charge,\n
system.
described
be noted that the circuitparameter
of two charged bodies with
equal
case
\ncapacitanceconcept can be
\nwith
q0 +
the
and
that
see
we
=
q0
value\n
*
Thus
system,
at any time
charge
distribution.
(Addison-Wesley
extended,
For
Publishing
however,
to
the
by
and
this
opposite
case of
holds
equation
charges.
The
several conductors
example, see Fowler, Introduction to
Co., Cambridge, 1953), pp. 73 ff.\n
Electric
the
given
Eq. 1-14
we rearrange
current,
to
be integrated
can
as\n
\342\200\224
idt\n
dq
which
Chap. 1\n
CIRCUIT CONCEPT\n
THE
OF
DEVELOPMENT
8\n
(1-17)\n
as\n
(1-18)\n
give\n
charge on the
The total
initial
the
\nof
related
are
\nage
more
once
the
by
be
to
seen
equal
by the
the charge deposited
to the relationship,
q
and
charge
Returning
system of platesis
\342\200\224
to the
sum
and
volt\302\254
current.\n
current
Cv,
equation\n
dq\n
(1-19)\n
dt\n
the
If
vary with time (or
C does not
capacitance
with
then\n
charge),
(1-20)\n
C is not
If, however,
/
\342\200\242d
l =
dt
\\
v-y
^
t;
the
time,
cur\302\254
=
i
S\n
dC\n
= C^{ft) +.
dt\n
the equation
with
starting
Similarly,
of
function
the general relationship\n
from
found
be
must
\nrent
constant but variesasa
=
v
V\n
we
Sq,
i
dt\n
(1-21)\n
dt\n
find
that\n
dt\n
(1-22)\n
/\n
1-19
Equations
\nsystem
Example
The
\nby
a
\nplates
and
1-22 relate
the
through
circuit
the voltage and
parameter
capacitive
C.\n
1\n
of Fig.
sketch
constant-speed
varies
according
1-5 (a) shows
motor
so that
to the
two plates, one
battery
the capacitance
of
which
is
driven
between the
two
equation\n
C(t) = Co(l
If the
current in the
\342\200\224
cos
potential remains constant
(1-23)\n
(at)
at
V
volts,
the
current
as
a\n
1-6
Art.
of time
function
may
be found from
=
i
^
(Cv) = V^-
CIRCUITCONCEPT
9\n
Eq. 1-21
as\n
=
sin
o)C0V
cot
(1-24)\n
of current is shownin Fig.
time variation
This
OF THE
DEVELOPMENT
l-5(c).\n
Fig.
1-6.
Variable
capacitance
system.\n
that the
cannot
Cv
q = Cv, it is seen
product
since an instantaneous
\nchange
instantaneously,
change in q would
infinite
which
an
is ruled out as a
\nmean
current,
phys\302\254
In
terms
of
the
time
\nical
system.
= U \342\200\224 in which
At
\ninterval
ti,
q or
amount
a finite
shown
\nCv
changes
At
cannot
be
zero.\n
\nin
Fig.
1-6,
of Cv shown
Instantaneous
change
1 is thus ruled out. Typical
\nas curve
are per\302\254
or q which
of
Cv
\nchanges
1_6,
of Cv wlth timeFig\342\200\230
Change
shown
as curves 2 and 3.
are
\nmitted
the
From
equation
possibilityina
another
\nFrom
the
approach,
q(t)
charge
\342\200\224
q0
is given
+\n
as\n
dt\n
(1-25)\n
/:*\n
by
1-18.
Eq.
\nvalue
in zero
portion of this
finite
with
time
i; that is,\n
The integral
i dt
lim
=
0,\n
equationcannot
have
i
9^
oo\n
a finite
(1-26)\n
\n<-\342\226\272
o\n
The
integration
\ninfinitesimal
\nto
ti
must
areas,
be
is illustrated
process
and dt
i in height
greater
than
in Fig. 1-7 as the summation
of
in width. The interval
zero for any area to be
from
summed.\n
t
=
0
OF
DEVELOPMENT
10\n
mathematical
These
the
\nthat
charge
time.
zero
\nin
THE CIRCUIT
cannot increase or decrease
or voltage can change
a capacitive
system
either
capacitance
However,
in
in\302\254
so
\nstantaneously
two
the
of
\nduct
\nconstant,
1-7.
of
Integration
the
where
conditions
apart (suchas
\nvanishingly small interval
pro\302\254
remains
quantities
=
C 2t>2\n
2 refer
a switch
after
and
times
at
existing
before
(1-27)\n
1 and
subscripts
current.\n
\nto
the
as
long
as\n
C\\V\\
Fig.
Chap.1\n
the requirement
in visualizing
aid
equations
CONCEPT\n
a
is
\nclosed).\n
\nnot
with
change
cannot
\nsystem
time
\nthe
\nthe needle
\ncarrying
In
\nnetism.
\ncurrent
i
and
\nbetween
\ntional
force
same
the
the important
discovery
an
\342\200\234action
is not
Ampere measured the force causedby the
in equation form. This mag\302\254
relationship
a distance\342\200\235
as in the case of the force
just
the
at
This
bodies.
charged
fact;!it
between
year,
expressed
effect'is
\nnetic
made
Oersted
two charged substances depended on
rate
of flow
of charge (the current). In Oersted\342\200\231s experiment,
of a compass was deflected by the presenceof a currentthat the effect was related to mag\302\254
indicating
conductor,
the
that
1820
parameter\n
phenomenon.
Physical
(1)
instantaneously.\n
change
The inductance
1-7.
\nin
the
to
\nsimplifies
does
considered, the capacitance of a network
time.
Under this condition, the above discussion
conclusion
that the voltage of a capacitive
important
to be
cases
most
In
\342\200\234action
at
a
distance\342\200\235
is
a basic
observa\302\254
from other knowledge.\n
deduced
(2) Field interpretation.The
in terms of the force per unit
\ninterpreted
phenomenon
magnetic
can
above
described
pole
at
all
be
points\n
(a)\n
Fig.
in
\nangles
\nof
Fig.
to
the
l-8(a),
The
magnetic
field.\n
that this force was directed at rightIn terms of the geometry
conductor.
current-carrying
described
a magnetic field density B, the force\n
Ampere
Oersted
space.
1-8.
discovered
Art.
1-7\n
per
unit
of
pole,
magnetic
CIRCUITCONCEPT\n
OF THE
DEVELOPMENT
11\n
value\n
a dl
cos
ni
_
(1-28)\n
\n4n-r2\n
the magnetic
n is
where
the
\nin which
magnetic
defined
are
\nquantities
1-9.
Pig.
of a
tion
conventions.\n
flux
By Eq. 1-28, the
conductor.
at a
constant
be
field and
Magnetic
current-carrying
will
\ndensity
permeability, which is a functionofthemedium
other
i is the currentin amperes,
and
on the figure. Figure 1-9(a) showsthe crosssec-\n
field exists,
magnetic
distance from the conductor.
constant
field
to indicate the direction
aid. These are magnetic field density lines or
\nof B\342\200\224as a
conceptual
For
force.\342\200\235
more
of
\n\342\200\234lines
than that shown in
complicated
geometries
the
of
the
lines
be
can
found
position
l-9(a),
\nFig.
by integratingEq.
or
\n1-28
with
lines
\nContinuous
by
experimentally
\nexisted) from place to
\nan
\ndensity
be drawn
may
a
moving
pole
\342\200\234point\342\200\235
magnetic
placein space.
A
would
compass
magnetic
sometimes convenient to replace the linesof magnetic
by lines of magnetic flux defined by the integralequation\n
$
where 0 is the
(if
one
give
of directions.\n
measure
approximate
It is
arrows
cos 0 dS
=
j'B
field
(1-29)\n
angle
between
the
surface
of integration
and the field
in
currents
each
of
N
\ndensity
conductors, representedin
are in such a direction that the fluxes add,
flux
N<f>
\nFig.
1-9(b),
to exist. If, however, <f>i lines
of flux
are said
link Ni con\302\254
\nlinkages*
link
lines
conductors
and
so
the
total
numberof
N2
<f>2
forth,
\nductors,
B.
the
If
then
\nflux
is
linkages
found
by
summation
algebraic
as\n
n\n
*
=
V
(1-30)\n
N^i
j=l\n
*
\nof
For
the
of some of the
flux linkages, see
a discussion
concept
of
\nmagnetic
fields,\342\200\235Am.
\nof
current
steady
J.
magnetic
Phys.,
problems
encountered
in the
use (and misuse)
in electric and
Joseph Slepian,
\342\200\234Topology
19, 87 (1951), and Keith McDonald,
Am.
fields,\342\200\235
\342\200\234Lines
of
J.
Phys.f
force
22, 586 (1954).\n
DEVELOPMENT OF
12\n
\nto
flux
give
linkages,
=
experimented
\nFaraday
found
He
\nimity.
a
\ninduced
voltage
\nof
a
\nconductor
moving
\nis
of
thought
\nthe
\nthought
in a
caused
\nfor
for
accounting
flux
\342\200\234changing
only
at a
action
distance.
\tCircuit interpretation.
in
current
v
is
law
Faraday\342\200\231s
Such
time.
with
we
\nseconds,and k = 1, then
and
linkages\342\200\235;
(as in a transformer)is
linkages\342\200\235
attach
not
a
means
is\n
=
O'32*\n
kfi
where k is a proportionalityconstant. In the
\nselected to make k have unit value:when
\nvoltage
flux
do
as conceptual aids so longas
to flux linkages whichare, afterall,
v
(3)
the
reducing
conductor
stationary
by
A
generator)
hence
and
flux
\342\200\234cutting
\nphysical significance
terms
a magnetic
in
valuable
are
\npictures
method of inducingvoltagein
of changes in flux linkages.
terms
field (as in the case of a
in
but
distance\342\200\235
as
as
of
prox\302\254
that a changing magnetic field produced by one
in the other circuit. The changing magnetic field
by
(1) a conductor moving in space or (2) a
induced
voltage
in spatial
circuits
conducting
not envision this
at
\342\200\234action
basic experimental discovery.
time.\n
did
Faraday
two
current
with
\nchanging
for the next
with
(1-31)\n
circuit
caused
be
\ncould
B cos 6 dS
N
J'
credit
goes
Faraday
1\n
as\n
*
To
Chap.
Eq. 1-29 may be modified
link all conductors,
lines
all
that
Assuming
CONCEPT\n
CIRCUIT
THE
inks
is
system
in
the
are
t is in
units
weber-turns,
volts.\n
in
To derive the circuit relationshipbetween
described in (2), we begin with
the system
law,\n
\nFaraday\342\200\231s
(1-33)\n
or
the
integral
equivalent
form\n
=
^
/ v dt
Note, incidentally, the similarity
this
of
\nin
terms
of
We see that ^
\nEq.
1-31,
by
is to
and
in turn,
Ampere\342\200\231s
as
voltage
Now
equations.
\nrent
expression
and
the one
for charge
current,\n
q =
\ntwo
(1-34)\n
law,
flux
charge
linkages
1-28.
is to
current,
are related
by comparing the
to the magnetic field
by
field density is relatedto
these substitutions,
with
Making
the magnetic
Eq.
(1-35)\n
jidt
the
cur\302\254
the\n
OF THE
DEVELOPMENT
1-7\n
Art.
i can
that
assumption
CIRCUITCONCEPT\n
be removed from
the integral,41
TJ
The
\ngeometries
the
\nto
\nance,
if a
However,
parameter
M
to
\nchanged
\342\200\224
Li
=
\nq
voltage
\nrelating
and current in
time,
law
Faraday\342\200\231s
relationship
an
gives
equation
a magnetic
circuit,\n
(1-39)\n
(Li)
L in appropriatecases).
(where M replaces
the
and
J-
with
(1-38)\n
of these equations
=
\302\273
\nvary
circuit,
letter symbolis
Mtiii
1-37 into
Eq.
Substituting
Cv.)
another
as\n
note the similarity
(Again,
(1-37)\n
current ii produces flux linkages
if/2 in
is one of mutual
inductance, and the
*2
=
d-8\302\256)\n
may be evaluated mathematically for simple
as the
or
be found
may
by measuring if/ and
i, is defined
the coefficient
of inductance).
If if/ end i refer
(or
parameter
is
defined
same
the
as self-induct\302\254
parameter
physical
system,
L as\n
symbolized
by the letter
if/
\nthe
f
M) AS]
which
term,
integral
\ninductance
have\n
we
\"
*\342\200\234[*/(/
13\n
does
inductance
the
If
not
1-39 becomes\n
Eq.
<w\302\260)\n
*
system is nonlinear,containing
the
coupled
magnetically
\nmedium, we may say that the flux
\nin all
other
linked
circuits,\n
If
in circuit
linkages
=
the voltage in
1-32,
Eq.
By
_
Vk ~
Each
\nThese
(ifa
=
~
dt
derivative
partial
terms
may
be
dii
,
i\"
djne
dii
is
some
saturating
of the
a function
currents
\342\200\242
\342\200\242.
fin)\n
circuit k is givenby
dii dt
A;
di2
dt
,
^
^,
term is evaluated with
defined as coefficients of
law
Faraday\342\200\231s
d\\pk
dik
dik
dt
,
i\"
'
as\n
, fyk din\n
\342\200\234l\"
dt\n
din
''
all other currentsheld constant.
inductance
so
that
the
voltage
\nbecomes\n
**
where
\nis
linear,
M
=
is used
this
Mkl
Tt
+ Mki
for mutual
equation
1ft +
Lkk
\342\200\242
\342\200\242
\342\200\242
+
W
+
Mkn
\342\200\242
\342\200\242
\342\200\242
+
It\n
inductance and L for self-inductance.When
to one which will later be written
as Eq.
reduces
a system
1-51.\n
\nletter
\nis
(1 /L)
quantity
The
gamma.
(1-41)\n
j
is sometimes symbolized by the uppercaseGreek
(after the American scientist Joseph Henry)
henry
for inductance.\n
unit
mks
the
dt
v
^
Chap.1\n
CONCEPT\n
give\n
=
i
The
to
be integrated
can
1-39
Equation
THE CIRCUIT
OF
DEVELOPMENT
14\n
and the
that
In the case of the capacitive
we
found
charge
system,
led to
be
might
\nproduct Cv could not change instantaneously.We
in
for
an
inductive
system
\nsuspect that there is a similarrelationship
that have been pointed out. Indeed there is such
of
the
\nview
analogies
in definite
\na relationship,
which
may be found with the help of Eq. 1-34,
form.\n
\nintegral
* =
to +
v
given in the last section
has zero value for t =
in this
\ngral
equation
closing of
\ninstantaneously\342\200\224say
by the
From
arguments
\na
same before and after the
interval
of time. In terms
the
be
\nmust
small
very
if/
which is to
of
\nciple
\njust
\nit
is
a
in
\nneously
say that
constant
before
our
altered,
(1-42)\n
= a
about capacitance,the
0. Thus, in a
altered
inte\302\254
system
a
systemis altered,but
of Eq.
(1-43)\n
be
cannot
linkages
=
changed
or
\\f/2
L\\i\\
=
\nin
an
to visualize
circuit.
electric
Newton\342\200\231s
F
where
\nknown
\ntaneously.
\nof
time,
\nconstant.
\nlaws:\n
F
as
is force,
M is mass,
the
momentum;
In
a system
must
velocity
We
see
that
changes
law
force
=
momentum
system
v
is
time
equations\n
(1-44)\n
to the
principle of
since
is helpful
in a mechanical
system than
is\n
Mv
4:
at\n
and
prin\302\254
after
LiU
flux
easier
sometimes
instanta\302\254
This conclusion
the
1
referto
the
If we let the subscript
The
is
for
only
constant
is similar
constant
linkages
\nconservationof momentumin mechanics.
analogy
The principle of
linkages
1-42,\n
same
system is altered and 2 refer to the
statements
can be summarized
by the
\\f/i
\nit
flux
switch\342\200\224the
is described as
linkages.
flux
the
to
flux
the
system.
given
=
dt
JQ
(1-45)\n
velocity.
of a system
The
Mv is
product
cannot change
instan\302\254
lostasa
function
such as a rocket where massis
in such a way that momentumremains
change
there are a number of analogous conservation
1-7\n
Art.
qi = 92 and
The
(2)
conservation
The
linkages:\n
\342\200\224
and
\nthe
of
Example
=
C%v2\n
=
L2i2\n
and
M\\Vi
linkages
=
M2v2\n
an important specialization of
results. In a fixed inductive
sys\302\254
instantaneously.\n
change
2\n
inductive
a certain
In
are
We
exists.
\n1-10
Liii
constant,
flux
constant
cannot
current
the
\ntem,
remains
inductance
principle
15\n
of momentum:\n
conservation
Pi = pa
When
CiVi
of flux
\\j/2
(3)
CIRCUITCONCEPT\n
of charge:\n
conservation
The
(1)
OF THE
DEVELOPMENT
system,
the current waveformshown in Fig.
the voltage that producesthis\n
to find
required
\t\n
1\n
assume
will
\nWe
t\n
waveform.\n
time.
associated
charge, both as functions of
The relationship v = L(di/dt)
L remains
constant.
the
and
waveform
current
4
3\n
Fig. 1-11. Voltage
waveform.\n
Current
1-10.
Fig.
1
2\n
1\n
that
can be found by differentiationof the currentand
The result is shown in Fig. 1-11.Charge
\nmultiplication
by
of the current to
be
found
\nmay
by integration
in Fig. 1-12.\n
result
shown
the
\ngive
It is important that we be able to apply the a 1
\nindicates
the voltage
a constant.
of
\nconcept
\nare
to
inductance
coupled.
magnetically
several
systems
A set
of three
which
coupled\n
Fig.
1-13. Set
coupled
magnetically
Fig.
is
i2 and
ii
Charge
waveform.\n
\ncoils.\n
system for the
of the
The
ii.
1-37
to
Eq.
in Fig. 1-13. To simplify the
zero and consider the effect
be
U
1 flux linkages, found from
produces
coils
\nlet
\nrent
1-12.
shown
=
Liii
(t*2 =
of\n
is =
moment,
current
cur\302\254
be\n
0)\n
(1-46)\n
16\n
OF THE
DEVELOPMENT
where
\ninductance).
\nflux
linkages
the
For
\neter.
self-inductance
the
is
L\\
In
each
by
the
other
circuit,
will
i\\
under
system
fa = Mziii (it
of subscripts for M requires
further
inductance
=
iz
=
some
\nthe
two
equations,
\nthe
flux
linkages
of
param\302\254
study,\n
fa = Afsit'i and
The order
the
number
some
produce
of the mutual
1\n
Chap.
(usually called just
parameter
proportionality
particular
CIRCUITCONCEPT\n
0)
(1-47)\n
attention.
From
be clear that the first subscriptrefersto
second
to the current. This particular con\302\254
it should
the
and
to the generalequations,
remembering this particular
\nvention
the order
if
we
\nassume
for our conceptual
scheme that current produces
In
the
to
general
case, there will be sources or loads connected
in Fig. 1-13 and no current
the
coils
shown
be
zero.
We
will
\nassume for the time being that the current
and
winding
are
\nsenses of the coils are such that all
linkages
additive,
postponing
more
case
for
2.
The
total
in coil 1
general
Chapter
linkages
be
made
up of flux linkages
produced by the current in coil1
\nflux
it and t3. In equation
linkages
produced
by currents
is
\nvention
a desired symmetry
to give
chosen
of study.
A crutch for
topic
is that
the subscripts
are in
next
\nour
con\302\254
\342\200\234effect,
cause,\342\200\235
flux.\n
each
\nof
will
directions
flux
flux
\nthe
\nwill
plus
form,\n
=
and
for
similarly
the
other
fa
^3
+ M\\%it
Liii
two
(1-48)\n
+ Mtziz
(1-49)\n
coils,\n
\342\200\224
=
+ Muiz
h%it
Mtiii
+
Mziii
+ Mztit -|-
Lziz
(1-50)\n
paragraph is now apparent.
inductance
mutual
coefficients
have subscripts
designating row
column
in the above
array of equations.\n
in
as a stepping
stone to
We are interested
linkages
only
in
The
induced
each coil is given by
law
as the
\nage.
voltage
rate
of change
of flux linkages. If the inductance parametersare
are readily
these
found by differentiation to
voltages
The
in the
discussed
symmetry
preceding
\nThe
\nand
flux
volt\302\254
Faraday\342\200\231s
\ntime
be\n
\nconstant,
Vl
=
In
Chapters
\nterms
,
g-
dx2
\342\226\240
dx3\n
Ti\302\243\342\226\240
=
1/
Uu
1/
Mn-s
di\\
.
di2
j
+ Lt
Si
H
dx 1
(1-51)\n
+Ml,H\n
.
+
*
+ Mn
Ttjr
dx2 . r
dx3\n
(1-52)\n
Ti\n
dx3
(1-53)\n
Ma-M+L,-s\n
3, we will consider the
equations will be negative.\n
2 and
in these
dx\\
Lllu+Mlt-di
=
\nv,
r
conditionsunder
which
some
resistance
The
1-8.
OF THE
DEVELOPMENT
1-8\n
Art.
CIRCUITCONCEPT\n
17\n
parameter\n
The passageof electronsthrougha
\nrial is not
without
collisions of the electrons
other
accomplished
\natomic particles.
these
collisions
are
not
and
Moreover,
elastic,
lost
in
each
collision.
This loss in energy per unit chargeis
in potential across the material. The amount energy
as a drop
\npreted
\nlost by the electrons is related to the physical
properties a particular
(1) Physical phenomenon.
mate\302\254
with
energy
\nis
inter\302\254
of
of
\nsubstance.\n
The German physicist Georg SimonOhm
there
is a relationship between the current
experimentally
\nin a
and
the potential
substance
drop. In terms of the fieldconcept,
in energy
\nthe
change
per unit of charge causes a changein the force
unit
electric
This
field.
effect
charge\342\200\224or
\nper
may be interpreted in
of a field
in the direction
of current through the conductingsub\302\254
\nterms
be stated
in terms of this fieldand the
Ohm\342\200\231sexperiment
\nstance.
may
Field
(2)
interpretation.
that
\nfound
\ncurrent
area
cross-sectional
unit
per
J
where, in mks
E
\nmeter,
\nand
<r
the
is
Circuit
an
\nhas
conductivity
<tE
(1-54)\n
current
in
density
amperes
per
square
the conducting substance in voltspermeter,
of the substance, which is a constant for each
material.*\n
\nparticular
(3)
units, J is the
field along
is the
=
as\n
If the
as that
interpretation.
idealized
geometry,
230(2>3\n
substance which carries the current
shown in Fig. l-14(b), it is possible\n
i\n
Fig.
\nIf
the
reduce
to
field form of Ohm\342\200\231s law
of the conductor is
related
are
\ndensity
by
the
where S is
*
Strictly
\nSimilarly
is uniform
Eq.
speaking,
independent
<r is
current
relate
to
and voltage.'
uniform, the currentand current
J cos
d
dS
=
the cross-sectionalarea. For
field
electric
Ohm\342\200\231s
law.\n
equation\n
i = J
\nthe
illustrating
section
cross
the
Conductors
1-14.
and
directed
JS
the
(1-55)\n
same
simple
geometry,
along the length of the
wire;\n
1-54 is a specialcasevalid
for isotropic
substances.
only
of the magnitude of E only for linear substances.\n
that
CIRCUIT CONCEPT\n
THE
OF
DEVELOPMENT
18\n
Chap. 1\n
is,\n
v
of the more
case
a special
as
v
and
El\n
(1-56)\n
general relationship\n
=
cos 6 dl\n
j E
(1-57)\n
the field form
1-56 into
of
Ohm\342\200\231s
law,
gives\n
1-54,
\nEq.
1-55
Eqs.
Substituting
=
v
=\n
(1-58)\n
for constant geometry the
is given
the name the resistance
the
\nconductor,
simply
and is symbolized
other
\nresistance,
by the letter R. For
the simple
one of Fig. 1-14(b), computation of the
\nthan
current
and
for a substance
will be more difficult.
\nrelating
voltage
measurement
of current
and voltage
can establish the value
resistance
the
and by-pass
the computation problem.
parameter
The
which
(1/<tS),
quantity
is a constant
of
parameter\342\200\224or
geometries
coefficient
\nHowever,
\nof
\nOhm\342\200\231slaw
be
may
written\n
v =
or, in terms
of
Ri
charge,\n
-
\342\200\242
The
v =
equation
Ri is
as
known
well
As
is
it
remember
\nand
law
\nhis
in
\naccepted.
\nand
voltage
\nbetween
body\342\200\235
\nby
laity
We
\nobserved
is
by
his
Gv
fellow
(1-61)\n
for conductance
children
(school
\342\200\234Vermont
scientists
mks
the
In
conductance.
with
association
form\n
is the
system,
mho.\n
are taught
=
the
Rhode
when he first
ideas
the law
Island\342\200\235),
announced
were
it was some 30 years beforehis
finally
of
of
current
We
must
that
the
concepts
remember,
course,
in his day, the first distinction
not well understood
were
been made by Ampere in 1820.
the
two
having
quantities
1826,
\nhis
1 -9.
law
f1-60*\n
and
can
such
statements
newspaper
\nReading
ohm and
Ohm\342\200\231s
by
ridiculed
was
\nOhm
the
i
written in the
=
l/R is known as the
resistance
for
\nunit
*
sometimes
i
where G =
(1-59)\n
convince
one
that
as
\342\200\234
10,000
the distinction
volts
passed
through
is not well understood
today.\n
Approximation
have
discussed
experimentally
of a system as a circuit\n
in which three electrical phenomena
the manner
in terms of circuit param-\n
be described
can
Art.
A
eters.
we must eventually
that
problem
CIRCUITCONCEPT\n
OF THE
DEVELOPMENT
1-9\n
19\n
face in makinguse
the
of
of
of representing a physicalsystemin terms
\nthese
can we draw a circuit that will
For example,
rep\302\254
parameters.
coil
of
a
trans\302\254
\nresent
a
a
an
electric
piezoelectric crystal,
wire,
motor,
a
few
to
name
but
\nmission
or
an
antenna,
systems?\n
line,
we examine
some arbitrary
physical system, looking for
Suppose
to be replaced by equivalent parameters. Pos\302\254
of
the
\nportions
system
of
the
effects
would be most easily recognized. A part
resistive
\nsibly
small
cross\nthe system
made of material of high resistivity, with
as equiv\302\254
\nsectional area and appreciable length, would be recognized
\ncircuit
\nalent
\neasily
large
be
distinguished
\nance.
could
and
resistance
to
of
\npart
is that
concept
another
from
of small resist\302\254
the
system
We
found
that there is a
have
between
effect
\ncapacitive
two
any
of
a system.
two parts
\nparts
\nconstitute a system capableof con\302\254
If the
of
\ncentration
electric
\nhigh
\nfor
small
and
storage
area
large
field\342\200\224say,
charge
a
producing
charge,
1-15.
Fig.
dis\302\254
for
\ncircuit
from
\ntance
\nis
that
of
\ntance
to
part
the
of
be
can
\nalent
can
we
of
that
taken
into
about
What
smaller
por\302\254
effects
or secondary
much
the same manner?
Just how
account
in representing a
systemby
in
recognized
answer
many
equiv\302\254
expect
by
a parameter.
\naccount
by
an
make
\npoint,
how
\nthat
the
they
\ncapacitors,
frequently
the
separate
We
must
be
electrical effects we can take into
stop somewhere. We must, at some
requires
engineering
one case will not
judgment.
An
be in another.In
approximation
many
practical
of connecting wires are so small
be neglected.
Likewise, in most cases of commercial
inductive
and resistive
parameters may be ignored. Much
resistance
of coils can be neglected.\n
the
and capacitance
resistance
may
many
good
approximation.\n
Approximation
\nwhich is valid
in
\ncases,
our questions only by asking another:just how
results
to be? The accuracy of our results will
the
\ndetermined
\nless
every
parameters?\n
We
\ndo
every currentbetween
then the inductance,selfor mutual,
large effects.
be
must
\neffects
crystal.\n
is large.\n
system
So much for
\nthat
a piezoelectric
system
one of which is carrying current. If thecon\302\254
in space in such a way
that
the
fields
magnetic
\nreinforce each other,
\ntion
of equivalent
antenna,
at least
located
are
\nductors
(b)
form
an end-excited
effect is associated with
an effect of mutual inductance
and
conductors
of
\npair
\nand
One
inductive
conductor,
\ncarrying
capaci\302\254
of the
an
Finally,
large.
part\342\200\224the
portion
(a)
and
inductance
Chap. 1\n
that
assume
will
we
have
of a system is given, all significant
parameters
into account.
Engineering judgment has been exercisedby
made
But when the student finally
who
up the problem.
a schematic
\nwhen
taken
\nbeen
individual
\nthe
CIRCUIT CONCEPT\n
in other chapters,
to follow
discussions
the
In
THE
OF
DEVELOPMENT
20\n
the
\napplies
\nthese
of analysis
techniques
with
associated
questions
to a problem
difficult
to
write
answers
\nis
usually
the
best
teacher.\n
must
approximation
to such questions
\nis
that he makes up,
be answered. It
in textbooks; experience
is not unique to circuit analysisby any means.
In
problems
by computing the electric and magneticfieldsfor all
will assuredly be approximations, either in
in space,
there
the
physical
system
by mathematical
equations or in
and analysis are bound together.
the
Approximation
equations.
of approximation
is to lack understanding ofthe
the problem
Approximation
\nsolving
\npositions
\nrepresenting
\nsolving
\nTo
ignore
of
\nresults
In
analysis.\n
\nresented
a
by
\ncombination
\nare
commonly
effects
\nwhich
as
\njudgment
\nisolated
\nthere
\nThis
1 -10.
an
such
is
Other
1-58,
\nthat
the
\narea.
\nare
some
circum\302\254
neither
these
necessary
not vary
the
assumptions
and
difficult
is lost
by
radiation.
section.\n
representation\n
Eqs.
parameters,
did
charge
different
in circuit
of the energy
equations for the circuit
was
it
with
If
in the next
approximations
In arriving at
\nand
part
interaction,
be discussed
will
\nvaried
It will, under
resistive
as the
long
all
In
rep\302\254
is useful.\n
approximation
cases we have assumed that the magnetic and electric
fields
are
that
there
is no interaction between the two fields.If
and
as
\nonly
be
to
system
effects. Such unwanted effects
capacitive
distinguished
by the name
parasitic. The decision of
must
be taken
into account involves the same engineering
discussed
The
effects can be ignored
earlier.
parasitic
and
exhibit
\nstances,
inductance.
a pure
unknown
commercial components in
labeled inductor, however,
with
A component
a circuit.
as
behave
start with an
instead
but
circuit,
forming
not
\nwill
we do not
cases,
many
to make simplifying
1-11,
1-36,
approximations: (1)
with dimensions, and (2) that the current
of the conductor nor the cross-sectional
length
do not hold, the values for the parameters
to
compute.\n
how current and charge might
with
space,
is made to flow for but a
that
the current
interval
of time,
\npose
that
this
flow is repeated at a periodic rate, a
large
pulsed
\nnumber
of times
each second.
Under such conditions, the current
will not be uniform
can
\nthe
charge
imagine
throughout the system.
of the system with charge and other portions
\nsome portions
This
the case, the general expressions must be used
being
\ncharge.
To illustrate
vary
sup\302\254
brief
\nand
very
and
We
without
in\n
the
evaluating
CIRCUITCONCEPT\n
21\n
and resistance parameters.
inductance,
or measured, will be different
values,
computed
with
uniform
current and charge in the
different
of a system
be computed for every
capacitance,
new
parameter
\nfrom
those
found
\nMust
the
parameters
\nThese
OF THE
DEVELOPMENT
1-10\n
Art.
system.
\ncurrent?\n
question is, again, a practicaloneof engineering
there will be conditions requiring someeffective
\njudgment.
Certainly,
\nvalue
of
the
parameters\342\200\224computed for a particular time waveform\342\200\224
\nto be
used.
But
in many cases, the approximation that parameter
\nvalues
are
found for nonvarying or static conditions
to those
equal
is strictly valid only in the
usable
This
results
\ngives
approximation
\ncases
in which
of current and charge is slow,theso-called
the variation
We
will assume
that we are operating in this
state.
\nquasi-stationary
in
We
\nstate
to
follow.
thus
assume
constant parametersfor
chapters
The
to this
answer
of
variations
\nchanging
current
and
charge.\n
that the parametersare
with
the
of the
of
and
This
is
a
\ntion
current.
charge
good
magnitude
in their nominal operating range.
for
most
elements
\ntion
system
of such elements is said to be linear.
will
assume
that
all
\ncomposed
a
re
linear.
\nsystems to be considered (unless otherwisespecified)
We further assume
varia\302\254
constant
approxima\302\254
A
We
We
\nthereby
exclude
\ntaining
dielectrics
nonlinear
change
elements
capacitance
and systems.
Some
with the quantity
systems
of charge in
con\302\254
in a magnetic system,the flux produced
related
to current because of saturation. Such resistive
\nis not
linearly
in a lamp bulb change resistanceas
the
carbon
filament
as
\nmaterials
of current.
It should be noted, however, that
of magnitude
\na function
linear under certain condi\302\254
can
be
considered
nonlinear
\nsome
systems
tubes
are nonlinear,
but for certain analyses may
Vacuum
\ntions.
be
linear
over a restricted
\nconsidered
range of operation.\n
the requirement
we will include
Besidesthe assumptionof linearity
in a system
be bilateral. In a bilateral system,the
all
elements
\nthat
between
current
exists for current flow\302\254
and voltage
\nsame
relationship
direction.
In
either
contrast, a unilateral system has different
\ning in
and
\nlaws
current
for the two possible directions of
voltage
relating
of unilateral
elements
are vacuum diodes, ger\302\254
\ncurrent.
Examples
\nthe
When
system.
\nmanium
Many
\nmission
iron
diodes,
crystal
electric
systems
line,
for
example,
is used
detectors,
selenium
are physically
may extend
rectifiers,
etc.\n
distributed in space. A trans\302\254
for hundreds of miles. When
a
to the transmission line,energyis trans\302\254
of light. Because of this finitevelocity,
\nported
nearly
velocity
\nall electrical
do not take place at the same instant of time.This
effects
the case,
the restrictions
discussed earlier apply in the computa\302\254
\nbeing
\ntion
of the
circuit
in\n
When a system is so concentrated
parameters.
\nsource
of energy
at
is connected
the
the
that
space
is
a good
\nconsider
only
\ntem
THE CIRCUIT
OF
DEVELOPMENT
22\n
Chap.1\n
CONCEPT\n
of simultaneous
actions through that
the system is said to be lumped.We
assumption
approximation,
sys\302\254
will
systems.\n
lumped
to the approximation
Our circuit approach
a
of
obscured
has
system
of electric and
As
that the
fields.
an approximation,
\nnetic
field
is associated
only with an inductive system and that the
field
is associated
with a capacitive
only
system. Fields
be so lumped.
The consequence of interaction of the fields
\nnot
actually
of electromagnetic
the
radiation
Open a switch in an
energy.
and
the effects will be observed as a noisein
radio
\ntive
system,
the ignition
\nreceivers.
Similarly,
spark of an automobile may affect
effect
\nan
of the interaction
we have assumed
terms
in
described
usually
\nmagnetic
mag\302\254
can\302\254
\nelectric
induc\302\254
\nis
nearby
\nnearby
television
\namount
of
be
\ncan
We
ignored.
have
will
\nsive
Sources
physical
elements
Energy and
1-11.
and
Energy
and
1-5
\nEqs.
inductive
an
and
capacitors.\n
are
given
power\n
which
is
spoken
\342\200\224
vi
=
\nsystem,
=
is spoken of as
\ncauses energy to be
= i
the
in
the energy is given by
which
by
\342\200\224
vi
dt\n
magnetic
field.\342\200\235
In
Prob.
(see
the
1-8)\n
(1-62)\n
the relationship
=
in
(see Prob.
joules
^
a
capacitive
1-9)\n
(1-63)\n
joules
\302\261Cv*
\342\200\234stored
j
has the value
t Li*
\342\200\234stored
Wc
w
and
energy
system,
of as
voltage and current
in terms of
are\n
WL
and
by different names
are distinguished
themselves
p
In
bilateral
and inductive elements are identifiedas pas\302\254
electric
energy are identified as active elements.
of
power
1-6,
approximation.\n
be lumped, linear, and
radiation.\n
inductors,
resistors,
as an approximation
thus
will
study
capacitive,
elements.
\nThe
\nas
shall
we
negligible
Resistive,
is small, and
by radiation
will make this
the
however,
conditions,
many
lost
energy
The systems
\nand
Under
receivers.
electric
field.\342\200\235
transformed into heat
or
in
Current
light.
The
a resistor
amount
of
\nenergy is\n
WK
The
equivalence
\nstrated
\nwhile
by
the
Joule.
resistor
of this
The
=
energy
inductor
is an
R f t*
dt
joules
(1-64)\n
to mechanical energy was first demon\302\254
and capacitor are energy storageelements,
energy sink.\n
\nin
a
given
\nin
a
system
THE CIRCUIT CONCEPT\n
23\n
quantity, always positive. The sum of the energy
can be found by algebraic summation. The power
system
is given
in terms of energy by the relationship\n
is a
Energy
scalar
P
Pr' is the
where
OF
DEVELOPMENT
1-11\n
Art.
=
W
= Pr' +
total powerof all
+
(1_65)\n
Wc>)
Jt(Wl\342\200\230'
elements
resistive
given
as\n
n\n
Pr'
=
=
^=
d^jf
3
and
is the
Wl
R,V
(1-66)\n
1\n
in all inductive
total energy stored
elements,\n
m\n
W.\n !
\342\200\224
h
(1-67)\n
L/ij*\n
^
3 = 1\n
and
is the
Wc
total energy stored
Wo'
=
*
in all capacitive
elements,\n
^=
3
where
n,
m,
and
p are
W
(1-68)\n
1\n
the total number of elementsof eachof
the
three
\nkinds.\n
FURTHER
READING\n
in further reading on the subjectsof this
should
consult
\nchapter
Elementary Electric-Circuit Theory (McGraw\nHill
New
Book
Co.,
York, 1945) by Richard H. Frazier, pp. 17-41.
\nMore
advanced
treatments
of these
concepts are given in Electric
\nCircuits
& Sons, Inc., New York, 1940) by the MIT
Wiley
(John
\nElectrical
Engineering
pp. 1-8, and in Linear Transient Analysis
Staff,
&
New
Sons, Inc.,
\n(John
Wiley
York, 1954) by Ernst Weber, pp. 1-14.
\nA discussion
of physical
in general is to be found in Chaps.1
systems
\nand
2 of Response
of Physical
Systems (John Wiley & Sons, Inc., New
John
D.
Trimmer.
Students interested in further
1950)
\nYork,
by
of
electric
and
fields as basic concepts should read
\nstudy
magnetic
\nThe
Fundamentals
of Electro-Magnetism
(The Macmillan
Company,
\nNew
1.
on
Maxwell\342\200\231s
1939)
York,
by Geoffrey
Cullwick,
starting
p.
in many libraries under the title, A Treatise
are found
\noriginal
writings
\non Electricity
and Magnetism
(Oxford Press, New York, 1892).\n
student
The
interested
PROBLEMS\n
A solid
1-1.
\ntrons
by
copper
a charging
of
101S elec\302\254
sphere 10 cm in diameteris deprived
is
the
of
scheme,
(a) What
charge the spherein\n
OF
DEVELOPMENT
26\n
THE CIRCUIT
CONCEPT\n
Chap.1\n
watt, (h) At what rate is energy being
as heat?
\ndissipated
(i) At what rate is the battery supplyingenergy?\n
1-16.
In the circuit
shown below, the capacitor is chargedto volt\302\254
in the cir\302\254
The
current
of 1 volt,
and at t = 0 the switch K is
\nage
is known
\ncuit
to be of the form i(t) = e-\342\200\230
(t > 0). At a certain
amp,
\ntime
the
has a value of 0.37 amp. (a) At what rate isthe
current
voltage\n
the
0.23
Answer:
inductor?
a
closed.
,
\t\n
K\n
+LV-1V\n
cr
c-it\n
Prob.
across the capacitor
\nthe
\n(d)
much
\nWhat
is
\nbeing
taken
At
the
what
changing? (b) What is
What
is the
the
value
of
the
charge
time rate of change of the product
on
Cvl
across the capacitor? Answer: 0.37 volt, (e)
in the electric field of the capacitor?(f)
is stored
energy
the resistor? (g) At what rate is energy
across
voltage
from
the electric
field of the capacitor? Answer:0.137watt,
is the
What
\nHow
\n(h)
(c)
capacitor?
1-16.\n
rate
voltage
is energy
being dissipated as
heat?\n
1\n
Chap.
the
CIRCUIT CONCEPT\n
THE
equation for energy, W
=
WL
\\Li% and WL = ^2/L.\n
the
From
1-8.
\nfor
OF
DEVELOPMENT
defining
inductance,
=
/ vi
25\n
dt show that,
for the
the equation for energy in Prob. 1-8,show
that
=
=
\ncapacitance, Wc
?Cv2 and Wc
?Sq2.\n
1-10. Assume that the inductance parameteris defined
as the
con\302\254
\nstant
stored
and the current squared by the equation
relating
energy
=
that
for
use of the relationship p = vi, show
\nWL
?Li2.
Making
\nconstant inductance
the voltage across the inductoris vL = L(di/dt).\n
in Prob.
1-11. Carry out a similar derivationto the one
suggested
=
to show
with
for a capacitive system, Wc
\n1-10
?Sq2
starting
energy
From
1-9.
constant
for
\nthat
S,\n
=
vc
\ntime:
RC;
(a)
= S
J i
dt\n
all have the dimension
the following quantities
(b) L/R]
(c) y/LC.\n
that
Show
1-12.
Sq
of
of
1-13. Show that (a) R2C has the
has
\ny/L/C has the dimension of resistance, (c) L/R2
dimension
\ncapacitance.\n
1-14.
the
\nin
of
\nrate
The current in
the
in
\npassed
through
weber-turn/sec,
\nreference
time).
i(t)
=
(1
inductor
(c)
0.5
switch
\342\200\224
e~l)
K
in the circuit
current
amp,
flux
flux\n
coulomb.\n
circuit shownabove,the
The
the
2 sec, (c) the quantity of chargehaving
after 1 sec. Answer: (a) 1 weber-turn,
after
system
the
1-16. In the
\ntion
shown
The current increases from t = 0 at
figure.
accompanying
1 amp/sec
seconds, at least). Find: (a) the
(for several
of
after 1 sec, (b) the time rate
in the
change
system
linkages
1
variation
the
follows
of
dimension
of
\nlinkages
\n(b)
a 1-henryinductor
(b)
inductance,
the
t
flowing
> 0. At
is closed
at t =
0 (the
is givenby the
equa\302\254
a certain time the current
has
a
What
At what rate is the currentchanging?
(b)
What
is
the
rate
of
the
value
flux linkages? (c)
change of
\nflux
is the voltage across the inductor? Answer:
What
linkages?
(d)
of
the
\n0.37
much energy is stored in the magneticfield
volt,
(e) How
is
the voltageacrossthe resis\302\254
\ninductor?
Answer: 0.20 joule, (f) What
in the magnetic
stored
\ntor?
field
of\n
At
is
what
rate
(g)
energy being
\nvalue
\nis
of 0.63
amp. (a)
of the total
FOR
CONVENTIONS
28\n
DESCRIBING
NETWORKS\n
Chap.
2\n
vacuum tube amplifier are
as ideal
of
that
can be approximated
practical
generators
sources
with
series
or parallel
passive elements. The current
an
of a circle but
is
also
represented
by the symbol
\nciated
arrow
rather
than
indicating the positive
markings
polarity
current
in Fig. 2-1
of
flow
as shown
For
both
the associated
sources,
symbol of a lower case letter or i
\ninfers a time-varying
source, while the upper case letter or / is used
The approximate
time variation of
denote
a time-invariant
source.
the circle. For example, the
is sometimes
sketched
within
~
or
in the circle indicates a source of sinusoidal
\nbol
placed
The
cell
photoelectric
the
and
pentode
\nexamples
\ncurrent
asso\302\254
with
\nsource
\ndirection
(c).\n
v
V
\nto
sym\302\254
\noutput
voltage
\ncurrent.\n
conventions\n
and voltage
Current
2-2.
within
the source
in the direc\302\254
A voltage source causes currentto flow
out
of the positive
to the positive terminal\342\200\224or
from
the
\ntion
negative
\nterminal and into the negative terminal. This particular
convention
Franklin in 1752. Franklin\342\200\231s
made
\nfollows
a decision
by Benjamin
was identified with the electron,
made
before
\nchoice
was
electricity
of charge were known. Actually,
or the nature
\nbefore
the
electron
to the positive terminal,
terminal
flow
from
the
\nelectrons
negative
direction
to that established by Franklin.
is
in
\nwhich
the
opposite
the flow of electrons is termed
the
two
\nTo
conventions,
distinguish
assumed
and
current
current
\nelectron
positive in the direction of Frank\302\254
is
\nlin\342\200\231s
convention
is
\nthis
current
the
conventional
called
we will
of
since
used
as
in measuring
a reference
the
poten\302\254
of a potential source, that voltageis con\302\254
positive
of as a voltage rise. Conversely,if the
is spoken
and
positive
is considered
to be the reference in measuring
the
terminal
terminal
of the voltage source, the voltage
of the
negative
terminal
the
\nsidered
\npositive
\npotential
considered
\nis
(or simply current,
use).\n
If the negativeterminalis
\ntial
current
rise\n
Voltage
0
is spoken
and
negative
Voltage
of as a voltagedrop.\n
b
e\n
a
d\n
drop\n
a\n
la)\n
2-2.
Fig.
lb)\n
Sign convention
\nsources.\n
In
\ncurrent
\nflowing
terms
to
in
for voltage
Fig.
2-3.
Direction
\nvoltage
of current
and
polarities.\n
causes
simple circuit of Fig. 2-3, the voltagesource
flow from a to 6 and around the circuita-b-c-d-a.
Current
the passive
resistive element is identified with energylossand\n
of the
CHAPTER
NETWORKS\n
DESCRIBING
FOR
CONVENTIONS
2\n
2-1. Active element conventions\n
Most
\nacteristics.
maintain
generators
practical
by their voltage-current char\302\254
approximately
with
voltage
electric
\nof
sources
of
\nof
current
with
into
relationships
\nvoltage-current
practical
account,
con\302\254
Still other types
constant output
than take actual
load current.
increasing
maintain
energy
approximately
terminal
increasing
voltage. Rather
terminal
\nstant
are classified
elements
circuit
Active
energy
sources are
ideal
sources
or
current
sources.
\napproximatedas either ideal
\nThese ideal sources are defined;
as
the
following
properties:\n
The Ideal Voltage Source. The ideal voltagesource
with
a given
time variation.
Neither voltage magnitude nor time
\nage
with
of output current. Thus the
changes
magnitude
\nminal
is assumed
to be maintained under all conditionsfrom
voltage
If in some manner the terminalvoltage
circuit
to short
circuit.
made
to zero, the source behaves as a short circuit.Theideal
equal
source
has
no resistive,
inductive, or capacitive effects. Most
voltage
having
volt\302\254
generates
\nvariation
ter\302\254
\nopen
\nis
\nvoltage
\nactual
generators
\nmated
as
with
\ncases,
an
a
represented
circle
as shown
volt\302\254
by
the
in Fig.\n
to
this
exception
prac\302\254
the
for
a
battery
symbol
in Fig.
2-1 (b). The polarity
a
(6)
(a)
One
(a).
\ntice
ideal
is
of
\nsymbol
2-1
The
inductor.
source
\nage
approxi\302\254
voltage source in
and in some
resistor,
ideal
an
\nseries
be
may
is
\nshown
\nmarks
+
and
\342\200\224
denote
and
positive
Fig.
2-1.
Symbols
(O\n
circuit
for active
\nelements.\n
terminals
negative
of the
source.\n
source generates
\nrent
time variation.
Neither current magnitude nor time
with
a given
This
with
load.
changes
output current is maintained for
and infinite resistance.* If the
load
zero
resistance
including
current
of the
ideal source is adjusted to be zero,the
is
\nput
in
caseof
an
circuit.
As
the
the
to
\nequivalent
open
voltagesource,
\nideal
has no resistive,
or
current
source
inductive,
capacitive
*
Of course,
this property of the ideal sourceto
into
an open
amperes
Ideal
The
Current
Source.
The
ideal current
cur\302\254
\nvariation
out\302\254
\nany
source
the
effects.\n
pump
\ncircuit
is
a poor
approximation
for actual
27\n
sources.\n
The
junction
formed
is
the
\ntogether
given
NETWORKS\n
or more elements being
Elements in series such
form a branch.
flows in them
A
\nparts
\nably,
\nthe schematic,
for
\nnode
branch
may
or mesh is
drawn
of
example.
Parts
pair.
called
are
coupled
\nmagnetically
find
in the
of
one
2-6.
\nword
are
there
\nBriefly,
node
\npendent
\nactive
of
of nodes.\n
Identification
The
inferred
meaning
in the next
are
as
inde\302\254
pairs.\n
the number of
network
num\302\254
num\302\254
will
\nnodes
of
the
by
chapter in moredetail.
currents in a system as there
unknown
unknown
voltages as there are
many
discussed
many
and
node
both
elements in the
(counting
but not mutual inductance). The
and passive elements,
is designated B. Quantities relatingto
in the network
branches
be
identified
by Nt for the number of nodes, J for the
different
node
pairs, and N for the number of independent
of independent loops, and finally,S is
L is the number
Also,
E be
Let
as
loops,
\nindependent
be
will
independent
and\n
(c)\n
of independent node pairs.
number
loops
branches)\n
(6)
Fig.
of
node\n
one
\n(four
different
quantities
different
(a)
Fig.\n
of the network.\n
parts
separate
isto
the
currents
Our objectivein circuitanalysis
Two
nodes.
\nbranches and the voltages at the
number
independent
\nimportance in this analysis are the
\nber
iden\302\254
a closedcontour
on
around
one or more window panes of the graph
a
Any two nodes in a network may be considered
not directly connected by wires but
of the network
A loop
circuit.
the
than
\nelements
\nber
connected
that
A network or circuit is formedby intercon\302\254
elements.
of a number
of branches or by coupling a numberof separate
We will use the words network and circuitinterchange\302\254
together.
is usually more complex, involving more
the network
that
except
\nnection
the
2\n
Chap.
active
\ninclude
2-4,
by two
name
node.
current
same
the
\ntically
DESCRIBING
FOR
CONVENTIONS
30\n
node
the
\npairs.
of
\nnumber
parts
separate
The number of
of the
network.\n
different node pairs is
given
by
the
topological
\nequation\n
J
We
are
usually
not
interested
= iN,(N,
- 1)
in all combinations
(2-1)\n
of nodes, but
is\n
FOR DESCRIBING
CONVENTIONS
2-3\n
Ait
29\n
NETWORKS\n
a
quantity
drop in potential from c to d. Voltageis scalar
of the two scalar quantitieswork
and
\ndefined
as the quotient
charge.
\nVoltage drops and voltage rises are distinguishedby a sign,positive
\nfor voltage
rise and negative
for voltage drop. For this polaritysignto
the
\nhave
reference
must be given or inferred. A
meaning,
voltage
is a
there
so
\ncertain
point
\nrespect
to
\nwith
a
\nwith
interconnecting
\nsuch
graphs,
\nwith
the
in
lines.
not
\ncles
\ndow
\ntwo
having
window
of each
panes.\n
lines,
other. For example, if
the number of lines is
cir\302\254
by
of
win\302\254
we have studied.\n
quantities
example
law
from
or
panes
the numberof
increased
one,
law
homely
\nlaws
window
more window pane.
of
Our discussion has beena
\nmathematics known as topology.The general
\nknown
and
circles,
Similarly,if the number
constant
is maintained
and one more circle is added,the
increase
lines
must
by one, a line having been dividedin
added
circle. There must exist some general
relating
of
\nthe three
and
constant
the
by
no lines
of
be one
panes
\nnumber
cir\302\254
all
definition,
By
independent
will
on
given graph,thenumber
remains
\nthere
ground
two
called
be
will
Now in any
\nare
the
called
relationship
rules
imposed:\n
least
\nclosed contours
\nwithin
be
will
lines must
and
circle,
(3) the graph
two
lines will
dimensions;
cross
\nnot
which
assumedto be
geometry)\n
terminate
must
at
every
be
\nwill
reference, it is
in Fig. 2-4 is made up of circles
that we wish to make a study of
of the number of circles and lines,
following
(2)
\njoin
with
b. Should the
shown
graph
lines.
Suppose
the
say
lines
\ncles,
of +50 volts
to point
respect
of zero potential
a point
two-dimensional
\tall
with
node.\n
datum
The
have a potential
without
given
Network topology (or
2-3.
(1)
be
point
to
respect
the
\nor
point
of
\npotential
a circuit
may
a and \342\200\22430volts
in
studies
facts
in
in
circuit
topology.
How can
an
branch
elegant
just
mentioned
of
is
we exploit these topological
analysis?\n
as
By. approximating a physical system by ideal circuit
we have eliminated consideration
in the first chapter,
\ndiscussed
in
favor
of a system of interconnected
\nthree-dimensional
systems
as a wiring diagram or a schematic.The
described
\nments
usually
to the topological
is equivalent
\nmatic
elements
graph of Fig. 2-4
of
Thus
the
laws
are
lines.
\nreplacing
topology
directly applicableto
Before
these
schematics.
giving
laws, we will define terms
elements,
of
ele\302\254
sche\302\254
with
\nnetwork
\nused
in
network
topology.\n
be
\nnot
chapter).
in this manner,
chosen
be
\nloops
In
independent.*
\nloops
is
\nsion
and
equal
of
\342\200\234window
at a
be determined
can
so
simple
number
the
to
Chap.
2\n
It is, however, not necessarythat the
will
but there is the risk that the loops
the number of independent
networks,
next
the
in
discussed
NETWORKS\n
DESCRIBING
FOR
CONVENTIONS
32\n
our
of
panes\342\200\231'
earlier
discus\302\254
glance.\n
1\n
Example
one
and
\nments,
in Fig. 2-7, there are five nodes, sevenele\302\254
Hence the number of independent loops is three,\n
shown
network
the
In
part.
of Example
Network
2-7.
Fig.
1.\n
since\n
=
L
=
\nN
The
1
=
node
independent
3
four, since
4\n
the figure,followingpaths
g is selected as the datum
node pairs is a-g}
b-g,
c-g,
d-g.\n
2\n
Example
of
Examination
=
Nt
\ncourse),
L
=
\nN
=
discussed
\nholtz
rule
\342\200\224
Nt
Nt
to
in
than
2-8
Fig.
\342\200\224
determine
Chap.
5
+
5 =
=
6
6
\342\200\224
2
independence
3, be
follows
6
+
2 =
2 independent
node
4 independent
requires
nonzero. It is usually
to experiment.\n
(M doesnot
count,
of
that\n
\342\200\224
=
E = 6
that
shows
and S = 2. It
6,
E
* The test
\nbe
of the
choice
\342\200\224
If
a-b-c-d-a.
and
suitable
a
\nnode,
5
as shown on
assigned
d-c-e-f-d,
\na-d-f-g-a,
\nand
be
may
loops
- S =
Nt
+ l=
7-5
node pairs is
of independent
number
the
\nand
+ S =
E-Nt
that
better
the
pairs\n
system
to
loops
follow
determinant, to
the von
Helm\302\254
\nof
unknown
given
voltages,
= Nt
=
\nof
elements
=
L
If
count
we
nodes
only
\nbranches
be B',
\nare related
as\n
then the number
2-2 is
if Eq.
=
these
equations,
or
L
\nquantity,
=
E
part.\n
given in terms of the
node pairs
E
number
as\n
- N
(2-3)\n
of branchesand let
of independentloops
and
-
of
number
that
branches
W
(2-4)\n
results\n
- Nt + S
(2-5)\n
determine for a given network which
of
the
two
enabling us to decide which
we may
is smaller,
N,
number
(2-2)\n
into Eq. 2-3, there
substituted
L
From
only one
E
at the ends
L
Finally,
is equal to the
S
of independent
loops is
of independent
number
and
number
The
for
1
31\n
\342\200\224
Nt
with
a network
\342\200\224
NETWORKS
as\n
N
or N
DESCRIBING
node pairs, which
of independent
number
the
FOR
CONVENTIONS
2-3
Art.
to analysis should be taken.\n
node pairs is known, there remains
be
which
N node pairs in the network will
\nthe
of selecting
problem
mem\302\254
\nused in analysis.
Analysis is simplified if one node is usedas one
to select the node
\nber
of each
of the N node pairs. It is conventional
as this
common
node and to designate it the datum
\nof zero
potential
the negative terminal of one of the active
node.
\nor reference
Usually,
\npossible approaches
so
is
\nsources
is a
There
\nindependent
Fig.
2-5.
\ning
The
the
rule
paths
chosen
\nsively
\nnot previously
\nrequired
\nthe
by
independence
of independent
number
the
Once
selected.\n
similar problem of choice in the caseof
loops once the number of such loopsis known
2-6.
Different independent
of von
Helmholtz
assigning
from
the
Eq.\n
loops in the samesystem.\n
may be used to advantagein designat\302\254
The rule requires that the loopsbe succes\302\254
that each new loop includes at least onenew
branch
in the path of a loop. If the loops
included
of
number
insure
to
Eq. 2-5 are so chosen, this processis sufficient
of the
be\n
loops (and of the voltage equations, as will
of the
such
loops.
the
of
of
scheme
\ntual
Current
(1)
to
\ndistance\342\200\235)
\nThe
magnitude
2\n
Chap.
that
Let us
dot. We will outline, step-by-step,our
into this
flows
\ncurrent
NETWORKS\n
with a dot.
marked
is shown
winding
DESCRIBING
FOR
CONVENTIONS
34\n
the
assume
concep\302\254
as a consequence of this current.\n
happens
in winding
1-1 causes a magnetic field (\342\200\234action at a
which
is concentrated
exist,
along the axis of the coil.
field can be computed from Ampere\342\200\231s law\n
of the
what
a
dl cos
nil
dB\n
(2-6)\n
\n4x r2\n
is a
There
(2)
flux
magnetic
the
with
associated
<t>
1-28.)\n
field
magnetic
value\n
a
\nhaving
1 in Eq.
in Chapter
defined
are
symbols
(These
4>
B cos 6
=
J
dA
(2-7)\n
\nApplying
determined
and given by the
experimentally
if the thumb
rule:
of the right-hand indicates the direction
the
the current-carrying conductor in
fingers
wrap around
of flux.
This flux is assumed confined to the magnetic
has
the property
of being a preferred path for the flux.
the
right-hand
rule, the flux is seen to have the direction
\nindicated
by
and
a direction
having
\nright-hand
\nof
current,
direction
\nthe
which
\ncore,
(3)
arrow
the
(clockwise).\n
Since winding 2-2 is on the
\nthe
flux
produced
\ncan
be
described
\ncause.\342\200\235
in
as
of flux
number
The
the
where
=
of
terms
In
1-1
\nterminal
law
Faraday\342\200\231s
age
order
the
have
flux
\342\200\234effect,
in winding 1-1 is\n
Ai$2i
be
\\pi can
(2-8)\n
from the voltage
computed
at
as\n
=
Eqs.
Combining
1-1,
winding
winding 2-2. This linking
subscripts
linkages
$i
as
core
magnetic
1-1 links
winding
</>2i,
same
2-8 and
/
Vi
2-9 gives the
dt
(2-9)\n
value of
flux
in
terms
of the
volt\302\254\n
t>i.\n
(2-10)\n
\tBecause
(4)
\ning
2-2
according
fai
is changing
to
with time, a
law.
Faraday\342\200\231s
The
voltage is inducedin
flux
linkages
wind\302\254
in winding 2-2
\nare\n
^ 2
and
02 has
the
=
Ni<j> 21\n
(2-11)\n
magnitude\n
(2-12)\n
2-4\n
Art.
The
When
convention
the
magnetic
for coupled
of Example
33\n
2.\n
circuits\n
field produced
by a changing current flowing
in
other coils,the coils said to be coupled,
If the details of transformer
a transformer.
constitute
windings
a
current
then
for
are
changing in one coil, it is
known,
and direction of the voltages
to
the
magnitude
compute
The necessity for cumbersome
in
all other
windings.
a voltage in
induces
\none coil
the
\nand
dot
Network
2-8.
rig.
2-4.
FOR DESCRIBING NE1WCR.S\n
CONVENTIONS
\nconstruction
\npossible
are
blue\302\254
\ninduced
is eliminated
construction
showing
\nprints
by two characterizing
fac\302\254
coefficient of mutual inductance, M (discussed
to details of construction in computing
\nin
1) is equivalent
Chapter
manufacturers
mark one end of
Most
of
induced
voltage.
\nmagnitude
with a dot (or some such symbol). The dot
transformer
\neach
winding
direction
is con\302\254
as far as voltage
of construction
to details
\nis equivalent
of
the
dot
we will discuss
In
this
\ncerned.
meaning
markings,
section,
and
their
are
\nhow
significance in cir\302\254
established,
experimentally
they
\ncuit
of the
value
The
\ntors.
analysis.\n
are shown
windings
sense is
the winding
Two
\nfigure,
Fig.
2-9.
A two-winding
of
(which
\nsecondary
\nto
winding
winding).
1-1
\nsource has the
In
winding
dot
primary
A time-varying
in series
and
and
winding)
source
with resistor Ri.
arrow
1-1\n
convention.\n
a
ii(t)
is increasing
with
is
vg(t)
instant,
given
current
2-2
winding
of voltage,
At
polarity shown,and the
\ndirectionshown by the
this
magnetic circuit used to establishmeaning\n
the
be called the
might
on a magnetic core in Fig. 2-9.
indicated for two windings,
is
(the
connected
the
voltage
in the
flowing
time.
The +
end\n
FOR
CONVENTIONS
36\n
a battery,
of
minal
the
of
\nend
\npositive,
with
measured
the negative terminal to the remaining
of winding 2-2 that momentarilygoes
is the terminal to be dottedin
a voltmeter,
with
dots,
the
generator
the
voltage
of
\nterminal
A
\n2-2.
step-by-step
current
\nincreasing
the
\ncauses
upper
to
of Fig. 2-9,
circuit
including\n
and resistor load interchanged. The
source is connected to the dotted end of
positive
winding
of this transformer will show that an
into
terminal of winding 2-2
the dotted
flowing
end of winding
1-1 to be positive and so to be the
analysis
expect, after all,
We would
2-2
Now
now establish,in
that dots established
should
agree with those established from 2-2 to
that
the voltage source of Fig. 2-11 has reverse
suppose
shown
and that an increasing current flows out of
or simply intuitive
step-by-step
analysis
reasoning will show
dotted
terminal
of winding
1-1 becomes negative under such
terminal.
\ndotted
from
1-1.\n
polarity
that
dot.
the
\nAnother
the
\nthat
2-11
Figure
\nanalysis?
the dots, which we can
the transformer
shows
are
value
what
Of
\nto
2\n
Chap.
2-2.\n
\nwinding
\n1-1
NETWORKS\n
connecting
The end
winding.
as
DESCRIBING
\nconditions.\n
with polarity markings (dots),
into
the
dot on one winding induces a voltage in the
flowing
is positive
which
at the dotted terminal; conversely,
winding
out
of a dotted
terminal
induces a voltage in the
flowing
\nond
which
is positive
at the undotted terminal. This important
winding
in Chapter
will
be
3 in formulating circuit equations.\n
applied
Thus far our discussion been limited
to a transformer
with two
In
with several windings, the same type of
a system
can
be carried
on for each pair of windings providingsome
\nysis
in the form
\ntion
of the dots is employed (suchas
to identify
between
each
In Chapter 3-, after
relationship
pair of windings.
of assumed
direction
of current is introduced, it
concept
positive
be
shown
that
the information
given by the pair of dots can
in the sign of the coefficient of mutual inductance. For a
\ngiven
this scheme
avoids the confusion of a large
\nwith
windings,
many
We
conclude
that,
for a
transformer
\ncurrent
\nsecond
\ncurrent
sec\302\254
\nrule
has
anal\302\254
\nwindings.
varia\302\254
\342\226\240
\342\226\262
\342\200\242
\342\231\246)
\nthe
\nthe
\nwill
be
system
num-\n
\nthe
under
system
inductance
flux
relate
to
introduced
\nwas
=\302\273
Mi.
in
written
be
may
V2
not
if M21 does
in
induced
\nis
of
rate
\ntime
(5)
of a
\nthe
transformer,
a
the
\noppose
\nto oppose
is
\nlaw
really
a magnitude
There remains the
an
2-2 can be
problem
the
found
the
of
the
that
voltage
a current in
in a coil
induced
the coil in a
</>2i
direction
application
another
<\302\24321,
to
direction
produced the voltage.The
a
2-9 and is increasing.To produce
flux
(by
requires
the
of
terms
In
in flux that
to aid
with
physicistLenz in 1834.
states
law
establishes
the
in
a flux
\nduced
Lem\342\200\231s
change
flow
us that a voltage
of Mu volts per unit
2-14 tells
Equation
by the German
as\n
(2-14)\n
in winding
voltage
in Fig.
upward
in
this increase
\ndirected
\ncurrent
of
flux
of
change
more useful) form
voltage.\n
law given
\naid
\nby
change
(but
Mn^
having
of current i\\.
winding
(2-13)\n
=
2-2
direction
\tThe
M
equivalent
time.
with
vary
of this
\ndirection
For
study\n
*2 =
and Eq. 2-12
35\n
NETWORKS\n
1, the coefficientof mutual
linkages with current as ^
of Chapter
discussion
the
In
FOR DESCRIBING
CONVENTIONS
2-4\n
Ait.
is
fax
flux
<t>\\2
rule) that the
right-hand
by the arrow (right to left). Lenz\342\200\231s
of conservation of energy, sinceif t'2 pro\302\254
current
would be induced in\n
increasing
shown
a vicious cycle to produceinfinitecurrent.\n
the
of current in winding 2-2 is established,
that
the direction
Now
and
so
marked
with
a
is
seen
to
be
of
the
end
winding
positive
\ntop
a time-varying voltage, the dotted terminalsare positive
\ndot. With
same time (and, of course, neg\302\254
\nat the
This action
at the same time).
\native
As shown,
in Fig. 2-10.
\nis illustrated
from
zero to a constant
increases
\nvg(t)
The
current
at time
\nvalue
ix and so
t\\.
with
time as shown
\nflux
<f>n increase
1-
\t1
on in
so
and
is
\nin
<t>21 in
induced
\nto
the
time
\nso
has
the
\n(c).
This
that
incidentally,
terms
Eq.\n
directly, since it
not apply
does
\ngives
The
Note,
(b).
2- \t10
of
vx
rather
than
vg.\n
v2 is proportional
voltage
rate
of change of ix and
time
example
variation
suggests
shown in
a simple
Fig.
2-10.
\nnetic
in
Waveforms
of Fig.
circuit
the
mag\302\254
2-9.\n
of
transformer
\nexperimental method for establishingthe dotted
On the winding selected as 1-1, arbitrarily mark oneend
\nwindings.
\nthe
with
a dot and to this terminal connectthe positive
winding
ends
of
ter-\n
CONVENTIONS
38
2-2.
Repeat
Prob.
2-1 for
FOR
DESCRIBING NETWORKS
the network shown
Chap.2\n
in the accompanying
\nfigure.\n
2-3.
In
the
accompanying
of a cube. For
figure,
a number of elements are arranged
this network, determine L, the
and N, the number of independent node
\nindependent
loops,
In the network
2-4.
of the figure, the paths for threeloops
as shown.
Are these three loops independent?
\nselected
in Li and L2 in terms
\nare the
currents
ii?)\n
\non
the
edges
of
number
pairs.\n
have
Why?
been
(What
of
Prob.
2-5.\n
The magnetic system shown in the figurehas three
\nmarked
1-1', 2-2', and 3-3'. Using three differentformsof
\nlish
for these
windings.\n
markings
polarity
2-6.
windings
dots,
estab\302\254
Prob.
shown
for
2-5 with
2-6. Place three windings on the core
terminals
in
(placed
\nwinding senses selected such that the following
1
in the figure for Prob. 2-5) have the samemark:(a)
order
\nthe
shown
3' and 1'.\n
\nand 2, 2 and
3',
3, 3 and 1, (b) 1' and2', and
2-7.
The
shows four windings on a magnetic flux-conducting\n
figure
2'
Prob.
2-7.\n
Art
2-4
ber
of
FOR DESCRIBINGNETWORKS
CONVENTIONS
are
both
and
\nproblems,
schemes have advantages for particular
used in electrical engineering literature.\n
Both
dots.
similar
ExampleS
the
In
,\n
JL/.'ll' P\n
in Fig. 2-12, the winding senseof eachcoilof
The polarity markings for each set of\n
indicated.
shown
system
is
transformer
\nthe
2-12.
Fig.
was
\nwinding-pair
then
was
\ndot
the
on
shown
are
coils
figure.
arbitrarily
in
Guillemin\342\200\231s
New
Inc.,
\nSons,
Matrix
\ninclude
&
\nWiley
of
\ntions
In each case, one of the dotsfor
selected and the position of the
Circuit
in Chap. 1.
by
theory\342\200\235
for
of the
and
network
dot convention
Circuits
Magnetic
especially
1950),
topology,\342\200\235
is
geometry)
Theory
other
treated
Wiley
(John
&
Other suggestedreferences
by LeCorbeiller (John
article
the
\342\200\234On
founda\302\254
Ingram
and
Cramlet
(1944).
See
also
Math. Phys., 23, 134-155
Electric
\nNew York,
network
Networks
of Electric
York, 1950) and the
network
electrical
subject
of
\nciples
New
Inc.,
Sons,
topology(or
1953)
York,
\n\342\200\234IREstandards
the
each
READING\n
Introductory
Analysis
\nappears in J.
\nOn
3.\n
determined.\n
The subject of network
detail
for Example
circuit
Magnetic
FURTHER
\nin
37\n
Proc.
IRE,
39,
which
the
article
27
(1951).
for mutual inductance,see
Prin\302\254
by Boast (Harper
Chaps. 15 and
&
Brothers,
16.\n
PROBLEMS\n
\nvalue
ent
\nvon
each
For
2-1.
of
node
the
quantities
pairs,
Helmholtz,
of the
the
circuits shown in the figure:(a) determine
t
he
select
N
independ\302\254\n
E, S, Nt, N, L, and J; (b)
using one datum
draw
L independent
node; and (c)
loops.\n
following
the
rule
of
CHAPTER
EQUATIONS\n
NETWORK
3-1.
Kirchlioff\342\200\231s
equations\n
network
Most
Kirchhoff
by
\ngiven
\ninstantaneous
\nloop
are
equations
in 1845.*
formulated
from two simple laws first
The first law relates to the sum of the
elements in a loop.
drops must equal the sum
of the
voltages
of the voltage
sum
the
3\n
This law follows from the scalarnature
of
voltage.
It statesthat in
of the voltage
To illustrate
any
rises.\n
the
\nconceptinvolved, consider another scalarquantity,
Suppose
a trip in an airplane, visiting a number
we
make
cities
\nthat
but
to our place of origin. At each stop,
will
returning
\neventually
elevation
and record the elevation increase or decrease.
the
\ndetermine
is completed,
we can be sure if
are
the
trip
sufficiently
sum
\nrate
that
the sum of the elevation increases will just equal
of
we would not be back at our
elevation
decreases.
Otherwise,
elevation.
of
we
we
\nWhen
accu\302\254
the
start\302\254
\nthe
elevation.\n
\ning
in a
we make a tour around someloop
network
fixed
instant
of time. At each node, we will measurethe volt\302\254
\nat some
to the previous node and record voltageincreases
and
with
\nage
respect
the
the
is
sum
decreases.
Once
of
completely traversed,
loop
\nvoltage
the
sum
must
of
the
decreases
\nthe
equal
(or drops)
voltage
voltage
in
are
other
the
same
fact
that
there
The
net\302\254
\nincreases
loops
rises).
(or
the
exist\302\254
\nwork
has
no effect
on the sum of the drops and rises,just
routes
does not affect the altitude change
\nence
alternate
airline
of
that
suppose
Next,
as
\nsummations.\n
law
second
KirchhofPs
relates
to the sum
of instantaneous currents
the sum of currentsflowingintothe node equals
\nthe sum
flowing out. In an analogous hydraulicsystem,the
of currents
out of a junction of pipes must equalthe
water
\nsum
of water
flowing
If
we
assume
no storage capacity at the junction.
\nflowing
in, assuming
at
the
a
\nno charge
nodes of network,then,just asin
storage
capacity
the currents
\nthe
into that node must equal thoseout.\n
hydraulic
system,
\nat
a
*
that
of
the
work of Kirchhoff closely followed the pioneerworks
and
electric induction, of Oersted in relating magnetism
describing
in 1820, of Ampere in relating force and
in
and of
current
1820-25
in
current
time
and
At
the
Kirchhoff
1826.
relating
published
voltage
in
\nelectricity
\nOhm
states
Historically,
\nFaraday
\nthe
It
node.
in
work
\nseveral
containing
sciences\342\200\224there
these
are
was 23 years of age. He.made
Kirchhoff
laws in other fields.\n
laws, he
other
40\n
contributions
in
oore.
FOR DESCRIBINGNETWORKS
CONVENTIONS
2
Chap.
different
Using
dots, establish polarity
shaped
39\n
markings for the
\nWindings.\n
with
\nsystem
the
legs
Show
\nmatic.
The
2-9.
the
\ndetermine
equivalent
LlK-My
lo-^qnnr\342\200\224wi\n
Lz\n
l'o-\n
(6)\n
Prob.
\nmary
\nand
\nweber,
Lz\n
IlVM7
(c)\n
A transformer
2-9.\n
has 100 turns
on the primary (terminals1-1')
the
secondary
(terminals 2-2'). A current
a magnetic
flux which links all turns of
causes
The flux decreases according to the
the
secondary.
200
on
turns
t
^ 0. Find:
(a) the
flux
linkages
\nsecondary, (b) the voltage inducedin the
secondary.\n
of
the
the
in
both the
when
-
systems,
l'o-\n
2-10.
trans\302\254
each show two inductors with
schematics
dot markings.
For each of the two
inductance
of the system at terminals 1-1'
lo\342\200\224i/qnnp\342\200\2242jqnnr\\--|
\nand
a
Draw
inductances.\n
combining
\nby
three coupled coils.
circuit of a
similar
accompanying
with
different
but
\ncoupling
shows the equivalent
to that shown for Prob.2-7andplacewindings
of the core in such a way
to the sche
as
to be equivalent
connections
between
the elements in the same drawing.\n
a core
with
\nformer
\non
schematic
accompanying
marks
on the
polarity
The
2-8.
pri\302\254
primary
law
primary
<f>
=
e~\342\200\230
and
EQUATIONS
NETWORK
42
the
With
\nlaws
as shown in
currents
loop
The
two
3-9,
are
of Kirchhoff
sets
- 11) +
since
course,
equations,
if h = Ia
are the
they
identical
voltage
\nelements\342\200\224the
routes
in
\ndirectly
of
\nterms
loop
than
\ngreater
terms
are
currents
Ri and
of
h
identical,
respectively.
R2,
and
12
as\n
- h
h
(3-10)\n
interested
ultimately
Positive directions for
3-3.
These
choice.\n
second
the
h.
and\n
by
or
\naddition
(3-9)\n
in determining currents in the
branch
currents.
has shown that there are
Our example
to
determine
the branch
currents:
(1) write equations
terms
of the branch
currents, or (2) write equationsin
currents
from
which the branch currents can be found
subtraction.
Since
the number
of branches is equal or
the
number
of loops, the advantage of simplicity is usually
we are
analysis,
\ntwo
(3-8)\n
3-8
currents flowing in
-
V
Eqs. 3-6 and 3-7and Eqs.
and /2 =
Ic
\nin
Kirchhoff
Rdi = 0
Ie is expressedin
3-5, we find that
\nBy Eq.
- It) =
+
-R3(h
In
Fig. 3-3, the
3\n
are\n
RJi
\nof
Chap.
currents\n
assigned the problem of counting cars travel\302\254
each
direction
on a busy street in a large city. Ourfirst
would
\ning
step
to distinguish
\nbe
cars traveling
in the two directions. We would
accom\302\254
this
on a positive direction of flow. With
decision
\nplish
by deciding
each
car
could
be considered
as moving in the positivedirection
\nmade,
\nor
to
the
direction
considered
opposite
positive (although handier
\nterms
such
as north
and south would likely be used).\n
were
we
that
Suppose
this
before
Similarly,
law,
\nvoltage
\nbe
for
assigned
the
in
\nCurrents
of
\narbitrary,
a
\nassigned
direction
but
course,
clockwise
\nmay
be
traversed
\nlaw.
If
the
\nfor
positive
is
loop
loop
equivalent
an
branch) and identified
establishes a positive or referencedirection.
to that considered positive are
opposite
sign. The direction to be assumed positiveis
for uniformity,
loop currents will usually be
(or for each
positive
Once the positive
\nthis
loop
a negative
with
\nmarked
each
a decision
Such
\narrow.
direction
a positive
in
based on Kirchhoff\342\200\231s
equations
of the loop (or branch) currentsmust
network
writing
with
direction.\n
direction for the loop
current
either
direction
in applying
is
assigned,
the
loop
the Kirchhoff voltage
in a direction opposite to that assigned
the Kirchhoff
equation is not changed, since
current,
to multiplying
all terms by \342\200\2241.\n
is traversed
across the three
\nvoltage drops
the
battery.
\nmust
Kirchhoff\342\200\231s
the
equal
A
The sum of
Fig. 3-1.
the
equals
currents
\nthe
attached
branch
flowing
Kirchhoff\342\200\231s
or
\nrents
\ncurrents,
assigned,
the
Kirchhoff
R^Ic
=
RJb
-H
node B, the
\nequation
\nequations
(3-2)\n
branch
equivalence of branch and loop
the
shown
in Fig. 3-3.
With
cur\302\254
currents
as\n
voltage
Ri
b
R*\n
loop
1)
(3-3)\n
loop
2)
(3-4)\n
Kirchhoff current
3-3.
Fig.
Two-loop
network.\n
is\n
Ia
This
or\n
0\n
(for
At
node,
V\n
(for
\342\200\224
Kirch\302\254
by using either
A
=
out,
Ii
13
are\n
Rzlc
+
that
flowing
be applied
may
network
the
by
currents\n
To show
currents.
loop
\nequations
loop
law
voltage
consider
R\\Ia
and
currents
Branch
node
the
node\n
law.\n
to a particular node.
must equal those
U=
+
into the
Currents
At
the node
into
of current\n
direction
equal thoseout of
by Kirch-
h
3-2.
3-2.
Fig.
hoff\342\200\231s
current
each
for
shown
the
with
law.
\nhoff\342\200\231s
voltage
(3-1)\n
Vg
Fig. 3-2
drops
voltage
voltage
battery
7, =
V, +
is shown in
network
of a
part
due
the voltage drops
that
states
law
or\n
+
Vi
are
there
that
see
We
passiveelementsanda voltagerise
voltage
rises,
voltage
41\n
3-1.
shown in Fig.
circuit
series
the
Consider
\nto
EQUATIONS\n
NETWORK
3-2\n
Art.
may
equation
of
Eqs.
=
and
3-4
RJa
\342\200\224
Ic =
or\n
la -
h
(3-5)\n
to eliminate Ic from the Kirchhoff
voltage
be used
3-3
Ic
lb +
f?3(/<.
+
as\n
- Ib) =
Rz(Ia
\342\200\224
Ib)
+
Rzlb
=
V\n
0\n
(3-6)\n
(3-7)\n
NETWORK
44\n
\nand
loops
u
positive
assigned
turn
in
in Fig. 3-6 with loop currentst\342\200\231i,
12,
three\n
the
as shown. Traversing
directions
is shown
network
three-loop
gives
three
the
Riii
3
\342\200\224
ii)
^\n
/\n
^\n
/\n
+
dt
C;\n
analysis
\342\200\224
(t'i
_1\n
Loop
Kirchhoff voltage
ii
(^3
j
\342\200\234
^2)
of circuits with
dt
-4*
dt
+
coupled
=
dt
U)
\342\226\240+\342\226\240
Ci\n
3-5.
3\n
8\n
Example
A
Chap.
EQUATIONS\n
q-
J
equations\n
(3-15)\n
v(t)\n
(tj
R2H =
\342\200\224
t*)
\342\200\224
0\n
dt
0\n
(3-16)\n
(3-17)\n
coils\n
2-4 regarding polarity of inducedvoltage
to dots can be used to advantage
\nand
current
direction
with
respect
\nin
with coupled coils. To apply the rule, (1) the
of circuits
analysis
for each pair of coupled coils\342\200\224or
(dots)
\npolarity
markings
equivalent
\ninformation\342\200\224must be given,
and (2) the positive direction of current
\nflow
must
be assumed
for each loop. A part of a circuitfulfilling
these
\ntwo
is shown
in Fig. 3-7(a). By the rule, current ii\n
requirements
The
rule
developed
in Art.
o3\n
04\n
Fig.
\ntion
3-7.
of
Coupled coils
current,
polarity
direc\302\254
illustrating the relationshipof assumed
and
of
induced
voltage.\n
markings,
polarity
1-2 and so willinducea
\nin winding
3-4
positive
terminal, terminal 3. The
thus
\nrent
induces
a voltage drop from 3 to 4, or a
rise
from
i\\
to
3. Similarly,
terminal\n
tj induces a voltage in winding1-2
enters
the
dotted
terminal
of winding
at the dotted
voltage
cur\302\254
voltage
\n4
with
Art.
3-4.
Formulatins
\nof
of
number
A
Example
3-4
examples
will illustrate
Kirchhoff\342\200\231s
basis\n
the formulation of
equations
law.\n
voltage
RLC circuit. It
a series
shows
is quiteclear
inspection
by
node
one loop, while there are 3 independent
pairs.
the
across
the
elements
must
voltage
equal
drops
passive
across
the\n
for
the
the active element.
voltages
Expressions
is but
there
voltage
to
due
\nrise
on the loop
43\n
1\n
Figure
\nThe
equations
using
equilibrium
\nthat
EQUATIONS\n
NETWORK
3-4\n
L\n
elements
passive
Kirchhoff\342\200\231s
\nsions,
1. In terms
in Chapter
derived
were
voltage
law
Ri
+ L
requires
+
jt
^
of these
expres\302\254
that\n
idt
j
=
(3-11)\n
v(t)
times. This is an integrodifferentialequation,which
\nchanged to a differential equation by differentiation to give\n
all
at
T
dH
+
\n\342\200\224
\nequations
_L
+
1
=
dt}^>\n
,\342\200\242
ct
(3-12)\n
~dT\n
been arranged in
descending
order.\n
3-5 has two independent loops, sinceL = E
4 +
and the two loop currents, ii and ii,
S = 5 \342\200\224
1=2,
directions
as shown. The equilibrium
assigned
positive
based on Kirchhoff\342\200\231s
are\n
of
the voltages,
law,
network
The
Nt
Rdt
be
2\n
Example
\nhave
have
derivatives
the
where
di
P
_|_
Ldi^
may
+
been
of Fig.
Riii
C
1\n
(i2
C\n
f\n
1\n
it)
+\n
/
\342\200\224
fc'i)
dt
+ L
-rjj
dt
=
(3-13)\n
v(t)\n
+ R2I2 =
0\n
(3-14)\n
Example
4\n
The
in
\nshown
into
taking
\nequations,
the
\ndots,
of Fig.
system
3 -10.
Kg.
We
3-9.
Fig.
flux-conducting material is
are required to write the Kirchhoffvoltage
With the aid of
account
inductance.
mutual
3-9 can be replaced by the equivalent
circuit\n
system of Fig. 3-9
Magnetic
\nand assumed
of
to
\nance
D
of
n
xt
\342\200\224
H)
\342\200\236\342\200\242
+
L\\
tut
i
~r M12
^
\342\200\242
\\
i
/\342\200\242
\342\200\224
Ii)
+
t
Li
d(ii
\342\200\224
d(iz
is)
n,
\342\200\224
M
^
*
(*,
-
\302\253
this
d(i2
dis\n
is\n
\342\200\224
Rziii
tj)
+
w
| M
^\342\200\224-
+ Mu
(ii
-
-
If-
\342\200\224
iz)
\302\253
U
particular
g
-
(\302\273,
ii)
problem,
-
M\342\200\236
jt
the
(i,
equations
- i,) +
i
(3-21)\n
diz\n
it-jj\n
Mzz
+
+
\342\200\224
t>(\302\243)
ii)
\302\247
+ U,z
\nIn
-
j,
M12
%\\)
^
+ Lt
L,
current.\n
are\n
d\"
R*(t2
markings
polarity
showing
direction
use a double subscript notationfor mutualinduct\302\254
the two coils being considered, the Kirchhoffvoltage
indicate
\nequations
K\\i\\
positive
If we
3-10.
Fig.
coils on a
of three
sense
winding
^^3
= 0
(3-22)\n
%\n
J
i,
dt
would have had
=
0
(3-23)
simpler
form\n
dotted
1\342\200\224the
terminal\342\200\224positive,
1 to
\nminal
\nthat
\nrent
is
\na
when
positive
terminal
and
in
positive,
winding
and
rise
from
terminal
Figure
3-8
Fig.
3-7(b)
\ndotted
voltage
2
\nminal
\nage
\nof
rules
\nhoff
voltage
to the
+
positive,
- M
^
polarity
terminals.
same
the
the
sign
conven\302\254
M.\n
first loop gives the equilibrium
equation\n
=
^
a voltage drop
the
with
for
coupled
two-loop
illustrating
Kirch-
the
applied
\ntion
A
8-8.
Tig.
\ncircuit
circuit. Applying
ti produces
across
\nrise
cur\302\254
it leaves the
induces
1-2 with ter\302\254
so with a volt\302\254
1 to 2.\n
discussed,
law
from
reversed
the coupled coils
into a
incorporated
just
current
\nwhen
has a positivedirection
hence
Riii
The
a voltage drop from ter\302\254
so with
and
shows
\ntwo-loop coupled
\nthe
45\n
2.\n
3-7 (b), the current it
Fig.
in Fig. 3-7(a). This
shown
In
EQUATIONS
NETWORK
3-5
Art.
markings
(3-18)\n
v(t)
across
but
L\\,
the
current
as shown induces
In the second
it,
a voltage
loop, the equilibrium
is\n
\nequation
+
**
-\302\260
<3-19>\n
a term of the
M (dit/dt)
a voltage drop if positive. long
are given
dots
along with the direction of positivecurrent,
polarity
is no
and the rule of Art. 2-4 can be applied
ambiguity,
use
of
to
all coupled
coils. If the number of coilsis large,
\nsively
it
is
\ndots of various shapes may become cumbersome.In this
to
or
M
\nmore convenient
minussignto to let
assign a plus
carry
in
the
formulation
rather
than
the
sign
equation
sign be
letting
these
In
equations
rise
a voltage
\ncates
the sign before
if negative and
indi\302\254
form
As
\nas
succes\302\254
\nthere
the
case,
M,
\nthe
of the
nature
the
by
\nspecified
as
just
\nfollow,
will
Consider the circuit
in
\nthe
coupled
coils
of
\ndrops
\nbe
\nas
erased,
\342\200\224
M.
these
either
By
are
loop.
provided
this
in
the
rise.
two
These
discussion
to
described
3-8,
by Eqs. 3-18 and 3-19.
that
the
equations
voltages induced by means of
rises
of
voltage
opposite polarity to the
With
the current directions given, the dots
of
observe
or
voltage\342\200\224drop
are used
both
\nWe
induced
and both
be
used
in the literature.\n
\nsystems are equivalent,
Fig.
voltage
can
a negative
system,
sign is
Eq. 3-18 is
identified with mutual inductance
written\n
(3-20)\n
EQUATIONS
NETWORK
48\n
\nmarked as
J4.
=
Iz
since
Now
or
h
(Vb
^MX 3
second
the
have
we
=
Iz
law
current
Kirchhoff\342\200\231s
There the
to node c.
our attention
turn
now
us
Let
-
requires
in
current
Ii
and
3\n
is
Rt
that\n
- Iz = 0
It
Ve)
Chap.
=
(3-29)\n
^-Vc
MX
(3-30)\n
4\n
equation,\n
equilibrium
n
v,\n
Vc\n
Rz
Rz\n
R\n
(3-31)\n
we
Have
\ncurrent
new
\nthis
=
\nIz
is
Iz
Iz,
Vc.\n
in choosing positive directions of current
nodes?
On the network under consideration, a new
In terms
marked
with an arrow such that Iz = \342\200\224
of
Iz.
this
Iz
+ I* =
with Eq. 3-29,
is identical
equation
is
law
current
Kirchhoff\342\200\231s
current,
\342\200\224
F6 and
solved simultaneously to give the
flexibility
any
different
the
\nfor
of
values
\nunknown
3-31 must be
and
3-28
Equations
0. But since
and so
with
Eq.
words, the positive direction of the branchcurrents
may
at
assumed
each
node independent
of previous designations. We
\nbe
have
two options:
\nthus
(1) Assume positive directions for branch cur\302\254
once
and
for all. (2) Assume new positive directions at eachnode,
\nrents
that
currents
flow out of the node for all passiveelements
\nfor example
in the
\nand
marked
direction
for active current sources.\n
of
As
a result
this discussion, we see that the stepsto be followed
in
other
In
\n3-31.
\nnode
the
are
analysis
following:\n
node
all
unknown
(1) Select a datum node and identify
voltages.\n
(2) Assume a positive direction for all branchcurrents.\n
law
Kirchhoff\342\200\231s
current
to each node of unknown voltage,
(3)
Apply
each
in terms of a node-to-node voltage
branch
current
\nwriting
\nand
It is
be
equivalent
\nthe
node
located
the
i(t)
circuit
this
for analysis.
source
into
a
math\302\254
In Fig. 3-12(a), let
of Fig.
of the
3-12(a)
v(t)
\nSolving
source
current
voltage
voltage source and vi(t) be the potentialof
between
the resistor
Ri and the rest of the network.
law
flows through
the resistor Ri. Kirchhoff\342\200\231s
voltage
potential
current
\nThe
\nfor
the
parameters.\n
convenient to change a
sometimes
\nematically
\nv(t)
circuit
appropriate
equation
for i(t)
is\n
=
+
Rii{t)
t>i(\302\243)
(3-32)\n
gives\n
v(t)
i(t)\n
Ri
_
Vi(t)\n
(3-33)\n
R\\\n
Art.
if the
v(t) and Ri
generator
\n(See
Prob.
3-6.
Formulating
of
of
each
the
three
loops.
of currents
that node.To
in
shown
\nnetwork
there
network
\nmarked
on the node basis\n
for formulating
the equilibrium equations for cir\302\254
the
Kirchhoff
law
of
that the sum
leaving
to the sum of currents entering
illustrate
in node analysis,
used
consider the simple resistive
procedures
\nthis
had been part
basis
is equal
node
\nthe
equations
use
makes
\ncuits
47\n
3-14.)\n
node
The
\na
EQUATIONS\n
NETWORK
3-6\n
d. Following
and
c,
6,
o,
3-11. For
Fig.
are four nodes,
\nconvention,the negativeterminal
of
active
\nthe
element,
three
\nthen
node.
datum
the
as
\nlected
to
\npotential
to
\nknown
\nunknown
\nwith
respect
\nnetwork
\naccomplished
\nbranch
\nb
marked
in
\n12,
and
know
with
the Fig. 3-11. Network illustrating pro\302\254
d. However,
in node analysis.\n
\ncedures
a to node
from
node
d is
be equal
to the battery voltage. There are
two
in the network:
the voltages of node b and node c
voltages
to the datum
node.\n
node
thus but
identified the unknown
Kirchhoff\342\200\231s
that\n
12 + ^3 =
I\\ +
What
write
At
the
\nwe
and c
voltages, our nexttask is to
in terms
of these unknown node voltages. This is
equations
in terms
of branch currents (never loop currents). Each
be assigned
current
must
a direction considered positive and
with
an arrow,
just as in the case of loopanalysis. node
network
of Fig. 3-11, the branch currents are markedas h,
current
out of the node. By
13, all directed
law,
Having
\nso
the
voltages,
node-pair
\npotentials of nodes a, b,
\nrespect
d, is se\302\254
There are
node
are
\nOhm\342\200\231s
law,
these
they
branch
0\n
(3-24)\n
in terms of the
currents
node voltages?
By
are\n
h
=
u
=
^
(Vb -
-
~
V)\n
(3-25)\n
o)\n
(3-26)\n
vc)\n
(3-27)\n
J\\>2\n
/.
Substituting
these
three
Vb
=
^
equations
V
,
Vb
(n
-
into Eq. 3-24
.
Vb
Vc\n
gives\n
EQUATIONS\n
NETWORK
50
of the
is
generator
in
v(t)
-
KOI
|\302\273'i
^
If
is
which
- 0
(3-35)\n
\nat\n
, f/i\342\200\231i r(f)\n
(it +
(\n
(it\n
lj\n
R\n
Kq. 3-31.\n Analysis
with
identical
+
/
/'
write\n
('(U''
<U
r\342\200\230\n
+\n
+
or\n
we may
3-13(a),
Fig.
source or the
fi\n
Eramplr
current
the
3\n
of
network
\nloop
of thiw-
circuit
Kquivalent
Hi*.
M.\n
2 are designated rt
1 and
and tv
work
net
The
\nis
\nthe
three-loop
\nFig.
3-0.
\nand
t he
\nAt
source.\n
current
\nequivalent
with
out
earritsl
be
may
\neither the voltage
-14.
3\n
Clup.
source
in Fig. 3-14
to
equivalent
network
in
shown
the datum node,
volt ages at nodes
unknown
= G\\ and\n
\\/Ri
1, setting
Node
mule
shown
3 is
\342\200\224
Gt,\n
\nl/Rt
(3-30)\n
at
and
node
2,\n
Oi
this
In
\342\200\224
+
O\342\200\231j \302\273\342\200\231i)
(\\
formulation
example,
\ndifferential
^
equations
than
4- G&t
\" 0
(3-37)\n
on the node basis has resultedin fewer
on the loop basis in Kxainplc 3. Ordinarily
simultaneous differential equations
in
of method of formulation, loopor
solving
also
on the objective
of analysis. In this example,
depends
\nthe
at
node
is
2
the
node method has the
voltage
desired,
\nover
the
But if it is the current flowingin capacitor
loop method.
\nthat
is to tie found,
we must weigh the relative advantages the two
\nmethods.
The loop currents can be assignedso that
one
loop
\nrent
flows
in Ct,
but three simultaneous
lie
must
solved.
equations
the
node
we might And the voltage at node2
and
method,
\nthen
determine
the current
in the capacitor from the
\nit
requires
less
\nthan
in solving two
three.
The choice
work
if
\nnode,
advantage
(\\
of
cur\302\254
only
first
\nUsing
equation\n
(3-38)\n
ic,\342\200\234C*W
The
second
ular
example.\n
method
appears to involve
less computationin this
partic\302\254\n
Art.
3-6\n
In
this
current
us
current
a
we will identify each separateterm.The
the current i(t) flowing into the network
is equal
minus
a current Vi(t)/Ri. This equation may
be
of\n
of the new network of Fig. 3-12(b)by means
that
v(t)/Ri
terms
in
\ninteipreted
3-12.
Fig.
\ncurrent
resistor
\nthe
the
\ntions,
Eqs.
Example
3-32
and
the
network
3-33,
giving
\ndescribed,
an
into
converted
and
node
the
were
voltage
\ncurrent
equation
course,
\nsource
differ\302\254
they are equivalent.\n
before
it
shown
Network
3-13.
the
for node
analysis.\n
currents are assigned to flow out of node 1.
in each of the elements in terms of
currents
in Chapter
the
1. By Kirchhoff\342\200\231s current
law,
given
is\n
iVi
Of
the current source.
all branch
for
expressions
\nthe
an equivalent
current flowing through
in Fig. 3-13(a). The voltagesourcemay
current source by the procedure just
equivalent
the\n
network
of Fig. 3-13 (b). Node 2 is designated
Fig.
datum
is the
5\n
Consider
\nThe
sources.\n
The
is the current flowing into the network.
Fig. 3-12 are described by the sameequa-
networks
two
of
v(t)/R i is from
current
Vi(t)/Ri
in shunt with
currents
of
two
these
\nSince
\nbe
connected
R\\
of
\nence
current
The
source.
Interchange
The
law.
current
Kirchhoff\342\200\231s
49\n
equation,
tells
\nequation
\nto
EQUATIONS\n
NETWORK
+
z
/
Vidt
+
=
C%1
It
(3_34)\n
to make the conversionto the current
network.
Since the voltage of the + terminal\n
is not
necessary
analyzing
the
50
EQUATIONS\n
NETWORK
of the
is
generator
i
in
v(t)
- KOI
[t>i
we may
3-13(a),
Fig.
Omp.3\n
=
+ C
dt
r,
+\n
write\n
0\n
(3-35)\n
l!\n
S
r(t)\n
dt\n
H\n
3-34. Analysis may
Eq.
the
either
2
\nj
i
(\n
V'+L/\"'\n
with
identical
is
which
, dr
dt +
or\n
net
The
the
is
3-14.
Fig.
circuit
Equivalent
of
network
loop
.
,
the
,
_.
of throe-
Fl*-
Fi\302\253.:M1.
and
1
are
2
and
=
\\/Rt
shown
work
in Fig.
3-14\n
source
current
node 1, setting
=
\\/R\\
G\\
and\n
Git\n
(?,!>,
at node
and
0\n
to\n
equivalent
net work shown in\n
three-loop
, ,
,\n
3'flNodc
,hc <latllm node.\n
3
the
unknown
at nodes\n
voltages
iq and vt. At
designated
the\n
source.\n
current
Bxamplr
ft;I>
>*,4=^
or
source
voltage
equivalent
with
out
carried
lx*
+
<?\342\226\240
^
+
-
<\342\200\235>
I(\"'
=
(3-36)\n
\342\200\231,)
s,\n
2,\n
Ct
^
(wj
\342\200\224
V\\)
+ G&t
+
=
0\n
(3-37)\n
basis has resultedin
than
on the loop basis in Example 3.
equations
less
in solving two simultaneous differential equations
work
requires
in
The choice of method of formulation, loop
three.
solving
on the objective
of analysis. In this example,
also
depends
at node
2 is desired, the node method has the
voltage
the
But if it is the current flowingin capacitor
loop method.
is to be found,
we must weigh the relative advantages the two
one
\nmethods.
The loop currents can be assigned so that
loop
must
be
flows
in Ct,
but three simultaneous
solved.
equations
the
node
we might And the voltage at node 2
method,
\nthen
in the capacitor from the
determine
the current
In
this
example,
formulation
on the node
fewer
\ndifferential
Ordinarily
\nit
\nthan
or
if
\nnode,
\nthe
advantage
\nover
Cj
\nthat
of
cur\302\254
only
\nrent
first
\nUsing
and
equation\n
ic,
The
second
ular
example.\n
method
appears
=
C,
to involve
(3-38)\n
^
less computation in this
partic\302\254\n
Art.
51\n
7\n
Example
\nAlthough
\nunknown
From
there
that
in
\nexamples
of other
source.
one
in Fig. 3-15 differs from the networks
is no series resistance with the voltage
has four independent
loops, there is but
that
at node
shown
network
The
\n2.
EQUATIONS\n
NETWORK
3-7\n
this
network
node
voltage,
C3\n
current
Kirchhoff\342\200\231s
we
law,
K\n
\nwrite\n
1\n
.
/OTTorv\342\200\224,\n
^AA/Vi\n
L\n
(---)
\302\253\342\200\242
\302\243
+\n
if
Gv
+
2\n
Ci
+
-\n
I\n
dv
2\n
= 0
W
as
before,
\nthat
Ci
does
not
This
\ntion.
\ntor
or
Ci
appear
is because
the
other
any
The
\nment.
O =
(3'39)\n
l/B. Note
in the equa\302\254
an ideal element), and
\naffecting the network equations.\n
3-7.
\npendent
\ndifferential
laws
The
chapter.
of
capaci\302\254
so Ci
may
be
without
removed
have been
statements
of
of a
all
the
word
larger
voltage
by
played
with
of identical behavior patterns in the
in network analysis. This similar\302\254
current
is termed the principle of duality.\n
pattern
and
implications,
(6)\n
(a)
Fig.
Consider
\nance
shown
\nlyzed
to
noted in the preceding
the two Kirchhofflaws
with voltage substituted for current, inde\302\254
for
node pair, etc. Likewise, the integroloop
independent
that
resulted
from the application of the Kirch\302\254
equations
have
been
similar in appearance. This repeated similarity
for
word
part
only
\nity,
this
of
almost
\nwere
\nroles
7
Example
1 is independent the
will
situations
analogous
\ndiscussions
\nis
o(
Network
Duality\n
Several
\nhoff
3\n
element. Capacitor Ci is an extraneousele\302\254
must maintain terminal voltage for any
load
source
voltage
Fie 3.18.
voltage at node
shunt
is not
\n(or it
^C2\n
^Ci
Lwhere,
R\n
dt\n
Vi)
(Vt~\n
3-16.
Dual
networks.\n
completely differentin physical
shows that the first mightbe
3-16. Inspection
basis.\n
on the loop basis and the other on the
the two networks
in
Fig.
advantage
appear\302\254
ana\302\254
node
The
equations
resulting
CW
two
+
Ri +
+
6v
+
i
J
L J
idt-v(f)
V
M
(3-40)\n
=
(3-41)\n
specify identical mathematical
equations
difference
\nonly
3\n
are\n
L~
These
Chap.
EQUATIONS\n
NETWORK
52\n
letter
in
being
solution
The
symbols.
operations,the
of one equation
roles
The two networks are duals.
\nof current
and
in the two networks have been interchanged.
voltage
a word
of caution,
one network is not the equivalent the other
that
the
sense
one can replace the
of the terms
An
of Eqs. 3-40 and 3-41 showsthat
inspection
\nis
the
also
other.
of the
solution
The
\nAs
of
\nin
other.\n
the
are
\nfollowing
quantities.\n
analogous
and
Ri
Gv\n
and\n
1\n
1\n
v
and\n
the
Evidently
are dual quantities.\n
pairs
following
R\n
and\n
G\n
L\n
and\n
C\n
and\n
v,
i\n
current,
loop
g\302\260rj\n
1
jidt)\n
loop\n
circuit\n
A simple
(1)
\tInside each loop
\nience.
Place
\nnetwork.
\non
\nNew
voltage\n
or\n
J v
dt\n
and\n
node
pair\n
and\n
open
circuit\n
may be followedin finding
the
dual
network.\n
\nof a
*
construction*
graphical
node-pair
| ^
and\n
short
dt\n
L\n /\n
C\n
Gardner
York,
an
Arrange
the
for
paper
and
1942),
Barnes,
pp.
46ff.\n
for
conven\302\254
place a node, givingit a
extra node, the datum node, external to
the
same
numbered
nodes on a separate space
number
the
construction
Transients
of the
dual.\n
in Linear Systems (John Wiley
it
Sons,
Inc.,
\ninal
traversing
network,
network,
original
chart
the
\nment\342\200\224from
the
in
traversed
\nelement
53\n
to node through the elementsin the orig\302\254
only one element at a time. For each
node
from
lines
Draw
(2)
EQUATIONS\n
NETWORK
3-8\n
Art.
the
above\342\200\224on
connect the dual ele\302\254
network
dual
being
\nconstructed.\n
(3) Continue this
processuntil thenumber
of
is exhausted.
\nsingle elements
\na
connecting
element
\ndual
wire
which
is an
open
is
construction
circuit.)\n
\nequations
on
\nother
the
be
may
is
dual
the
network.
by writing the differential
one on the loop basis and the
systems,
basis.\n
node
manner
checked
two
the
for
through
(Should you slip and go through
assumed to be a short circuit,the
(4) The network constructedin this
\nThis
paths
possible
3-17.
Networks
This graphical construction is illustrated in Fig.
\nare not
(that is, cannot be shown schematicallyin
planar
do not have duals.\n
\nwith
no wires
crossing)
that
one
plane
zC\n
(b)\n
3-17.
Fig.
Graphical
procedures for
\nnal;
General
3-8.
far
Thus
\nto
successively
\nour
approach
L
\nwhere
\nin
3-18.
Fig.
\nbeen
replaced
\ninductance
Consider
\ncapacitance
is
network
(b)
finding
dual
of network:
(a)
origi\302\254
dual.\n
equations\n
from analysis of very simple networks
To systematize
network
more
configurations.
complex
of networks, consider an L-loopnetwork,
to the analysis
A
of such a networkis shown
number.
representation
any
the circuit elements in each branchhave
In this diagram,
line for simplicity. The effectsof mutual
by a straight
are
indicated
but are assumed to be present.\n
not
1. This
loop
loop may contain resistance, inductance, and
in any
one or all of the branches that make up the loop
we have
progressed
\nLet\n
Rube
resistance
in loop
the
total
in loop 1.\n
total inductance
total elastance of loop 1.\n
Ln
be
the
Su
be
the
1.\n
EQUATIONS\n
NETWORK
54\n
\nadd
a series
for
directly
J_
i
=
\nin
on
\nspecialize
4\342\200\224in
loop
3,
loop
drops in loop
be voltage
will
i ++ i+
c,
c,+
^+
c,
e\342\200\236
There
all
fact,
two
\nthese
\nj;
\nand
Skj
j;
(h)\n
(a)\n
(a)\n
(A)\n
Csi\n
(s)\n
(h)\n
(a)\n
\ndrop
this
\nby
letting
elastance
total
+
Lkj
implies
symbolism
Rkj,
by
by
\nmultiplication
in
The
total
\nsidering
loop
\nsources
voltage
within
+ Skj
J
symbol
J dt^
ij dt
(3-42)\n
ij =
form
(3-43)\n
dkjij
ij is operated upon
Lkj and differentiation, and
All three operations are
the variable
multiplication
by
and
integration.
Skj
by
mul\302\254
finally,
sum\302\254
ajy.\n
k will be found by successively
con\302\254
flowing in every other loop. Math\302\254
is done
by letting j have all values from 1 to L. This
must
be equal to the total voltage rise from active
drop
volt-\n
we write as vk. Then
loop
k, which
by Kirchhoff\342\200\231s
voltage
k and
this
\nematically
\ntotal
the
that
^
k
voltage
adopt a special notation for equationsof this
be the equivalent of Eq. 3-42.\n
following
equation
+
\nmarized
The
we will
the
and
is\n
+ Skj
Lkj
For
to loops
common
common to loops k and j.
by current ij
k produced
point,
\ntiplication
mutual)
(including
(^Rkj
This
on\n
network.\n
L-loop
inductance
Rk/ij
At
loop
from
1 to L.
k, where j and k are any integers
=
let Ry
the total resistance common to loopsk
the
in loop
general
currents in the jth
(Ri\n
total
2,
loop
case. Rather than
G)\n
loops,
=
the
in
(s!\n
loop
\342\200\224
the
Lkj
in
flow
(*\342\226\240)\n
3-18.
in
c.\n
1 produced by current
loops
Fig.
voltage
+i\n
+
the effect of
1, consider
loop
3\n
of capacitance here because elastance terms
circuit, while capacitance terms combine as\n
instead
elastance
use
We
Chap.
drop in loop
the currents
NETWORK
Ait.
3-8\n
age
law,
EQUATIONS\n
55\n
have\n
we
(3-44)\n
to repeat this process all loops, by
1 to L. Thus the most generalformfor
There remains only
from
values
all
\nhave
Kirchhoff\342\200\231s
network
an L-loop
for
law
\nvoltage
is\n
=
k
The
diiii
+
ffluti
...,
1, 2,
L\n
(3-45)\n
set of
concise equation is the following
of this
expansion
k
letting
for
022^2
\n\342\200\234b
*b ciiiiz
+
... +
023^3
\342\200\234b
+
\342\200\242
\342\200\242
\342\200\242
+
=
flihiv
=
oul^l
equations.\n
V\\
t>2\n
(3-46)\n
~b Oi2*2
Oilii
...
+ Oi3*3 +
is
\nchart
a
in the
Sucha
emphasized.
1\n
Hi\n
2\n
v2\n
L\n
vL\n
the
currents
loop
\ntion,
clockwise
\na,jk(j
k)
\ntwo
\nage
Z5
\342\200\242
\342\200\242
\342\200\242
011,\n
an\n
an\n
^13\n
an\n
fliB\n
Cl 21\n
a 22\n
0 23\n
024\n
026\n
. .
. . .\n
. . .\n
. . .\n
. . .\n
. . .\n
. . .\n
CIli\n
0,L2\n
0
0 L4\n
0 L5\n
L3\n
Z jj\n
.\n
02Z\n
. . .\n
dLL\n
are all assumed positive in the samepath direc\302\254
then all an terms are positive and all
example,
In actual problems, of course, many
negative.
are
zero.\n
8\n
Example
A
are
terms
a-coefficients
the
\nof
for
of\n
Z$
%2
%\\
Voltage\n
Eq.\n
of
form
below.\n
shown
Coefficient
If
Vl\n
to arrange these equations given above
in which the a-coefficientsare
(or schedule)
is helpful
It
\nchart
=
\302\256ii*i
-b
of
sources
law
in Fig. 3-19. In this network
there
are
and no mutual inductance.
The Kirchhoff volt\302\254
is shown
network
two-loop
voltage
is\n
2\n
Q'kj'ij
j
883 1\n
\342\200\224
Via\n
k
=
1,
2\n
56\n
3-19.
Fig.
expanded
=
=
da
=
a 12
the
Similarly,
found
-f- R2)
(Ri
+ Ri) +
=
Ri
Li
=
for
\nfound
as
analysis,
loop
a
Consider
\nduality.
for
equations
+
(<Si
+
+
{Si
+
\342\200\224
Si
^
network
to
=
Vi
v,\n
<*>\n
node
two
and
\nto
give
an
=
Ci
-f
Ci
j
appearing
and k. The
in
\ntentials
and
be
combined
replaced
+
...; (2) an
may
\nlent
by
equivalent
system
node.
of the
parameters
the
between
For
elements
are
po\302\254
ele\302\254
nodes
con\302\254
3-20, the ele\302\254
be replaced
by an equiva\302\254
made
up as follows: (1) all
shown
\nments
out of the
the
considered.
as
current
of
terms
connected
\nnected
Kirchhoff\342\200\231s
node, in turn,
of node-to-node
the
into
\nwritten
\nbeing
system.\n
equivalent
capacitances
\nparallel
\nCu,
nodes,
L\342\200\231s
may
terms
in
law
\ndirected into and
A\n
\nments
\nC\342\200\231a,
G's,
from
written
is
\nCurrents
\nbetween
be\n
analysis will be similar to those
be expected from the principle of
might
with Nt nodes and only one part such\n
that
there
are N = Nt \342\200\224
1 independ\302\254
\nent node
Now each of the N
pairs.
node
Elements
dt\n
j
-vb\n
\ncurrent
3-20.
Si)
dt\n
j
dt\n
\nequations
Fig.
S2)
follows,\n
j
are recognized
terms
voltage
general
{Li + Li) ^
\342\200\224 \342\200\224
=\n
of the network as
+ (Li -f- L2)
(i2i
dn
-f- <222^2
dull
V\\f
by inspection
t>i
The
c3\n
network.\n
Two-loop
=
\"t* duti
are
a-coefficients
an
3\n
form,\n
ffllt'l
The
L3
R3
Ri
or in
Chap.
EQUATIONS\n
NETWORK
an equivalent
resistance
in Fig.
of value
capacitance
found
by
adding\n
Art.
value
of
\ninductance
network
Ly,
where
to
simplification
=
\nfromy
- l/Rkj =
as (?*,
conductances
\nthis
EQUATIONS\n
NETWORK
3-8\n
= N,
1 toy
Gi +
Gt
57\n
and (3) an
...;
+
equivalent
1 /Lkj
+ 1/1/2 + ....
l/Li
the circuits from node k to all othernodes
\342\200\224
we have the
Applying
equation\n
N\n
2,
lb-1,
(3-47)\n
...,N
>-x\n
which
written
be
may
concisely
as\n
N\n
(3-48)\n
i-i\n
The
\nthe
network
\nthe
sum
of
the
of
\nsum
=
1/R.
for
by
i\342\200\231s
by
6\342\200\231s,
and
v\342\200\231s,
the
v\342\200\231s
by
i\342\200\231a.\n
for
inductance
inverse
can
elements
are
1/L
and for conductance
be found by inspection by
thus
\342\200\234hanging
on\342\200\235
or
\342\200\234hanging
simply
between\342\200\235
the
nodes.\n
\nvarious
If the
all
\nlating
\nwhen
o\342\200\231s
replaced
Coefficients
which
\nnoting
same form as the expansion
to networks, it is not necessary
to simplify
elements.
At node j, the capacitanceCjjis
by combining
the capacitance
connected to node j. The valueof Ckj is the
connected
between node j and node k. Similar
capacitances
hold
\ninstructions
k
same convention for positivecurrentis maintained
node
for a network, the sign of bkj
equations
=
j, and negative when k j* j.\n
in
will
be
formu\302\254
positive
9\n
Example
A
3-48 has the
Eq. 3-46, with
this equation
applying
In
\nFor
of Eq.
expansion
\nloop case,
\nG
the operations\n
bkj summarize
letting
by
network
this
with
network,
two
independent
current
Kirchhoff\342\200\231s
-=\302\261=-Datum
\nFig. 3-21.
node pairs is
law
is\n
node
Two-node
network.\n
shown in Fig. 3-21.
58\n
=
hM
or\n
\nform
the
for
Values
&i2^2 =
+
bnVi
k
ik,\n
=
1,
Current
1\n
Gl
solution
of
\ntion.
The
+
Cl7t
v2\n
(k+t)
1\n
the
\n^
1
by
+
+6r2
type
with straight line
dt\n
use for systematic
derived in the last
sec\302\254
brackets on either
d 12\n
Ol3\n
\302\256i\302\273\n
\302\25621\n
&22\n
a23\n
tt2\302\273\n
dnl\n
On
dni
2\n
+ C2
tools we will
of the
equations
of quantities
(Ci
determinants\n
mathematical
simultaneous
array
jdt\n
H\n
w\n **
\342\200\242ere\n
of equations
are
Determinants
\nsolution
+
l\n
ib\n
The
3-9.
in chart
of\n
Vi
ia\n
2\n
*2\n
follows.\n
as
Coefficient
Eq.
=
summarized
be
i\342\200\231s
may
3\n
2\n
b22V2
&21^l
i\\j\n
and the
6-coefficients
Chap.
EQUATIONS\n
NETWORK
\342\200\242
\342\200\242
\342\200\242\n
dfin\n
side,\n
(3-50)\n
determinant of order n. Quantitiesin horizontal
lines
form
in vertical lines form columns. Such a deter\302\254
\nrows, and quantities
n rows and n columns. Each of the n2 quan\302\254
\nminant
is square,
having
is known as an element. Element position
\ntities
in
the
determinant
in
\nthe
determinant
is identified
by a double subscript, the first subscript
column (numbered from the
row
and
the second
indicating
\nindicating
the sloping line extending
left-hand
Elements
corner).
along
\nupper
of the determinant.\n
\nfrom
to a\342\200\236\342\200\236
the
form
au
principal
diagonal
A
has
a value
which is a function of the valuesof its
determinant
\nelements.
In finding
this value, we must make use of rulesfor expan\302\254
of the
\nsion
determinant
in terms of the elements. Second- and third\norder
determinants
have
that are familiar from studies in
expansions
for determinants
of order higher than
\nelementary
algebra.
Expansions
third
are
in
\nthe
made
terms
of
minors.\n
conveniently
The
minor
of any element
of a determinant qp is the determinant\n
is known
as a
=\n
A
for
minor
the
for example,
an,
deleted.
and row containing a#, are
determinant,\n
third-order
of the
terms
59\n
column
the
when
remains
which
\nIn
EQUATIONS
NETWORK
3-9
Art.
012\n
021\n
022\n
023\n
081\n
032\n
033\n
O13\n
(3-51)\n
is\n
=\n
An
Oil\n
022\n
023\n
032\n
033\n
(3-52)\n
of the element a,* multipliedby (\342\200\2241),+*
The
cofactor sign is thus foundby raising
\ncofactor.
the
row and the column,j
\nfound
+ k as\n
adding
by
A minor
=
(cofactor)
(
is
(
the
to
1)
name
the
given
\342\200\224
power
\342\200\224
l),+fc(minor)
(3-53)\n
rule, the cofactor signsalternatealongany row
the proper
cofactor
\nor column,
sign can be determined by \342\200\234counting\342\200\235
a
from
etc.)
positive an position to any element,
minus,
plus,
\n(plus,
or vertical path.\n
along
any horizontal
\nproceeding
in terms of minors (or cofactors)con\302\254
of a determinant
Expansion
reduction
of determinant
order. A determinant
of
\nsists
of successive
to the sum of the product of the elementsof any
row
\norder to is equal
to this
according
Since,
\nor column
rule to the
\nApplying
this
\nalong the
first column
\342\200\224
022\n
023\n
O32\n
O33\n
or
the
\nby
the
\nas
2to equivalent
The
to columns.
are
\nrows
rule
same
of to!
sum
and the
\ntial
in
solving
that
\nsimilar
have
resulted
equations
-|-
process
Ol3\n
032\n
O33\n
+
a 12
Ois\n
\302\25631\n
\n022\n
O23\n
about the
to
be expanded
continued until the valueof
given
A
is
factors.\n
equations
+
Oi2*2
013*3 +
o 1,2*2 \342\200\234IOisi\342\200\231s
H-
from
from
3-51
(3-64)\n
a 12\n
determinantsthat
\"I\"
On*\342\200\231i
Eq.
O31A31\n
\342\200\224
a 21\n
simultaneous
Oli*i
+
expansions of the determinant
minor determinants can, in turn,
product
The facts about
cofactors.
order
gives\n
an\n
There
1)
(to
of
TO21A21
flxiAn
\342\200\224
corresponding
expansion of the determinant
\342\200\224
\342\200\224
A
their
by
multiplied
application
the
Kirchhoff
have
we
of the
reviewed
just
are
essen\302\254
form\n
... + au.iL
\342\200\224
Vi\n
...
of
-f\"
a Lift
l
Kirchhoff\342\200\231s
current
\342\200\224
Vl\n
voltage
law
(and
law). The solution
to\n
60\n
such
simultaneous
is given
equations
A is
the system determinant
=\n
A
be
must
which
a-coefficients
\nof
With
\nneous
. . .
au,\n
0> 21\n
\302\25622\n
. . .
a-tL\n
aLi\n
Oz,2\n
...
Oll\n
=\n
ii
or\n
(3-56)\n
...,
solutions ix, iit
zero for the
vit Vz,
method
...,
for
=\n
by minors,
can be solved.
For
\342\200\224
column
simulta\302\254
third-
a
V3A31\n
4\"
V2A21
(3-57)\n
A
ii
be
is\n
t\342\200\230i
ViAn
Di
jth
to
i\302\273
vn.\n
of expansion
of Eq. 3-46
form
the solution
\norder equation,
(3-55)\n
as\n
12\n
the
and
of the
equations
given
a
column
rule
Cramer\342\200\231s
Dm\n
the
the
by
3\n
as\n
determinant formed by replacing
Dj is the
and
\nunique,
rule
\342\200\224\n
\342\226\240
lL
flu\n
from
different
Cramer\342\200\231s
=\n
12
T\n
where
by
D,\n
^1\n
\342\200\224
\342\200\234
i
Chap.
EQUATIONS\n
NETWORK
A\n
An
1
T
A2i
\342\200\234
+
,
Asi\n
\342\200\235*\n
T
\"a
(3-58)\n
Similarly,\n
=\n
ii
A21
,
\342\200\234
+
A_\342\200\2351
and so on.
\nexcept
Example
For
one
\"A
are
zero,
all
if
simplified
greatly
to only one
corresponding
(3-59)\n
V*~TV*\n
The form of these equationsis
driving voltage
source.\n
10\n
a certain
three-loop
network, the following equations are
5ii
-
2i%
\n-2it 4-
4\302\273,
\342\200\224
It*
\342\200\2243ti
From
A23\n
A22
rule
Cramer\342\200\231s
we
4
10\n
\n-1
write
-
3t> =
-
1 it
-|-
-1
6\n
5
0\n
=
0\n
-3
-2
6\n
-2
-3
4
-1
as\n
4-0\n
\n-1
\n-2
\n-3
=
-3
-2
-0\n
10
for t'i
solution
the
given.\n
\n4
-1\n
230\n
43\n
-1
6\n
v\342\200\231s
Art. 3-10
EQUATIONS\n
NETWORK
61\n
\nSimilarly,\n
-2\n
-(+10)\n
-2\n
-1\n
+
-3\n
150
6\n
\n43
A\n
network
Resistive
3-10.
in
\na-coefficients
this
we can
restriction,
\noperations of integration
\nKirchhoff\342\200\231s
law
voltage
terms
as\n
\342\200\224\342\226\272
Gkj\n
postpone our questions relating to
and differentiation.
The
for the
equations
include
which
6-coefficients
and
a-coefficients
of
\nmanipulation
elements, the
Rkj\n
bkj
Under
43\n
A\n
conductance
become
6-coefficients
the
140\n
\342\200\231\n
\342\200\224>
ay
and
-1\n
to contain
only resistive
become resistance terms as\n
3-46
Eq.
4\n
-3\n
analysis\n
restricted
networks
For
(10)\n
the
of
form
general
resistive case
the
is\n
L\n
Rujij =
k
vk,
=
1, 2,
^
..., L
(3-60)\n
j-i\n
where
\nin
resistance in loop j, and Rkj is
and
loop j
loop k. If the loopsare
total
the
R,-,- is
common
to
the
all
VW\n
I\342\200\224VW\n
drawn
resistance
total
in
the
same\n
\342\226\240AAA/\342\200\224i\n
()l
volts*''\n
-AAA/\342\200\224\n
1 volt\n
|*
0
0\n
-AAA/\342\200\224\n
0\n
VW\n
Fig. 3-22.
Resistive network analyzed
in
direction
\nan
example,
\nKirchhoff
consider
voltage
example:
values
of resistance\n
As
Rjj is positive and Rkj is negative.
network
of Fig. 3-22. For this example,
are summarized
in the following chart,\n
then
clockwise),
(say
in
ohms.\n
the
equations
the
\342\200\224
\342\200\224
0
ij
4i\\
Otj
+
\342\200\224
tz\n
1\n
0\n
=\n
2\n
1\n
=\n
3\n
0\n
=\n
-1\n
-1\n
-1\n
-1\n
0\n
4\n
0\n
0\n
5\n
0\n
=\n
0\n
6\n
0\n
=\n
0\n
0\n
7\n
1\n
=\n
0\n
0\n
0\n
8\n
0\n
=\n
0\n
0\n
0\n
0\n
9\n
0\n
=\n
0\n
0\n
0\n
0\n
-1\n
0\n
0\n
0\n
0\n
0\n
0\n
0\n
0\n
0\n
0\n
0\n
0\n
-1\n
-1\n
-1\n
0\n
0\n
0\n
4\n
-1\n
0\n
-1\n
-1\n
0\n
5\n
0\n
0\n
4\n
-1\n
0\n
-1\n
0\n
-1\n
0\n
4\n
can be made from this
are the elements of the
(2)
\npal
diagonal
\ntry
about
elements
are
positive;
the
principal
\nsymmetry and the
\napply
when
\nmon
direction.
\nof
on
\nanalyzed
\n\342\226\274aloes
of
basis
node
network
\ntive
in ohms.\n
resistance
\nlyzed
3-23.
\nFig.
\nis
chart
The
for
node:
Current
vm
p\302\273
sign rule
drawn
always
in a
com\302\254
is
characteristic
A
the
for
equations
network
node
resis\302\254
to be
ana\302\254
is shown in
basis
node-pair voltageequations
6\n
0\n
c\n
0\n
0\n
d\n
0\n
0\n
0\n
0\n
0\n
5/2\n
-1\n
1\n
0\n
-1/2\n
of\n
v,
ve
-1\n
0\n
f\n
node
Coefficient
a\n
e\n
This
diagonal.
example will illustrate
networks.
on
symme\302\254
below.\n
shown
Eq.
deter\302\254
of the princi\302\254
all others are
equations.\n
checking
of
of the six
equivalent
This
second
\nformation
in example:
are
loops
in
value
The
Resistive
chart.\n
system
The
\nnegativeor zero. (3) Thereisa
3-23.
-1\n
5\n
importance
\nminant.
rig.
-1\n
0\n
-1\n
0\n
0\n
*9\n
is\n
*7\n
-1\n
0\n
-1\n
-1\n
chart
of the
elements
\tThe
(1)
of
observations
Several
-1\n
0\n
0\n
*\302\253\n
0\n
5\n
-1\n
of\n
*5\n
-1\n
0\n
5\n
0\n
=\n
-1\n
-1\n
4\n
4\" 0i>\n
\342\226\240+\342\226\240
Oit
0t7
0*\302\253-I-
*4\n
0\n
2\n
-1\n
-1\n
5/2\n
0\n
-1/2\n
9/\n
0\n
0\n
0\n
0\n
-1\n
-1\n
2\n
0\n
3\n
equation\n
Coefficient
**\n
4\n
H-
~l\342\200\234
Ots
Voltage
Eq.
of the
is the equivalent
row
first
the
where
Chap.
EQUATIONS
NETWORK
62\n
-1\n
0\n
-1\n
5/2\n
0\n
-1/2\n
-1/2\n
0\n
-1/2\n
0\n
-1/2\n
3/2\n
Chap.
EQUATIONS\n
NETWORK
3\n
READING\n
FURTHER
of the formulation of equilibriumequations
for
For discussions
see
\nworks,
\nMIT
\n25-49,
New
\nInc.,
\nin
the
\nmay
\npp.
\n(John
many
Mathematics
\nApplied
\ning
in
found
Inc.,
New
for
Wiley
principle of duality is discussed
Gardner and
coupled
equations for magnetically
Corcoran,
and
Circuits
Alternating-Current
New York, 1951),
pp. 222-230,or LePage
Book Co., Inc.,
Analysis
(McGraw-Hill
102-110.
Further discussion of determinants
texts in mathematics, for example in Pipes\342\200\231
and Physicists (McGraw-Hill Book
Engineers
pp. 69-76, in Wylie\342\200\231s Advanced
Book Co., Inc., New
(McGraw-Hill
York,
Mathematics
573-579,
Inc.,
pp.
1952),
York,
be
Wiley & Sons,
Network
General
Seely,
\nNew
\nCo.,
& Sons,
Wiley
\nand
and
Kerchner
see
\n(John
(John
by Johnson and by
those
example
of writing circuit
subject
\ncircuits,
2. The
Analysis
above.\n
cited
On
for
texts,
Transient
Chap.
1954),
York,
many
\nBarnes
Engi\302\254
New York, 1944),
More
Systems
in Weber\342\200\231s Linear
and
Inc.,
Engineering Staff at
& Sons, Inc., New York, 1940),pp. 112-138.
are contained
in Gardner and Barnes, Transients
(John
Wiley & Sons, Inc., New York, 1942),pp.
treatments
Linear
\nin
Co.,
of
Principles
Physical
net\302\254
by the Electrical
Circuits
Electric
Wiley
(John
\nadvanced
Book
(McGraw-Hill
or
45-67,
\npp.
and
Mathematical
Johnson\342\200\231s
Analysis
\nneering
63\n
in
& Sons,
1946),
Guilleman\342\200\231s
Inc.,
The
Mathematics
York, 1951),
Circuit
of
New York, 1949), Chap.
Engineer\302\254
1.\n
Analysis
NETWORK
64\n
Chap.3\n
EQUATIONS\n
PROBLEMS\n
3-1. For the
\nKirchhoff
part
d, select
four appropriate
the
formulate
figures,
voltage
for
\ncated;
four networks shown in the
For parts a, b, and c,
equations.
the
use
loops
indi\302\254
loops.\n
*2\n
\nmit
we are to write equationsthat will
currents
in the inductors to be found. Howmany
simultaneous
the
to describe
are required
equations
system? Write
Discuss.\n
on the loop basis.
equations
the
In
3-2.
the
\ndifferential
\nequilibrium
network
L2
L\\
shown,
\302\24313\n
v2\n
\302\2731\n
pOp\n
ic!
-\n
3-3.
In
the
\nthrough
\nare
\non
required
the
loop
the
t\302\2733\n
pOp\n
pOp\n
\342\200\234\n
-c3
^C2
Prob.
per\302\254
-\n
3-3.\n
the
current
shown, the problem is to
How many simultaneous differentialequations
capacitors.
to describe
the system?
Write the equilibrium equations
basis.
conclusions with those found for Prob.
Compare
the
network
find
3-2.\n
\nin
the
figure.\n
3-5.
The
shown in the
network
network
a ladder
as
known
is
h\n
the
describe
to
\nfigure
a set of node equa\302\254
shown
network
Formulate
3-4.
\ntions
65\n
EQUATIONS\n
NETWORK
3\n
Chap.
\n(becauseof its physical appearance).
the
on
\ntions
extended
is
adding
alternately
\nby
indefinitely
inductors
and
\ncapacitors. Compare the
\nloops
shown
\nnetwork
a
\nmulate
The
3-7.
The
3-8.
\nwork
of
of the parameters are
is called
a Butterworth
properlyselected,the
low-pass filter.
For\302\254
equations on the basis (loopor node)that
smaller
of simultaneous differential equations.\n
the
number
the
network
shown in the figure is of a type designed
by
insertion-loss
\nDarlington
to\n
of differential
set
in
\nresults
above
number of
addition
each
values
the
When
3-6.
for
nodes
and
equa\302\254
Suppose that
basis.
loop
ladder
\nthe
of differential
set
a
\nFormulate
method.
Repeat
shown in the figure
vacuum
tube amplifiers.
network
some
\nnetwork.\n
Prob.
3-8.\n
Prob. 3-6 for
this
network.\n
represents the interstage
Repeat Prob. 3-6 for this
net\302\254
NETWORK
66\n
network
shown
two-stage
vacuum
The
3-9.
a
of
\nwork
Chap.
EQUATIONS\n
in the figure represents
tube amplifier. Repeat
3\n
the equivalent
Prob. 3-6 for this
net\302\254
\nnetwork.\n
\nwork
(the
network.\n
this
\nfor
forms
inductor
problem represents a bridged-Tfilternet\302\254
the bridge across the T). Repeat Prob.3-6
of this
network
The
3-10.
3-11. For the double-T network shownin the
\n3-12. The network shown in the figureis
Prob.
3-6 for this network.\n
\nRepeat
a
figure,
repeat
lattice
symmetrical
3-6.
Prob.
filter.
La\n
\ncoil.
\nfluxes
\nof
M
two
Consider
3-13.
Show
aid,
is
if the
that
the
sign
negative.\n
magnetically
currents
of M is
coupled
are in such
coils with current
a direction that
positive, while if the
fluxes
oppose,
in
each
the
two
the
sign
Chap. 3\n
3-14.
\n4,
\nFor
Fig.
this
\nCompare
\nfound
EQUATIONS\n
NETWORK
67\n
shown below is identical to that usedin Example
the loops are chosen differently than in the example.
but
3-10,
the differential equations on the loopbasis.
network
formulate
those
the
number
of terms in the equations that result with
The
circuit
in Example
4.\n
R i\n
3-16.
A network
\ncoil winding
\nequations
3-16.
\nin
the
for
Write
sense
this
the
accompanying
the
mutual inductance is shown below,with
indicated
dots.
Write
Kirchhoff
the
voltage
by
network.
Note that Mu = 0.\n
with
loop basis
network equations for the systemshown
figure.\n
Prob.
3-16.\n
EQUATIONS\n
NETWORK
68\n
to
duals
3-17.\n
Prob.
3-18.
Find the dual
3-19.
Find
the
dual
3-20.
Find
the
dual
3-21.
Find
the
dual
3-22. The
it
\nease
crossover
(a)
\tDraw
\nplanar
3-5.\n
3-9.\n
of
3-12
Prob.
to be nonplanar
appears
(in
this particular network,
be removed
so that the network is
network, (b) Find the dual
planar
a dual). For
have
can
point
the
of the networkof Prob.
of the network of Prob.
of the network of Prob.
of the network of Prob. 3-10.\n
3-8.\n
network
not
does
\nthe
give
coefficients.\n
identical
\nhave
3\n
dual and (b)
element
make the network equationsfor the
(a) find the
shown,
schematic
the
on
\nvalues
network
the
For
3-17.
Chap.
which
however,
planar,\n
equivalent
the
of
network.\n
3-23.
system of equationsfor ilf
Solve the following
i2,
and
i3, using
\ndeterminants.\n
- 2i2 +
3ti
9u-
+
\n-2ix
\342\200\224
Answers.
3-24.
\nknowns,
=
ix
Solve
it,
ii,
=
of
system
7it
\342\200\224
+
\n\342\200\2245ix
0i2
+
i2 =
-1\n
0\n
0\n
3\n
-2\n
-2\n
O\n
0\n
3\n
-1\n
=
Oij
-1\n
(a)
+9,
\342\200\22410
i, = -445/342.\n
-615/342,
1\n
(b)\n
-2\n
-1\n
4\n
(b) +133.\n
3\n
4\n
1\n
0\n
1\n
1\n
3\n
4\n
2\n
0\n
4\n
2\n
2\n
minors.\n
by
3\n
Answers,
5\n
= -10\n
Hi,
determinants
following
for the three
equations
3i2- 5is=
+
-295/342,
10\n
i, = 270/159.\n
i, by determinants.\n
-1\n
0\n
=
and
the
0\n
9i,
following
3-26. Evaluate
2\n
= 0
the
\342\200\2243ii
t\\
4i,
i2 = 210/159,
405/159,
8ii -
Answers.
+
4i2
\nOil
Oi, = 5
0\n
-2\n
1\n
-1\n
3\n
-2\n
-1\n
1\n
un\302\254
3\n
Chap.
3-26.
Consider
the equations\n
3x
\nx
\n4x
Is
(a)
(4,
2, 3)
the
regarding
writing
Answer.\n
6
-3
3y
+
circuit
the
Oy
2 =
+
values
7
-4
6\n
Answers.\n
Repeat
Prob.
3-27 for
What
(b)
can
these
Can
you
con\302\254
by these equations?\n
the
write
equations),
are in ohms and
}\n
solution?
(c)
Why?
in
networks
-3
3-28.
1\n
\342\200\224
a
\342\200\2241)
1,
\342\200\2241,
Answer.\n
-3\n
1
\342\200\224 =
5\302\253
represented
geometry
Element
\ndeterminant.
y
\342\200\224
determinants?
3-27. By inspectingthe
\n(without
\342\200\224 \342\200\224 \302\253\342\226\240
1
3z
Is (
a solution?
\nequations be solved by
\nclude
69\n
EQUATIONS\n
NETWORK
-4
11
-3
accompanying
the loop basis
volts.\n
-3\n
-3\n
8\n
the networks shown in the figure.\n
figure
system
3-29.
the
inspecting
By
write
\nequations),
the
shown (without writing
system determinant.\n
network
basis
node
node
node
1
Prob.
Answer.\n
9\n
-1\n
5\n
\ntains
1-ohm
a
source.
\nvoltage
\nof
equations.
number
\nThe
the
In
network
3\n
the circuit
2\n
3-29.\n
20\n
3-30.
Chap.
EQUATIONS
NETWORK
70
graph shown in
-1\n
5\n
17\n
10\n
the figure,each
branch
con\302\254
as shown, contain a 1-volt
branches,
on
the
this network
loop basis to obtaina set
Analyze
in any one equation.
like
terms
by combining
Simplify
is the loop number.\n
inside
each
square
resistor.
Four
2h
2h\n
graph shown in the figure, each interiorbranch
an
inductor
of 1 henry
\ncontains
and each exterior branch an inductor
\nof
A 1-volt
The
2 henrys.
source is located in one branch as shown.
\nnumber
inside
each
is the loop number. Analyze the network
square
\non the
loop basis to obtain a set of equations. Simplifyby
combining
in any one equation.\n
\nlike terms
3-31.
In
the
network
a current
contains
\nbranch
in
any
a
obtain
to
basis
\nterms
network
the
In
3-32.
\nnode
EQUATIONS\n
NETWORK
3\n
Chap.
one
graph
shown,
\nohms
and
interior
\342\200\224
Ri
\342\200\224
Analyze
1
ohm,
one
and
on the
network
combininglike
equation.\n
R\n
Prob.
Repeat
R
source of 1 amp.
the
set of equations. Simplify by
R
3-33.
71\n
3-32.\n
3-32 with external branch
of Ri = 1 ohm.\n
resistors
branch
Prob.
resistors
of
R
=
2
CHAPTER
4-1. Definitions for differential
this
of
\ntion
equations\n
study a number
we will
chapter,
the
differential
simplest
equations,
a\302\260
^
This
is of
equation
derivative
\nordered
be
\nmay
equations
An
contain.
they
of techniquesfor the
those of first order such
solu\302\254
= 0
+ axi
first order because
differential
Thus
first.
the
\nis
EQUATIONS\n
DIFFERENTIAL
FIRST-ORDER
In
4\n
as\n
(4-1)\n
the highest-orderedderivative
are classified by the highestnth order differential equation
written\n
dni
3F
for
equations
\nto
which
of the
the
dn~H\n
+
dF7'\n
di\n
. .
+
Un-l
\342\200\242
+
first degree. The
+ dni
degreeof an
derivative
highest-ordered
^
\342\200\224
is
equation
appears
(4-2)\n
V(t)\n
after
the
power
all possible
reduction.\n
\nalgebraic
is the dependent variable and t is the independent
var\302\254
an energy source, it is known as theforcing
\niable.
When
v(t) represents
The
variable
\nfunction.
dependent
i, which is to be found,is calledthe
or solution.
The differential
\nresponse
equation is linear if the dependent
\nvariable
and
all its derivatives
are of first degree. All other equations
\nare
A differential
is said to be ordinary if it con\302\254
nonlinear.
equation
\ntains
not partial)
derivatives. For the type of circuits
total
(and
only
in
\nassumed
Chapter
equations that describe net\302\254
1, the differential
\nworks
will
all
be ordinary,
linear differential equations with constant
It
should
be remembered
that the techniques we will
\ncoefficients.
\ndiscuss
will
in general,
not,
apply to nonlinear differential equations.\n
is
to
be
4-2
said
Equation
homogeneous when v(t) = 0; if v(t) is not
In
Eq.
the
\nzero
4-2, i
is nonhomogeneous.\n
equation
4-2. General and
In
\nknown
electrical
state
particular
solutions\n
the
problems,
with
all voltages
network is assumed to be initiallyin a
and currents fixed. At an instant of
that
the system is altered in a manner
the
of
one
or
or
more
switches.
\nrepresented
by
opening
closing
of
is
to
obtain
mathematicalequationsfor
\nobjective
analysis
\ntime
designated
t
* 0,
72\n
be
can
The
current,\n
Art.
altered
was
\nlibrium
1 to
taken
\nage
equation
+
\ntial
can
It
be
can
\nables
in
4-3
switching
the
(4-3)\n
linear
differen\302\254
constant
if the
solved
be
coeffi\302\254
may be accomplished
This
separated.
vari\302\254
by rearranging
variables
the
(4-4)\n
L\n
-
-
i
+
t
to
can be integrated
the equation
separated,
In
give\n
K
(4-5)\n
\302\243
that the logarithmisto
e = 2.718
base
this equation, the constant K is redefined
in
where In designates
the
\nof
form
the
simplify
of
constant
of another
logarithm
4-5 may then
Equation
be
=
the
\nAlso, from
of a
definition
In
Eq. 4-7 may
so that
be
the
equation
In k
(4-6)\n
\\nk
logarithm,
In e* =
(4-7)\n
x, or logio
=
10*
x.
that\n
y
+
In z
i
=
In (ke~M/L)
\342\200\224
In
(4-8)\n
yz
written\n
In
With
as\n
In e~Rt'L +
we know
logarithms
terms
written\n
i =
In
by
To\n
the
K
since,
Eq.\n
form\n
t
With
posi\302\254
volt\302\254
0\n
with
equation
\ncients.
the
JK\n
first-order
a
is
equi\302\254
is\n
La
This
from
is changed
reference time t * 0.* After the
Kirchhoff
the
place,
from the instant
K
switch
the
2 at
73\n
switching.\n
shown in Fig.4-1,
position
\nhas
the
by
In the network
\ntion
of time measured
in terms
etc.
charge,
voltage,
EQUATIONS\n
DIFFERENTIAL
FIRST-ORDER
4-2\n
in this
(4-9)\n
form, the antilogarithm may
be taken to
\ngive,\n
=
i
*It is assumed that
\ntransition
\nrent
i.\n
from
position
the
1 to
switch
position
is
ke~Rt/L
a
(4-10)\n
\342\200\234make-before-break\342\200\235
2 does not
cause
an
type
interruption
that
and
of
the
the
cur\302\254
EQUATIONS\n
DIFFERENTIAL
FIRST-ORDER
74\n
Chap.
4\n
the logarithm of another constanthas
the
form
of the
solution.
4-10 is the network
\nsimplified
Equation
or solution.
This
solution
is free of derivatives and expresses
\nresponse
\nthe
between
the
variables.
and independent
relationship
dependent
\nThat
it is the
solution
can be verified by substituting Eq. 4-10into
K as
constant
the
Redefining
4-3.\n
\nEq.
the solution is
of Eq. 4-10,
form
the
In
known
as
the
solution.
general
is evaluated, the solution is a 'particular
The
to any number of situations. A
\nsolution.
applies
solution
fits the
of a particular problem.\n
specifications
\nparticular
new about the
To evaluatethe constant
we
must
know
something
k,
In
of i and t.
this particular prob\302\254
\nproblem,such as any pair of values
know
that
has
taken placemustbe
we
the
current
after
switching
\nlem,
as before switching because of the inductor in the
the
same
cir\302\254
\njust
of integration
solution
general
constant
the
\nIf
at
Thus
\ncuit.*
t =
that the current has
0, we know
=
i(0)
as the initial
is known
This
value
\nthis
required
into
condition
Z
|
=
(4-11)\n
condition of the circuit.
4-10
Eq.
value\n
the
Substituting
gives\n
jceo =
k
(4-12)\n
XI\n
The
solution
particular
of this
example becomes\n
i = Z
e~Rt/L
(4-13)\n
MX\n
The
4-3.
integrating
factor\n
a nonhomogeneous
Consider
I +
\nvariable
\302\243
or
by
\nmultiplied
\nthe
constant
is a
P
where
a
constant.
the
same
ept, which
quantity
written\n
equation
\302\253-\302\253\n
and Q may
The
factor.
be a
equation
Suppose
will be known
of
function
the
independent
is not altered if every term is
that we multiply Eq. 4-14by
as an integratingfactor,
f
There
\nresults\n
di\n
ept
jt
*
\nArt.
This
is an
application
+ Piept =
of the principle of
Qept
constant
(4-15)\n
flux
linkages
discussed
in
1-7.\n
f If
P is
a function of time, the properintegrating
factor
is
efPdt.
See
Prob.
4-5.\n
this
That
factor
by a
multiplication
EQUATIONS\n
DIFFERENTIAL
FIRST-ORDER
4-3\n
Art.
of Eq. 4-14 can be recognized
solution
by
of a product:\n
for the derivative
the
\npossible
\nequation
d(xy) = xdy +
letting
By
the
of
out
\342\200\234pulled
=
x\n
y =
i and
ept, we
y
75\n
made
hat\342\200\235has
the
recalling
dx\n
(4-16)\n
have\n
(4-17)\n
is
which
4-15; thus we
of Eq.
side
left-hand
the\n
have\n
(4-18)\n
This
=
iept
first
\nular
\nmentary
\nby
a
\nor
=
(4-19)\n
/ Qept dt
e~pt
Ke~pt\n
(4-20)\n
comple\302\254
network
any
network
\ntary
\nthat
is\n
function
Q.\n
a positive constant determined
and Q will be either the forcingfunction
In the limit, the
function.
P is a positive constant;
zero, because
P will be
problem,
parameters,
the forcing
must
on the forcing
depend
of
derivative
function
not
does
function
the
K\n
partic\302\254
integral
For
dt +
the
is
\nsecond
/ Qept
give\n
+
in Eq. 4-20 is known as the particularintegral;
term
as the complementary
known
function. Note that the
does
not contain
the arbitrary
constant, and the
i
or\n
The
to
be integrated
may
equation
complemen\302\254
approach
=
Ke~pt
lim
0
(4-21)\n
oo\n
t\342\200\224\342\226\272
Thus
i(
the
When
\nvalue
at
particular
t = \302\253is
\nwould
reduce
\nmost
frequently
to
=
oo)
lim i(t)
t
> 00
of
spoken
as
contain
encountered
i = A sin
+
{wt
of
the
\ncomplementary
solution,
function;
engineering,
have
the
forms\n
<j>)
and
i
4-20
iP be the
letting
dt
(4-22)\n
00\n
\342\226\272
not approach zero in the limit,its
the steady-state
value. For this case, the
no exponential
factor or otherwise it
Let the general solutionof Eq.
\nparts
Qept
is\n
does
In electrical
zero.
infinity
= lim e~Pt
J
t
integral
must
integral
\nparticular
as time approaches
of i
value
the
be
=
written
the steady-state values
a constant
as the
(4-23)\n
sum of the
particular integral and ic
be
two
the
thus\n
\342\200\242\342\200\242
\342\200\242\n
t
% ==
Zp
I\342\204\242
%c\n
(4-24)\n
If
\nstate
\ning
\nthis
=
i
\nand
the
\nthe
in
\ntron
\nor
the
\nto
a
by convention; the
purely
way of knowing whether it
of the current.\n
division
has no
individual
is in
the
elec\302\254
transient
and steady-state
portions of the solution
the network
of Fig. 4-1 with the switchmoved
consider
=
1 at t
2 to position
0. The Kirchhoffvoltageequation
position
division
by
L,\n
V
.
R
di
\ndt+ L%
this
Comparing
to Eq.
equation
t>
\nP
R
=
this
to
solution
L\n
4-14, we see
n
and a
Q
j-
The
0,
transient
the
problem,
\nis, after
=
1\n
illustrate
\nfrom
t
to account for
is made
solution
steady-state
To
as havingbeenestablished
at
adjust
itself, mathematically,
0 and all other times. This is an arbitrary
division
which nevertheless
has utility as a conceptualaid. The
a current
Example
(4-25)\n
t =
at
of
it
must
solution
\ndivision
i\302\273+
value is regarded
transient
response
the
\nof
steady-
convention
has been established
for call\302\254
the
transient portion of the solution.By
ic
is
made
response
up of two separate parts:\n
the
convention,
steady-state
as a
written
term
remaining
The
be
may
Chap. 4\n
A
i\342\200\236.
designated
value,
the
4-23, it
forms of Eq.
of the
either
has
iP
EQUATIONS\n
DIFFERENTIAL
FIRST-ORDER
76\n
is given as
equation
that\n
V
\342\200\224\n
Eq. 4-20
which
becomes
for
\nthis problem,\n
i =
r-Rt/L\n
eRt/L
e\n
fa
+
Ke~Rt/L\n
(\n
the
Evaluating
integral,
we obtain\n
i =
as
the
general
\nis
zero
before
\nof
the
inductor.
^ +
Ke~Rt/L\n
If the current in the networkbeingconsidered
it must be zero afterward because
switching
action,
that i(0) = 0 leads to the particular
The requirement
solution.
the
\nsolution\n
=
\342\200\242
^
The
steady-state
and
transient
(1
-
(4-26)\n
e~Ri/L)
divisions
of this current
are
shown
in\n
Art.
4-4\n
Fig.
4-2
=
0,
t =
\nrent
at
4*4.
Time
The
\na
constants\n
of Eq. 4-3
be written
in
solution
4-13
Eq.
by
cur\302\254
0.\n
particular
\ngiven
is zero
there
that
such
\nadjusted
steady-state\n
is
term
transient
the
and
The
current.
actual
77\n
at
established
is
(V/R)
\nportion
\nt
sum or the
their
with
along
EQUATIONS
differential
first-order
may
form
nondimensional
as\n
Fig. 4-2. Transient
=
Y
\n1
where
\ntime
7o
\nhomogeneous
\ndifferent
\nattached
(4-27)
0\n
of
parts
\nExample
the
initial
t
When
\nengineering.
\342\200\224
T,
by
Eq.
=
e-1
=
i(T)
= 0.3770
In other words, the current
to
or
0.37\n
(4-28)\n
constant.
A
to
plot
Fig.
37%
of its
a similar
By
decreases
(4-29)\n
initial value in one
can
it
be
shown that the
computation,
initial
of
its
value in four time
2%
approximately
in
70
is
shown
if
against t/T
Fig.4-3.\n
decreases
\nconstants.
of
4-27,\n
70\n
\ncurrent
solution
1.\n
value
t\\T)\n
\ntime
and steady-
of current at t = 0 and T = (L/R)
is the
of the
The form of Eq. 4-27 is the solutionof all
system.
first-order
differential
where 7o and T have
equations,
for
values
different
The physical
problems.
significance
the
time
to
constant
is of great importance in electrical
is the
constant
e~t/T
\nstate
of
4-3.
Normalized
exponential
curve for
<r\342\200\230/r\n
FIRST-ORDER
78\n
The
a
\nwith
solution
of the
constant
forcing
Chap.
4\n
nonhomogeneous differential equation
is of the form of Eq. 4-26,which
may
first-order
function
form
nondimensional
in
written
\nbe
EQUATIONS\n
DIFFERENTIAL
as\n
\342\200\242\n
i-
l
=
i(T)
final
its
Fig. 4-4.
\ntransient
\nat
disappears
this
least,
requires
function
\nexponential
\nto
another.
of
that
\nwith
its
of
63%
for
\nstandard
final
(1
Normalized
time
\nvalue
\n-
0.1
\nT1000
(4-31)\n
comparison.
constant
of 100
Msec.
for
curve
(1
\342\200\224
e
t/T).\n
is conveniently measured and usedas a
As an example,
consider a series RC circuit
solution,\n
i =
The
T,\n
- 0.63/o
- 0.37)/o
exponential
value)
a general
has
\nwhich
When t =
is useful in comparing the behaviorof one system
the
It is not possible to compare times at which
its steady state) since, mathematically
(or reaches
infinite
time. However, the time interval for an
to 37 % of its initial value (orincrease
to decrease
constant
time
The
(4.30)\n
time
reached 63% of its steady-state value in one
will increase to approximately 98%
the current
Similarly,
in four time constants.\n
value
\nconstant.
\nof
e-t/r
has
current
the
or
1 _
in Fig. 4-4.
is plotted
function
This
=
o\n
circuit is T
for the
ohms and C
However,
hfi~i/RC\n
is 1000
jujuf;
= RC. Supposethat
then
=
T
if R = 1000 megohms
sec, or 17 min.
For onecombination
of
R
100 X 1000 X
and C
R
and
\302\2731
C, the
the
has
/if,
10\342\200\234\342\200\234
then
current\n
would
decrease
to
other,
the
the
\nfor
\n37
%
of the initial value in the smalltime
current
would require about 17 min to
value.\n
In experimentally recording a transient, the
this
of the order of 1 or 2%. For
\nment is often
\nsometimes assumed to have disappearedwhen
value
\nfinal
\nbe
the
case
\nnential)
is
\nin
an
of
reaches
2%
it is
dis\302\254
constants.\n
time
four
is
of the
expo\302\254
constants,
a transient
that
in
\nappears
(or 98%
increasing
time
four
assumed
\noften
final
of the
2%
value
Since
determined).
\nreach
the
can
time to
measure\302\254
a transient
reason,
it
^sec;
decreaseto
of
accuracy
0.1
of
as
accurately
(as
79\n
37 %
initial
of the
EQUATIONS\n
DIFFERENTIAL
FIRST-ORDER
4-5\n
Ait.
This basis is sometimes used to meas\302\254
the
time
constant
of a system.\n
\nure
4-5. The principleof
To
di\n
^
...
+
\nname
\ndoes
the
of
\nnature
not
forcing
of the
\nfree
response.
The
and
\nparts\342\200\224forced
of
the
+
(t>i
function
the
the
henry. By the Kirchhoffvoltage
t>2(t)
+
... +
+
(4-32)\n
vn(t)
of a first-order nonhomogeneous
=
vi
4-20, P = R, a constant,
Q
is\n
\342\200\242
\342\200\242
\342\200\242
+
vn)eRt
dt
+ Ke~Rt
(4-33)\n
integral depends on the
particular
voltages and, for this reason, is giventhe
other hand, the complementary function
the
function (except that K is fixedby the
conditions
function at t \342\200\224
circuit
exist\302\254
0 and
forcing
function is given the name
complementary
total
can be thought of as made up of two
response
We now have three sets of terms definingthe
free.
solution
of the differential equation.\n
particular integral
\nsteady-state
\nforced
t>2
before,
forcing
The
time).
that
\ning
/
On
on
depend
at
parts
e~Rt
+
by Eq.
solution
the
and
response.
forced
\nmagnitude
\ntwo
n
super\302\254
theorem.\n
\nposition
solution
given
observed
have
we
with
the
differ\302\254
V\\{t)
general
vn
=
i
As
+ Ri =
equation,
+
t>2
the
of
terms
\ndifferential
\n+
4-5.\n
series circuit
illustrating
have\n
we
In
sources
\nvoltage
to be 1
taken
is
L
equation,
\nlaw,
of the
form
the
simplify
\nential
in Fig.
is shown
sources
\nvoltage
Fig. 4-5. RL
n series
with
circuit
RL
series
A
superposition\n
response
solution
and
complementary
and
transient
and
free
function
solution
response\n
\nused
text
the
thioughout
Chap. 4\n
be
engineering literature, and
our preference will be for the last
used in electrical
are
sets
three
All
EQUATIONS\n
DIFFERENTIAL
FIRST-ORDER
80\n
will
although
\nset.\n
circuits.
short
by
\nreplaced
=
ix
this
If
the
4-5,
\nthat
the
\nmay
be
(4-34)\n
kxe~Ht
of
Suppose
sum
written\n
=
in
\342\200\242
\342\200\242
\342\200\242
+
R is
dt]
dt
+
+
(kx
/
\302\253
+ k2 +
we
... +
+
k2
+
kx
... +
(4-35)\n
kn)e~Rt
set\n
may
kn
(4-36)\n
in Eq. 4-35
the integral terms
a constant,
dt\n
v<&Rl
arbitrary constant,
K
Because
VieRt
vneRt
so far an
k is
each
e~Rt [J
...+/
+
Since
vtf*1 dt +
each generator of the circuit
to that given in Eq. 4-34.
in this manner are added together. This
response
currents
found
+ ia +
ix
e~Rt J
is\n
is repeated
for
will be similar
experiment
\nFig.
except Vi(t), are removed and
under this condition
all voltage sources
The response
that
assume
Next,
be
may
combined
\nto give\n
+ ia +
ix
This
then
\na
rule
general
for
\nholds
is
\nand
known
great
hold
as the
... +
+
principle
systems
in more complex
\n1951)
in
READING\n
by
York,
the
pp.
Transients
\npp.
that it
differential equations are discussed Fich in Transient
in Electrical Engineering (Prentice-Hall, Inc., New
Electric
15-30.\n
networks
systems.\n
\nTransients,\342\200\235
\nto
Ke~Rt\n
network theory. The fact
is the root of the greatdifficulty
First-order
under
+
applicationof
of superposition. This principle
located
in linear
FURTHER
\nAnalysis
dt
vn)eRt
This is the
responses.
arbitrarily
importance
for nonlinear
such
\nanalyzing
individual
sources
voltage
of
not
\ndoes
the
summing
t>2
+
(vx
with Eq. 4-33 which was found for the
In summary, the total response of a linear
found by considering each voltage source
removed
and replaced by short circuits
sources
other
all
with
\nalone
e~Rt f
that
to
identical
is
=
functions.
forcing
\nnetwork
in
is identical
equation
\ncombined
\nand
\342\200\242..
+
of
heading
36-66.
See
(John
\342\200\234Classical
also
Wiley
Kurtz
Solution
and
& Sons,
of
Corcoran\342\200\231s
Inc.,
Single-Energy
Introduction
New York, 1945),
Chap.
DIFFERENTIAL
FIRST-ORDER
4\n
EQUATIONS\n
81\n
PROBLEMS\n
shown in the figure, the switchis changed
from
=
1 to position
2 at t
having
pre\302\254
0, a steady-statecurrent
been
in the RL circuit.
established
Find the particular solution
=
current
in the circuit.
Answer, i
(V/J?1)e~(Rl+B*)</z'\n
In
4-1.
\nposition
\nviously
the
\nfor
4-2.
\nthe
2, assuming
position
source while the switch was in position1. (b) Write
as\n
for the charge under the same conditions
equation
the
for the charge
as a function of time and evaluate
\nthe differential
Solve
(c)
(a),
In
4-3.
\n7o
at
and
of
\ntion
to (c). q =
Answer
constant.
\narbitrary
in Fig. 4-1 with a capacitor, (a) Write
in the system after the switch
the current
that the capacitor was charged to a voltage
of the
that
to
for
equation
integral
\nequal
inductor
the
Replace
in
\nis
circuit
the
circuit
=
t
0 the
the
CVert/RC\n
the capacitor Ci
shown,
is chargedto a
Solvefor the
switch is closed.
charge
as
voltage
a
func\302\254
time.\n
\342\226\240AAAr\n
K\n
R\n
:C2\n
v0ipcx\n
4-8.\n
Prob.
In the circuit
4-4.
\nfrom
no
\n4-5.
position
on the capacitor.
charge
We
1 at t
2 to
position
\nwas
of Prob. 4-2, supposethat
an
\ntion
\ngrating
\nsolution
derivative
the
equals
is
factor
to
the
R
factor\342\200\235
\342\200\234integrating
R
P(t)i
such
d(Ri)/dt.
while
=
that
switch
is
in position
changed
2 there
as a function
of
time.
equation\n
<2(fl\n
the
left-hand
side of the
(a) Show that the
equa\302\254
required
inte\302\254
(b) Using this integrating factor, find the
that corresponds
to Eq. 4-20.\n
equation
= eSFdt.
differential
that
Find the charge
+
jt
by
and
the differential
to multiply
wish
\342\200\224
0
the
is
\nK
\nrent
the current assuming
a
when
=
that a
R/L.
follows the
closed at
(b)
t
for
Solve
use of
Make
Suggestion:
R/L.
switch
time.\n
shown, the voltage source
a is a constant.
The switch is
where
Ve~at,
\nSolve for
figure, the
steady-state having previously been attained.
the circuit as a function of
in the accompanying
circuit
the
In
4-7.
\n=
in
current
the
for
\nSolve
0, a
at
closed
shown
circuit
t =
the
In
4-6.
Chap. 4\n
EQUATIONS\n
DIFFERENTIAL
FIRST-ORDER
82\n
law
=
v(t)
0.
the
(a)
cur\302\254
for
rule
1\342\200\231Hospital\342\200\231s
forms.\n
\nindeterminant
L\n
4-7.\n
Prob.
\na
source
voltage
\ntem,
solve
for
v(t)
the
response
Show
that
\nintersects the time
constant.
axis at t
\342\200\224
\n70(1
e~t/T)
Similarly,
at
t =
circuit.
0
connecting
For this
sys\302\254
4-8.\n
the tangent to
\342\200\224
T.
\ndecreased
at the initialrate, it
\ntime
RL
=
t
i(t).\n
Prob.
4-9.
closedat
shown, the switch is
= V sin u>t to a series
circuit
the
In
4-8.
show
0 intersects
the curve i
would
that
be
to
reduced
the
that line i
70
t =
0
if the current
zero value in one
to the curve i =
tangent
=
at
7oe_t/r
that
show
will
This
=
at
time
t =
T.\n
FIRST-ORDER
4\n
Chap.
network
the
In
4-10.
shown,
is observed
\ninitial
value
of the
current
is
in
0.1
sec. Find
\ndisappears
\nof
and
the
of
(c)
C,
equation
=
=
2.5 Mf, (c) i
\nC
10-2e~40*.\n
waveform
\ncurrent
EQUATIONS\n
DIFFERENTIAL
is closedat = 0. The
with a cathode ray oscillograph.The
measured to be 10 ma. The transient
value
(a) the value of R, (b)
=
104ohms,
i(t). Answers, (a) R
the switch K
t
the
(b)
4-11. The circuit shown in
accompanying
\nresistor and a relay with inductanceL.
figure
the
The
\nit
is
\nwhen
\nhenrys,
4-12.
\nV
with
\nto the
\nof
t
of
\ntion
time,
is closed
K
\nswitch
=
current
=
at t
0, and
when
actuated
0.1 sec.
i(t)
with
i =
the
Find:
all
83\n
through
is
relay
(a) the inductance L
coefficients
evaluated.
of
the
the
of a
so that
adjusted
the coil is
it is observedthat
consists
0.008 amp. The
is actuated
relay
coil,
(b)
the
equa\302\254
Answers, (a) L = 620
\342\200\224
e-1**)
0.01(1
amp.\n
of
A switch
is closed at t = 0, connectinga battery
voltage
delivered
a series
RC circuit,
(a) Determine the ratio of energy
to the total energy supplied by the sourceas function
capacitor
\302\260\302\260.\n
that this ratio approaches 0.50 as t \342\200\224>
Show
(b)
(b)
a
CHAPTER
5\n
IN
CONDITIONS
INITIAL
5-1. Initial conditions in
individual
NETWORKS\n
elements\n
chapter, we found that the generalsolutionof a first\norder differential
equation contained an unknown designatedan arbi\302\254
For
constant.
differential
of higher order, the pattern
\ntrary
equations
\nwill
that
the number
of arbitrary constants equals the equation
develop
If
\norder.
the
unknown
constants
are to be evaluated for
arbitrary
other
must be known about the network
\nparticular
solutions,
things
the
differential
We must form a set of simulta\302\254
\ndescribed
by
equation.
one
of which is the general solution, with additional
\nneous
equations,
to total
the number
of unknowns. The additional equations
\nequations
as values
of voltage, current, charge, etc., or
\nare
conveniently
given
these
at
the instant
\nderivatives
of
network equilibrium is
quantities
=
\naltered
t
0.
Conditions
by
switching
action,
existing at this instant
the
last
known
as
In
\nare
that alters network equilibrium,
of the
network
might have voltages across their terminalsor
them
as a consequence
of past history of the
through
in
network.
To evaluate
initial voltages or currents,
the
each
and current changes when the
determine
how
voltage
the switching action
Before
\nelements
\ncurrents
\nforces
conditions.\n
initial
the
driving
we
\nmust
net\302\254
is
\nwork
altered.\n
assumed to exist before
in turn, established
\naction
takes
by switching actionat
place
were,
in the past.
Such voltages and currents in the
remote
time
\nwork
are
said
to be in the steady
act in zero time. To differentiate
switches
We
assume
that
before
and immediately
after the closing of a
time
\nthe
immediately
Thus conditions existing just
and
we
will
use
+
signs.
will be designated as
the
switch
is operated
etc.,
In
problems,
many
conditions
switching
net\302\254
\nsome
state.\n
between
\342\200\224
\nswitch,
\nbefore
t(0\342\200\224),
after
\nconditions
as
i\342\200\230(0+),
v(0+),
v(0\342\200\224)
etc.\n
in networks, we will study the
at the instant equilibrium is
of
each
different
element
The
In the ideal resistor, current and voltage are related
Resistor.
in Fig.
If a step input of voltage,
v = Ri.
5-1,
law,
to a resistor
network, the current will have the same
applied
altered
(1 /R). The current througha
by the scale factor
\nform,
Before
analyzing
initial
conditions
\naction
\nby
altered.\n
shown
Ohm\342\200\231s
wave\302\254
\nis
resistor\n
84\n
Art.
if the
instantaneously
change
may
NETWORKS\n
voltage
instantaneously
may
change
was
concluded
voltage
\nSimilarly,
IN
CONDITIONS
INITIAL
5-1\n
85\n
changes instantaneously.
if current
changes
\ninstantaneously.\n
act
at the initial instant, and the inductorwill
terminals.
an open circuit independent of the voltageat the
of value
Io flows in the inductor at the instantswitching
to flow. For the initial instant,
will continue
that
current
of as a current source of Io amp.\n
can
be thought
to flow
current
\ncause
\nas
if
it were
\nIf
a
current
\ntakes
connect
to
switch
a
\nclosing
in Art. 1-7, that the currentcannot
a system
of constant inductance. Thus
not
an inductor to a source of energywill
in
instantaneously
\nchange
place,
inductor
\nthe
It
Inductor.
The
Vv\n
b\342\200\231b\n
V\n
(V7R)\n
O'
and
Current
6-1.
Tig.
6'\n
relationships in a
voltage
resistive\n
purely
element.\n
The
\nnot
a short
(or
\nvoltage
is
\ncapacitor
in
capacitor
and
for
5-3.
Fig.
are summarized in
the
can\302\254
voltage
of fixed capacitance. If an
energy source, a current will
will be equivalent to a short
charge are proportional
source of value
a voltage
to
offered that the
in a
so that zero charge corresponds to zero
With an initial charge in the system,the
q/C,
charge.\n
conclusions
conditions
\nshown
a system
to an
voltage
circuit).
equivalent
initial
the
These
\nfinal
v =
system,
\ncapacitive
the
because
follows
This
connected
since
instantaneously,
\ncircuit.
is
is
capacitor
\nuncharged
\nqo
in
instantaneously
change
\nflow
1-6, proof was
Art.
In
Capacitor.
case
special
These
equivalent
Fig. 5-2.
of constant
Vo
A
=
qo/C,
similar
where
chart
of
voltage sources is
circuits are derived
from the
\nrelationships\n
y
Vl
d%
\342\200\224
L \342\200\224r~
and
derivatives
\nsources.
The
\nopposite
to
It
\na
is
network
zero
having
those
not
the
yy\n
C -r:
in each case for invariant voltage
for L and C are
for final conditions
conditions for these elements.\n
initial
possible to interrupt a current instantaneouslyin
a switch.
If an attempt is made to open
a switch\n
opening
always
by
for
=
value
circuits
equivalent
.
ic
at\n
\ndt
the
J
the
\nof
from a voltage source, an arc will
switch
to permit the current to flow until
field is spent.\n
the
across
\nlished
magnetic
Element
initial
land
Chap.
inductor
an
disconnect
to
NETWORKS\n
IN
CONDITIONS
INITIAL
86\n
circuit\n
Equivalent
at t
condition)
estab\302\254
the
energy
Equivalent
- 0+\n
Element\n
\ncircuit at
t-co
voltage
sources
\n[for
R
be
5\n
R\n
\nof
constant
potential)\n
R\n
sc\n
-o\n
o
o-\n
O
oc\n
o\n
0\n
lQ\302\261
V\302\260\n
Vo-g-
condition
Initial
5-2.
Tig.
the
for
\ncircuits
equivalent
Fig.
elements.\n
6-3.
\nsources of
Geometrical
5-2.
to
a constant
for
elements
constant
potential.\n
equation that describesan
circuit
RL
may
equation
show
\nderivative
\nthe
\nbe
circuit
the
of
current.
is
closed
From
zero.
that
relationship
Eq.
If
the
at t =
5-2,
the
(5-1)\n
in the
be arranged
a
to
con\302\254
source:\n
voltage
Ljt+Ri=V
This
voltage
of derivatives\n
interpretation
Consider the differential
\nnected
Final condition equivalent
the
for
\ncircuits
i(F
form\n
-iR)
must exist between
switch
connecting
(5-2)\n
current and the
time
the voltage sourceto
0, the current in the systemat
initial value of the derivative
=
t
0 must
is\n
=
It(0+)
Now
the
\nfunction
\nhas
quantity
of time.
a magnitude
\napproximate
<M>\n
~L
of the requiredplot current
as a
and
Equation 5-3 tells us that this slopeis
For some small interval of time, this
must
V/L.
actual
curve found by solving Eq. 5-1. Assume
di/dt
is the slope
of
positive
slope
the
that\n
INITIAL
Art.
5-3\n
the
current
second
A
\nt\\.
increases
\ntime
5-2
Eq.
\nusing
-
differential
a
equation.
more
first
the
curvature
\nrepresents
Fig.
\nz(0+)
Curves
0,
di/dt(
The
= 0,
dH/dt*(0+)
initial
of an
Approximation
\ncurve
tangents
by
to typical
corresponding
=
0+)
0, dH/dP(0+) =
= + K,
to the
actual
curve.\n
=
=
initial conditions:
(a)
K;
0,
*(0+)
(b)
= K,
t'(0+)
(c)
di/dt(
= 0,
di/dt(0+)\n
=
0+)
+Kh\n
-K2.\n
shows several combinations
\nsponding
5-4.
Fig.
dH/dt*(0+) = 0;
=
dH/dt*(0+)
0; (d) z(0+)
\ndi/dt(0+)
ure 5-5
a graph\302\254
solution
the
6-6.
=
illus\302\254
curve approach the actual curve.\n
approximate
derivative
represents slope so the secondderivative
or the
rate of change of the slope with time. Fig-\n
will
closely
as
Just
of
(5-4)\n
intervals are chosen,
\nsmaller the time
\nthe
process,
provides
of the
5-4,
Fig.
interpretation
\nical
\nof
this
of
in
as a function
point by
- hR)
(V
J
Continuation
\ntrated
time
i\\ at
value
as\n
\302\253.)
%
87\n
NETWORKS\n
at the rate V/L to a new
to the curve of current
linearly
approximation
made
at this
be
may
IN
CONDITIONS
slope
and
of initial conditions,
with
the
corre\302\254
curvature.\n
5-3. A procedure for evaluating initialconditions\n
that must be followed in solvingfor
is no unique
There
procedure
it is usually
wise practice to solve first
\ninitial
conditions.
However,
\nfor
the
initial
values
of the
variables\342\200\224currents
or
voltages\342\200\224and then
The first step is essentiallyroutineand based
circuits
for t = 0+ given in Fig. 5-2. In the
\non the
second,
equivalent
and order of manipulation will be differentfor eachdifferent\n
\nthe details
\nsolve for
derivatives.
A successful
network.
Chap.
will not be obvious at all, a fact
a challenge in the solution of initial
5\n
that
approach
offers
and
interest
\nadds
IN NETWORKS\n
CONDITIONS
INITIAL
88\n
value
\nproblems.\n
voltage may be found directly
For each element in the
schematic.
of the network
determine
just what will happen when the switchingactiontakes
a new schematic of an equivalent
From
this
analysis,
= 0+
may be constructed according to these
\nstudy
a
from
we
network,
\nmust
network
\nplace.
t
\nfor
of current or
values
Initial
rules:\n
by open circuits or by current generators
of current
the
value
flowing at t = 0+.\n
of
all capacitors
by short circuits or by a voltagesource
=
is
an
initial
Vo
charge.\n
qo/C if there
are left in the network without change.\n
\nhaving
Replace
(2)
\nvalue
Resistors
(3)
the
Consider
is
switch
\nthe
6-6.
Tig.
inductors
all
Replace
(1)
network
passive
t = 0, no
solution
illustrating
(b) equivalent
network;
time.
to that
prior
Supposethat
voltage having been appliedto
Network
\nloop
in Fig. 5-6(a).
shown
network
two-loop
at
closed
for initial conditions:(a)
network at t = 0+.\n
Since there is
the\n
two-
no initial
voltage
be replaced by a short circuit;similarly,
the
be replaced
\ninductor
by an open circuit, there being no initial
may
The resulting
network is shown as (b) in
\nvalue
of current.
equivalent
In this
\nthe
particular
figure.
case, there is no need to write equations
network.
the initial values of thecurrents
\nfor
resistor
the
By inspection
=
and t*2(0+) = 0 because the secondloopis open.\n
\nare
V/Ri
ti(O-f-)
the
in solving initial values of derivatives is to write
The
first
step
\non
the
capacitor,
\nIn
loop
terms
or
from
equations
\nintegrodifferential
\nthe
it may
node
of the
basis
Kirchhoff\342\200\231s
employing
laws,
either
give the required quantities moredirectly.
of Fig. 5-6(a), the Kirchhoff voltageequations
as will
network
\nare\n
i
j
R,(\302\273.
i,di + R,(i,
+
\302\253,)
Bit!
-
+ L
U)
^
=
=
V
0\n
(5-5)\n
(5-6)\n
Ait
these
Since
they hold at
in general,
hold
equations
NETWORKS\n
IN
CONDITIONS
INITIAL
5-3\n
the
it are known at t = 0+. Also
value
at t = 0+, since this term is the
\nhas
a known
capacitor
\ncapacitor, which is known to be zero sincethe
\ncircuit.
the
inductor.)\n
\nterms
involving
\nically
solving
only
for
(1/L)
basis,
v dt
\342\200\234
The
as
t
for
\nonly
Neither
5-5,
\nEq.
current
(5-7)\n
=
K!)0]
(t
Z
=
0+)
(8_8)\n
as (general) or (t = 0+) is
equations
against
differentiating
equations that hold
Eq. 5-5 nor
Eq. 5-6 containa
in
holds
which
if
However,
and manipulated
is differentiated
general,
=
dii\n
R^-Ri^
dt\n
+
C\n
ix
for
(5-10)\n
(general)\n
RXC\n
known
are
(5-9)\n
(general)\n
%j\n
dt
dt
0\n
dt\n
di%
d%\\
Both dit/dt and
term.
(dix/dt)
results\n
there
%
t =
so that (dii/dt) may
0+,
be
as\n
\nevaluated
((
it is required
that
Suppose
of
point
practical
of
=
W
5-7,
Eq.
at
(dHt/dt2)
t =
0+.
From
Z [Rl Tt
be applied
in solving for all derivatives.
gives\n
~ {R' +
=
1&^
to evaluate
can
manipulation
\nDifferentiation
(5-11)\n
0+)\n
and higher-order derivatives are less
the first derivative
in the solution of differ\302\254
the procedure
of continued differentiation
However,
equations.
algebraic
=
second-
view,
than
required
\nfrequently
\nand
short
marking
safeguard
i_i\n
\nential
as a
= 0+.\n
\nalgebraically,
\na
the
(general)
~ (Bl +
Z [Sl Ti
of
a
+
(fil
precaution
\nsuggested
acts
across
[***\302\273
\"
(0+)
voltage
gives\n
(dit/dt)
Yt=Z
W
J ix dt
(1/C)
represents
similarly
\302\247
term
the
Now
to
Eq. 5-6 contains a derivativetermin addition
=
ix and ta, which are known at t
0+. Algebra\302\254
that
observe
We
node
the
(On
\nthrough
t = 0-K
and
of ti
\nvalues
89\n
-v{r/lc
+
(general)
t)\n
(t =
0+)\n
<5-12>\n
(5-13)\n
each
In
(network
\nshould
not
and
be
given
NETWORKS\n
5\n
Chap.
have been given in terms
solutions to problems
parameters);
driving
in terms of integral or derivative
conditions
initial
the
case,
\nstants
IN
CONDITIONS
INITIAL
90\n
of
con\302\254
force
expressions.\n
1\n
Example
=
C
and
\nhenry,
in Fig. 5-7,
shown
circuit
the
In
Let it
10 pf.
be\n
=
V
find
to
\nrequired
\nand
dH/dt2(0+).
\nhoff
voltage
ohms, L = 1
di/dt(0+),
t(0+),
Kirch-
the
From
law,\n
=
v
R = 10
10 volts,
+ m +
Lft
vIidl\n
(general)
of Example
Network
5-7.
Pig.
1.\n
for t
values
element
equivalent
of the
because
that
shows
in terms
circuit
the
Analyzing
\342\226\240
\n= 0
(5-14)\n
of
open\n
\ncircuit,\n
=
\302\273(
0+)
last term in
The
\nthe
capacitor,
becomes
5-14
Eq. 5-14, (1/C) J
=
To find
=
i
the second derivative,
+
LU
\nt
0+)\n
the
represents
dt,
Eq.
5-15,
0+;
general expressionin
(t =
+ 0
values
RIt
=
10^
Eq. 5-14must
+ b
=
for the second
\302\260
be
(5-15)\n
(general)
and third
\nthe
at
known
are
terms
-100\n
amp\n
sec2\n
2\n
network
the
switch
find
as\n
differentiated
thus\n
Example
\nto
0+)\n
0+>\n
3?\342\200\231(\302\260+)\342\200\234-ll<0+)
\nthe
Eq.\n
<\342\200\230
=
In
across
voltage
which\n
=
=
i
+ R 0
(0+)
Ljt
f(0+)
In
=
(t
which
is zero at t = 0. The
the following
for t = 0+.\n
V
from
0\n
K open, and
the
initial
value
various
currents
in Figure
shown
at t
and
the
\342\200\224
0
of all three
voltages
5-8, a steady
switch
state is
is closed.
with
reached
Let it be required
loop currents. We must
in the network at t =
find
first
before\n
0\342\200\224,
5-3
the
switch
=
4,(0-)
current
The
is closed.
will
that
Ri;
+
=
0-)\n
Ri\n
(5-16)\n
+
R\\
Rz\n
the
on the
charge
divide
will
capacitors
(7x7c, =
q\\ = q2 or
have
\nseries we
capacitors
must be equal
2.\n
in
connected
when
the
Hence
CzVCt.
of Example
Network
6-8.
Fig.
\nthe
be\n
will
is,\n
=\n
VCt
Since
and L
voltage across the capaci\302\254
be
the
same
as the drop
The total
FCl
flowing through Rz, Rh
Rz\n
-|\342\200\234
(t
\nacross
91\n
=\n
4(0-)
Ri
\ntors
IN NETWORKS
CONDITIONS
INITIAL
Art.
voltage
across
as\n
rc,
Cz
_
_
Si\n
(general)\n
Cl
VCt
(5-17)\n
Sz\n
and\n
y C|
\342\200\224
Ri
I\"
y-
+ Sz]
+ R2 Ui
y
\342\200\231
'
Sz\n
\342\200\224
\t\n
#i
S i
Rz\n
+\n
+
Sz\n
(5-18)\n
4 at t
find
To
(not
\nloop
= 0+, apply
the
on
drawn
V
\342\200\224
\342\200\224
7c,
7c,
that\n
since the current
\nlast
two
4 cannot
shows
equations
\nacross
consider
the
the
4(0+)
(5-20)\n
\nV\n
R i
0-0+)
+
(5-21)\n
Rz\n
Comparing
instantaneously.
=
4(0+)
= 4(0+)
0\n
the
(5-22)\n
in the resistor Rx.
Since the voltage
instantaneously,\n
change
\342\200\224
or\n
0+)
=\n
change
flowing
cannot
4(0+)
=
that\n
current
capacitor
(5-19)\n
+ Rz
Rz\n
4(0+)
Next,
outside
write\n
V
\n(t
+
= 4(0+)
- 4(0+)
4(0+)
Now,
R i
the
Rz\n
R i
=\n
4(0+)
around
this loop, we
=\n
V\n
so
law
voltage
Traversing
diagram).
=
hRz
Kirchhoff\342\200\231s
=
-
(t =
0+)\n
(5-23)\n
0
0+)\n
(5-24)\n
=
IN
CONDITIONS
INITIAL
92\n
5\n
Chap.
NETWORKS\n
that\n
so
V
Si
^2(0+)\n
Ri
Si
R2
\342\200\234h
V\n
V
(t
R 2
\342\200\234I-
\nRi
be
\nto
in
found
Weber\342\200\231s
=
0+)
(5-26)\n
\342\200\234IS2\n
of the method
discussion
(5-25)\n
READING\n
FURTHER
A good
Si
0+)
R2\n
-f-
R\\
S2
=\n
i2(0+)
Finally,
=
01
-f- S2
of evaluatinginitial
Transient
Linear
is
conditions
\nSons, Inc., New York, 1954),pp. 42-45,and
\nTransients
in Linear Systems (John Wiley &
in
and
Gardner
Barnes,
New
Inc.,
Sons,
&
Wiley
(John
Analysis
York,
\n1942), pp. 26-34.\n
PROBLEMS\n
5-1. In the
of
\nvalues
\n1000
\nFind
\nL
=
In
the
circuit
In
the
circuit
at t = 0+,
1 yi. Answers.
0.1,
closed at t =
when
\342\200\224100,
V
=
0. Findthe
100
R =
volts,
100,000.\n
\nhenry,
5-4.
C
=
from
\nchanged
at
\342\200\224
100
t
from
a
position
in
circuit
position
0+,
=
1
\342\200\224100,
\342\200\224
volts.
the switch K
when
10/xf,
the
In
shown,
state
Find
a.
\nposition
is changed
t = 0, having already established a steady
= 0+,
L
i, di/dt, dH/dt2, and dH/dt3 at t
and V = 100 volts. Answers. 0,
0, 107.\n
b at
\nto position
\nV
dH/dt2
K is
switch
at
of the figure, the switch K is closed
t = 0.
= 10
the
values
of i, di/dt,
R
dH/dt2 at t = 0+, when
ohms,
=
1 henry,
and V
100 volts. Answers. 0, 100,\342\200\2241000.\n
5-3.
\ndH/dt2
C =
and
the
shown,
and
di/dt,
i,
ohms,
5-2.
circuit
when
Answers.
of the accompanying figure,
o to position b at t = 0. Solve
\\ henry,
R = 1000 ohms, L \342\200\224
0.1,
\342\200\224100,
0.\n
the
for
i,
C
K
switch
di/dt,
\342\200\224
1
yi,
is
and
and
INITIAL
5\n
Chap.
\nohms,
for
solve
\nt \302\273
0+,
shown, the switch K is
v, dv/dt, and dh)/dt2, when I
circuit
the
In
'6-5.
for
\nohms, and
\tVi
(a)
and
\nR2Vj
circuit
t =
\nK
is
\nlished
t
at
R/).
0.
=
R
At
1000
\342\200\224
0+.
100,
vx and
is
t =
at
opened
= 1 amp,
=
R
0.
100
\342\200\224104, 10\302\256.\n
the switch K
(b)
5-6.\n
is closedat
v2 at
t =
t
\302\253.
=
(c)
0.
Solve
dvi/dt
for:\n
and
(d) dh)2/dt2 at t = 0+. Answers,(a) 0, 0.
dh2/dt2
(c) dvjdt = V/CRU dv2/dt = 0. (d)
(b)
=
R\\LC.\n
\\j\n
5-8.
Answers.
shown,
0+.
+
RiV/iRj.
\n0,
of the figure, the switch K
and
at t = 0+, when 7
dh/dt2
henry.
t>2
at
\ndo2/dt
Prob.
\342\200\224
1
the
In
6-7.
L
=
t
\342\200\2241010.\n
6-5.
dv/dt,
v,
93-\n
opened at
= 10 amp,
circuit
the
In
\nSolve
Answers. 0, 107,
and C = 1 nf.
Prob.
6-6.
NETWORKS\n
IN
CONDITIONS
In
changed
at
shown in the accompanying figure, the
been
a to b at t = 0 (a steady state having
=
a). Show that at t
0+,\n
network
the
from
position
V\n
i\\
1%
\342\200\224
\"p
tl\\
v
-J-
Prob.
n
It2
T
~t~
5-8.\n
p\342\200\235'
rtz\n
^
\302\256\n
switch
estab\302\254
\nt
=
\nt
=
=
R2
\nvolts,
given network,
K is closed at
switch
the
\nand
the
In
6-9.
1 megohm,
C\\ = 10
6-10. In the circuit
0
an
connecting
\nswitch
x
=
C2
open
to
charged
Fo
voltage
2 megohms,
Fo = 1000
20/ttf, solve
for dS^/dt* at
=
amp/sec2.\n
the
the
figure,
F0 sin
voltage,
alternating
K is closed at
switch
t
the
to
cot,
and (b) di2/dtat
network shown, a steady state is
=
with F = 100 volts,
10 ohms,
In the
K
is
C\\
5\n
Chap.
Find (a) dii/dt
\nRL-RC circuit.
6-11.
and
*tf,
in
shown
NETWORKS.\n
the capacitor
=
t
0. When R
1.41 X 10-6
Answer.
0+.
IN
CONDITIONS
INITIAL
94\n
=
parallel
0+.\n
the
with
reached
R2 = 20
R\\
ohms,\n
L = 1 henry, and
R
ohms,
1 /if. At time t = 0, the switch
\n(7 =
the integrodiffer\nis closed,
(a) Write
for the network
\nential
after
equations
\nthe
switch
is closed,
(b) What is the
C before the switch
Fo across
\nvoltage
3 =
20
\nAnswer. 66.7 volts,
of
value
\ninitial
i\\
for
the
\342\200\22483,300.
(e)
\nSolve
\ndi2/dt
\ndii/dt
at
t
at
t =
The
6-12.
\npairs.
\nat
t
If
the
\342\200\224
0+:
0-f.
Answer.
33.3,
values
What
polarity?
(c) Solve
the
for
and
\nAnswer. 3.33 amp,
\342\200\224
its
is
What
closed?
\nis
i2
= 0+).
(t
1.67 amp.
(d)
of dii/dt and
is the value of
oo ?\n
shown
network
switch
(a)
K is
t>i, (b)
v2f
in the figure
opened at t
(c)
dvx/dt,
Prob.
\342\200\224
0,
(d)
has two independent
find
dv2/dt.\n
6-12.\n
node
the
following
quantities
Chap.5\n
INITIAL
In the network
6-13.
NETWORKS\n
shown, the switch K is
95\n
at
closed
0.
t\n
Find\n
t = 0+.\n
at
dvi/dt
IN
CONDITIONS
Vl\n
0\342\200\224W\\r\n
1\342\200\224'CHKT\342\200\2241\n
K\n
L1\n
<
<r2\n
*>2
~cx\n
1\n
)\n
5-13.\n
Prob.
is
\nrium
an
shown in the accompanying
=
is
Find
K
at t
opened.
0, switch
In the network
6-14.
and
reached,
figure,
equilib\302\254
initial\n
the
V -=-\n
5-14.\n
Prob.
voltage across the switch
\nacross the switch.\n
and the initial
of
derivative
time
the
voltage
^3\n
\nthe
the
In
6-16.
t = 0,
instant
Let
\nsource.
network
Mu
=
shown
connecting
0. Determine
the switch K is
at
an unenergized system to a voltage
in the figure,
the values
and
of\n
(0+)\n
fdt\n
^\342\200\230(0+)
closed
Answer.\n
dii
dt
dU
(L\\
/a
K
dt
F(L2
_
/rti'v
1
\\
+
Lz
+ 2M13) (L2 +
V(Lz
__
'ri
(Lx
+ Lz
\t\n
-f- Lz -f- 2M23)
Lz -f- 2Mzz)
+
M\\z
+ 2Ml3)(L2 + Lz +
+
\342\200\224
(Lz
-(- M13
+
M23)2\n
(Lz
+
+
Mzz)*\n
M%3)\n
2M2Z)-
Mxz
INITIAL
96\n
The
6-16.
At
\nitor.
\nv(t)
=
consists
of two coupled
a
closedconnecting
t = 0, the switch K is
Determine
V sin (t/\\/MC).
^(0+),
the values
and
Chap.
coils and a
of
generator
5\n
capac\302\254
voltage,
of\n
^(0+)\n
0, dva/dt(0+) = (V/L) \\/M/C,
Answer. vo(0+) =
d2va/dt2(0+)
0.\n
the
\nafter
Find
network
the
In
6-17.
(a)
IN NETWORKS\n
network
given
t>.(0+),
\n=
CONDITIONS
has
network
an
expression
of the figure, the switch K is opened
at
t = 0
attained
a steady state with the switchclosed.\n
for the voltage across the switch at t = 0+.\n
=
are adjusted such that
1 and di/dt
(0+)
is the value of the derivative of the voltage
the
1, what
Answers,
Ri.\n
(a) VR1/R2, (b) l/C
dvK/dt
(0+)?
\nswitch,
in
In the
network
shown
the figure, the switchK is
6-18.
at
the
with
an
0 connecting
battery
unenergized system, (a) Find
=
at
at
t
across
0+. (b) Find the
voltage
point
a, Va
(b)
\n=
If the
parameters
t\342\200\230(0+)
\342\200\224
across
\342\200\224
closed
\nt \342\200\224
\nthe
voltage
\ncapacitor
t =
C\\ at
Prob.
6-19.
For
5-38.\n
the network
\342\200\2241
(0-f)
dt2
\302\273, Fc,(\302\260\302\260).\n
=
of the figure, show
that\n
-\n
Rx\n
dv{t)
\"I
\n\"l\"
\ndt
\\\n
d2v(t)\\
dt2
)\n
CHAPTER
DIFFERENTIAL
6\n
CONTINUED\n
EQUATIONS,
Differential
limited
to linear
equations studied in Chapter 4 were
In
\nequations of the first order with constant coefficients.
chapter,
\nwe will
continue
our
of
differential
with
the
same
study
equations
\nrestrictions
as to linearity
and constant
coefficients but of higher order.
\nThe
in these two chapters are included
mathematical
procedures
given
\nunder
of the classical
method of solution. As we will see,
the
heading
\nthe
classical
method
affords
a better
insight into the interpretation of
and
of a solution. Aside from
\ndifferential
the requirements
equations
this
the
advantages,
\nconceptual
\ntransformationis better suitedto
more
but
\nmethod
be
will
\nformation
use.
our
differential
homogeneous
differential
second-order
in
written
\nbe
the
general
of this
equation\n
form\n
+
solution
treat\302\254
with constant coefficients may
equation
\342\200\234\302\2603*
The
reason,
Laplace
our
Topics
6-1. Solutionof a second-order
A
this
For
the
using
covered using the classical
ordinarily
with the aid of the Laplacetrans\302\254
easily
developed
for the next chapter.\n
reserved
brief.
be
will
\nment
method
operational
differential
+ a'i =
a'7t
equation
0
(6'1)\n
must be of such
the
that
form
and its secondderivative,
add to zero. To satisfy this
a constant
\ntiplied
by
coefficient,
in their
the
three
terms
must be of the same form,
\nment,
only
Is there such a function? By whatevermethod search,
\ncoefficients.
the search always leads to the
trying
\nperhaps
possible
functions,
\nsolution
itself,
its first derivative,
each
mul\302\254
require\302\254
differing
we
\nexponential*\n
i(t)
where- k and
\ninto
Eq.
6-1
shown
(6-2)\n
the
exponential
solution
gives\n
+ aimkemt +
atkemt
=
0
(6-3)\n
two at a time, the sine and the cosineor the hyperbolic
sine
the
and
cosine satisfy the requirement; however, the exponentialsolutionwill
\nhyperbolic
\nbe
he1\"*
m are constants. Substituting
aom2kemt
* Taken
=
to
simplify
to these
forms.\n
97\n
DIFFERENTIAL
98\n
the
requirement
\nas
the
characteristic
\nroots
for
the
mh m2 =
\nsolution
ii
if
Now,
Zd o\n
\ninto
=
.
d2
(ii
o\302\260
^
.
d2ii
+ ai
w
V\342\200\234\302\260
or
0
two
(6-5)\n
forms of the
exponential
(6-6)\n
the differentialequation
of
Eq.
i\\
ii
+
(6-7)\n
be shown
6-7
of Eq.
substitution
direct
by
giving\n
6-1,
/
4a0a2
i2 = k2ema
and
kiemit
a solution. This may
Eq.
\342\200\224
y/ai1
there are two
iz =
is also
by the
of these solutions,\n
sum
\n6-1, the
\302\261
each solutions of
i2 are
and
ii
^
Zd o
is known
are\n
they
kemt;
~
\342\200\224
that
discovered
have
now
We
It is satisfied
equation.
formula\n
quadratic
(6-4)\n
This equation
auxiliary)
(or
by
given
0
solution.
the
be
to
kemt
Chap. 6\n
t,\n
+ a2 =
aim
+
a0m2
as
finite
be zero for
never
can
kemt
since
or,
CONTINUED\n
EQUATIONS,
+ 0 = 0.
.
+ ii) +
dii
n
d\n
ai ^
+
(t\342\200\231i
/
. \\
+
a2{i\\ + ii)
+
ii)
dH2
w
atH)+
The general solution of the
ii+
\\a\302\260
=
i(t)
kiemit
+
(6-8)\n
. =0
di2 .
+01
= 0
.
_
\\
a,Iv
differential
is
equation
.\n
{M)\n
k2emit
thus\n
(6-10)\n
coefficients
in Eq. 6-1 determines the form
of the characteristic
\nof
the
roots
equation. In Eq. 6-5, the radical
\342\200\224
be
4a<ja2
\n\302\261
may
real,
y/ai2
zero, or imaginary
depending on the
with
of
\nvalue
The
4aoa2.
forms
of
the solutionsforthese
ai2
compared
will
be given
three
cases
\nthree
by
simple examples.\n
The
of
magnitude
the
1\n
Example
The
\nrent
in
differential
equation
the
of Fig.
circuit
law
\nKirchhoff\342\200\231s
Kg. 6-1.
for
Circuit
\n1
and
and
Differentiating
\nFig.
6-1
+
Examples
Ljt
3.\n
using
numerical
values
for the
cur\302\254
6-1 is givenby
as\n
R*
+
fifidt
for R, L,
= v
and C
(6-11)\n
in
shown
gives\n
dH
. 0 di
,\n
(6-12)\n
Art.
DIFFERENTIAL
6-1\n
characteristic
The
\nand m
for
emt or
(di/dt);
solution
\neral
+
\nat
t
=
\nneously
\nterms
\342\200\224
1
=
fcx
the
that
so
\342\200\2242,
gen\302\254
(6-14)\n
for a specific prob\302\254
be evaluated
k2 can
and
=
Jt
initial
two
gives
the
]r
zero becauseit is
=1
+
&2
=
i(t)
of the separate
in
Pig.
6-2.
voltage
into the general
solution, Eq.
equations,\n
\342\200\224
2*2
\342\200\224ki
0,
The solution of these equations is
to Eq. 6-12 is\n
\nparticular solution
discussed
initial
the
amp/sec\n
substituted
conditions,
&i
\nAs
=
m2
+ ktf-*
fcie-*
f i dt being
Hence\n
0 and
(l/C)
\nacross the capacitor.
A plot
and
(6-13)\n
initial conditions. If the switchK is closed
=
t(0+)
0, then
0, because current cannot changeinstanta\302\254
in the
inductor.
In Eq. 6-11, the second and third voltage
are
at the instant
of switching, 7fo'(0+) being zero because
zero
=
\n6-14,
0
of the
a knowledge
\ni(0+)
The
(dH/dt2),
is\n
The arbitrary constants
by
+ 2 =
3m
mx =
roots
the
i(t)
\nlem
for
m2
trial
thus\n
has
equation
of substituting
by the equivalent
m2
This
99\n
by substituting the
be found
can
equation
=
i
\nsolution
CONTINUED\n
EQUATIONS,
Chapter
=
ki
e~*
=
+1
and
=
1
(6-15)\n
k2 =
\342\200\224
1;
\342\200\224
e~n
(6-16)\n
is
in
parts and their combination
4, the total current may be thought
Current as a
the
hence
shown
of
functionof time for the circuitof
Example
Fig.
6-2.
made\n
as
1.\n
DIFFERENTIAL
100
two
of
up
\nway
Example
The
basis
node
the
on
shown in Fig.
network
the
for
equation
a
such
6-3
for\302\254
is\n
+
Gv
+
CJt
vdt = 1
lJ
(6-17)\n
differentiation,\n
by
or,
0 and combinein
2\n
equilibrium
\nmulated
Chap. 6\n
the initial conditions.\n
to satisfy
as
exist from t =
which
components
CONTINUED
EQUATIONS,
<S-i8>\n
cS+4<+z=\302\260
as given in
this equation
into
values
numerical
Substituting
Fig.
6-3\n
va\n
for Example
Circuit
6-3.
Tig.
2.\n
gives\n
+
2&
The
characteristic
corresponding
the
into
\nstituting
mi =
roots
as
has
which
\nthe
\ntain
\nin
\nnew
=
K
where
+
two
arbitrary
manner
solution
to
to
differential
successive
or
\342\200\2242,
roots.
repeated
gives\n
solution
The
=
v
of repeated
for the condition
be v = ye~2t, where y is a factor
the
at
arrive
we
(6-21)\n
form of the solution,since
differential equation must con\302\254
a second-order
6-19,
Sub\302\254
a complete
is not
roots. If
to
be modified
must
ke~2t
we
assume
the
and
y satisfy
be determined,
requirement
that
equation\n
\"
S
Two
=
(6-20)\n
solution, Eq. 6-10,
/c2e-21 = Ke~2t
+
kxe~2t
constants.
\nsubstitute into Eq.
\nthe
=
This
fc2.
solution
general
some
ki
m2
0
form of the
general
v(t)
and
^\n
is\n
+ 8 =
8m
\342\200\224
2
*-\302\260
equation
+
2m2
+
8i
integrations
of the
y
-
0\n
equation
ki +
kit\n
lead to the
solution\n
(6-23)\n
Art.
6-1\n
DIFFERENTIAL
to our problem with repeatedroots
the solution
Thus
101\n
CONTINUED\n
EQUATIONS,
becomes\n
=
v(t)
(6-24)\n
solution for this problem require
knowledge
two
initial
From
conditions.
the network of Fig. 6-3,
must
since the capacitor
acts as a short circuitat
initial
\nequal
zero,
In Eq. 6-17, the second and third terms are equalto
\ninstant.
the
=
\nfirst because
0 and the second becausethere is
current
t>(0+)
=
at the initial instant.
inductor
the
Then, by Eq. 6-17,
(0+)
for this
network.
these initial
volt/sec
\n1/C
\\
Substituting
=
=
6-24 leads to the result that
\ntions
into
The
0 and
Eq.
a particular
To obtain
will
\nof
v(0-\\~)
the
zero,
not
\nin
dv/dt
\342\200\224
condi\302\254
kx
solution
\ndesired particular
is\n
v(t)
6-4.
(6-25)\n
function of time for
as a
Voltage
the circuitof
2.
Example
3\n
\nExample
this
For
use the network of Fig. 6-1with
values:
V = 1 volt, L = 1 henry,R
we will
example,
network
parameter
C =
farad.
The characteristic equation
m2
+ 2m + 2 = 0
the
follow\302\254
\342\200\224
2
ohms,
becomes\n
(6-26)\n
roots*\n
with
mh
The
This
=
We
letter
\ntextbooks
will
equivalent
the letter
j in textbooks
of mathematics
use
values for m,
+
e-\342\200\230(M\342\200\230
becomes\n
k2e~*)\n
for
of the solution is not convenient
interpreta\302\254
with
Euler\342\200\231s
form may be found starting
equation,\n
form
e\302\261i'4=
*
\342\200\224
1 \302\261
jl\n
=
particular
An
\ntion.
m2 =
6-10, with these
Eq.
solution,
general
i(t)
\nThe
$te~2t
6-4.\n
Fig.
\nand
=
this solution is shownin Fig.
A plot of
\ning
k%
cos
t +
j sin t
the
letter
j for the operator\\A-1 to reserve
of electrical engineering is equivalentto
and
physics.\n
(6-27)\n
i for
i
current.
\342\200\224
\\/
in
\342\200\2241
DIFFERENTIAL
102
=
i(t)
where
=
kt
ki
to
equal
kt
=
conditions
initial
The
kt).
t)\n
= 0 and di/dt (0+)
=
1
are the
Sub\302\254
amp/sec.
have\n
^4 sin 0)
+
0
cos
e~\302\260(ki
=
(0+)
1. The
0
cos
fc\302\253(e\342\200\234\302\260
particular
i(t)
of the two
A plot
ki sin
=
kt\n
zero,\n
^
whence
0 =
=
t(0+)
\342\200\224
j(kt
1: t(0+)
Example
into
the solution,
we
\nstituting
With
=
and
fcj
form\n
cos l -f-
e~*(ki
in
as
\nsame
+
ki
to the
solution
our
reduces
which
Chap.6\n
CONTINUED
EQUATIONS,
\342\200\224
solution
\342\200\224
e~\342\200\230
sin
sin
0
c~\302\260)
=
1\n
is\n
t
(6-28)\n
factors in thissolutionand
their
product
is shown
6-5.\n
\nFig.
Fig.
6-6.
Current
as a
(a)
standard
The
6-2.
the
Consider
\ncircuit
on
the
Form of
KirchhofT
loop
basis\n
function of time for the circuitof Example
(b)
e~\342\200\230;
sin
l; (c)
e~( sin
the solution of second-orderdifferential
voltage
equation
3:\n
t.\n
that describes a
equations\n
series
RLC
in
Ait
6-2\n
DIFFERENTIAL
+ Ri +
Ljt
If
v(t)
\nto
a
or a
zero
either
is
dH
.
+
di
dT*+Ldt
\nfound
of the
roots
two
The
the
by
-
as
vanish
\nto
critical
the
=
LCi
A\n
\302\260\n
(6-30)\n
equation may be
be\n
i
(6\"31>\n
\302\261/(\302\243)-\302\243
standard form,
will
define
the
radical term in the above equation
Rcr. This value is found
solving
the
causes
that
resistance
of
\nvalue
.
6-30 to a
Eq.
converting
begin
to
formula
quadratic
1
thus\n
characteristic
corresponding
=
To
order;
(6-29)\n
reduces this equation
differentiation
R
.
= v(t)
idt
J
of second
equation
homogeneous
\302\261
constant,
103\n
CONTINUED\n
EQUATIONS,
resistance,
we
by
\nthe equation\n
fey
w\n
=
Rcr
or
We will next
2
Vl/C
(6-33)\n
introduce two definitions;
we
define
f =
dimensionless
the
as
The
\nzeta.)
damping
of
value
\ncritical
damping
ratio
is
quantity\n
t\302\256-34)\n
ratio,
the
The
resistance.
the
(f is the
lower-case Greek letter
ratio of the actual
other definition is\n
resistanceto
the
=
(6-35>\n
The
for
\nreason
3
Case
\nof
of
\nsions
\nit
con is
quantity
as
a
Now
such
o>\342\200\236
giving
in
this
are
o)\342\200\236
section.
(time)-1
natural angular frequency.The
will be discussed under the heading
a name
For the time being, we note that the
it does not seem unreasonable to define
so that
the
undamped
dimen\302\254
frequency.\n
the
has the
product
value\n
R\n
2fa>\342\200\236\n
2
\342\200\224\n
2
\\L
(6-36)\n
y/LC\n
L\n
1\n
and\n
(6-37)\n
to*\n
LC\n
DIFFERENTIAL EQUATIONS, CONTINUED
104
these
Substituting
\nard
The
form.
characteristic
m2
Before
varies
\n\302\243
=
R
\nto
If
this
from
zero
oo).
There
1: f
> 1, the
Case
=
\nwill
\nfor
\302\243
roots are
1, the
roots
<
\302\243
1, the
roots
conjugates
a locus
real.\n
m2
with,
(6-42)\n
\302\261j\302\253\342\200\236
=
m2
\342\200\224
\302\261
\302\243o>\302\273
j(>>n
are complex numbers
parts
imaginary
of m
<
\302\243
1, the
roots
are
com\302\254
for
\\/l
<
\302\243
\342\200\224
1,
=
<r
let
us define
\342\226\240+\342\200\242
jo)
Greek letter sigma). The roots of the equation
real
the complex
m plane as shown in Fig. 6-6. The
lower-case
in
(6-43)\n
f2
imaginary
part
the real
as\n
<r
the
=
imaginary. For
m
and
we
\302\260ot
as\n
roots
the
plotted
to R =
roots for a variation of \302\243
from
0 to
of roots in the complex plane. To start
mi,
\nbe
(6-41)\n
are real and repeated.\n
are complex and conjugates.\n
roots are purely
\nplex
the
written\n
0,\n
is, the
(<r is
(6-40)\n
of the
form
the
that
\nand
1
roots:\n
mi,
Since
\342\200\224
0) to infinity(corresponding
evidently three different forms for the
\302\243
recognize
=
equation
are
2:
follow
(6-39)\n
let us examine the behaviorof the
as the dimensionless damping ratio
solution,
(corresponding
Case
we
0
+ K2e[ \342\200\234f\"\"-\342\200\234-Vcr
characteristic
Case 3:
=
\302\261
w\342\200\236
-\302\243co\342\200\236
\\/f2
=
simplifying
the
of
=
m2
may now be
solution
i
\nroots
wb!
is\n
are\n
mi,
general
-f-
2\302\243u}nm
equation
stand\302\254
of the characteristicequation
and the roots
The
-|-
C5-3*)\n
equation is called the
differential
corresponding
gives\n
= \302\260
\342\200\242\342\226\240\342\200\231*
2f\342\200\234\"
second-order
of the
form
This
| +
+
%
Eq. 6-30
into
relationships
6\n
Chap.
=
\342\200\224
\302\243w\302\273
(6-44)\n
may
part
(6-45)\n
is\n
a)\n
Vl
\302\261\302\253\302\273
~
\302\2432\n
(6-46)\n
is\n
DIFFERENTIAL
6-2
Art.
other
In
two roots have the same real part, and the
by their sign. Since\n
only
differ
<T2
that
it follows
\nof
radius
wn
\nThis
locus
\nerty
of
the
W2 =
+
r2Wn2 _|_ Wn2(!
_ f2)
the locus of the roots in the complex
this locus is formed by
and
that
geometry
\ninverse
from the
measured
root
either
to
=
inverse tangent
\nthe
term
is
co\342\200\236
\nthe
angle
is\n
the
\ndently,
f
circle
to 1.
is a
plane
0
from
varying
Locus
\342\200\224
of roots
with
f.\n
of the
in terms
is given
<r axis
tan-
=
cos\"1
real
Since
part.
and the denominator,
f*
V'1~
(6-49)\n
f and y/1
has unit value,
hypotenuse
the
over
the numerator
both
sides
the
dmi
(<M8)\n
of the imaginarypart
to
common
having
triangle
m
(6-47)\n
tan-1
=
A
Wn2
as\n
tangent
or the
\302\253
6-7.
Fig.
angle
imag\302\254
6-7. It is interesting to noteanother
prop\302\254
m plane for these second-orderroots. The\n
in Fig.
of the
shown
is
105\n
the
words,
parts
\ninary
CONTINUED
EQUATIONS,
\342\200\224
f2
is
shown
in
Fig.
6-8.
Evi\302\254
and\n
(6-50)\n
f\n
'A-f2\n
\nare
of
lines
\ninary
\nhave
constant
ratio
\ndamping
part
the
the origin of the m plane
When the
f as con is varied.
=
value
the
f
1, the imag\302\254
from
lines
radial
Thus
of
same
has
the
roots
vanishes,
two
Fig.
6-8.
Right
triangle
\nrelationships.\n
value,\n
(mi,
These
and the roots
superimposed
values
m2) =
(6-51)\n
-\302\253\302\273\n
are shown in
Fig. 6-7. For f
> 1,
the\n
DIFFERENTIAL
106\n
Chap.
6\n
become\n
roots
two
CONTINUED\n
EQUATIONS,
= ( \342\200\224f\302\261Vi\"2
mh m2
-
1) \"n
(6-52)\n
of these roots are real. As f becomes
1 is small
such
that
large
\342\200\224
with
\ncompared
f2, the rootsapproach 2fton and 0. The locus of roots
in
\nshown
6-7 illustrates
this separation
of the real roots as f
Fig.
Both
\nincreases.\n
\nof
f is determined by circuit parameters.With
served
as an example in deriving the relation\302\254
RLC
circuit
that
of
the resistance R will vary the rootsin
a simple
adjustment
for
the
m
three
cases. This general solution,Eq.\n
plane
complex
forms for each of the three cases.We
reduces
to different
will
this algebraic
reduction
for each of the three cases.\n
investigate
The
f.
\nthe
\nships,
\nthe
6-41,
\nnext
Case
in
and
K\\
K-i
=
The
constants
arbitrary
to evaluate
convenient
the
x
of x is
sine
hyperbolic
equivalent
these
\nsubtracting
i =
equations;
or
i =
=
=
or
is,\n
x
cosh
by successivelyadding
+ cosh x
x
sinh
(6-55)\n
e~*)
\342\200\224
(6-56)\n
x
sinh
(6-57)\n
may be used to convertEq.
6-53
terms
to
involv\302\254
thus\n
functions;
+
\342\200\224
\302\243(e*
that
two identities
e-r\"\"\342\200\230{IiLi[cosh
=
(6-54)\n
as\n
two
e-*
hyperbolic
+ e~x)
\\{ex
can be obtained
and
\ning
results\n
as\n
relationship
6*
These
=
defined
x
sinh
An
there
equation,
of integration. This equation
in terms of hyperbolicfunc\302\254
is defined
of x
cosine
hyperbolic
cosh
and
this
from
factored
is
e~i<ant
is as
form
exponential
e-^C(6-53)\n
are
more
sometimes
\ntions.
If
6-41.
Eq.
case, the solutionin
1. For this
>
f
i
where
ratio
damping
1,
\ngiven
\nis
for the values
the three possibilities
illustrated
has
discussion
This
\\/f2
(o)n
Kt[cosh
~
(\302\253. V?2
e~iant[Kz cosh
y/t2
(co\342\200\236
1 t)
~
+ sinh
- sinh
11)
\342\200\224
1
t)
+
\342\200\224
1
0]\n
\\/f2
(co\342\200\236
\342\200\234
(\302\253nVf2
Ka sinh
1
(6-58)\n
Oil
\342\200\224
(w\302\253y/t*
12)]\n
(6-59)\n
where
Kz
= Ki
and
Ka
=
This
equation
is the
equivalent
+ K.2
- K2
of Eq. 6-53.
(6-60)\n
(6-61)\n
Each has
two
arbitrary\n
DIFFERENTIAL
6-2
Art.
which are usually evaluatedto
constants,
initial
the
of
\nterms
\nbecome
identical.
\nis
by
For this
With
6-24,
Eq.
of
limit
this
If
\nrule.
in
solution
particular
of the equation
solution
the
roots,
two roots
the
that
shown
have
we
case,
repeated
giving\n
i
The
a
find
107\n
conditions.\n
= 1.
Case 2, f
given
CONTINUED
EQUATIONS,
-
(Kt + K2t)e^'1
the
te~Unt
quantity
is written
quantity
be
may
(6-62)\n
by
investigated
1*
Hospital\342\200\231s
as\n
t\n
(6-63)\n
\302\243+\302\253\342\200\242*\n
of
differentiation
with respect to t
denominator
and
numerator
shows
\nthat\n
Case
1.
Case
For
i
3, the roots
in terms
Euler\342\200\231s
e-fcos
which
again,
are,
be
written
~
Vl
constants
form by
=
K
Kt = K
the
y)
=
\nsinusoid
\nsame
3 reduces
VT^T*
(\302\253\342\200\236
K< = j(K1 -
K2)
to\n
i)]
(6-67)\n
(6-68)\n
This equation
of integration.
defining\n
sin 0
(6-69)\n
cos 0
(6-70)\n
sin
x cos
y + sin
y
cos
x
(6-71)\n
6-67 becomes\n
i = Ke~sin
These
(6-66)\n
identity\n
trigonometric
sin (x +
Eq.
quantities\n
x
sin
r21) + K< sin
K6
Using
\302\261
j
+ K2 and
arbitrary
in different
cosine
and
sine
for Case
solution
(\302\253n
Kh =
where
\nmay
the
equation,
x
of
6-27.\n
Eq.
= cos
=
become complex,and Eq. 6-41
(6-65)\n
equation may be written
by making use of Euler\342\200\231s equation,
i
(6-64)\n
=
This
Using
0
00\n
written\n
be
\nmay
<
\302\243
8,
=
te~Unt
lim
<->
algebraic
manipulations
to
equivalent
frequency.
In
the
(wn
~
f21
+
<t>)
(6-72)\n
equation of one
sinusoids of the
the two arbitrary constants
in an
resulted
which contains two
have
6-67,
revised
form,
Eq.
\\/l
are\n
DIFFERENTIAL
108\n
6-70.
\nand
the
Summing
=
\nand
6-69
Eq.
+
VK62
6\n
6-69
of Eqs.
gives the
relationship\n
1C\302\2532
(6-73)\n
equation for
6-70 gives an
by Eq.
Chap.
means
by
Ka
of K6 and
squares
K
Dividing
to Kt and
may be related
0, which
and
K
CONTINUED\n
EQUATIONS,
<f>
in
of K6
terms
:\n
K<
=
0
~
tan-1
(6-74)\n
As\n
\nThe
appeared in these solutions are givennames.
the name damping factor. The product
e~*Unt is given
decrement
factor or attenuation factor. The three cases
expression
the
is
called
\n$\342\226\240&>\342\200\236
\nare
the
given
Case
Case
the
names,\n
1. the overdamped case\n
2. the critically
damped case\n
3. the underdamped
(or oscillatory)
Case
In
have
that
terms
Several
sin
\342\200\224
where
w\342\200\236
-\\A
f2
\ning
R
no
to
\342\200\224 or
Of
to
\nreduces
f2
f21)\n
f = 0,
When
term in the
radical
correspond\302\254
angular
frequency
basis, the following definitions are
=
actual
=
(on
\342\200\224
frequency.
the
damping,
\342\200\224
y/\\
(\302\253\342\200\236
angular
On this
unity.
\\/l
o)n
an
is
of the following form appear:\n
expressions
case,
underdamped
case\n
the
frequency\n
angular
natural
undamped
made:\n
To illustrate these definitions,let us
to
return
is
which
frequency\n
angular
the
series
described
RLC
by
the
of
Fig.
circuit\n
original
\ndifferential equation of this section.
the
\nLet
capacitor
\nchargedto a voltage
\nt
=
RLC
6-9.
\nrespectto the
circuit.\n
\nbe
reduced
\nFor
the
\nThe
in
possibilities
With
term
damped,
critically
overdamped,
\nthree
R >
or
circuit
(1/C)
K be closed.
resistance
R with
the
critical resistance
Rer
the
whether
system
Consider these
underdamped.
turn.\n
is overdamped.The
solution
can
solution
for a given set of initial conditions.
Rer, the system
to
V0,
6-9 be
at time
switch
determine
\nwill
\nis
of
value
\nThe
Fig.
the
let
0
and
particular
in
shown
j idt
general
Fig. 6-9,
=
\342\200\224
Vo
t\342\200\230(0+)
at
=
t = 0
0 because
of the
inductance.
(the initial voltage
on
the\n
Art.
DIFFERENTIAL
6-2\n
such
capacitor)
109\n
CONTINUED\n
EQUATIONS,
that\n
=
+T\n
|'(\302\260+)
\nzero
that
value;
0 at t = 0,
that i(t) =
The requirement
=
cosh
- 11)
KVC2
sinh
+
\nbolic
term
sine
term
cosine
hyperbolic
approaches
r
\nA
=
i
co\342\200\236L
V
The
shape
general
in
\nshown
Fig.
- 1
(6-76)\n
\342\200\224>
0,
and
the
hyper\302\254
-
1
(6-77)\n
1\302\260\n
Vo
,\n
(6-78)\n
-
Vf2
1\n
case thus
overdamped
Vf2
(o>\342\200\236
becomes\n
\342\200\224
1
(6-79)\n
0
1\n
current
against time curve for this equationis
6-10(a).\n
6-10. Network response
Fig.
(a)
of the
as\n
VF^l\n
sinh
~
r
o>.
Vf -
for the
solution
particular
condition
\342\200\242
=
4
(6-75)\n
initial
approaches
unity as t
zero as t \342\200\224\302\273
0.
Hence\n
uJL
The
1t
second
(w, Vf!
=
| (0+)
and\n
-
o>\302\253
Vf2
the
from
jt
The
sinh
Kte-!\"'*
can be evaluated
=
6-59 has
Eq.
is,\n
i
Constant
K% in
that
means
(b)
overdamped;
critically
correspondingto the three
cases:\n
and (c) underdamped
damped;
or
\noscillatory.\n
For
the
critically
case, R =
damped
*
subject to the
=
(Ki
same initial
conditions
Rcrand the solutionin
general
+ Kit)e-ant
as
the
is\n
(6-80)\n
overdamped
case.
The\n
DIFFERENTIAL EQUATIONS,
110\n
is differentiated
6-80
Eq.
\ncondition,
reduce
not
does
\nequation
that
implies
to zero at
condition
current
initial
=
Chap.
6\n
since otherwisethis
To apply the
K\\ = 0,
t = 0.
derivative
as\n
Kt[Ur\"'*{-
ft
CONTINUED\n
+
o>\342\200\236)
e\342\200\224\342\200\2301]
(6-81)\n
Hence\n
\"
\342\200\234
S(0+)
the
and
damped case
for the critically
solution
particular
T\n
is\n
i\n
of Fig.
\nas that
For the underdamped,
\nis
given
by
\nK& be
zero,
Eq. 6-67.
so that
the
of
\native
=
=
current;
sin
K6e~iUnt
cos
fan
\nsuch that
=
(0+)
V*
=
i
-
vi
ujj
\\/l
K%wn
solution for the
particular
the solution
and
the currentrequires
that
(6-84)\n
initial condition
the
of
deriv\302\254
vT^f\"21)\n
(a>n
\342\200\224
The
appearance
\342\200\224
\302\243H)
-y/l
(o>\342\200\236
by using the
Rcr,
same
thus\n
K6e-^Wn
jt
<
R
evaluated
K6 is
constant
or oscillatorycase,
The initial condition for
the solution can be written\n
i
The
and has muchthe
in Fig. 6-10(b)
6-10(a).\n
is shown
curve
This
(6-83)\n
sin
\342\200\224
T2
(coft \\/l
T2
(6-85)
<)]
=
oscillatory case
e-f\"-4 sin
\342\200\224
(6-86)\n
is\n
-
(\302\253*VI
f21)
(6-87)\n
r2\n
in
with time for the oscillatory caseis shown
is the product of the dampingfactor
Since
the current
\nFig.
6-10(c).
or
\nand
the
oscillatory
term, the damping factor represents an envelope
factor
determines
The attenuation
curve
for the oscillations.
\nboundary
of current
variation
The
\nhow
the
rapidly
become
\noscillations
undamped,
The physical meaning
\npreted
in
element
\nstorage
\nAfter
terms
the
switch
are
oscillations
of an
(C)
is
damped.
and
sustained
As R approaches
oscillations
of this mathematical
result
might
zero, the
result.\n
be
inter\302\254
of energy between the electricenergy
and
the magnetic
energy storage element (L).
the energy which is stored in the electric\n
closed,
interchange
DIFFERENTIAL
6-3\n
Art.
to the inductoras
field is transferred
\nrent
begins
\nfrom
the
\nrent
will
\nent
(and
\nto
current
and the total
with
each
all the energy will be
cycle.
Eventually
current
will be reduced to zero. If a scheme
can
the energy that is lost in each cycle,
oscillations
This is accomplished in the vacuum-tubeoscillator
decrease
the
and
\ndissipated
to
be
supply
the
sustained.
\ncan be
the
indefinitely.
However, if there is resistance pres\302\254
there
is in any practical circuit) the
flow
always
will cause energy to be dissipated,
the
resistor
will
\ndevised
cur\302\254
sustained
be
\nthrough
\nenergy
the
When
energy.
magnetic
energy
This
decrease,
field.
magnetic
If
111\n
is being returned to the electricfield
continues as long as any
interchange
resistance
has zero value, the oscillatory cur\302\254
to
remains.
\nenergy
CONTINUED\n
EQUATIONS,
audio
produce
or radio
frequency
frequency
power.\n
6-3. Higher-order homogeneousdifferential
The
method
of solution
discussed
for first- and second-order
\nential
be followed in the solution of higher-order
equations
may
\ntions.
For
an nth
order differential
equation, the characteristic
equations\n
differ\302\254
equa\302\254
equa\302\254
will
\ntion
be\n
... +
aimn_1 +
+
amn
+
an-im
roots.
n
\342\200\224
\ntion.
Thus,
\ntions
is
of
the
\342\200\224
=
wi\342\200\236)
form
kiemit
(6-88)\n
a
(6-88).\n
0
in the
higher-order
homogeneous
matter
of finding the roots
n
order
of
(6-89)\n
The
solution.
of the differential
the solution
constitutes
of
solution
primarily
mi).
to a factor
factors
such
all
.. (m
\342\200\224
(m
raj)
root gives rise
of
\nsum
0
equation
a0(m
Each
=
theorem of algebra states that an
These
roots can be found by factoring Eq.
A fundamental
\nhas
on
differential
equa\302\254
equa\302\254
of the characteristic
\nequation.\n
coefficients
the
\nbecause
because
follows
\nThis
\nreal
are
There
roots,
\ninary
way
only
(the
and
L,
R,
\nparameters,
three
and
in finding these roots
simplification
of Eq. 6-88 are 'positive and real coefficients.
these
coefficients
are made up of the system
C. And since R, L, and C mustbe positive
and
is
there
Fortunately,
some
appear in nature), so must the a-coefficients.\n
forms for the roots: (1) real roots,(2) imag\302\254
possible
roots. For the first-order characteristic
(3) complex
they
\nequation\n
am
the root is m
\nare
always
=
positive
\342\200\224
di/a0,
and
is negative
which
real.
am2
ax =
+
P or a
+
0
(6-90)\n
and real because
do and a}
second-order characteristic
axm
+ ai =
0\n
equation\n
(6-91)\n
112\n
DIFFERENTIAL
roots
the
CONTINUED\n
EQUATIONS,
\342\200\224
Pi\n
m2\n
the
2^
complex.
this
coeffi\302\254
Thus,
they
in
occur
\nmust
or
imaginary,
forms\342\200\224real,
they occur in conjugate pairs, since
roots can combine to give positivereal
way complex
for
characteristic
roots to be complex,
equation
only
\ncients.
(6-92)\n
are complex,
roots
the
if
the
\nis
any
4aoa2\n
on the a-coefficients,these roots
positive
have
\nBut
real restrictions
of the three possible
~
\\/pi2
\302\261\n
2ao\n
\nmay
6\n
are\n
Wli,
With
Chap.
pairs.\n
conjugate
next a third-order characteristic equation. In this case,
of the
rule just given for complex roots, at least oneroot
must
The
other
two may be both real or a conjugatepairof complex
For
a fourth-order
characteristic
equation, there are morepos\302\254
four
real
two real roots and a conjugate complex
roots,
pair,
sets
of conjugate
complex roots. The general pattern is thus
rules may be given:\n
and
the
following
Consider
\nbecause
real.
\nbe
\nroots.*
\nsibilities:
two
\nor
\nestablished
occur
in conjugate
they
(1) If the roots complex,
pairs.\n
at least one root is
(2) If the characteristicequationis odd order,
be
real
or
occur
in conjugate
The
roots
may
remaining
are
of
\nreal.
pairs.\n
\ncomplex
(3)
occur
or
\nreal
this
Summarizing
its
\ninto
equation is of
in conjugate
complex
characteristic
the
If
\ndifferential
have
which
\ntions
the
as
equation
pairs.\n
of any order can befactored
the solution of the homogeneous
of first-order (or second-order)
determine
sum
solu\302\254
considered.\n
been
already
An example will illustrate the methodof solution
of higher
\nhomogeneous differential equations. The differential equation\n
dH
dH
\342\200\236
.
+
S?
6^
a characteristic
has
(m
,
+
\342\200\236
_ dH
+
l)(m
.
dH
.
+ 28S? +
17d?
equation
root,
the
\nUsing
\342\200\236. \342\200\236\n
=
0
-
(Ki
see
+ 1)(m
as\n
+ 2)(m2+
+
2m
=0
4)
one
+ KiQe-* +
derived
already
the solution
is\n
and
f =
0.5.
and second-order
for first-order
Kie-v + e-*(Kisin V31 +
(6-94)\n
nonrepeated
2
<an
that
(6-93)\n
which may be factored
there are two repeated real roots,
=
one conjugate
complex pair with
equations
we
\nsystems,
*
and
order
.
+
24S 8,
^. di
In this equation,
\nreal
be
may
an equation
discussion,
the roots
and
roots,
even order, the roots
Kt
cos
V31)\n
(6-95)\n
* In
\nroots.\n
this discussion,imaginary roots are
considered
as
a special
case of
complex
DIFFERENTIAL
6-4
Art.
\nthe
equations\n
the
nonhomogeneous
equation
is
\nferential
equation
a simple
by
+6
This equation has as roots
\nm2
the
\342\200\2243.Thus
ic
=
Suppose that some function
\nthe
a
also
add
\nwords,
part
to
the
to
analogy
\nby
the
solution
the
\nis
nothing
of
The
\njunction.
\noperations
\nintegral.
=
v(t)
0
\342\200\224
2
we
and
is\n
(6-97)\n
will
find, satisfies
presently
given above
since
either kxe~u or k2e~3tinto Eq. 6-96
substituting
to the right-hand
side of the equation. In other
a
solution
of
nonhomogeneous differential equation
the homogeneous
differential equation. That part,
discussion
in Art. 4-3, is termed the complementary
solution,
\nwould
mx =
equation,
for the case
which
Eq.
equation,
nonhomogeneous
\nis
(e-96)\n
+ k2e~3t
kxe~u
iP}
equation\n
=
solution
complete
consider the
characteristic
its
of
example,
+6f
a
a\302\243
=
of
but
zero,
illustrate
To
\nequation.
the right-hand side
equation,
equal to the forcing function or some
forcing
v(t). In studying such equations, we
function,
the solution
to the corresponding homogeneous dif\302\254
is a part
of the solution of the nonhomogeneous
that
observe
\nfirst
differential
not
the
of
\nderivative
113\n
nonhomogeneousdifferential
6-4. Solution of
In
CONTINUED
EQUATIONS,
of the
part
remaining
iP plus ic
Then
6-96.
the
make
to
solution\342\200\224needed
of the differential equation add to v(t)\342\200\224is the particular
of the
Thus we write the total solution as the sum
of two
parts
\nsolution\n
%
Since we can
6-5.
The
particular
the
analysis
In
\na
forms
\nematical
\nterms
\nthe
\nforms
\nis
sin at, kt,
are
the
trial
of
constant),
driving
by only a
or
products
force.
few
As
math\302\254
of these
to give
our
use.\n
of undetermined
method
functions
e~at,
the
differential
square waves, pulses, etc.). We
of such functions as
physical
generators
mathematical
methods
are available for deter\302\254
If only driving forces of the practical
integral.
the method of undetermined coefficients
considered,
to
suited
Ordinarily,
derivative
V (a
particular
mentioned
iP.\n
v(t)
are represented
Several
section,
coefficients\n
forces
ordinarily
the
integral
driving
encounter
particularly
\nselecting
like
last
the
in
discussed
of undetermined
in
the
circuits, the term
force or a
combinations
tangent.
\nmining
of electric
linear
(or
not
\ndo
(6-98)\n
by the method
integral
matter,
practical
ic
only the particular
found
be
is the driving
\nequation
ip -f-
find ic for any equation,as
to
remains
\nthere
=
of
all
possible
coefficients is applied
forms that might satisfy
by
the\n
DIFFERENTIAL
114\n
differential
Chap.
is assigned an
function
trial
Each
equation.
CONTINUED\n
EQUATIONS,
6\n
undetermined
trial functions is substituted into the differ\302\254
\nential
a set of linear algebraic equations is formedby
and
equation,
coefficients
of like
functions
in the equation resulting from
\nequating
\nthis
The
undetermined
substitution.
coefficients
are thus determined
If
solution
of
this
set
of
trial
function
is not a solu\302\254
\nby
equations.
any
It is not
of
\nforms
necessary to study rules
for
is given in
is suggested:\n
functions
trial
procedure
\nfollowing
the
\nof
\ngiven
\nof
in
V
2.
aitn\n
3.
a2ert\n
4.
CL3
cos
5.
a4
sin
6.
astnert
cos
7.
a6tneTt
sin\n
choice
\nparticular
constant)\n
(a
part
The rules
functions haveterms
6-1\n
Necessary
\nv(t)*\n
1.
the form of v(t).
with
TABLE
Factor
each
Compare
form.\n
mathematical
same
the
the
table,
modified if these two
6-1 are
Table
required
Table 6-1. In usingthis
function
complementary
in
are considering. The
complementaryfunctionic.
(1) Determine the
for the
functions
trial
selecting
v(t) we
function
force
driving
the
of
\nform
be zero.\n
will
coefficient
its
\ntion,
of the
sum
The
\ncoefficient.
the
for
integralf\n
A\n
+ ...
+ B\\tn~l
B0tn
+
-f-
Bn\n
Cert\n
cd\n
D
<d
cos
+ E sin
cot\n
ut\n
...
cat\n
(F\302\261tn 4\342\200\235
4~
Fn\342\200\224it4~
cos
Fn)ert
4~
(G\\tn
cot\n
...
4\342\200\234
4\342\200\234
Gri)crt
sin
cot\n
* When
inte\302\254
v(t) consists of a sum of severalterms, the
particular
terms
individually.\n
\ngral is the sum of the particular integrals corresponding
listed in this columnis alreadya
a term in any of the trial integrals
t Whenever
to modify
of
the
function
of
the
\npart
given equation, it necessary
complementary
\nthe indicated
it. If such a term appears
choice by multiplying it by t before
using
appropriate
to these
is
\nr
times
\nby
in
the
function,
complementary
the indicated choiceshould be
multiplied
tr-\n
(By
\n1951.
(2)
McGraw-Hill
Write
\nEach
\ncoefficient,
Advanced Engineering
Book Co., Inc.)\n
from
permission
the
trial
form
trial
different
and
all
Mathematics by
of the particular
solution
similar
should
functions
Wylie.
Copyright,
integral, usingTable6-1.
be assigned
should
a different letter
be combined.\n
DIFFERENTIAL
6-5\n
Art.
CONTINUED\n
EQUATIONS,
the trial solution into the
coefficients
of all like terms, form
(3) Substitute
differential
\nequating
and
(4) Solve for the undetermined
\nintegral. These coefficientsmust be in
of
terms
a
the particular
and
circuit
driv\302\254
in the
the total solution may be
function
to the particular integral.
the complementary
by
adding
the arbitrary
constants of ic can be
is required,
solution
particular
of the initial conditions. As a precaution,
from
a knowledge
be applied to the total
initial
conditions
must
always
alone unless iP = 0 [when
function
to the
complementary
the
determined
Having
\nIf
algebraic
integral.\n
\nparticular
\nfound
of
constants
no arbitrary
are
There
parameters.
a set
so find
coefficients
force
By
equation.
coefficients.\n
undetermined
the
in
\nequations
\ning
115\n
integral,
particular
\nevaluated
solution\342\200\224
\nthe
\nnever
\n=
v(t)
0].\n
Example
4\n
=
\nv(t)
\nlaw, the
V and
where
Ve~at,
with the driving
RL circuit
a series
Consider
differential
equation
force voltageof the
a are constants. By
is, after division by
Kirchhoff\342\200\231s
voltage
L,\n
di
V
.
R
,
\342\200\224at\n
(6-99)\n
\ndi+Li=Le\n
characteristic
The
function
\nmentary
is m
equation
=
Table 6-1, the
\nthis
a
^
R/L,
trial
A is the
where
A
is the
i
The
arbitrary
=
constant
ke~Rt/L
=
the
comple\302\254
(6-100)\n
sum
~B
K
Ae~*\342\200\230
+
=
R
aL
e~at
+
ic,
^
gives\n
e\342\200\224\342\200\230
a 5* y-
f,
\342\200\224
\342\200\224\342\200\224f
\342\200\224
can
Ae~at =
^
of iP and
ocJj
equation
V
D
(6-101)\n
undetermined coefficient.Substituting
differential
the
into
solution
or
solution
that
be\n
-aAe~at
The
= 0, so
trial solution should
iP
if
+ (R/L)
is\n
ic
From
form
(6-102)\n
(6-103)\n
L\n
or\n
Ke~Rt/L,
be evaluated
a
L/\n
(6-104)\n
from knowledge of the
initial\n
DIFFERENTIAL
116
=
Ate\342\200\2241
=
A
or
The solution for this caseis
Example
a
As
V\n
=
TIj\n e~at
(6-106)\n
~
(6-107)\n
+
=
a
Ke-\302\253t/L,
(6-108)\n
j-
5\n
second
force
\ndriving
gives\n
thus\n
~ te~at
=
equation
<xAte~at
+
e~at)
A(\342\200\224ate+
i
(6-105)\n
into the differential
solution
this
be\n
should
iP
Substituting
Chap. 6\n
form of the trialsolution
= R/L, the
a
If
conditions.
CONTINUED
EQUATIONS,
example,
v(t)
voltage
consider a series RC circuit with a sinusoidal
= V sin cot. The Kirchhoff
voltage equation
\nis\n
+
Ri
V
sin
<ot\n
(6-109)\n
^\n
and
differentiating
or,
di
dt
Table
\ncosine term,
=
\nand
of
coefficients
\nlinear
+
for A and B
a
\nA
some
(6-110)\n
the sum
of
a
sine
and
a
cot -J- B
A cos
are
sin
cot
(6-111)\n
into the differential equation
the following system of
equated,
these
A
=
\342\200\236B
-
\302\273A
=
0
(6-112)\n
yields\n
wCT
_
1
\nafter
cos\342\200\234<\n
iP should be
functions
like
A
Substituting
~R
results.\n
equations
Solving
coV\n
is substituted
solution
assumed
this
=
as\n
iP
If
R,\n
ml
the assumed
6-1,
.
1
,
+
From
by
dividing
+
u'RC'V
a,2ft2C2\342\200\231\n
\n1
+
into the assumed
values
a>2tf2C2\n
solution, there
(6-113)\n
results,
simplification,\n
%p
\342\200\224\n
i/tPC*
+
fi,
cos 0,1 +
sin\n
(6-114)\n
6-5\n
Ait
This
DIFFERENTIAL
\ntity
use
of
cosine
the
for
cos
=\n
tp
+1
Vfl2
where\n
tan-1
(6-115)\n
0)\n
o)RC\n
(6-116)\n
be added ic =
of iP must
value
this
To
Finally,\n
A>2C2\n
=
0
iden\302\254
trigonometric
\342\200\224
{(tit
*\302\273
1/ooC
defining
by
sin 0, and making of the
the difference of two angles.
R = K
0 and
cos
\nK
117\n
CONTINUED\n
to a single sinusoid
be reduced
can
equation
EQUATIONS,
complete\n
the
for
Ke~,/RC
\nsolution.\n
Example
6\n
is
\nvoltages
important
Kirchhoff
\nequation for this
di\n
Ri = V
+
L^
equiv\302\254
system
sin
(<tit
in
shown
system
The
6-11.
\nFig.
a
such
of
\nalent
circuit
the
Consider
\nsystems.
with sinusoidal driving force
generation and distribution
of systems
response
in studies
of power
the
of
Knowledge
voltage
is\n
+
6)
\nat\n
(6-117)\n
\nin
for
method
The
last
the
result
The
example.
integral is like that illustrated
the particular
finding
is\n
V\n
ip
=\n
sin
V#2
To this
result must be
4
\nExample
+
\302\2532L2\n
+
Vfl2
=
from
which
((at
(6-119)\n
+ 6
\342\200\224
+
tan'\n
(6-120)\n
^\n
t
=
V-USin
has zero value
current
initial
the
0,
requiring
inductor,
that\n
(\302\256
+
Ke\302\260
=
0
\342\200\234tan\"
vWT^E'-sin
(6-121)\n
tan\"Tf)
K = ~
angle 0, which
Ke~Rt/L
\342\200\234>2\302\2432\n
switch is closedat
the
Ke~Rt/L
\342\200\234
If the
(6-118)\n
R)\n
(\n
thus becomes
sin
fin
or
tan-1\n
function,
\nV\n
of
i^\\\n
\342\200\224
is\n
The total solution
\nbecause
6
+
added the complementary
ic
Now if the
(o)t
(*
represents the angle
of
(6_122)\n
Tt)
the
sinusoid
at the
time
the\n
DIFFERENTIAL
118\n
is
switch
has the
closed,
CONTINUED\n
EQUATIONS,
Chap.
6\n
value\n
6
=
tan-1
(6-123)\n
K\n
other
In
\nvanish.
transient. The same conclusioncan be reached
network
but not for an RLC network.\n
\nRC series
6-6.
RLC series
Capacitor
charge
in an
circuit
shown
in Fig.
The
=
\nat t
words,
zero value, and the transient term ic will
if the switch is closed at the proper
instant,
be no
will
\nthere
have
will
K
constant
the
0. It
is desired to
circuit\n
by closingthe
6-12 is energized
find chargeon
the
1
C-ZrZ\n
after
is,
division
by
of
law,\n
voltage
for the charge
equation
r\342\200\235j\302\243
+\n
\n+1
K
switch
a function
Kirchhoff\342\200\231s
differential
#\342\200\224the
as
capacitor
time, q(t). By
\nv
the
for
L,\n
9c(t)\n
series
RLC
6-12.
Fig.
\nequation, and the
\nlar
or
integral
circuit.\n
of
solution
=
q3\342\200\236
Case
q
=
f
1,
qt
+
>
=
q*8
\302\260
+
-^A
f =
2,
q
=
CV
f <
The
parts.
particu\302\254
be\n
(6-125)\n
gives\n
equation
A =
or
=j/\n
the
underdamped,
CV\n
(6-126)\n
corresponding, to overdamped,
solutions
are:\n
1.\n
CV
cosh
+ e-^^Ki
\342\200\224
Vf2
(co\342\200\236
11)
l\302\273L2(sinh
<on
\t\n
\342\200\224
1 \302\243)]
\\/f2
(6-127)\n
1.\n
q
Case 3,
differential
the differential
+
Case
two
(6_124)\n
(a constant)
studied,
previously
or
damped
\ncritically
+
cases
three
the
For
A
this solution into
0
will
I
a nonhomogeneous
solutionwillbe composed
steady-state
Substituting
is
This
=
+
+
^
1\n
=
CV
\342\200\234H
(Ki
(6-128)\n
K%t)e~<*nt
1.\n
+ e-t^Ki
cos
(<on
\\/l
\342\200\224
f21)
+
K2 sin
(\302\253nV1
\342\200\224
f2
<)]\n
(6-129)\n
Each
\nwill
not
be
two arbitrary
constants, Ki
same for the three cases, but
has
solution
the
and Ki. Theseconstants
they can
be
evaluated
from\n
Ait
DIFFERENTIAL
6-6\n
CONTINUED\n
EQUATIONS,
119\n
conditions. As initial conditions, we will
that
assume
at
= 0
\342\200\224
\342\200\224 so
initial
the
and
on
\nt
ic 0,
0,
qc
(no
charge
capacitor)
= 0 (no initial current because of the inductor).
will
We
carry
\ndqc/dt
two
of the constants in some detailfor Case3. The
\nout the evaluation
value
follow a similar (and easier) pattern. Usingthe initial
\nother cases
initial
the
\nof
condition
q
gives\n
0 =
or
=
Kx
The derivative of q
dq
K2
CV + e*(Kx +
0)\n
-CV
with respect to
(6-130)\n
is\n
time
_\n
dt
\t\n
\342\200\224
\\/l
(jin
+
\342\200\224
sin
r2
\\/l
X2e~\"\"\342\200\230[a)\342\200\236
\342\200\224
T2
\342\200\224
cos
\342\200\224
f21)
(w\302\253a/1
\342\200\224
(w\302\273a/I
f2
cos
\342\200\224
0
-\\/l
(ion
sin
(<on \\/l
\342\200\224
f2
\342\200\224
f2
<)]\n
0]\n
(6-131)\n
At < =
= 0,
0,
0
Hence
The
g(<)
K,
=
-
CV
^
Jl
\342\200\224
K\\{
fan)
(6-132)\n
r2)
(6-133)\n
^=L=g
is\n
\342\200\234
*\342\200\234nt=
[Vl
r2
cos
may also be written in
=
q(t)
CF
^=%=
+
This equation
\342\200\224
K2(a)n \\/l
+
= -
= K,
solution
particular
=
so\n
-
cv
Vi
(o>\342\200\236
f sin
0\n
(c0n
vr^T*
*
+ o
^i
8
where
For
=
tan-1
2, the
Case
q(t)\n
for
f2/f.\n
=
solution
particular
1, the
Case
=
- (1 +
C7[l
particular
will be found to
solution
\342\200\224
CV
e
Jl
(6-135)\n
j
\342\200\224
\\/l
q(t)
and
(6-134)\n
(u.VW\342\200\231O]\n
form\n
the
sin
- f2
(\302\253\342\200\236
\\/f2
be\n
Me-\"\"4]
(6-1.36/\n
is\n
\342\200\224
11)\n
j^cosh
sinh
+\n
Vt2
-1\n
(o)n \\/{2
\342\200\224
1
0\n
(6-137)\n
DIFFERENTIAL
120\n
three
These
the
\nSince
an
\nby
ratio
6-13.
Kg.
=
= 0.2),
(f
are
\ntions
\ninitial
circuit as the
for
valid
the
and
ratio
damping
changes.
These
solu\302\254
system with these particular
networks but also in mechanical
in servomechanisms).\n
example,
second-order
only in electric
any
not
conditions,
electromechanical
\nand
values of f illustrating
(f = 1.0),
overdamped
cases.\n
1.4)
charge in the
of the
behavior\n
response for three
critically damped
System
\nunderdamped
\n(f
merely
of the
adjustment
Chap.6\n
the three conditions of damping.
as the ratio of actual resistance
three
conditions
could be obtained
resistor R. Figure 6-13 shows the
the
resistance,
CONTINUED\n
give q(t) for
is determined
equations
damping
critical
\nto
EQUATIONS,
(for
systems
FURTHER READING\n
Solutionofdifferential
of
equations
\nis
treated
concisely
Book
\n(McGraw-Hill
4,
\nChap.
\nin
Fich,
\nInc.,
\nreading
by
New
titled
Transient
York,
include:
Wylie
Inc.,
Co.,
\342\200\234Classical
Analysis
1951).
Skilling,
the
type
discussed
in this chapter
in Advanced Engineering Mathematics
New York, 1951), pp. 1\342\200\22445.See also
Analysis
of
Electrical
Double-Energy
Transients\342\200\235
Engineering (Prentice-Hall,
texts recommended for supplementary
Currents
Transient
Electric
(McGraw-Hill\n
in
Other
Chap. 6\n
DIFFERENTIAL
and
\nmatical
\nBook
Mathe\302\254
Johnson,
of Engineering Analysis (McGraw-Hill
1944), Chap. 6; Salvadori and Schwarz,
Problems
Engineering
Inc.,
(Prentice-Hall,
3 and 4.\n
Physical
Principles
New
Inc.,
York,
Co.,
in
Equations
\nDifferential
\nNew
121\n
York, 1952), Chaps.3 and 4;
Co., Inc., New
Book
CONTINUED\n
EQUATIONS,
Chaps.
1954),
York,
PROBLEMS\n
that i =
Show
6-1.
ke~2tand i =
of the
solutions
are
ke~l
differential
\nequation\n
dS
+
\nential
i
that
Show
6-2.
+ 2i =
3ft
\ndt2\n
\342\200\224
ke~\342\200\230
and
i
=
0\n
kte~* are
of the
solutions
differ\302\254
equation\n
dH
+2s+i=0\n
\ndt2\n
<*>
the general solution
Find
6-3.
+
3?
+
5S
+
+
6s'\342\200\234\302\260
\302\260\n
+
<d>3?
+
5;l
Find
6-3(a)
particular
and
Prob.
Repeat
=
2e~*
subject
6-3(b)
=
e~2t,
Write
the
characteristic
\nfollowing
(a)
(m
(b)
(m
(c)
(m
i + 2*
1,
i
=
+
\302\260\n
+
3e-2<
=
(0+)
j(
2, ^
solution
4f
=
0\n
4t\342\200\231
of
0\n
\342\200\224
2e~3t.\n
6-4
general
0\n
=
of the differential equation
to the initial conditions:\n
solution
\342\200\224
+ 6x =
0\n
variable (0+) =
6-6.
+
B
7t
for the differential equations
to the initial conditions\n
(h) subject
Prob.
through
(c)
w
<h)3F
the
i
Answers,
\nparts
+
= 0\n
4\342\200\231
i(\302\260+)
6-6.
<\342\200\242>
W
+ 12i =
7m
\n^\302\247
\nProb.
2'
equations:\n
following
3\302\256
(b)S
6-4.
=
+
each of the
(0+)]
[variable
of the differential
equations:\n
+ l)(m + 2)(m2 + m -f 1) =
+ l)2(m
+ 5) = 0\n
\342\200\224
+ a){m -j- b)(m \342\200\224
c) (m
d)
0\n
=
0\n
=
of Prob. 6-3,
1\n
equations with the
DIFFERENTIAL
122\n
=
\n(m
<r
+
\tR =
(b)
roots
the
Plot
6-7.
L
for
ju)
1000
CONTINUED\n
EQUATIONS,
Chap.
equation in the ra-plane
=
= 1 henry, C
1
and
R = 500
(a)
R =
(c) R = 2000 ohms, (d) = 3000
of the characteristic
ohms,\n
jff,
ohms,
6\n
(e)
ohms,
\n4000 ohms.\n
not
constants.\n
arbitrary
In a certain
6-9.
6-7
find
to the
\nsolution
\nthe
=
R
(b)
\nohms,
= 500
and
for
described in Prob.
(a) R
the
2000
general
ohms, and (c) R = 4000 ohms,
but
differential equation. Evaluate all coefficients
For the system
6-8.
network, it isfoundthat
is given
current
the
by the
\nexpression\n
=
i
\nShow that
\342\200\224
Kie~ait
>
t
K2e~att,
maximum value at
i(t) reaches a
a.2
\302\253i>
0,
time\n
1
t
=\n
ln\n
\342\200\224
ai
6-10.
The
graph
CX2K2\n
<*2
shown is a
record of the
current
as
of
a function
when
a switch
is closed at t = 0, connectinga
resulting
\nto a network.
more than a cycle is shownin
Only
slightly
record,
Determine
the
current
zero
value,
the
\nbut
reaches
(a)
eventually
of
for
waveform,
the
current
f and
(b) Write the equation
current
as a function
of time with all coefficients
\ntime
battery
the
\nvalues
<o\342\200\236
evaluated.\n
\nof
Prob.
6-11.
\nK is
\nthe
\ntion
In the
opened
at t
\342\200\224
is opened.
switch
of
network
time.
Answer.
0
6-10.\n
shown in the
with
the current
u(<) = 10 cos
Prob.
the
conditions
equilibrium
Find
accompanyingfigure,
existing
switch
before the
through the inductor as a
316L\n
6-11.\n
func\302\254
DIFFERENTIAL EQUATIONS, CONTINUED
Chap. 6
that
Show
6-12.
\342\200\224
Ke~t<iKt
i
the
Give
the
Solve
6-13.
and
for K
values
cos
<t>
0)\n
and K6 of
Eq.
differential
nonhomogeneous
following
+
f21
of K6
terms
in
\342\200\224
\\/l
(o>\302\273
form\n
6-67.\n
equations.\n
+ ai+*-1\n
M$
+
+
i =
Answer,
*-\302\253\n
\302\273s
o\302\273>\302\243
\342\200\224
\302\247\302\243
kxe~l
+
(c) S
+3 s
+
(d)
+
+ 69 =
^
Answer,
^
5l
=
q
(e)
written in the
6-67 can be
Eq.
123\n
v =
Answer.
!| +
(fCi
A special
= t
6-14.
=
10
(to
=
10i\n
kie-2*+
+
e-\342\200\234
+ t)e~2t +
generator
sin
fe_\342\200\230\n
74\n e-< +
^2
+ 5
2<
+ kgr2*.\n
ktfru\n
5e-\342\200\234\n
(K2
\342\200\224
bt)e~u.\n
variation given by the
has a voltage
equa\302\254
where t is the time in seconds.This generator
is
v(t)
volts,
=
=
to an RL series circuit, where R
1
2 ohms
L
\nconnected
=
Find
0 by the closingof a switch.
the equation
for
\nhenry, at time t
\342\200\224
=
of time i(t). Answer. i(t)
as a function
\nthe
current
x + tC_2\342\200\230-\n
\ntion
and
of lightning
having
a transmission
\nas v(t)
te~* strikes
\nand
inductance
L = 0.1 henry (the
A bolt
6-15.
=
An
\nnegligible).
What
\ndiagram.
base
\ntime
is
\nat
t
=
of
normalized.)
Prob.
6-16.
ohm
line-to-line capacitance is assumed
is shown
in the accompanying
network
equivalent
is the form
of the current as a function time?
(This
in amperes per unit volt
the
likewise
the
lightning;
of
will be
\ncurrent
a waveform which is approximated
line having resistance R = 0.1
Answer.
i(t)
=
5t2e~\342\200\230.\n
Prob.
6-15.
In the circuit
0. Answer,
i =
shown in the figure,
solve
10~4e~\342\200\230
(cos
t
\342\200\224
sin
t)
for
\342\200\224
6-16.\n
i(t)
10-4e~104*.\n
if K
is closed
DIFFERENTIAL
124\n
network
the
In
6-17.
CONTINUED\n
EQUATIONS,
is closedat
the switch K
shown,
having been established previousto
\nsteady state
time
\nof
\nsin
\nin
force
6-19.
=
*
2L
\nthe
of
frequency
w =
(1)
=
o)
(2)
total
the
Find
is,
the
con
natural
undamped
\342\200\224
natural
(the
f2
\nis,
\nvalue
\nwhich
\nis
value
current,
which
the
as
a
of time. Show the
wR
and
Wl
in
peak
(a)
approach
of
accompanying
angular
angular
separate
frequency)\n
frequency)\n
experiments.
when
0, and
(b)
the
steadycase (that
is the maximum
(b) In
greater?
is, which frequency)
value
of the steady-state
(that
maximum
diagram,
is\n
transient
the
case
<
=
two
the
In which
frequency)
of
at
closed
maximum
\nstate
current
transient
is
switch
\nthe
that
values
force voltage
driving
(the
oon \\/\\
the
of
\nthe
field
energy supplied by the voltage sourcein
These frequencies are applied
\nexperiment we measure (a) the
\nvalue
the
magnetic
a function
wL as
RLC circuit shownin
In the series
6-21.
a
state.\n
steady
\nthe
that
asymptotes,
oo.
* \342\200\224*\342\226\240
(d)
\nas
V
as
2 RJ\n
energy in the
an
\nsteady-state
resistor
e\342\200\2242Rt/L\n
\n2 R\n
for the
expression
of time,
(c) Sketch wR and
\nfunction
of voltage
energy in the
3L\\\n
L_
,-Rt/L\n
\nR\n
=?(\342\226\240+\n
\tFind
partic\302\254
is\n
wR\n
(b)
the
3.\n
a battery
0 connecting
(a) Show that the
circuit,
RL
of time
\nfunction
at
is closed
switch
series
a
\nwith
time. Prove
Eq. 6-137is
given for Case
conditions
initial
the
with
A
6-20.
a sinusoidal
with
at an appropriate
with Eq. 6-127,verify that
Starting
solution
\nular
vanish
cannot
is correct.\n
statement
this
\nthat
term
in the
transient
circuit
series
a switch
closing
by
merely
in the dia\302\254
as a function
i = (20.3
1414* + 1485
amp.\n
the
solution
the
\nRLC
\ndriving
10-3
that
made
6-17.\n
Prob.
the\n
For
On page 118, the statement
6-18.
\nis
cos
0.8
X
1414*)
a
with
0,
shown
Answer,
i(t).
-
377*
\ncos
=
*
time.
current
the
find
\ngram,
this
values
parameter
rt-HPO\n
Chap. 6\n
current greater?\n
In
each
CHAPTER
TRANSFORMATION\n
LAPLACE
THE
7\n
7-1. Introduction\n
Laplacetransformation
The forerunner of the
\nferential
the
equations,
\nliant
dif\302\254
solving
was invented by the
calculus,
Heaviside
Heaviside
(1850-1925).
operational
Oliver
engineer
English
of
method
bril\302\254
was
practical solution of electric
rather
than
of his methods. He
careful
problems
justification
with
an insight
into physical problems that enabled him to
gifted
the
correct
solution
from a number of alternatives. This heuristic
of
view
drew
bitter
and perpetual
criticism from the leading
of
his
In the years that followed publication of
time.
the
was supplied
work,
rigor
by such men as Bromwich,
and
the work of
others.
The basis for substantiating
Carson,
in the writings
was
found
of Laplace in 1780. As the years
the
structural
of the framework of
members
passed,
calculus
have
been
\noperational
replaced,
piece by piece, by new
This transformation has
\nbers
derived
transformation.
by the
Laplace
\na
was in the
interest
his
and
man
practical
\ncircuit
\nwas
\npick
\npoint
\nmathematicians
\nHeaviside\342\200\231s
\nGiorgi,
\nHeaviside
Heaviside\342\200\231s
\nhave
mem\302\254
The
in
discussed
\nwere
of advantages
4 and 6.
Chapters
(1) The solution
of
differential
no
impor\302\254
results.\n
Heaviside\342\200\231s
for solving
differential
over the classical
number
a
methods;
operational
method
transformation
Laplace
offers
\ntions
in
discovered
been
have
errors
\ntant
of the
substantiation
rigorous
\nprovided
equa\302\254
methods that
For example:\n
is routine
equations
and progresses
\nsystematically.\n
(2) The method givesthe total
solution\342\200\224the
\nthe
(3)
problem
\nsuch
\ning
\nto
step.\n
we
as
operations
to
quantities
have
The logarithm is
powers.
the
division,
accuracy
of the
trans\302\254
extracting
roots, and
rais\302\254
we have two numbers,given
we arc required
to find the product,
given numbers. Rather than just
Suppose
and
an example of a
past. Logarithms greatly simplify
used in the
multiplication,
accuracy,
seven-place
\nmaintaining
operation.\n
transformed
a transformation?
that
\nformation
and
integral
are automatically specifiedin the
are incorporated into
the initial
conditions
Further,
as one of the first steps rather than as the last
\nequations.
is
one
conditions
Initial
\nthe
What
function\342\200\224in
complementary
particular
that
mul-\n
125\n
THE
126\n
TRANSFORMATION\n
LAPLACE
Chap. 7\n
we transform these numbers by
are added (or subtracted in
their
These
\ntaking
logarithm.
logarithms
of division).
sum itself has little meaning. How\302\254
\nthe
case
The resulting
if we find the antilog\302\254
if we
an inverse transformation,
\never,
perform
then
we have the desired numerical result. The directdivision
\narithm,
has been that the use
\nlooks
more
but our experience
straightforward,
\nof the
often
saves time. If the simple problem of multiply\302\254
logarithm
consider evaluating (1437)\302\260-1328
two
is not convincing,
numbers
\ning
numbers
two
tiplying
the
\nwithout
logarithms!\n
together,
Logarithm\n
Logarithms\n
Numbers\n
of numbers\n
1\n
1\n
Direct\n
Addition
multiplication
\nor
\nof
division\n
numbers\n
1\n
Product
\nor
Sum of
Antilogarithm\n
\nlogarithms\n
quotient\n
(a)\n
Initial\n
Conditions\n
Integra-\n
Laplace\n
differential\n
Transform\n
transformation\n
equation\n
I\n
I\n
Classical\n
Algebraic\n
solution\n
manipulation\n
\\\n
Inverse\n
Solution\n
transfo\n
Time domain
Revised\n
Laplace\n
Transform\n
rmation\n
\342\200\242\n
\342\226\240
Frequency
domain\n
(6)\n
Fig.
7-1.
Comparison
of logarithms and
the
Laplace
transformation.\n
a product
operation of using logarithms to find
in
The
are:
individual steps
is shown
\nof
a quotient
(1)find
Fig. 7-1.
of the
\nthe
numbers,
(2) add or subtract the num\302\254
separate
logarithm
\nbers
to obtain
the sum of logarithms, and (3) take the antilogarithm
the
or quotient.
This is roundabout comparedwith
\nto obtain
product
or division,
\ndirect
yet we use logarithms to advantage,
multiplication
when
a good
table of logarithms is available.\n
\nparticularly
The flow sheet idea may be used to illustrate what we willdoin using
\nthe
transformation
to solve a differential
Laplace
equation. The flow
a\n
for
the
\nsheet
is also shown in Fig. 7-1with
transformation
Laplace
A
flow
sheet
of the
7-2\n
Art.
block
TRANSFORMATION\n
LAPLACE
THE
to
corresponding
of the logarithm
block
every
127\n
flow sheet
con\302\254
follows. (1) Start
an
integro\ndifferential
and
find
the
equation
corresponding Laplace transform.
is a mathematical
but
there are tables of transforms just
process,
there
are
tables
of logarithms
(and one is included in this chapter).
The
transform
is manipulated
after the initial
algebraically
\nditions
are
The
is a revised transform. As step (3),
inserted.
result
to give us the solution. In
an
inverse
transformation
\nperform
Laplace
use
we also can use a table of transforms, just as
the table
step,
for
in the
step
logarithms. The flow sheet
logarithms
corresponding
the
classical solution. It looks
is
another
that
there
us
way:
it is for simple problems). For complicated
sometimes
direct
(and
an
will be found for the Laplace transformation,
advantage
was found for the use of
as an
advantage
The
above.
\nsidered
steps
will be as
with
\nThis
\nas
con\302\254
\n(2)
we
\nthis
we
\nof
\nreminds
\nmore
\nproblems,
logarithms.\n
\njust
The
7-2.
transformation\n
Laplace
a Laplace transform
To construct
first
\nwe
\nThis
is
the
\nDenoting
L
reserve
\nto
mathematical
\nthe
function
time
of
f(t),
a + j<a.
f(t)
e~ft, where s is a complexnumber,s
with respect to time from zero to infinity.
is integrated
transform
of /(<), which is designatedF(s).
the Laplace
transformation
order
by the script letter <\302\243
(in
Laplace
the Laplace transformation is given by
for inductance),
multiply
result
given
=
by
product
\nThe
for a
expression\n
(7-1)\n
letter
The
\nin
above
the
Although
\nthe
actual
be
\302\243
can
by the
replaced
expression.\n
once
\nFurthermore,
of F(s)
transform
the
\342\200\234theLaplace
transform
of\342\200\235
formidableappearingintegral,
given f(t) is usually not
a function is found, it neednot
can be tabulated. The time
is a rather
this equation
evaluation
words
for a
of
difficult.
be
func\302\254
problem but
\ntion
and
the transform
F(s) of this function are calledtransform
f(t)
A table
of transform
pairs is given on page 146.\n
\npairs.
The
which
a function of s back to a function
of
operation
changes
is called
\ntime
the inverse Laplace transformation. This operation is
\nfound
again
for
as
\nsymbolized
a new
<\302\243-1.
Then
by
definition,\n
jb-'WWD
The
inverse
Laplace
transformation
=\302\243-[?(\302\253)]=/(()\n
is given by
(7-2)\n
the complexinversion
\nintegral,\n
(7-3)\n
TRANSFORMATION\n
LAPLACE
THE
128\n
Chap. 7\n
This integral is seldom used becauseof the
of the Laplace transformation that there isa one\nuniqueness
property
and inverse transforms. In
\nto-one
between
the direct
correspondence
to a given
a table
\nother
can be used to find the/(\302\243) corresponding
words,
as well
as an F(s) for a given f(t).\n
\nF(s)
Not
all functions
can be integrated
to find the Laplacetransforms.
contin\302\254
\nThe
usual
for an f(t) are that it be (1) piecewise
requirements
and
\nuous
of exponential
order. These requirements are discussed
(2)
to say
detail
in references
\nin
cited at the end of this chapter. Suffice
c is
where
existence
the
\nfor
have the propertiesrequired
of engineering interest
of the transform.\n
functions
all
\nthat
a constant.
As an example
of the evaluation
of
\302\243[1]
=\n
= 1;
f(t)
then\n
\342\200\2248t\n
e\n
=\n
dt
e~at
let
7-1,
Eq.
(7-4)\n
8\n
Because
\ndoes
\nf(t)
\nt
<
limit of the integral, the value of f(t) for t < 0
Thus the Laplace transformfor
not
enter
into the final solution.
\342\200\224
for
1 is the
same
as that of a special function havingzerovalue
called
a unit
unit value for t > 0. Such a functionwas
0 and
step\n
Heaviside.
We will
by
function
use
the
u(t) for a function\n
symbol
\nu(tj
lower
the
of
1
described
m
u(t)
\342\200\2351\302\260+*
in
Fig.
7-2.
Such
a switch
closing
at
network
Vu(t);
Unit
7-2.
Fig.
that
a function
t>Q
.
(7-5)\n
.\n
of
time\n
to
0 by closing a switch, that voltagecanbe described
zero
for t < 0.
is, V times unity for t > 0 and V times
a second
\nwhere a
as
!
a function is the mathematical equivalentof physically\n
at t = 0. If a battery of voltageV is connected
for
Vu(t)
=
of a transform,let
Substituting into Eq. 7-1,
is a constant.
as\n
The\n
(7-6)\n
^\n
o\n
of the calculation
example
a\n
is\n
\302\243lVu(t)]
As
f\302\260
t =
transform
Laplace
shown
and
< <0\n
=
function.
step
as\n
mathematically
we
g-(.-o)t
dt\n
=
f(t)
eat)
have\n
*>a
dt\n
(7-7)\n
L\n
Thus
For
e\302\260\342\200\230
and
one
l/(s
further
\ndefining equation,
\342\200\224
a)
example,
we
a transform
constitute
have\n
let
f(t)
\342\200\224
sin
pair.\n
<at.
Substituting
into
the
7-3
Art.
129\n
TRANSFORMATION\n
LAPLACE
THE
-
4
+;/\342\200\224\342\226\240]*\n
f\342\200\231
o>\n
(7-8)\n
\302\2532
+
two
These
\npairs
can form
computations
as shown
tt2\n
the beginning of a table of
transform
below.\n
of Transform
Table
Pairs\n
m\n
F(s)\n
uit)\n
1/s\n
1\n
e\302\260*\n
s
\342\200\224
a\n
CO\n
sin
ut\n
s2
More
\nto
be
in reference
found
Transforms
(1)
of
\nfunctions
time
Laplace
Linear
of
and
a and
+
\302\243[a/i(<)
the
\nthat
\nof
is
theorem
This
the
\302\243[afi(f)
integral
that
terms;
+
bf2{t)]
transformation\n
If fi{t) and
Combinations.
b are constants,
= aF1(s)
bf2(t)]
sum of terms
=
[afi(t) +
a
J
+ bF2(s)
is equalto
We will make use
(2)
\nLaplace
sum
of
this
of derivative
Transforms
transformation,
(7-9)\n
follows from the
the
of
sum
the
fact
integrals
dt\n
bf2(t)]e-\342\200\230
fi{t)e~tl dt + b
fi{t)e~H
j
=
the
two
is,\n
=
\nof
f2(t) are
then\n
Eq. 7-1. It
with
established
of a
<a2\n
as they are computed. Exhaustivetablesare
books cited at the end of the chapter.\n
for the
theorems
Basic
7-3.
be added
can
pairs
+
terms
appearing
From
of Derivatives.
we
+ bF2(s)
(7-10)\n
the Laplace transformation
in taking
theorem
aF^s)
dt\n
in a differential
the defining equation for
equation.\n
the
write\n
s-tf(i)e~',dt
4itf(t>}=r
This equation may be integrated
by
u = e~tl
and
by
parts
dv
=
letting\n
df(t)\n
(7-n)\n
THE LAPLACE TRANSFORMATION
130
in
the
equation\n
fb
=
udv
l6
uv
fb\n
\342\200\224
I
J
Then
=
\302\243
(7-12)\n
du\n
v =
and
-se\342\200\224'dt
+
e-\342\200\230f(t)]\"
f(t)\n
becomes\n
of a derivative
transform
the
v
J
=
du
that
so
7\n
Chap.
=
dt
\302\253
[\"/(()\302\253-
sF(s)
- /(0+)
(7-\n
13)\n
[|/(0]
=
lim
provided
0, which
follows from
pro-\n
1\342\200\231Hospital\342\200\231s
rule,
\302\2530\n
t\342\200\224\342\226\272
all its derivatives are not infiniteat t
transform
of the second derivative, we follow
Since\n
but make use of the result of Eq.
that
vided
f(t)
the
find
To
and
=
<\302\273.\n
similar
a
7-13.
\nprocedure
(7-14)\n
then\n
-
=
=
this
In
\nevaluated
The
=
^
-
for an
-
*-F(\302\273)
\302\253/(0+)
^
(7-15)\n
(0+)\n
(0+)\n
the derivative
after
switchingactionis
immediately
expression
general
ft
df/dt (0+) is
the quantity
expression,
=
at t
0+ (the time
\ninitiated).
\302\243
-
a!F(s)
-
/(0+)]
\302\253(<#(\302\273)
nth order derivative
*-* % (0+)
s\342\200\224/(0+)
of
-
... -
/(f)
is\n
1^/(0+)\n
(7-16)\n
\nsion,
of
\tTransforms
(S)
is
found
by
transform
The
Integrals.
for an integral
expres\302\254
with the equation\n
starting
dt
me-1
= F(s)\n
(7-17)\n
/.\n
by parts to
and integrating
F(s)
=
e\342\200\2241
give\n
+ 8
f(t)
j
The
JQ [J
quantity\n
j
m
^](\n
fit)
dt
e\342\200\224*\n
(7-18)\n
the
\nof
the
at
zero
becomes
upper
evaluated
integral
TRANSFORMATION\n
LAPLACE
THE
7-4\n
Art.
limit, and at the lower limit has the
at t = 0, usually written by the notation\n
m
-
dt\n
the
s times
\nnized
as
7-18,
there
value
(7-19)\n
/<~1)(0+)\n
t-o+\n
/\n
where /(-1) indicates
131\n
integration. The last
transform
Laplace
of
term
of / f(t)
is
7-18
Eq.
recog\302\254
dt. Rearranging
Eq.\n
results\n
=
rn+f^f\302\261i\n
*[jmdt]
is found
it
Similarly,
(7-20)\n
that\n
-\n
m
,
/(\n -\342\200\230>(0+)
0+)\n
/<-\302\273(
r
.9
I\n
-9
(7-21)\n
_\n
[//*H
the
In
the
\nand
on the loop basis, f(t) is often
a current
the current
is the charge q(t). Equation20then
of networks
analysis
of
integral
i(t)
has
\nthe form\n
= rn
\302\243[fmdt]
where
charge (say on the
is the
q(0+)
+
rn\302\261i\n
plates
of
a
(7-22)\n
time
at the
capacitor)
\nt \342\200\224
0-!-.\n
If
is a
f{t)
voltage,
then\n
(7-23)\n
\342\226\240[/<\302\253]\342\200\242\302\273*\302\253\n
the
since
7-4.
of the
Examples
to
yet,
now
are
we
\nsection,
is flux linkage
of problems
if/.\n
with
the
of transforms that
has
theorems
basic
three
the
\nas
solution
the short table
With
\nand
of voltage
integral
equipped
Laplace
been
transformation\n
on
given
129
page
that have been derived in the previous
to solve a network problem (elementary
using the Laplace
be sure)
\ntransformation.\n
K
Example
this
For
example,
\nthe
Kirchhoff
voltage
\nRC
network
shown
\nwill
be
\nis
closed
\nof
the
R\n
1\n
that
assumed
at
network
t =
will write
law for a series
in Fig. 7-3. It
K
the switch
we
0. This information
equations
by
writing
Tig.
7-3.
RC series
circuit.\n
will be includedin
formation
the voltage expression as
the
Vu(t).\n
TRANSFORMATION\n
LAPLACE
THE
132\n
7\n
Chap.
Hence\n
=
dt + Ri
the
is
This
\nfirst term,
M
of the
terms
\nand
may
the
terms
\342\200\236
v-.i\n
\302\253/(,)
J
the
taken
as\n
(7-25)\n
8\n
have
the
Laplace
and there has resulted a
conditions
are automatically
trans\302\254
transform
specified
second step (rather than as the final
step
classical
Now
methods).
by
equations
g(0+)
the capacitor
at t = 0+. If the capacitoris initially
= 0 and the last equation reduces to the form\n
solved
on
?(0+)
\nuncharged,
for
7-20
as the
inserted
charge
8
initial
required
differential
\nas in
\nis
be
+
\302\253(\302\260\302\261>1
equation
integral
The
\nexpression.
+
8
flow chart of Fig. 7-1,we
the
of
\nformation
solve. UsingEq.
the transforms of the linearcombination
of
i[\n
In
we wish to
equation
integral
we take
(7-24)\n
Vu(t)\n
V\n
=\n
m\n
R)
Or*+
again according to the flow chart, is algebraic
manipula\302\254
of
this
is
to
solve
for
objective
manipulation
/($). This
by s and dividing by R to give\n
by multiplying
next
The
\ntion.
step,
The
\naccomplished
\nchart
is a
is
\342\200\234revised
to
perform
That
\nsolution.
s
transform\342\200\235
the
(7-27)\n
1 IRC\n
+
expression.
next
The
step
Laplace transformation
inverse
is
V/R\n
=\n
m
which
(7-26)\n
8\n
on our
flow
and obtainthe
is,\n
-
\302\243-(/(\302\273)]
=
\302\243-
P-28)\n
(fZfTfUc)
the second transform
Using
pair of our shorttable,
Of)
is
This
\nzero
the
\342\202\254*\302\273\n
(7-29)\n
solution
(the steady state in this
constant
arbitrary
emerges evaluated
case
being
of
(and has the
IB\n
Example
7-4.
is\n
V/R).\n
\nmagnitude
As
=
complete
The
value).
solution
the
second
our
As
in
example,
Example
1, the
consider
the RL series
circuit shownin
switch is closed at t = 0.
The
differential\n
Fig.\n
is, by KirchhofTs
circuit
the
for
equation
The corresponding transform
-
L[\302\253/(\302\253)
condition
initial
\nrent
after
\nOur
equation
\nnew
+
\n[1/(8
This suggests that
\nwere
some
form
terms
\nlet
be
could
\nation
us
L
this
us
simplify
senes
clrcult-\n
Ki\n
(7-33)\n
s
R/L\n
coefficients.As the first
putting all terms overa
K i are unknown
by
common
= Ko
(*+
of like
coefficients
r)
+
functions,
KiS\n
we obtain a set
of linear
equations:\n
\nalgebraic
R
K
\nKo
_
L
From
prod\302\254
attempt to performthisoperation,
s
R/L)\n
RL
7\"4*
rifir*
K0 ,
= Ko,\n
equation
r
equating
the
of
oper\302\254
the trans-
and
the
something
We
Then\n
\ndenominator.
By
R/L)\n
if there
-j-
K0
equation
let
\nstep,
(7-32)\n
\302\253(s+
in our shorttable. need
that this term is madeup
V/L
In
0.
terms.\n
the inverse
s(s
=
r\n
into several parts. As an
the following
expansion:\n
try
cur\302\254
of
performed
to break
way
the
1\n
=
individual
these
of
\302\253(0+),
/(\302\253);
transformation
Laplace
\neach
(7-31)\n
Notice
(1/s) and the term
know
the in\302\254
We
R/L)].
\nverse
-
t\342\200\230(0+)
table).
term
the
of
\nuct
is\n
by the last
however, is not
a larger
(or
(7-30)\n
+ RI(8) =
*(0+>]
m
This transform,
law,\n
equation is
Because of the inductance,
to solve for
thus\n
manipulated
be
now
may
133\n
Vu(t)
equation
specified
is closed.
switch
the
Ri=
+
Ljt
The
TRANSFORMATION
LAPLACE
THE
7-4
Art.
these
two
Ko
+
Ki
=
0\n
L\342\200\231\n
we find the
equations,
Ko
V
=
^
and
required values for
Kx =
\342\200\224
^\n
Ko
and
K\\\\\n
THE LAPLACE TRANSFORMATION\n
134
has permittedEq.
This algebraic manipulation
1
V
/(\342\200\242)\n
V\\l\n
R
(7-34)\n
s
[s\n
R/L\n
-(-
transform pairs corresponding to each of
\nThe current
as a function of time isfoundby taking
\ntransformation of the individual expressions;thus\n
We have
-
=
i(t)
is
This
\nexpand
\nwe
of 'partial
heading
Laplace
(7-35)\n
-
-|-
R/L\n )\n
(7-36)\n
erRt/L)\n
solution. The method we
(time-domain)
a transform
into the sum of several separate parts
to
used
is
known
is this subject
fraction expansion. It
that
next.\n
study
fraction expansion\n
Partial
7-5.
expressions.
inverse
final
the
the
\nunder
the
1\n
-i\n
(1
^
these
\302\243\n
s
or\n
written\n
be
to
7-32
1\n
=
+ R/L)
L s(s
Chap.7\n
The examples of the
in
\ncedure
last
the
applying
the
suggested
general
pro\302\254
to the solution of
of the general
equation
transformation
Laplace
differential
A
equations.
\nintegrodifferential
have
section
\nform\n
dni
o0
as
becomes,
which
\ntion
,
+ ai
^
^n-1 +
\nLet
\nand
+
pairs,
\never,
the
transform
any
by
expression
\ndenominator
QCs)
an algebraic
equa\302\254
terms
+ an~iS +
is a quotient
polynomials
of
(7-38)\n
a\342\200\236\n
in
polynomials
=
(7-39)\n
can now be foundin a
table
of
trans\302\254
i(t) can be written directly. In general,how\302\254
for I(s) must be broken into simpler
expression
transform
table can be used. To simplify
practical
fraction
we find the roots of the
partial
expansion,
polynomial\n
=
aos\"
+
s.
be designated P(s)
solution
the
before
\nterms
\nthe
term P(s)/Q(s)
transform
\nform
(7-37)\n
as\n
/\302\253
If the
^
.. .
ais\342\200\235-1 +
denominator
and
respectively,
Q(s),
...\n
= v(t)
+ ani
condition
initial
of this equation
form
numerator
the
+
\302\243[t)(Q]
=\n
\na<>sn
general
on_i
solved
be
m
The
.
di
.
... +
of the Laplace transformation,
for the unknown as\n
a result
may
.
dn~H
ax\302\256\"-1
+
...
+
On
=
a0(s
+
\302\2530...
+
(\302\253
s\342\200\236)(7-40)\n
Art.
7-5\n
coefficients
the
Because
TRANSFORMATION\n
LAPLACE
THE
...,
do, di,
are
d\342\200\236
\nfunctions of network parameters R, L,
\nrestricted to be: (1) simple and real,(2)
The
rules
the
three
(1)
\tIf
\nof
for
possibilities
the
roots
are
all
\nfraction
+
Si)
repeated), then the partial
Kn\n
_Kj_
82) .
+
(\302\253
terms
is\n
expansion
P(a)
(s
or\n
6-3.\n
fractions are givenin
(that is, not
simple
pairs,
complex
in Art.
I(s) by partial
just mentioned:\n
expanding
(being
of Q(s) are
roots
the
C),
real
and
positive
conjugate
as discussed
nonsimple)
(or
\trepeated
(3)
and
135\n
. . (s
+
8
Sn)
+ Sj
8
*1
+
j
8+
8n\n
(7-41)\n
\tIf a
(2)
r times, the partial
is repeated
one (repeated)
root is\n
this
root
to
\nsponding
-f-
(8
conjugate
complex
(7-42)\n
(8 +
(* + Si)2
Si
Si)r\n
terms for every other repeatedroot.\n
rule may be given for two rootswhich
form
special
For this case, the partial fraction expansion
pair.
\tAn important
(3)
\na
Si)r
Klr\n
be similar
will
there
and
8
corre\302\254
Kxi
Kxx
_
P(s)
fractionexpansion
\nis\n
P(s)
\302\253
+
+
Qi(\302\253)(\302\253
^\n
a
+
j<a)(s
\342\200\224
ju>)\n
K
Kx\n
(s
where
Kx*
\nroots
\nAn
are
expansion
the
As
+ b)(s2+ cs +
+
In an expansion of a
\nof
the
necessary
examples
K\342\200\231b.\n
quotient
any
\t\n
(s
\nSeveral
(7-43)\n
to expand second-
as\n
As
be
...
+
jta)\n
convenient
more
sometimes
is
terms
denominator
Qi(s)(s2 + as
may
\342\200\224
sufficientto expand
three rules are
above
it
PW
\nit
a
roots.\n
polynomials,
\norder
+
(s
jui)\n
the
complex conjugate of Kx. In other words,when
fraction
so are the partial
expansioncoefficients.
conjugates,
is
shown
above
of the
necessary for each pair of
type
Although
\nof
a +
!*\n
is the
conjugate
\ncomplex
+
+\n
+
ai)2
s2
d)\n
B\n
as
+
Cs
B\n
+
+ b
+
D\n
as)2
+
\n+
&>i2
quotient
of
(s
+
polynomials
+\n
by
2
Cs
+
D\n
+
cs
+
+
...
+\n
d\n
(7-44)\n
\302\25322\n
partial
fractions,
a combination of the three rulesgiven
above.
will illustrate
the expansion and the determination
to use
136\n
TRANSFORMATION\n
LAPLACE
THE
the
Consider
of polynomials,\n
quotient
I(s)
first
appropriate
+
(s
denominator
\ncommon
\nor
\nlinear
equations:
\ntions,
we
\ntion
-f- 3
2s
+
3 =
=
Kx
Kx =
that
2s
may
expansion
fraction
Example
+ 2)K!
K2\n
s+
s+ 1
2\n
Placing eachfactor
the
over
\342\200\224(s -1-
l)iCj
(2Ki +
K2)\n
we obtain the
functions,
+ K2}3 = 2Ki +
1 and K2 = 1. The
the
of
result
followingset
two
these
From
K2.
+ 3
2
+
1
=
of
equa\302\254
partial
frac\302\254
1\n
s +
1
< -f
2\n
by combining the
be checked
two terms
of
the
expansion.\n
of polynomials with repeated
a quotient
consider
example,
roots:\n
\ndenominator
s
(s
is required
form
This
K2
4\n
this
For
2\n
\342\200\234I-
(Ki + K2)s +
3s
-|\342\200\234
s2
\npartial
3
is thus\n
expansion
The
3s
=
= (s
of like
2
find
-I-
+ 2)
l)(s
2s
coefficients
Equating
+
and not repeated.
of the equation gives\n
are real
roots
the
\ns2
3
+
2s
since
2s
=\n
factor the denominator polynomialand thenexpand
rule. For this example, the expansionis\n
is to
step
the
\nby
7\n
S\n
Example
The
Chap.
+
2
+
l)2
.
Kn
_-
s+ 1
K\\2\n
(s +
l)2\n
Multiplying the equation
by rule (2).
by
(s
+
l)2
\ngives\n
s
The
\n=
set
resulting
2,
or
Ki2
=
this
\nfirst
in
+
algebraic
+
can
expansion
the expansion
2
l)2
+
(s +
1. The resulting
\n(s
term
2 =
of linear
8
Again,
+
K\\2\n
equations is Kn
\342\200\224
1,
Kn
+
Kn
partial fractionexpansion
is\n
1
1
\"
s
+ 1
\"*\342\226\240
be checked, in
by (s + l)/(s
(s
+
1)S\n
this case by multiplying
the
+
1).\n
7-5\n
Art.
Example
This
where
\nsider
the
denominator
the
quotient\n
1
_
+ 5
2s
+
the
\ndenominator,
- j2)Kx+
(1
two
\nThis
1-
+
(1
+
(a
1
+
=
be solved
to put all terms
+
Kx*(8
1
-\\-j2)\n
1
Kx + Kx* =
and
= -n\n
\nbetween
and
Kx
for
solve
\nfirst
of
equation
is found by any method,it is not necessary
If the values for Kx and Kx*aresubstitutedintothe
this example,
there results\n
1\n
use
\npleting
3x\n
square.
2s
+
(s2
+ 5) =
1
the
is
the
imaginary
The three
\nto
the
cases
denominator
\nthe
\nnecessary
\nan
form
general
expansion
to
+
2s
1) +
+
[(s + o)2
5
1
fit)\n
4
=
(s
+
l)2 +
by
com\302\254
root,
and
22\n
1
_
that\n
\n82
\nb
(s2 + 2s +
-Jr\n
should be revised
tables, such terms
In this example,\n
transform
some
the
+\n
'
(s + 1+ j2) (s +
5
+ 2s +
s2
the conjugate relationship
If Ki
Ki*.
K\\*.
use of
make
not
did
we
In
0\n
for Kx and Kx*.
simultaneously
and
In this development,
so
common
a
over
k'*-th\n
To
j2)\n
gives\n
=
\nto
-
terms,\n
j2)Kx*
may
equations
Kx*\n
results:\n
j2)
of like
coefficients
the
^
(a + 1 + j2)
equation
following
1 = Kx(e+
Equating
Kx
by a factor
is multiplied
term
These
the expansion of a quotient of polyno\302\254
roots are a complex conjugate pair. Con\302\254
illustrate
will
example
each
137\n
6\n
\nmials
If
TRANSFORMATION\n
LAPLACE
THE
(s + l)2
+ 62],a is
the
+
real
22\n
part
of
the
part.\n
rules just given for partial fractionexpansion
are
restricted
in which the order of the numerator polynomial
is lessthan
If this
condition
is not fulfilled, it is
polynomial.
first
divide
the denominator
into the numerator to obtain
of
the
form\n
(7-45)\n
n is
where
\nand
TRANSFORMATION\n
LAPLACE
THE
138\n
Chap. 7\n
of the numerator, d is the order
of the
denominator,
of polynomials with the orderof the
is a new quotient
than that of the denominator such that theusualrules
the order
Pi(s)/Q(s)
less
\nnumerator
\napply.\n
an
As
example,
let\n
P(s)
direct
+
s
\nQ(s)
By
2
+ 2s +
s2
=
1\n
division,\n
s2 + 2s +
+ 1)
s
S2
+
2
+
(s
1\n
3\n
s
2
\342\200\224|\342\200\224
\ns
+
1
\n1\n
s2
+ 2
2s
+
=
or\n
in Eq.
that
so
1, Bi =
expansion
theorem\n
to the problem
Let us return
of
+
(s
first
the
As
+
(2s
common
canceling
3)(s
s.
\nof
\ns
=
+
s
\342\200\224
1, the
\nfollow the
+
\ns
(7-48)\n
2\n
+
the coefficient K x is not multiplied
an algebraic factor that can
is merely
of K2 reduces to zero and we
coefficient
same
result
+
2
as found previously.
set s =
3\n
=
s
K!\n
1\n
\342\200\2242
in
+
by any
have
can
function
value.
any
for Ki
solve
= 1
^ri
,\n
multiply Eq. 7-46 by
same pattern,
Ki} we
3
+
\302\253
+
evaluate
(7-47)\n
2\n
s_\302\2611
2\n
2s+
To
+
\"^2s
Als+1
+ 3
\ns
the
1
factors,\n
2s
is
(7-46)\n
2)\n
equation,
Now
written\n
Ki\n
(s +1)
2)
K,\n
which
.
1).\n
\342\200\234
\ns
this
was
3 which
Example
+
l/(s
as\n
2s
In
1\n
(s +
the equation by (s + 1)
s+
+ 1) _ K s + 1 .
\n(s+l)(s
or,
+
=
K\\
+ 2)
l)(s
multiply
step,
1\n
+\n
1, andPi(s)/Q(s)
-j- 3
2s
s
s
7-45, Bo =
Heaviside's
7-6.
+
1\n
+
\n3
1
K2
2)
as\n
P-49)\n
To evaluate
(s +
If
to
and
to
obtain\n
2\n
+
Ki\n
(7-50)\n
3+1\n
order
to
reduce
the coefficient of Ki\n
7-6\n
Ait.
to zero.
TRANSFORMATION\n
LAPLACE
THE
139\n
Then\n
+
2s
3\n
4
+
3\n
2
+
1\n
K,\n
1\n
\342\200\242
+
This
\nHeaviside's
'partial
when
\nexample
expansion
has
Q(s)
S
Q(s)
then
of the
any
\ncoefficient
In
form
given
=
in
this
(s
by
Multiplying
s =
and when
the
follow
+ l)2
+
\342\200\224
1,
same
+ 2
+
l)2
In other words,to
to
the
find
this
If,
in
2 =
\nbe
evaluated.
s =
(s + l)tf
and
we
\nindeterminant
are
+
now
K\342\200\230~\302\260V\n
l)2
problem
0 =
Kn
evaluated
s+2
(s + l)2
Kn
(7-55)\n
as Kn =
1. If
we
attempt
is,\n
<7-56>\n
sTl
becomes
infinite
and Kn cannot
can be resolved if
we return to
to s. (This is a reasonable
thing
before to remove trouble with
with
Differentiating
1
constants
term
respect
differentiation
forms.)
(s +
Kn, trouble develops. That
+
with
used
have
one
1,
the
differentiate
Kn\n
+
\342\200\236
evaluated
\342\200\224
However,
Kn
s + 1^
to evaluate
pattern
equation,
theorem
expansion
gives\n
is readily
Kn
_\"
-
The
by
and setting s
coefficient
Heaviside\342\200\231s
7T1
do:
be evaluated
example,\n
(s
s
\nto
Sn\n
+(7-53)\n
equation,
s
Q(s)
7-55
(7-52)\n
+
with nonrepeated denominator roots. To
case of repeated roots, considerthe
P(s)
\nEq.
can
Kn
...,
K%,
of that
[(s
of the
discussion
our
*'
to functions
only
\napplied
\nto
S3
of the denominator.
root
S
Kj,\n
the
\nstart
K1, Kif
denominator
the
the
of
value
Si
+
Kn\n
,
S+
S + 82
coefficients
by
\nmultiplying
,Kt
K2
K,
P(s)
\nthe
example
manipulation
\nalgebraic
(7-51)\n
has shown will eliminate muchof the
is known as
the coefficients,
of
evaluating
will work as in this
The
method
method.
roots. In general, if\n
no repeated
the
which
method,
--2\n
\342\200\242
+ 0
or Kn =
and the partial
1,1\n
s + 1^
to
respect
(s +
s,\n
1\n
fraction expansion
is\n
(7-57)\n
l)2\n
140
LAPLACE
THE
general case
To consider a
P(S)
R(8)
=
Kn
=\n
+
(s
Q(\302\253)
of
Kj2
(8
8i\n
+
+
,
Kin\n
...\n
\342\200\242
\342\200\242
4\342\200\234
+
(\302\253
8y)2\n
\302\253;)\342\200\231
\nKir\n
...
\n+
n is
+
partial fraction expansionand
term in the
any
(7-58)\n
+\n
(a
where
7\n
Chap.
let\n
roots,
r-repeated
,
8 +
8y)r
TRANSFORMATION\n
8i)r\n
is
R(s)
defined
\nas\n
-
B(,)
this
+
Kn(s
\342\200\224
sy,
+ Kir
to be usedto
in the
terms
all
...
+
sy)r~2
visualize the method
s =
we let
If
gives\n
+ Kiiia +
3y)r_1
we can
equation,
coefficient.
\neach
(s + a,)r
by
\342\200\224
R(&)
From
7-58
Eq.
Multiplying
(7-59)\n
W)(*+s,)r\n
differentiate
\nexceptK^, which can be evaluated.
\nonce with respect to s. The term
but
vanish,
\nwithout a multiplying function of 8. Again,
the
Next,
iCy,
\nby
letting
7-60
(r
\342\200\224
n)
\342\200\224
s,.
let 8
and
times
the
find
To
be
can
remain
evaluated
differentiate
/\302\243,\302\273,
Eq.\n
\342\200\224
\342\200\224
then\n
\302\253y;
1
dr-^Ria)\n
=\n
Kin
term
general
equation
fCy,r_i will
ICy,r_i
8 =
evaluate
disappear
equation
will
(7-60)\n
\342\200\224
(r
d,8r\n
n)l\n
I*
(7-61)\n
-
-\342\200\242*\n
or\n
(7-62)\n
actual
The
\nplexity
of
use of this
this
38
+
+\n
'
+
(\302\253
the
Multiplying
2 8*
this
\nFrom
+
38 + 2
l)2
(8+
(s + 1)*,
by
equation
= Kn(a +
'
+\n
(8+1)
1)\302\273
we
1)*+
2s2 + 38 +
+
Kn(8
2
- 3 +
\342\200\242--i\n
we
Next,
with
differentiate
48
so
that\n
(8 +
(7-63)\n
1)\302\273\n
have\n
=
2\n
+
K\\a
to 8 to
respect
3 =
2Kn(8
= 48 +
obtain\n
+ 1) +
Ku\n
-1\n
3\n
\342\200\242--i\n
com\302\254
K\n 13\n
1)
+
Ku
equation,\n
Kl% =
the
consider\n
K\n 12\n
K\n ii\n
2\n
from
appear
might
For example,
equation.
general
2\302\253*+
idea is easier than
2=
1\n
THE
7-6\n
Art.
=
4
\nThe
fraction
partial
2
+ 2
3s
rules
\nboth
i
used.
the
of
+
(s
In
this
be
\nmay
s
-p
(7-65)\n
+ 3)
is\n
+
1\n
'
+\n
l)2
2\n
Ki\n
s +
(7-66)\n
3\n
1\n
-3\n
l)2, we
by (s +
7-66
Eq.
of
K\n 12\n
\302\245TT)i\n
Multiplying
roots, a combination
then\n
=\n
K2
i)*\n
evaluated by Eq. 7-53and Kn and Kn
7-62;
Eq.
(7-64)\n
(* +
+\n
K2 may be
from
l)2(s
\302\253
+
1\n
2\n
K\n n\n
+ 3)
l)2(s
+
expansion
2\n
\342\200\234H
expansion,
found
fraction
partial
s
+
(s
Q(s)
form
.
+ i)>+
(*
s
POO\n
\nThe
-1
+\n
'
simple and repeated
As an example, let\n
both
be
may
give\n
= 2
Kn
,
\302\253
+
contains
to
141\n
is\n
(*+i)*
Q(s)
or
2Kn
expansion
2\302\2532+
If
TRANSFORMATION\n
last equation
the
differentiate
we
Again,
LAPLACE
4\n
have\n
(7-67)\n
Constant
is evaluated
Kn
s
Kn
=\n
-{- 2
Kn
be found
will
+
(s
1
-
+
\n(\302\253
and
this
\nthe
d_
[
ds
|_
term
\nentiation
\ndifferentiation
terms
+ 1)21
(s
s
-f-
vanishes
contains
multiplying
vanishes
all
to
common
\nremains
(s + 2)
(s
factor
3
thus\n
2\n
1
+
K\"
K'l\n
\ns
1 because
when s = \342\200\224
in the numerator. In
(s + 3)2(8 +
\342\200\2241:\n
(s + l)2
\342\200\242
1) -
(s
+
-f-
an
the
l)2
3\n
(s
+ 1) term
example\n
\342\200\242
1\n
(s -J- 3)2\n
J
when
1;
Eq. 7-67beforelettings =
3)2\n
of K2
coefficient
The
\342\200\242
-I\n
by differentiating
\342\200\242
3)
\342\200\224
1\n
3\n
\ns \342\200\234j-
and
=
by letting s
directly
s =
\342\200\224
1,
because
each
term
in the
differ\302\254
order
the
+ 1). This is always the case,since
(s + Sj)T is higher than the numberof times
is required.\n
of
THE LAPLACE
142\n
be
can
\nexpansion
Chap.
(7-68)\n
F(s)\n
for
roots
simple
and
(nonrepeated)
as\n
Kjk\n
=\n
F(s)
a single
be
\nnow
\342\200\224
root,
Sj,
the
for
found,
of
\ntransformation
The
f(t) may
corresponding
by taking the inverse Laplace
case,
general
Sj)k\n
times.
r
repeated
(7-69)\n
+
(s
for
7\n
the coefficients of the partialfraction
transform equation can be written\n
all
methods,
and the
found
these
using
By
TRANSFORMATION\n
as\n
F(s)
+
+
M\n
S.)
\342\200\242'
(7-70)\n
Q(s)\n
\342\226\240?
-1\n
the
as
for simple roots.
solution
time-domain
Likewise,for repeated
\nroots,\n
=
f(t)
1
e-'1
n
^\n
=
dsr~n
n)!
\n(r
tn~l
dr~nR(\342\200\224Sj)
\342\200\224
(7-71)\n
\342\200\224
(n
1)!\n
1\n
this equation, is the root that is repeatedr times.By
for the case of both simple and repeatedroots,
a
equations
\nsolution is obtained
in the form originally givenas Heaviside\342\200\231s
Sj, in
where
\nboth
\nsion
using
general
expan\302\254
theorem.\n
be
used
partial fraction expansion
\nto give
for finding the inverse transform the
a simplified
procedure
\nterms
for a conjugate
roots
complex pair of roots. Supposethat
\nhave
a real
is
part a and an imaginary part, co.The first
Heaviside
of the
method
The
may
of
these
coefficient
\nevaluated
the
by
procedure,\n
P(s)\n
(s
\342\200\224
a
-f-
=
jta)\n
and
the
second
Serf\n
as\n
K''
=
=
W){\302\260-a-ju)\n
The
inverse
(7-72)\n
a \342\200\224jta\n
8\342\200\224
Q(s)\n
of these
transformation
h(f)
=
Rei9e{a+Mi
two terms
+
Re~i6e(a~iu)t\n
Re+*9\n
(7-73)\n
gives\n
(7-74)\n
Art.
LAPLACE
THE
7-7\n
to the
may be rearranged
This equation
143\n
TRANSFORMATION\n
form\n
i
p\342\200\224/(\302\253f+0)\"]\n
-
IT
=
R
The
factors
\nthe
magnitude
7-7.
and
0 in this
and
phase
(<at
J\n
+ 0)
(7-75)\n
equation are easily foundin
of the coefficient
angle
by the Laplace
solutions
total
of
Examples
cos
2Reat
as
6-72
Eq.
K\\.\n
transformation\n
6\n
Example
an
As
\nfraction
been
have
expansion
solution, now that the methodsof partial
consider the differential
reviewed,
total
of the
example
\nequation\n
The
-
-
\302\253\342\200\231/(\302\253)
\302\273(0+)
\nin
this
\ndi/dt
equation.
*\342\200\242(0+)
=
+
/(s)(s2
or\n
s2
W
s(s
To
evaluate
be expanded
may
6s
+
+ 2 +jl)(s
Ki,
evaluate
K2l
_
~
multiply
_
s2 + 6s +
s2 + 4s +
the
-s\n
found:\n
-
(0+)
2\n
the transform equation
+ s+
+ 4s +
s
5
5\n
equation
to\n
6\n
5)\n
as\n
K2
\"*\342\226\240
s
2
+
by s and
the equation
\n1
are
by partial fractions
+ 2 - jl)
multiply
p-
To
+ 5
Jt
+ 5)=-
/(.).s*+6s+5\n
equation
is\n
conditions are automatically specified
from the physical system, \302\253(0+) and
s(s!
This
equation
- i(0+)] + 5I(s)=
simplifies
4s
(7\"76)\n
differential
and
1
conditions
initial
these
Inserting
^
the following values
Suppose
(0+).
\342\200\234
\302\256*
+ 4[s/(s)
(0+)l
Jt
initial
required
We must
know,
the
that
Notice
+
<5
of this
transformation
Laplace
4
+
3P
=
^
+jl
let s
=
0.
K2*\n
s+ 2
-
jl\n
Then\n
1
\n\342\200\242\302\253*o\n
by (s +
2 + jl)
and
let\n
THE
144
s =
\342\200\224
\342\200\2242
jl
s2
\n+
The
+ 5
obtain
fraction
- i
6
\nand
of
\342\200\224
90\302\260for
second
the
2-
' s
+
+\n
+ jl
jl\n
equation, we take the
the first term and use Eq. 7-75
and third terms to give the
transform
this
transformation
\nLaplace
2
+
=\n
becomes\n
expansion
+\n
= j
-j2\n
(-2-jl)(-j2)
-2-,i
from
i(t)
2\n
j2
partial
m
To
\342\200\224
\342\200\2244
2\342\200\224jl)\n
complete
7\n
Chap.
as\n
6s
+
=\n
Kt
TRANSFORMATION\n
LAPLACE
with
inverse
=
R
1
solution\n
=
i(t)
+ 2e_2< sin t
1
(7-77)\n
7\n
Example
this
For
circuit with the
a series RLC
consider
example,
\nVo
differential
current
\nthe
Fig.
series
RLC
+
Lf
circuit,
voltage
in
indicated
as
\nThe
7-5.
to
charged
initially
capacitor\n
7-5.
Fig.
for
equation
is\n
i(t)
JK +
=
\302\260\n
<?/ld\342\200\230
(7-78)\n
the
\nand
corresponding
- i(0+)]
L[sl(s)
The
\nL
\nand
1 henry.
if
C
The
is initially
equation
[I(s) + g(0+)]-
initial
as C =
specified
farad,
=
R
of
charged
The
0\n
2 ohms,
and
inductor,
polarity),\n
Vo\n
_
~
Cs
volt.
|
current t(0+) = 0 because the
to voltage V0 (with the given
ff(0+)
or\342\200\224l/sifVo=l
is\n
+ RI(s) +
been
have
parameters
=
transform
s\n
transform
for /(s) then
equation
becomes\n
=
s2
or,
completing
the
+ 2s
2\n
+
i
square,\n
/(s)
=
+
(\302\253
\nUsing transform
+
pair 15 of page
i(t)
-
i)2
146,\n
\302\273
er*
\302\243-\302\273/(\302\253)
sin
t\n
(7-79)\n
Art.
7-8\n
Example
In
\nWith
network
shown
the
network
parameter
Kirchhoff
7-6, the switch
in Fig.
the
the
is closedat
=
t
0.
values
voltage
rAAAr-nnnp\n.4
100
equa\302\254
lh
I
lh\n
are\n
\ntions
di,\n
+
^
-
are
initially
+
10ti
are
we
that
0\n
-\\- 20
s
\342\200\224\302\273
partial
fraction
inverse
s +
The initial
7-8.
The
and final value
theorem
value
initial
\nin
network
analysis.
To
\nto
approach
infinity
in the
lim
In
\nare
\nexists.
this
writing
/
and
the
two
-
40s
+
300)\n
3.33
5
s
s+
1.67\n
+
10
gives iz(t)
5e-10\342\200\230+
is\n
s
+
30\n
as\n
1.67e-30t\n
theorems\n
and final value theorem find frequentuse
derive the initial value theorem, we allow
s
for the transform
equation
f'(t)er\302\253
dt
=
lim
[sF(s)
of a derivative
- /(OH-)]
as\n
(7-80)\n
we assume that f(t) and its first derivative
that
the limit of sF(s) as s approachesinfinity
has zero value for s\342\200\224>\302\253, and
is\n
integral
/(0+)
equation,
transformable
Since
3.33
+
s(s2
of this equation
+ 30)
=
*2
of
20\n
transformation
Laplace
0\n
1000\n
-10\n
1000
The
=
as\n
0
expansion
s(8 + 10) (s
and
100/s\n
-10
The
+ 20) J,(s)
(s
to find the current *2 as a function
may be found from the last
/2(s)
20
+
\ns
+
-10/i(s)
s\n
-10
I2(s)\n
8.\n
before the switch is closed,both ii
transform equations may be written\n
determinants
by
of Example
Network
7-6.
rtg.
current
equations
\nalgebraic
=
required
transform
The
100u(<),\n
- 10/2(s) =
20)/1(s)
Suppose
\ntime.
KH'a =
is unenergized
and the
zero,
network
the
(s
20\302\273i
+ 20*2-
^
\n*2
145\n
8\n
\nshown,
If
TRANSFORMATION\n
LAPLACE
THE
TRANSFORMATION\n
LAPLACE
THE
146\n
op
Table
Chap.
Transforms\n
m\n
1. u(t) or
m\n
1\n
1\n
2.
t\n
s2\n
n\342\200\224
1\n
1\n
\t\n
3
=
n
-
integer\n
sn\n
1)!\n
(\302\273
1\n
4.
e\302\260\342\200\230\n
\342\200\224
a
s
\n1\n
5.
teat\n
\342\200\224
(s
a)2
\n1\n
-
fn\342\200\224lpat\n
e\n
\342\200\224
a)n
(s
(n-l)r
\n1\n
7.\n
'\n
\342\200\224
a
\342\200\224
bK
a)
(s
\342\200\224
b)\n
(s
1
e-at\n
8.\n
\342\200\224
(b
(c
a)
a)\n
\t\n
+ b)(s +
(s + a)(s
\342\200\224
o\342\200\224bt\n
+\n
-
(a
b) (c
-
b)\n
o\342\200\224ct\n
+\n
\342\200\224
(a
9.
1
-
c)(b
\342\200\224
c)\n
\342\200\224a\n
e+ot\n
\342\200\224\n
s(s
1\n
10.
\342\200\224
cat
sin
(a2
+
s2
\n(a\n
\ns\n
11.
COS
(at\n
+
S2
12.
1
\342\200\224
COS
(at\n
\ns
sin
(cat
+
cos
((at +
+ (a cos
0
sin
&)\n
+
S2
s
14.
(a2)
+
s(s2
13.
(a2\n
cos
0)\n
(a2\n
\342\200\224
0
co sin
+
\ns2
m2\n
CO\n
15.
e~at
sin
cat\n
(s
+
a)2
S
16.
e~at
cos
+
(at\n
(s
+
a)2
a\n
17.
sinh
at\n
S2
\342\200\224
a2
\nS\n
18.
cosh
at\n
0\n
+
co2\n
a\n
+
co2\n
0
c)\n
7\n
7-8\n
Art.
THE
of
independent
TRANSFORMATION\n
LAPLACE
a,\n
lim
=
lim
(7-81)\n
which we conclude
from
f{t),
= 0
- /(0+)]
[aF(s)
oo\n
8\342\200\224\342\226\272
But/(0+)
147\n
that\n
tr->0-f*\n
lim
=
sF(s)
lim f(t)
*\342\200\224\302\273\302\273
In deriving the
is
0+
infinity.\n
start
we
dt =
fit)e~9t
f
lim [sF(s)
8\342\200\224>0
JO
t are
and
s
to
equal
final value theorem
lim
Since
Equation 7-82 shows
the limit of the product
previously.
from
the
but let s approach zero.
theorem,
derivative
are transformable,
we write
first
its
and
\nf{t)
\342\200\224
value
initial
the
\nas
at t
s approaches
as
\nsF(s)
mentioned
of f(t)
value
the
\nthat
restrictions
the
to
subject
(7-82)\n
<\342\200\224\302\273o+\n
8\342\200\224>0\n
equation
Assumingthat
.\n
- /(0+)]
(and e~9t\342\200\224>
1,
independent
same
as
s
(7-83)\n
\342\200\224>
0)
the
integral
\nbecomes\n
/'(<)
0
dt
=
/.\n
This
be equated
may
expression
\nlim
lim /(f)
00\n
t-+
to Eq. 7-83 to
-
[sF(s)J
- /(0+)\n
= lim [/(()]
(7-84)\n
give
-\n
(7-85)\n
00\n
t\342\200\224\342\226\272
from
we conclude
which
that\n
lim
=
[sF(s)]
8\342\200\224\342\226\272O
lim
[/(<)]\n
oo\n
t\342\200\224+
(7-86)\n
final value theorem. This result holdsprovided
\nall roots
of the denominator
of sF(s) have negative real parts. Because
the final value theorem does not apply in the case
\nof this
restriction,
sinusoidal
\nof
because
the denominator
roots of the trans\302\254
excitation,
are purely imaginary.\n
\nform
sinusoid
of the
which
is
as the
known
READING\n
FURTHER
An
historical
interesting
\nHeaviside\342\200\231\342\200\231
by
\n1929),
pp.
is contained
Behrend
\nside\342\200\231s
Operational
173-208.
summary
Calculus
(McGraw-Hill
Heaviside\342\200\231s
original
titled
\342\200\234The
as an appendix
Book
writings
of
Work
in
Oliver
Berg\342\200\231s
Heavi\302\254
Co., Inc., New
York,
have
recently
been
\nreprinted as Electromagnetic Theory (Dover Publications,
York,
contain
an extensive presentation of his method. For
\n1960) and
New
a\n
more
\nneering
\nChap.
\nWiley
\nModem
\nCo.,
in
found
be
New
Controls
in Linear Systems
Transients
Barnes,
pp. 334-356 and in
New York, 1942),
Inc.,
of Automatic
\nPrinciples
and
tables
transform
Extensive
1950).
York,
Gardner
& Sons,
Wiley
\n(John
\npp.
7\n
Laplace transformation than is offered
to the following: Wylie, Advanced
the
student
is referred
Engi\302\254
Book
New
Mathematics
(McGraw-Hill
Co., Inc.,
York, 1951),
in Linear Systems (John
Gardner
and
Transients
6;
Barnes,
&
New
York,
1942), beginning on p. 93; Churchill,
Sons,
Inc.,
in Engineering
Mathematics
Book
(McGraw-Hill
Operational
New
1951) and Thomson,
Laplace Transformation
Inc.,
York,
\n(Prentice-Hall,
Inc.,
\nmay
Chap.
of the
treatment
complete
\nhere,
TRANSFORMATION\n
LAPLACE
THE
148\n
Nixon,
Inc., New York, 1953),
(Prentice-Hall,
371-399.\n
PROBLEMS\n
7-1.
\nof
f(t)
into
following transform
7-1 and integrating.\n
Eq.
\302\243[cos
In each of the
\nfor
the
pair by substituting
the
Verify
^^5\n
the
of Prob.
procedure
of the following
pairs
value
=
problemsthat follow,repeat
transform
various
orf]
the
7-1
table.\n
F{*)\n
m\n
2\n
7-2.\n
t*\n
s*\n
a\n
sinh
7-3.\n
at\n
s2
-
s2
-
a2\n
8\n
cosh
7-4.\n
at\n
a2\n
ta\n
e~at
7-5.\n
sin
tat\n
(s
+
a)2
s
er**
7-6.\n
cos
s
sin
7-7.\n
(tat
+
(tat
-f
sin
a)2
to
prove
7-10.
\ncharged
\nAnswer.
Letting
Eq.
= f(t)
7-13.\n
u
+
<o
\ns2
+
\302\2532\n
the
voltage
(Vf2R)e-t/BC.\n
cos
0
cos
0
\342\200\224
ta
sin
0
0)\n
and
Rework Example 1,
to
ta2\n
6
\ns2
7-9.\n
+
0)\n
s
cos
7-8.\n
+
to2\n
a\n
&it\n
(s
+
V/2
dv\n
=
6~*\342\200\230
dt,
+
\302\2532\n
integrate
Eq.
7-1 by
is
assuming that the capacitor
with the upper plate positiveat t
parts\n
originally
\342\200\224
0.
Chap.
THE
7\n
In the network
7-11.
\nLaplace
7-12.
\nfrom
\nthe
0.
In the network
shown in the
to
a
at
established
Laplace
at t
is closed
K
0,
a
state
steady
is moved
having
pre\302\254
(V/Ri)e~<'Rl+Rt)t/L.\n
to
charged
current
the
for
Solve
K
the current i(t), using
shown, C is initially
\342\200\224
0.
Answer.
\ntransformation.
switch
the
\342\200\224
Answer.
In the network
7-13.
figure,
a. Solve for
position
transformation.
i(t) using the
(V/R)e~t/RC.\n
b at t
position
Vo, and
to
charged
current
the
for
Solve
is
C
figure,
Answer.
been
\nswitch
\342\200\224
transformation.
position
\nviously
shown in the
at t
closed
is
K
switch
\nthe
149\n
TRANSFORMATION\n
LAPLACE
sin
(V/y/L/C)
i(t), using the
V0.
The
Laplace
(t/y/LC).\n
L\n
Prob.
to
error.\n
2s
s2
\nrw\\
Cc')
^
s3
1
_
- 1
2
2s
+ 3s2 +
^\342\200\224
s2
,
1\n
s+l+s-l
+
+ 5s +
6
=
(d)4-^r
(t/y/LC).\n
the following partial fraction expansionsby expanding
of polynomials in partial fractions. Two of the set
quotient
7s
w
positiona
Check
\nthe given
/_%
^
shown, the switch K is movedfrom position
=
t
0 (a steady state existingin
before
the current
i(t), using the Laplace transformation.
cos
(V/R)
7-16.
in
for
Solve
\nAnswer.
\nare
b at
position
\nt \342\200\224
0).
7-14.\n
network
the
In
7-14.
\na
Prob.
7-13.\n
r^r
= 1
s
2
\342\200\224
-
2
F+\"2 +
,
3\n
s + 3^s
+ s +
p
i\n
+
2\n
\342\200\2243\n
7+3\n
+
1) _
+
1
2(s
(e)\n
s2
s2
4s
+
(f)\n
+
s(s
\n3s8
(g)\n
GO\n
7-16.
\nthe
2\n
+
(s + l)2 s
2
- 1+
2
+ s2 + s
+ 9= 1 ,
,
1_
\342\200\224j
1\n
-
(s
l)2\n
+j\n
,
s
s + jZ
s2
s
\342\200\224
j\342\200\2303\n
by partial fractions and find
transformation,
f{t) = \302\243rlF(s).\n
functions
following
inverse
corresponding
1
s
the
Expand
jl\n
1
3s + 2
l)2
5s2 + 9s
s2(s2 + 9)
s2(s
-
88
Laplace
3s\n
Answer.
F(s)\n
(a)
+
(s2
F(s)\n
(b)
\ns+
1\n
+
2s\n
s2
(c) F(s)
sin
2<)].\n
(d)
F(s)\n
\n+
Answer.
-
2
2s
=
f(t)
+
i(l
Answer.
+
\342\200\224
cos
f(t)
+ 4)
l)(s2
=
f(t)
1) (s
+
Answer.
+ 2)2
s8(s2
+
2) (s2
(s2 +
l)s\n
(s
W\n
(g)
the
Solve
\nmation
\nd2i\n
y
d?v\n
7-18.
Answer.
+
given
+ 4v
= sin t
21
\342\200\224
\\t
+
7-19.
ST
q
=
+
H\342\200\230
\nthat
\nilarity
\nteristic
\nroutine
2er>
in
the
equation
as
+
any
cos
2
\342\200\224
e-2\342\200\230(l+
\342\200\224
\342\200\224
f(t)
= \\e~2t
=
f(t)
1
t2/2
21\n
+
t).\n
cosh
+
f.\n
+ |sin2L\n
2t
cos
\302\243
cos
21.
L\n
+ K2e~t
v =
Answer,
transfor\302\254
(where specified).
=
i
+ * sin
by the Laplace
conditions
Answer,
\342\200\224
^t cos t
equations
initial
e2*.
sin
~ <! +
%
- 25 +
Ki sin 2t
+
is24.\n
K2\n
t.\n
=
\302\273.7(0+)
= -2.
(0+)
+
Answer.\n
2.\n
6-13 (d), using
transformation
Laplace
the
in
+ -J
Prob.
Solve
7-20.
2t
sin
\342\200\224
?et(
at1\n
cos
2t.\n
cos
4)\n
differential
25 +
e~l
=
1\n
- t =
at*\n
\302\243[1 +
=
f{t)
f(t)
Answer.
following
to the
subject
7-17.
1)
2s +
+
\ns2
F(s)\n
(f)
Answer.
-
F(s)\n
\342\200\224
5)\n
\n1\n
(e)
t
e~2t).\n
1\n
(s
7\n
\342\200\224
1+
s +
l)2
s2 -
-
s
1
__
Chap.
1 -jl\n
1 + j1
s + j'l
+ 1
TRANSFORMATION\n
LAPLACE
THE
150\n
of the driving
form
give
other
no
force
concern\342\200\224the
problem.\n
the Laplace transformation. Note
method, special conditions of
sim\302\254
v(<)
and
solution
the
of
roots
such
of the
charac\302\254
a problem
is as
Chap. 7\n
TRANSFORMATION\n
LAPLACE
THE
151\n
In the series RLC circuit shown below,the
applied
sin t. For the parameter values specified,findi(t) if the
2 sin t \342\200\224
e~*
closed
at
t +
t = 0. Answer. i{t) \342\200\224
\302\243(cos
7-21.
\nis
v(t)
\nK
is
sin
\n36\"*
V
0, a switch is closed, connectinga voltage
a series RL circuit. By the methodof the
show that the current is given by the equation\n
t =
At
\342\200\224
cat to
sin
\ntransformation,
=
i
\\
=
Z
where
sin
\342\200\224
(\302\253<
4)
in
the
figure,
Prob.
the
\nof
mechanism
is represented
\nthe
advantage
studied
(to
equation
for
\neffects
\nform
to
\nnormalized
7-24.
\nDr.
e~Rt/L\n
<f>
=
tan-1
^\n
K\n
the following quantities
are
duals:\n
7-23.\n
This
etc.). The concentration of the
and the voltage va(t) at node a is the dual
(kidney,
as V0
stream. The analognetworkhas
elements
may be readily changed and the
of the saving of cats). Find the
say nothing
Va(s) with the coefficient of the highest-orderterm
of drug
that
the
amount
the
Laplace
volume of drug-containing fluid, R\\ is the \342\200\234resist\302\254
of the drug from the pool to the
C2
stream,
passage
of the blood stream, and R2 is equivalent
to the
volume
\nbody\342\200\231sexcretion
dose
source
blood
the
\nrepresents
\ndrug
\342\200\224
the
represents
\nance\342\200\235
to
+
+ (a>L)2 and
\\/R2
shown
network
Ci
t
of the University of Utah Schoolof Med\302\254
Woodbury
In
use of an electrical analog in studies of convulsions.
made
has
\nthe
cos
L. A.
Dr.
7-23.
\nicine
switch
t).\n
7-22.
\nv
voltage
=
in the blood
trans\302\254
unity.\n
problem
Woodbury\342\200\231s
analog.
is a continuation
The
following
of Prob. 7-23 concerning
constants
for the network
are\n
Ci
selected:
\nohms.
\nvA(t),
a
\nas
R2 = 5
and
megohms,
Vo
\342\200\224
100
meg\302\254
is closed at / = 0, solve
of drug in the bloodstream,
volts
for
1286\342\200\230.\n
the time C when the concentration of drug in
Find
for Prob.
7-24, is a maximum. (This information is desired
second
dose may be given at that time to buildup
16.7
to the point
where a convulsion is induced.)
sec.\n
If a second dose (the voltageequivalent
a magnitude
blood
\n(Note:
having
volts)
is injected
tm
in
found
as
the maximum
that the total
giving
the
made.)
sec.\n
\n25
K
\nswitch
\nthe
network
\nFor
the
\nthe
With
\nattains
\nFind
equilibrium.
the
voltage
across
time
t = 0,
(a)
find
on
(b) find
shown
values
element
diagram:
ii(t),
#\342\200\230.\n
shownin
the switch is movedto
the
position
R2 a$ a function of
Prob.
unenergized.
previously
\342\200\224
4.54e_28
(a).
ix = 3.33 + 1.21e~#,36\342\200\230
switch
K in position a, the network
At
at
closed
is
shown, the
t = 0 with
network
the
In
7-27.
Answer
voltage\n
plus the voltage
at the time the addi\302\254
14 volts in
Answer.
plates
is
\ntion
obtained?
vA
volts
100
then
\non
7-28.
7-25, what will be
Prob.
of time, and what will be
the second dose we will assume
a function
In
at t =
is
\nit(t).
concen\302\254
Answer.
7-26.
as
9
the
\ntration
\nvA
=
Rx
7\n
so
a
100
ni,
the
\nstream
\nof
C2 = 8
1
Chap.
and
the switch
the
of
concentration
the
equivalent
\342\200\224
Answer. va(t) = 13e_002l5\342\200\230
of time.
13e-0
function
If
7-25.
\nthat
=
TRANSFORMATION\n
LAPLACE
THE
152\n
figure
b.
time.\n
7-28.\n
0 with
resulting from closing the switch at t \342\200\224
the
The
=
\ncircuit
circuit
constants
are: Li 1
unenergized.
previously
\342\200\224
= 4 henrys,
=
=
M
2
1
L*
R*
henrys,
\nhenry,
ohm, V = 1 volt.\n
7-29.
Find
ii(t)
Prob.
7-29.\n
CHAPTER
The time-domain
in
considered
\nstudies
will
\nsources
and
\ndescribed
\nin
by
8-1.
Fig.
forces has
cases: (1) single pulses
which recur a finite
functions
(2) time-varying
waveforms,
which cannot be
waveforms
and
recurring
(3)
times,
of such waveforms are shown
a single
Examples
equation.
of networks subjected to
The
transient
response
of
\nnumber
driving
specializing
to the following
by
sources)
voltage
related
\nand
various
In this chapter, time-domain
the driving forces (current
chapters.
previous
extended
be
to
networks
of
response
\nbeen
DOMAIN\n
FREQUENCY
THE
\nAND
DOMAIN
TIME
THE
IN
TOPICS
8\n
these\n
V\n
t\n
(a)\n
8-1.
Fig.
force
Driving
forces
using
\nstudies,
8-1. The
The
unit
unit
using the Laplacetransformation.
will be studied,
Fourier
series
step
function\n
step
function
The
followed
be
will
studies
\ntime-domain
wave.\n
square
(c)
driving
(a) pulse; (b) sectionofsinewave;\n
waveforms:
the Fourier
and
is defined
|
frequency-domain
integral.\n
as\n
-
for
related
by
0,\n
t
>
0
\nt
<
0\n
(8-1)\n
at
which changes abruptly
from zero to unit value
This
be
0.
generalizedby the definition\n
expression may
a function
t =
\ntime
-
-
\342\200\234>
u
for
\ngeneral,
a
function
step
the
step
which
function
153\n
(8-2)\n
{<:
changes abruptly at the. time t =
has unit value when the quantity
the
+a. In
(t
\342\200\224
a)\n
AND FREQUENCYDOMAIN\n
TIME DOMAIN
154\n
a
has
will
definition
\nThis
\nfunction
u(t
function
the
\nSimilarly,
\nvalue
apply
\342\200\224
u(a
t)
at
time)
increasing
(with
changes from zerotounit
a) is one that
+
is
that
one
uU)\n
a
t\n
Unit step
8-2.
t\n
functions:\n
.
t >
fl,
a)
+
\nu(l
functions\n
u{a-t)\n
|\n
Fig.
zero
t\n
a
^\n
\342\200\224a.
ult-a)\n
t\n
a\n
from unit to
changes
1\n
1\n
uW+a)
t =
at
value
the time t = a.* These two
1\n
1\n
8\n
has zero value when (t \342\200\224
is negative.
a)
for any form of the variable. Hencethe
and
value,
positive
Chap.
t <
-a;
\n\342\200\234(o^
_a\n
_
II,
\\0, t >
V\n
V\n
ult\n
u(t-o)\n
<a
t
-a)
-
o\n
u[t-\n \342\226\240b)\n
\342\200\224\342\200\224\n
\342\200\224
\342\200\224\342\200\224
\342\200\234
\342\200\224\342\200\234
r
l\n
l\n
*\n
1
-ult*-b)\n
V\n
V\n
u(6-t)
u(h-<)\n
L\n
} t\n
f_.J\n
:
i
-1\n
-
u(c-\n
c\nl
i\n
t\n
\342\200\242t)\n
t\n)
t\n
\342\200\224
u{a-f)\n
V\n
V\n
b
a
t\n
a\n
uit-a)\n
ulb-t)\n
u[b-t)>u[t-\n
a)\n
1\n
1\n
1\n
b
a
are
represented
as
\ntions
Construction
8-8.
Fig.
illustrated
a
t\n
in Fig. 8-2.
in these
of pulse
l\n
>
t\n
from unit step functions.\n
defini\302\254
The use of unit step functionswith
examples will make it possibleto represent
shifted.
Further, the unit step functions
\ncan
be
used
as building
blocks
to represent other time-varying func\302\254
such
as
a pulse.
\ntions
The construction
of a pulse from two unit step
\342\200\224
is
illustrated
in
\nfunctions
and
Fig. 8-3. A unit step function
u{t
a)
\nfunctions
\nm
Note
that
time
the
with
u(a
\342\200\224
t)
*
axis
\342\200\224
\342\200\224u(l
a).\n
unit
a
function
step
\342\200\224
u(t
two
a
\nunit
be
may
pulse
\342\200\224
a)
\342\200\224
u(t
\342\200\224
(8-3)\n
b)
a to
t
=
same
The
b.
step functionbuilding
as\n
\nblocks
=
v(t)
or
\342\200\224 \342\200\224
u(b
\342\200\224
u(a
t)
\342\200\224
v(t)
These
u(b
u(t
t)
A
wave.
square
constructing
of
representation
a starting at t = 0 is
of width
pulse
functionsin
mathematical
the
consider
waveforms,
\ntime-varying
(8-5)\n
a)
are illustrated in Fig. 8-3.\n
of the use of unit step
example
another
(8-4)\n
t)
\342\200\224 \342\200\242 \342\200\224
operations
As
\na
By taking the
figure.
of unit amplitude from t =
in terms of the unit
formed
is formed
pulse
the
155\n
functions,\n
step
v(t) = u(t
in
shown
are
b)
these
between
\ndifference
AND FREQUENCYDOMAIN
DOMAIN
TIME
8-1
Art.
the
by
given
\nequation\n
\342\200\224
u(t)
that
\nsuppose
\342\200\224
2u(t
The
subtracted.
is
a)
V\n
V\n
+
+1\n
a\n
t\n
step
\nzero
value
for
in
\ntrated
all
time
8-4(b)
Fig.
\nof
of a
of 4-1 to
\342\200\224
the
to
2a)
\342\200\224
2
u(t
square
1
at
function,
\342\200\224
a)
a. By next adding the
the waveform assumes
2a. This constructionis
+
u(t
\342\200\224
in this pattern, any square wave
(square
only
but
known
as a square wave for any value of a) form
can be synthesized.
It is clear that a square
waveform
course,
\nvariation
\ncontinues
infinitely
=
u(t)
This
waveform
\ntion
will
\nshown
be
that
\342\200\224
2u(t
\342\200\224
is shown
found
given by the
long in duration is
to be
the Laplace
a)
+
2u(t
\342\200\224
2a)
\342\200\224
2u(t
3a)
more convenient than
time
of
which
series\n
infinite
\342\200\224
= 1,
if a
+
...
This infiniteseries
it appears
transformation reduces to a closed
in Fig. 8-4 (c).
illus\302\254
(8-7)\n
2a)
following
v{t)
5 a\n
4 a\n
wave.\n
t =
\342\200\224
greater than t =
for the function\n
v(t) = u(t)
By
3 a\n
(c)\n
Evolution
value
u(t
will\n
-1\n
8-4.
the
from
u(t),
t\n
(6)
function
\nunit
2 a\n
a\n
-1\n
change
from
waveform
resulting
t\n
Ia)
then
a)
V\n
1\n
Fig.
\342\200\224
4-1\n
2a\n
a
(8-6)\n
a)
Instead of subtractingu(t
in Fig. 8-4(a).
as illustrated
\342\200\224
u(t
(8-8)\n
representa\302\254
when
form.
it
is
As\n
a
third
\nwaveform
has
of T is
a period
with
times from t =
for all other
value
zero
wave
\nsine
given
y
this
\nof
time
a
follow
\nwill
(1)
t
(2)
\342\200\224
is
wave
time
3)
t
sin
\342\200\224
ir(t
\342\200\224
1)
\342\200\224
u(t
times
a similar
after t = 3.
first
the
Hence
3).
represented
product
all
time.\n
waveform
all
this
\342\200\224
cancels
3
v(t
same reasoning, the
resented
cycle
\342\200\224
5)
sin
3 and
to
5)
by\n
\342\200\224
3)
sin
\342\200\224
r(t
second cycle
the
5 to 7,
waveform
\342\200\224
u(t
is the sum
function
each
because
\nHence
\342\200\224
r(t
waveform
total
follows
(8-10)\n
3)
wave
sine
of
is
rep\302\254\n
\342\200\224
7)
sin
of the
\342\200\224
ir(t
(8-11)\n
7)
two
This\n
expressions.
is defined only
for its
interval
respectively) and is zero all other time.
of Fig. 8-5 may be represented the
for
by
\nequation\n
t>(<)
- tift
of
as\n
u(t
The
sin
1)
inter\302\254
all other
eliminates
from
Subtracting
\342\200\224
u(t
completely
1)
the
By
\n(1
to
\342\200\224
u(t
the
u(t
by
1)
\342\200\224
in the
shown
also exists for
waveform
the
\342\200\224
1.
is
product
\nsine
We
second.
a function to
constructing
waveform
the
has
1)
\342\200\224
t
shifted
\342\200\224
but
r(t
before
\nproduct
(4)
3,
sin
times
\nThis
(3)
t
to
1
\342\200\224
Multiplying
\nat
in
procedure
1 unit
by
for
sin x(<
The function
\nval
shifted
waveform.\n
this
\nrepresent
(8-9)\n
and the time axisis
the
and by 5 units of time
step-by-step
A
+\302\273.
as\n
T = 2
example,
first wave
particular
for
the
< =
\342\200\224
\302\253to
t
sin
In
8\n
Chap.
of time-varying waveforms,con\302\254
representation
of representing
the waveform shown in Fig.8-5.The
problem
is sinusoidal
from
t = 1 to t = 3 and fromt = 5 to t = 7.\n
the
It
DOMAIN\n
of the
example
\nsider
FREQUENCY
AND
DOMAIN
TIME
156\n
\342\200\224
1)
+
sin
x(<
-
u\302\253
\342\200\224
1)
\342\200\224
5) sin r(t
u(t
\342\200\224
3)
- 5) -
sin
u(t
r(t
-
\342\200\224
3)\n
7) sin
x(<
-
7)
(8-12)\n
8-2
TIME
By
following
similar
unit
FREQUENCYDOMAIN
time function can
any
patterns,
and
series
time
a
\nby
AND
DOMAIN
Art.
157\n
be represented
step functions.
0\n
^
\\\n
been described mathematically in the
in this chapter to describe driving forces
networks
of one or more electric elements. The
consisting
for the individual
elements are\n
relationships
to
\napplied
doublet\n
\nvoltage-current
used
be
will
section
\nlast
and
have
which
waveforms
The
the impulse, ramp,
functions:
unit
Other
8-2.
di
,
II\n
<2\n
*\n
Jo\n
vL\n
\n=
and\n
Vc
=
Ldt\342\200\231\n
1\n
i dt
C\n
/\n
&\n
II\n
or\n
ic\n
\342\200\242S\n
II\n
II\n
and\n
Q\n
\nv
su
the
Whether
this
\nfrom
\nstep
\nthe
\nunit
voltage
were
derivative
of
and
integrals
step
function
to an inductor,
applied
be the integral of the
would
function
\nboth
a step
were
waveform
this
\nIf
Consider
waveforms.
of
\natives
volt\302\254
is a voltage source or a currentsource,
in the network
are described by integrals and deriv\302\254
force
driving
currents
and
\nages
dt\n
/\n
voltage driving force.
the current resulting
voltage. If a
to a single capacitor, the current wouldbe
applied
we shall be concernedwith
this
Evidently
voltage.
of driving-force functions. For the
the derivatives
such
function,
Fig.
8-6.
waveforms
Derivative
are
illustrated
and integral
voltage
by Fig. 8-6.
The\n
functions.\n
function varies linearly with time, and is known
as
a
The
of
the
function
linear ramp).
derivative
\na
(or
step
function
ramp
at
of
the
there
\nhas
value
the
function:
a nonzero
only
beginning
step
for all other values of time the value zero.
This
\nthe
is infinite,
value
with only one nonzero value (and that infinite)
function
\nrather
unusual
integral
of the
step
is
\nis
known
as
an
impulse
function.\n
the
If
function
function.
\nstep
In general,
\nnitude
of
the
unit
step
impulse
modified
\nthe
ramp
to the
the slope of the ramp functionis equal
mag\302\254
from which it is derived.\n
is not defined so easily as the unit ramp. Consider
shown in Fig. 8-7. This function is linear\n
function
t
=
\342\200\224
blot
interval
\nThe time
c and
(c
of unit
Derivation
8-7.
Fig.
from
Chap. 8\n
has unit magnitude, the slope of the correspond\302\254
is unity
since the ramp function is the integralof the
as
A ramp
a unit
function
with unity slope is known
\nramp.
The
DOMAIN\n
function
step
ramp
\ning
FREQUENCY
AND
DOMAIN
TIME
158\n
then has a
\342\200\224
b)
is
impulse.\n
constant value of unity
for
The derivative
as a.
defined
time.
all
of this
mod\302\254
pulse of width a as shownin the figure.
(Con\302\254
linear
the
function
is
the
of
If
the
pulse
ramp.) the
integral
\nversely,
as the variable i, the pulsehasa magnitude
is designated
function
\nramp
of the ramp. The slope of the ramp the
distance
1
the
slope
\ndi/dt,
\ndivided
a; that is,\n
by the distance
\nified
is a
function
ramp
is
cU
1
=
(8-13)\n
a\n
\ndt
of the
area
the
Now
pulse is o
X
l/o
=
1, for
any
value of a.
As
a
the modified
ramp function approaches a unit step
zero,
\napproaches
the
same
At
time, the pulse approaches infinite height
constant at unity. In the limit,
with
the area remaining
\nzero
width
This
as a unit impulse, and is designated 6(t
is known
\nfunction
b).*
which is zero when t j*b and
a function
indicates
\nsymbolism
since the total area under the curveis
\nwhen
t = b. Also,
and
\nfunction.
this
\342\200\224
infinite
unity,\n
S(t
\nt
=
the
same
value for any
l
(8-14)\n
limits which boundthe
time
b.)\n
From
is
\ntion
*
has
integral
(The
=
-b)dt
In
this
discussion,
a unit
mathematical
impulse.
physics,
that the derivative of a unitstep func\302\254
The same process of reasoning mightbe used\n
we see
this function
is called a Diracdeltafunction.\n
Art.
8-2\n
to
find
the
\nshown
in
8-8\342\200\224a
\nthen
d
let
\342\200\242\n
l\n
waveform
pulse with two
limit, first let a
zero, i assumes
a approaches
As
\342\200\224>
0.
of a
up
the mathematical
to visualize
order
In
\nfunctions.
made
trapezoid
159\n
DOMAIN\n
impulse. Considerthe
of the unit
derivative
Fig.
FREQUENCY
AND
DOMAIN
TIME
the form of a
ramp
\342\200\224>
0,
pulse\n
f\302\253o\n
Unit
impulse\n
y\n
A
i\n
4\n i a\n
t\n
a
-*\342\200\224<*\342\200\224-\n U\n
1\n
1
i\n
i\n
i\n
i\n
i\n
i\n
di
dt\n
\n!
.1.\n
Unit
doublet\n
Had\n
ki\n
6\n
t\n
1-00\n
T\n
and
is
function
\nresulting
going.
negative
this
Of
\njunctions.
\nunit
and
impulse
see
us
what
basic
These
time, are known
func\302\254
as singular
step function is an old friend.The
seems friendly enough. But the
too familiar,
doublet
are rather terrifying! To break the ice,
unit
when the impulse is applied to ordinaryele\302\254
happens
and
\nments\342\200\224inductance
The
d, one
two
give a unit triplet.
with
behavior
the
family,
while
not
function,
\nramp
be continued to
discontinuous
with
\ntions,
\nlet
might
process
by the distance
As d approaches
of
\nimpulse.\n
This
\342\200\224*\342\226\240
0.\n
zero,
form
of
infinite
di/dt
impulse;
the two impulses superimpose at t = b. This
called
a unit doublet. It is the derivative a unit
but
impulses,
\ngoing
d
remains in the
a unit
approaches
separated
impulses,
other
the
going,
\npositive
\ni
two
becomes
di/dt
doublet in the limit, a\342\200\224*
0,
of unit
Derivation
8-8.
Fig.
t\n
Had\n
equation
unit
capacitance.\n
relating
current and voltage in an
is\n
inductance
(8-i5)
i=Uvdt
(C\n
For this problem,
v(t)
indicating
\nfind
the
a unit
current,
=
|\n
S(t
\342\200\224
(8-16)
a)
at
impulse
we obtain\n
8(t
Fig.
time t = a.
\342\200\224
a)
dt
=\n
8-9.\n
Substituting into Eq. 8-15to
^
for
b
>
a\n
0
for
b
<
a\n
Li\n
(8-17)\n
TIME
160\n
That
application
to start
is,
current
\nof
AND
DOMAIN
unit impulse at t = a causesa
flowing at t = a. The current is\n
of a
=
i{t)
is a
\nafter
all,
\nimpulse
step
function
\342\200\224
u(t
8\n
Chap.
(8-18)\n
a)
rather unusual behavior for the conservative
inductance,
it was hit by a rather unusual drivingforce.Another
of the unit impulse is that
this
unusual
property
will cause a current of 1 amp to be established
of L units
This
\nstating
DOMAIN\n
FREQUENCY
inductance
but
a
\ncapacitor.
in
an\n
be found for a
may
of
source
current
a
Let
voltage
A similar
immediately.
\nrelationship
of
way
Vtf)\n
\nvalue\n
=
i(t)
\nthe
is given
capacitance
Fig.
a capacitance
8-10.
The
voltage
to
(8-19)\n
as shown
across
i(t) dt
=
^
v{t)
j
=
^
u(t
(8-20)\n
the result
is\n
\342\200\224
a)
(8-21)\n
a unit impulse of current appliedto a capacitance
1 /C
on the capacitance
volts
to appear
instantaneously
delivers
a
unit
a unit
of
current
charge.\n
impulse
of
a unit impulsein
of
the
illustrate
the
concept
application
consider
the second-order differential
of
a familiar
problem,
other
In
\nin
as before and
is evaluated
integral
applied
- a)
basic relationship\n
by the
v(t)
This
be
8(t
\ncauses
\nbecause
To
\nterms
words,
\nequation\n
+
LW\342\200\230
If
the
\nthe
of the
solution
in
\nstudied
Chapter
\nonce
as\n
This
equation
\niables
voltage
driving-force
di/dt
\nfunction,
and
equation
6. Now
RIt
of
gi
=
W\n
\302\273W
is taken as a unit step function
v(t)
is the familiar solution for an RLC
suppose that the
is exactly the same
have replaced
dv/dt
the derivative
+
v(t)
with
equation is
in form as Eq.
i and v. Now if
respect
to
time
v(t)
u(t),
circuit
differentiated
the
where
8-22
=
is
a unit
is a unit
var\302\254
step
impulse\n
\nunit
function
step
driving
\nimpulse
\nfor
found
be
\ncan
In
8-22.
Eq.
and
\nwork
by
161\n
a
with
for i(t) of Eq. 8-22 known
driving
force, the solution for Eq. 8-23with a unit
can be found by simply differentiating i(t) found
force
other words, the unit impulse response of a network
for the unit step function responseof the net\302\254
solving
solution
the
With
8(t).
function,
FREQUENCY DOMAIN\n
AND
DOMAIN
TIME
8-3\n
Art.
it.\n
differentiating
in reverse. The step functionresponse
be
the impulse response. Likewise, the ramp
found
\nmay
by integrating
be found
the step function
\nfunction
may
by integrating
response
all the singular
functions
are related by differentiation
Since
\nresponse.
the solution
for one singular function is known,
\nand
once
integration,
is readily found by simple
solution
for
other
functions
\nthe
singular
same
The
or
\ndifferentiation
work
will
process
is an
This
integration.
important
property of
sin\302\254
functions.\n
\ngular
The
8-3.
and singular
sections
of this chapter have been devoted to
previous
of shifted
and singular functions.
In this
representation
will
consider
the derivation
of the transforms of these
two
The
transform for shifted
Laplace
\nmathematical
we
\nsection,
functions\n
\nfunctions.\n
unit
The
function
step
\nshown
in
\nThe
Laplace
has been
8-2(6),
of this
transform
Fig.
(where a is a
represented by the notation u(t a).
function may be computedfrom the
beginning
at t = a
constant),
\342\200\224
equation,\n
\ndefining
=
F(s)
f{t)e~\342\200\230l
dt
j
\nFor
the
case,
= u(t
f(t)
\342\200\224
a),\n
\342\200\224
\302\243u(t
le~at
a)\n
dt\n
\342\200\224
(8-24)\n
a)\n
\302\243u(t
factors:
the
factor
up of the product of two
is
of the unit step function beginning at the time
\n1/s
is a function which effectively\342\200\234shifts\342\200\235
\nt =
the
trans\302\254
0; the term
\nform
from
one
at t = 0 to one beginning at t = a.\n
beginning
The
for a unit step function may be generalizedfor
example
given
This
\nany
\ntime,
is
equation
the
transform
time
t =
function
a.
Such
made
f(t)
which
a time shifted
f(t
To
find the
transform
in its beginning to some
is delayed
other
function is represented
as\n
\342\200\224
a)u(t
\342\200\224
of this equation,
a)\n
we
(8-25)\n
write
the
defining
equation\n
TIME DOMAIN AND
162\n
in
new variable,
of a
terms
that
t\342\200\231;
is,\n
(8-26)\n
=t
defined as t'
t' be
variable
the
8\n
Chap.
=\n
m
Let
DOMAIN\n
FREQUENCY
\342\200\224
a
that
such
the
defining
\nequation becomes\n
=\n
m
fit
or\n
f{t
factor
constant
The
\nthe
of the
lower
limit
\342\200\224
thus\n
a);
\nuit
=
\342\200\224
fit
e\302\260*
the integral and
\342\200\224
\342\200\224
a)uit
(8-28)\n
within
from
\342\200\224
(8-27)\n
dt\n
a)e-*t(ea<')
changed to 0 if fit
integral
Fis)
dt\n
a)e~(t~a)t
removed
be
e\302\260*
may
\342\200\224
a)e~H
a)
is multiplied
dt
by
(8-29)\n
J
The
\342\200\224
\342\200\224
a)uit
\ntion/^
is recognized as the transformof
expression
integral
a),
time
func\302\254
that\n
so
\342\200\224
\302\243fit
\342\200\224
a)
a)uit
=
(8-30)\n
e~\302\260*\302\243fit)
conversely,\n
or,
equations
at the
begin
to
\ndelayed
function
the
number
A
In
\ntions.
tell
two
last
These
time t
a)uit -
-
a)
us that the transform of
= a is equalto
the
times
\302\243-le~\302\260s\302\243fit)
\nof
the
=
fit
(8-31)\n
function
any
e-04
transform
at the time t =
beginning
will illustrate the use of the last
of examples
unit
shown in Fig. 8-11, a pulse
network
amplitude\n
0.\n
two
the
equa\302\254
of
v\n
*R-1
ohm\n
1\n
[L-l
0
henry\n
a\n
(o)\n
Fig.
and
\nthe
\nof
width
current
two
unit
a is
applied
flowing
step
in
8-11.
Pulsed
RL
circuit.\n
to a series RL circuit. Let it be required
to
the network.
The pulse is givenby thedifference
functions
find
as\n
v(t)
=
uit)
\342\200\224
uit
\342\200\224
a)\n
(8-32)\n
From
AND
TIME DOMAIN
8-3
Art.
of this voltage
the transform
8-30
Eq.
of V
value
this
Substituting
-
L[al(8)
the
Substituting
is\n
(8-33)\n
e\342\200\224)
into the transform
+ RI(a) =
t(0+)]
equation,\n
- (1 -
(8-34)\n
e\342\200\224)
condition, t(0+) =
and the initial
values
parameter
-
- (1
8\n
=
F(s)
163\n
DOMAIN
FREQUENCY
0
\ngives\n
\342\200\224
(1
-
m
e~at)\n
(8-35)\n
*+1)\n
This expressionmay be writtenasa
of
sum
1
terms,\n
e~\n
=\n
m
(8-36)\n
+
\302\273(\302\253
1)\n
\nto
term
first
The
+
1)\n
is easily expanded
of this equation
by
fractions
partial
give\n
(M7)\n
-rrn
In terms of this expansion,
Eq.
r,
1
x
8-36
written\n
be
may
1\n
+\n
m-J\"7+J\n
\nin
this
Laplace
to
equation
\302\2431I(s)
The
third
waveform
(8-38)\n
1\n
out term by term
give\n
\342\200\224
\342\200\224
\342\200\224
1
e~\342\200\230
i(t)
\342\200\224
u(t
terms of this
and fourth
\nsecond only in
\nThe
=
+
may be carried
transformation
inverse
The
8
a)
n*.8
by
-12.
this
from
differ
time
equation
Response
of RL
and
are
the
first
opposite
is plotted as
circuit.\n
(8-39)\n
a)
expression
that they are shiftedin
represented
\342\200\224
+
Fig.
and
in sign.
8-12.\n
DOMAIN\n
FREQUENCY
AND
DOMAIN
TIME
164\n
Chap.8\n
obtained is the same as wouldbe found
by
using
\ntwo voltage
sources
and the principle of superposition. The resulting
\ncurrent
shown
in Fig. 8-12, is the summation of the two
waveform,
of the
circuit
caused
\nresponses
by the superimposed voltage sources
The
we have
result
make
\nthat
the
up
of
square
\nperiodic
pulse.\n
example, consider the problem
wave shown in Fig. 8-13 by a
As a second
been
has
3 a\n
2a\n
Periodic
8-13.
Fig.
wave
The
transform.
a\n
wave.\n
sum of step functions
by an infinite
represented
square\n
5 a\n t\n
4 a\n
square
the
representing
of
the
\nform\n
=
v(t)
The
\nterm
infinite
\nfollowing
\342\200\224
\342\200\224
2a)
\342\200\224
3a)
2u(t
+
...
(8-40)\n
this expressionterm
be applied to
that
series
2
h
s
-
-
from
2
by
3a\302\253\n
p\342\200\224
2
s
+\n
tr-
becomes\n
-
+
\342\200\224
e~\302\260*+
e-2\302\260*
...)]
e-*\342\200\234\342\200\242
+
in this equation may
the binomial
theorem,\n
= 1
(8-41)\n
s\n
terms, the equation
2e~\302\260*(l
appearing
expansion
V (s)
2a*\n
o\342\200\224
\342\200\224
common
- [1
=
\342\200\224
e_8\302\260*
(8-42)\n
be identified
by
+\n
the
(8-43)\n
e\n
becomes\n
^
or,
2u(t
may
s
1+
such
-1\n
=
out
factoring
V(s)
The
+
a)
give\n
V(s)
By
\342\200\224
2u(t
transformation
Laplace
to
\342\200\224
u(t)
-
=
K1
\342\200\234
T&)
(8^>\n
Kr+^-)
finally,\n
V(s)
\nw
\342\200\224
tanh
*
^2\n
(8-45)\n
Art.
The
be
outlined in the example
applied
The transform of any such function period
procedure
may
\nperiodic function.
=
F(s)
dt =
\n0
T, we
to
f(t)e-\342\200\234
dt
+
...
dt +
Me-*
/
Jt\n
any
T
is\n
(8-46)\n
where
n is the
each transform term by
to make the limits of the integral expression
shifting
shifts
of
\nnumber
/
Jo
Jo
By successively
to
of
Me-*
/
165\n
DOMAIN\n
FREQUENCY
AND
DOMAIN
TIME
8-3\n
e~n*T
necessary
have\n
rr\n
=
F(s)
(1
now
\342\200\224
b).
\n8(t
terms
series,\n
dt\n
r-'e-r*
of
limit\n
=
\342\200\224
8(t
b)
-
lim
&\n
a\342\200\224>0
-
=
b)
\302\2438(t
of this
transform
Laplace
[u(t
\342\200\224
\342\200\224\342\200\224
\342\200\224
b)
b
u(t
limit equation
l/o\n
r-\n
as\n
I\n
\nresult
is\n
-l/o\n
b) =
\302\243S(t
b
where
\n(that
is,
is
the
the
time
impulse
e~b\342\200\242
of appearance
occurs at / =
\nvoltage
result
has
ratio
transfer
significance
function
of the unit
0), we
=
Fin(s)\n
=
1\n
G(s)\n
The
unit
impulse.\n
impulse.When 6 = 0
have\n
in terms of
of a network
Fout(s)\n
8-14.
Fig.
(8-51)\n
\302\2438(t)
This
t\n
The
l\342\200\231Hospital\342\200\231srule.
of
\342\226\240i\302\260h\n
by the
found
be
may
\napplication
(8-49)\n
is\n
(8-50)\n
limit
a)]\n
lim\n
a\342\200\224>0\n
This
(8-48)\n
Jo
may
the
as
\ndefined
The
(8-47)\n
compute the transform any periodic
and requires
one
only
integration.\n
turn
our
attention
to the transform of the unit impulse
The
of this function were discussed on page 158.
properties
sketch
of the
be
shown as Fig. 8-14, the unit impulse
\nwaveform,
\nIn
dt
e-*f(t)
be used to
may
equation
We
I
\342\200\242.)
=
F(s)
This
.
to identify the
theorem
binomial
\nUsing the
+ e-T + e-*'T+
(8-52)\n
a transfer function. The
is\n
(8-53)\n
TIME DOMAIN AND
166\n
If
=
Vin(t)
impulse, then Fin(s)
a unit
S(t),
=
Fout(s)
is, the output
that
= 1,
and\n
(8-54)\n
G(s)
voltage from a unit
8\n
Chap.
DOMAIN\n
FREQUENCY
is
voltage
input
impulse
deter\302\254
of the network. Such a response
characterizes
the
network.
We will exploit this fact in our study
\ntruly
\nof the
convolution
in the next section.\n
integral
Consider next the unit doublet
and
the
transform
of this function.
\nThe
that
for the unit impulse. The unit
procedure
parallels
given
\nmined
by
solely
is
\ndoublet
defined
function
transfer
the
the
by
lim
limit\n
\342\200\224
2u(t
\\
[w(\302\243)
\302\256\n
a\342\200\224>0
The
term
transform,
Laplace
Z
rule
l\342\200\231Hospital\342\200\231s
may
\nnumeratorand denominator
lim
(
JT\342\200\234
+
u(t
\342\200\224
(8-55)\n
2a)]\n
of this limit
is\n
+\n
(8-56)\n
sa2\n
a\342\200\224>0
Again,
a)
by term,
<!
lim
\342\200\224
be
used.
The
second differentiation of
yields\n
\342\200\224se\342\200\234\302\260*
+
2se_2\302\260*)
=
s
(8-57)\n
a\342\200\224\342\226\272
0\n
\nbeen
\nunit
\nsponds
\nby
\nshifted
has the transform s. This result mighthave
from
the fact that the doublet is the derivativeof the
anticipated
are ignored, differentiation corre\302\254
conditions
If initial
impulse.
to multiplication
by s, while integration corresponds to division
the family of singular functions (not
The
among
relationship
as follows:\n
from
t = 0) is tabulated
the
Thus
s.
doublet
unit
Function
Laplace
transform\n
Unit
ramp
Unit
step
Unit
impulse
1\n
Unit
doublet
s\n
Unit
triplet
1/s2\n
1/s\n
s2\n
either direction (up or
\nAs was
if the response to any of the singular
suggested
earlier,
the response
for any other singular function may befound
known,
\ndifferentiation
or integration.
from one known response
Further,
the
the
\nquently
impulse
response),
response for any other driving
found
be
the
use
of
the
convolution integral to be
in
\nmay
by
The
table
might
be further
extended
down).
functions
\nis
by
(fre\302\254
force
discussed
\nthe
next
section.\n
Art.
The
8-4.
An
DOMAIN
convolution
integral\n
\nwhere
where
=
\nconvolution
integral*
\ntransform
/2($)\n
as the
expression is known
This
and /2(f) are convolvedto
give g(t)
find
the
and
pairs
=
\302\253\302\243_1F2(s)
(8-58)
by
applica\302\254
convolution
the
of
XV,(X) d\\
In this section, we will study the
the use of this equation to
integral:
the
use of this equation to
response
convolution.
of
process
f/.\302\253
and
fx(t)
of integration.
in which
fi(t)
variable
a
is
X
=
- Sr'FfrW)
\302\243-1Fi(s)
\ntions
in network theory
frequently
appears
167\n
form\n
g(l)
\nthe
that
DOMAIN\n
FREQUENCY
AND
expression
integral
the
\nhas
TIME
8-4\n
find
new
of
\nnetworks for
complicated inputs.\n
= 1/s
to F(s)
As an example, suppose that the /(<)\342\200\231scorresponding
= l/(s
transform
\nand
to
+ 1) are known, and that the inverse
F(s)
=
=
as Fi(s)
\nfor
+ 1) is to be found.If we designate
Fi(s)
F(s)
l/s(s
= 1 or
then
\n1/s,
fi(t)
u(t), and similarly, if F2(s) = l/(s + 1), then
= e~l. From
\nfi(t)
Eq. 8-58, fi(t) and /2(<) may be convolvedto give
\ng(t)
follows:\n
as
1\n
=\n
g(t)
+
s(s
1)\n
dk\n
X)e-X
of this
evaluation
The
in
terms
\nthe
in
\nshown
\ntial
the
Fig.
is
e-x
X.
The
has
unit
negative
\ntion
u(t
and
zero value for
function
\nstep
unit
>
X
n{t
X)
t,
has
the
from
=
g(t)
positive
<
X
*
Proof
\nEquations
\nalso
\nNew
of this
in
in
Wylie,
York,
1951),
p.
188.\n
convolution
in the
integral.\n
t\n
discussed
was
= -e~x
e-x dk
Problems
Advanced
involved
in Art. 8-1. Since the unit
unit
value
over the limits of integration, it
integral
expression to give\n
as
equation can be
Engineering
the
of
\nevaluation
for
was given by
result
same
Functions
func\302\254
Jq
The
8-16.
Fig.
step
value
\342\200\224
removed
be
\nmay
both
for
\nand
X)
of
which are
integral
8-15.
The exponen\302\254
shown
\342\200\224
ex\302\254
interpretation
requires
\npression
integral
Engineering
= 1
j*
\342\200\224
e~*
partial fractions in the last
found
in
Salvadori
(Prentice-Hall,
Mathematics
and
Schwarz,
(8-59)\n
chapter.
It\n
Differential
Inc., New York, 1964),p. 214;
(McGraw-Hill Book Co., Inc.,
be
should
and
\nfi(t)
Consider
DOMAIN\n
FREQUENCY
AND
DOMAIN
TIME
168\n
functions to be
the
out that the choice of
pointed
is arbitrary
and does not affect
f2(t)
next
a two-terminal-pair
network,
a
ratio
voltage
is
\nform
Vi(s)
input
In
terms
transform.
\nage
=
V2(s)
is the
equation\n
G(s)7,(s)
(8-60)\n
voltage transform and F2(s) is the output
of the convolution
integral, let\n
=
Fi(s)
and
the
by
trans\302\254
of the input
terms
in
8-16. Two-terminal-pair
where
voltage
output
transform
\nnetwork.\n
designated
result.\n
given
\nvoltage
=
F2(s)
8\n
shown in Fig. 8-16,with\n
function G(s).*
transfer
the
that
\nAssume
Fig.
Chap.
7i(s)
and
fx(t) =
G(s)
and
f2(t) = h(t)
volt\302\254
(8-61)\n
vx(t)
(8-62)\n
as the transfer function G(s); f2(t) = k(t)
\nis the
related
time-domain
From the discussion of the last
response.
it will
be recognized
that h(t) is the unit impulseresponse
of
\nsection,
with
a
\nthe
network
transfer
function
G(s). For a unit impulseinput,
\nthe
is
determined
the
inverse
output
by
Laplace transform of the
Function
is identified
F2(s)
=
git)
=
\302\243,-'[Fi(s)F2(s)]
=
g(t)
=
g(t) is identified as the output voltage
\nconvolution integral has the form\n
-
v2(t)
the
only
\nknown,
the
\nmine
unit
\nthe
impulse
of
terms
\nIn
\nand
when
\nmeasured
*
Transfer
\nQ(s) is an
get a
term,
better
functions
deter\302\254
with
of
X).
An
arbitrary
from any
will be
algebraic function
t\\
that
is,
of the convolution
meaning
each term in
us examine
\342\200\224
the
plot, what is X? When t
\342\200\224 *=
=
vx(t
X)
t>i(0).
X, then
backward
(8-65)\n
unit impulse response,is
specified in orderto
other words, any input convolved
of this
t
d\\
\\)h(\\)
The
domain.
time
if h(t), the
Vi(t) need be
picture
let
Vi(t
the
in
gives the output.\n
response
8-65,
Eq.
the
\nconsider
voltage
voltage 1 In
input
output
In order to
\nintegral
that
indicates
equation
-
vi(t
(8-64)\n
v2(t)\n
JB\342\200\2351?^\302\256)
Thus
This
(8-63)\n
\302\243-'{Vi(s)G(s)]\n
8-60,\n
Eq.
By
From the convolution integral,\n
is h(t).
function
This
function.
\ntransfer
Vi(t)
the expression.First,
* 0, then
Evidently
it is a time
studied in Chap.
in Fig. 8-17.
is shown
\342\200\224
X)
vx(t
X
is
a
interval
t>i(\342\200\224X)
quantity
measured\n
10. Forthe present,
relating Vi(s) and Ft(\302\253).\n
=
assume
that
8-4\n
Ait.
\nthe
The
figure.
\nwhich
the
8-17.
be
of
response.
\342\200\224
\nall
history
past
of
and
h(\\)
(that is,
\342\200\224
Vi(t
X)
must
input,
of
i>i (t),
to
find
repeated
very
simple
for
have
present
very*little
effect on the
output is determined
by
weighted by the unit input response.For
it may be necessary to use numericalor
v2 (t) from the
-V\\Ar\n
the
Further,
be
the
by recent values
influenced
mainly
input
speaking,
of the
integral.
must
product
integration\342\200\224is
integration
\ngraphical
of
response.
old\342\200\235values
forms
\nconvolution
The
to t = 0
t = t
by
Strictly
\ncomplicated
from
increasing
response.\n
to give the output
This
process
as weighting
all past values of the input
the
unit
Since h(k) is usually small for large the output
at
by
\342\200\234Very
output.
X
Impulse
X,
time\342\200\224found
\npresent
Fig. 8-18.
from 0 to t
thought
input.
8-18).
Fig.
integrated
\nimpulse
\ntion
\342\200\224
Vi(t
voltage.\n
with
8-17
Fig.
from
\nbackward
\nnext be
input
Arbitrary
on
imposed
\nof
0 to t,
from
super-\n
Fig.
\nany
time t. This is illustratedin
from
0 to t, the limits of the
the
the
\ncan
vary
169\n
t
\nWhat
\nto
can
X
DOMAIN\n
then
X)
ranges
through
values
of the input
is
assumed
start
at
to
(the input
0).
about
and
?
The
is
transient
h(t)
h(\\)
quantity h(t)
response
unit
Its
form
on
exact
the
transfer
impulse.
depends
function,
we
have
not yet specified.
It might have an appearancesimilar
shown
in Fig. 8-18. A plot of h(\\) could be
waveform
past
\nto
reference
specified
quantity
varies
+X
As
\nintegration.
\nall
some
from
negatively
FREQUENCY
AND
DOMAIN
TIME
integra\302\254
1
value of t
each
ohm\n
1 farad\n
interest.\n
\nof
As
\nconcept,
\ndriving
a
suppose
force
Vi
of this
that
the response
from the
= e~2t is
for the
required
application
\ntwo-terminal-pairnetwork shownin Fig.
\ntransfer function for the voltage ratio
8-19.
8-19.
Fig.
For
this
RC
network.\n
the
network,
is*\n
1\n
GOO
-\n
(8-66)\n
5+1\n
a table
From
of transforms,
the corresponding h(t) is foundto
h(t)
*
The
computation
of this
transfer
=
er*
functions is given in
be\n
(8-67)\n
Chap.
10.\n
convolution
the
from
Then,
integral,\n
=
0%{t)\n
Chap.8\n
DOMAIN\n
FREQUENCY
AND
DOMAIN
TIME
170\n
d\\\n
g-S\302\253-Me-X
(8-68)\n
\302\243
t\n
=
this
For
For
more
be
used
to
Thus
of
\nterms
\nresponse
\nintegration
\nthe
step
\nvolution
be
\nThe
-
=
the
\nof
step
to
system
to
extended
by letting u
=
J'
input
any
any
vx
\ntion
force.
driving
network
to
to
of the
=
(8-71)\n
\342\200\224
vx(t
-
k(\\)vi(t
fixed
and
X)
and
X)
dk
*= h(\\)
do
dk.
(8-72)\n
be determined
can
for the unit
family of singular
by
con\302\254
impulse and unit step
can
functions.\n
turning point in
the response of a
our
network
studies
The
a sinusoidal
time-domain
driving
study.
to
Behind
are
a given
time-
concern the response
force of variable frequencyin
yet to come
of
addi\302\254
topics.\n
electrical engineer must be bilingual
He must speak the language of the
response.
The modem
\nnetwork
is
made
section marks the
\nstudies in the time domain,
\na
+
fc(<)\302\273i(0)
This
\nvarying
X)h(X) d\\
series\n
Fourier
8-5.
-
is\n
statements
The
\nvolution.
\nbe
is more
integral
unit step function response of the system.
We
note
within
the
of h(k) to give k(k) is compensated
integral
to
time.\n
of vx with
respect
a useful property of convolution.If
illustrates
equation
of a system is determined, the response
function
response
last
unit
(8-70)\n
is the
integration
\nthe
\302\273i(t
f*
by parts
equation
k(t)
This
-
e-\342\200\230 e~2t\n
by partial fractions
expansion
forms of input, the convolution
complicated
integrated
differentiation
\nby
=
1)
advantage.\n
t>*(0
\nthat
0\n
is\n
resulting
where
(8-69)\n
j
e-*\342\200\230(e\342\200\230
vt(t)
can
e~2tex\n
in
far the application of the convolutionintegralhas been
the unit
impulse
impulse response of a system. If the unit
can
be
found
unit
function
the
is known,
response
by
step
in the last section. In some cases,however,
as discussed
is more conveniently
recorded. The con\302\254
function
response
be put in another form for this case. Equation\n
can
integral
which
8-65,
=
d\\
e2Xe-x
example,
particular
\ndirect.
\ncan
=
Vt(t)\n
Finally,\n
e~u
when
of
speaking
time
domain\n
8-5\n
and
must
at a
other
mechanical,
\npurely
two
the
To
begin
\ninclude
the
\nrecurring
\nnext
study
moment\342\200\231s
\342\200\234large
notice.
bandwidth\342\200\235
frequencydomain.
able to translate
He
from
translation
The
may be
process
\342\200\234desirable
equals
171\n
function
step
based on an understanding of the concepts
in terms of a common root or origin.\n
domains
we have now extended our time-domain studies to
with,
of a network
to (1) a nonrecurring pulse and (2)
response
and
waveforms
such as the square wave. We
will
periodic
these
two classes of driving force functions in terms of
it
Or
\nresponse.\342\200\235
\nof
either
in
the
to
\none
in the language of the
but he must be
language,
be trained
also
think
\nmay
FREQUENCY DOMAIN\n
AND
DOMAIN
TIME
Art.
be
may
\nsinusoids.\n
we
do
What
wave
sine
\nfamiliar
after
(at
of
\nvalue
sin
=
61
by
sin
has
function
the
since
a periodic function (or
If represented as sin (at,
is periodic.
=
cat
0, then for a sine wave,\n
mean
n = any
(2nx + 0i),
identical
form from
(at
=
waveform)?The
if di
and
is some
integer\n
0
to
cat =
2ir, from
4ir, etc. Similarly, any function is periodicin (at if
=
+ 2nx) and the period is 2r.\n
\nf(di)
/(0i
were studied
functions
Such
by the French mathematician
periodic
was the first to show that periodic functions
who
\nFourier
(1768-1830)
form
in terms of harmonically related
\ncould
in
series
be
expanded
\ncat
=
(at =
to
2ir
as\n
\nsinusoids
\342\200\224
do
f((at)
+
ax cos
(at + a2 cos 2(at + ...
...
-f*
-|- b\\ sin (at
+ an cos ruat\n
bn
sin
tuat
...
-|-
(8-73)\n
series, and the process represent\302\254
function
problem
by such a series is Fourieranalysis.
\ning a periodic
of the
of
the
coefficients
of
the
values
is
determination
analysis
time
a given
a2, ..., bh b2, ...,
series,
a0, ah
of the Fourier series such
coefficients
the
we
select
Suppose
=
=
coefficients are equal to zero. The
all other
&2
1, and
1, and
of the two functions with t is shownin
the
combination
of
the
and
resulting
f((at) has small amplitude from 0 to and
large
8-20,
from
to
This
distorted
waveform
resulted
from
the
2ir.
\namplitude
two
of
related
terms. It seems quite
harmonically
merely
that
function
could be synthesized with the
any
periodic
number
of terms
that are available in the
\nnite
for
the
coefficients
of the Fourier series for use in
Equations
found
are
\nysis
by the mathematical
procedure of (1) multiplication of
series
a suitable
by
(2) integration of the resulting
factor,
This
series
is known
as the Fourier
of
The
\nof
for
\nFourier
function/(\302\253<).\n
that\n
\342\200\224
Fig.\n
\nplot
ir
7r
\ncombination
infi\302\254
\npossible
series.\n
anal\302\254
\nthe
equations\n
172\n
TIME
H*.
8-20.
Waveform
by
term
over the period,
DOMAIN
from the addition
resulting
(
term
FREQUENCY
AND
DOMAIN
\342\200\224
sin
of
Chap.8\n
<at)
(cos
and\n
2col).\n
and (3) simplification
by
the
of the
use
\nfollowing definite integrals.\n
ruat
d<at =
\nsin
ruat
deal
\nsin
mat
cos
/.
\ns>\n
cos
=
nat
0
n
0
\302\273^0\n
(8-75)\n
m 7*
m
9* n\n
(8-76)\n
n
(8-77)\n
dat
= 0
f>\n
t6
m
r-\n
L
7*
cos
mat
cos
cos2
ruat
d<at
sin2
ruat
d<at
ruat dat
=
x\n
= 0
m
m
n
t*
t*
0\n
(8-74)\n
0\n
n\n
(8-78)\n
n\n
(8-79)\n
0\n
(8-80)\n
0\n
(8-81)\n
\n/.\n
These
equations
\nlimits
of the
also
hold
integrals
can
\342\200\224
x\n
n
period, 0i to 0i + 2t,
be replaced by these moregeneral
for any other
and
terms.\n
the
In
no
\302\253o,
evaluating
\nIntegration
r2x
term
multiplying
=
dut
f(ut)
+
... +
+
nut
cos
6 x
such
that
8-82,
\nand
8-75.
si\n
sin
/
reduces
equation
dut
=
\nEq.
all
which
\nsimplifies to
Eq.\n
8-74
Eqs.
to\n
(8-83)\n
dut\n
(8-84)\n
give\n
the
over
nut
cos
6n coefficient
the
\nintegrating
In
is
On\n
Similarly,
(8-82)\n
zero, each term in the Fourier series
from 0 to 2ir. In the resulting
cos nut
and integrated
will
vanish
except the one of the form of
integrals
The
thus
is the integral
with a coefficient a\342\200\236.
equation
by
8-80,
...
+
than
other
n
for
a\342\200\236
\nexpression,
dut
Oo(2t)\n
ut)
\nmultiplied
nut
is permitted.
integration
or\n
find
dut\n
by
total
ut)
To
ut
right except the first have zerovalue
the
Hence
do\302\273t\n
Jo\n
term-by-term
on the
terms
all
2(at
cos
I
a,2
2r\n
sin
assuming
2r\n
Jo\n
r
[2t
I
Jo
an
+\n
cos ut doit+
Jo
Jo
...
r
dut + fli /
do
Jo
required.
gives\n
/*2ir
r2*\342\200\242
/
series
173\n
as step 1 is
specified
Fourier
of the
term
each
of
FREQUENCY DOMAIN
AND
DOMAIN
TIME
8-5
Art.
period
is evaluated by
0 to 2x, giving\n
dut\n
(8-85)\n
multiplying by sin nutand
1\n
nut
sin
bn\n
dut\n
(8-86)\n
T\n
These
three
\nThese
integrals
at
\nwith
\nber
of
f(ut)
to
The
\ncan
be
must
The practical
be valid.
is
without
the
that
represents
num\302\254
of maxima and minima and a finite
in every
finite interval. These are the Dirichlet
be satisfied
for the Fourier series representation
discontinuities
which
f(ut)
of the Fourier series.
a finite periodic function
all coefficients
number
a finite
\napplication
\nfunctions
when
hold
most
\nconditions,
\nof
determine
equations
Fourier
consequence in terms of engineering
series can be written for engineering
concern.\n
amount
of labor involved
reduced
when
there
in the evaluation
is symmetry
of
with respect to
the
coefficients
the axisin
the\n
of/(\302\253<).
plot
\nfor
\nhave
symmetry
\nSuch
a function
the
\nfunction,
DOMAIN\n
Chap.8\n
shows a plot of sine and cosinefunctions
values
of at. The cosine function is seento
8-21 (a)
Figure
and
positive
FREQUENCY
AND
DOMAIN
TIME
174\n
negative
the
is said
to be an
of the
value
same value for
/ axis, the
about
even
\342\200\224
at
\342\200\224
ut.
and
of the sine
of that for\n
negative
f\n
wave\n
Square
-w
the
is
case
the
In
function.
for
function
+ad
t\n
+ut\n
)\n
8-21.
Kg.
Even
(a)
and
\nfunction
vice
may
and
functions\342\200\224cosine
and
\nfunctions\342\200\224sine
wave:
square
(b)
odd
waves.\n
triangular
Such a function is an oddfunction.Any
general
described
as odd or even when it meetsthe following
versa.
be
\nconditions.\n
function:
Even
=
/(&><)
function:
Odd
=
f(<at)
/(
\342\200\224
(at)\n
\342\200\224/(\342\200\224\302\253<)\n
are shown in Fig. 8-21(b), square
wave
The
wave
is
an
even
triangular
wave,
respectively.
square
\ntion
it
be
made
odd
(although
might
axis).
by shifting the
Being
term
in
its
Fourier
series
even
every
representation
function,
odd
term
\nalso
be
a
would
the
even
even;
single
destroy
\nThe
same
can be applied
to the triangular
in that
its
argument
\nFourier series must contain only odd terms. These
of this
even
An
and
an odd
function
a
\nand
func\302\254
cit
\nan
must
symmetry.
wave
conclusions
can
\ndiscussion
be verified
The equation for the
mathematically.*\n
ao
Oo =
This
*
\np.
integral
For
359.\n
example,
represents
see Wylie,
of
coefficient
1
4K
Fourier
the
series
is\n
r2'\n
/
/(\302\253<)
doit\n
Jo\n
the area under
op. cit., p.
the
/(\302\253<)
curve
122,or Salvadoriand
from
Schwarz,
0 to
op.
2x.\n
cit.,
Art.
TIME
8-5\n
of
Fig.\n
of
value
in the following
are summarized
conclusions
three
These
zero.
is
\nao
of the square wave and the triangularwave
is as much positive area as negative area, the
there
8-21(b),
175\n
case
the
in
as
If,
DOMAIN\n
FREQUENCY
AND
DOMAIN
\ntable.\n
of
Simplification
\nFourier
Condition\n
=
/(\302\253<)
=
f(<at)
negative\n
waveform
the
under
areas
all
0,
=
an
0,
=
0\n
ao
-f(-<at)\n
and
positive
Equal
bn =
f(-wt)\n
series\n
n\n
all n including
a0\n
over\n
one cycle.\n
Example
such
\nany
\nare
a
From
series.
Fourier
that
a square
shows
8-22
Figure
\nby
1\n
v(<at)
cycle
determined
adds
=
wave function which we wish to represent
the figure it is seen that the symmetryis\n
coefficient
the
to
zero,
by
evaluating
=
a\342\200\236
the
1
voltage
v(<at)
has
values
may
be
an
Jo\n
to
set of values over one
cycle:\n
v{<at)\n
V\n
7t/2\n
37t/2 to
These
The coefficients
v(<at) cos mat dtat
/
Interval\n
7t/2
ao is zero.
total area under
integral\n
the following
0 to
0. Since the
P\342\200\231\n
-
\n*
The
6n =
so
and
v(\342\200\224<at),
37t/2\n
2tt\n
substituted
-V\n
V\n
into the integral
equation to
give\n
AND
DOMAIN
TIME
176\n
FREQUENCY DOMAIN\n
Chap.8\n
as\n
a\302\273
=\n
On
Sir/2
dut
cos nut
^
V
y
^F
\nV
=
\342\200\224
^Vsl\n
The
term
\nzero
for
0
even
|3r/2\n )\n
of
=
a.
l
(cos
\neach varying
8-23.
Fig.
of n,
values
and
1, 5,
9, ...\n
to
=
3, 7,
11,
\nn
=
even
integers\n
ut
...
is\n
\342\200\224
\\
cos
IT\n
a Fourier
By
\nin
series
\342\200\224
odd
=
\342\200\2244F
\n0,\n
t>(w<) =
for
to
\ntot
Fourier
4
\302\261
hence\n
values;
TO?r\n
the
)\n
sin\n
has the value
+4F\n
Thus
nw<
|2ir\n
+
|\302\273/2
in parenthesis
cos
/
J&w/2\n
nul
sin
nut\n
F
+
\n/2
I3t/2
\342\200\242\n
sin
nul
cos
J
(
/*2ir\n
\342\200\224
Sut
+
i cos
we have shown
expansion,
5a>/
\342\200\224
y
7a><
cos
that a sum
of
+ ...)\n
terms
voltage
as illustrated
is equivalent to a squarewave,
sinusoidally
of superposition, the responseof each\n
By the principle
(U)\n
Two equivalent
8 -28.
Fig.
\ntime-varying
can
generator
\nand
\nfound
\ning
\njust
\npute
total
the
(as
the
one
the
systems: the Fourier seriesexpansion
driving-force
of
a
function.\n
with all other generators short-circuited,
will be the sum of the individualresponses
so
be determined
response
discussed
problem
being
response
in
rather
required.
as a
appear to be
than simplifying it, many solutionsinstead
to
We have yet to show that it is
function of frequency by complexalgebra
Chapter
4). This may
complicat\302\254
of
easy
com\302\254
and\n
FREQUENCY DOMAIN\n
AND
DOMAIN
TIME
8-5\n
Art.
177\n
so interested in the actual
as
\nwe are
In
frequency
response required for a given
we have a choice between the solution the
\nsolving a problem,
\nsient
time
and the solution of responsein terms
domain)
problem
(the
in
that
many
in the
we are not
cases
response
waveform.
of
\nof
sinusoidal
a
Example
A
saw-tooth)
(or
8-24.
Fig.
the
\nby
\nstraight
for
\nand
odd
an
is
function
including
is shown in
waveform
with an equal
function
Triangular
(or sawtooth)
0
to zero for all
to
7r
function.\n
to
\342\200\224
2V\n
Z
(cat)
\nv(o)t)
=
+
2V\n
TT\n
2V\n
\342\200\224
0
2t\n
((at)
-
4V\n
TT\n
\n-g-
the
,\n
ir\n
Sir\n
3w
out
a
=\n
2toT\n
Carrying
n\n
v((at)\n
2\n
integration,
f sin
as in Example
\342\200\224
o\302\273t
^
sin
Scot
+
1, gives
^
sin
5(at\n
)\n
3\n
Example
In
some
\nof
mathematical
\nrectangles
Fig. 8-24. This
0. To
2V
\nintegration.
domain).\n
find the ^coefficients, we represent the waveform
each derived in terms of the equationfor
following
equations,
y = mx + 6, where m is the slope,6 is the y intercept
line,
= v and x = t.\n
this
y
problem,
=
n
Interval\n
\ncases,
(frequency
frequency
2\n
triangular
\nvoltage
of variable
generator
tran\302\254
the waveforms are not known in the form
but
rather
as recorded graphs. In such
equations
coefficients
be evaluated
may
by approximate graphical
As
shown
in Fig. 8-25, the waveform is divided into m
of
width
Awt and
height f(<at). In terms of a summation\n
practical
the
problems,
A recorded
8-25.
Fig.
\nwaveform
than
rather
AND
DOMAIN
TIME
178\n
Chap.8\n
FREQUENCY DOMAIN\n
waveform. The Fourier seriesequivalentof
may be found by graphical integration.\n
the
integration,
for the Fourier
equations
approximate
this
become\n
\ncoefficients
Aoit\n
do\n
(8-87)\n
i-i\n
m\n
i
=
2x^
^
^
cosn
(8-88)\n
Acot\n
2x^
y-i\n
m\n
i
=
2x^
^
sin n
2x^
(8-89)\n
Acot\n
l\n
2x\n
where\n
Acot
\342\200\224\n
(8-90)\n
m\n
The
\nform,
are
required
as follows.\n
summations
for
example,
3\n
3-
(360\302\260)
=
\302\273
3
=
cos
e
of
for Calculation
Table
\nm\n
carried out in tabular
most conveniently
Aut =
15\302\260
m
=
o\302\273\n
24\n
\302\273
cos
f(6)
(jt
360\302\260)\n
nO\n
/(m360\342\200\234)\n
1\n
15\302\260\n
cos
45\302\260=
0.707\n
1.52\n
1.07\n
2\n
30\302\260\n
cos
90\302\260=
0\n
1.77\n
0\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
\342\200\242\n
0\n
\342\200\242\n
\342\200\242\n
m\n
2\n
The coefficient has
the
FREQUENCY DOMAIN\n
AND
DOMAIN
TIME
8-6\n
Art
179\n
value\n
8-6. Complexexponential
form
o8
=
of
the
-
2
(8-91)\n
series\n
Fourier
in
series studied in the last sectioncan be expressed
form in terms of exponential quantities. Supposethat the
\nequivalent
in the
\nterms
series
are grouped
together by harmonic number as\n
Fourier
The
=
f(o)t)
+
o0
+
mat
cos
(u\342\200\236
^
sin
6\342\200\236
(8-92)\n
mat)
\302\273=i\n
6-2.
Art.
in
\nlearned
sine may be expressed in exponentialform,as we
with Euler\342\200\231s equation,\n
Starting
and
cosine
the
Now
=
e\302\261iut
the cosine is
found in terms
forms
exponential
\nnegative
cat =
is found
sine
the
these
<at
sin
(8-93)\n
positive
adding
by
exponentials
+ e~iat)
v(eiat
and
=
(eiat
into
equations
(8-94)\n
these quantities
by subtracting
(at
sin
Substituting
of
\302\261
j
as\n
cos
Similarly,
(at
cos
-
as\n
(8-95)\n
e~iat)\n
Eq. 8-92, there
results\n
oo\n
'
\\
=
+2,
In
that
\nNoting
pi nut
I
=
l/j
+
^
n =
nut
p\342\200\224jnut
our
\342\200\224j,
p\342\200\224jnut\\\n
*>\302\273
j
(\342\200\234\342\200\242
1\n
f
)
(8_96)\n
like exponential terms are grouped.
this equation,
to simplify
order
/
\302\253o
becomes\n
equation
oo\n
- a, +
/M
\302\253*\302\243
1\n
n=
To
this
simplify
the
\nreplace
a and
we
expression,
b coefficients.
\302\256n
\342\200\236
\342\200\224
The
new
jbn
sn
0
form
for Eq.
/M)
(8-97)\n
next introduce
By definition,\n
dn
y
8-97
=
+
[(~^)
co
^
\342\200\224
c\342\200\224n
\"I\342\200\235
jbn
)
pr
and
a new coefficientto
\342\200\236
\t\n
\342\200\224\n
CLq\n
Co
(8-98)\n
is\n
+
^=
n
(cne^nt
1\n
+
c_\342\200\236e-\302\273'\"n<)\n
(8-99)\n
are
\nwe
have
\noo
this
in
\n(including
Chap.8\n
to understand better all the
been through.
just
Letting n range through valuesfrom 1 to
oo
to + oo
is equivalent
to letting n range from
equation
zero) in a compact
in a
now
We
FREQUENCY DOMAIN\n
AND
DOMAIN
TIME
180\n
position
maneuvering
\342\200\224
equation,\n
=\n
(8-100)\n
/(\302\253fl
we
Here
can
c\342\200\236
\ncients
The
be evaluated
easily
coeffi\302\254
already
bn,
Then\n
\nknew.
1
f2r
r-
Cn
\342\200\234
2tt\n
hr
/
7T
2m
H*\n
1
=
f (cat)
/
Jo\n
f2r\n
7
f(<at) cos
\342\200\224
dcat
ruat
2tt
Jo
l
\342\200\224
(cos neat
\342\200\224
sin
j
J
sin
f(cat)
/
mat
dcat\n
)\n
neat)
d<at\n
f2*\n
do)t\n
2r
This
form of the Fourier series.
which
we
in terms of an and
the exponential
have
for
equation
(8-101)\n
Jo
cn holds
n is positive as
whether
have
we
assumed,
same
or zero, as can be shown by exactlythe
procedure.
\nnegative,
series?
Fourier
\nHas this form any advantage over the otherform of the
be\n
sine and cosine form usually may
\nIn
the
coefficients,
computing
But in discussing
used
to advantage.
\nthe
of frequency
concepts
\nthis
Fig.
\nform
8-26.
used
in
of the
Sweep
voltage
a cathode
ray
Example
oscillo\302\254
The
\nin
one
cycle
by the equation
\ncn-coefficients,defined by Eq.
8-\n
sweep
f2r
8-26
straight
y\n
\342\200\224
<ate~ino>t
hj\n
_
jv\n
2 TIT\n
\no
2t\n
n
9*
V\n
O
*\n
voltage
II\n
\302\243\n
o\n
we need
0\n
waveform
shown
be represented
may
=
v
line,
(V/2ic)<at. The
are\n
\n01,
next,
4\n
Fig.
a
\nif
study
form.\n
exponential
\ngraph.\n
over
will
we
\nwhich
and
Fourier integral,
the
\nintroducing
spectra
dcat
the
Hence
...
=
v(at)
\342\200\224
~
4r
\342\200\224
e~iZot
jf-
6ir
~
e~iwt
2ir
^
+
reducethis
If we wish to
definitions
the
\nfrom
Cn
\302\256n
these
From
o\342\200\236
- ( sin ut
v(o)t)\n
+
The
sinusoidal
of
\nber
\nFourier
phase,*
\nspectrum
+
\nvalues
is
plot
the
\nfar),
\nas
values
are
related
\ncosine
*
not
do
in
term
The
(8-103)\n
\342\226\240\342\226\240\342\226\240)]\n
can
We
example
visualize
last
the
of
a large
num\302\254
voltage.
of exponential voltage
picturing
of
harmonics
(merely
the
function.
phase
discrete
a plot is
Such
plotted
\ncies
+\n
appropriate
frequency
n.
\nnegative
We
3<at
series of our
interpreted.
for different
actually
is for
plot
to
\nalent
\na
the
fundamental
\nthe
sin
^
of voltage
as specified by the
in series to produce a sweep
connected
difficulty
frequency,
of
a
the
and
\342\200\224V/mr,
waveforms\n
The plot of the magnitudeand phaseof
Actually,
\nrequires special interpretation.
\nSuch
\342\200\224
Co\n
generators
in
contain
Eq. 8-102, but the coefficients
appearing
in Eq. 8-103. This information is con\302\254
information
as those
of the magnitude of cn, and sometimesof
in a plot
displayed
Such a plot showsthefrequency
as a function
of frequency.
waveform.\n
to a particular
corresponding
\nveniently
\nthe
series,
the equations which follow
form
the
same
\nthe
all
some
of
\nterms
of Fourier
generators
coefficient,
have
\nWe
form
and bn =
F/2,
2o>t
sin
Fourier
of the
form
the most easily
\nsection is
=
of periodic
spectra
frequency
second
The
(8-102)\n
C\342\200\224n),
^
-ds-K-\n
8-7.
j(pn
ao
0,
...
e,2\342\200\234*
+
becomes\n
series
\nFourier
=
equations,
^
\342\200\224
bn
d\342\200\235
C\342\200\224n,
is\n
waveform
^\n
2\n
alternate
may be found from
of Eq. 8-98.\n
b coefficients
and
a
\nthe
e*\342\200\230
+
the
to
result
+
\342\200\224
e~i2at
181\n
series for this
of the Fourier
form
exponential
FREQUENCY DOMAIN
AND
DOMAIN
TIME
8-7
Art.
of
n or
as
cn
of frequency
a function
has
c\342\200\236
values
only
for discrete
fundamental
frequency.
values of n, or if we identify
<o0 as
to in the equations
derived thus
of a>, the
values of the
ratio
\302\253/wo,
which
is
equiv\302\254
shown in Fig. 8-27for both positiveand
co/W\n
significance to negative frequencies
We do note that these
the
spectrum.
frequency
such
to exponential
factors of the form
a term
and
form e~inat may be added to be equivalentto a
or
Thus
a positive frequency and a negative
attach
any particular
frequen\302\254
einat;
sine
frequency\n
angle
of
is abbreviated
c\342\200\236
as Ang
c\302\273.\n
182\n
the
form
to
combine
AND
DOMAIN
TIME
frequency
FREQUENCY
DOMAIN\n
associated
with actual sinusoidal
\ngenerators.\n
the
\nfor
wave
square
other
two
shows
8-28
Figure
8-27.
Tig.
\nof
8-26.
Fig.
of
\n(Ang)
spectra\n
V\n
2r\n
1
I\n
1\n
-5\n
-6
CO\n
1\n
CM\n
1\n
0
line spectrum
plots are
Separate
The
1\n
CM\n
that
\nterms
\nfor
c\302\253.\n
The
8-28.
shown
of the
line
(6)\n
for (a) a
spectrum
to
\naddition
\ntains
\nwaveform.\n
in Fig. 8-27, we
harmonics of the
different
the
many
6\n
corresponding to the sweepvoltage
made for magnitude and phaseangle
\nangular
to
1\n
5
CO\n
(a)
Fig.
8-5
wave. Comparingthese
i\n
i\n
V\n
2w\n
I
for the examples of Art.
spectra,
and the triangular
Chap. 8\n
the
waveforms.
fundamental,
harmonic
terms
square
wave,
while
of larger
(b)
a
tri\302\254
wave.\n
see that the
Fourier seriesmust
The
and
in
distribution
amplitude
be
quite
different
wave contains little in
triangular
the sweep voltage
magnitude
waveform
than for the triangular
con\302\254
Fourier
The
=
\nwot
\342\200\224
to
7r
of
of cB are known as
magnitude
continuous-frequency
line\n
spectra\n
waveform
of a periodic pulse of magnitudeV
marked on the figure as T, extendsfrom
period,
the
shows
The
a.
duration
\nand
and
integral
8-29
Figure
of the
plots
183\n
for discrete values
of frequency
components
such
only,
\nfrequency
8-8.
are
there
Because
FREQUENCY DOMAIN\n
AND
DOMAIN
TIME
8-8\n
Ait
03at =
following the practice
+*-, where,
established
in\n
volts\n
V\n
V\n
V\n
-1\n r\n
7I\n
r\n
last
the
rather
series
\nFourier
may be
series
of the
form
\nnential
cn
Since
the
\n(o/2)
and
1
=
^
(\342\200\224a/2),
the
-v
expo\302\254
Eq.\n
1\n
]+o/2
-a/2
jn\n
/(coo<)e_,n\"0<
do)ot
r\n
Ve-\342\200\231nmt
y-
^gjnaM/2
nir
\\
g\342\200\224jnwoa/2^\n
)\n
particular
sin
problem, the ratio
function
This analysis brings
cn
\twhat
\nnonrecurring
cooa
as
change
happens
pulse?\n
the
may
if the
(8-106)\n
/\n
written\n
be
(ntooa/2)
no)
(8-107)\n
od/2\n
a/T will be fixed,and
will
c\342\200\236
vary
(sin x)/x.\n
up a number
of
of the pulse
period becomes
width
(no)oa/2)\\\n
2tt ^sin
\\ no}0a/2
finally
\nT
the
y
= 2x/co0,the equation
mathematical
(8-105)\n
do>ot\n
2j
a
For any
except between the limits
J_\302\243\n
_
Now since T
(8-104)\n
becomes\n
-i
\342\200\224\n
2t
(2)
of
f+x\n
I
integral
2ir
\ndoes
T.\n
of the fundamental the
coefficientsfor the
computed for this problemfrom
has zero value
waveform
voltage
Cn
\nas
period
written\n
8-101
cn
+Q)t\n
used as the frequency
than to. The Fourier
too is
section,
2\n
pulse of duration a and
A recurrent
8-29.
a\n
2\n
\t\n
*-r
Fig.
a\n
questions
of
interest:
(1)
how
or the ratio a/T changes,
infinite, leaving us with one
and\n
the
answer
=
(T/a)
To
help
for
\none
AND
DOMAIN
TIME
184\n
FREQUENCY
\nof
there
view,
the
second
the
second
w0
is
of the
amplitude
frequency
+<i)t
\n2w,
etc.\n
We
\nas
begin
what
to
\napproaches
\nwill
\nthe
be
added.
Fourier
second
pulse.
shorter
the
pulse,
is smaller.\n
\342\200\234(t)t\n
+wt\n
16)
\302\243-5\n
(6)
\302\243-5\n
pulses with differentvalues of (T/a) and
holding a constant. The envelopeof
spectra
Because
is of the general form (sin x/x).
|cn| is plotted,
a- to
always
positive whereas (sin x/x) is negativefrom
Two
recurrent
line
corresponding
envelope
ampli\302\254
another point
the
for
\302\243-2\n
-80.
\nthe
(2) the
plots
From
components
\302\243-2
line
in two
V\n
(a)
\nthe
differ
1\n
\342\200\224cot\n
\nthe
8-30,
and
plot.
u\n
Tig.S
two
smaller
spectra
is
a trend that should help answeroursecond
question
As
the
an
in
the
case
of
period
aperiodic pulse.
happens
of
smaller
more
amplitude
components
infinity,
frequency
for
this limit in terms of the expressions
To accomplish
we begin with Eq. 8-100, which is\n
series,*
to see
=\n
/(\302\253oO
(8-108)\n
c\342\200\236e,w\n
n\n
the
substitute
and
f(<aot)
equation
for
giving\n
c\302\273,
=\n
/(wo0e-,w
dco01
e\342\200\231\342\204\242*1\n
(8-109)\n
]\n
* The
\nfor
the
following
Fourier
8\n
made in Fig.
plot
1\n
la)
Chap.
components are required to makeup
\nMore frequency
\nbut
= 5. The
2 and one for (T/a)
in the
by a factor of \302\243
are more lines because
smaller
is
\ntude
two plots are
first question,
\nrespects: (1) there are more linesin
DOMAIN\n
discussion
integr\302\253d
theorem.
is intended to
It
is not
providea
a rigorous
heuristic
derivation.\n
proof
or
motivation
Art.
8-8\n
We
next
let
coo =
Aw
as
of
n
the
as
limit,
i
=
new
variable
by
than
(8-110)\n
a\n
\342\200\224
\302\253\n
0,
a process of
becomes
summation
the
\342\200\224>
Aw
=
m
This
integra\302\254
the
g(a),
equation
(8-ui)\n
r
form
a slightly
in
written
be
\nmay
one
is
equation
that\n
as a Fourierintegral.It
calling the bracket term
=
dt\n
^
=
fit)
two
these
and
of what is known
different form by
as\n
g(u)
so
a
introduce
and\n
\ntion,
\nin
\302\273 and
T\342\200\224\342\226\272
185\n
write Eq. 8-109 in terms of fit) rather
integration
change to give the followingexpression.\n
w-
In
FREQUENCY DOMAIN\n
AND
If we
w.
limits
the
\n/(woO>
DOMAIN
=
nAu
\nletting
TIME
be
\nchlet
conditions,
/
f{t)e~iat
(8-112)\n
^
gio>)elul
da\n
(8-113)\n
a Fourier transform pair. These
used
to represent
f(t) provided f(t) satisfies the Diriin Art. 8-5, and if the integral\n
mentioned
constitute
equations
can
\nequations
j
(8-114)\n
is finite.\n
|cn| determined
Now
is
\nsion
\ncourse,
is
\ng(a)
term
The
series.
\nrier
since
da
finite
and
to
corresponding
of
The
da.
g(a)
the frequency spectrum
in
c\342\200\236
in the case
of
Fourier
the
integral
is
g{<x>)
vanishingly
amplitude
is an infinitesimal
However,
quantity.
in magnitude
and phase as
is plotted
da
Fou-
the
expres\302\254
small,
of
the function
the
frequency
corresponding
aperiodic
fit). No longer is this spectrum
\nspectrum
for
discrete
values
of frequency.
The function g(a) is a
only
\ngiven
function
\ncontinuous
for
all a. Because
of this difference in spectra,
is
sometimes
called
a continuous
while |cn| is a line
spectrum,
\n|gf(<o)|
In
terms
of the synthesis
of a pulse by addition of frequency
\nspectrum.
the
continuous
\ncomponents,
spectrum
requires all frequencies com\302\254
to
\nbined
as
required
by
Eq.
an
8-113.\n
For the single pulse shownin Fig.8-29,
\nto
infinity
in
opposite
directions,
the
the
other
frequency
two
having
spectrum
moved
may
be\n
DOMAIN
TIME
186
8-112
found from Eq.
AND
FREQUENCY DOMAIN\n
8\n
as\n
If
2ir J \342\200\224a/2
Q(u)\n
Chap.
*<!!$}\n
(coa/2)\n
2x
(8-115)\n
is similar in form to Eq. 8-107
for
recurring
pulses,
given
\nbut
is a continuous
rather
than a line spectrum. The plots of the math\302\254
\nematical
shown in Fig. 8-30 constitute the enve\302\254
expression
|(sin
x)/x\\
of the
\nlope
spectrum
|gr(co)| for the single pulse.\n
The
Fourier
transform
Eqs. 8-112 and 8-113,
pair of equations,
\nserves
to
illustrate
the relationship
of the time-domain function f(t)
\nand
the
the frequency
spectrum, g(ui).
frequency-domain
quantity,
a given
we can find the corresponding g(<a). And for a given
\nFor
f(t),
can
find the corresponding
similarly
f(t). The Fourier trans\302\254
\ng((\302\273), we
us
street
with which we can
with a two-way
\nform
provide
equations
domain
time
to frequency domain or vice versa. The same
\ngo from
exists
for the Fourier
series and the associatedconcept
street
\ntwo-way
and the time domain.\n
line
\nof the
spectrum
a single
can be synthesized
of
We know that the waveform
pulse
\nfrom
specified
by Eq. 8-115. Suppose that we
frequency
components
transmit
to the input of a system whichdoesnot
a single
pulse
\napply
of
this
The
\nthe higher-frequency
system will no
output
components.
function
be a square pulse but will be someothertime-domain
\nlonger
from Eq. 8-113 using the g(<*>) of the output of
\nthat could
be computed
As we change frequency
\nthe
system.
response, we\n
frequency-selective
This
equation
8-81.
Fig.
and
Input
waveforms
output
\nfrequency-selective
also
\na
frequency
Fourier
8-9.
Fourier
The
\nused
\ndomain
\ntransform
in
sensitive
An input and the correspondingoutputfor
in Fig. 8-31.\n
are illustrated
system
transforms
and their
time
change
the
and
for a two-terminal-pair
network.\n
response.
and 8-113,have been
to illustrate
the relationship of frequency
We
domain
have not yet discussedthe
concepts.
of these
which can be used in
equations
transforms,
last
section
time
property
relationship to Laplace transforms\n
defined
by Eqs. 8-112
solving\n
Art.
8-9\n
TIME
circuit equations in
\ntion
been
has
used.
FREQUENCY DOMAIN\n
AND
DOMAIN
much the same
as
manner
For
the
Laplace
Eqs. 8-112 and
reference,
187\n
transforma\302\254
8-113 are repeated
\nhere.\n
=
gM
gj
/(() =
\nof
the
gMe<-
J_
of these
character
transform
The
J'j(t)e-*\342\200\234dt
equations
=
%f(t)
Eqs.
four
\nThese
\nequations
and
\n7-1
the use
(8-118)\n
9(<\302\273)
(8-H9)\n
8-116 and 8-118 define the directFourier
transformation,
8-117 and 8-119 define the inverseFourier transformation.
are similar
in appearance
to the corresponding
equations
for
the
Laplace
transformation,
given in Chapter 7 as Eqs.
which
7-3,
are\n
m\n
=
m\n
=
\302\243f(t)\n
\302\243-'F(s)\n
Comparison
=
=
r* +y \302\260\302\260\n
jr\342\200\224.
/
\n2x7
and\n
(8-120)\n
f{t)e~atdt\n
1
\nences:
is emphasized by
=m
fc-Wco)
\nwhile
(8-117)\n
notation.\n
following
Equations
dw
(8-116)\n
F(s)elt
ds
(8-121)\n
>-y\302\273\n
F(s)\n
(8-122)\n
m\n
(8-123)\n
of these two sets of equations revealsseveral
differ\302\254
jo) in the Fourier transform occupies the sameposition
(1)
The
the
Laplace
(2) The letters F and g signifyfunctions
roles.
\nwith
similar
(3) The limits of integration in Eqs. 8-116and
\342\200\224
oo
in the
Fourier
transform
\nare
different,
corresponding to 0 in the
constants
transform.
(4) The
multiplying
(l/2x) and (1/2vj)
\nLaplace
this
is
a
matter
of
convention since
different
positions
although
\noccupy
in Eq.
8-111. Since we
with either f(t) or g(u>)
be associated
\nl/2x
may
of the two systems of transforms, the fac\302\254
\nare
the
similarities
stressing
for
to Eq.
8-117
\ntor
will
be shifted
from Eq. 8-116 for g(u>)
for/(\302\243)
1/27T
\nthe remainder
of this discussion.\n
\nas
s in
transform.
8-120
an
will
we
To illustrate the consequencesof thesedifferences,
study
circuits
\nexample of the computation of a Fourier transform.In most
\nstudied
interest has centered on what happensin a\n
in past
chapters,
\nof
an
after
circuit
This
a switch.
\ntime,
t
\342\200\224
0.
A
Chap.8\n
DOMAIN\n
FREQUENCY
AND
DOMAIN
TIME
188\n
of time corresponding to the openingor closing
of time is conveniently taken as the reference
instant
of
function
that has been used to denote the closing
instant
a
function
circuit is the unit
use of
the Fourier transform of u(t),
make
\nu(t). To determine
In this
to
8-116.
changed
case, the lower limit of the integral
\nzero, since u(t) has zero value for all negativet. In this
in which
\nsituation
f(t) has zero value before t = 0, Eq. 8-116
as the
\nwritten
with
the lower limit of zero, and is then
unilateral
out
the
have
\nFourier transformation.
Carrying
operations
just
from
we
removed
with
l/2ir
Eq.
\ndescribed,
have,
force to a
a driving
connect
to
\nswitch
step
we
Eq.\n
be
may
circuit
usual
be
may
known
we
8-116\n
r
=
dt
f(t)e~iat
*\n
iat
e~>'
I
dt\n
(8-124)\n
/oo
=
g(a>)
\n-1\n
e~>wt
-r-^
cot
(cos
no meaning, since neither
cot. This difficulty can be
factor
defined in the equation\n
has
equation
for infinite
\ndefined
\nconvergence
j
sin
(8-125)\n
(at)\n
3<\302\273\n
\n3<*\n
This
\342\200\224
=
Mt)
the sine nor
avoided
is
cosine
the
a
introducing
by
(8-126)\n
e-'W)\n
is a modified function and is real and positive.
This
difficulty
\nprocedure provides the convergence necessary to avoid
of
the
Fourier
and
\nin evaluating
computation
permits
Eq. 8-125,
= e~atu{t) into
as
0.
limit
\nform
as the
fi(t)
Substituting
Eq. 8-116
where
fi(t)
<r
the
trans\302\254
<r
factor
the
\nwithout
g{u>)
and\n
\ng(co)
this
From
\342\200\224\342\226\272
l/2ir
=
lim
=
lim
gives\n
[
/\n
a\342\200\224>0
JO\n
equation,
fi{t)e~iot dt =
e-(<r+;\302\253)<
we have
_
fa
lim /
e-*te-iot
1
lim\n
<r
<r\342\200\224>0
+
^
(8-127)\n
1\n
(8-128)\n
ja
jco\n
the Fourier transform
of u(t)
<r =
with
0
\nas\n
5\302\253(<)
=
(8-129)\n
^\n
Thus
a real
\nducing
\ntion
difficulty has been avoided
the
part to be added to j<a. Since s
Laplace
=
as a complex
s
we
+ ja,
number,
the convergence
is
defined
by
effectively
of
<r
intro\302\254
transforma\302\254
see
that
the\n
Ait
automatically has the
Laplace transformation
\never,
the
\nform
of
incorporating
transform
by
\nconvergence
Laplace
the
the
study
transform
Fourier
\nthe
transformation
Laplace
conveys
is conceptually
tie-in
this
\nseries,
\nthe
Laplace
transformation
is
\nthe
Fourier
transformation
and
\nextensive
of
compilation
the Fourier
more
physical
is more
extensively used.\n
an example being
are available,
excitations
READING\n
York,
Thomson,
\nby
Linear
Problems
Book
Inc.,
Co.,
Fourier
on
Wiley
(John
and Schwarz,
Inc.,
(Prentice-Hall,
and Wylie,
\n214r-219;
\nreading
response
of a system
to
step
Systems
Salvadori
228-241;
pp.
\nEngineering
\nHill
the
impulse, ramp function, square
function,
(Prentice-Hall, Inc.,
Laplace Transformation
Thomson,
is also discussed
The
convolution
23-26.
integral
pp.
1950),
and in such references as Gardner and Barnes,
pp. 37-38,
the
as
in
\nTransients
\n1942),
of
subject
see
etc.,
\nwave,
\nNew
the
the
Foster.*\n
and
FURTHER
\nsuch
derivation
Fourier integral and Fourier
Aside from this advantage,
important.
a more powerful mathematical tool than
Campbell
For further reading on
a heuristic
transformation. Since
meaning than the Laplace
from
transforms
Fourier
of
Tables
as
regarded
unifies two important
out of the
does
it
as
arising
\ntransform,
How\302\254
(8-130)\n
The preceding discussionmightbe
\nof
stronger
factor.
\342\200\234built-in\342\200\235
convergence
relationship of transforms
of electric circuits.\n
of this
in
\ntopics
of
=
\302\243f(t)
Recognition
advantage
of f(t) is identical with the Fouriertrans\302\254
by the convergence factor e~**;that is,\n
multiplied
f(t)
a
189\n
DOMAIN\n
FREQUENCY
AND
DOMAIN
TIME
8-9\n
Advanced
New
series,
Engineering
York,
see
& Sons, Inc., New York,
Differential Equations in
New
York,
1954), pp.
Mathematics
(McGraw-
1951), pp. 188-197. For further
Kerchner
and Corcoran, Alternating-
New York, 1951),Chap. 6.
series and integral is given
very
complete
The
Mathematics
of Circuit Analysis
Guillemin,
(John Wiley
New
\nSons,
York,
Inc.,
1949), Chap. 7. Chapter 5 of Wylie (op.cit.)
concise
on
these
subjects and is especially recommended. Two
very
information
on the Fourier
references
valuable
containing
and
are Fich, Transient Analysis in
spectra
frequency
\ntrical
New York, 1951), pp. 199-214;
(Prentice-Hall,
Inc.,
Engineering
and
General
Network Analysis (McGraw-Hill Book
Seely,
LePage
Circuits
\nCurrent
\nA
Wiley
discussion
(John
& Sons, Inc.,
of the Fourier
&
\nby
\nis
\nadditional
Elec\302\254
\nintegral
\nand
\nCo.,
Inc.,
New
York,
1952),
pp. 444-463.\n
* Campbell and Foster, Fourier
\nNostrand
Company,
Inc.,
New
Integrals
York,
1950).\n
for
Practical
Applications
(D. Van
TIME DOMAIN AND
190\n
Chap.
DOMAIN\n
FREQUENCY
8\n
PROBLEMS\n
8-1.
\nthe
Write
an
equation
in
terms
of unit
figure
for the nonrecurring waveform shown in
step functions.\n
1\n
1\n
1\t\n
1
1\n
2
4
J\n
Prob.
an
Write
8-2.
unit
of
\nterms
step
6
5
f
i\n
*\n
i
9
t\n
8-1.\n
for the nonrecurring waveform shown in
equation
functions.\n
10\n
8
12
3
4\n
(Jl\n
9\n
t\n
Ol\n
-10\n
Prob.
the
In
8-3.
\nincreases
to
\nnentially
to
waveform
\nThe
Write
\nnitudes.
\nfunctions.
8-4.
\nis
applied
\nMhenry.
A
waveform shown, the function suddenly
nonrecurring
b at the time t = 0 and then decreases
a value
expo\302\254
a at t = c before decreasingsuddenlyto zero.
a value
then
the same cycle with negativemag\302\254
goes through
an
for this waveform,
expression
using unit step
v =
Partial
answer,
voltage
an
to
pulse of 10 volts
RL series circuit
Plot
8-2.\n
the
waveform
of
be~[ln<-a/b)]t/cu(t)
\342\200\224
.\n
magnitude and of 5
duration
Msec
ohms
and
where R
2
the current as a function of
\342\200\224
time.\n
L =
10
A
8-6.
\nis
applied
=
R
\nwhere
and
\nrent
plot
a
\nthe
of
frequency
One
RL
series
a radio
Prob.
transmit\302\254
is
8-5.\n
shown\n
the equation for this
repeated,
(b) Suppose that
circuit
with R = 1 ohm and L
the
a
duration
shift
to
used
is
of staircase
cycle
figure,
(a)
it is not
\nassuming
\nto
/usee
waveform known
\342\200\234staircase\342\200\235
\nter.
and of 5
of time.\n
A voltage
8-6.
\nas
10 volts magnitude
circuit
=
ohms and C
0.05
pulse of
RC series
191\n
for the cur\302\254
equation
the current
waveform
function
a
\nas
100
the
Find
\nnf.
in
voltage
an
to
DOMAIN\n
FREQUENCY
AND
DOMAIN
TIME
8\n
Chap.
Write
voltage
this voltageis
= 1
Sketch
waveform
henry.
v(t),
applied
the\n
volts\n
6
waveform
to scale on
approximately
sec\n
8-6.\n
Prob.
current
f,
the same coordinatesas
\342\200\234staircase\342\200\235voltage.\n
\nthe
Show
8-7.
that
the
of the square wave
transform
is\n
1\n
F(s)
=\n
s(l
The
8-8.
\ntified
waveform
The
voltage.
\n\342\200\224sin
t
from
it
to
shown
equation
2x,
etc.
-f-
e~\302\260s)\n
in the figure is that of a full-wave
for the waveform is sin t from
Show that the
transform of this
function
rec\302\254
0 to t,
is\n
cathode
a
in
\nbeam
DOMAIN\n
FREQUENCY
Chap.8\n
is a sweep voltage used to deflectthe
Show that the transform of this
oscilloscope.
shown
waveform
The
8-9.
AND
DOMAIN
TIME
192\n
ray
is\n
\nfunction
pi
\\
1\n
~
as*
-
s(l
\302\253r*4)\n
the
waveform shown in
of the voltage
transform
the
Find
8-10.
/
{s)
\nfigure.\n
8-11.
=
F(s)
\ne0*,
(a)
-
^
8-12.
a).\n
(c)
a)2
+
(\302\253
\342\200\224
(s
^
\342\200\224
a)(s
find
\342\200\224
a)(s
the
\ni(t)
=
\ninput
\noutput
\nas for
8-14.
Laplace transformation
inverse
of
+
$\n
(b)
l)2
on a
Tests
\342\200\2242e~l
+
at
current
of
previous
The
(s + l)(s2
+
1)\n
certain network showed that
t
i(t)
\342\200\224
an
same
Answer,
of a half-wave
cos
cot,\n
v
=
\342\200\224
4e\342\200\230 3.\n
rectifier is given by
0
sis Cot
<
^\n
v(wt)
was
output
a unit
test?
output
the current
voltage was suddenly applied to the
0.
must
be applied to give
What
voltage
= 2e~l if the network remains in the
form
when
4e-8*
terminals
the
\342\200\224
c)\n
b)(s
1
8-13.
b)\n
functions.\n
following
(s2
+
o)(\302\253
\342\200\224
(s
6)
convolution,
By
the time functions corresponding to the
with the transform pair f(t) =
starting
\342\200\224
1 /(s
(\302\253
\nthe
functions
transform
\nfollowing
find
convolution,
By
~\n
T
<r
S*\n
t
o,\n
cos
at,\n
y
^
<*t
g
2x\n
the
equation\n
this
that
Show
FREQUENCY DOMAIN
AND
DOMAIN
TIME
8
Chap.
can be represented
waveform
periodic
193\n
by the Fourier
\nseries\n
t
1
v(ut)\n
Find
8-15.
8-16.
Draw
8-17.
The
of
\nfunction
2
the
Fourier
ut)
other
in
\342\200\224f(ut);
positive
line
words,
4ut
cos
+\n
jg
loop from r
negative
f(ut)\n
ut\n
0\n
85.0\n
120\n
77.9\n
30\302\260\n
75\n
135\n
77.8\n
45\302\260\n
77.8\n
150\n
75\n
60\302\260\n
77.9\n
165\n
49.7\n
75\302\260\n
85.0\n
180\n
0\n
90\302\260\n
90\n
coefficients for the
for this waveform.\n
Fourier
and
|fir(w)| and
sketch
the
to 2ir
/(\302\2530\n
105\302\260\n
49.7\n
spectra
spectrum
the
v.\n
8-18. For the waveform shownin
\nuous
2\n
\342\200\224
15\302\260\n
the
Determine
the
g
cos 2ut
loop from 0 to
0\n
\nDraw
2
line
the
ut\n
(a)
+
of the trapazoidal wave\302\254
series representation
8-10. Draw the line spectrum for this waveform.\n
Prob.
the
to
similar
cos
for the waveform of Prob. 8-14.\n
spectrum
table
following
gives the ordinates of a waveform as a
The
values
for r to 2?r are defined by the relationship
<at.
=
+
\n/(*\342\200\242
\nis
a\n
for
shown
\nform
+
first five harmonics, (b)
the
determine
figure,
contin\302\254
Ang g(u).\n
V\n
V\n
~
a\n
2\n
1\n
*
-V\n
Prob.
The
8-19.
defined
\nwave
8-18.\n
aperiodic function shown in
\342\200\224
to
only from
v/2
+ir/2.
\nspectrumand sketch
8-20.
An aperiodic
v(t)
and
\ntrum
represents
for
this
and
|gr(w)|
Ang
the figureis part
Determine
a damped
function
and
Ve~at
of
a
cosine
continuous
g{u).\n
function is defined by the
=
the
sin uot,
t ^
equation\n
0\n
Determine the continuous
oscillation.
sketch both |(7(w)| and Ang g{u).\n
spec\302\254
CHAPTER
the operationalmethodstudied
be used to introduce the concepts
In this chapter,
will
\ntransformation
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
9\n
the
by
of
Laplace
and
impedance
\nadmittance.\n
9-1. The concept
for networks has
differential
equations
functions
of the form\n
the
time-domain
to
\nrise
frequency\n
of
solution
The
of complex
Kne-t
where
a
is
s\342\200\236
=
Sn
the
Here
equation,
(or
\nquency
(9-2)\n
has been interpreted as radian fre\302\254
and it appears in time-domain equations
frequency)
angular
+ jun
<Tn
of
part
imaginary
co\342\200\236,
forms\n
the
sin
Radian
T, in
period
of frequency,
terms
in
cos unt
or
ccnt
(9-3)\n
of radians per secondand
in
fn,
cycles per second,orin
the dimensions
has
frequency
\nexpressed
\nthe
of the characteristic
as\n
\nexpressed
\nin
(9-1)\n
a root
number,
complex
given
be
may
of
terms
seconds, by the
equation\n
2_\n
=
=
co\342\200\236 2tr/\342\200\236
By
we see
9-2
Eq.
\ndimensionof
\n(being length
that
and
<rn
is (time)-1,since radian
of arc per length radius).
time.
unit
Since
I\n
that
\nsion
that the
is
<r the
\342\200\234something\342\200\235
unit.
logarithmic
for
of
quantity
<rn
The
usual
I
oe*nt\n
(9-5)\n
T\n
(9-6)\n
per
70\n
unit
unit
time
This
194\n
should
for the natural
unit is commonly
neper.
neper
per second.\n
the
be
must
<rn\n
evident
\nlogarithm
dimensionless
dimension
factor,\n
tln
\nsional
The
The
in dimensions.
an appears as an exponential
1
it is
is a
the
of
\n\342\200\234something\342\200\235 per
such
identical
be
must
co\342\200\236
(9-4)\n
y
be a nondimen-
(or Naperian)
used making the
dimen\302\254
AND
IMPEDANCE
9-1\n
Art.
The complex
ADMITTANCE
195\n
FUNCTIONS\n
sum\n
-
(9-7)\n
\302\253\302\273
<r\342\200\236
+
jun
as the complexfrequency. The imaginary
part of the complex
is the radian frequency (or real frequency),and the
real
\nfrequency
part
\nof
is neper
(rather than the misleading
complex
frequency
frequency*
\nterm
The
of complex
interpretation
\342\200\234imaginary
frequency\342\200\235).
physical
in
will
be
the
e,Ht
studied
function
\nfrequency
appearing
by considering
\na number
of special
cases for the value of \302\253\302\273.\n
= <rn +
and negative
Let
\302\253\342\200\236
(1)
positive,
zero,
jO and let <rn have
an
\nvalues.
The
function
of Eq. 9-1 becomes Kne9nt,
expo\302\254
exponential
is defined
<rn = 0,
When
so that
-
a
time-invariant
\nin
terms
\342\200\234direct
real
for
the
for
are
s\342\200\236
(2) Let
=
8n
of
\nterms
\na
unit
\nthe
the
sign.
A
frequency).\n
neper
9-1.\n
this
In
only).
frequency
(radian
case, the
becomes\n
=
The
equation.
o)nt
\302\261
j
sin
e\302\261,w
is
(9-9)\n
uj)
in
interpreted
usually
no actual physical significance)of
f the direction of rotation being determinedby
phasor,
positive
Kn(cos
exponential
model
physical
rotating
\n(a
\342\200\224
with time
of e9i
Variation
9-1.
possibili\302\254
in Fig.
Kne\302\261iUnt
Euler\342\200\231s
The
Jig.
three
\302\261
join
factor
\nexponential
is
voltage
current.\342\200\235
shown
0
(9-8)\n
which
and
current
of
variation
by
decreases
0 and
>
+ jO,
= Kn
Kneot
quantity
\ndescribedas
\nties
<rn
< 0.\n
<r\342\200\236
=
s\342\200\236 0
Kne\342\200\230*\342\200\230
\ntime
for
exponentially
becomes\n
term
\nthe
for
exponentially
decays)
\n(or
increases
which
function
\nnential
sign,
e+jWnt
(with
implies
counterclockwise
positive)
(or
clockwise
\nrotation, while a negative sign, e~iUnt
(or
implies
negative)
\nrotation. For positive rotation, the real part of
the
(or
projection
\non
the
real
varies
as
the
of
while
cosine
the
axis)
o)nt,
imaginary
part
on
the
varies
sine
as
the
of
This
o>nt.
\n(or
projection
axis)
imaginary
is
illustrated
9-2.
The
variation
of
the
\nconcept
by Fig.
exponential
with
is sinusoidal
\nfunction
time
and corresponds to the case of the
\nsinusoidal
\342\200\242The
terms
\n\342\200\234The
potential
state.\n
steady
used
radian
analog
magnitude
and
Air
by
W.
Force
H.
Huggins,
Cambridge
Report No. E5066, March 1951.\n
the word vector
in place
of phasor.
A phasor is characterized
with respect to a reference.\n
phase
\nResearch Laboratories,
texts used
t Many
\nby
and neper frequency were
frequency
in network
synthesis and analysis,\342\200\235
9-2.
fig.
ADMITTANCE
AND
IMPEDANCE
196\n
and imaginary and
phasor
Rotating
(sine
\tLet
(3)
=
s\302\273
this
For
\ncomplex).
jun
shows
expression
\nthe
of the
product
\nrepresented
real
9\n
Chap.
axis
projections\n
cosine).\n
the frequencyis
general case and
case,\n
= Kne(9n+iUn)t =
Kne'nt
This
and
is the
(this
<r\342\200\236
+
FUNCTIONS\n
for
result
by the rotating
(9-10)\n
has a time variation
such a term
that
Kne9nteiUnt
for
=
s\342\200\236
\302\261ju>n.
phasor model, the other
term
\na
and
in Fig. 9-3.
The
of this
projections
imaginary
a
Such
time.
with
changes
\nphasor is illustrated
\nreal
func\302\254
be thought of as
with
a magnitude
phasor
rotating
\nwhich
expo\302\254
can
result
This
an
by
is
term
One
or decreasing
increasing
\nnentially
\ntion.
is
which
=
and
s\342\200\236 <r\342\200\236
\nphasor are\n
Re(e**0
\nand
9-8.
Fig.
\ning
in
Rotating
magnitude
phasor
with
Im(e*\"\342\200\230)
=
e~9nt cos
=
e~9nt
decreas\302\254
for
time.\n
a
phasor
rotating
sin
o>nt
(9-11)
<ant
(9-12)\n
in the positive
\ndirection and negative a. These
9-4. Such waveforms have been classified
pro\302\254
are
\njections
\nas
\nconcept
\nreal
this
part
Fig.
of
there is nothingreally
in the
frequency. The imaginary part of complex
(or real) frequency corresponds to oscillations.
neper frequency, corresponds to
frequency,
discussion,
of complex
the
\nquency,
in
sinusoids.\n
damped
From
shown
radian
complex
we see that
new
fre\302\254
The
expo-\n
nential
or
decay
two
the
of
\342\200\234kinds\342\200\235
\nquences
are
\ncomplex
frequency.\n
should
We
work
the
\nof
or
ghostlike)
words
the
\nborrowed
\nas
quantity
in the sense
often
\nwe
\nematician\342\200\231s
9-2.
\nnext
determine
\neach
of
the
as
is
the
mathematicians
(which
about
connotation
have
We
or function
phase). The
the
given
by
corresponding
to the transform
of
parameters.\n
expression for the
The time-domain
Ohm\342\200\231s
in
law
v(t) = Ri(t)
The
from
invis\302\254
math\302\254
physical
transform
V(s)
=
the
or
equations
RI(s)
a
current
generalized)
impedance. The reciprocal
(or generalized) admittance. We will
for the impedance and admittance for
expression
network
Resistance.
\nresistor
frequency.\n
admittance\n
transform,
(or
the transform
an
the
no
name\342\200\224
imaginary (that is
parts of a quantity
of magnitude and
of a voltage
transform
of the
defined
is
complex
not physicallyreal.
\342\200\234imaginary\342\200\235
carries
and
impedance
as
defined
\nis
\nratio
under one
conse\302\254
us!\n
Transform
ratio
that it is
terms
\342\200\234imaginary\342\200\235
about
The
part
the
even though
semantic difficulties in the use
of a complex quantity. The
is not physically
distinct
in
reinterpret
\nuniverse
one
\342\200\234real\342\200\235
and
two
of
designations
s is
e~\342\200\242*
where
against
as
of a
part
\nimaginary
\nible
\342\200\234imaginary\342\200\235
same,
the two concepts
unify
guard
constantly
the
is
Time variation of
9-4.
Tig.
(depending on sign) or to no
We have talked about such expo\302\254
of the time constant. Since the role
frequency
we
different,
197\n
increase
frequency.
in terms
neper
before
functions
\nnential
FUNCTIONS\n
ADMITTANCE
exponential
zero
for
\nvariation
\nof
AND
IMPEDANCE
9-2\n
Art.
or
voltageacrossa
forms\n
i{t) =
Gv(t)
(9-13)\n
GV(s)\n
(9-14)\n
are\n
I(s)
=
m\n
the
Following
transform
\nand
definitions
given
admittance,
we
Z(s) is the
F(s) is the
transform impedance,
transform
\nlent
(9-15)\n
and\n
_
Y(s)
the actual
be replaced
Two
such
quantities.
R
G
(9-16)\n
admittance.\n
shows
can
current
and
\nvoltage
transform
which
schematic
The
=
Z\302\253
_
where
impedance
have\n
=
where
for the transform
above
Chap.9\n
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
by
time-domain
a diagram to represent
resistor and the
equiva\302\254
are shown in
diagrams
(a)
Fig.
9-5.\n
(6)\n
Resistor
9-5.
Fig.
and admittance.\n
impedance
of
the
actual
The time-domain schematic is a representation
physical
element
diagram is composed of time-domain
\nsystem. The transform
actual
element is replaced
but
letter
for the
the
symbol
\nrepresentations,
an
\nby
or admittance
impedance
symbol.\n
The time-domain relationship betweenvoltage
inductor
is expressed
by the following
Inductance.
in
\nrent
an
and
equations.\n
v(t)
The
=
terms,
we
=
this
\nLi(0+)
L[sl(s)
^
(9-17)\n
expression
Designating
Fj(s)
=
F(s)
is\n
- i(0+)]
(9-18)\n
=
7(s)
+ Li( 0+)
(9-19)\n
transform of the appliedvoltage,
resulting from the initial current in the
across Z(s) as Fi(s),
transform
the
voltage
+ Li{0+), the transform impedance
expression,
V(s) is the
is a transform
voltage
\ninductor.
\nwhere
j
have\n
Lsl(s)
In
Jj
for the voltage
equation
F(s)
the
~
and
L
transform
equivalent
Regrouping
cur\302\254
and
becomes\n
Vx(s)\n
\342\200\224
Z(s)\n
m\n
=
Ls\n
transform
The
imped\302\254
equivalence
in Fig.
for the current is\n
is shown
schematic
time-domain
\nto the
transform
source
a voltage
and
199\n
diagram thus containsa
due to. the initial current. This
The equivalent transform
\nance
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
M\n
Ait
equation
9-6.\n
(9-20)\n
initial-value
The
Li
\nlinkages
can be evaluated
w-1(0+)
integral
in terms of
as\n
tr'(o+)\n
be
for 7(s) may
The equation
=
dt\n
(t)
Lt*(0+)\n
(9-21)\n
1-0+\n
-/\342\200\242\n
rewritten\n
iEW+S2\302\261l\n
s
L
n
this
(9-22)\n
s\n
- m
F(s)
or\n
In
flux
-\n
\302\273\342\200\242(o+)\n
(9-23)\n
source
t'(0-f )/s is an equivalenttransformcurrent
initial
from
the
current in the inductor. Designatingthe
in
current
as Ii(s) = /($) \342\200\224
the
transform\n
Y(s)
becomes
....
r..\n
equation,
\nresulting
\ntransform
admittance
=
=
F(s)
m
^\n
rs
The equivalenttransform
diagram
\nthus
contains
and
\nvalue
1/Ls
\nrent
source
\nThis
equivalent
\ntime
domain
\nFig.
admittance
an
equivalent
in
defined
Z(s)
of
cur\302\254
note
9-23.
Eq.
for
schematic
the
in
is shown
diagram
We
9-6.
an
that,\n
=
=
U
(6)\n
(9-25)\n
9-6.
Fig.
Capacitance.
\nrelationship
time-domain
The
between
i(t)
L
and current for
voltage
=
and
C
(a)
Impedance
diagram
for
\302\273dmi\342\200\234ancediagram
\342\200\242(b)
a capacitoris
i(t)
v(t)\n
for\n
L.
as\n
given
dt\n
(9-26)\n
\342\226\240H\n
The
equivalent
transform
for the voltage
equation
V(s)
=
i
cL
[
^
1\n
\302\253\n
*
J\n
expression
is\n
(9-27)\n
where
initial voltage of the
is the
q(0+)/C
is
\ncharge polarity,
\342\200\224
This
Vo.
i
the
of
\nratio
transform
the
Designating
transform
-
/(\302\253)
voltage
the
Vo/\302\253,
is\n
=
7(s) '
_L\n
(9-29)\n
Cs\n
an equivalent
charge thus has
an initial
with
(9-28)\n
^
Vi(s) = V(s) +
to the transform current
7(s)
capacitor
be written\n
of Z(s) as
voltage
the
to
due
which,
+
F(\302\253)
=
The
capacitor
may
equation
Chap.9\n
FUNCTIONS
ADMITTANCE
AND
IMPEDANCE
200
in
\nshown
\ncurrent
my\n
9-7.\n
Fig.
of
=
for the
equation
expression
/(\302\253)
is
combination
this
transform
The
The
\342\200\224v(0+)/\302\253.
of
\nschematic
V0zfcC\n
source
voltage
a transform
\nhaving
1/Cs
impedance
a
with
scries
\nin
an
with
diagram
itf).\n
transform\n
9-26
Eq.
C[\302\253F(\302\253)
-
is\n
K0+)]\n
(9-30)\n
lrc\n
or\n
-
CeV(s)
-
1(a)
CF,\n
(9-31)\n
9-7.
Fig.
\nC,
(b)
for
diagram
to
transform
-
^
the
-
an
-
the
C70,
becomes\n
(9-32)\n
transform
equivalent
of value Cs
admittance
/(\302\253)
Ca
CVo. This schematic
of value
source
\ncurrent
\nFor
of
representation
/,(\302\253)
voltage
with an initial chargehasan
The capacitor
\nmatic
Y(a)
as
V(s)
current
transform
the
Designating
C.\n
\nin
current
transform
of
\nratio
diagram for
(a) Impedance
admittance
in parallel
sche\302\254
with
a
is shown in Fig.
9-7(b).
capacitor,\n
*w
(9^>\n
-mmn
9-3. Seriesand parallel
combinations\n
In
\nseries,
\nTo
\nreserved
this
we will
section,
and
parallel
series-parallel
schematic
simplify
for
the
consider
resistor
diagrams,
together
the impedance and
combinations
we
with
will
admittance
of
of different elements.
the
use
symbol normally
the letters Z(a)
or
Y(e)
to
des\302\254\n
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
9-3\n
Ait
201\n
We will not consistentlyuse
a transform
broken
\nthe
zigzag
impedance symbol, but either
element symbol depending on which is most
or the actual
\nthis
symbol
the series combination
Consider
of ele\302\254
and
\nconvenient
descriptive.
and the equivalent transform impedance
in Fig.
\nments
shown
9-8(a)
In Fig. 9-8(a), the same current i(t)\n
in Fig.
shown
9-8(b).
\ndiagram
a transform
ign&te
or admittance.
impedance
line
for
\t\n
nnro^nnr
\342\200\242
\342\200\242
\342\200\242
1>\\
Rt
R\\
1(\342\200\224|(
\342\200\242
\342\200\242
\342\200\242
Ci
\342\200\242\n
\342\200\242
\342\200\242
Cj
v(t)\n
(\320\260)
\nila)\n
*
'\302\273\342\226\240
V\\ArAAAr-\342\200\224WWW-\342\200\224-WWW\342\200\224*\n
Zci
Zi,
ZLl
Zr,
Zr,
zc,\n
Vb)\n
\342\226\240
'\342\226\240
I\n
(\320\261)\n
Fig.
\nments.
elements,
sum
=
V(s)
this
Dividing
of
\nvoltage
and
so in Fig.
VRl(s)
Z(s)
to v(t).
+
ele\302\254
of
is,\n
... + ViM +
by
element
each
that
V(s);
equation
I(s)
... +
+
VcM
and recognizing that
by the current
divided
\342\200\242
\342\200\242
\342\200\242
(9-34)\n
the ratio
for that
of
the
is
element
have\n
we
\nimpedance,
to
networks.\n
9-8(b), J(s) is commonto all
the sum of the drops of voltage
law,
Hence the transform of all voltages
voltage
is equal
elements
elements
\nthe
Impedance
Kirchhoff\342\200\231s
By
all
\nfor
all
in
flows
of series
9-8.
= Zr,(s)
+
\342\200\242
\342\200\242.
+
+
Zl^8)
.
\342\200\242
\342\200\242
+
Zct(s)
+
...
(9-35)\n
n\n
or
Z(\302\253)
=
V
(9-36)\n
Z\342\200\236(s)
1\n
in
\nelements
It
where n is
the total number
of
series.\n
be recognized that in performing such a summation,
the
are not being combined.
Rather, only a characteristicfea\302\254
the
element
is being summed and added to
(its impedance)
of
a
of
\ncharacteristic
Consider
9-9(a)
of elements,
should
\nelements
\nture
combination
series
a
for
and
another
element.\n
next the parallel combination of elementsshown
in equivalent
transform form in Fig. 9-9(b).In
in
this
Fig.\n
net-\n
\nis
the
Chap. 9\n
is the same across all elements,
andso F(s)
the sum of\n
From Kirchhoff\342\200\231s
current
law,
drop v(t)
voltage
for all elements.
same
the
work,
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
202\n
(h)\n
the
currents
\nthe
network;
that
the
and
If
this
\nof
the
=
\342\200\242
\342\200\242
\342\200\242
+
*\302\243,(<)
transform
(9-37)\n
is\n
\342\200\242
\342\200\242.
+
+
/c,(s)
...
(9-38)\n
it is recognizedthat
the voltage transform is transform
to
transform
equation
\342\200\242
\342\200\242.
by F(s) and
is divided
current
+ icXt) +
...
+
+ ILl(s)+
...
+
/o,(s)
equation
ratio
the
admit\302\254
results\n
there
\ntance,
+
iai(0
corresponding
I(s)
to
supplied
is,\n
=
i(t)
networks.\n
the total current
is equal to
elements
the
in
of parallel
Admittance
9-9.
Fig.
=
Y(s)
Yai(s) +
\342\200\242
\342\200\242
\342\200\242
+
+
FLl(s)
... +
YCl(s)
\342\200\242
\342\200\242
\342\200\242
(9-39)\n
+
W\n
or
=
Y(s)
Yk(s)
^
(9-40)\n
k~\\\n
For a series-parallelnetwork,rules
for
admittance
of
\nand
\nsingle
equivalent
with
\nillustrated
Example
In
\ntion
the
a until
where n is the total
of
number
in parallel.\n
of elements
\nall kinds
of elements,
combination
a parallel
for
be
can
used
a number
successively
or
impedance
the
admittance.
of impedance
combination
to reduce a networkto a
This procedure will be
of examples.\n
1\n
series
such
circuit
a time
shown
that
switchK is
a current 70 flowsin the
in Fig. 9-10, the
inductor
held
in
and
posi\302\254
the\n
Art.
9-3\n
is
capacitor
to
\nthrown
\nThe
to
charged
I(s)
transform
\nmarked with
RLC
9-10.
Fig.
find
to
is
problem
HJNUIUNb\n
203\n
Vo. At that instant, the switch is
the circuit to a voltage sourcev(t).
voltage
6, connecting
position
AINU:
AUMIII
AND
IMPEDANCE
and so i(t).
An
circuit
equivalent
The
circuit.\n
Fig.
9-11.
the
and
values
of
derivations
\nthe
\nrevised
The
\nOhm\342\200\231s
law.
V(S)
iy_
current
I(s)
_
\nw
Z(s)
This
transform
\nthe
corresponding
\nthe
current, may be found
by
the total transformvoltage
total transform impedance.
Then\n
i(t)
and
+
+
L/o\302\253
Rs
-
F0
/n
^
}\n
+ l/C
be expanded by partial fractions to find
the inverse
Laplace transformation. This
can
by
found
been
system,
a transform
is given as
V1(s) + LIo - Vo/s\342\200\234
sVris)
R + Ls + 1/Cs
Ls2
equation
has
\nsolution
by the
divided
network
the
\nin
I($),
for
9-10.\n
voltage source values are taken from
given on pages 199 and 200. In this
equivalent
this
section
current
the
form,
imped-\n
diagram
Equivalent
\nimpedanceof Fig.
ance
diagram
impedances is shown in Fig.9-11.
without
automatically
the differential equation of
initial
the required
incorporates
writing
\nconditions.\n
2\n
Example
network
of the
dual
The
the switch
\nnetwork,
of Example 1 is shownin Fig.9-12.
K i is opened
Fig. 9-12.
rent
\ni
=
\nthe
\nalent
at an instant
when
the
In
this
cur-\n
inductor
Parallel RLC network.\n
is charged to F0. At the same
capacitor
the
transform
of
A2 is closed. It is requiredto find
0, the switch
can
be
node
so that v(t)
determined. Fromthe equiv\302\254
voltage
F(s)
is
70
and
admittance
instant,
the
diagram
shown
in Fig.
9-13, the transform
voltage\n
is found
F(s)
Cs
Y(s)
9-13.
\ndomain
CT0s
+
Cs2+
1/Ls
h
/\302\253
^on\n
'\n
^
+
Gs
1/L
of Eq. 9-41
(and
for admittance
diagram
\342\200\224
of
could\n
9-12.\n
Fig.
by inspection). The correspondingtimecan be found by taking the inverse Laplacetrans\302\254
above
transform
has been expanded by partial
the
after
\nformation
s-^i(s)
written
v(t)
voltage,
_
of the transform
Equivalent
been
have
+ G+
dual
is the
transform
therefore
\342\200\224
^\302\260/s
^i(s)
_
w
Fig.
Chap.9\n
as\n
t/y\302\253a \342\200\224
This
FUNCTIONS
ADMITTANCE
AND
IMPEDANCE
204
\nfractions.\n
3\n
Example
In
\ntion
\nmine
this
make use of the lawsfor the series
combina\302\254
and the parallel combination of admittance to deter\302\254
shown in Fig. 9-14, it will
the network
be
assumed\n
we will
example,
of impedance
current.
In
is initially relaxed (no current, no charge)and
that
=
0. It is requiredto find the current in the
was closed at t
\nthe switch
The imped\302\254
I(s).
\ngenerator i{t) by finding the transformof this current
the
1-ohm
of the
branch
resistor
and
\nance
2-henryinductoris\n
containing
that
the
network
=
Z(s)
This
\nThe
impedance
is in
admittances
be
-
+ 2s
(9-43)\n
2/s of the
with the impedance
parallel
may
1
added
directly.
+
2s2
r.
impedance
from
s
+ 2
+\n
(9-44)\n
\n2(2s
The
capacitor.
Thus\n
a to
Zobis')
&
=\n
is
the
1
Yab(s)
reciprocal
=
2 (2s
2 s2
of the
+
+
1)\n
admittance;
thus\n
1)\n
s -+- 2\n
(9-45)\n
total
The
resistor.
i
-z
total
This
FUNCTIONS\n
ADMITTANCE
is now found by adding to
Then the total impedance
impedance
1-ohm
the
\nof
AND
IMPEDANCE
9-4\n
Art.
+
in series
impedance
impedance
is\n
5s +
4\n
(9-46)\n
2s2 +
+ a + 2
2s2
the
\302\243\342\200\236&(\302\253)
_ 2s2 +
+1)
2(2s
i
1
205\n
a
2\n
+
of the voltage
with the transform
source\n
\342\226\240AAAr\n
2**+\302\253+2\n
9-16.
Fig.
of
\npedance
in
shown
is
transform
\nfor
Fig.
that
9-4.
several
When
by
Ohm\342\200\231s
(9-47)\n
4)\n
at t = 0
more
somewhat
the
if
or
compli\302\254
theorem\n
in a network,
sources are present
\nvoltage
as far as
sources,
one
by
due
\ntheorem
law
are involved.\n
Norton\342\200\231s
or current
+
or
branch
the current in
at one node are concerned, may be taken into account
due
to ThGvenin*
(and the dual of this theorem
of all
effect
net
\nthe
and
voltage
problemis
sources
voltage
theorem
Thevenin\342\200\231s
flowing through it
at t = 0, the
several
since
[(\302\253+
a current
has
is charged
\ncapacitor
\ncated,
Z(s)
diagram of
9-15.\n
now be found
2(2s2 + s + 2)\n
l)2 + 4](2s2 + 5s
V(s)
~
inductor
the
If
Equivalent
is,\n
'
.
9 -16.
\nFig.
The current may
quantities;
1(
IW
Fig.
9-14.\n
9-16.
Fig.
for im
diagram
Equivalent
the
a
to
\nNorton).\n
The network
\n(less
one
One
branch)
\nactive and
passive
\nany
\nsources\n
Fig.
that
Suppose
The
\nnetwork.
*
shown
are
\nbox
\nNorton
Comptes
of
the
Arbitrary
elements\n
network.\n
in the current in one branchof a
interested
and the remainder of the network as
single branch
in Fig.
9-17. We will assume that the remainderof the\n
we
are
was first
theorem
This
\njournal,
9 -17.
branch
\ncontaining
rendu*,
Bell
in
Telephone
a
proposed
by
M.
L.
The dual of
Laboratories.\n
1883.
Thdvenin
Thgvenin\342\200\231s
in the
theorem
French scientific
is due to E. L.
IMPEDANCE AND
206\n
is
network
of
\nnumber
a
\nIf
\nthat
voltage
the
current
source
may
as far as
\nentire network
\nalent
\nbeing
considered.
If
affecting
the
\nwithout
to the net effect
this onebranchis
in the
connected
generator
it contains an arbitrary
waveform as a function of time.
that
in the branch under consideration
andis
in this branch is equal to zero at all time,
be said to be an equivalentvoltage
in
source
it is identical
that
\nthe sense
9\n
Chap.
inserted
is now
until
\nadjusted
of arbitrary
sources
voltage
generator
and
complicated
arbitrarily
FUNCTIONS\n
ADMITTANCE
current
no
network.
of
all
this
With
concerned.
the
in
generators
equiv\302\254
circuit, no current flows in the branch
flows, the branch could be broken
The voltage across the network termi\302\254
removed is the voltage the
Thus the equivalent generator
for polarity.
\nalent
generator
except
same
as the voltage measured by removing the
is
the
circuit
the
\nand
considering
voltage but of opposite
open
Now
if the
equivalent
voltage generator is placed in the loop
with
all active sources within the
\nconsidered
polarity
reversed,
them with short circuits, and the
could
be removed
\nwork
by replacing
will be the same as in the original
in the
branch
The
is
of an equivalent
\nconcept
voltage source for a singlebranch,
in Fig. 9-18. This
is illustrated
\nthe basis
of
theorem,
in the
to that
of Fig. 9-17 as far as the
is equivalent
one
\nwork
and generator
branch
the
with
\nnals
of
equiv\302\254
branch
\nvoltage
polarity.\n
being
net\302\254
network.
\ncurrent
which
Th6venin\342\200\231s
net\302\254
current
\nbranch is
concerned.\n
network
The
\n(less one
\nall
active
One branch
branch)
sources
\ncontaining
\nany elements\n
\nshort-circuited\n
Fig.
To
time-domain.
\nare
\nonly
to
find
the
currents,
the time domain.
The
and the resulting
to the frequency
convert
Laplace
network.\n
equivalent
far apply to
time-domain
are
considered
\nrents
Th^venin\342\200\231s
thus
made
statements
The
9-18.
transforms
domain it is
cur\302\254
voltages
necessary
for all time-domain quantities
\ninvolved.\n
we can say that by Th^venin\342\200\231s
\ntheorem
we
network
shown in Fig. 9-18with
one
have
the equivalent
one passive
branch
\nvoltage
(although it is not necessarythat
source,
\nit
be
and
with
a network
passive)
containing passive elements only.
\nWe
in terms of an equivalent transform
consider
this
network
may
and
of network
elements. These impedances
\nvoltage
by
impedances
be
combined
rules for series and parallel combinationpre\302\254
\nmay
by the
To
\nviously
summarize
discussed.
our
Finally,
discussion,
the
entire
passive
network
may be
made\n
let
and
\nZtq(s)
branch is given as\n
in this
current
the
Then
207\n
Let this impedance be
impedance.
the branch being consideredbe Zfcr(s).\n
transform
a single
of
the Impedance
to
equivalent
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
9-4\n
Art.
AAAr\n
Z^B)\n
Vtq(s)\n
=\n
m
(9-48)\n
+
Z\302\253q(s)
\nalent
for
The equiv\302\254
in Fig. 9-19, does not
the one
than
other
branch.
given
shown
circuit,
\nhold
branch
any
and
over
start
As far
as the current
network
\nthe
as
\n(1)
the
\nalent
\nof
transform
a
replaced
open-circuit
and
(2) as
branch
\nimpedancefor current
of
dual
Thevenin\342\200\231s
the
concerned,
of
remainder
theorem
is
Norton\342\200\231s
9-17. The
theorem.
in
current
Once
more
the
single
being
placing a current
may
in parallel
with
the branch
and adjusting the current until the
across
the
branch
is zero. The voltage across the branch is
the
current
from the network is just balanced by an
\nsource
\nvoltage
\nzero
because
\nsite
current
the
from
\nall
parallel
current
source.
one
active
One
branch)
sources
\342\200\242ejlfl\n
Fig.
source
\nall
sources
\nand
the
Norton\342\200\231s
equivalent
branch
elements\n
network.\n
a short circuit may be placedin parallel
with
the
the network. The current in the shortcircuit
affecting
be zero,
because
there will be a current from the equiv\302\254
which
cancels
the current from the network. If
exactly
is zero,
without
actually
9 -20.
across\n
\ncontaining
\nany
\nremoved\n
branch
Since the voltage
network
The
\nbranch
to zero by
oppo\302\254
\n(less
\nalent
be
sources.\n
be reduced
considered
\nbranch
\nwill
can
theorem
by an equivalent network having:
source, the transform of that voltage
terminals
resulting from the removal
shown in Fig.
network
\nconsider the
the
is necessary
series transform impedance, an equiv\302\254
of the network from the terminals
to that
equal
with
all energy sources replaced by their internal
for
and infinite
sources
voltage
impedance
branch,
\nimpedances\342\200\224zero
The
equiv\302\254
network.\n
is needed, it
Th&venin\342\200\231s
in one branch is
voltage
the
at
impedance
the
be
may
\nappearing
\nof
Th&venin\342\200\231s
transform
\nalent
follows:\n
as
\nstated
theorem.
the
reapply
9-19.
rig.
current
branch
If another
consideration.
\nunder
\nto
to a
only
\napplies
that this analysis
be recognized
must
It
Zbr(s)\n
within
equivalent
the
network
current
are replaced by their internal impedance
source which caused the voltage across the\n
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
208\n
Chap. 9\n
(but with opposite direction) is placedin
with
the
then this network, shown in Fig. 9-20,is
\nparallel
branch,
to the
network.
This equivalent current source has
\nequivalent
original
\nthe
value
and
is found by short-circuiting the branch
direction
that
\nin
consideration
and
the current in the short circuit. In sum\302\254
measuring
can
be stated
as follows:\n
Norton\342\200\231s
theorem
\nmary,
to
reduce
to
branch
zero
As far as the voltageacross
network
the
\nof
transform
a
\nin
short
circuit
an
\nadmittance,
and
sources
\nage
transform
net\302\254
equivalent
internal
their
by
\nreplaced
across
terminals
the
from
\nwork
an equivalent network having:
the transform of that current
the branch, and (2) as parallel
to that of the
admittance
equal
of the branch with all energy sources
be replaced by
current
source,
may
a
as
\n(1)
infinite
in
shown
two
admittances.
is given
voltage
=\n
7(s)
node basis network
transform
parallel
unknown
/\342\200\242,(\302\253)\n
as a
9-21
Fig.
The\n
as\n
m\n
+
YU*)
Y(%)
1VW\n
volt\302\254
sources.\n
for current
impedance
and
source
current
a
for
impedances\342\200\224zero impedance
The equivalent networkis
\nwith
the remainder
is concerned,
branch
any
YU*)\n
(9-49)\n
Norton\342\200\231s
network.\n
\nform
\ncuit,
\nThese
will
operations
Example
The
\nswitch
reduce
\nto
an
form
the
equivalent
series
simple
illustrated
of the current
by two examples.\n
is unenergized
until the instant t
be
network
of any
can be
cir\302\254
found.
4\n
shown
network
K
\nto
the transform
circuit
this
from
and
trans\302\254
equivalent
will allow us
theorems
two
These
9-21.
Fig.
It is required
is closed.
/?!
to find the
current
= 0,
the
when
flowing
ii(t)
in\n
L]\n
\302\253\342\200\224VA\342\200\224\n
10
1
ohms
henry\n
jr*1\n
V _=.
100 volts\n
10
Fig.
the
resistor
\nthe
ohm,
\nelement
I2\302\273. The
the
henry,
impedances
9-22.
values
and
Two-loop
given
in Fig.
ohms\n
network.\n
on the
the farad.
is shown
10
ohms\n
schematic
have the units of
An equivalent schematic
9-23.
Th&venin\342\200\231s
theorem
showing
will
be\n
Ait
\nthe
10-ohm
impedance
9-23.
Fig.
the
the
of
\nance
and
network
series
resistor;
\ncircuited
R%.
The
imped\302\254
thus\n
10(100/s)
1000\n
_
s(s +
+ 10
10 + s
(9-50)\n
20)\n
with the voltage
network
source 100/s short-
is\n
+
10(s
10)
(9-51)\n
Zeq{$)\n
\ns
The
net\302\254
9-22.\n
by 10 ohms,the
this current
multiplying
of the
impedance
for Fig.
schematic
Impedance
VoeKS)
The
and
L%
simple series network, and the voltageacross
will be the open-circuit voltage for the Th6venin
is found by finding the current in\n
This
voltage
network.
\nequivalent
containing
209\n
is a
remains
that
\nwork
branch
the
disconnecting
by
applied
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
9-4\n
current
T/
20\n
is\n
1000/s(s
Vqc(&)
\\
W
which
9-48,
by Eq.
transform,
+
10(s + 10)/(s +
Zeg(s) + ZUs)
20)
\t\n
+
(s
20)
+
10)\n
to\n
simplifies
=
s($2
This
may
equation
s(s!
With
Ki,
Ki,
+
and
40s
-
js ^ (s +
time-domain
-5
\nformation
=\n
s + 10
s
current
i(t)
+
10)
as\n
K,\n
(s +
the current transform
Kz evaluated,
3.33
The
K,
- g.
+ 300)
m
^9\342\200\23453)\n
300)
by partial fractions
be expanded
1000
+ 40s +
{
30)
becomes\n
1.67\n
+ s
+
is found by the
(9-55)\n
30\n
inverse Laplace
trans\302\254
as\n
i(t)
=
3.33
-
5e-10\342\200\230
-f-
1.67e-*0<\n
(9-56)\n
'\n
a
As
this
check,
ADMITTANCE
AND
IMPEDANCE
210\n
reduces to the
equation
9\n
Chap.
FUNCTIONS\n
correct valuesfor initial
and
conditions.\n
\nfinal
5\n
Example
in Fig. 9-24, it is requiredto find
the
current
\nin the
resistor
R%. The equivalent
impedance schematic is shownin Fig.
It is assumed
that the capacitor C2 is initially unchargedandthat
\n9-25.
at t = 0. Thevenin\342\200\231s
theorem
is applied
at ter\302\254
switch
K is closed
\nthe
\nminals
shown
network
the
In
the
and
a-a',
and equivalent voltage at\n
impedance
equivalent
AAA/
1\342\200\224\n
Ao\n
\342\200\224
-fe.\n
*1
9-24.
Fig.
Fig. 9-25.
network.
Two-loop
Th6venin\342\200\231s
will be
terminals
these
the
of
\ncombination
The equivalent impedance is a
found.
of two
impedance
+
(Ri
=\n
Z\342\200\236(\302\253)
Ri
current
The
/*(\302\253)
Ri
E2
through
parallel
thus\n
1/Cis)l/Czs\n
+ I/C2S\n
1/CiS
(9-57)\n
(Vo/s)(l/C2s)\n
=\n
Us)
and\n
branches;
of\n
equivalent
9-24.\n
Fig.
(9-58)\n
1/C2s\n
\342\200\234b
1/CiS
is\n
Voc(s)\n
=\n
R2\n
Zeq(s)
V0/C1\n
-f-
RiRts2
\nC2
\nWith
the
that
Suppose
\342\200\224
8
pf,
these
Ri
(R1/C2
following
= 9
parameter
-|-
\342\200\234IR2/C1
network:Ci
=
= 75
5 megohms, and
R*
megohms,
Vo
Eq.
values,
9-59 reduces
0.208
This
h(s)
\nThe
inverse
Laplace
+ 0.045) (s
be expanded
can
equation
=
X
transformation
+
volts.
(9-60)\n
0.0077)\n
to
1
0.0077
gives
8 pf,
10^6\n
by partial fractions
10~#^\ns +
=
to\n
1
5.55\n X
^\n
-f- I/C1C2
are given for the
values
(s
R%/Ci)s
give\n
1\n
s + 0.045.
\302\273*(<) as\n
(9-61)\n
Art.
AND
IMPEDANCE
9*4\n
U{t) = 5.55 X
which
the
is
it
\nrequired
current.
to start
required
is
necessary
211\n
FUNCTIONS\n
ADMITTANCE
-
10-#(e-\302\260-0077*
e\"0
(9-62)\n
046')\n
If the current
in any otherbranchis
over, applying
Th^venin\342\200\231s
theorem.\n
READING\n
FURTHER
of the concept of complexfrequency,the
and Seely, General Network Analysis
to LePage
\nstudent
is
referred
Book
Co.,
Inc., New York, 1952), pp. 189-193 and to
\n(McGraw-Hill
Feedback
Network
and
Amplifier
Design
(D. Van
Analysis
\nBode,
\nNostrand
New
1945), pp. 18-30. For a discussionof
York,
Inc.,
Co.,
in
\nthe
direct
use
of transforms
solving equations for a network,read
in Linear
Transients
\nGardner
and
Systems (John Wiley &
Barnes,
New
1942), pp. 176-214.\n
\nSons,
Inc.,
York,
For
discussion
further
PROBLEMS\n
that
differential
the
For systems described
equations
follow,
in
the
and
\ndetermine the complex frequencies that
appear
solution,
\ndesignate whether these frequencies are natural frequencies
of the system or frequenciesdetermined
\nby the
passive
parameters
\nthe
nature
of the
force. Call these two kinds of frequencies
driving
by
will
determined
by
\n\342\200\234free\342\200\235
and
& *
/
\\
^
9-1\342\200\230
\342\200\234forced,\342\200\235
respectively.\n
dt2
M
+
1~
dt +
+
4p 4-
\t(p2
(b)
9-2-
.di
dH
%
+
^
+
\302\253-al
9-4.
9-5.
\nfigure.
^
+
=
Be~*
sin
t,
p =
d/dt\n
*\342\200\242-\302\253\302\273<\n
+
25?
4- i\\
De~2t 4* E sin
5i=
4-
\342\200\224
3i2
the
Consider
Given
5)(p2 4- 2p 4* 5)y
\302\253!'
di\n
(b)
.\n
Ae\n
that
+
25
=
two
vx{t)
sin
=
,\342\200\231
1
2
t,
+
71\n
e\342\200\234I\342\200\230sin3<\n
+ it
^
\342\200\224
=
3t\342\200\230i
1\n
series circuits shown in the
= sin
for t
10% v2(t) = e-looo<
L\n
tfeU)\n
(a)\n
ib)\n
Prob.
9-6.\n
accompanying
> 0,
and
C
=\n
AND
IMPEDANCE
212\n
1 id.
the
Determine
\n(b)
the
\nDiscuss
\nfrequencies
two
the
9-6. Two black
is known
It
\ntical.
the
that
\nShow
complex
as (b) with
Zin(s) = V'\302\273n(s)//,-n(s)
shown
network
impedance,
input
\nthe
the
If
9-7.
\nrent
for
J(s)
be
may
,,
Prob.
9-9.
\nnetwork
In
the
reaches
purely
etc.\n
flows
Answer.\n
v
9-7 for
+
l)(s2
s
+
1)\n
+ 2s +
the network shown
+
s(s
(s2+
+
10(s2
\\
w
Yf
both\n
cur\302\254
(s2
Repeat
for
9-6.\n
examined,
w
9-8.
R
in
uncharged and no current
determine the transform of the generator
shown in the accompanying figure.
network
the
(a)
y/L/C.
are initially
capacitors
at t = 0,
inductors
=
of distinguishing the
measurements
may be made, initial
external
Any
conditions
final
\nand
R
and
(a)
the possibility
Investigate
network.
\nresistive
as
=
(6)\n
Prob.
(b)
iden\302\254
externally
shown
(a)
networks,
(c)
of
circuits.\n
each
the
0.
t >
all
for
t2(l)
L for (a) to hold,
this problem in terms of the
boxes with two terminals
are
that one box contains the network
contains
other
\nthe
meaning
series
=
Chap.9\n
of R and
values
required
physical
of
FUNCTIONS\n
to have ii(t)
it is possible
that
Show
(a)
ADMITTANCE
4s
+
2) (5s
13) (10s2
2)\n
in the figure.
+ 6)\t\n
+ 18s +
the switch K is in
network,
a steady
state. Then at t \342\200\224
0, the
given
Answer.\n
4)\n
positiona
switch
the
until
K is
moved\n
\nthe
the
\nin
of Example
4,\n
modified
as shown
\nstate
in
current
using
Rit
find
10
9-9.\n
Prob.
Th&venin\342\200\231s
with
result
this
Compare
steady
existed,
previously
having
\ntheorem.
at t = 0, a
closed
is
K
the
If
figure.
accompanying
\nswitch
\nthe
been
has
0-22,
Fig.
network
The
9-10.
capac\302\254
theorem.\n
Thevenin\342\200\231s
using
\nitor,
0.5-farad
the
across
voltage
213\n
transform of
b. Find the
to position
FUNCTIONS\n
ADMITTANCE
AND
IMPEDANCE
9\n
Chap.
10
Q\n
Eq.
9-56.\n
9\n
I\n
ioov^-y\n
10
Q\n
*3 <109\n
i\n
Prob.
9-11. The network
in
\nstudied
\nand
the
9-10.\n
shown in the
is
14). The input voltage Vi(Z)
have the value R
and load resistors
of
the transform
that
show
theorem,
input
F,(S)
(LC)*'*
9-12. In
\nage
\nalso a
of
pulse,
+
[\302\253(\302\253*
network
the
chosen
are
pulse
a
1-volt
and
such
shown
that
amplitude
L
4
s*
y/iJLC
in
\342\200\224
CRf
and
find its amplitude
the
filter
low-pass
unit
a
Chap.
\nThevenin\342\200\231s
\nments
is
figure
=
step
y/L/C-
(to
be
function,
By
using
the output voltage
+ 8s/LC +
is\n
8/(LC)*'1\n
sketch, the ele\302\254
R*. If Vx(t) is a volt\302\254
accompanying
and
R\\
=
T sec duration,
and time
duration.\n
show that
Vj(f)
is
10\n
CHAPTER
FUNCTIONS\n
NETWORK
the concept of transform
and
transform
which was introduced
\nadmittance
in the last chapter
be
studied
\nand extended.
a function relating currents or voltagesat
Further,
of the network,
called a transfer function, will be
parts
similar
\nto be
to the transform
mathematically
impedance function.
In this chapter,
impedance
will
\ndifferent
found
two
\nThese
10-1. Terminalsand
an
Consider
\nTo
\nthe
pairs\n
made up entirely
network
nature
general
of a
symbol
\nnode in
terminal
arbitrary
the
indicate
network functions.\n
called
are
functions
of the network,
(or a box). If
rectangle
and brought out
the network
of passiveelements.
let it be represented
by
a conductoris
of the
box
for
to
fastened
access,
any
of
end
the
Terminals are required
for connecting some other
to the network,
forces
\nconnecting
driving
The
or for making measurements.
a load),
(say
the
are
is
two.
terminals
of terminals
that
useful
\nnumber
Further,
in pairs,
one pair for a driving force, another pair
the
\nassociated
are given the name terminal
terminals
etc.
Two
associated
\nload,
In Fig.
10-1 (a) is shown a symbolic representation of a
network.
The terminal
pair is customarily
two-terminal)
(or
\npair
to a driving
force and so is sometimes given the
\nnected
driving\n
\nthis
conductor
is
designated
as a
terminal.
for
minimum
\nnetwork
are
for
pair.\n
one-terminal-
con\302\254
name
(6)
la)
Fig.
point.
Figure
10-1.
Network
10-1(b) shows a
(c)\n
representations.\n
two-terminal-pair
network.
The
ter\302\254
connected to a driving force
pair
designated
usually
while
terminal
the
\ninput)
pair marked 2 is usually connectedto a
The number
of terminal pairs in a networkcan
\n(as an
output).
\nwithout
limit:
10-1 (c) shows a representation of an n-terminalFig.
network.
All the discussion
in this chapter, however, will be
1 is
\nminal
(or
load
increase
con\302\254
\npair
\ncerned
with
one-
and
two-terminal-pair
214\n
networks.\n
10-2. Driving-point
transform
The
FUNCTIONS\n
NETWORK
10-2\n
Art.
immittances\n
current
the
defined as the
that is,\n
has been
impedance
to
\ntransform
215\n
transform;
voltage
V(8)\n
-\n
Z(s)
ratio of the
(10-1)\n
m\n
admittance is definedas
Similarly, the transform
m\n
=\n
Y(s)
ratio\n
the
(10-2)\n
V(s)\n
\nimpedance
\nterminals.
transform
The
impedance
a driving-point
Because of the
\navoid
writing
driving-point
terms
\nimpedance
\nof
bosm
In this
of
\nbe
+
his\342\200\235*-1-|-
\nis
a
\norder
order
will
we
the
two
Figure
bm\n
integers).\n
numerator
m
and
polynomial
The polynomials may
\302\260\342\200\224
\342\226\240AAA\342\200\224annr\342\200\224,\n
R
in
Ls\n
2(sl\n
Cs\342\200\242\n
o \t\n
\nform
impedances
marked
\nance
is\n
Z(s)
=
~J\n
an RLC series
10-2 shows
network
or\n
there
that
\302\253\\\n
al\302\254
polynomials.
one-terminal-pair
Z(s)
are
m
polynomial.
1
Example
of
quotient
an
+ bm-is +
order of the
difference
the
in
..
and
zero,
including
later
show
restriction
.
denominator
the
of
#any
of
combination
in the form of a
fln-is
of s (n
n is the
equation,
order
\nthough
admittance.\n
This algebraic
function
flis\342\200\235-1
function
a rational
is
the
1/Cs)
-f-
apsn
\nis
and
or an
found by combining
or admittance terms (Cs, G, and
or dividing.
immittance
in an
are
quantities
as\n
\npolynomials
which
and
to
(and
of impedance
(a
multiplying,
results
terms
two
admittance\342\200\235), the
and
R,
(Ls,
adding,
by
\n1/Ls)
admittance
and
impedance
combination
is thus an impedance
immittance
of a network is
immittance
An
The
of
immittance
name,
\nadmittance).
(or admittance).\n
impedance
similarity
\342\200\234impedance
one
\nassigned
admittance
or
transform
relate to the same pair of
found at a given terminal
must
admittance
and
called
is
\npair
that define
and current transform
transform
The voltage
with
,
r\302\273,
R
r
+ Ls +
z(\302\253)
=
element.
each
for
L
^1^7^
nrtwork.
trans\302\254
*.+
1
Jr
=
LCs2
The driving-point
+ RCs
\t\n+
jr
imped\302\254
1\n
(10-4)\n
R*Jk\302\261}lMl\n
S\n
(10-5)\n
216\n
polynomial for this
The numerator
\nond
is of first
polynomial
of
sec\302\254
order.\n
shunted
network
n/
\\
w
this
complicated network consisting of a series
a capacitor.
The driving-point impedance is\n
a more
shows
10-3
Figure
In
is
driving-pointimpedance
2\n
Example
\nRL
denominator
the
while
order,
Chap.10\n
FUNCTIONS\n
NETWORK
by
1
_
\342\200\234
+
Cs
+ Ls)
1 /{R
_\" 1
C s2 +
s -I- R/L
...\302\273\n
U
+
Rs/L
the numerator is of first
function,
second order. The driving-pointadmit\302\254
is the reciprocal of Eq. 10-6.\n
this network
driving-point
impedance
is of
the denominator
\norder
and
\ntance
function
Y(s)
for
2\n
1\n
The
\nAlthough
\nminal
pair
and
\ntion
terminal
of
a quantity
(1)
several
ratio
The
The
\nfer
(3)
The
ratio
having
at one
of one
voltage
or the
to another voltage,
of one current
to another current, or
ratio
of one current
voltage
current
the
to another voltage or
trans\302\254
\nterminal-pair, and terminal
computing
one
to
voltage
current.\n
transfer function for a voltage or current
10-4
\nsymbol G(s). If terminal pair 2 of Fig.
In
networks
with
Such
The
\nsero.\n
network.\n
ratio.\n
\nanother
*
Two-terminal-pair
ratio.\n
\ntransfer
(2)
10-4.
terminal pair to the transform ofanother
terminal
pair is given the name transferfunction.*
forms
for transfer
functions in electric networks:\n
another
at
are
\nThere
Fig.
a network is shown in Fig. 10-4.
pairs.
immittances
at terminal pair 1 and ter\302\254
the
driving-point
we are also interested in theratioof excita\302\254
2 are of interest,
for the two terminal
response
pairs. The function relatingthe
two
\ntransform
\nquantity
2.\n
function is identified
of a transfer
concept
least
of Example
functions\n
Transfer
10-3.
\nat
Network
10-3.
Fig.
'\n
1/LC
the
transfer
pair 1 is
designated
function, all initial
ratio
is
the output
is designated
the
input,
conditionsare
the
assigned
assumed
then
the\n
to
be
10-3\n
Art.
ru~\\
GW
a ratio
such
\nterminals,
transfer
The
\nmhos.
immittance
\ndriving-point
\nFor
since
but
\nimmittance,
^2(5)\n
VIW
VdW\n
(10-7)\n
or current to voltageis
are not measured at the same
the two quantities
a transfer immittance in ohmsor
is designated
imfhittance
is given the same symbol as the
with
to identify
the terminals.
subscripts
dimensionally
example,\n
=
Zls(s)
identifies
\nsecond
\ndetermined
\nas
This
vider.
the
61s\"*-1
+
\n60s\"
has the same
+
...
+
... +
-|- an
(10-9)\n
bm\n
general form as the
driving-
8\n
network
two-terminal-pair
in
as
1
\342\200\234I<Zis\"
function.\n
immittance
Example
\nFi(s)
thus
function
transfer
\nshown
numerator
the
ctoSn
=\n
Q(S)\n
The
(10-8)\n
^
of polynomials,\n
G(s)
\npoint
r\342\200\236(\302\253)
quantity and the
The transfer function is
quantity.
and can always be reduced to
immittances
network
the
by
quotient
The
=
and
identifies
subscript
denominator
the
first
the
where
\na
VU8)
is\n
to current
of voltage
ratio
217\n
output to the input
ratio of the
voltage transfer
The
FUNCTIONS\n
NETWORK
the
10-5
Fig.
a voltage
in the
current
no
Two-terminal-pair
net-\n
work,\n
di-
as
acts
10-6.
transform.
voltage
network
marked
and F2(s)
voltage
input
output
With
has
terminals, the voltage equations
output
\nare\n
+
ltl(s)
~
/(\302\253)
=
(10-10)\n
F^s)
->,(.)\n
^/(s)
(10-11)\n
The ratio of these equations
is\n
G(s)
=\n
V*(s)
\nV,
(l/Cs)/(s)
(.R
(s)\n
+
l/Cs)/(s)
1
or\n
G(S)
_
=\n
\ns
+
1\n
RCs +
1\n
(10-12)\n
/RC
1 IRC\n
(10-13)\n
FUNCTIONS
NETWORK
218
and
order
zero
\nof
Example
order.\n
4\n
The
Example
\nIt
is
the resistor has been
to write Kirchhoff\342\200\231s
equations
that
1 except
not
necessary
this network is
\ntransfer function, since
the
for
function
transfer
=\n
G(s)
or\n
Ls
is
of
+
above
as
a
essentially
inductor.
divider.
voltage
LCs2
+
(10-14)\n
1\n
\nl/LC\n
s2
the
to find
1\n
1/Cs
=\n
G(s)
+
(10-15)\n
l/LC\n
is of zero
The numerator polynomial
replaced by an
1/Cs
F2(s)
is similarto that
ratio becomes\n
voltage
Fi(s)
\nnomial
in Fig. 10-6
shown
network
two-terminal-pair
\nof
\nThe
has a numerator polynomial
polynomial of first
function
transfer
a denominator
This
network.
this
for
Chap. 10\n
the
and
order,
denominator
poly\302\254
order.\n
second
\342\226\240Kr\n
o\t\n
\342\200\224ORRP
4\342\200\234\n
i\342\200\224\342\200\224o
1\n
1\n
Cs\n
1\n
0\n
o\342\200\224l
same
The
one
can
network
voltage-divider
current
shows
10-7
\nFigure
be
loop
such
\n1
v2w\n
1
O\n
1-\n
Network of Example5.
combined
can be used with
concept
the
transfer
Ri\n
=\n
Cs
=
G(s)
(10-16)\n
1 / Ri\n
\342\200\224f\342\200\224
Ri
=\n
RiRiCs
4-
RiRiCs
may
be
reduced
G(s)
1\n
Ri\n
Vt(8)\n
Fl(s)
\nG(s)
4-
RiCs
becomes\n
function
or\n
more
the network by using network reduction.
a network.
The transform impedances Ri and
into an equivalent
impedance having the value\n
1
which
1
in
Zeq(s)
Then
|\n
>R2
\n1
\t\n
Fig. 10-7.
network.
Two-terminal-pair
>
5\n
\nExample
\n1/Cs
VXW
\n1\n
1
10-6.
Rl\n
1
\n1\n
o\342\200\224\n
4-\n
\nthan
V^(s)
0\n
h\n
1\n
|
1\n
Fig.
1
i\n
1\n
__
Fife)\n
\t\n
O
\ni\n
Ls\n
+ Zeq(s)\n
4Ri
(10-17)\n
Ri\n
-t-
(10-18)\n
Ri\n
to\n
=\n
s
s -|- (Ri
4~
4~
1/RiC
Ri)/RiRiG\n
\t\n
(10-19)\n
10-4\n
Art.
219\n
FUNCTIONS\n
NETWORK
transfer function, the order of the numerator
and
\ndenominator are the same. This particular network
finds
\nin servomechanisms
where
it is known as a \342\200\234lead\342\200\235
network.\n
In this
\ndenominator
\nbe
1
+
Cl\\8n
+
\342\200\242
\342\200\242
\342\200\242
+
\nboSm
+
biSm~l
+
... + bm-i8+
the
in
(s
(s
\n8i,
values
\nthe
Si,
-
s2,
...,
are
zeros
\ntwo
equations
its
\nby
are
that
and
zeros,
function becomes infinite.These
called
poles of the network function. Poles
in network
theory; a comparison of the last
function is completely specified
a network
the scale factor.\n
possibilitythat
of
roots
to
corresponding
roots,
\nmultiple
Such complex
s has the
When
sm, the network
important
There is the
s has
variable
the
function.
network
the
roots
the
and
factor,
vanishes.
function
network
the
s\342\200\236,
shows
poles,
(10-21)\n
When
...,
sb,
\nand
scale
... are complexfrequencies.
frequencies
\ncomplex
Sn)\n
known as the
of
So,
-<>)...(*-
\342\200\224
\342\200\224
8m)\n
s\342\200\236)(s
\302\253*)...(\302\253
\nfrequenciesare called zeros
\nvalues
into its n roots, and the
its m roots, the equation can
8b,
s0,
...,
\302\2532,
- Si)(a
is a constant
= a0/60
H
(10-20)\n
bm\n
form\n
H
where
into
is factored
polynomial
written
fln-lS
is factored
as\n
+ On
gpsn
polynomial
of polynomials
of a quotient
the form
have
functions
numerator
the
application
zeros\n
network
All
If
and
Poles
10-4.
of the
order
10-21
Eq.
may
of the form
a factor
Such
coincide.
(s
\342\200\224
sq)r,
are
or zeros (depending on location in the numerator
or
=
that
r
\ndenominator)
of order r. For a nonrepeated root, such
1, the
or
said
zero
is
to be simple.\n
\npole
Both zero and infinitevaluesof s are possible
pole or zero locations.
\nFrom
10-21
it is seen that:\n
Eq.
as poles
\ndescribed
s=
(1) When n > m,
(2)
When
(3)
When
<
=
n
n
\302\253is
a
pole
of order
is
a zero
of order
m, s =
=
co is neither a zero
m, s
<\302\273
n
m
\342\200\224
m.\n
\342\200\224
n.\n
a
nor
pole
but
an
ordinary
\npoint.\n
for
If,
any
are
\ninfinity
\nthe
\nthe
total
number
network
into
account
of zeros
is equal
taken
function,
poles and zeros at zero and
in addition to finite poles and zeros,
network
rational
to the total number of poles.Forexample,
function\n
(s
+ l)(s +
\ns3(s
2 + jl)(s
+
'+
5)\n
3\(\302\253") +
2
-j
1)
(10-22)\n
five
has
\nS3
=
10\n
\342\200\224
= \342\200\2242
five poles. The zeros are at Si = \342\200\224
1, s2
jl,
and s* = s8 = \302\273. The
occur
at
the
fre\302\254
poles
complex
jl,
=
=
Sb
on
the
s\302\253
plotted
Chap.
and
zeros
\342\200\2242
+
\nquencies
\nare
FUNCTIONS\n
NETWORK
220\n
sc =
sg
0, Sd \342\200\224
\342\200\2243,
s
complex
plane (s =
=
<r
\342\200\2245.These
+
ju)
in Fig.
poles
and
zeros
10-8. The
real\n
JO)\n
2 zeros\n
/\n
00 <\342\200\224
\302\256\n
>1\n
3 poles\n
+<r\n
-4
-5
\342\200\2422
-1
-3\n
\no\n
Fig.
The
10-9.
\nplex
is
part
The
\naxis.
\nzero.
the location of a
zeros
becomes
\nfunction
At
other
plane,
the <r axis,
and
the imaginary
part along the ja
is
used
to
the
location
of a zero and the
O
designate
symbol
and
Poles
network function plottedinthe com\302\254
showing two poles and one zero.\n
of a
along
plotted
x for
\nsymbol
magnitude
frequency
-A\n
Poles and zerosin the s plane.\n
io-8.
ng.
-
frequencies. At poles the network
at zeros the network function becomes
while
the network function has a finite,\n
frequencies,
designate
infinite,
complex
pole.\n
critical
10-4\n
Art.
\nFig. 10-9,
three-dimensional
for one
represented
\nplane
in
shown
Fig.
finite
one
\npoles,
at
zero
\norder
The
x\n
function
zero
represents
the
which
at
frequency
or
up\342\200\235
\342\200\234blowing
10-10.
Fig.
opposite
place: the network
\nbehavior takes
\nEither
Ju\n
a frequency
network
the
\n\342\200\234blows up.\342\200\235The
\na
complex
infinity.\n
pole represents
which
\nat
oo \342\200\242*o3 zeros
a third-
and
zero,
finite
in
shown
..\n
*
particufour
the
of
- ni\302\273n\302\253
is
10-9
Fig.
has
function
network
of complex frequency is
portion
This
10-10.
of the magnitude
of the s plane. The
quadrant
in
221\n
representation
function
as a
function
transfer
the
lar
A
value.
nonzero
\nof
FUNCTIONS\n
NETWORK
for Fig.
function becomes
sounds
like
10-9.\n
at
nothing
nothing\342\200\235
\342\200\234becoming
s plane
all.
drastic
rather
network function.VWemight
if
it would
not
avoid poles and zeros, to select networkfunctions
wise
to completely
or zeros.
Such
is not the case at all. Polesand
are
poles
\nthe lifeblood
of a function; without poles and zerosthe
reduces
\nbehavior for the
wonder
\nbe
\nwithout
zeros
function
a
\nto
\nunder
grubby
drab,
dull,
any
of
\nrepresentation
\nmathematical
\nzeros
desert\342\200\224absolutely
springs
concepts
\ns
the
Consider
of spectacular peaks (elevation: \302\253) and
(elevation:
0). This picture will becomecleareras we
with the aid of poles and zeros
of network
behavior
function
for a voltage ratio\n
transfer
a land
have
we
and
\nbeautiful
\nstudy
network
the
not
change
and
the
three-dimensional
poles
zeros,
function
becomes a tedious expanse of
flat.
But
add a few poles and a few
Without
conditions.
does
which
function
constant\342\200\224a
= GW
(10-23)\n
G(s)Vin(s)
(10-24)\n
tSkt
which may be
written\n
=
Vout(s)
In the usual
\nthe
network.
\nlast
equation
fraction
\npartial
\nno
repeated
problem,
vin(t)
is
specified,
and
problem
\nFtn(s)-
from
is to
V\n
l
p
computed
find the response, vout(t).
When
the
is expanded
by partial fractions, the denominator of each
term
a pole of either G(s) or Vin(s);that is, with
gives
roots
in the denominator
of Vout(s)\n
The
P
where
can be
G(s)
is the
Perfornung
number
the
of poles of
inverse
^
\302\243
*=1\n
G(s), and
Laplace
v
is
(10-25)\n
s
the
transformation
number
of poles
of this
equation\n
of
FUNCTIONS\n
NETWORK
222\n
Chap.
10\n
gives\n
p
=
Voul(t)
=
\302\243,~'G(s)Vin(s)
V\n
+
^
the
the natural complex frequenciescorrespond\302\254
The frequencies sk are the driving-forcecomplex
The poles therefore
oscillations.
to forced
Sj are
frequencies
oscillations.
to free
\ning
corresponding
\nfrequencies
\ndeterminethe waveformof the
since
\nresponse,
response,
the
out\302\254
of each part of the
the magnitude
of K, and Kk in the par\302\254
the magnitude
determine
as we shall see.\n
they
fraction
\ntial
of the
variation
time
determine
zeros
The
voltage.
\nput
(10-26)\n
*-i\n
y-1\n
Thus
Kke^
\302\243
expansion,
immittances,
poles and zeros have easily
=
a pole of Z(s) implies
Since
\nvisualized
Z(s)
V(s)/I(s),
meanings.
which means an open circuit.A zero
for
a finite voltage,
\nzero
current
or
a
other hand, means no voltage for a finitecurrent,
\nof Z(s),
on the
an
network is
Thus a one-terminal-pair
\nshort circuit.
open circuitfor
for
zero
a
circuit
and
short
frequencies. This can be
frequencies
\npole
of driving-point
terms
In
\nvisualized
\nthe
the
\nat
\n= Ls
at
circuit
\nshort
zero
s
1
=
at s =
and
frequency
as an
behaves
a short circuitat
the driving-point
\302\273)
function
as an open
open circuit
infinite
fre\302\254
impedance
element
this
and
a capacitor,
network
This
/Cs.
<\302\273.It
0) and as
inductor,
s = 0, pole
at
(zero
an
for
Likewise,
\nquency.
(to =
frequency
pole
\342\200\224
Z{s)
a zero at
= 0 and
at s
a pole
\nhas
is
impedance
driving-point
element networks. For
of single
terms
in
easily
circuit at
Z(s)
as a
behaves
infinite
\nfrequency.\n
on
Restrictions
10-5.
The
and
poles
in
\nlocation
the
in
\nterms
of
\nnation
\n(the
only
\nered
are
\npoles
\nfunctions,
and
R,
way
made
zeros
and
L,
to
the
of the
aoSn
+
(a<>,
Oi,
each
because
\nfollows
OiSn
...,
of these
form\n
1
-f-
...
numerator
\ntance
functions
on\342\200\224is~h
an
(10-27)\n
which are positive
coefficients is determined
on)
and real. This
by some
combi\302\254
and
real
C, and these parameters must be positive
(2) The networksbeingconsid\302\254
they appear in nature).
elements only. The rules for location
of
up of passive
are different
for driving-point
functions and for transfer
so will be considered
separately.\n
and
Driving-Point Immittance Functions.
\nboth
s-plane\n
functions have limitationsas their
These restrictions follow from twofacts:(1)
polynomials
coefficients
have
in
zeros of network
s plane.
the
and zero locations
pole
and
are
denominator
positive
and
(1) Since
the
coefficients
of
polynomials of driving-point immit\302\254
real, poles and zeros are eitherreal\n
10-5
Art.
or
in
occur
\nArt.
This
pairs.
conjugate
223\n
was discussed in
more detail in
6-3.\n
All
(2)
a pole
\ntive,
\342\200\224
sa),
where
s\342\200\236
If
and imaginary part, sa = <r\302\253.
+ jo>\342\200\236<rB is posi\302\254
give rise to a time-domain factor (by findingthe
will
pole
\ninverse
(s
a real
having
this
factor
a denominator
Consider
parts.
immittance functions have
of driving-point
zeros
and
poles
real
\nnegative
\nis
FUNCTIONS
NETWORK
of the
transformation)
Laplace
form\n
=
Kae,at
(10-28)\n
term (e*at) increases exponentiallyas t increases.For
in Z(s),
a pole
the voltage would increase without limit for any
\nsuch
for such a pole in Y(s) the current would
\ncurrent
and
increase
input,
limit
for any voltage input. Since this cannot happenphys\302\254
\nwithout
in the network, the poles and zeros
with
elements
\nically
only
passive
a
immittance
function
\nof
have negative real parts. In
driving-point
in
and
zero
location
the
of
s plane, all polesand zeros
\nterms
must
pole
half
and
in
the
left
can
never
occur
in
\nbe
the
half
plane (LHP)
right
and
Poles
zeros
can
be
on
the
boundary
(the
\nplane (RHP).
jot axis)
limitations
we
to
the
discuss
next.\n
\nsubject
and
zeros
on the j<a axis of the s plane (corresponding
Poles
to
(3)
The exponential
radian
\nreal
the
\nconsider
made
a network
for
\npossible
limit
without
increase
\nterms
listed in (2). Multiple polesgiveriseto
type (t cos ut), (t sin ut), etc., and such
as t increases.
Such an increase is not
up of passive elements only. For example,
transform
following
pair.\n
^
w2)2
(s2
transform
wt
to two poles
corresponds
expression
sin
=
\302\243_1
The
reason for this
be simple. The
of the
functions
domain
always
as that
same
the
is
\nrestriction
\ntime
will
frequency)
(10-29)\n
at
time domain factor of the transformpairis
\n+jo). The
a
\342\200\224
jca
at
two
and
increas\302\254
linearly
sinusoid.\n
\ning
poles
Multiple
of
\nhalf
s plane,
the
the
having
\ntne~at,
and
zeros are permitted
since such poles
required
at other locationsin
give rise to
zero
limit
since\n
lim
tne~al
=
terms
of
the
the
left
form
0\n
oo\n
t\342\200\224>
n by
finite
for
(4)
\nnomial
\nunity.
\nalgebraic
The
for
If
rule.\n
l\342\200\231Hospital\342\200\231s
polynomial and denominator
function can differ at most
driving
point immittance
immittance
function
is found without
driving-point
this
restriction
will always be observed. It can,
order
a
the
error,
of the
numerator
poly\302\254
by
how\302\254\n
NETWORK
224\n
under
\nlisted
\ninfinite
above.
(2)
We
frequency.
visualize
usually
sinusoidal
\nincreasing
Starting
frequency.
\ngenerators, we next
Chap.
10\n
a requirement which follows from the restriction
we must further discuss the meaningof
First,
to be
shown
be
ever,
FUNCTIONS\n
high frequency in
with
conventional
visualizean audio
radio
a
oscillator,
terms of
an
60-cycle
frequency
frequency oscillator, and then, somewhere
This
is a rather nebulous and
infinite
frequency.
\nbeyond,
but
it is the best we have. In terms of the s plane,
have
\nconcept,
\nfollowed
It
the jo> axis from a value near the originon
to infinity.
axis.
not
to follow the
Following
any other
necessary
path in the
will
lead us to infinity, and once
there
we are
get
plane
eventually
the
same
as if we had followed the ju axis. In
place
words,
one frequency, and is reached by
\ninfinite
is just
frequency
direction
from
the origin
of the s plane. Infinity is a
\nany
point
not be infinity).
We can say that the s planeis
not
\n(or it would
to
earth.
the
at
is
similar
the
Let
north pole
a sphere,
plane
at
the
the
s
the
north
of
plane. Standing
origin
pole, s
not
too
unreasonable
since
it
is of
looks
which
is
really
\nplhne
flat,
of infinite
But if you go far enoughin
radius.
direction
sphere
\nfrom
the
north
south
which
pole of the s plane, you end up at
pole,
from the north pole of thes
far removed
one
point
infinitely
if infinity
Now
is only one point in the s plane,it
the
ja
date line in the s world).
\naxis (sort
of an international
item
by
(3),
zeros
on
the
axis
mustbe
and
and
zeros
poles
\npoles
jo
simple.
immittance
for
a driving-point
function must be simple.*
infinity
at
we can get a simple pole or a simple
is to
only
way
infinity
exceed that of the denominator
the
order
of the numerator
and have the order of the
a pole
at infinity,
for
\nunity
\nexceed
that
of the numerator
This
by unity for a zero at
when you compute the driving-point
is satisfied
\nrule
automatically
is
If it
function.
not, you have made an algebraic
The restrictions
Functions
for transfer
Transfer
for Output/Input.
are
not
as those for driving-point immittances,
so rigid
is determined
the
transfer
function
as the ratio oftwo
in one network. The restrictions
at different
be
\nquantities
points
to those for driving-point immittance functions
\ngiven
by analogy
a microwave
\noscillator,
far
lies
hazy
we
out
\nis
jn>
we
\ns
other
\nat
traveling
unique
really
all\342\200\224it
\na
the
\nrepresent
part
\na
any
the
\nis
plane.\n
includes
But
Hence
\nat
\nThe
zero
\nhave
by
denominator
infinity.
\nimmittance
mistake.\n
\nfunctions
\nbecause
different
will
and
same
the
\nin
(1) This
\nzeros
The
*
this
This
also holds for transfer functions.
and
real or occur in conjugate
of a transfer function must have negativereal
does not hold for the zeros.
zeros
with
network
restriction
either
are
(2)
\nbut
order.\n
poles
restriction
is intended
Poles
pairs.\n
parts,
A
to be
only a suggestiveor heuristic
proof.\n
in\n
Art.
10-6\n
the
left
half
the
\nin
This
(3)
the
on
For
(4)
\ntor
order
this
In
transfer
its
\ndriving-forces.
\ncurrent
J(s),
pole and zero plot\n
from the
behavior
show that the time-domain behaviorof a
from the s plane plot of the poles and zeros
be determined
of the transform of the active-source
function
and those
transform
that
the
of some variable, say a
Suppose
and the poles and zeros are determinedas\n
is found,
we will
section,
can
\nsystem
order
may
Time-domain
10-6.
the order of the numera\302\254
functions,
of the denominator by one. However,the
be any value less than that of the denominator.\n
transfer
the
exceed
\nnumerator
\nof
zeros
only is classified as minimumphase;thosewith
half plane
are nonminimum phase.\n
for transfer
restriction
holds
function poles. Poles and
will
be
axis
simple.\n
fa
always
(output/input)
may
225\n
plane
right
\nzeros
FUNCTIONS\n
NETWORK
=
m
P(s)
=
where\n
(s
H
(s
\nQ(s)
(10-30)\n
7(a)\n
\342\200\224
Si)
-
(s
\342\200\224
.(s
s2)..
\342\200\224
sn)
(10-31)\n
\342\200\224
s\342\200\236)(s
sm)\n
shown in Art. 10-4 that the polesof this function
\ntime-domain behavior of i(t). It was suggestedthat
of each of the terms of i(t). In
the
\nmine
magnitude
It was
the
determine
zeros
the
this
deter\302\254
we
section,
these
\nwill amplify
by showing how i(t) can be
concepts
of the poles, the zeros, and the scalefactor
a knowledge
\nfrom
ratio f and the undampednatural
of the damping
In
terms
discussed
on page
as
104, the poles and zeros of the last
\nquency,
determined
H.\n
fre\302\254
co\342\200\236
the
have
will
\nequation
*1, Si =
\nin
\nlines
\nare
s plane,
the
through
\nof
fan\n
\302\261
fan
Vl
\342\200\224
f*\n
-
r
<
i\n
(10-32)\n
r
>
i\n
(10-33)\n
=
i\n
(10-34)\n
=
o\n
(10-35)\n
Si,
Si
=
\342\200\224
Si,
Si
=
\342\200\2240)n\n
r
Si,
Si
=
\302\261fan\n
r
fan\n
\302\261
o>\302\273
Vf2
\342\226\240\n
are
circles
on page 105 that contoursof constant
\302\253\342\200\236
that
contours
of constant damping ratio are straight
the
and that contours of constant damping {fan)
origin,
lines
straight
\nparallel
\342\200\224
also shown
was
It
forms.\n
following
to the
oscillation,
parallel
a axis of the
to the
ju axis of
s planearelines
\342\200\224
Vl
\302\253\342\200\236
\302\243l. These
of
facts
are
the s plane.
constant
summarized
Further,
actual
lines
frequency
in Fig. 10-11.\n
226
NETWORK
the
\nof
s plane can be
in terms of f and
response
poles in the
of the
location
The
time-domain
general
=
i(t)
w\302\273.\n
(10-36)\n
+
the use of the contoursof
To illustrate
terms
in
interpreted
A',e(_r\"\"+w\"v/f7:r3)\342\200\230
10\n
Chap.
FUNCTIONS
consider
10-11,
Fig.
s-plane\n
the array
of\n
jej\n
(d I\n
<*,;
\nf
and
+
un
2fw\302\253a
in
shown
of
\npair
actual
Fig.
10-12
sa*
and
Typical
plane.\n
poles in the
real
part
and
se
\nof
decreasing
for
of
\nmately
the
same
corre\302\254
So* is higher than that
just as the damping (or rate
is less for s\302\253
and
amplitude)
sc and sc*. The natural fre\302\254
sc*,
\nquency
oscillation
the
two
same,
radius
pole pairs is approxi\302\254
since they are on about
from
the origin. The
\ndifferencein actual frequency
to a lower damping ratio for sa and
and
different from the conjugate pairs just
ss are quite
to the overdamped
They
correspond
case, and have
of
\ntion
is
The
\nconsidered.
due
poles
of
\302\261
\302\253\342\200\236
\\/l~\342\200\224J*-
8a and
to
\nof
\nthe
of
frequency
\ns0* than
\n8
lines u =
(zeros have been omitted for clarity). The
the pair se and sc* correspondto oscillatory\n
in
the
time
domain. The
expressions
\nsponding
10-12.
(-\\/l
any
\342\200\224
\302\243*/\302\243);
second-order characteristic equation
by the
\nactual
Fig.
radius
constant
\342\204\242
0.\n
\302\253,*
+
and
s\302\253
poles
(a)
\342\200\224
tan-1
\342\200\224
fw\342\200\236
(or
*=
<r
of oscillation
frequency
defined
are
line
damping
\302\253
plane:
line 0
ratio
damping
negative
constant
(d)
\na*
poles
(b)
constant
\n(c)
\na);
constant
in the
contours
Constant
10-11.
Fig.
\n=\342\226\240
oscilla\302\254
sa*.\n
s\302\273
an\n
Ait.
10*6\n
\nthe
for
than
sti
Sb is
pole
sb. From another
than that
greater
to each pole
\ncorresponding
for
for
\npoles
tit)
As
\ngive
is
found
=
Kae+
usual,
the
factor.
each
for
by
adding
Fig. 10-13
terms
response
an
for
the s
arbitrary\n
plane\n
amplitudes).\n
The total response correspondingto
each of the individual factors as\n
Ka*e>\302\260'\342\200\230
+
+ K*e^
Kce^
+
Kbe^
+
to conjugate pairs
corresponding
sinusoidal
damped
time-domain
s\302\253j.
Typical
is shown in
(arbitrary
amplitude
damping is greater
point of view, the time constant
Response comparison for various polesin
10-13.
Fig.
227\n
time domain. The
in the
form
decay
exponential
\nfor
FUNCTIONS\n
NETWORK
Kde*\302\253
these
(10-37)\n
will combineto
expressions.\n
remains the problem of determiningthe multiplying
constant
for each of the terms (or modes). The startingpoint
\n(or magnitude)
10-31.
To find the time-domain response correspondingto
\nEq.
\ntransform
we expand
equation,
by partial fractions. Hence\n
- There
I(s) =
Any
the
of
~
Kr
Ka
+
8
...
\342\200\224+
S
Sa
Sb
say
A-coefficients,
+
S
-4Sf
Ar, can be
.... +
is
this
Km\n
5
(10-38)\n
$tn\n
found by the Heaviside
as\n
\nmethod
v
'
-
\342\200\224
u
*i)(*
(\302\253
\"
Is
\342\200\224
So).
-\302\253\302\273)...(\302\253
. . (S-\342\200\224-S;) . . . (s
-
s\302\273)\n
(10-39)\n
\342\200\224
Sm)\n
+\302\253r\n
NETWORK
228
Eq. 10-38 gives the
s in
sT for
Substituting
(Sr
This
both
\nwhere
\ntwo
in
\nwritten
form
polar
for
Sn)\n
-
(10-40)\n
Sm)\n
(sr
as\n
8n) =
-
(10-41)\n
is
where Mnris the magnitude the phasor
and
(ar
\nangle of the same phasor. The differenceof the
complex
and
in Fig. 10-14 (other poles and zeros
8h is illustrated
\342\200\224
*\342\200\236),
of
the
<f>\342\200\236r
two
for
again
\nthe phase
\nrespect
angle
to the
<t>
10-40
are
10-40,
the
V
=*
The
easily found.
of Kr is
value
\342\200\224
If
r
as
interpreted
and
magnitude
and so all
In terms of
seen to
a
phasor
with
measured
\302\253r,
phase of the
factor
terms of this general
type
M and
<t>
for
each
factor
in
become\n
. .Mnr
MirMuMlr.
and
the distance from s\342\200\236
to sr;
of the linefromsnto
measured,
easily
is
Mnr is
magnitude
line.
0
\342\200\224
\302\253*)
(sr
is the angle
thus
\nEq.
The
to
#\342\200\236
sr.
\342\200\224
are
*\302\273)
Eq.
term
The
clarity).
\ndirected from
\nin
omitted\n
and phase of (\302\253r\342\200\224
\302\253\302\253)
(other
poles
diagram.\n
(a) polar diagram; (b) string
10-14. Magnitude
\nzeros omitted):
phase
quantities
are
\n\302\253r
\n(sr
Kr.\n
of
(*r
nr
10\n
of the general form
sn),
known complex numbers. The difference
is another
number which may be
complex
numbers
complex
followingvalue
of factors
is composed
sn are
8, and
equation
-
\342\200\224
So)(\302\253r
Chap.
82) . . ,j(ST
Sb). . . (Sr
$l)(&r
(&r
ij
jy
FUNCTIONS
..)\n
(10-42)\n
MarMbrMCT...Mnre\n
This
equation
\noperations
\nuated.
\nby
a
(1)
gives Kr as a
indicated
by this
the
Determining
\tPlot
\ncomplex
the
poles
8
plane.\n
By
which
and
performing
the
the constant Kr can be
in Eq. 10-42 is readily accomplished
may be outlined
equation,
quantities
procedure
graphical
magnitude and phase.
zeros of
eval\302\254
as:\n
/(\302\253)
=
P(s)/Q(s)
to scale
on the
Art. 10-6
Measure
(2)
and
Measure
and
An
\ndure.
Suppose
/(\302\253)
evaluate
Kr.\n
\342\200\224
\342\200\224
1
origin, and H
transform has
current
The
5.
as
given
10-42 and so
Eq.
proce\302\254
s
poles
at the
a zero
\342\200\2243
and
\nand
this
has
of the other finite
sr.\n
into
quantities
illustrate
will
example
\nis
these
Substitute
(4)
pole
given
other finite
\302\253f.\n
angle from each
the
compute)
zeros
to a
(or
\npoles
pole
given
229\n
from each of the
distance
the
compute)
zeros
to a
(or
\npoles
(3)
FUNCTIONS
NETWORK
0r\n
\342\200\224M-\n
-1\n
-3\n
form\n
\nthe
\342\200\235
Pole-zero
10-15.
Kg.
\342\200\224|\342\200\224
\342\200\224|\342\200\224
3)
l)(s
configu\302\254\n
ration.\n
\ntial
\nto
is easily
can
but
function
This
fractions,
is seen
it
10-15,
Fig.
by par\302\254
expanded
be evaluated as
also
that\n
M oie,>#l
More*\"
=
Ki
=
H\n
= le,l80,
=
ieiHl
\nMz
Hence\n
outlined above. Referring
(10-44)\n
(10-45)\n
2e,\342\200\2300,\n
5 X
=\n
ie+'180\302\260
(10-46)\n
-2.5\n
\nMz ief*n\n
Similarly,\n
3g/18\302\260*\n
=
Kz
5
H\n
X\n
Since the poles determinethe
the
for
write
\nwe
since
\nand
zero
have
we
locations,
(in
case
this
neper
(10-48)\n
evaluated from a
as a
of
knowledge
the
pole
solution,\n
particular
i(t) = 2.5e_< + 7.5e~3t
this discussion and with the aid Eq.
\342\200\224
From
frequency),
= Ktf-' + Kze~u
Kz have been
K\\ and
(10-47)\n
7.5\n
solution,\n
general
i(t)
and
frequency
=
2^nos\n
Mize\342\200\231*1*\n
of
(10-49)\n
10-40,
the
of
influence
can be visualized. Consider one
response
If
all
other
say
Fig.
poles and zerosin the s plane
\npole,
sr,
\nremain
in position
and the zero s\342\200\236
the
fixed
is
of a
moved,
proximity
is seen by Eq. 10-42 to reducethe magnitude
\nzero
to
a pole
of
the
\n^-coefficient
associated
with the complex frequency of the pole sr.
from
of a pole to sr is seen to have
the
Eq.
\nAgain,
10-42,
proximity
\na
time-domain
the
on
zero
10-14.
in
\nopposite
\nand
\ncoefficient
effect\342\200\224since
of
proximity
Kr.
pole
another
When
the
magnitudes
pole
zero
in
appear
to sr increases
is
s\342\200\236
moved
so
the
denominator\342\200\224
the magnitude of the
close
to
sT that
they\n
230\n
the
coincide,
\nular
to
Kr
reduce the value
zero caned and
and
pole
zero.\n
Chap.
FUNCTIONS\n
NETWORK
of
the
10\n
partic\302\254
of the ^-coefficient corresponding to a particular
determined
\npole
by the proximity of both poles and zeros.If,
can
be
\nin the
of a network, the position of the polesand zeros
design
pattern:\n
they should be selected according to the following
\nselected,
The
magnitude
thus
is
Select
(1)
\nin
pole
locations
to give the
complex
frequencies.\n
of
terms
Fix the position of the zerosin
\nmagnitudes of the various K
(2)
required time behavior.Do
this
the
to
plane
complex
the
adjust
coefficients.\n
\nof
the graphical interpretation of the position
for the case of nonrepeated (orsimple)
and
zeros
was discussed
In
the
case of multiple
poles, it is suggested that expansion
by
a
rather
than
modification
of
be followed
the
fractions
seeking
discussed
to fit the new case.\n
that
have
been
be
should
It
poles
\npoles.
\npartial
\nprocedures
that
network functions for
for finding
Procedure
10-7.
noted
general
two-terminal-pair
\nnetworks\n
For
the
networks,
two-terminal-pair
complicated
of
computation
become
\ntransfer functions and driving-point immittances
quite
In this section, we will discuss systematic procedures
\ninvolved.
may
for
functions.\n
network
such
\nfinding
of
network can be thought of as made
\nnumber of one-terminal-pair networks. There
up
Any
of a
combination
the
is no
unique
rule
for\n
Z3\n
10-16.
ng.
Grouping
\nalternate
the
dividing
\nponents.
\nthe
\ncases.
The
arbitrary
of certain
network
in
impedances
However,
voltage
of elements in
network
such
series
a
and
to
form
admittances
in
network
a system
of
parallel.\n
into elementary one-terminal-pair com\302\254
in
is made on the basis of interest
a division
nodes
of Fig.
in certain branchesin
10-16, for example, is groupedintoa
or the current
many
num\302\254\n
Art.
10-7\n
ber
of
transform
the
can
\nFig.
10-17.
and
combination
parallel
each of the one-terminal-pair
Z(a) or the transform admittance
impedance
This
computed.
Such
combination
be
\nY(s)
in
for a number of examples
is accomplished
by the usual rules for
discussed earlier.\n
of immittances
is illustrated
Several two-terminal-pairnetworksoccurso
\nthat
are
given
special
a series
impedance,
they
to
\nreduces
231\n
For
networks.
one-terminal-pair
\nnetworks,
\nseries
FUNCTIONS\n
NETWORK
in
often
networks
useful
names. The general network sometimes
a parallel impedance, and another series\n
l
Zla)
1\n
or
Yfs)\n
i\302\260\342\200\224m\342\200\224\302\2602\n
2o\n
lo\n
Z{s) or
Y(\302\253)
\nlo-^VW\342\200\224\302\2602\n
2o\n
rVWi\n
Z{a)or
Y(s)
io-^wv-L_|j_J\n
\nlo\342\200\224VW\342\200\224\302\2602\n
Immittance
10-17.
Fig.
10-18.
Fig.
as shown in
impedance
\nT
network.
=
\nZi
\nshows
\nis
Zz,
Zc
\nwork
\nnetwork,
\nIf
T
network
other
two
network,
\nnetwork
of
If
network
the
a v
\nwhen
if
Further,
=
and
Fig.
of one-terminal-pair
Network
configurations.\n
Fig. 10-18(a).Sucha
the
series
networks.\n
impedances
is designated
configurations.
network
is
are equal, that
Zd and Za
two terminals
a
is if
as a symmetricalT. Figure10-18
The network of Fig. 10-18(b)
Yi = F3, the networkis a symmetrical
or
a symmetrical
10-18(c) is a lattice network
with
as
known
ir.
The
lattice
\342\200\224
Zb.\n
of a terminal pair of a two-terminal-pair
net\302\254
are
connected
to a terminal
pair of another two-terminal-pair
the
networks
are said to be connected in tandem or cascade.
or ir networks
networks
are connected in cascade, the resulting\n
the
FUNCTIONS\n
NETWORK
232\n
is an
network
of
\nber
\nwith
important
and
sections
=
Zi(s)
begin
as shown in
Fig. 10-19,or
10\n
num\302\254
begin
may
in
manner
in
com\302\254
are computed as impedances, and the
are computed
as admittances.\n
immittances
immittances
shunt)
(or
\nparallel
structure.
same thing may be said about the
ends (at terminal pair 2). For convenience
series
the
\nputation,
It may contain any
network
0. The
network
\nwhich the
may
Chap.
\302\260\342\200\224VW\n
Zi(s)\n
1\n
2\n
Ladder
10-19.
Fig.
network.\n
any network can be found by
or
node
If the network can be made into a
\nwriting
loop
equations.
it is possible
to find the driving-point immittance
\n\342\200\234ladder structure/'
and
combination
of immittances
without writing loop
series
\nby
parallel
Assume
that
the
ladder
\nor
node
network of Fig.\n
directly.
equations
of
the
six
immittances
shown.
is made
10-19
up
Combiningimmit\302\254
The
is
\nimmittance
this
\nNext
sum
required,
is inverted
the
until
continued
\nbe
terminals
the
at
\ntances
of
immittance
driving-point
for which the driving-point
is first inverted
and combined with Z6(s).
F6(s)
and combined with F4(s). This pattern
network
reduces to a single immittance. In
opposite
those
may
\nsummary,\n
=
Z*M
Zi(s)
+
i
Y*(*) +
Z*(\302\273)
(10-50)\n
\t\n
\t\n
+
=
-T-\n
+
F4(\302\253)
+
which is read from the
to
bottom
the
the pattern of com\302\254
to give
top
F^i)\n
\nbraic
is
configuration
\ntinued
fraction
an
Such
immittances.
\nbining
or a
alge\302\254
a
called
con\302\254
continued
Stieltjes
\nfraction. Forming a continued
frac\302\254
\na
10-20.
Two-terminal-pair
net-\n
\nwork.\n
\nfor
\nwhen
\nWhen
the
the
network
transfer
driving-point
The
function
of interest
function
is desired,
ladder
network
procedure
systematic
\nthe
ng.\n
a
for
\ntion
use
network
provides
for
finding
immittance.\n
of a continued
reduction
fraction
applies
only
is a driving-point immittance.
a different procedure must be\n
10-7\n
Art.
In
followed.
several
\nwith
current
\nvoltage
\302\253
-
(10-51)\n
r4 + zji\n
(10-52)\n
Theseequationsdescribethe
(10-54)\n
F, + Zj/1\n
network
and
\342\200\224
and
(10-55)\n
(10-56)\n
=
the usual
contain
they
Z4i(s)
V4(s)/Vi(a),
equation
Ztlt\n
-
F,
first
(10-53)\n
- U + r,F,\n
h
the
Y4Vt\n
\342\226\240
F\302\273+
Fj
\nwith
com\302\254
f.f4\n
/.-/.+
(?(*)
that
requires
opposite the driving-point terminals.
write the following equations for Kirchhoff's
Vi
quantities,
redrawn
As in finding
designated.
procedure
following
It
\nfer
10-19is
terminals
we may
laws:\n
network,
and
the
at
begin
\nputations
the
the
immittance,
currents
and
voltages
pertinent
833\n
network of Fig.
the ladder
10-20,
Fig.
\ndriving-point
\nFor
FUNCTIONS\n
NETWORK
V4(s)//i(\302\253), etc.
Starting
second equation
it into the
substituting
trans\302\254
\ngives\n
-
V,
In
turn,
this
to
\nequation
(10-57)\n
V\302\273Zt)V4
with Eq. 10-51into
may be substituted
equation
the
next
give\n
/,
=
Vi
=
(10-58)\n
YMIV,
+
r\302\253(l
pattern,\n
+ z$ir, +
yj>)
+
{(i
+
[K,
to this
according
Continuing
(1 +
y4(i + ym)\\v4
(10-59)\n
/l-\n
([F,+
r4(l
+ Y#>)) +
r,|(l + YM+Z4Y.+
1
Y4(
+
YJM\\)V4\n
(10-60)\n
This equation gives the transfer
which is\n
\nadmittance
F^s),
y\342\200\236(.)
-
-
5*1
\nthis procedure
(io-6i)\n
r.\302\253]
) eZ4
(10-62)\n
voltage ratio may be foundby carrying
one step further by substituting into Eq. 10-56;that is,\n
for the
function
transfer
+
of the
inverse
the
+ z,(Y. +
r, +
where
The
impedanceZ4i(s)as
Vfr)\n
0(b)\n
V4(b)\n
6
+ Z\\(Y %
+
r\302\2535)
+
%\\\\Y%
+
Y46 +
r,(\302\253
+
Z,(K*
+
r4*)J)\n
(10-63)\n
234\n
FUNCTIONS\n
NETWORK
The
transfer
\nThe
method
\342\200\234
rKKK\\
Z-8\n
Vi
\t\n
Z-8\n
\"
1
1
v2\n
and
s,
\nnetwork,
\nvoltage
\nin the
inductances
series
of
shunt
the
net\302\254
is made
which
10-21,
Fig.
(1 henry)
net\302\254
\nand
The
(1 farad).
capacitors
series elements
\nimpedance of the
of the shunt elements is s. Forthis
likewise,
the
and the transfer function of the
driving-point
impedance
will
ratio
be found.
Other currents and voltages that will aid
on the
computation
(but not appear in the solution) are shown
admittance
the
The
\nfigure.
of
\nwork
o\n
Two-terminal-pair
with a
\nspecific example, consider
farad\n
\nwork.\n
\nis
pat\302\254
by this
method
the
illustrate
To
Y-\302\253P\n-
\nup
10-21.
Fig.
the\n
\nexample.\n
henry\n
\t\n
o
general
established
been
has
\ntern
10\n
given.
The
network.
ladder
1 henry\n
Y-8~\n
1 farad\n
of G(s) as
of sectionsin
is the reciprocal
F4(s)/Fi(s)
above
holds for any number
function
illustrated
ft
Chap.
in continued-fraction
written
impedance,
driving-point
is\n
\nform,
=
Zdp(s)
S
+\n
(10-64)\n
1\n
+\n
+
*
~8\n
This
\nto the
be reduced,
can
equation
at the bottom
starting
=
(10-65)\n
2~\342\200\224
\342\200\224+
find
\nand
the
F\342\200\236(s)
/i(\302\253)
7i(s)
=
=
=
Vt
/*
+
+
F\342\200\236
Zh
=
YVt
= aVi
+
ItZ
- (s* +
YVa =
=
+
(\302\253*
Vt(s)
s4
particular
terminals
as
behaves
a low-pass
additional
to
discussion
Gardner
and
(10-66)\n
l)Vt
+
s(s\342\200\231
+
1)F,
(10-67)\n
(10-68)\n
l)]Vt
+
\302\253(\302\253*
2s)
Vt
(10-69)\n
becomes\n
+ 3s*
We will show
network.
+
(10-70)\n
1\n
in another chapterthat
this
filter.\n
READING\n
FURTHER
\nis referred
V%
1\n
=
F,(s)
For
+
[\302\253
ratio transfer function thus
The voltage
\nnetwork
the
as\n
proceed
this
function start at
transfer
ratio
voltage
J,(s)
for
up,
working
form\n
ZdpW
To
and
relating
Barnes,
to network functions, the
Transients in Linear Systems
reader
(John\n
FUNCTIONS\n
NETWORK
10\n
Chap.
235\n
1942), Chap. 5; to ValleyandWallman,
\nVacuum
Tube
18 of the Radiation Laboratory series,
Amplifiers
(Vol.
\nMcGraw-Hill
Book
Co., Inc., New York, 1948), pp. 42-53; to Tuttle,
2
\nNetwork
vols.
in
Synthesis,
(John
Wiley & Sons, Inc., New
York,
and
to
\npreparation);
Guillemin, Communications Networks, Vol. I
&
\n(John
Wiley
Sons,
Inc., New York, 1932). It should be notedthat
\nthe
s is equivalent
to p and X as used by some authors.\n
quantity
A Sons,
Wiley
New
Inc.,
York,
PROBLEMS\n
10-1. Find the
the
\nin
2s2
+
\ns3
s3
o
normalized
term
+
+ fs
+
fs\n
r000>
2
10\n
p\n -
$f
+,
s3
the
Find
*
+
the
10-2.\n
shown
network
in
the
figure.\n
s\n
for the network shown
with the
the polynomials
Arrange
to unity
coefficient.
Answer. Z{s)\n
admittance
figure.
+
V+
10\n
+ * \342\200\242\n
driving-point
normalized
term
+
t _?
s2
+
2_f
accompanying
,
lh<\n
Prob.
Prob. 10-1 for
\nhighest-ordered
_
Hf-\n
If\n
Z(s)\n
10-1.\n
Z(S) =
Anmer.
Ht-\n
O \t\n
10-2. Repeat
the
-
lfP\n
\t\n
Prob.
\nin
o\342\200\224\n
1\n
h\n
If\n
Zt\302\273)
10-3.
of this function with the high\302\254
polynomials
to unity
coefficient. Answer. Z(s) =
1\n
Vv\\r\342\200\224i\n
O\342\200\224
shown
network
the
for
the
Arrange
figure.
ordered
\nest
driving-pointimpedance
f\n
+
\302\245\302\253\302\273
*)\342\226\240\n
O \t\n
Hf\n
if\n
GHHP\342\200\224|\n
h\n
3
>\342\200\224
\342\200\224yOOP
\n2 h\n
yis)\n
Y(s)
=:
if:\n
-\n
3
h\n
if-\n
0\342\200\224\342\200\224\n
10-4.
=\n
+
13S2
+
fs(2s2
Find
10-5.
\nratio,
for
the
10-3
Prob.
Repeat
S4
Y(s)
Prob.
10-3.\n
Prob.
the
+
the network shown
above.
Answer.\n
5\n
'\n
1)
transfer
networks
for
10-4.\n
the output voltage to input voltage
function,
in the figure. Arrange the polynomials\n
shown
236\n
FUNCTIONS\n
NETWORK
the
with
/n
\302\261
w
/y(
to unity coefficient.
normalized
term
highest-ordered
Chap.
\t\n
I/R1R2C1C2
\\
S2
+ RlCi
+
+ R2C2)s/R1R2C1C2 +
0\342\200\224VW\n
0\342\200\224vw\n
Answer\n
l/RJl&Ci\n
\342\200\224Wv\342\200\2241\n
O\n
r2\n
Ri\n
\342\200\234\n
-\n
-Cl
u,
0
10\n
^C2
\t\n
\n\t
V2
0\n
10-5.\n
Prob.
O\342\200\224wv\n
Ri\n
Ci\n
LAA/V\342\200\2241
\n*1\n
v2\n
Vl\n
U2\n
<>1\n
(6)\n
la)\n
10-6.\n
Prob.
10-6.
s2 -(s2
Prob.
Repeat
+
(R1C1
(R1C1
10-7.
Show
+ RiC2 +
that
both
shown.
for the networks
-l-
\342\200\234IR2C2)s/RiR2GiC2
X/R\\R2CiC2
Answer
to
(b).\n
1\n
R2C2)s/RiR2CiC2 +
of the networks of the figurehave the
(s + l)(s + 3) I
\\/RiR2CiC2\\\n
Z(s)\n
impedance
driving-point
10-5
\ns(s
+
2)(s
+ 4)
same\n
J\n
jo\n
10-8.\n
Prob.
10-8.
c|riving-point
Show
that
the network of
impedance
Z(s)\n
the accompanying
figure
s4 + 18s2 +
7s*
4-
12s\n
24\n
has
the\n
Chap.
10\n
that the total number of poles is equalto
of zeros for any network function having the form
of
Prove
10-9.
\nnumber
\nof
237\n
FUNCTIONS\n
NETWORK
rational
total
the
quotient
polynomials.\n
shown below is known to have the
the figure.
In addition, it is known that
network
The
10-10.
\nconfiguration
pole-zero
in
shown
the\n
(a)\n
10-10.\n
Prob.
\nis
Z(0)
\nAnswer.
10-11.
\nis
is
(s = 0 or direct current) 1 ohm, that
frequency
= 1. Determine
the values of R, L, and C in the
network.
=
=
=
L
C
0.1
farad.\n
1 ohm,
R
\302\243
henry,
the
that
It is known
response in the time domainof a system
zero
at
impedance
of
summation
the
terms
f = 0.5 (second
= 0. Plot
1, f
=
\n\302\253\342\200\236
2,
(c)
=
\t(on
10-12.
is
A transient
order); (b) T (the time
the poles of this systemin
found to be of the
a
=
-
2e~*
constant)
the
s
3
(a)
sec;\n
plane.\n
le~6t\n
for 7(s) that
configuration
pole-zero
=
form\n
i(t)
Find
the following characteristics:
having
gives this time-domain
\nresponse.\n
is found
A transient
10-13.
a
pole-zero
-
=
i(t)
Find
to be of the
2e-8\342\200\230
+\n
for I(s)
configuration
form\n
that gives this
time domain
\nresponse.\n
Given
10-14.
\nthe
value
10
and
that
that
of a current transformI(s) has
the following finite poles and zerosdescribe
the
scale factor
the
\nsystem:\n
Zeros\n
Poles
\342\200\224
2
\302\261
j
1
none\n
-5\n
Find
\ni(t)
=
the time-domain response corresponding to
e~6t + \\/l0
cos (t \342\200\224
108.4\302\260).\n
this 7(s).
Answer.
NETWORK
238\n
and
5.0
value
\nthe
Chap.
10\n
of a current transformI(s)
the following finite poles and zerosdescribe
the
that
Given
10-15.
FUNCTIONS\n
that
scale factor
has
the
\nsystem.\n
Zeros\n
Poles
-1
-6\n
\302\261j2
-3\n
response i(t) corresponding to this J(s).\n
the
10-16.
For
each of the networks shown in the figure,find
transfer
=
ratio
and
the
transfer
func\302\254
voltage
Z<r(s)
V0ut(s)/Iin(s)
\nimpedance
Find
\ntion
the
G(s)
time-domain
=
Vout(s)/Vin(s).\n
(81\n
Prob.
10-16.\n
/il\302\253)
\no\t\n
\342\226\240o\n
-c\n
o
\342\226\240
\342\200\224O\n
-\n
Prob.
10-17.
\nJi.
The
output
=
\nZti(s)
The network
F2(s)//i(s).
voltage
V^(\302\253)\n
shown in the
10-17.\n
figureis
driven
by
a current
source
W (a) Find the transfer impedance,
(b) Show the pole-zero configuration for Zsi(s).\n
is
FUNCTIONS\n
NETWORK
10\n
Chap.
For a given
10-18.
network, it is
=
Z21
V
n
2/11
_
239\n
that\n
known
\342\200\234
*i)(\302\253
(\302\253
\t\n
-
Si)\n
s\n
where
Si,
Si
10-19.
\nat
=
(a)
\342\200\224
1 \302\261
jlO.
For
\nG(s)
=
Vi(s)/Vi(s)
Zn(s)
for this
\342\200\224
e~0
ii(t)
network
the
1 is
terminal-pair
If
6t,
function
network.\n
are
\nregion
\nthe
0
fan
specifications
given
Vz(t).\n
shown, show that the input impedance
= 1 ohm. (b) Find the transfer
10-20. Two conjugate complexpoles
\nious
find
For
belqw.
each
required
to
specification,
the
var\302\254
sketch the
crosshatching for identification) that
be
located,
^
^
may
(a) f ^ 0.707,
poles
1,
(b)\n
actual
1
^
=
^
^
t
frequency
2, fan negative,
0.5,
2.5.\n
g
^ 0.5. (d) 0.5 ^ f < 0.866,
in
the
s plane
(using
\342\200\2244.
\302\253\302\273
^
meet
(c)
con
4,\n
CHAPTER
11\n
CONFIGURATIONS\n
POLE-ZERO
\nFROM
There is somethingdistinctiveabout
sinusoidal
\na
in
\nonly
the
in
\nsinusoidal
steady
state,
and
phase
amplitude
waveform.
sinusoidal
If
to a network of linear passive
in that network will be
current
every
differing from the driving-forcesinusoid
angle. This property follows from two
and
voltage
every
the
is applied
force
driving
\nelements,
ANALYSIS
STEADY-STATE
SINUSOIDAL
\nfacts:\n
(1)
The
sinusoid
\nstill
be
sinusoid
a
(2) The sum
of a
of
number
and
amplitudes
\ntrary
be repeatedly differentiated
of the same frequency.\n
may
sinusoids
phase
angles
of
one
or integrated and
frequency
is a sinusoid
with
arbi\302\254
of the same
\nfrequency.\n
In
distinction, the sinusoid is gen\302\254
in nature:
a bottle bobbing in the water,
a
rather
commonly
on a wheel\342\200\224all
these
of a crank-handle
devices
the
shadow
A sinusoidal voltage is generatedby a con\302\254
motion.
sinusoidal
in a circular path at right anglesto a
to
move
constrained
to
addition
\nerated
\npendulum,
\ndescribe
\nductor
this
mathematical
field.\n
\nmagnetic
the
under
Analysis
assumption
of a sinusoidal
driving forceand a
fields as electronics, network theory,and
the driving forces are sel\302\254
In these
\nservomechanisms.
fields, however,
We
\ndom
sinusoidal.
rightfully
question how valid such analysis
might
of this method stems from Fourieranalysis:
\nis.
Part
of the justification
be approximated
can
waveforms
by a finite sum of sinusoids.
\nperiodic
waveforms
can be expressed
(and
nonrecurring)
\nFurther,
nonperiodic
\nin
terms
of
of sinusoids
by use of the Fourier integral. This concept
state
\nsteady
\nanalysis
\nresponse
\nfrom
In
\nsolution
\nstate.
\nof
is
used
in such
allows the
frequency
components
of
a network
to a nonsinusoidal
waveform to be predicted
a known
as
a
function
of
response
frequency.\n
this
we
will
the
section,
develop
relationship between the general
of a network
and the solution for the sinusoidal
problem
steady
This
will be accomplished
in terms of the pole-zeroconfiguration
network
in
terms
of
harmonic
functions.\n
240\n
Radian
11-1,
241\n
sinusoid\n
and the
frequency
ANALYSIS
STEADY-STATE
SINUSOIDAL
11-1
Art.
cosine
The term sinusoidincludesthe sine
or either
wave,
the
or
cosine
with a phase angle. The transforms of the
the
wave,
\nthe cosine
are\n
=
wt
\302\243
sin
The
the
on
appear
11-1,
w2
ju
s-plane\n
s-plane\n
9\n
(\n5
(a)
11-1.
Pole-zero
(6)\n
for sinusoids: (a)
configurations
(b)P
radian
(11-1)\n
w2\n
+ \342\200\2245
j<a\n
\302\253
Hg.
s2
for these transform equations, as shownin Fig.\n
j<a axis. Such frequencies have been definedas\n
zeros
and
poles
<at =#
cos
\302\243
\342\200\224r>
+
s2.
sine
wave;\n
wave.\n
cosine
described by positions on the jca axis
Frequencies
represent
pure radian frequencies such as occurin the
state
and correspond
to the time-domain factors eiot
frequencies.
s plane
the
\nof
and
sine
\nsine
\nsinusoidal
steady
e~iut.\n
\nand
and
Sine
cosine
to exponential factors by the
are related
functions
\nequations\n
sin
(at
cos (at
=
pjut
p\342\200\224jitit\n
(H-2)\n
jp
ei<*t
e
jut
\342\200\224\n
(11-3)\n
\n2\n
term
The
\nphasor*
likewise
e\342\200\231at
is
rotating
is
*
For
in
the
interpreted
direction.
\n(clockwise)
those
interpreted
commonly
who prefer,
positive
as a
The
terms
of a unit
(or counterclockwise)
unit rotating
unit
in
phasors
rotating
direction;
e~\342\200\231at\n
phasor rotating in the negative
in Fig.
are illustrated
the term phasor may be read as vector,\n
11-2,\n
the
Now
\nof
two
\nby
the
\nunit
unit
rotating
factor
Fig.
\n(1
is
illustrated
11-3.
The
exponentials
and
phasor
/j)
\nzero
when
0 to
1
The
11\n
of a sine
in
of these
terms
Fig. 11-3. The combination of
wave
the\n
from rotating
wave
sine
on the
a phasor
gives
of
rotation
a negative
phasors.\n
ja axis. The factor
90\302\260 (\342\200\224x/2
radians).
is a real number (on the axisof reals);
it has a value of
at = 0, and a value of unity when at \342\200\224
As at increases
x/2.
the
limits
2x, the sine function is seen to have valuesbetween
of at
sine
\nof
to
corresponds
\342\200\224j
\nThe
\nfrom
in
(\342\200\224e~iat/2)
(e,\342\200\231\"</2)
=
phasors,
The construction
(2j).
Chap.
to Eq. 11-2, is made up of the difference
in opposite directions, divided
rotating
according
sinusoid,
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
242\n
and
\342\200\2241.\n
as
factors
\nnential
similarly constructed in terms of expo\302\254
in Fig. 11-4. The cosine is alsoa real\n
may be
is illustrated
function
cosine
eiwt+g-jut\n
-cos
t\n
2\n
Fig.
number
having
The
11-4.
cosine wave
from +1 to
variation
a total
\ncosine has a value of unity;
from rotating
at
when
=
x/2,
phasors.\n
\342\200\2241.When
the cosine has
at
=
0, the
zero value.
sine are generated by two \342\200\234frequencies\342\200\235:
\nand
This
is also
shown from the pole locations of Fig. 11-1.\n
\342\200\224ja.
The
factors
to the cosine or sinetermscan
exponential
corresponding
in computing
\nbe used
for the sinusoidal steady state. Con\302\254
impedance
a series
with a cosine driving force givenas\n
RL circuit
\nsider
\nBoth
the
cosine
and
the
F
cos
=
\302\253i
r
+\n
(^-\342\200\230
(U-4)\n
STEADY-STATE ANALYSIS
SINUSOIDAL
Art. 11-1
cosine
The
single
\nbe
equivalent
to
\ngenerating
the
Using
(7/2)e-,\342\200\230\"\342\200\230.
\nconsider
the
\ncurrents
to
\nferential
equation
forces
driving
\nof
Ri
the resulting
generator, the
Aeiut.
i\342\200\236(t)
=\n
dif\302\254
(11-5)\n
(the particular integral) will be
this solution into the equation
solution
of the
part
=
form
other
becomes\n
steady-state
the
to
seen
we may
of superposition,
For the first
solution.
the
(F/2)e,\342\200\231\"<,
and then combine
separately
final
the
obtain
cos at is thus
generating
principle
+
The
v(t) = 7
generating
generator
two generators,
one
243\n
Substituting
\ngives\n
or
-
V/2
=
A
A
= 7/2
RA
+
jaLA
ZZ?\n
R+jaL
\nwe
is
Z
where
let
may
\nreplacing
Z\n
steady state. Similarly,
e~iat
solution of Eq. 11-5
for the sinusoidal
the
impedance
Be~iat
be the
eio>t to
steady-state
with
give\n
7/2
for the
solution
total
7/2\n
_
- jaL
R
The
(11-6)\n
(11-7)\n
Z*\n
state becomes\n
steady
(11-8)\n
If
7/Z
is defined
equation may be writtenin
as 7, this
=
*..(<)
In this equationI
\nnumbers
to
relating
the
I*
number,
exponential
cosine and
form\n
(11-9)\n
+ I*e~iat)
v(Ieiut
complex
the
and
number,
\ncomplex
is a
the
is the
of this
conjugate
eiat and e~lulare
factors
complex
sine functions according to
Euler\342\200\231s
\nequation,\n
e\302\261jo>t =
If we let 7 =
a + jb,
i$*(t)
11-9
Eq.
=
a cos
= Re
where
\nimaginary
the
letters
part
Re mean
of\342\200\235). But
to the
reduces
at
\342\200\224
b
sin
[(a + jb)(cos
\342\200\234the
real
(a
sjn
+ j
cos
part
(11-10)\n
form\n
at
(11-11)\n
at + j sin
of\342\200\235
(similarly,
+ jb) = I
&><)]
(11-12)\n
Im
means
and (cosat +sin
\342\200\234the
at)
is\n
defined
Euler\342\200\231s
by
=
i\342\200\236(t)
=
i\342\200\236(t)
\nthe
equation
sinusoid
for
Re
f
that
\ncircuit
in the
written
be
forms\n
=
we
(11-15)\n
Ve\342\200\231at
(11-16)\n
of the exponential equivalent of the sinusoid,
consider
the differential
equation for an RLC series
as\n
given
of the
and
\nferentiation
idt
+ Ri +
The sinusoidalimpedanceis
is given
= R +
j-
this
real part
\nV\n
Re\n
U
= Re
1/
f
=
j (uL
\342\200\224
(11-19)\n
=
of
^
this
e,w
(11-20)\n
is taken;
expression
=
Re
that
is,\n
(11-21)\n
(\302\245
example,\n
i(t)
i(t)
(11-18)\n
as\n
i(t)
For
dif\302\254
as\n
defined
i(t)
provided only the
required
R+]L)l
=
Z(j(a)
(11-17)\n
= v
+
current
VeP*
gives\n
integration
(j\342\200\236L
The
=
^ J
solution must be Ieiat. Performingthe
jt
form
Ie,0,t
use
the
L
The
the
differentiated
v(t) =
\nsuppose
of
place
take
i(t)
illustrate
(11-14)\n
and
may
To
ey\342\200\234M\n
^
the exponential is easily
method
of the solution is convenient. Providedit is
that
the real part has meaning, currents
only
understood
\nvoltages
11\n
(11-13)\n
in
may use the exponential
solution providing we
only
the
this
integrated,
\nalways
(Ie\342\200\231at)\n
steady-state
the
solution.
Since
of
\nreal part
\nand
Re
us that we
tells
last
This
Chap.
Hence\n
etut.
as
equation
or\n
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
244\n
+
(J-
|^-R
-
+j(aL
r
i/ggp [R
~
/
(11-22)\n
1/coC)\n
-
i\n
+irin\342\200\234\n i)\n
i)](cos\342\200\234'
(11-23)\n
Ait
Z
Since
\342\200\224
R
\342\200\224
=
ut
cos
+
ANALYSIS\n
245\n
of i(t) in
part
\342\200\224
<at
sin
(wL
t.hia equation is
(11-24)\n
j
finally,\n
=
i(t)
same
This
\na
real
the
1/toC),
j(\302\253L
\ni(0
or,
STEADY-STATE
SINUSOIDAL
11-2\n
j^|
for
example,
exponential
this
form,
aL
tan\"1
(11-25)\n
to represent a sinusoid
at
let\n
t>(\302\243)
In
-
may be used
factor
exponential
angle;
phase
cos (ad
\342\200\224
V
COS (bit
-f-
(11-26)\n
<f>)\n
becomes\n
equation
(11-27)\n
\342\200\224V-\342\200\224J\n
or
=
v(t)
The
be
will
\nwhich
a
(Ve\342\200\231+)is
quantity
represented
by
\ning as
=
+
This exponentialfactor
result
\nsame
to
\nnecessary
that
\nment
In this
\nto
with
less
carry
the
only
the
s plane.\n
\nin
s;
in
1 =
long
as the
kept
require\302\254
in
mind.\n
correspond
frequencies located on the
complex
j<a
functions\n
network
as a quotient
be written
may
sinusoidal
P(s)
Q(s)
j
\nZ(s)
the
so
of polynomials
form,\n
general
G(s)
For
phase of
functions
network
have
and
11-2. Magnitude and
All
manipulation.
the
factors
of
11-26 to give the
Again, it is not
of Eq.
place
the result has meaningis
we have seen that the sineand
cosine
section,
reason\302\254
(11-30)\n
of\342\200\235
notation
part
same
the
by
real part of
the
exponential
\naxis
in
mathematical
\342\200\234real
(11-29)\n
V*e-i\342\200\234t)
(Ve**)
used
be
may
phase angle 0,
Then\n
to\n
= Re
v(t)
V.
as Eq. 11-9,and
is equivalent
given
previously
V and
magnitude
notation
the
the same form
is of
equation
of
phasor
v(t)
This
(11-28)\n
+
^[(Fe,\342\200\231*)e,\342\200\231\"\342\200\230
(Ve~,'+)e~\342\200\231at]\n
steady
=
apsT
+
ais\"-1
bosm
+
blSm~l\n
state,
+
\342\226\240\342\200\242\342\226\240+\302\253\342\200\236
(11-31)\n
s = ja,
and the
network
function\n
\nbecomes\n
G(ja)
\nZ(ju)
J
j
=
P(j<*)
_ ao(j<a)n +
bo(joi)m
+
aipo?)\"-1 +
...
-J- ...\n
(11-32)\n
this
In
the
n
whether
\non
will alternately
be real and imaginary.Which
are
real
and which imaginary depends
general
expression
and m are even or odd. However, in every
case
it will be
the quotient of polynomials
in
\npossible to write
where\n
A(a>)
Z(j\302\273)\n
\nC(\302\253)
A(w)
A(a>),
P(j<a)
and
=
Re
Q(jo>)
and
the
=
and
C(w),
phase
form\n
+ jX(
(11-33)\n
<*)\n
P(j<a)
(11-34)\n
Q(ju)\n
by rationalizingthe
Phase
D(a>).
in terms
defined
are
R( co)
B(u>) = Im
D(co) = Im
X(w) are found
function
for
equation
\ning
jB(a))
the
jD(\302\253)\n
Re
B(\302\253),
network
\ngeneral
+
=
R(o>) and
quantities
\ninvolving
-f-
G(i\302\253)\n
\nC(o))
The
Chap. 11\n
terms
equation,
of
\nterms
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
246\n
and
expression
of the
magnitude
of R and X.
The
defin\302\254
is\n
=
*M
ten\"
(11-35)\n
IM\n
equation for magnitude
and the defining
is\n
=
M(\302\253)
The complex variable,
V[X(o>))*
-f
R(o))
jX(oj),
+
[^(co)]2\n
is
thus
defined
(11-36)\n
in polar
coor\302\254
as\n
\ndinates
(11-37)\n
the
Alternately,
\nbe
magnitude
from
directly
computed
and phase of the networkfunction
may
form of Eq. 11-33as\n
the quotient
=\n
[Af(w)]2
of
a network
\nimportant
in
magnitude
network
\nquantities
are
\nand
phase
The
\nquency.
[B(\302\253)]\302\273
\n[C( a,)]2
+
[D(<o)]2\n
(11-38)\n
tan-1^\n
A
(11-39)\n
C(\302\253)\n
((a)\n
that have been describedthe
function
may be found as a function
and phase
characteristics
of networks
theory,
partly because measurements of these
methods
of the
either
+
tan-1\n
<!>(<*)\n
By
[A(q,)P
magnitude
of
fre\302\254
are
made.\n
easily
of
problem before us is the computationand plotting
magnitude
a
as
function
of
The
amount
of
phase
frequency.
computations
The
\nand
be reduced
\ncan frequently
\nof
functions
these
\ngiven
as
Eq.
in terms
11-32.\n
Asymptotes.
High-Frequency
\nbeing
considered
by first considering the asymptoticvalues
of the original quotient of polynomial
form
is
a transfer
Assume
function
the network function
G(jw). For large values of
that
\302\273,\n
Art.
11-2\n
only
the
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
247\n
%\n
are
\nG(ju>)
that
significant;
lim
is,\n
=
G(jo>)
\nthat
..
+
.
+
...\n
(11-40)\n
H\n
00\n
co\342\200\224\342\226\272
\n(jo>)n
=
function
(j<\302\273)n
lim
oo
co\342\200\224\342\226\272
Thelimit of this
and denominator of
numerator
in the
terms
highest-ordered
lim H(joi)n~m\n
which
on
depends
(11-41)\n
n or m,
number,
is larger;
is,\n
00 g/(n-m)T/2
lim
\n\342\200\242
0\n
=\n
G(j<j})
n
>
m
\nn
<
m\n
n
=
m\n
00\n
CO\342\200\224>
H\n
The
limiting
\nThe
angles
value
in
each
case
is zero, infinity, or a
magnitude
are some multiple of w/2 radians.\n
is
determined
polynomials.
The
\nwork
function
\ntient
of
case
The
Asymptotes.
Low-Frequency
\nthis
of the
(11-42)\n
net\302\254
terms in the
quo\302\254
of the network
part
important
of the
behavior
low-frequency
the lowest-ordered
by
constantH.
function for
is\n
..
.
~b
an-i(ju>)
o>n
(11-43)\n
\342\200\242
\342\200\242
\n\342\200\242
\342\200\235(\"
bm\342\200\224l(j\"oj)
If
neither
nor
a\342\200\236
bm is
zero,
the low-frequency
lim
G(ju>)
=H^
co\342\200\224>0
function
network
\nthen the
may be
+
...
\\a0'(j<a)\302\273
where
p may
small
\nbecomes
be positive
of this
that
\nnegative;
while
In
\nways:
a
polar
etc.,
+ a/1
(11-45)\n
as
function
\302\253\n
function
(j\302\273)\302\273\n
on whether p
depends
is positive,
zero,
or
is,\n
p >
0
p<
p
0\n
=
(11-46)\n
0\n
a
of the magnitude is zero, infinity,or con\302\254
the
is some multiple of (w/2) radians.\n
angle
the phase
and magnitude information is plotted in two
coordinate
plot, and separate plots of M and <f> against\n
limiting
practice,
\302\253n-i,
H(a\342\200\236\342\200\231/bJ)\n
{oo H(a.\342\200\231/bJ),
\nstant,
an,
G{ju>)\n
0e~ivT/i,
the
bm-i,
bm,
bm'\\\n
e~ipT/2,
Again,
as
... +
or negative. The limit of this
+
LW(jo,)m
co\342\200\224\342\226\272
0\n
limit
(11-44)\n
is\n
lim
The
is\n
written\n
H
\n(i\302\253)p
asymptote
Vm\n
are zero such
terms
or more
one
if
However,
\342\200\234f\342\200\234
bm\n
value
part
\ninary
real
and
types
made
be
can
plot
\npolar
two
These
frequency.
part
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
248\n
of plots are
in terms of
Chap, n\n
illustrated in Fig. 11-5.
either M(u) and
The
the
or
<f>(<a)
imag\302\254
of\n
U)\n
Tig.
11-6.
of phase
Plotting
and magnitude,
\noo\n
11-6.
Fig.
High
frequency
asymptotes of G(jw)f wherem
n
and
denominator
=\n
order
of
=
order
of\n
numerator.\n
00\n
j Im
G{ju>)\n
Jl\n
P-3/\n
fpm~1
'~2\n
*\"73
Re
P-0
\\p--3
G{ju)\n
1\n
oo\n
Tig.
Plots
on
the
lla7*
M(<a)
\ncurves. The quantity
phasor
represented
\nfusion,
only
\nof
the
phasor
the
by
Low-frequency
and
<\302\243(co)
asymptotes
coordinates
of\n
are
made
however, is usuallythought
an arrow as shown in Fig. 11-5.To
of
the
is plotted.
The locus of the
phasor
as the phasor locus of the
function.
of
avoid
\342\200\234tip\342\200\235
is known
as continuous
network
as
a\n
con\302\254
\342\200\234tip\342\200\235
The\n
11-8\n
Ait
in
shown
Fig.
Thus the low-frequency
and
of
\ntask
the
Example
points.
in
shown
is
made
network
Vt{s)
R
o
-\n
low
\nhigh
\nand
\nis
\nthe
the
frequencies,
\ncomes
X
especially
magnitude
\nsummarized
other
One
for
convenient
while
for
value
be\302\254
the phase is
11 -
8.
pair
RC
1/RC.
At
\342\200\22445\302\260.
This
Two\n
=
*W\n
5-
-terminal-\n
network.\n
this
frequency,
is
information
11-1\n
GW\n
locus
M(w)
and
Pig.
0
<\302\243(o>)plots
11-9.
at
is that of a
RC
are
network
0\302\260\n
at
0.707
oo\n
phasor
at
1
0\n
1 /RC\n
\nequivalent
0\n
in Table 11-1.\n
(O\n
complete
-\n
Pig.
a) =
TABLE
The
\342\200\224
\342\226\240
V\\Ar\342\200\224\n
frequency
computation:
and
0.707
is
1,
is, zero magnitude
angle.
letting
by
Vi (s)
asymptotic
that
e-,T/2;
\342\200\224
90\302\260
phase
is found
(11-48)\n
\342\200\224>
G(j<a)
frequencies
0
1\n
R\n
jo> + 1/RC\n
1
+
jaRC
(11-47)\n
RCs +
1 /RC\n
1
is\n
1\n
~
jo)', thus\n
0(j\302\273)
transfer function
sinusoidal steady state
for the
function
transfer
l/Cs
+ l/Cs
_
}
and one
of one resistor
up
The voltage ratio
11-8.
Fig.
Fx(S)
For
a network
There remains the tedious
A number
of examples will
by inspection.
two-terminal-pair
_
=
of
behavior
procedure.\n
general
\nK
\n8
frequency
1\n
\ncapacitor
The
for low and high
high-frequency
intermediate
computing
\nillustrate
A
determined
be
can
\nfunction
249\n
ANALYSIS\n
of the network functions
11-6 and Fig. 11-7.\n
values
asymptotic
\nare
STEADY-STATE
SINUSOIDAL
-45\n
-90\302\260\n
circle as shownin Fig.11-9.
The
also
shown
in the
characteristics.\n
figure.\n
SINUSOIDAL STEADY-STATEANALYSIS\n
250\n
2\n
Example
1
\nple
are
interchanged,
in
\nshown
the capacitor of the RC networkusedfor
there
results the two-terminal-pair
and
resistor
the
If
11\n
Chap.
\nstudied.
Fig.
11-10.
This
function
the voltage ratio
Again,
the
has
RCs\n
s\n
s+ 1
RCs +1
1/Cs
(11-49)\n
/RC\n
at s
\342\200\224
=
1 /RC.
at s = 0 and a pole
Letting\n
s = jo: gives the transfer functionfor the
a zero
to
corresponding
+
be
to
value\n
R\n
R
network
transfer functionis
=\n
0(8)
Exam\302\254
ft\n
\nsinusoidal
state
steady
as\n
Cs\n
V2(s)\n
Vi(s)\n
0(i\302\253)
=\n
\nbe
applied).
\nand
90\302\260
to
\nreduce
=
\nG(ju)
\nmarized
phase
Again,
angle.
an
simple
especially
In
eir/i.
0.707
as
asymptotes.
high-frequency
becomes
large,
the
(alternately,
tabular
form:
co
=
1 /RC.
1/RC can be
term
For
can
rule
^Hospital\342\200\231s
of co, G(ju>) approaches
causes
one frequency
values
small
co
unity
approaches
a,ndG(ja>)
For
and
\nquency
\nAs
\nneglected,
low-fre\302\254
Ex\302\254
2.\n
\nample
will be examinedfor
This function
of
Network
11-10.
Fig.
(11-50)\n
1 /RC\n
+
jdi
zero magnitude
the function to
that
form, these computations
frequency,
be
may
sum\302\254
follows:\n
TABLE
11-2\n
co
0
1
/RC
oo
G(ju>)\n
0 at
0.707
+90\302\260\n
at
1 at
+45\302\260\n
0\302\260\n
computation. In orderto
other
several
values will have to be found. The
plot,
\ncomplete
locus
is that of a circle, as shown in Fig. 11-11.
phasor
and
Af(co)
<f>(co) plots for this same function are also
The two plotting
\nfigure.
systems display the same information.
it will
be possible to visualize one form from an
These
values
serve
as a
guide to the
make
the
com\302\254
The
\nplete
\nalent
equiv\302\254
in
shown
the
With
\npractice,
the
\nof
inspection
other.\n
of Example 1 and Example2,itis
all
provides
positive phase shift (or phase lead)
while
the
latter
\nquencies,
provides
negative
phase shift (or phase
\nfor
all
at
For the first, the output per unit input is
frequencies.
\nlow frequencies
and
low at high frequencies. The opposite behavior
in the
second
network.
This result can be correlated
place
Comparing
the
first
\nthat
the two
networks
seen
for
fre\302\254
lag)
high
\ntakes
with\n
STEADY-STATE
SINUSOIDAL
Art.
11-2\n
the
behavior
of
\nquencies
and
251\n
elements of the two networks.In the
acts as an open circuitat low fre\302\254
11-8, the capacitor
Fig.
the same voltage
hence
appears on the output terminals\n
individual
of the
\nnetwprk
ANALYSIS\n
characteristics.\n
RC network
11-11.
Fig.
the
as appears on the input terminals.At high frequencies,
however,
the
circuit
and
short
output
voltage
approaches
\ncapacitor acts as a
the
For
the net\302\254
the
across
\nzero
capacitor).
drop
magnitude
(being
low
fre\302\254
\nwork
of Fig.
11-10, the capacitor acts as an open circuitfor
is no output voltage; however, at highfrequen\302\254
so that
there
\nquencies,
as a short cir\302\254
behaves
\ncies
the
capacitor
\nage
to
on the
appear
the
on
\nappears
same
the
approximately
causing
\ncuit,
output
\342\226\240AAAr\n
volt\302\254
R\n
terminals as
Z[s)
terminals.\n
input
and
3\n
Example
11-12.
Fig.
As
Ls\n
Yts)\n
the
\npoint
third
consider
example,
of
immittance
a series
the
have
functions
\nimmittance
=
Z(s)
the drivingRL network shown in
RL
Fig. 11-12.The
forms\n
R -J-
*
Ls\n
+
-H\n
y(s) = W>
the impedance function has a
(11-51)\n
f)
=
Thus
\ninfinity,
state,
admittance
become
(poles
\nfiguration
\nsteady
the
while
s =
zeros;
function
zeros
s
+
+
Z(\302\273=i(>
(11-52)\n
R/L
at
zero
s =
\342\200\224R/L
has the opposite
become poles). In
j<a, and the immittance
r(M
network.\n
functions
=
B(l+>|)
f)
and
at
a pole
pole-zero
the sinusoidal
con\302\254
become\n
(11-53)\n
1/R\n
=\n
jo)L/
R
-j-
(11-54)\n
1\n
STEADY-STATE
SINUSOIDAL
252\n
ANALYSIS\n
Chap. 11\n
real part of Z(ju) is constant and the imaginary
part
form
with frequency.
11-54
has
the
same
\nincreases
Eq.\n
Equation
aji
and\n
has
circular
a
locus.
The
locifor
which
impedance
phasor
11-48,
the
11-53
Eq.
By
or
G(\302\253)
characteristics
Immittance
11-13.
Fig.
Locus
Z(s)\n
j Im
Magnitude
plot\n
G.\n
for RL
network.\n
Phase
plot\n
plot\n
ReG\n
0\n
1-
\342\200\242\n
1\"\n
00\n
j Im
G\n
I
00
It\n
\302\2532+a*+1\n
IZJ.\n
j Im
O
\noo
Re
G\n
Re
G\n
\302\253(as+1)\n
j Im
G
a\n
\noo
1\n
M\n
\342\200\242*(\302\253\302\253+!)\n
Fig.
admittance
11-14
\nFig.
11-3.
As
\nwaveforms
shown
are
for
given
was
pointed
in
any
in Fig. 11-13.
transfer
network
Sinusoidal
11-14.\n
functions
functions
Several other plots
are
or immittance
in terms
of poles and
shown
in
functions.\n
zeros\n
earlier in this chapter, all voltage and current
linear network
are sinusoidal in the steady stateif\n
out
11-3
the
network
\nare
STEADY-STATE
SINUSOIDAL
Ait.
V(jot)
=
/(j\302\253)
not interested in
we are
waveform
\nthe
\nis:
(1)
V(jot)?
to
\nnecessary
the
will
to
\nV(ju>)
\nof
of
give
an
that
of
V
between
\nship
if V is
this ratio is
\nThe
term
\nratio
of
\nwith
Y
of
\n(and,
\nmined
\nand
to
=
the magnitude
the
all
relation\302\254
values
of
Y(jot),\n
\\Y(jo,)\\e\302\273\342\204\242
(11-56)\n
=
(11-57)\n
YW
complex
ratio
of
to voltage.
current
thus
the
are
in radian frequency) is deter\302\254
specializing
that is, by immittance functions
ratios;
now
complex
functions.\n
same
arguments
given in the
because
equations
typical
V(jot)
V2
for
(jot)
=
previous paragraphapplyto
of their
to Eq. 11-55.\n
similarity
(11-58)\n
G(jot)Vx(jo>)
= Z(jot)I(jo>)
=
the
'
(11-59)\n
(11-60)\n
Ztr(jo>)I(jot)
is the transfer impedance).Thusit is
the
that
in
to
network
function in the
general
apply
Y(jot)
any
and so on (Ztr
\nsinusoidal
of
phase
form\n
F,(j\302\253)
\ncussions
relates
10
V of
amp,
implies not only the magnitude of the
but also the phase of the onequantity
to another
other.
Network behavior as a function of frequency
by
entirely
\nfollowing
in the
ratio
we
course,
Y (jot)
and phase of
magnitude
quantity
transfer
The
likewise,
(jot);
often described as the
respect
1
of
I (jot). In the caseof the last equation,
and I (jot) is given completely
(for
(jot)
complex
one
1 volt to give I
the
I (jot) is
is written
11-55
(jot) is linearlydependent
10 amp. The quantity that relates
Y(jot)
Eq.
of I
magnitude
V(jot):
I of
by the
\nfrequency)
If
know
We
I(jot).
of
that
to
F(jcj)
the
linear,
magnitude
\nvolts
of
waveform
a sinusoid. The informationwe do need
of the voltage V(jw), what is the mag\302\254
it is
magnitude
are
\nconsideration
\non
(11-55)\n
I (jot)
of
in terms
(2) what is the phaserelationship
we are interested only in magnitudeand
words,
and
I (jot). To find this information,it is not
V(jot)
know
the magnitude
of V(jot). Since the networksunder
relating
\nphase
we
I (jo>), and
In
other
of
\nnitude
\nof
a given
for
reason,if
Y(jot)V(jo>)
solving for the
advance:
in
253\n
waveforms. For this
by sinusoidal
in this equation,\n
is driven
gjven
ANALYSIS
steady
seen
state.\n
dis\302\254
SINUSOIDAL STEADY-STATEANALYSIS
254
which
\nnomials
may
-
(8
either a pole
where sr is
\ns
=
the
and
ju,
typical
or
the
plane,
complex
\ndifference of these
in
\nis,
complex;
general,
Direction
11-16.
Tig.
poly\302\254
(11-61)\n
sinusoidal
the
state,
steady
term becomes\n
ju and sr are
the
(11-62)\n
\302\253r)
phasors.
in the
interested
are
We
The
a phasor.
also
is
phasors\342\200\224which
two phasors
These
\nju axis.
two
of
form\n
In
zero.
a
quotient
Sr)
(ju In
form of a
of the
Y(s) has the
into roots
be factored
function
admittance
The
11\n
Chap.
phasor s,
phasor ju is purely imaginary and is
and their difference are shownin Fig.
the
on
11-15.\n
of the phasor (ju
(b) string diagram.\n
\342\200\224
\302\253,):
(a)
diagram;\n
polar
the phasors with respect to the s plane
origin.
shows
the
The
\342\200\234string\342\200\235
phasor
diagram.
ll-15(b)
equivalent
\nFigure
to be a phasor directed from sr to
is seen
Sr)
\nphasor difference
(ju \342\200\224
of ju changes\342\200\224always
from 0 to <\302\273,the
As
<o changes
position
\nju.
of several of thesephasors
The combination
on
the ju axis.
\nremaining
network
functions. This will be
sinusoidal
\ncan
be
used
to determine
11-15
Figure
\nillustrated
(a)
shows
a number
by
of examples.\n
first the admittance
Consider
of a
is
\nof
in
shown
this
11-12.
Fig.
network
a
Such
circuit.
RL
series
The impedance
is\n
=
Z(s)\n
(11-63)\n
*,(\302\253+!)\n
and so the
admittance has the
y<*>
Thus
\nand
11-16.
a
As
u
has
form\n
-
(1MH)\n
i jjTFm
a finite
pole at s
=
\342\200\224
R/L
This pole-zero configuration is shownin Fig.\n
infinity.
increases
from zero to infinity, the phasorohanges
position\n
at
aero
Y(a)
circuit\n
in
shown
\nfound
Phasor
11-17.
Fig.
as
STEADY-STATE
SINUSOIDAL
11-4\n
Art.
+
Then the admittance
be
may
It can be
frequency.\n
variation of the impedance
is
the
from
found
(11-65)\n
equation\n
from
changes
l/R
variation
for
\nvoltage
has
\nlinearly
for
as
0\302\260
o>
changes
\342\200\22490\302\260
with
frequency
\nfrequency
to 0
as
similarly, the phase changes from to
The polar coordinate representation of
zero to infinity.
in Fig. 11-18. The variation of admittance
is shown
is exactly
the same as the variation of current
of voltage of l/R volts. If the
a constant
magnitude
the current will increase or decrease
different
magnitude,
to infinity;
from
\n\302\253
varies
= M^^*M
f)
seen that the magnitude
\nfrom zero
\nwith
255\n
writing\n
by
\nthis
changing with
diagram
11-17. The frequency
Fig.
ANALYSIS\n
a
of frequency.\n
values
all
AA/V\n
Ls\n
R\n
1 JCs-Ztz\n
Y[s)\n
o
Fig. 11-18. Variation of
\nwith
11-4.
\ncuit
\nwork
circuit
circuit
shown
as
RLC
circuit.\n
in
Q, and
bandwidth\n
by means of the study
networks.
Considera series
other
applies to
Fig. 11-19. The driving-point impedance for this
that
method
\nRL series
Series
frequency.\n
Resonance,
The
phasor
\t\n
Fig.11-19.
has
been illustrated
of
RLC
the
cir\302\254
net\302\254
is\n
Z(\302\253)
=
Ls
+
R +
^\n
(11-67)\n
the
and
is the
admittance
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
256\n
of Z(s); that
reciprocal
Chap.
is\n
=
or\n
L
(s2
\\s2 +
\nL
((B
Y(s)\n
(11-68)\n
1/Lc)\n
Rs/L
Its/L + X/LC,
-
\302\253.)(\302\253
11\n
(11-69)\n
\302\253.*))\n
where\n
Sa,
=
V\n
1vr:rT\n
\302\251\342\200\230
(11-70)\n
In this expression,
is
wn
natural
the
undamped
and
is
f
consider
we
If
ratio.
\ning
of the
frequency
dimensionless
the
system,\n
damp\302\254
the
only
\nunderdamped (or oscillatory) case
of the
< 1, the variation
f
of Y(s) for constant
of
the
\ntion
poles
and
variable
f is shown in Fig. 11-20,
\nwhere
posi\302\254
\ncon
Pole-zero
11-20.
\ntion
for
In
F(s).\n
\n(s
\nthe
\nallowing co
\na
frequency
as
quantities
a
to
over
vary
range
variation
are shown
such
system,
Fig.
plete
magnitude
\ncies,
the
\nmagnitude
phase
starts
11-21.
and
s
\nthe
configura\302\254
Frequency
has
Y(s)
\ntion,
Fig.
locus
the
\nwhere
=
|/(j'<o)|,
is
In
addi\302\254
at the origin of
a zero
plane.\n
sinusoidal
the
the
ju>),
\\
Y(ju>)\\
steady
frequency
may
etc.,
state
response
of
be found
by
of frequencies. Several steps in such
in Fig. 11-21, together with the
com-\n
response of an RLC
network.\n
Over the
characteristic.
0\302\260to
from +90\302\260, through
phase
changes
from zero
a circle.
value, attains a
range of
frequen\302\254
while
\342\200\22490\302\260,
maximum value,and
the
then\n
phase
The
maximum
\nhave a
\napproach
\nchanges
\nslowly.
the
\342\200\224
jco
of magnitude
terms
in
of Y(joo)
value
Since I (jo) varies just as Y(jo),the
the frequency of maximum I (jo).\n
of resonance.
is also
resonance
of
The
of
magnitude
this
from
in the
may be written
Y(jo)
1
-\n
l*WI
\n(1
The function
frequencies.
a
form\n
frequency
\nquency
and
257\n
evidently takes place near the frequency
value. The frequency to cause Ma to
has a minimum
Ma
near
to the point
of closest
minimum
value is a frequency very
In
that
to
one
of the conjugate
frequencyrange,Ma
poles.
and
same
time
and
at
the
Mo are changing
Ma*
very
rapidly,
Y
to
a
maximum
is
defined
The
(jo)
corresponding
frequency
which
\nas
ANALYSIS\n
high
letting
11-69,
Eq.
the
in
factors
\nand
\nat
from
found
is
for
zero
approaches
asymptotically
\nY(ja)
STEADY-STATE
SINUSOIDAL
11-4\n
Art.
form\n
\t\n
(11-72)\n
\342\200\224
1/oC)2\n
(oL
\342\226\240\\/R2\342\226\240+\342\200\242
it is seen
equation
fre\302\254
that F(jco)has a
of
value
maximum
when\n
/R)
(11-73)\n
the
opposite
\nAn
that resonance occursat
is to say
that
is
\nphase
angles
\naxis
of
\n4>a*
is
point\n
various
reso\302\254
The
phasor
jon is along the jco
has a constant
phase
+90\302\260.
equal
at the
not
(and
to
zero
thus
and
on
11-22. The
poles to jon
and
<f>a
the
Fig.
the
from
marked
\nangle
in
=
jco axis).
the
condition of
shown
\nnance
\nfrom
of
view
enlarged
\nphasors for the
\nare
on the
pole
o
to
of
sum
The
<f>a
and
the
90\302\260because
tri\302\254
is a right
ABC
triangle (being
\nangle
\ninscribedin a semicircle). The total
\nphase
0o
angle,
\342\200\224
<l>a
which
\342\200\224
<f>a*
is\n
\342\200\224
+90\302\260\n
\342\200\224
(+90\302\260)
=
0\302\260
Fig.
(11-74)
11-22.
Phasors
drawn for
reso\302\254\n
nance.\n
thus
has
zero
value.
The
phase\n
at
angle of Y(jco)is zerodegrees
\nresonanceis (1/R), and the
magnitude
The
resonance.
of
I (jco)
magnitude
at resonance
of
F(jo)
at
is (V/R).\n
Q or
circuit
The
\342\200\234
(R/2L) is
\ncircuit
defined
is
Q
1
_
1
\302\253\342\200\236
_
2
2
the
Thus the
11-70.
Eq.
by
the
length
the
length
O A\n
(11-76)\n
EB\n
11-22. The circuit
be
thus
can
Q
of
from a scale plot of the polesandzeros
an RLC circuit. The circuit Q can alternately
\ntaken directly
\nfunction for
\nin
X
shown in Fig.
of quantities
terms
as
same
the
as\n
^
in
(11-75)\n
2 R/2L\n
R
11\n
as\n
defined
1
oanL ~_
_
q
\nV
Chap.
RLC seriescircuitis
Q for an
simply
Now the quantity
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
258\n
immittance
the
written
be
form\n
(11-77)\n
1\n
the
0 is
where
these
From
11-22.
from the
angle
last
9\n
the
to
\342\200\224a
axis
three
(11-78)\n
cos
2
of
OB (or OD)
line
Fig.\n
be
several conclusions can
equations,
\nwritten:\n
(1)
\nQ.
(2)
poles
(This
R
sa*
the jta axis,
to
are
A
of
\nvalue
and
s\302\253
EB.)\n
value
The
\nlow
the higher the
follows
since
Q varies inversely with the distance
of Q varies inversely with damping ratio, f.
high
circuit
with
Q infers a low value of damping ratio.
closer
The
A
has
thus
high
Q.\n
of
Plots
various
\nfor
\nin
Fig.
of Y(jw)
values
of Q are shown
11-23.
The circuit
Q is an
the
magnitude
\nimportantfactor in circuits
the
(of
for
used
considered)
being
\ntype
\nselectors
(filters).\n
means
Another
Fig.
11-23.
Variation
of | Y(jut)
]
with\n
\ncircuit Q is
Q.\n
\nof
the
shown,
current
\ncurrent has the
at
resonance
\nhas
\nhalf-power
has
been
specification in terms
As
points.
When
V/R.
magnitude
the\n
V\n
=\n
(11-79)\n
y/2
power
the
the
magnitude\n
I
the
half-power
of specifying
half of that at
the magnitude
points,
will be
R\n
resonance(beingequalto
of the admittance
Y(j<a)
I*R).
is
At
{\\/y/2R)\\\n
the
Art.
11*4\n
SINUSOIDAL
this requires
that\n
+
Vtf2
-
(\302\253L
to the
reduces
equation
II\n
(11-80)\n
H-\n
(11-81)\n
form\n
.
,
= V2R\n
1/coC)2
I\n
or\n
This
259\n
ANALYSIS\n
STEADY-STATE
1
R
n\n
(11-82)\n
\"^\302\261r\342\200\234_Ec\342\200\234\302\260\n
values
The
that
a
of\n
this
satisfy
=
\302\273
\302\261
of dampingratioand
or, in terms
=
the
\ncondition,
last
form,\n
positive
are the
approximation
be
frequency
\npower
=
\302\261
w\302\273
frequencies defined by\n
half-power frequencies. Let the highest half\302\254
is defined
which
designated
\302\253i,
=
=
\302\2532
the
is
(f\302\253\342\200\236)
quantity
Fig.
\npoints
The
+
<o\342\200\236
as\n
(11-86)\n
fwn
the smaller half-power frequency be \302\2532be
and let
from
(11-85)\n
frequencies).
OJl
The
(11-84)\n
damping
that
only
(considering
frequency,\n
\302\253.(\302\261f \302\261
w
this
natural
undamped
ratio
networks used as selectors,the
f is
with
is
Under
this
f2
negligible compared
unity.
to
an
reduces
especially simple
equation
practical
small,
\nvery
the
in
the
11-24.
as\n
given
\342\200\224
\302\243\302\253)\342\200\236
EB
distance
Bandwidth
to the pole sa (or
s plane is shown in
jco axis
(11-83)\n
1/LC\n
+
VF+l)
\302\253
so
are\n
\302\261V(*/2t)*
gg
in most
equation
(11-87)\n
of Fig. 11-22,
on the
The
\302\253\342\200\236*).
s
or the
distance\n
plane.\n
location
Fig. 11-24.
A
circle
of the
of
half-power
radius
(fa>\342\200\236)\n
\nthese
The
11-79.
\nEq.
the magnitude
given as (coi
of frequencies
range
At
has
Y(jo>)
frequencies,
half-power
11\n
at the half-powerpoints.
the ja axis
j<an crosses
at
centered
and
Chap.
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
260\n
A
by
as the
defined
is
to2)
with Q.
varies inversely
\nbandwidth. Bandwidth
to a selective network.\n
\ncorresponds
0.707(1/#)
\342\200\224
bandwidth
small
can
thus
be
The concepts of resonance,circuitQ, and
for the RLC
\nvisualized in terms of the pole-zeroconfiguration Y(s)
\ncircuit. These concepts are easily visualizedand do
depend
upon
The
definitions
of
numbers.
\nalgebraic
manipulation
specific
complex
in
this
section do not
circuit
and
bandwidth
Q,
given
resonance,
For example, resonance
network
to
all possible
configurations.
the
sense
of a maximum
impedance or admittance does not coincide
of unity
the
power factor for most networks.
frequency
can be visualized in terms of phasor magnitude
\nall these
quantities
can be accomplished
and
design
by means of simple graphical
\nphase,
bandwidth
of
not
\nof
\napply
\nin
\nwith
However,
and
\nconstructions.number
some
In
\nSince
the
\nthe
analysis
RLC networks are used as selectors.
is the dual of the seriesRLC
RLC network
network,
in this section applies in terms of impedance
and
parallel
applications,
parallel
given
of
instead
\nvoltage
\nand
magnitude
with
and driving-point
functions
of the
factors
frequency
which in the sinusoidal
-
\nlarger,
\ncompared
become\n
(11-89)\n
of to, this factor can be approximatedas (\342\200\224So)*1,
As
to becomes
point sa to the origin of the s
small
values
from
the
with the phasor
as
is valid. But
an approximation
no such
(11-88)\n
- So)*1
O'to
phasor
up
s,,)*1
state
steady
\na
immittances are made
form\n
(s
very
of poles
terms
in
frequency
zeros\n
transfer
Both
current.\n
and
admittance
11-5. Asymptotic change of
For
analysis
concepts in
these
illustrate
14.\n
\nChapter
\nof
of network
examples
further
will
configuration
pole-zero
\nby
additional
of
A
plane.
to
becomes
sa, the frequencyfactorcan
large
very
be
represented
\nas\n
0\342\200\230to)*1
For
large
to, the
\nor
inversely
with
\nor
\342\200\22490\302\260
depending
on
'
asymptotic
whether
(11-90)\n
(\302\253)\302\2611e\302\261fr/l
of this factor
magnitude
to. The
=
the
changes either
phase angle is
factor is a zero or
linearly
either
+90\302\260
a pole.
This\n
11-5
Ait
of
behavior
and
magnitude
ANALYSIS
STEADY-STATE
SINUSOIDAL
with
phase
261\n
is illustrated in
frequenoy
11-25.\n
\nFig.
To
\nfer
we are interested
that
suppose
illustrate,
in a voltageratio
trans\302\254
function,\n
7*(j\302\253)\n
w\n
(11-91)\n
VrU<o)\n
sinusoidal
the
Assume
that
\nand that
for frequencies
magnitude
|s0| the
to
\nof
=
the
is of
Vi(jo>)
voltage
\nif
G(s)
result
form\n
the
had
unity
(11-92)\n
V2 would
of
variation
of
form\n
CO\n
Such
magnitude
in excess
of the\n
-
-
|F,(j\302\273)|
voltage has a
input
1\n
=\n
0(a)
is
provided
a
\nto
The
unity.
s
(11-93)\n
-f-
a\n
small
very
=
|<?0\302\273)|
The
\nof
logarithmic
\nvoltage
is
but
powers
i\n
be\n
(11-94)\n
Cl)\n
decibel, was originally defined for a ratio
used for voltage and current ratios. The
in decibels
is defined by the equation\n
ratio
amplitude
For the example
=
I
20 log,.
(11-95)\n
being considered,the
db) has the
\ndecibels (abbreviated
20
=
\302\253
function would then
the
unit,
often
now
|G(\302\273
When
\nfrequency.\n
compared
of the transfer
magnitude
with
variation
Phasor
11-25.
Fig.
1, G(j<a)
in
form\n
\342\200\224
=
logio
ratio
amplitude
voltage
o)
\342\200\22420
logio
(11-96)\n
CO\n
has a
value of 0 db, and when
logic
2 =
<w
=
2, G(jo))
has the
\nvalue\n
-20
Two frequencies
\noctave.
In
\ndecibels.
\nan
additional
one
In
having a ratio
the
octave,
another
of 2:1
(to
The
w
=
said
are
db
-6
\302\253
to
be
of this example
magnitude
octave
6 decibels.
X 0.301
-20
4),
magnitude
the
magnitude
(11-97)\n
separated
by an
has decreased6
would decrease
is thus changing at
the rate
of\n
decibels\n
-6\n
octave\n
octave\n
(11-98)\n
SINUSOIDAL STEADY-STATEANALYSIS\n
262\n
G(s) given in
pole in
to one
due
Eq. 11-93.*Had
=
G(s)
of
\nasymptotic
(11-99)\n
11-90.\n
\nEq.
form
the
of
\ntors
that
considered;
G(s)
-
\nasymptotic
\nfrequency limit, these
of
an
example,
(11-100)\n
Sm)\n
transfer
frequency
function
of the
of finite
=
transfer
ji>> and
function
for large
in
cancel
pole-zero
of one zero). The net asymptotic
will thus be the number of finite
effect
the
db per
octave.\n
network having a
volt\302\254
form\n
s\n
H\n
\342\200\224
(s
=
~
. . . (s
consider
G(s)
This
Sb)
poles times 6
a two-terminal-pair
number
the
ratio
\nage
Sn)
\342\200\224
sa)(s
will
with
magnitude
less
As
\342\200\224
\342\200\242
\342\200\242
\342\200\242
(s
S2)
increases and decreases
cancels
pole
(one
\nzeros
fac\302\254
will cause an asymptotic increasein the
expression
of G(s)
of +6 db per octave, while each pole will
cause
an
\342\200\224
in magnitude
of G(s) of 6 db per octave.
decrease
In the
\nmagnitude
\nchange
of
number
this
in
zero
\npairs
\342\200\224
(s
Si)
H\n
up of a
is,\n
\342\200\224
(s
=
\n(s
Each
will be made
in general
function
transfer
A
\ns
+ a
form\n
in G(s), the asymptotic change in the mag\302\254
the
ratio
function would have been an
transfer
voltage
6 decibels
This case would correspond to the
per octave.
form
from
the choice of the positive sign in
resulting
of
\nincrease
of the
been
G(s)
zero
one
to
corresponding
\nnitude
s
11\n
Chap.
sa)(s
the
o>,
0(j\302\273)
s6)(s
=
(11-101)\n
\342\200\224
sc)\n
finite poles and
has three
values of
-
one finite
of G(jw)
magnitude
For
zero.
has the
H\302\261\n
form\n
(11-102)\n
8 =JO)\n
The
output
\ndetermined
\nular
network
\nput
compared
network will fall off at a rate
two-terminal-pair
excess
the
of poles over zeros. The rate for thispartic\302\254
by
octave.
The
is \342\200\22412db
per
asymptotic
phase of the out\302\254
will be 180\302\260 as given
to the input
by Eq. 11-90.\n
of this
The RLC selector
\nadmittance function\n
network studied in
1
Y(s)\n
corresponding
\nthis
*
\nper
selective
Quantities
octave
to two
network
in the
is equivalent
L
previous
section
has
the
1\n
(S
poles and one
for
the
constant
\342\200\224
Sa)(s
(11-103)\n
\342\200\224
So*)\n
zero. The current passing
will vary with
voltage
through
frequency\n
beseparated
ratio of 10:1 are saidto
to 20 decibels per decade.\n
by
a decade.
Six decibels
the
as
with
\ndecrease
of Fig.
curve
M(w)
frequency
11-6. An application:
The
\nzero
The
11-26.
as
configuration,
shall
we
symmetrical
The lattice structurecanbeput in a
Assume
network.
bridge
in
\nfer
the
function,
\nare
\nof
the
and
the
network
\nKirchhoff\342\200\231s
If
of
these
the
determinants;
Fig.
The
load current
ZaI
+ ZL(I
Vi
=
zj
+ zbr
Vi
=
thus\n
ZaI
and I'
been
has
is a function
in the
are arranged
(2Za
(11-104)\n
(n-105)\n
shown
quantities
=
is marked as 12. From
- 1') + ZaI
notation
functional
Fi
sym\302\254
lattice.\n
write\n
=
equations
Bridge form of the
11-27.
\nmetrical
Vt
the unknown currents I
\nof
Fig.\n
symmetry
we
law,
the
s.
trans\302\254
configuration.
In theseequations,
\nquency
load
marked
marked I'
of the
voltage
\nplicity\342\200\224each
\342\200\234unwrap\302\254
F2(s)/Fi(s).\n
currents
two
because
equal
a
identified in
currents
two
The
11-27.
the
ratio
voltage
=
G(s)
Several currents are
\n/
by
and that we are
\nimpedance ZJs)
\ninterested
that
in
terminated
is
\nnetwork
form
familiar
more
of the pole-
11-27 as
in Fig.
shown
as
it
\nping\342\200\235
\na
lattice.\n
easily visualized in terms
show in this section.\n
are
network
this
of
shown in Fig. 11-26is a two-terminalfor phase correction. The\n
application
frequent
Fiff.
properties
lattice\n
symmetrical
network
finds
that
network
\npair
263\n
11-21. For large frequency, the currentwill
at the rate of \342\200\224
6 db
per octave.\n
the
lattice
symmetrical
ANALYSIS
STEADY-STATE
SINUSOIDAL
11-6
Art.
+ ZL)I
for
omitted
of the complex
fre\302\254
forms\n
- ZlF
(11-106)\n
-f- ZbIf
may
be
sim\302\254
(11-107)\n
found
conveniently
by
the
use
-ZL\n
Vi
Vi
zb\n
-ZL
+ ZL)
(2 Za
\nZ
a
Vi\n
Vl
+
T2
7
=
The
Vi
IiZl
_~
lattice
\nments
are
these
such
selected
\342\200\224
properties when the
=
\nVi
and
R
ZaZb
-
Za)
(Zb
-\\- {Za
Za)(R
{R
Za){R
(11-112)\n
-
2 RZa\n
\342\200\224
R
Za\n
(11-113)\n
R -f-
Z^)
Z02
-}- Za2
R2
\342\200\234H
Za)
{R
\342\200\224
R^\n
fi2
=
\342\200\234IZb)
\342\200\224
Za\n
to a specificnetwork,let
impedance
in
the
LC
network
shown
by
Fig.
parallel
this derivation
apply
\nrepresented
ele\302\254
network
becomes\n
the equation
2R
(11-111)\n
zb)\n
that\n
restrictions,
Vi =
To
ratio
ZL(Zb Za)\n
2ZaZb + ZL{Za+
has very useful
network
Zl
With
_
~
~V\\
Vl
The
ZaZL\n
becomes\n
function
\ntransfer
(11-110)\n
+
is IiZL\\ hence the voltage
load impedance
the
across
voltage
as\n
Za)\n
+ ZL)
Zb{2Za
/'
and
I
of
(11-109)\n
ZaZL\n
\342\200\224
Vi(Zb
=
\342\200\224
T'
+
ZL)
found in terms
/2 may be
current
load
Zl)
+
(Za
\nZb(2Za
The
(11-108)\n
ZaZL\n
Zl)
\nZb(2Za
Zb\n
+ ZL)
(2 Za
ZL)
+
Vi(Zb
Za be
the
11-28.
o\n
The
net-\n
o\342\200\224nnnp\342\200\2241(\342\200\224o\n
^
C2\n
zb\n
work
\nseries
\nfor the
Zb must
for Zb is
to
represent
LC network
parallel
Networks
11-28.
Fig.
/
aW
and
Zb.\n
of that representing
also shown in Fig. 11-28.
be the dual
LC network
y
for Za
Za.
The
The
impedance
is\n
1
\\
Cis
+ 1/Lis
\"
L\\S\n
LiCiS2 +
(11-114)\n
1\n
11-6
Art.
series
the
while
STEADY-STATE
LC network
has impedance given
-
z*\302\253
+
*
the
=
\nL\\
such
\nother
\nthe
=
Li
frequency
by\n
o1-118)\n
^r1
C2, and R must be selectedto satisfy
One
set of parameters
that satisfies the requirements is
\342\200\224 =
1 henry,
1 farad,
and 5 = 1 ohm. There are
C\\
C%
these
are selected for their simplicity in illustrating
values;
behavior
of the lattice
network. With this choice of
the
\nparameters,
z. =
L2, C\\,
Lh
parameters
11-112.
\nEq.
265\n
\302\243
Now
ANALYSIS
SINUSOIDAL
become\n
functions
impedance
Zi. =
s^r-[;
Z.Z,
transfer function is given
\nassignedparameters, this transfer function
The voltage ratio
=
R*
(11-116)\n
With
11-113.
Eq.
by
=
1
the
becomes\n
72(s)
_
Vi(s)
F2(s)
or\n
s2
=
S2
Fl(s)
The
Si,
and the
poles have the
1
+
+
1)\n
s/(*2 +
1)\n
s/(s2
(11-117)\n
(s -
Si)(s
(s
Sa)(s
\342\200\224
Si*)\n
(11-118)\n
-
\342\200\224
8a*)\n
transfer function have the
8!*
*
=
I
+
1
2
.\n
1
\302\261
J
values\n
(11-119)\n
-~2~\n
values\n
Sa,
The pole-zero configuration
The poles and
\nFig. 11-30.
1
\342\200\224
- 8+ 1
+ S + 1
of the voltage
zeros
two
1
*
=
s\342\200\236*
-
1
2
.
-\\/3\n
\302\261
J
(11-120)\n
~2~\n
is
for the network of Fig. 11-29
about
zeros are located on a unit
in
shown
circle
the\n
farad\n
jo>\n
\"V1\n
/A\n
/\n
/\n
N\n
\\\n
\\\n
t\n
1\n
l\n
\\\n
1\n
/\n
\\\n
\\\n
/\n
/\n
/\n
Fig.
11-29.
Symmetrical
values
\nelement
of
origin
\nthe
axis
lattice
assigned.\n
with
Pig.
11-30.
1 \t\n\\\n
\\\n
\342\200\242\n
b*\n
Pole-zero
s plane. They are symmetrically located with
reals and axis of imaginaries. The two
two
the
of
configura\302\254
\ntion.\n
polesand
respectto
zeros
so\n
\nare
selected
\nand
zeros
are known as a quad. If otherparametervalues
the requirements
of Eq. 11-112, the poles
satisfying
still be located
with respect to the axisof
symmetrically
The
zeros
will always be located in the right half plane
will
in the left half plane, and both polesand zeros
occur\n
either
in conjugate
pairs or on the
will
the
poles
axis.\n
\nreal
the
\nfinding
response\342\200\224the
frequency
the
of
\nvariation
angle
\nwith
frequency.
and
magnitude
function
transfer
of the
\nphase
problem of
to the
come
now
We
in
outlined
As
\nprevious sections, the frequency
different
\nto
\nfrequencybehavior of the
Vi{joy)
function
{joy
_
the
from
\342\200\224
is
Si*)
\n{joy
last
\nthe
V2
of
magnitude
always
we
equation
\342\200\224
\342\200\224
\342\200\224
\342\200\224
is
angle
and
For
V\\.
sa)
Si*)
{joy
(11-121)\n
sa*)\n
the magnitude
we see that
figure,
the
to
\nequal
11-31
positive
examine the
us
let
relating
Si)(j(o
{joy
\nVi(ja)
But
respect
First
transfer
phase
to the
A
11-118 becomes\n
Eq.
joy,
with
axis.
\na
Each
joyi.
\nfound
\nputation.\n
\ns
s
\nfor
com\302\254
response
Frequency
in Fig.
is shown
graph
\342\200\224
joy axis.
the
on
re\302\254
phasors
drawing
by
points
\ntypical
Fig. 11-31.
found
is
\nsponse
=
11\n
still
\nimaginaries.
\nand
Chap.
s plane
the
in
arranged
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
266\n
\342\200\224
of
\342\200\224
equal
to the
have
discovered
magnitude
v*U*)\n
of
Si)
the
likewise,
sa);
{joy
{joy
of
magnitude
\342\200\224
s\302\253*).
{joy
is always
In
of
terms
that\n
=
1\n
(11-122)\n
Vi{joy)\n
other
In
\nthis
network,
of
\nnitude
the
four
\ninductors,
There
\ntion
as
\ngiven
as\n
resistor,
and yet it has the same
with purely resistive net\302\254
must
be
else of interest
this
far.
from
something
Let us examine
in this network
the phase
after
the
of
transfer
func\302\254
of frequency.\n
In computing the
and
one
associated
a function
\npositive
and
characteristic
come
have
\nwe
capacitors,
invariant
\nfrequency
\nworks.
we have arrived at the remarkableconclusion
for
that
of the output is always equalto the mag\302\254
the magnitude
Our network
is made up of four
any
frequency.
input\342\200\224for
words,
phase,
pole
terms
we
regard
the
as negative.
phase
When
from
<0
=
zero terms as
0 the phase is
6
=
STEADY-STATE
SINUSOIDAL
11-6\n
Art.
-
0i* ~ 0.
+
=
<t>a*
+
240\302\260
ANALYSIS\n
-
120\302\260
267\n
-
300\302\260
60\302\260 =
0\302\260\n
(11-123)\n
There
\nof
is no
phase
11-29
that
Fig.
inductor
the
\nwith
the
As
\ntogether.)
\nbecomes
frequency
The
Tig.
and
phase
circuits.
\nfound
zeros
\ntransfer
functions,
neither
\nfunction;
\nfor
Fig.\n
network
in\n
symmetrical\n
network.\n
that for the
first time
we
have
last
in the right half plane. As discussedin the
in the right half planefor (output/input)
are permitted
but
are not. This is true only for the transfer
poles
nor zeros are permitted in the right half plane
poles
located
zeros
\nchapter,
of a
characteristics
incidentally,
Note,
are shown in
as a phase-shifting
magnitude
lattice
telephone
becomes
frequency
characteristics
application
and
Phase
11-32.
the
as
\342\200\224360\302\260
magnitude
finds
network
This
11-32.
directly connected
two-terminal-pairs
the phase of V/V
increases,
approaching
negative,
\ninfinite.
the
thus
and
circuit,
\nopen
the
shift at zero frequency. (It is seenfrom
network
and output are identical at zerofrequency,
the input
as a short circuit, the capacitor acting as an
acting
driving-point
immittances.\n
READING\n
FURTHER
in
state
reading on analysis in the sinusoidal
Network
\nterms of poles and zeros, see LePage and Seely,
Analy\302\254
Book
(McGraw-Hill
Co., Inc., New York, 1952), pp. 8-12, 193-196;
Circuit
Theory (John Wiley & Sons, Inc.,
Guillemin,
Introductory
6. See also D. F. Tuttle, Jr.,
York,
1953),
Chap.
2 vols.
\nthesist,
(John
Wiley & Sons, Inc., New York, in
For further
steady
General
\nsis
\nand
Network
\nNew
Syn\302\254
preparation).\n
PROBLEMS\n
11-1.
By manipulating
unit phasors as
sin2
(That
is,
start
\nphasors to prove
with
the
<at
+
phasors
cos2
eiut
tat
in Art. 11-1,
show
=
1\n
e~iat
and
the identity givenabove
by
that\n
and
a graphical
manipulate
these
construction.)\n
SINUSOIDAL STEADY-STATEANALYSIS
268
the
Find
11-2.
+
Ri(t)
a sinusoidal
for
the
Find
11-8.
^
i(t) dt
j
steady-state
^
a
sinusoidal
j
i(t) dt =
equation\n
v(t)\n
have
v(t)
the
exponential
Ve^K\n
\nform
of
\ncomponent
network
the
For
11-4.
when
i(t)
a function
as
of
<o
coordinates
rectangular
\n(b)
Prob.
For the network
11-5.
(ju>)
shown in the figure, find the
v(t) = V sin cat, by using Eq. 11-14.\n
11-4.
Prob.
\nVi
v(t)\n
v(t), by letting
force
driving
\342\200\224
form\n
L
for
equation\n
v(t) have the
solution of the differential
by letting
v(t),
of the differential
solution
steady-state
Chap.11\n
shown in the
for
(a)
coordinates
polar
M vs w
sketch
figure,
and
<f>
\nindicatethe low- and high-frequency
Prob.
11-5 for the network
11-6.
Repeat
vs
co.
steady-state
11-5.\n
=
G(j<a)
M(ca)
On the
and
V2(ja)/
0(\302\253),
plots,
and
clearly
asymptotes.\n
Prob.
\nas
(a)
11-7.
Repeat
11-8.
Show
Fig.
11-9
11-9.
For
\tthe
\ndinates
as
Prob.
11-6.
Prob. 11-6 for
the network
shown.\n
the
impedance
admittance
in
11-7.\n
of Eq. 11-48given
phasor locus representation
=
is a semicircle
centered at Re G(jta)
0.5.\n
the one-terminal-pair
network
shown in the figure,sketch:\n
that
driving-point
\ndriving-point
shown.\n
Fig.
11-13.
Z(jta)
as a function
Y(ju) as a function
The sketches should
of
o>,
of
using
a>,
and
the
(b)
coor\302\254
polar
have one point
located\n
11\n
Chap.
low- and high-frequency asymptotes clearly
the
and
accurately
269\n
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
\nindicated.\n
Wv\n
\342\226\240AAAr\n
1
R\n
and
Z{ju)
Yiju)\n
and
\n10\n
=t=lf\n
Yiju)\n
11-10.\n
Prob.
11-9.\n
Prob.
\342\226\240AAA/\342\200\224i
Q\n
network
Repeat Prob. 11-9 for the one-terminal-pair
11-9 for the oneProb.
Repeat
11-10.
11-11.
\nterminal-pairnetwork
shown.\n
shown.\n
1
pole\n
+J4\n
L\n
and
Z{ju)
Yl/w)\n
+>3\n
11-11.\n
Prob.
+J2\n
11-12. The
the
in
\nshown
in
\nshown
determine:
\nconfiguration,
\nnatural
\nf,
frequency
\nthe
the
(f)
\nsponse,
\ntransient
of
of
L
if the
values
of
\nnitude
the
admittance
frequency
(to
cannot
of
frequency,
u\n
-l
-2\n
fre\302\254
re\302\254
the
of
-;2\n
otherwise be
Sketch
| Y (ju)
If the
the
| as a
mag\302\254
func\302\254
\342\200\234>3\n
frequency
\nscale
is
of 1000, how
magnified
\ndo
the
values
of the parameters,
R, L, and
\nC
Answers,
change?
4.04,
(b) 0.124,\n
(a)
(c) \t4.04,
(d)
1.0,
(e) 4.0, (f) 0.5, (g) 4.04,
=
=
R
C
\n(h)
L,
0.061/L.\n
\ntion
1\n
1 zero\n
parameter values (in
(i)
determined),
\nuniquely
transient
factor
of
the
(g)
\nresonance, (h) the
\nterms
the
damping
response,
actual
the
(e)
oscillation
of
\nquency
the bandwidth
points),
half-power
undamped
ratio
damping
the
Q, (d)
+j
the
(a)
w\302\273,(b)
circuit
the
(c)
From
11-19.
admit\302\254
circuit
the pole-zero
the
for
Fig.
the
RLC
represents
series
figure
function
\ntance
configuration
pole-zero
(j)
by a factor
1 pole\n
g \t\n
-;4\n
-;5\n
Prob.
11-19.\n
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
270\n
Chap.11\n
is
The passive network shown in the accompanying
figure
net\302\254
as a double-tuned
circuit. It consists of two parallelRLC
with
a capacitor
Cc. For a certain combination of\n
coupled
11-13.
\nknown
\nworks
A
,c
\t\n
-0.5+;
2.0
x\n
-0.5+7
1.5
x\n
(Scale
factor
3 zerosv\n
^\n
^\n
h
1
-0.5-7
1.5
-O.5-7
(61\n
Prob.
the
\nand
(b)
\npole-zero
\ntransfer
11-14.
\ngiven
for
the
transfer
impedance
The
network,
in the
=
Vi(s)/Ii(s).
impedance, Z2i(s)
(accurately
configuration
as
frequency
Draw
11-13.\n
is as shown
configuration
pole-zero
*\n
x\n
-0.5->2
(0)\n
parameters,
c\n
^
*\n
x\n
-0.5->2
j(jJ\n
1 x\n
3 zeros
*
x\n
-1)\n
-0.5+7
factor-1)\n
(Scale
2
-0.5+7
jo)
a function
response
a pole-zero
plotted),
of frequency
as
figure
From
(a)
the
plot the magnitude of the
to.\n
in the figure is observed
a
for
that can give this\n
configuration
shown
is no unique
(Note: there
must be such
response.
\nsolution
as to meet
and
\nhigh frequencies,
a
the
since
\nhave,
\nlem
can
how
\nstate,
oscillate
will
\nbox
=
\nG(s)
the
Draw
11-18.
-f-
l/(s2
varies
an RLC series
phasor
1/L, where L is
locus corresponding
the
Draw
locus
phasor
the
identify
Carefully
1).
\n+
your
as vacuum
tube
range.
the
product
of
|
inductance.\n
to the transfer function,
the high- and low-frequency
identify
Carefully
a).
the
network
\nasymptotes.\n
11-19.
In
you would perform.\n
with the circuit
inversely
experiment
B
bandwidth
B equals
bandwidth
the
\nand
the
Show that for
11-17.
will oscillate
circuit.\n
RLC
series
a
for
compo\302\254
frequency
generators\342\200\224any
the
Describe
11-16. Show that
\nQ
no
fre\302\254
oscillation.
of
\nquency
but
Circuit\342\200\235
prob\302\254
is
face
you
frequencies,
cathode ray oscillograph. You are
detect oscillation with the instruments you
you could
of oscillation
may be very high. The
frequency
made in the sinusoidalsteady
this: with measurements
the
determine
whether
current through the black
you
is closed and what will be the
when
the switch
that
certain
every
no adequate
have
you
\nHowever,
\nnot
wave
sine
ammeters,
\nvoltmeters,
Series
RLC
test equipment such
standard
have
you
\nlaboratory,
low
are seeking a circuit that
box by the closingof a switch.
You
to the
is connected
battery
but
problem,
the requirements
at
is marked u
indicated.
are
values
\nnent
solution to the
resonance.)\n
11-16. A black box
\nif
271\n
ANALYSIS\n
STEADY-STATE
SINUSOIDAL
11\n
Chap,
for the function
G(s) = l/s(s2 +
and
high-frequency
as
low-frequency
\nasymptotes.\n
zero shown in the s planeof the accom\302\254
function of a two-terminal-pair
sketch
are
for the transfer
=
zero
is on the
The
G(s)
F2(s)/Fi(s).
at a position
to correspond with the
The poles have
of
the
part
poles.
=
to f
0.707(0 = 45\302\260);
corresponding
\npanying
\nnetwork,
axis
\nreal
real
\nsame
\npositions
the
of
\nthe
modified
chapter
as
=
0
to
the
of
\nbandwidth
by computations
(a) The bandwidth
zero,
is
\nsystem
the origin to the pole as
we will investigate
the
problem,
finite
zero
the
without
\nand
\n<o
this
In
\nshown.
and
poles
from
distance
the
is
\nw\342\200\236
\neffect
two
The
11-20.
the
point.
system
\nconfigurationshown above;
with
\nwidth
\nand
by
what
the
zero
factor?
from
removed.
Compute
the
with
in
given
of frequencies
range
halfpower
the
definition
the
from
with
of the
the
pole-zero
computethe
band\302\254
In which case
A graphical
construction
is the bandwidthgreater
is suggested, (b)
We\n
the
define
per
STEADY-STATE
cent overshoot
in response to
\342\200\224
value
\nmaximum
the per cent
is
case
\nwhich
\nof
\ntively
G(*)-V2(*)/Vi(\302\253)\n
\n<r
axis
\nfrom
x\n
-1+/2
\nto
(1)
Test\n
-4
-5
zero\n
*\n
\"
9\n
the
For
11-21.
\nbandwidth
\na
function
\n<r
=
zero
Show
0.
in
defined
(as
of
pole-zero
configuration
figure, compute a
\nshown in the
any
11-21.\n
Prob.
In
\nbandwidth.\n
I
-1-/2x\n
in (a).
described
cases
two
By what factor? Check point:\n
the zero, (c) Discuss qualita\302\254
without
of another
the
effect
pole on the real
ten
with
a
times further
but
position
than
the zero with respect
the
origin
and (2) the
transient
the
response
%
jta\n
zeros
and
100%\n
greater?
4.3
Poles
value\n
final
overshoot for the
overshoot
the
Chap. 11\n
a step functioninputas
value\n
final
Compute
ANALYSIS
SINUSOIDAL
272
position
changes
of
curve
Prob. 11-20) as
from a = \342\200\2244
to
in
curvature
\ncarefully.\n
11-22.
\nthe
\nof
the
Consider
\342\200\234test zero.\342\200\235
a zero
at
<r
=
To
of Prob. 11-21 without
configuration
is added
a so-called \342\200\234dipole\342\200\235
configuration
pole-zero
the
\342\200\2240.1 and
a pole
at
=
<r
\342\200\2240.105.
\342\226\240AAAr
\nR\n
Yl\302\253)\n
Prob.
11-28.\n
Show
that
this
dipole\n
not
does
\nto
a
\nof
network
relating
\nthe
to
respect
Plot
driving-point
0
(with
to
RL
series
the
current
values)
of the
admittance
10
the
waveform
coordinate
RL circuit.\n
30
20
Frequency
Prob.
response
to change certaincharacteristics
bandwidth
is described in the
as lag or integral compensation.)\n
of the voltage waveform
phase
is measured and plotted in the
the pole-zero configuration for
of a dipole
circuit,
273\n
ANALYSIS\n
or the transient
bandwidth
without
changing
servomechanisms
response
11-23.For a
\nfigure.
use
(The
input.
step
the
affect
appreciably
\nliterature
\nwith
STEADY-STATE
SINUSOIDAL
Ckp.11\n
40
50
60\n
in cycles/sec.\n
11-24.\n
11-24. The curve of the accompanying
the
current
figure
represents
as
a
with
constant
for
function
of
\nmagnitude
inputvoltage
frequency
\nan RLC
series
From this plot, determine the locationsof the
circuit.
under
\npoles and zeros in the s plane for the network
study.\n
CHAPTER
12\n
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
12-1. Reactive networks\n
The
\nways.
\nand
\nments,
\nnetworks
in this chapter will be restrictedin two
will be assumed to be made up of inductances
The
networks
(1)
capacitances
only. Since these networks contain no resistiveele\302\254
are said to be dissipationless.
(2) Only one-terminal-pair
they
The appropriate
will
be considered.
network function for the
be studied
to
networks
network is
\none-terminal-pair
\nimpedance
the
immittance
driving-point
V(s)\n
=\n
Z(s)
=\n
Y(s)
the voltage and
where F(s) is
terminals.\n
respectively,
\ndriving-point
\nadmittance
for
expressions
and
inductance
are found by
immittances
Driving-point
m\n
7(s) is
of
capacitance
Admittance
Ls
of
\ncombination
=
Zt(s)
Zt(s)
= (Li +
=
In this
\nulated
be broken into parts con\302\254
of elements. For a series
and capacitances, the total
can
combinations
of inductances
is\n
\nimpedance
or
Cs\n
network
number
any
l/Ls\n
1/Cs
complicated
and
parallel
series
or
impedance
in the network. These expressions
in the following table.\n
are summarized
Capacitance
\nsisting
the
elements
\nInductance
arbitrarily
at
current
the
combining
Impedance
Any
(12-1)\n
V(s)\n
I(s)\342\200\231\n
\nalent
admit\302\254
are\n
\ntance
\nfor
(either
The driving-point impedance and
or admittance).
algebraically
Li -j-
L%
-f-
...)\302\253
+
\342\200\242..
+
+
(12-2)\n
Zn(s)
~
\342\200\224h
^
Leq is the
of
(12-3)\n
(12-4)\n
+
L\342\200\236s
expression
capacitance
+ Zt(s)
Zi(s)
series
the
to
equivalent inductanceand
the
is
the
12-4 may be
equiv\302\254
manip\302\254
form\n
S2
Zt(s)
Equation
system.
Ceq
\342\200\224
+
1 /LeqCeq
Leq\n
\n8\n
8\n
274\n
(12-5)\n
and
inductances
of
\nnumber
-
Yt(s)
or
=
(Ci
=
Ceqs
F)(s)
of a parallel
admittance
total
the
Similarly,
capacitances
+
Y1(s)
275\n
combination of any
is\n
F,(\302\253)
+ Yn(s)
...
+
...)s +
+ C3 +
4- C2
NETWORKS
REACTIVE
ONE-TERMINAL-PAIR
12-1
Art.
(12-6)\n
\342\200\224
..
\342\200\224|- \342\200\224f-
(12-7)\n
+ -=J-
(12-8)\n
LieqS\n
where
be
may
\nequation
elements
those
from
different
symbolized
in
rearranged
system.
Yt and
Zt
for
\ntions
Ceq
+ 1/LegCag
(12-9)\n
s\n
only difference in
The
a parallel
thus\n
**
=
Yt(s)
for
capacitance and inductanceof the
(hence this set of equivalent values
in Eq. 12-4). The last
identically
a form similar to Eq. 12-5;
the equivalent
of
combination
\nparallel
\nis
Leq are
and
Ceq
two
equa\302\254
is the multiplying constant.\n
o
o\342\200\224
L2
L1
the form of the
1(
1(
Cl
C2
\342\200\224
Leq
\342\200\224|(-o\n
Ceq\n
la)\n
0
l\n
:
ill\n
l2
i
Ci
Ln
=\n
*\342\200\224J
Leg
0\n
O\n
Ceq\n
1\n
?\n
\302\243\n
.
l\n
\342\200\224
\342\226\240
si\n
lb)\n
12-1.
Fig.
function
immittance
Equivalent
representations.\n
The driving-point immittance for a complex
network
found
is
by
and the reciprocals of admittances, or admittances
\nadding impedances
\nand
the
of impedances.
Since all networks can be arbi\302\254
reciprocals
and parallel networks, the
into
a
divided
number
of
series
\ntrarily
of terms
for
immittance
will be a combination
driving-point
\nexpression
of Eqs. 12-5 and 12-9, and
\nof the
form
\ntheir
\nwill
Several
reciprocals.
such
illustrate
Example
1\n
The
network
\nis
seen
\nLC
network
\nLC
network.
to
in
shown
made
be
in
The
examples
combinations.\n
series
Fig.
12-2
of a series
with a parallel
up
impedance
driving-point
Zae
=
Zab
+
is given
y\342\200\224\n
1 be\n
as\n
(12-10)\n
REACTIVE
ONE-TERMINAL-PAIR
276
=
Zac
This
+
*2
=
Zac
L\\S -f-
+\n
t=\342\200\224-
U
(12-11)\n
C%8
C\\S
\342\200\234f\"
I/L2S\n
in the form
be rearranged
may
equation
+ (1/LxCx +
s4
r
_
V
1
\nac
of Eqs. 12-5and12-9
as\n
s\n
l/L\302\261Cl
(12-12)\n
+\n
+
C2(s2
or\n
12\n
Chap.
element immittances,\n
of individual
terms
in
or,
NETWORKS\n
1/L2C2)\n
1/L2C2 + 1/LiC2)s2+
+
S8
1/UUCrC,
s/L2C2\n
(12-13)\n
Example
2\n
values
\nelement
2\n
1 farad
2
.
7^.\n
henry\n
\nseries and
1 farad
\342\200\242\n
\nThe
2*\n
\nb
found
parallel
from
impedance
combinations.
1S-S.
Tig.
The
of
\ntance
the
1-farad
=
Y& can be
impedance
capacitor;
(12-15)\n
3s
+
(12-16)\n
+
\n2(sl
impedance
admit\302\254
2<\302\273* +1).\n
2s8
\nthe
combinedwith the
*+
=\n
F22'(s)
driving-point
(12-14)\n
s\n
thus\n
K-+h.=
or\n
node
^
^
s
=
The
+ - =
2s
network.\n
of this
reciprocal
a to
node
is\n
Za6(s)
LC ladder
this\n
several
into
-o-\n
l'o-\n
specific
for
by grouping
elements
network
\nthe
with
admittance
be
will
network
lo\342\200\224nflflT'-\n
1 henry
network
in Fig. 12-3 is a ladder
The
designated.
driving-point
shown
network
The
1)\n
1-1' is found
at terminals
impedance
of the 1-henry
inductance
by
combining
with the reciprocal of
Ytr)
\nthus\n
7
Zn'(s)
-
f\302\273\\
_
.
7
+ J_
Zla
ytr
=\n
The driving-point
\nby
the
admittance
(12-17)\n
3s\n
+ 2
(12-18)\n
is the reciprocalof
Zw
and
so
is given
expression\n
Yn<>)
or\n
1)\n
3s\n
+
\n2s8
+
2s8 +
5s*
+
2s4
Zn'(s)
Simplifying,\n
2(^
, +\342\226\240
Fn'(s)
=
=\n
\ns4
5s* +
+
2s4
2\n
(12-19)\n
1.5s
s8
+
+
2.5s*
+
(12-20)\n
1\n
A
a number
show
will
\nimmittance
the
equations\342\200\224Eqs.
\nintermediate
steps\342\200\224are
All
\nconstant
NETWORKS
derived expressions for
of common features:\n
various
the
of
comparison
(1)
REACTIVE
ONE-TERMINAL-PAIR
1J-1
Art.
12-9,
12-5,
quotients
The
\nin
as well as
a
with
\302\253
multiplier.\n
one
never
polynomials
even powers of s or odd powers
of
s
if the numerator polynomial
Further,
have
polynomials
any
driving-point
12-13, and 12-20,
of polynomials
in
(2) The order of the numeratorand denominator
\ndiffers by more than unity.\n
(3)
277\n
only
polynomial.
the denominator polynomial always
\nhas
even
and vice versa.\n
only
powers
In
a polynomial
with even powers of a, no even term of degree
\nless than
the term of maximum degree can be missing.
The
\nsame
condition
holds
for odd terms in odd polynomials.\n
\nhas
(4)
only
odd
powers
of a,
of a,
whatever the
is by definition
an even polynomial. Similarly, a
may
be)
with
odd powers
of a is by definition an oddpolynomial.
only
in different words as: the
statement
be
(3)
may
expressed
immittance
functions
are all odd to even or even to odd
a
Now
\niable
with
polynomial
even
only
powers
of a (or
var\302\254
poly\302\254
\nnomial
\nHence
\ndriving-point
of
\nquotients
polynomials.\n
illustrate
To
odd polynomials,
of even and
the concept
further
the
\nequation\n
=
Px(a)
even
an
is
the
\nilarly,
a8a8
since
polynomial,
odd
an
a\302\253a# +
cua4
+
+ 02a2
it contains only
aoa\302\260
(12-21)\n
even powers
a.
of
Sim\302\254
equation\n
P2(s)
is
+
= a7a7 + a6a6 + a3a3 +
containing
polynomial
01a
only odd powers
(12-22)\n
of a. The
equation\n
aia +
both
and odd powers of a and so is neitheran
contains
even
nor
an
and
an
both
even
odd
but
has
polynomial,
a
four
The
statements
only
just given are made on the basis
of examples. However, these statements are
\nlimited number
for LC networks in advanced textbooks.*
true
in general
be
statements
(2) are true for the driving-pointimmittance
(1) and
or the general RLC. Only statements (3)
network\342\200\224RL, RC,
any
in the case of the dissipationless LC
only
apply
P3(s)
=
ats*
+ a3a* + a2a2 +
a0
(12-23)\n
even
odd
\nan
part.\n
of
shown
\nto
Fur\302\254
\nther,
\nof
\nand
network.\n
(4)
\342\231\246See
Guillemin,
\nNew
\nSons,
York,
Inc.,
1935),
New
Communications
Network,
or Tuttle,
pp.
in
preparation).\n
York,
184 ff.,
Network
Vol. II
Synthesis,
(John
Wiley
2 vols.
A
Sons,
(John
Inc.,
Wiley
A
278\n
\npoint
\neven
to
\norder
of
\nfor
the
than
admittance)
+ a2n-2S2n~2
at nS2n
=\n
\nto
even
\nbe
an
...
+
+
qp\n
(12-24)\n
blS\n
= 0 such that
even to odd polynomial, and with
ao
factored
out of both numerator and denominator,an odd
be
The
polynomial.
in s2 which
equation
be
\nequation
can
\nequation
becomes\n
7(n\\
W
numerator
into its n roots.
the denominator of the last
1
roots
+
*32).
S22)(S2 +
+
=
H
where\n
\342\200\224
\302\256l2)(s2
\342\200\224
ff
S(S2
the
After
the
equation,
into n
factored
may be consideredto
polynomial
may be factored
from
factored
s is
\ncommon
+
c^s2
\342\200\242
\342\200\242
\342\200\242
+
of an
a quotient
s may
+
62n-3S2n_3
+
&2n-lS2n_1
\nan
of
polynomials
Z(s)
as
driving-
are quotients
networks
(LC)
dissipationless
12\n
or odd to even polynomials. Further, the
and denominator
numerator
polynomials will never differ
hold
Such a general impedance (and the samewill
unity.
in the form\n
can be written
odd
more
\nby
for
immittances
Chap.
will assumethat
above discussion, we
of the
basis
the
On
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
,a2n
s2.
. .
(S2 +
+
\342\200\242
\342\226\240
\342\200\242
\302\25342)
(S2
\302\2532n-l2)\n
(12-25)\n
\302\2532n-22)\n
constant
a
>
form, the
In factored
in
(12-26)\n
t>2n-l\n
roots
\ntwo
previously
\nthis
of
\nof
the
the
will
\nequation
=
s =
-s:2;
form\n
the
if Oo
= 0 in
Eq.
(s2
+
S (s2 +
for
becomes\n
o>12)(s2 + o>32)...\n
+ 0)42) . . .\n
G>22)
(s2
(12-29)\n
12-24,\n
=
Hs\n
(s2
+ 0)22)(s2 +
+
\n(\302\2562
reason
(12-28)\n
\302\261jan
expression
impedance
driving-point
Z(s)
\nlater.\n
of
numbers
H
The
(12-27)\n
impedance
Z(s)\n
or,
\302\261js i
purely imaginary.
s =
Then
into
factors
emphasize
identification,
\nterms
which
Si2),
Such imaginary values roots
been
associated
with radian frequency. To
will
we
notation
at this point by letting
change
form
Hence typical roots of the
sn become
in pairs as purely imaginary
have
roots
occurring
are thus
roots
\nhave
(s2+
as\n
s2
The
in Eq. 12-25 is
factors
of the
form
Typical
the particular
0)12)(s2
+
0>42)
. . .
(12-30)\n
0)32)...\n
choice of subscripts
for
to
will
be
discussed
REACTIVE
ONE-TERMINAL-PAIR
12-2\n
Art.
279\n
NETWORKS\n
considerationsto the special
consequence of
in
terms the
point
on, we will restrict our
of
state. The mathematical
the sinusoidal
\ncase
steady
=
restriction
is that
s
\nthis
j<a and that the s2
For
the
two cases of the
become
s2 = \342\200\224
w2.
\nequations
this
From
\ntions,
s =
substitution
the
jta
H
-
(w2
The
are
state, the general
steady
of the
and
complex
or
=
R(a)
=
\nB(o))
driving-point
admittance),
\nterms
of
the
form
is
H
constant
fl(\302\253)
-
account for there being
than the denominator
<?(\302\253)
=
jX
(\302\253)
Since the impedance function
\nspoken of as reactive networks.\n
the
When
reactance
with
differentiated
Pig.
is
always
+
12-4.
positive;
and
(12-33)\n
jX(u)
(12-34)\n
jB(\302\273)
Y(j<*)
function
respect
(12-35)\n
jB(u)
LC
networks
are
networks\n
X(u>) discussed in the previoussection
to radian
frequency w, the resultant func-\n
of positive
Geometry
that
=
is purely reactive,
for reactive
property
Separation
+
= reactance, G(a) = conductance,
to Eqs. 12-31 and 12-32 written for
According
of the same form for the drivingimpedance
(and
is purely
This follows because
Z(jw)
imaginary.
\342\200\224
are
co\342\200\2362)
real, and likewise the multiply\302\254
(<o2
always
always real as well as positive. Thusfor LC networks,\n
Z0\302\253)
12-2.
driving-pointimmittance
X(w)
resistance,
susceptance.
\npoint
tion
(12-32)\n
form\n
=
F(j\302\253)
where
\nis
W42)\n
\342\200\224
\302\253J2)\n
is introduced to
from
the numerator
Z(j\302\273)
\ning
(12-31)\n
\342\200\224
\302\261j\302\273H
\342\200\224
Wi2)(o)2
\302\243\302\243\n
equations
sinusoidal
the
\nfunctions
\nthe
&>22)(tt2
.\n
versa.\n
vice
In
0>82)
\342\200\224
\302\25342).\n
\342\200\224
&>22)(\302\2532
=
removed
(\342\200\2241)factors
\nmore
\nor
these
in
\302\261
sign
(j\302\273)
\342\200\224
\302\253i2)(tt2
-
fa2\n
\nj<0
z
equa\302\254
gives\n
ZU\\n")
or\n
previous
last two
and negative slope.\n
is,\n
dX\n
>
dco\n
0\n
(12-36)\n
\nfrequency,
\nof
the
postpone
\nexpansion
be
\nmust
a
\nmust
increase,
12-5.
At
finally
the
frequency
of
\nvalue
before
a
the
zero
which result in a zerovalue
(of frequency). Frequencies resultingin
are poles (of frequency). The zerofre\302\254
of as resonant frequencies (frequen\302\254
spoken
the pole frequencies
are called
antiresonant
property
derivative
network
always
must
the
and
poles.
by
The
be positive.
again
of reactive networks that the
be positive,
the poles and zeros of the reactive
alternate.
The poles must be separated by zeros
This is referred to as the separation
for
of the
\nfunction
must
reactance).\n
(infinite
Because
\nzeros
and
reactance),
\nfrequencies
\ndX/dco
reactance
sometimes
also
are
of
\ncies
reac\302\254
property
networks.\n
\ntive
The
and
poles
located
be
\ncan
\nthe form
\nco
of
magnitude
\nquencies
zeros
are
reactance
the
slope
of frequency
values
Those
changes.
\ninfinite
\342\200\224
\302\260o,the
frequency.\n
to increase to zero value, thenceincrease
to
infinite
cycle of change is repeated. The reactancefunction
which
from zero value to infinite valueasfre\302\254
varies
magnitude
\nquency
=
X(<a)
function of
is shown
X
\ncurve
\nfor
at
again
as a
Reactance
12-6.
Fig.
Starting
X
coi, then as frequency increases,
frequency
value. This is illustrated in Fig.\n
to an infinite
of infinite reactance, the sign of X changes.\n
some
at
Xx
reactance,
\nhas
Chap. 12\n
proof for this statement until the partialfraction
In terms of a plot of X as function
of
of X(co)
is studied.
the
of the curve must always be positive;that is,
slope
co. If we start with a givenvalue
with
increasing
increasing
will
We
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
280\n
=
zeros of the
inspection
by
(co2
\342\200\224
in
w\342\200\2362)
\nlocatestwo poles at
at
zeros
\nand
\nstudy
The
\nthe
of these
of
\nhavior
zero
elements
reactance
equation\n
co
=
12-5
A
at
term
a
Similarly,
\302\261con.
the
reactance function illustratedin Fig.
of
term
of the reactance function.
of X(co) locates two zeros
numerator
(to2
\302\261
com.
frequency
\342\200\224
There
com2)
remain
and at infinite
lower and upper
in
the
to consider the
only
frequency.
limit frequencies
and
combinations
of an
inductance
of X(a)
denominator
by
We
will
considering
= 0 and
of elements at
varies with frequency as
co
poles
start
the
co =
given
our
be\302\254
<*.\n
by
Art.
with
\ndirectly
w =
at
zero
a
has
XL
\302\253.The
0 and a
for
expression
281\n
= (oL
XL
Thus
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
12-2\n
(12-37)\n
pole at
=
w
reactance
the
\302\273, since
of a capacitance\n
X\n
\nfrequency
so
\nquency
inductance
\nfor
varies
Xc
Xc\n
XL\n
0\n
pole\n
zero\n
00\n
zero\n
pole\n
CO\n
\nthe
terms
\nof
reactance
pole
does
\ncapacitance
inductance
an
\nilarly,
an open circuit at zero frequency(direct
are not in physical contact. Sim\302\254
capacitor
plates
to be an open circuit at very high
appears
to be
appear
the
since
\ncurrent)
zero
\nreasoning,
a
\nreactance
(and
word
\342\200\234choke\342\200\235
is
to
applied
high reactance at high frequency. By dual
means zero value of reactance. Zero
of reactance
since there is no resistance pres\302\254
zero impedance,
of this
because
inductor
The
frequencies.
infinite)
\n(approaching
\nthe
LC networks, we can attach a physical
to
significance
and
zero. A pole of reactance means an infinitevalue
as an open circuit. The
we
which
physically
interpret
case of
the
In
(12-38)\n
at zero
inversely with w, Xc is infinite
that
zero frequency
is a pole and is zero at infinite
so
fre\302\254
that
infinite
defines a zero. These relationships
frequency
and
are summarized
below.\n
capacitance
since
that
showing
varies
XL
thus
is interpreted as a shortcircuit.
considered)
to be a short circuit at zero frequency(direct
\nAn
inductance
appears
= 0 and
no voltage appears across the
because
\ncurrent)
(d/dt)(Li)
the
to be a short circuit at
\ninductance.
capacitance
appears
Likewise,
This
of poles and zeros at
\ninfinite
interpretation
physical
frequency.
=
oo allows
the zero and infinite poles and zerosof a net\302\254
\n<o =
0 and \302\253
\nent
\nwork
networks
the
in
to
be
found
being
Several examples will illustrate this
by inspection.
\nfeature.\n
o-\n
o\342\200\224\342\200\224If
\nL
c\n
Z-+\n
Z\n
o \t\n
L2\n
o-\n
(a)\n
Fig.
12-6.
Figure 12-6(a) shows a
\n(direct
\ndriving-point
current)
the
impedance
for examples.\n
LC networks
simpleseries
LC
for
this
At
zero
frequency
as an open circuit. Hence
has a pole at zero. At infinite\n
network
acts
capacitor
network.
the
in
\nshown
12-6(b).
Fig.
the
at infinity. A more
At zero frequency, C\\
a pole
has
also
\nfunction
\nmaking the behavior of any other
The
Irrelevant.
\nzero
\nand
to
\nterminal
\nof
a
has
\nance
12-6
Fig.
of
can
\nfrequencies
the
place of j<a to simplify notation,the
as Eq. 12-24 may be rewritten as\n
Un$n
_
is even
be
need
\npolynomials
dp
+
... +
bi$\n
(12-39)\n
s = jcoapproachesa
is to say
This
considered.
=
Z(s)
large
very
value,
and denominator
numerator
of the
term
lim
...
6m_2sOT-2
and m odd. As
highest-ordered
~l~
2
dn-2Sn
~f~
-f-
\nbmsm
\nonly
short circuits,
impedance at zero and infinite
the mathematical
form of the
from
given
\\
yr
the
has a pole at
s in
Using
impedance
\ndriving-point
n
thus
impedance
and C3 behave as
driving-point
determined
be
function.
\nimpedance
where
open circuit,
at zero fre\302\254
network
from
open circuits. There is a zero impedance
path
terminal
Ci-Cz-Cz. Then the driving-point imped\302\254
through
at infinite frequency.
zero
We conclude that the network
(b) has a pole at zero and a zeroat infinity.\n
behavior
The
C2,
C1,
<*>,
an
as
acts
the
is
network
L2 as
and
L?
=
\302\253
At
frequency.
driving-point
impedance
complicated
in
elements
\nquency
Chap.12\n
acts as an opencircuit.Thus
the inductance
frequency,
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
282\n
that\n
lim\n
(12-40)\n
bmsm\n
at most by unity and
never
as
equal,
\ndiscussed in the last section. Hence as s approaches
Z(s)
either
zero
or
on
m
is
whether
\napproaches
infinity,
depending
\nthan
than
m.
In
either
worn
larger
case, because n and can differ
the pole or zero at infinity will be
at most,
(not
unity
If the order of the numerator polynomial
is
. In summary:
\ntiple)
\nthan
of the denominator
the
order
polynomial, there will be a
at
If the converse is true, there will be a
at infinity.
zero
\npole
m can differ
and
n
Now
are
infinity,
larger
m
\nby
mul\302\254
simple
greater
simple
simple
\ninfinity.\n
the
For
low-frequency
of
\npolynomials
the
12-39,
the
case, only the
may be ignored
terms
higher-order
=\n
Z(\302\253)
two
The
can
\npolynomials
\nin
this
cases
possible
equation:
be
taken
(1) do
...
-f- d2s2
account
^ 0, and (2)
by
do
=
the
Eq.\n
as\n
-|- do
\n... -f- 6js* -|-
of an even-to-odd
into
be considered. In
need
function
impedance
lowest-ordered terms in
(12-41)\n
bi$\n
or odd-to-evenquotient
considering two possibilities
of
0.
For
case
(1),\n
REACTIVE
ONE-TERMINAL-PAIR
12-2\n
Art.
lim
=
Z(s)
lim
NETWORKS\n
283\n
+ oo
\342\200\224
=
(12-42)\n
\n\342\200\242\342\200\224\302\273o
\302\253->0 \302\253\n
is a
there
and
For case
at zero frequency.
pole
lim
=
Z(s)
(2),\n
= 0
lim Hs
(12-43)\n
8\342\200\224>0\n
at
zero
\nple
term
lowest-ordered
the
\nthan
From this discussion, see that
is of higher order
there will be a
term
converse is true, there will be a
at
pole
zero frequency.
lowest-ordered
the
\nwhen
zero at
is a
there
and
If the
zero.
we
of the numerator
of the denominator,
sim\302\254
simple
\nzero.\n
all
In
networks.
LC
\nfor
\nTwo
Example
\nby
the
these
conclusions.\n
S\n
this
For
zeros
these
Further,
will
illustrate
examples
(frequency) are either polesor
poles and zeros are always simple.
and infinity
zero
cases,
the driving-point impedance is given
that
suppose
example,
expression\n
Z(s)
(s2
=
(12-44)\n
\ns(s2
\ninator
\ninfinity
\nof
\nform
the
4)\n
of the
x at
s-plane\n
at
to =
12-7.
\nand
different
The
s plane
12-7.
=
\302\253
\302\2611 and
and reactance plot of Example3.\n
\302\2613,
and
a finite
pole at
reactance
values
of
o>
=
\302\2612.
The
pole-
in
of this reactancefunction shown
function
X(<a) may be found from Eq. 12-44
and a plot
\nzero configuration
\nFig.
oo\n
ju\n
Fig.
zeros
+
numerator
polynomial is 4, and that of the denom\302\254
is 3.
the rule for infinite frequency stated on page282,
Applying
is a pole,
since
the order of the numerator is greaterthanthat
denominator.
Since
for small values of s, Z(s) approachesthe
zero
is also a pole. By inspection, there are finite\n
H/s,
frequency
order
The
+ 9)
1 )(s2
+
2\n
\302\253
substituted
are
into
this
equation
to make the
plot.\n
Example
this
consider
example,
denominator
the
greater
an impedance
than that of the
-
In this equation,
4.
is
\ndenominator
of
there
that
numerator.
at
zero
finite
that of the
shows that both zeroand
this equation
are finite poles at =
and
go
\nreactance function plot
four
reactance
been
shown
The
12-3.
has
It
\nfrequencies
are
two
possible
\nthe
=
w~3
\nance
two
but
\nare
+
\n(\302\253*
an2)
Ccue
1.
\n8
in
the
With
representation
and
the
12-8.\n
=
CO
0\n
=
1\n
pole\n
pole\n
2\n
zero\n
zero\n
3\n
pole\n
zero\n
4\n
zero\n
pole\n
of finding the
of factors
forms
and
s-plane
forms\n
<o
corresponding
admittance)
(or
4.\n
in the previous section that zero and infinite
either
always
poles or zeros. The four possibilitiesfor
conditions
at the two frequencies are tabulatedbelow.\n
the task
remains
The
to-oo\n
Example
are shownin Fig.
Case\n
There
\342\200\224
\302\2613\n
*\342\200\224\342\200\224<}>\n
\302\2612.
function
go
\302\2611
e
w
Plots of
12-8.
<o
(m3)\n
is 3 and
numerator
the
|
one
Let\n
9)
\302\253\342\200\2341
w\342\200\2340
\302\253-2
Fig.
order
function with the
(* + l)tA
of
Analysis
and
zeros
are
\ninfinity
3
order
the
and
12\n
4\n
For
\nof
Chap.
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
284\n
form of the
to each
00\n
driving-point
of these four
for the numerator
imped\302\254
cases.
There
and the denominator:
8.\n
a pole
denominator
at both
and
one
zero and infinity, there must
in
factor
more (a2 +
type
be
s\302\2732)
an
the\n
ONE-TERMINAL-PAIR
18-3
Art
is\n
impedance
\npoint
H
8
where\n
\302\2731<
Fig. 12-9.
This
#.
it
infinity,
additional
has
impedance
(12-47)\n
\302\253\302\273\n
=
the
same
the
plot
12-47.
the
Case
for
1.\n
zero
at
both
and
and
numerator
For Case 2, the
form\n
+
(a2
&>22)...
He\n
relationship
The
reactance
in
12-9.\n
Reactance
\n(a2
Eq.
<
is the
Z(a)
\nin
\302\253n-l
(12-46)\n
tt\302\273_i2)\n
inverse of Case 1. With a zero
that there be an s term in the
is necessary
in the denominator.
term
(a2 + a\302\2732)type
\ndriving-point
where
\342\200\242
. .
in Fig.
shown
is
function
\nreaction
\nan
<
\302\2533<
a>n2)\n
last finite critical frequencies (poles or zeros)are
of this equation.
The general form of the corresponding
\nnumerator
\nat
+
(a2
+ ttj2) . . . (a2 +
(S2
\302\2732<
a>x\302\273)..,
and
first
Case
+
(a2
Z(8)\n
The
285\n
The general form for thindriving-
denominator.
in the
than
numerator
NETWORKS
REACTIVE
wi2)...
+
+ w\302\273-i2)
(a2 + \302\253\342\200\2362)\n
(a2
this
of
exists for the to\342\200\231s
function plot for Case
(12-48)\n
as given
equation
2 is
in
shown
Fig.\n
12-10.\n
Case S.
\nfor
the
To have a
impedance
\norder of
is already
(a2
+
a\342\200\2362)
type
an a in
terms
in
H/s
multiplier
zero at infinity, total
be greater than that the numerator.
For there
must
an
requires
frequency
the denominator
\nSince there
\nmany
pole at zero
function.
to be a
the
of
the denominator,
there
the
numerator
must
as in the
be
denominator.\n
just
as
286
7(K)
H
s
+
(s2
S
(s2
. . (s2 + C0n-12)\n
+ o>22)...(s2 + g>\342\200\2362)\n
fa?l2) .
Fig.12-11.
Reactance
Case
\n(s2
\nan
(Hs)
for
\nance
4
=
Cs2
Hs\n
\n(s2
Case 4 reactance
From
\nCase 3
that
COi2)
for
^
>
.\n
positive
+
imped\302\254
0)n2)\n
(12-50)\n
Wn-12)\n
12-12.\n
it is seen that
in
the
for
function
_
sinusoidal
Case
1 and
(\302\2532-mi2)(\302\2532-<032)...\n
+H
0)
=
0>22)(w2
4, the form
(to2
i
case the
slope
Ho)\n
dX/du
0>22)
\342\200\242
\342\200\242\n
is\n
(t>)2
\342\200\224
\302\253l2) (ft)2
\342\200\224
0>42)
\342\200\224
Wa2)
\342\200\242
\342\200\242
\342\200\242
(12-52)\n
\342\200\242
\342\200\242\n
\342\200\242
to give
reactance function mustbe
(see Prob. 12-7). In order to havethis
sign of the
to
\342\200\224
\302\25342).
\342\200\224
(w2
\n(tt2
\na
(S2
form of the reactance
2 and Case
Case
X(\302\253)
each
+
. (s2
\342\200\224
In
addition,
is\n
y,
and
+
discussed,
just
(s = ju) the
state
\nsteady
+ W)
function is plotted in Fig.
cases
four
the
of
number
is\n
Z(s)
The
3.\n
The driving-point
numerator.
in the
factor
multiplying
Case
Case
must also be
there
case,
in
sn2) factors
+
for
an equal
the numerator and denominatorwith, in
this
For
4-
plot
(12-49)\n
Fig. 12-11.\n
is plotted in
function
3 reactance
Case
12\n
Chap.
impedance becomes\n
The driving-point
The
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
selected
posi\302\254\n
sign
of X
Consider
the
The
\ninfinity.
zero.
a
X at
the value of
tive slope,
\nor
REACTIVE
ONE-TERMINAL-PAIR
12-4\n
Art.
=
<a
each time the
changes
287\n
be zero or
either
must
0
NETWORKS\n
negative
frequency passesa pole
factor\n
-
the
alternate
\nmust
(the
the
pole
\nfunction
zero
and
above,
given
to
changing successively
negative,
In
of the reactance
the
two
forms
frequencies.
the pole and zero frequencies must satisfy the
from
alternates
\nfunction
\nat
of the
sign
\n6)\342\200\236,
the sign of the factor is negative.When
<a exceeds
factor becomes positive. Since the polesand zeros
the sign of the reactance
separation
property),
than
a less
For
(12-53)\n
\302\253n2)
(\302\2532
positive
\ncondition\n
<
0
\nis
the
function.\n
susceptance
H which appears
The factor
\nDoubling
for
function
\nance
<
...
(12-54)\n
in all the
as the
reactance
multiplying
is
equations
or scale factor.
a
The
posi\302\254
func\302\254
reactance.
of the reactance is to fix the scale
of H, for example, doubles the values of the react\302\254
Thus H fixes the levelof
all values
of frequency.
value
the
<04
of the
in terms
H
of
\ntion
known
constant
real
\ntive
<
section for the reactance functionX(u)
admittance
case, where Y(ja>) = jB(co),and B(w)
the
to
directly
\napply
<03
in this
made
statements
The
0)2 <
COi <
\nimpedance.\n
12-4.
for reactance
Specifications
functions\n
and number quantities
\nwhich
known to completely specify the impedancefunction
an LC network.
The term critical frequency will be definedto
From the last section, know that
a pole
or a zero frequency.
\nzero frequency
and infinite frequency are always critical frequencies.
and
zeros
at zero and infinity (frequency) are defined as
we will
this
In
section,
must
be
discuss the nature
of
mean
\nfor
\neither
we
\nThese
poles
critical
\nexternal
\ncies
are
Poles
frequencies.
defined
as
critical
frequencies
\nlar pole-zero
\ntified
\nthe
\nas
in
poles
frequency
for a
Fig.
and
12-13.
zeros
increases
at finite, nonzero
frequen\302\254
Internal\n
exter\302\254
particu\302\254
are
configuration
zeros
critical
internal
\nfrequencies. The internal and
\nnal
and
iden\302\254
We know that
alternate
of the
because
must
Fig.
12-13.
Designations
for critical
\nfrequencies.\n
critical frequencies\n
\nseparationproperty for LC networks, If the internal
as
or
there
\nare
remains
no
choice
for the external
zeros,
specified
poles
must
be the opposite of the nearest finite(or
\ncritical
They
frequencies.
in order
critical
that the poles and zeros alternate\n
\ninternal)
frequency
as
In
required.
is
\ncies
\nThe
critical
internal
of
specification
Chap.
12\n
If the nature of the internal criticalfrequen\302\254
of the external critical frequenciesis fixed.
summary:
the nature
specified,
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
288\n
specifies all critical
frequencies
\nfrequencies.\n
next
We
reactance
the
\nof
remains
\nthere
only
determine
to
\nin order
or
\nfrequency,
\nsome
In
a value
(b)
than
other
frequency
the driving-point
impedance function
of the value of H, an equivalent
spec\302\254
place
value of the reactance at somenoncritical
of the slope of the reactancecurvedX/duat
This information is sum\302\254
a pole frequency.
(a) a
either
be
would
\nification
Eqs.
completely
networks.
reactive
\nfor
forms
12-51 and 12-52, the onlypossible
if all critical frequencies are known,
that
function,
the scale factor H (or the equivalent) to be specified
from
observe,
below.\n
\nmarized
A. The internal
One
B.
\nto be
computed.
value
of
(b)
the
value
of X
(c)
the
value
two
at a noncritical frequency, or\n
at some nonpole frequency.\n
of dX/du
of specification
types
=
\302\253
first
investigated
\nby
\nThe
\nwhich
study
Foster
form of
partial
fraction
considering
Case
Eq.\n
study the problemof designing
net\302\254
the
in
now
classified
are
thus
chapter
far
Tel\302\254
by R. M. Foster, then of the Bell
at Brooklyn
Institute.
Polytechnic
1924
reactive
under
the
heading
of
Foster\342\200\231s
networks\n
functions may bestudied
specializing to the other three
function
for Case 1 is given in Eq. 1246,
expansion
1 and then
impedance
driving-point
of reactance
cases,
is\n
nt
where
by
theorem.\n
\nreactance
The
found
the type studied
but
\nFeaturesof this
12-5.
in
Laboratories
\nephone
are made, Z(s) can be
of the form of
the equations
Z(s) specification.\n
Reactance functions of
\nwere
in
s/j
We will next
12-52.
Eq.
the
to meet
or
\nworks
H,\n
substitution
the
H
are:\n
the
\nmaking
12-51
of information
to give H or to allow
The three most common forms of this second
(a)
these
frequencies.\n
bit
additional
\nspecification
When
critical
Networks\n
Reactive
for
Specifications
there
is one
\\
_
H
(s2
+ o>i2)(a2 + q>s2).\342\231\246.\n
*
(s*
+
more factor
\302\2532*)(s2
+
of the form
W4*) .
(s24-
.
coj2)
.\n
in
the
numerator\n
12*5
Art.
the
in
than
denominator.
the
of
terms
\nequation,
NETWORKS
REACTIVE
ONE-TERMINAL-PAIR
fraction expressionof this
In the partial
type (s2 + \302\253j2) will
as\n
expand
g.\302\253
m
+
(\302\253*
^-coefficients
\nthe
are
roots
\ninary
\nKt* are
k2*\n
+
1
\nCase
in
\ninfinity
Case
=
\342\200\2241
+
Z(s)
\nis
philosophy
to
identify
\nple network
\nin
to
series
,
t2*''*
+ 0)22
+
,2f48
S2
+ 0)42\n,
expansion, Hs, is necessary to give
be
1. The H value of the coefficient
may
or
rule
^Hospital\342\200\231s
=
last
the
\ncapacitor.
Z,(\302\253)
+
Zz(s)
+
the
at
pole
either
verified
division.\n
direct
by
(12-58)\n
total
impedance
... + Zn(s)
will
be
(12-59)\n
design procedure to arrive at a Fosternetwork
a sim\302\254
of
each term in the last equation as the impedance
will then be combined
These
configurations
configuration.
give
a composite
this
equation
philosophy,
with
the
network having the
required driving-point
L
=
same
H henrys.
and
form
The
this
associate
we will
term K0/s in
the impedance Zi
Eq. 12-58. Thus Zi(s)
=
in
K0/s
of value C = l/2Co. Similarlyassociating
Zn(s)
conclusion
that Zn(s) represents an inductorof
All other terms in the impedanceexpression
are
represent
of
impedance
Z^
Comparing
+
Zi(s)
a capacitor
\nrepresents
\nand
Hs
leads
to the
\nof
... +Hs
+
of the
Following
\nvalue
for
12-55
Eq.
terms,
Z(s).\n
\nimpedance
\nthe
0)22\n
o>22)
the
For a series combination of impedances,
that is,\n
\nthe
sum
of the series
impedances;
The
(12-57)\n
+
82
for the (s2 +
82
8
of
application
\nby
JO) 2)\n
in the
term
last
\342\200\224
(s
2)
as\n
expands
Z(\302\253)
The
JO)
of expansion
form
texts that
that\n
(8
this
JO) 2)\n
fraction expansion of terms with imag\302\254
and real. This being the case,K2 and
K2
Using
\342\200\224
in advanced
shown
is
It
f*L2.
(8
JO) 2)
+
(8
positive
always
so
equal,
(s
(12-56)\n
I
\342\200\224
of
conjugate
in the partial
the
is
where
+
(s
\302\25322)
289\n
equation
a parallel combination
such a network is\n
=
Cs
+ l/Ls
with,
=
s2Vl/LC
for example,
of inductorand
(12-60)\n
Eq. 12-57, gives
values\n
290
and inductors in the
for the capacitors
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
12\n
Chap.
as\n
network
1\n
(12-61)\n
Cn\n
2 Kn
2
\n1\n
Kn\n
Ln\n
(12-62)\n
0>n Cn\n
the
\nare
summarized
fraction expansion.These
in the partial
term
nth
for
in
conclusions
12-14.\n
Fig.
values\n
Element
Network\n
Term\n
K0\n
**\n
C0~fara(J\n
8\n
Co\n
Hs\n
henry\n
L0mH
L0\n
C.-
r-Wn\n
2K\342\200\236b\n
farad\n
2Jt_
Ln\n
\302\2532+\302\253i\n
1/\n
11\n
henry\n
Ln-~^~
Cn\n
12-14.
Fig.
for
\nexpression
\nis
of
realization
The
the
\nas
Impedance
Z(s)
network
in Fig.
shown
networks.\n
to the
corresponding
12-15. This realization
first
1.
figure
at
Pole
12-16.
Fig.
Before
cases.
element
each
Pole
origin\n
Network
of the
this
undertaking
in the
in
network
first Foster
study,
other\n
I\342\200\224\342\200\224\n
c\342\200\236
\nof
known
form (series impedances). The network of the
There
remains
the problem of specializing to the
\342\200\224
three
expanded
is
Foster
Case
for
is
the
and equivalent
expressions
at
/\n
infinity\n
type.\n
let us consider the
terms of the known
role
pole-zero
\nconfiguration.\n
Capacitor
(1)
term
\nthe
\nto
\na
Kq/s
a pole
pole
(2)
\nance
at
at the
the
Inductor
expression
in the network because
in the partial fraction expansion. This term corresponds
being
origin. Hence the presence of Codepends there
Co.
Capacitor
Co appears
of
on
origin.\n
La. Inductor L0 appears in the network
fraction expansion.
Hs in the partial
from
the
imped\302\254
This term is
pres-\n
Art.
12-5
ent
because
a pole at
\nthere being
\tParallel
(3)
\ndenominator
each
of
\no>n
LC
\nual
\nthe
networks
gives
function
increases
of a
reactance
Lo
on
depends
in
\302\253\302\2732)
the
parallelLCnetwork.The
Thus antiresonance in the
rise to the poles of Z(s). The
any specific elements. In terms the
of the parallel circuits changessignas
frequency
individ\302\254
of
of
At some frequency,
antiresonance.
through
of parallel
group
type (a2 +
zeros
with
X(<o), each
associated
be
frequency
\nthe
of the
term
Each
of Z(s) gives rise to a
term
is a pole frequency.
parallel
\nreactance
of
291\n
infinity.\n
network.
L\302\273Cn
cannot
\nZ(s)
Hence the presence
pole at infinity.
of the
NETWORKS
REACTIVE
ONE-TERMINAL-PAIR
circuits is equal to and
in
opposite
of all remaining parallel circuits. Underthiscon\302\254
There will be as many
zeros
as there
are
\ndition
is a zero of Z(s).
there
These zeros can be thought of
because
of the
property.
separation
\npoles
first
network
res\302\254
the
caused
\nas
parallel
being in \342\200\234series\342\200\235
by
being
with the rest of the network,
of zero impedance)
\nonance
resonance
(the
the
and second
\nthen
the
first
parallel networks in resonance with
and so on, until finally the last parallelnetwork
network
\nremaining
\nsign to
the reactance
combined
the
with
\nresonates
network.\n
preceding
features of the
Since the distinguishing
cases
four
of
reactive
net\302\254
the poles and zeros at zero and infinity,it follows
4
can be specialized from Case1 simply
\nthat
Cases
by
leaving
2, 3, and
of Co and Lo. For example,if thereis no s term in the
\nout
either
or both
term
in the partial
fraction
\ndenominator of Z(s), there will be no K0/s
are
considered
\nworks
is
polynomial
\nexpansion. Similarly, if the order of the denominator
in
polynomial, there will be no Hs factor
\ngreater than the numerator
and consequently, in terms of the phys\302\254
\nthe partial
fraction
expansion,
in the circuit. The nature of the \342\200\234end
inductor
\nical elements,
no series
summarized
is
below.\n
four
cases
for
the
\nelements\342\200\235
CO
Case\n
Values
Element
End
=
=
CO
0\n
\302\260\302\260\n
present\n
short-circuited\n
pole\n
short-circuited\n
zero\n
3\n
pole\n
4\n
zero\n
\nit
is
possible
found
\nNetworks
\nof
the
second
Consider
\nand
1924 paper, Foster
a given
to realize
infinity.
by
Foster
present\n
pointed out that
present\n
if
the
admittance
corre\302\254
function Y(s) = 1/Z(s) is determined,
impedance
a physical network for the admittance function.
an admittance
expansion are known as networks
form.\n
an admittance
This
Lo\n
Co\n
zero\n
pole\n
zero\n
to
Networks\n
Foster
present\n
short-circuited\n
pole\n
\nsponding
First
short-circuited\n
1\n
2\n
In his
in
function
function of Case 1
a pole
at both zero
has the same form as Eq. 12-55
Y(s)\n
with
with
Chap. 12\n
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
292\n
expansion for F(s) is the sameas
remains
the problem
12-58.
There
of identifying individual terms
\nEq.
\nin
the
fraction
as admittance
expansion
expressions for LC
partial
of parallel net\302\254
The
admittance
\nnetwork
driving-point
configurations.
of the networks in parallel;that is,\n
is the
sum of the admittances
\nworks
The
Z{&).
replacing
=
F(\302\253)
fraction
partial
+
Fx(s)
... +
+ F3(s) +
F2(s)
(12-63)\n
F\342\200\236(s)
Eq. 12-58, we see that the termKo/s
so that Fx(s) = K0/s. ThenFx(s)
to
admittance,
\ncorresponds
of an inductor
of value L = l/K0 henry.
admittance
the
\nrepresents
=
of a capacitor ofH farad
Hs is the admittance
value.
F\342\200\236(s)
\nSimilarly
in the admittance
\nAll other
terms
expression are the sameandrepresent
and capacitor (the dual of the net\302\254
of a series inductor
admittance
\nthe
the first Foster
for
\nwork
found
form). The admittance of the series
this
Comparing
network
\nLC
with
equation
the first
is\n
this
\nrequired
\npole
of
Ls
values
equation
of L and
admittance
as\n
Comparing
1
-\n
Y(s)
+
with
_
s/L\n
s2 + 1/LC\n
l/Cs
(12-64)\n
12-57 permits identification of the
Eq.
network correspondingto the
C for the
nth
1\n
Ln\n
(12-65)\n
2 Kn\n
1
_
Cn\n
2Kn\n
(12-66)\n
co\342\200\2362\n
L\342\200\2360)\342\200\2362
These
summarized
are
conclusions
in Fig. 12-16.\n
Kp\n
L0-
t\n
|f
\342\200\224
henry\n
Ap\n
Lq\n
Ha\n
values\n
Element
Network\n
Term\n
\302\260\n
Cq-H
farad\n
Co\n
2K\342\200\236t\n
\342\226\240\342\226\240
o
o\342\200\224\342\200\224\342\200\242
ir
uuu
\342\200\242L\302\273'dbhenry\n
\302\2732+\302\253S\n
\nCn\n
Fig.
12-16.
Admittance
expressions
C\342\200\236farad\n
and equivalent
networks.\n
shown in Fig. 12-16
is shownin Fig.
\nconform
with
Eq. 12-63 for parallel admittances
\nThe
network
is for Case 1 but specializes to the other
cases
shown
each
as in the
first Foster type of network. The role
type
\njust
The
combination
of
networks
of the types
to
12-17.
three
of
of\n
network
is
function
admittance
\npoint
terms
in
configuration
(1) Inductor Lo. Inductor
\nthe
which
origin
poles and zeros
summarized
as follows:\n
of the
Lo
293\n
of the drivingpole
(of Y) at
the partial fraction
Kq/s term in
the
the
of
because
appears
to
rise
gives
NETWORKS
REACTIVE
ONE-TERMINAL-PAIR
12-5
Art.
\nexpansion.\n
Co.
Capacitor
(2)
\nat
infinity
of the pole (of
the partial fraction
because
Co appears
Hs term to appear in
Capacitor
the
causes
which
Y)
\nexpansion.\n
(3)
LnCn
network.
type
of each
(s2
such
Series
the
of
\nfactor
<an
\nquency
+
in
\302\253\302\2732)
of
denominator
the
is a pole
term
corresponds to a
series LC network
Each
frequency
F(s).
is
which
The
fre\302\254
associated\n
^C0\n
Pole at
Pole
\norigin\n
resonance
with
\ninfinity\n
second Foster type.\n
in the series impedance
to antiresonance
contrast
(in
of the
Network
12-17.
Fig.
at
Thus all poles can be associated
in the sense that resonance within
elements
with
\ndirectly
specific
\nindividual
cause
the entire
network to have a pole of admit\302\254
networks
\ntance.
cannot
As in the
case of the first Foster form, the zerosof Y(s)
\nbe
so
element
contributes
identified.
some part to the condi\302\254
Every
\ntions
with
a zero of admittance,
in much the sameway as
associated
\ndiscussed
for the first Foster form.\n
The
\342\200\234end
elements\342\200\235 (that
is, L0 and Co) are associated with the poles
\nof
F(\302\253)
\nplace
of
\ntions
for
a
pole
the
End
thus
infinity
(frequency).
infers the absence of
for
<o
1\n
pole\n
2\n
a zero in
networks
condi\302\254
sum\302\254
functions.\n
in Second Foster
Values
\342\200\224
0\n
The presence of
an end element.End
the second Foster type of
are
impedance
Element
Case\n
of
cases
four
below
\nmarized
and
zero
at
networks.
LC
the
of
\nrealization)
co
=
oo\n
Networks\n
Co\n
Lo\n
zero\n
zero\n
absent\n
present\n
absent\n
3\n
pole\n
zero\n
present\n
absent\n
4\n
zero\n
pole\n
absent\n
present\n
this discussion,
From
equivalent
\npletely
\nform
is
a
series
networks
impedance
pole\n
present\n
we see that
from
any
specifications
two
com\302\254
be designed.
The first Foster network
realization
of Z(s); the second Foster net\302\254\n
can
then
and
\n1/Z(s),
with
identified
\nare
\nthat
and
Z(s)
\nset of
Y(s)
\ninternal
\n1
/Z(s).
as
a given
If
is:
Z(s)
The
2
\nCase
1
\nCase
3
\nCase
4
\nCase
4\n
\nCase
3\n
Foster
networks
for
+
(s2
=
Z(s)
9)\n
(12-67)\n
-f-
Case 1 and
is of
function
impedance
+
l)(s2
2\n
s(s2
This
first and
procedure for finding the
a given Z(s). Consider the impedance\n
the
illustrate
will
is:
\nCase 2
\nCase
example
\nsecond
F(s)
corresponding
\nCase 1
An
12\n
a function interchanges poles and zeros,
Inverting
will never
have the same number of internal poles
Z(s)
case
will differ for Z(s) and Y(s) =
the
designation
zeros,
conclusions
can easily be verified.\n
The
following
specifications.
since
\nand
Chap.
=
by forming Y(s) by inverting Z(s), as F(s)
as parallel admittance functions which
expanding
network
forms. It is important to observe
specific
will never be of the same case classifications
for
one
found
is
form
work
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
294\n
4)\n
the partial fraction
expansion\n
is\n
\342\200\234
T
+
s-T72
i
farad\n
\nbe
2
farad\n
Hs\n
(12-68)\n
-j2\n
of this
constants
evaluated
the
by
equation
may
Heaviside
rule
henry\n
'~\\SlQSLrJ
Zls)\n
+
i\n
The
Mr\n
&
k2*\n
+
\nas
follows.\n
\n$ henry\n
2
9
X
=\n
Ao
=
\n1
First
12-18.
Fig.
network of
Foster
K2
2(s2
=\n
H
since
by
= ^\n
2(158/4)\n
+
s2
+ 4
Using the equationsdisplayedin Fig.
form
Foster
To
\nfunction
is
found
are found
network
the
determine
4\n
-}2\n
inspection,\n
*(.)
\nfirst
15\n
*=
i2)\n
\n\302\253(\302\253
\342\200\224
2
2\n
+ 9)
l)(s2
~
\nexample.\n
and
+
9
\342\200\235
the
12-14,
as shown in
form
Foster
second
+ZS\n
(12-69)\n
element
values
and
Fig. 12-18.\n
of network,
the admittance
as\n
Y(s)W
^
=
i
2 (s*
Z(s)
2KiS
_\n\342\200\234
s2
+
1
+
s(s2 + 4)\n
+ l)(s2 +
(12-70)\n
9)\n
2Kzs
s2 +
32\n
(12-71)\n
constants
two
The
REACTIVE NETWORKS
ONE-TERMINAL-PAIR
Art. 12-5
92+
=
Kl
\nelement
values
are
Fig.
12-19.
in
\nshown
5_\n
-
l)(s
two
32\n
\302\273--j3\n
jZ)\n
the network configurationand
as
O
1\n
equiv\302\254
<\n
3
c\n
henry
number
same
the
have
networks
\nalent
4)\n
determined
The
32\n
\302\273-ji\n
9)\n
of Fig. 12-16,
chart
the
of
use
Making
+
+
2(s2
z_\n
+
s(s2
=\n
as\n
4)\n
2(s-il)(s2
\nXs
by the Heavisidemethod
K* are found
and
Ki
\nof
equivalent
\nto
consider
unfinished
the
Second
12-19.
Tig.
\nwork
\na
for
proof
the
Now that we
\ntance
\nof,
have completed the partial
fraction
we
functions,
at
most,
three
net\302\254
immit-
for
function is composed
of terms:\n
types
2
Kns\n
Hs\n
S2
s\n
reactance
Foster
example.\n
expansion
any immittance
that
know
Ko
The
of
property.\n
separation
farad\"\n
\t\n
o
us digress
of
business
let
networks,
=:
farad\"\n
h
other forms
two
consider
we
Before
\302\253\n
henry
Zis)
elements.\n
\nof
295\n
+
terms
for these
expressions
i*l\n H\n
II\n
(12-72)\n
0>\302\2732\n
are\n
1\n
(12-73)\n
*1\302\256\n
Y
2Knu\n
\342\200\234
\342\200\235
*\n CO\nII\n
(Susceptance
\ntives
of
these\n
have
expressions
\nthree
(12-75)\n
_
values
\ndriving-point
of
09\n
II\n
typical terms is
(and
deriva-\n
+ 2EW
(12-77)\n
the
+\n iS\n
positive for both
(12-78)\n
positive
and
neg\302\254
the reactance function for the
same thing might be said for the sus-\n
co. Since
frequency
terminals
The
(12-76)\n
\342\200\224
(co\342\200\2362 co2)2\n
ft\n
\native
form.)
co2\n
+2tfnco2
\ndco
three
same
,Ko
\ndo)
of the
the
are\n
dXi _
Each
0>\342\200\2362\n
3\n
exactly
expressions
dXn
(12-74)\n
+
-CO2
296
ONE-TERMINAL-PAIR
sum
is given as a
function)
ceptance
NETWORKS\n
REACTIVE
=
X
\342\200\242
of
12\n
reactances,\n
Xi
^
Chap.
(12-79)\n
y-i\n
which
by term to
term
differentiated
be
may
;=
y
give\n
dxj\n
(12-80)\n
do\n
A/
/-1\n
Then
is positive,\n
dX/dw
ft\n
8-1
both
for
and
positive
the
\nclusion,
o\n
(12-81)\n
values of frequency
negative
\302\253.From
this
con\302\254
follows.\n
property
separation
V\n
12-6. Cauer form of reactive networks\n
An
\nin
could
function
\ntions
a continued
by
of
form
basic
\nThe
W.
by
Germany
\nance
Foster reactance theorem was made
in 1927. He first pointed out that the react\302\254
Cauer*
be represented
by two different network configura\302\254
fraction
expansion of the driving-point impedance.
for the Cauer realization is the ladder\n
the network
to the
extension
important
o
VW\n
\342\200\224AAA/
Zi\n
^7\n
Z{s)\n
n\n
12-20.
Fig.
with
shown
\nof
a
and
impedance
of
impedance
\ndriving-point
fraction
continued
Z(s)
-
Zx
Ladder
network.\n
admittance
designations in Fig.
such
a network
may be written
+\n
+\n
z%
+\n
Yt
Cauer,
Arch.
The
form
(12-82)\n
+\n
f4
W.
12-20.
in the
as\n
Z*
*
\t\n
Elektroteck.,
17, 365 (1927).\n
+\n
Art.
12-6\n
Let
us
\nat
our
restrict
the
In
infinity.
general
\nit
This
must).
procedure
the
Consider
\nexample.
12s4
division
\n6s8
as
6s8
3s)
+
+
=\n
proceeds
that
for forming the
+ 1
12s2
+ 12s2 +
12s4
+
+
1\n
(12-85)\n
+
3g\n
and dividing
6s8
gives\n
+ 3s
(s\n
s
+
6s8
(2s\n
1\n
+
+
1)
1
6s2\n
Qg3
+
6s2
(12-84)\n
3s\n
+
12s4
=2s
term
remainder
the
(12-83)\n
follows.\n
Z(s)
Inverting
...\n
procedure
6s2
so
+
function\n
impedance
Z(s)
Direct
...
divide,
this
continue
and
\ndivide,
The
m.
4~
pole
then invert and divide, invert and
until the expansion terminates (as
process
can best be illustrated with a numerical
to
is
fraction
\ncontinued
bm-iSm~2
+
than
greater
2
an_2Sn
Q>\302\2738n+
=\n
\nbmsm
n is
297\n
to an impedance function with a
for the driving-point impedance\n
expression
discusssion
Z(8)
this means that
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
\n2s\n
or
=
ZW
28
+
\302\273
+
Continuing
the
invert
and
divide
2s)
+
2V(6s>
(12-86)\n
1)\n
procedure,\n
6s2 + 1
(3s\n
6s2\n
1\n
such
that,
finally,\n
Z(s)
\342\200\224
2s
(12-87)\n
-|-\n
s
+\n
3s
+
2i\n
Comparing
\nresents
\n1
\na
farad,
capacitor
with Eq. 12-82, we see that: Z\\ = 2s rep\302\254
of 2 henrys,
a capacitor
of
inductor
s represents
an
Yz \342\200\224
= 3s
an inductor of 3 henrys, Y* = 2s represents
Z\302\273
represents
The network
2 farads.
of
configuration for this specific\n
this
expression
\nfirst
12-22. For this case, a poleat
will always be of the form of Fig.
network
Cauer
and
\n\342\200\234series\342\200\235
inductors
\342\200\234parallel\342\200\235
o\342\200\224<innrL-\n
i-W-n\n
Lx\n
Li\n
L%\n
\342\200\234\n
~'C2
\nbe
will
\nexpansion
this
with
equation
Zi
Eq.
+\n
2 henry\n
3 henry\n
case. If
is
Fig.
In
12-23.
\nfor
Eq.
form
\nwork
network
Cauer
then
of
\nior
12-84.\n
the
the
\nthat
the
first
first
A
element
is\n
l3\n
Let
\nthat
us
we
focus
have
our
studied
-\n
-c4
^pC4
form
General
of
first
=:c8
^C8
Cauer
requires
requires
r\n
-\n
-C6
^C6
is shown
the first or
l9\n
L?\n
-\n
=:C2
net\302\254
Cauer net\302\254
is determined
by the behav\302\254
network
function
at infinite
at
infinity
pole
\nfrequency.
be a series inductor.
at infinity
zero
be a parallel (or shunt) capacitor.\n
element
Tig. 12-28.
Cauer
first
summary,
the
first
in
\342\200\224'T5W'1\342\200\224i\n
\342\200\224nm'\342\200\224|\n
\342\200\224\342\200\224i\nr-^tRKP
Zls)
the first\n
from Fig.
absent
A
\nthat
differ\302\254
Li,
at infinity
zero
a
with
\n\342\200\234end\342\200\235
element
First
0,
farad:\n
\nin
Fig.12-22.
that the only
=
Z\\
inductor,
\nwork
2
+\n
\n12-22.The form of the
1\n
farad:\n
Z 6
12-82, we see
series
1
the
+\n
in the second
= 0
-nm'
Zls)\n
of
+\n
f4
that
form
(12-88)\n
Z*
is
by
can
=\n
i%
\nence
n
exceeds
m
be\n
m
Comparing
*>).\n
expansion
it
made,
infinity. For this case,
12-83, and before the continuedfraction
is necessary
to invert the polynomial. The
in Eq.
\nunity
^Cq\n
first Cauer network (poleat
form of
a zero at
considered:
be
to
case
-\n
-C6
-c4
General
other\n
one
only
as
far
\342\200\224'TRHP\342\200\2241\n
l3\n
~\n
12-21.
with
12-21
as
is
There
12\n
the
infinity,
extending
capacitors
continued fractionexpansion.
\nrequired by the
Tig.
Chap.
in Fig.
shown
is
example
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
298\n
network
^\n
^Cio\n
(zero at
\302\253).\n
on the way the Cauernetworkendsnow
the factor controlling
the way it begins.There\n
attention
Art.
12-6\n
are
but
ways
possible
in Fig.
as shown
12-24
in
\nshown
can end. Oneis
other is
with
a
an
with
as in
capacitor
networks
attached to the general form
any number of elements in a ladder arrangement)
(with
12-22 and 12-23. With the inductor as the last element,
in a path from one terminal to the other.
inductors
At\n
Figs.
series
are
\nthere
12-24 (a); the
299\n
these
Imagine
(b).
network
the
\nof
NETWORKS\n
the ladder network
two
\ninductor
\nFig.
REACTIVE
ONE-TERMINAL-PAIR
\342\200\224i\n
(a)
zero
(direct
frequency
element
Last
12-24.
lig.
(6)\n
forms for first Cauer network.\n
these
current),
inductors
offer zero impedance
zero of impedance. Alternately, if the
in
from ter\302\254
the
there is a \342\200\234break\342\200\235
path
\nminal
to terminal
and at zero frequency there is a poleof impedance.
\nIn
The
is
last (or far end) element in a first Cauernetwork
summary:
\ndetermined
of the impedance function at zerofrequency.
by the nature
\nA zero
at zero
at
that the last element be an inductor.A pole
requires
\nzero
a capacitor
as the last element. These conclusions are
requires
is a
zero
\nsuch
that
\nlast
element
frequency
is a capacitor,
below.\n
\nsummarized
w
Case\n
1\n
=
=
03
0\n
00\n
Last
element\n
First
pole\n
pole\n
Elements\n
End
Network
Cauer
First
element\n
L\n
C\n
2\n
zero\n
zero\n
C\n
L\n
3\n
pole\n
zero\n
C\n
C\n
4\n
zero\n
pole\n
L\n
L\n
basic
The
but
\nnetwork,
\nthe
\nsecond
a
first
the
with
Fig.
Such
for
form
ladder
Second
12-26.
is
or
Cauer
is illustrated
network
element
(parallel
second Cauer network is again the ladder
of capacitor and inductor interchanged.\n
position
the
Zi(s)
shunt)
= l/Cis;
element
network
form (pole at
0).\n
in Fig. 12-25. The
impedance
the
admittance
of the
similarly,
is F2
= 1/L2s. The
of
continued
frac\302\254\n
tion
of this network
realization
the
for
expansion
=
zw
NETWORKS
REACTIVE
ONE-TERMINAL-PAIR
300
form must
be\n
1
+
c^
Chap. 12\n
-1
,
1\n
1
L%s
^
,\n
\342\200\242
\342\200\242
*\n
CiS
\ndifferent
An
a function
of such
example
of
case
the
consider
\nzero.
that
than
procedure
\nillustrate,
fraction expansion will requirea
used for the first Cauer network. To
an impedance function with a poleat
of continued
form
this
obtain
To
is\n
-
(1M0)\n
w-6
To
expand
\nfor
end
this
of Eq. 12-89,
in the form
function
we
=
\nis
\342\200\234invert-and-divide\342\200\235
the
and
inverted
then
3s
to obtain
numerator
division
+
\n6
2
3s2
that
Z(s)
+
s4)
+
3s
+ 2s8
\n3s
+
=
f
s4\n
(12-92)\n
2s8\n
gives\n
(1/s
s8\n
1\n
(12-93)\n
+\n
i
+\n
3s2
s
final
+
is\n
Z(s)
The
s4\n
and dividing
term
remainder
+
3s
s
the
example,\n
+ s4 (2/s
3s2
,
+
\302\260r
Inverting
We divide
+4s2\n
3s2
*
started.
one term; the remainder
In our
repeated.
6 + 7s2
2s8)
be
now
may
procedure
the
into
denominator
such
it end
(12-91>\n
2s1
+
3\302\253
\nthe
turn
as\n
m
The
first
\342\200\234invert-and-divide\342\200\235
step
s8)
3s2+
+
gives\n
s4
\n3s2\n
<*\n
(3/s
s4\n
Art.
12-6\n
such
that
REACTIVE
ONE-TERMINAL-PAIR
of the continued
form
final
the
-
Z(s)
+
-s
NETWORKS\n
fraction
301\n
is\n
(12-94)\n
j
J-j\342\200\224
+
s
3,1
1
\n\302\253\"*\"
\n\302\253\n
this
Comparing
2\n
-
=
s\n
-
=
r*(s)
a capacitor
represents
an inductor of 1 henry,
3\n
=
=
Yi(s)
a capacitor of
- represents
\ns\n
an inductor of 1 henry.
\nFig.
must
\nnetworks
\nto-even
farad.
second Cauer form,
is shown in
function
for LC
impedance
i farad
be an even-to-odd
or odd1
\nZ(s)
The
12-26.
^
of the
configuration
last
equation,
the
to
\nalent
of
quotient
the following
farad.\n
\302\243
- represents
\ns\n
network
The
of
represents
\ns\n
Zz(s)
12-82 and 12-89,
Eqs.
made:\n
are
\nidentifications
Zi(\302\253)
with
expression
L
henry\302\256\n
the
For
polynomials.
which is
equiv\302\254
i farad
\n1
henry*\n
\nprocedurejust shown by exampleto work,
\nthe
\nat
\nto
invert
\nof
the
the
If
zero.
poly\302\254
first
=
-j
ES
j
+ T-
c.s+
of Eq.
Comparison
first
\nThe
first
is
element
element
function
\nimpedance
\nfunction
Second
12-26.
Cauer
net\302\254
form\n
Z(\302\253)
\nthe
Fig.
that
for example.\n
\nwork
pole
function
has a zero at zero, it is necessary
impedance
before
such that the continued fraction will be
dividing
to saying
equivalent
must
have a
expanded
function
\nthe
even-to-odd
is
This
\nnomial.
be an
must
quotient
has
a zero
(12-95)\n
\t\n
1
j_\n
a zero
at zero
12-89showsthat
and that C\\ is not present. In summary:
an inductor
in the second
Cauer network is a capacitor the
the impedance
has a pole at zero; it is an
12-95 and Eq.
with
if
inductor
at zero.\n
if
By
the
\nof
\nsible
network
the
networks
\nlast element
the
general
(6)\n
element
network
being
forms for second Cauer network.\n
of Fig. 12-25, we
the network has
is a capacitor,
there
\nfrequency,
Last
12-27.
Fig.
to
12\n
other
(a)
networks
Chap.
Cauer case, the infinite frequencybehavior
is determined
by the way the network ends; that is,
last
is an inductor or a capacitor. The two
element
pos\302\254
are
shown
in Fig. 12-27. Attaching these terminating\n
the
to
analogy
\nwhether
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
302\n
a short-circuited
can see that
zero impedanceat
path
if
the
infinite
from terminal to
ter\302\254
element is an inductor,
at infinite frequency. In summary:The
\nwork
has
a pole
of impedance
second
in the
Cauer network is an inductor if the
element
has
a pole at infinity; it is a capacitorif the
\nance
function
These conclusions are summarized
at infinity.
a zero
\nfunction
has
\nminal.
On
other
the
if the last
hand,
the
\nlast
net\302\254
imped\302\254
impedance
\nbelow.\n
=
\302\253
Case\n
O)
0\n
End
Network
Cauer
Second
=
CO\n
First
Elements\n
element\n
Last
element\n
1\n
pole\n
pole\n
C\n
L\n
2\n
zero\n
zero\n
L\n
C\n
3\n
pole\n
zero\n
C\n
C\n
4\n
zero\n
pole\n
L\n
L\n
12-7.
Choice
of network
realizations\n
network
to the
In discussing the specificationsfor a reactive
leading
of (1) the
specification
\nsummary on page 288 it was found that the
or some
\ninternal critical frequencies and (2) the scalefactor
equiv\302\254
H,
fix
the
was
sufficient
to
\nalent
driving-point impedance
specification,
be
shown
can
In
last
two
we
have
that a network
the
\nZ(s).
sections,
\nrealized
in four basic forms from the driving-pointimpedance
only.
\nIn
are
network
there
are
as
unknowns
as
there
many
elements,
any
\nand
these
unknowns
must
be specified by Z(s). In the solutionof any
as there are
of equations,
there must be as many specifications
\nsystem
\nunknowns.
The
of specifications
total
number
that we have found
\nsufficient
is one
more
than the number of internal critical frequencies.
total
number
of unknowns
\nThe
equals the total number of elements.
\nFrom
the
equality
of specifications
and
unknowns,
we conclude
that\n
REACTIVE
ONE-TERMINAL-PAIR
12-7\n
Ait
NETWORKS\n
303\n
in a network realization is one more
the
number
of internal
critical frequencies.
It should be noted
\nthan
in this
each pair of conjugate critical frequenciescount
statement
\nthat
and
critical
We do not distinguish
between \302\253\302\273
\nas one
\342\200\224\302\253\302\273.\n
frequency.
number
A knowledge
of the
of elements required to realize a net\302\254
of end elements with
with
our association
\nwork
along
specification
\nnetwork
at zero and infinite
behavior
frequency can be used in drawing
the
minimum
\nthe
four
of elements
number
network
possible
configurations
As
configuration.
\npole-zero
example
directly
will illustrate
by inspection of the
the procedure.\n
Example
5\n
the
Consider
critical
\nnal
specifications
of two
consist
pole-zero
frequencies
00
s-plane\n
given in Fig.
12-28. The
inter\302\254
poles and one zero for Z(s). The\n
Q\n
JO\302\273\n
<>\n
<T\n
-6-\n
X
O
-H\n
Q> *0\n
ci)\n
o\n
0)\n
critical
external
\nerty
be
must
\nthere
Since
at
two
\nthe
poles
12-29
\nFig.
by the separation
prop\302\254
a time.\n
infinity
network
The
12-15.
\nFig.
5.\n
are
both
network. Because zero and
zeros,
Foster
form
of
in
the
basic
are
missing
must have two parallel LC networks to give
network
The network is shown in
frequencies).
(antiresonant
elements
end
constrained
of Example
are three internal critical frequencies,
in each network realization. Considerthe
(1) First Foster
\nthe
configuration
there
elements
four
one
\nrealizations
are
frequencies
zeros.
be
to
Pole-zero
12-28.
Fig.
(a).\n
for\n
network. In finding the critical frequencies
versa.
the poles of Z(s) become zeros of F(s) and vice
Y(s)
1/Z(s),
are
\nSince both
zero and infinity are poles, both end elements
present.
in the
network
series
LC
\nThe one internal
pole is caused by a single
network
in Fig. 12-17. The four-element network is shown
\nbasic
shown
Foster
Second
(2)
=
\nin
Fig.
(3)
\nfirst
\na
zero
\nsince
12-29
First
Cauer
of
there
\342\200\242\n
(b).
Cauer
network.
network.
Z(s)
at
The
Referring
infinity
is a zero
is a capacitor.
the
at zero,
last elementis
means
of Z(s)
end elements will first be foundfor the
to the table of page 299, we see that
the first element
Also
An
inductor.\n
REACTIVE
ONE-TERMINAL-PAIR
304\n
There
in the network, specification of the
element
determines
the schematic shown in Fig. 12-29
(c).\n
Ca/uer
table
of
network.
The
page 302 may be usedto
the
second Cauer network form. From the\n
investigating
last
\tSecond
(4)
in
\nadvantage
Chap. 12\n
12-29.
Fig.
elements
four
only
being
and
\nfirst
NETWORKS\n
The
four network
realizations of Example 5.\n
implies that the first elementis an inductor.The
identifies
the last element as a capacitor. The fourat infinity
\nzero
is shown
in Fig. 12-29 (d).\n
realization
\nelement
the value of H must be given,andthe partial
find
element
To
values,
must be completed.\n
fraction
\nor continued
expansions
in selecting a net\302\254
matters
involved
There are a numberof practical
four
forms for a specific application. These
\nwork
the
from
possible
at zero
a zero
table,
\ninclude:\n
Element
(1)
\ncapacitors
with
\nEconomic
factors
(2)
\nout
reasonable
thus
may
size
the
This
\nrealization.
the
when
Use
com\302\254
For example, 1-farad
quantities.
voltage
rating are hard to come by.
one network form the advantage over
give
to construct inductors
into account
This
capacitance
the parallel capacitor in the first Foster
It is impossible
with\302\254
of
vacuum
important
is high.\n
frequency
tube circuits. Vacuum
capacitors
of normalized
of
when specifications are rigid
in
interstage
tube circuits frequently
coupling
\nrequirementmay specify whichnetworkmustbe
Use
by
form
is especially
operating
in
blocking
\nrequire
12-8.
of
can be taken
capacitance.
\nreducing
(3)
any
capacitance.
Stray
stray
\nor
limited range
three.\n
other
\nthe
are but a
in commercial
available
sizes
\nponent
There
values.
component
networks.
This
used.\n
frequency\n
made
The examples given in this chapterhave
\ncritical frequency values to advantagein simplifying
use
of small
numerical
integer
opera-\n
Ait
problems, the criticalfrequencies in
radians
per second. In such cases, the arithmetic
to some value that makes
normalizing
frequency
lions.
In many practical
\nsands
or
\ncan
be
\nthe
critical
by
simplified
small
\342\200\224
Xl
values,
OioEaet
thou\302\254
frequency
let\n
-
\342\200\224
CoLaet
To normalize
integers.
on element
effect
the
are
values
frequency
observe
\nand
of
millions
305\n
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
12-8\n
)
= LnW*'
(12-96)\n
y\n\n>\302\253o/\n
and\n
=
\342\200\224
Be
<oCaet
MoCact
= Cner^o'
-)
(12-97)\n
^\n\n.\"0/\n
where\n
=
Laet
inductance,\n
=
the
normalized
inductance
=
=
the
normalized
capacitance
=
=
the
normalized
frequency
Lnom
Cnorm
w'
the
\nfrom
can be found
values
element
actual
The
Co* = the
actual
the
=
actual
capacitance\n
<o0\302\243/\302\253\302\253* (12-98)\n
u>oCaci
(12-99)\n
(12-100)\n
o\302\273/oto
from the normalizedvalues
equations\n
Laet
=
L^i
(12-101)\n
COo\n
=
Cact
(12-102)\n
0)0\n
With
can
\ninductance
the
\nillustrate
expansion.
an equation
An example will
in
function
reactance
frequency.\n
the
per second. Fromthe data,
X
c\302\2602
+
1QI)*TC
fa)[-w2 + (6r X
frequency
normalizing
=
o)o
Several other choices
of
=
\302\253
\nSubstituting
w
X(w)
at 1000and 4000
is\n
[
=
at 3000 cycles
a pole
and
to have zeros
is known
impedance
second
per
X(<o)
\nLet
fraction
in normalizing
procedure
driving-point
\ncycles
\nthe
actual values of capacitance and
the normalized values found in the
6\n
Example
A
from
continued
or
fraction
\npartial
the
normalized,
be
found
frequency
wow'
\342\200\236
_
X 1Q3)2]
103)2]
(12-103)\n
be\n
2v
X 10s
normalizing
into
+a^1T
Eq.
-Ho,o(-co/2
w'
(12-104)\n
radians/sec\n
frequency
12-103
might
have
been made.
gives\n
+
+
l)(-\302\253'2
(-fa)'2+
16)\n
(12-105)\n
9)\n
108
X
radians/sec
into Eq. 12-105 fixes the valueof
\nthese values
normalized
\nthe
^
\nKS)
+
\nthis
of the elements
values can be
values
normalized
After
actual
the
equation,
-
\302\253'=
2.
at
Huo
l)(s'2+
_ 1000(s'2+
36s'(s'2
^
when
Substituting
so that
1000/36
becomes\n
function
impedance
Chap. 12\n
be 100 ohms
that the reactance
to
corresponding
is required
it
that
Suppose
\na) =
4tt
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
306\n
16)
(12-106)\n
9)\n
are found by the
found by division
of
expansion
<12-107)\n
2^00
=
C\342\200\234\342\200\230
2ir
on
\nE.
A.
\nD.
F.
\nYork,
M.
\nR.
New
in
\n(1924),
\n17, 355
and
one-terminal-pair
II
Vol.
Networks,
Wiley
(John
5, is recommended,
as
&
as
well
New
Synthesis, 2 vols. (John Wiley& Sons,
Source material may be found in articlesby
\342\200\234A
reactance
W.
Chap.
networks,
Network
preparation).
Foster,
\nstanden
1935),
York,
Jr.,
Tuttle,
of
topic
Communications
Guillemin\342\200\231s
Inc.,
\nSons,
the
(12-108)\n
1000
X
READING\n
FURTHER
For further study
as\n
w0
by
Cauer,
theorem,\342\200\235
Bell
Tech.
J.,
3, 259
Wechselstromwider-
von
\342\200\234Die Verwirklichung
vorgeschriebener
System
Arch.
Frequenzabhangigkeit,\342\200\235
Elektrotech.,
(1927).\n
PROBLEMS\n
For the networks shown in the
as a quotient of polynomials.
\nimpedance
12-1.
figure,
Prob.
polynomials.
Compare
\npolynomials.
Answer.
(s4
+
4s*
the
Identify the
driving-point
even
and
odd\n
12-1.\n
the order of
(a)
find
the numerator and
+ l)/(3s* + s); (b) 3s/(s*+
denominator
1).\n
12-2.
shown
3(s\302\256+
\n(b)
NETWORKS\n
307\n
find the driving-point admittance of
Answer,
below.
+ 1/6)/(s* + s/2);
+ lls2/6
(a) (\302\2534
+ 14s2 + 30).\n
Prob.
Repeat
networks
\nthe
REACTIVE
ONE-TERMINAL-PAIR
12
Chap.
7s/2)/(s4
but
12-1
1 \302\256-\n
io\342\200\224nmrHt\n
2f\n
3h<\n
lh\n
lo-\n
3z__T\n
lo-\n
(a)\n
(6)\n
Prob.
A certain
12-3.
\nance
\n1
with
and
\nquotient
\nunity.
12-4.
3
poles
LC network is known to havea driving-point
imped\302\254
at the frequencies of 0 and 2 radians/sec
and zeros at
of
for
\nimpedances
\302\26173
X\n
-
\n(co2
(co2
(c)
\302\261jo) X\n
(w2
Answer,
(a)
-
No\342\200\224no
In
4)(co2
16)
-
l)(a>2
pole
case,
- 25)
(co2
3)\n
each
or
why?\n
10
(b)\n
\n\302\261
X
(co2
- 25) (co2
(co2
jco
64)\n
-
(co2
(d)
driving-point
\302\261j'5co
X\n
5)\n
-
at
\302\253
;
(b)
4)(co2
No\342\200\224separation
36)
49)\n
16) (co2
\n(co2
zero
-
-
-
25)
-
9)\n
property.\n
shown in the
determine:
\nwhether zero frequency
whether
represents a pole or a zero,and
a pole or a zero of the
\ninfinity
(frequency)
represents
Do this
of the networks (and not
by inspection
determining
12-5.
For
a
is
(s4
networks?
LC
(co2
(a)
+ 10s2 + 9)/(s3 + 4s).\n
the following functions may represent
=
Z(s)
of
driving-point
impedance as
the multiplying factor,
if ff,
(expanded)
polynomials
Which
the
Determine
radians/sec.
Answer.
12-2.\n
the LC networks
figure,
(a)
(b)
impedance
\ntion.
\nthe driving-point
by
impedance).\n
Prob.
12-5.\n
func\302\254
12-6.
For
the
following
(a)
sketch
the
\nworks),
the
of
\nExamples
12\n
net\302\254
configuration
X as a function
desired are given
of sketches
type
Chap.
(representing LC
in the s plane, and
functions
network
pole-zero
function
reactance
the
sketch
\n(b)
NETWORKS\n
REACTIVE
ONE-TERMINAL-PAIR
308\n
of frequency
w.
in Fig. 12-7
in
and
12-8.\n
\nFig.
M
w
+ 4)
(s2
l)(s2
-1-
(s2
+
s(s2
that for case 1
be negative and
should
\nequation
+
(s2
+
36)\n
+
64)\n
+
64)
+
(s2
81)
+
(s2
144)\n
100)\n
will consider which sign shouldbe used
that for Case 2 and Case3 the sign of
we
problem,
12-52. Show
and
12-51
10
W s
(s2 9)(s2
,
(s2+ 25)
w
25) (s2 + 81)
(s2 +
this
In
12-7.
\nEqs.
...
(s2 + 36) (s2 + 100)
(s2 + 16) (s2 + 81)
+ 5 )(s2 + 49)
s(s2
.
W
be
should
12-8.
\nat
=
a)
the
sign\n
positive.\n
A network function
4 and a zero at o> =
internal
only
is required
that
the
are
\ntwo
4 the
Case
and
in
has a pole
These
10.
critical
fre\302\254
the magni\302\254
be 100 ohms at <a = 6
\ntude
of reactance
Determine
(a) the sche\302\254
\nradians/sec.
It
\nquencies.
\nmatic
\nworks
two Foster
the
and (b) the
specifications,
net\302\254
element values
networks.\n
two
the
\nfor
these
to
corresponding
of
diagrams
to serve as the load for a
12-9. A reactive network is to be designed
for
\nvacuum tube amplifier. The following specificationsare given
the
\nLC
function.
impedance
\n1000
of
\nslope
a frequency
at
\nifications:
(a)
Sketch
the
Determine
(b)
each
Determine
(c)
the
of
\nWhich
two
the
\naccount
12-10.
A
stray
spec\302\254
configuration
pole-zero
schematic
for the
diagrams
networks
factors
of coils,
capacitance
impedance
driving-point
impedance
(b)
\nconfiguration,
\n(c)
Find
the
nature
and the
X(u>)
two Foster
curve.\n
networks.\n
value in the two networks of part (b). (d)
would you select for a practical applica\302\254
as estimated
cost of elements, taking into
element
,
this
internal
kilo-
\nZW
For
The
the
such
Consider
\ntion?
vs.
reactance
the
\ncycle/sec
zero),
(a
cycles/sec
critical frequencies are:
3000 cycles/sec,
4000 cycles/sec. (2) The
must
curve
be 100 ohmsper
frequency
of 1000 cycles per second. From these
(1)
function:
Determine
of the
_
15s6
7.5s4
etc.\n
is given by
+ 29s3 +
+
7s2
+
the
equation,\n
6s
1\n
Cauer
in the
external critical frequencies (that is,
the first
(a) Determine
the value
of each element
network
network,
does\n
12
zero
frequency
\nzero
REACTIVE
ONE-TERMINAL-PAIR
Chap.
for
configuration
=\n
\n12s8
\nconfiguration,
Find
the
nature
\tInvestigate
(c)
or
\npoles
the
For
the
of
\ndiagrams
element
\nmine
Plots
*
-O
>
0
f
external critical frequencies (are they
the pole-zero configuration for Z(s).\n
configuration
shown, draw the schematic
and two Cauer networks. (Do not deter\302\254
are of impedance unless otherwise noted.\n
pole-zero
two Foster
values.)
58\n
of the
Sketch
(d)
zeros?),
12-12.
the
(b)
+
the second Cauer network
(a) Determine
for each element in the network,\n
value
function:
impedance
as\n
13s2 + 5
+
2 s*
this
etc.), (d) Sketchthe pole-
impedance is given
Z(s)
For
^
X
w-0
0\342\200\224x
<\302\253>\"oo\n
w -0
(!)
<*>\n
For the pole-zero
\ndiagrams of the two Foster
element
12-14.
12-13.\n
Prob.
12-12.
12-13.
\nmine
\302\253\342\200\224if\n
<f
<*>\"\302\253>
Prob.
309\n
Z(s).\n
A driving-point
12-11.
or a zero,
a pole
represent
NETWORKS
the
configuration shown,
and two Cauer networks.
draw
(Do
schematic
not
deter\302\254
values.)\n
12-13 for the
Prob.
Repeat
pole-zero configuration
of
the
\nfigure.\n
1
co *0
<*>
\342\231\246\n
\342\200\242
oo\n
<*>\n
12-14.\n
Prob.
12-15.
two
the
Draw
\nzero
with
Starting
12-16.
Eq. 12-103, verify Eq. 12-106.\n
and two Cauer networks for
Foster
in the
shown
configuration
co
the pole-
figure.\n
*\n
*
\"0
co\n
CO\n
Prob.
first
the
(b) the
\nat
1
\nthat
\nnumber
\nment
are
following
specifications
must be a capacitor in
element
The
12-17.
(a)
network
of
values.
avoid
<a
the
\\Z(jl)\\
made for an
series(to
must have zero impedanceat = 2
radian/sec,
is,
12-16.\n
=
impedance
LC
network:\n
a d-c
radians/sec,
path),\n
(c)
have a magnitude of 10 ohms;
must have the smallestpossible
schematic
and indicate
the network
farad.\n
henrys, C =
must
10, (d) the network
elements.
Answer.
Draw
L
ele\302\254
=
-rtr
13\n
CHAPTER
REACTIVE
TWO-TERMINAL-PAIR
\nNETWORKS
(FILTERS)\n
The discussionsin Chapter12
confined
to networks
with one
\nterminal pair (the driving-point terminals). In this
we
will
networks.
One of the terminal pairs
be
\nstudy
tow-terminal-pair
\nidentified
as the input,
the other as the output. Our ultimate
to
networks
to give a specified relationship between voltages
design
currents
at one terminal
pair and voltages or currents at the
were
chapter,
will
objective
\nis
\nor
other.\n
The
in this
concepts
\nnature
well
contrast
in
\ndriving-point
Fig. 13-1.
net
\nnetwork
\nwork.\n
work
\nthe
\nmarked 1-1 will
the
\nwith
Most
structure
\nladder
\nin
our
of
constructing
for
\ndeveloped
be identified
\nBy
convention,
\narate
\nthe
\nvalues
T
input
specifically
12.
A
rep\302\254
two-terminal-pair
in Fig. 13-1.In
the two terminals
follow,
and the two marked 2-2
noted.\n
the ladder network structure.The
is important
it was the first structure used
historically;
a design
for filters. In addition, the concepts
procedure
the
ladder
structure
can be applied to other
will concern
studies
structures\n
13-3.
lattice.
the
A
Standard
standard
impedance
ladder
ladder
of all
network designations.\n
network
is shown
series elements is
in Fig. 13-2.
Zi(s) and
of
all
to sep\302\254
For our studies, it will be convenient
the
network into two other network structures:
standard
ladder
section
and
the t section. This separation and the resulting
is illustrated in Fig. 13-3.\n
for
the
series
and shunt impedances
elements
\nshunt
otherwise
is shown
to
to exclusively
Chapter
a
in
driving-point
network\n
Fig.
such as the
the
with
unless
outpilt
The ladder
13-1.
of
\nresentation
Two-terminal-pair
as
as
in
\ntransfer
are thus
chapter
is Zs($).
310\n
\nance
13-3 (c), together with the specific
imped\302\254
will
be adopted
as a standard T section.The net\302\254
r section.
(d) will likewise be adopted as a standard
shall
for T or w sections
will
refer
to these
develop
The
T and ir section building blocks can\n
networks.
designations,
of
\nwork
\nAll
13-3
Fig.
we
equations
standard
\nspecific
(a)
\nan
L
Evolution
13-3.
Fig.
\nof
(b) x
T sections;
of the ladder network into tandemnetworks
a x section;
and (e)
(d)
sections; (c) a T section;
section.\n
VWtAA/V
AAA/\342\200\224\302\260\n
o\342\200\224A/W-\n
\nZJ2\n
Zj/2\n
Zx/2\n
2 Z2\n
2 Z2\n
'2Zz\n
\342\226\240o\n
o\n
lb)\n
(a)\n
13-4.
Fig.
(a)
from two L
T section
L
13-3
\nFig.
\nsection
the
\nin
13-4.
to
the
term
These
follow.\n
three
x section
(b)
from
into a more primitive
is designated as a
network
seem
ir section
network
two\n
shown
network
similarity).
standard
sections;
sections.\n
\342\200\234inverted
might
geometrical
and
\nstudies
primitive
end-for-end,
of
\nsection
Fig.
This
(e).
(although
\nturned
\nview
step further
one
divided
be
311\n
NETWORKS\n
in Fig.
shown
network
The
REACTIVE
TWO-TERMINAL-PAIR
13-1\n
Art.
more
The
or
L,\342\200\235
gamma
appropriate
construction
in
standardL
network
when
from the point of
of the standard T
from the primitive L sectionis shown
will form the basis of the
structures
REACTIVE
TWO-TERMINAL-PAIR
312\n
13-2.
NETWORKS\n
13\n
Chap.
impedance\n
Image
in the
T section and t section just discussed
are
symmetrical
L sec\302\254
1-1 and 2-2 could be interchanged.The
\nsense that terminals
In order to derive an equation that
is unsymmetrical.
\ntion,
however,
\nwill
a symmetrical
or an unsymmetrical network, con\302\254
to either
apply
T section
\nsider
the
shown in Fig. 13-5. Let the gen\302\254
unsymmetrical
and
the
load
be ZL. We will also\n
\nerator
be Z\342\200\236
impedance
impedance
The
at terminals 1-1 with ZL connected
(and
as
the
at
and
terminals
2-2
Za*
similarly
impedance
When the impedances Zg
ZL disconnected).
(but
the
as
Zu
define
impedance
disconnected),
\nZ,
\nwith
connected
Zg
\nand
are
ZL
adjusted
such
Zg
that\n
*
\nZu, the
Ztt
(13-1)\n
match is said
and
empha\302\254
will
Ztt
\nwritten
2.\n
The
the
\nUnder
specified
1-1 is
\nterminals
the
conditions,
the same as
that
\nimpedance) which views the
the
when
of the
\nimpedance
seen
impedance
a mirror
\342\200\234seen\342\200\235
in
For
the
\nwritten
network
in terms
An
the
their
of Fig.
of Zh Zt,
=
13-5, expressions for
and
Zi
These
Z%.
Z\302\273(Z
=
i
+
Z%
-f-
+\n
Z\\
\nalgebraic
+
Zt(Z\\
Zt
equations
+\n
\nZt
+
Z%
Zu and
can
Zu
be
are\n
Zu)
+ Zu\n
(13-2)\n
Z\\i)\n
+
(13-3)\n
Zu\n
two equations in two unknowns, Zu and Z\302\253.
it is possible to solve for the unknowns.
operation,
here
match
image
image
impedance is the same as
if terminals 2-2 are also terminatedin
generator
Zu\n
in\342\200\235
at
to see
(constructed
driving-point
at\n
have
\342\200\234looking
generatorimpedance.
impedance.\n
\nimage
We
\342\200\224
Zi,
To add
2-2.
to exist at terminals1-1
be
defined by Eq. 13-1, Zn
written
impedance
special
at terminal pair 1, and similarly will be
image impedance
at terminal pair
Zw, the image impedance
for using
the word image is suggested by Eq. 13-1.
reason
an image
the
\nsis to
\nexists
and
Z\\\\
By
routine
However,\n
a more
useful
\ncircuit
and
=
\nZu
313\n
of
by solving for Zu and Zuin terms
openLet
impedances.
at terminal-pair
1 with terminal-pair 2 open,
at terminal-pair
1 with terminal-pair 2 short-
short-circuit
the
NETWORKS
is found
result
impedance
\342\200\224
the
\nZu
REACTIVE
TWO-TERMINAL-PAIR
13-2
Art.
impedance
\ncircuited,\n
Zu
\nZu
=
the
impedance
=
the
impedance
2 with terminal-pair 1 open,
2 with terminal-pair 1 short-
at terminal-pair
at terminal-pair
\ncircuited.\n
terminal
for
\nances
of Fig.
network
the
For
pair
13-5, the open-circuit
1 have the values\n
\342\200\224
Z\\
Z\\o
\342\200\224
Z
i
Zu
+
(13-4)\n
z%z%\n
+\n
=
Zu
two
For
\nlow.
\nFor
this
the
case,
the
are
equations
symmetrical
the notation
Several examples, important
given
Image
13-3 (c)
that\n
(13-6)\n
VZioZu
(13-7)\n
=
Zu
= Zi
in terms
(13-8)\n
to
discussion
the
of
follow,
will
in
Fig.\n
next.\n
Section. For the T section
of the T
Impedance
shown
the image impedance
=
Z*
=
7
ZiT
ir
the
\342\200\224
VZioZu 7
.%/7
ZiT
+ ZxZ2
The
Section.
(13-9)\n
(13-10)\n
is shown in
v section
Fig.\n
as follows:\n
is found
-\342\200\224/
(I + z)
Mfz)
V^i2/4
impedance
image
+
V(t
Image Impedanceof
The
is\n
=
vz^zr.
ZiT
13-3(d).
it is found
of much of the analysisto fol\302\254
foundation
the image impedances are equal.
network,
will be simplified by letting\n
Zu
\nbe
Zs\n
Zu = s/Z^Zu
Similarly,
These
(13-5)\n
+
last four equations,
of the
manipulation
algebraic
imped\302\254
Zz\n
Z%
By
and short-circuit
+
2Z2(Zi
y/2z2 + 2Zi
2Zj)
2ZiZj\n
+ Zi (Zi +
(13-11)\n
2Zj)
Z\\Zi\n
=\n
\\/
Zi2/4
+ Z \\Zi\n
(13-12)\n
is
13-3(e)
L Section. The L section shownin Fig.\n
below as Fig. 13-6. This network is unsymmetmust be computed for each terminalpair.\n
impedance
At
terminals
1-1, the image impedance is\n
reproduced
the
and
\nrical,
image
-o2\n
1\302\260\342\200\224AAA/\n
Zi/2\n
\342\200\224
Zul
+
y/(Zi/2
2Z,\n
-o2\n
with
\nComparison
\nZul
+ ZxZi (13-13)
13-10
Eq.
L section.\n
13-6.
2Zz)Zi/2\n
= v%2/4
lo-\n
Fig.
Chap. 13\n
the
of
Impedance
Image
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
314\n
=
(13-14)\n
ZiT
the image impedance of the L sectionat terminals
of the T section. At the otherterminal
\nimage impedance
or that
Z
2Zs(Zi/2)\n
(2 Zi)\n
*2iL\n
2Z2
this
of
Comparison
+
with
equation
Zul
\nir
The
section.
\nanother
point
combined
be
\ncan
one
in
\nlooking
impedance
\nimage
y/z
i2/4 +
Eq. 13-12 shows
is the
1-1
pair\n
2\n
\\Z
(13-15)\n
ZxZz\n
that\n
\342\200\224
(13-16)\n
Zir\n
image impedanceis that the symmetrical
of the L section appears as a T
impedance
image
From
and as a ir section looking in the
direction,
of view this conclusion seems reasonable. Two L sections
an image
with
match to form a T section,
ZiT on each end as shown in Fig. 13-7.
pair 2 the
at terminal
Thus
Zi/2
that
shows
of
section
other.
the
with
Similarly,\n
o\n
\342\200\242o
\302\260AAA/\n
A/W-0---0-VW\n
AAA/-\302\260\n
-Zir\n
Zt.\n
ZiT\n
ZiT\n
o
o\n
(c)\n
of image
Combination
13-7.
Fig.
\nT
L
both
\nat
with
combine
sections
terminal
pairs
section,
and
matched L sections to form
(b)
the
an image match
as required.\n
7r
(a)
the
section.\n
to form a
ir
section
with
Z\342\200\236
13-3. Image transferfunction\n
It
\nance
\nis
\none
a
concept
in
driving-point
terminal
we need something
that
is evident
pair
the
two-terminal-pair
We need
concept.
to
the
other
in addition to the
problem.
image
imped\302\254
The image impedance
a function to relate variablesat
terminal pair. For the time being,we\n
REACTIVE
TWO-TERMINAL-PAIR
13-3\n
Ait.
ratio
function to the
the transfer
will restrict
NETWORKS\n
of
315\n
thus\n
currents;
h(s)\n
(13-17)\n
-<?(\342\200\242)\n
/*(\342\200\242)\n
and
input current, 12 is the output (or load)
current,
In the sinusoidal steady state, the trans\302\254
transfer
function.
a complex
number which may be expressedas a
becomes
is the
h
where
the
is
\nG(s)
function
\nfer
and
\nmagnitude
angle
phase
as\n
=
G(ju>)
In practice, the
=
e\342\200\234
\\G(j<\302\273)\\
The
/3, so
is designated
of G(j<a)
angle
=
G(jco)
where a =
a logarithmic
in
measured
is
|Cr(jw)|
(13-18)\n
unit
that\n
so
neper)
\n(the
magnitude
a(ia)
\\G(ja)\\eiAna
(13-19)\n
that\n
= ey
eae*
the attenuation
=
/S
(nepers),
(13-20)\n
the
phase
shift
(radians),
As a practical matter, the mostcom\302\254
function.
transfer
\n7
image
\nmon
unit
for attenuation
is the decibel, abbreviated db, even though
the ratio of powers\342\200\224not
\nthe
involves
definition
for
the
decibel
voltages
=
the
currents\342\200\224as
\nor
follows.\n
oidb =
If the powerratio is
10 logio
to
related
Ei\n
Pi\n
P2\n
Then\n
otdb
the
2\n
Pi\n
or
=\n
K
Ei\n
or
logxo\n
that
\nrestriction
last
20
adb in
\n8.686 (the
should
It
\nthe
\nto
decibels
be
input
the
=
20 logio
magnitude
e\" =
8.686 a
is found by multiplying
emphasized
attenuation
current
(13-23)\n
It\n
the
under
neper,
neper being the
The
\nnature.
I_i\n
logxo\n
ratio
current
as\n
adb
Thus
(13-22)\n
of decibels correspondingto a
Eq. 13-22 applies, we substitute for the
equation
as\n
2\n
the number
To find
the
Ii
ratio
It\n
E2\n
\nin
(13-21)\n
or current
voltage
Ej2\n
\342\200\224
20
db
(P1/P2)
(which
of the
larger
db
(13-24)\n
a in nepers
the
by
factor
unit).\n
in
that the quantities a and /9 are transfer
of
is a measure
of the ratio of the magnitude
must be sinusoidal for a to have meaning)
output
current.
The phase angle
/3
is
the
phase\n
TWO-TERMINAL-PAIR
316\n
the
of
the
\nKnowing
input
with respect to the output sinusoid.
and the image transfer function, the out\302\254
when the network is terminated in the image
current
is determined
current
\nput
13\n
Chap.
measured
sinusoid
input
REACTIVE NETWORKS\n
\nimpedance.\n
13-8.
Tig.
the
\nconsider
the
\nby
the
find
To
for computing
Network
function for a symmetrical network,
in Fig. 13-8. The currents are related
transfer
image
T section
shown
equation\n
Ii
\n7-
this
Solving
=
^
ey =
Zi + Zi/2
for Zit
equation
=
This image impedanceis
Eq.
by
given
the
Eq.
\nthere
13-26
after
results,
Z2V7
e27
of the
=
(13-25)\n
(13-27)\n
this squared equation
to Eq. 13-27,
are canceled,\n
2ey + 1)
\342\200\224
2e7
-I-
-
=
ZiZ2ey
1
0
(13-28)\n
Z\\\n
(13-29)\n
Zl\n
e7
This
\n?(ey
equation
+
e~y)
can be put in hyperbolic
= cosh 7, so that finally\n
cosh
\nthe
/10
T section which
ZlZi\n
1\342\200\234
or\n
A
-7-
is\n
terms
-
Zi
(13-26)\n
impedance
and equating
common
Zi
|
ot~ +
1)-^]
image
ZiT2
Squaring
= 1
1 +,
there results\n
Z,Uer-
which
13-10,
4- Zi
7
Z<
\nis
image transfer function.\n
similar
expression
for the
y
=
1 +
hyperbolic
form by recognizing
that
(13-30)\n
^
sine is found by making
use
identity\n
cosh
7
sinh 7 =
e7\n
(13-31)\n
of
REACTIVE
TWO-TERMINAL-PAIR
Art. 13-4
this
Comparing
with
equation
Dividing Eq. 13-32 by
\ntion
the
for
found
is
13-25, it is
Eq.
sinh
13-30and
seen
317\n
that\n
=
t
(13-32)\n
the
canceling
tangent
hyperbolic
NETWORKS\n
common
of 7, which
Z2,
an
=
tonh
f13-33*\n
\302\273
zjhz>
This equation can be expressedin terms
\nparametersby making use of Eq. 13-6.\n
=
Zi
(which could be
is
\nnetwork
two
these
expression
\ntion,
\nto
which
This
follow.
\ntion.
It
7
=
and
l\342\200\231s
by
short-circuit
2*s,
the
since
for the open-circuit
equation
for any
holds
Zl0
imped\302\254
(13-35)\n
written\n
(13-36)\n
\342\226\240%/Zu/Z\\o
equation for the
the basis of much of the discussion
is more general than is implied by our
passive reciprocal network. (See Prob. 13-2,
useful form of the
with
13-34 forms
Eq.
most
is a
the
13-33 may be
Eq.
identities,
tanh
This
open-
(13-34)\n
replacing
by
Zi/2 + Z2 =
Using
the
T network.\n
the
for
\nance
the
and
symmetrical)
of
y/ZuZu
written for side2
equa\302\254
is\n
func\302\254
image-transfer
deriva\302\254
for
\nexample.)\n
13-4.
In
a
purely
\nurements
LC
networks
12,
is
purely
class
we
is approximate.
however,
well,
sufficiently
\nChapter
\nwork
of two-terminal-pair networks, all the
and capacitors. Since all
the
network
are inductors
have
any result based on the assumption\n
resistance,
elements
\npractical
networks\n
important
very
within
\nelements
of
to LC
Application
know
reactive
The results conform to
to be of engineering value.
meas\302\254
From
the driving-point impedance of an LC net\302\254
(and the admittance
purely susceptive) such\n
that
TWO-TERMINAL-PAIR
318\n
Chap.
NETWORKS\n
REACTIVE
13\n
that\n
Z
=
0\342\200\230co)
foundation
The
\nthese
given
equations
networks
reactive
\nFor
and
impedance
\nimage
From our knowledge
the
of
\nwith
for
frequency
\na
of
As
\nmagnitude
Xu
and
Case\n
2\n
Xu
\nreal or
con\302\254
tanh
Zi\n
+\n
\342\200\224\n
real\n
an
,
,
both
cosh
For
real.
is
investigation
y =
Ri\n
imaginary\n
opposite
the
image
and tanh
Z, is real
signs,
is
impedance
y
only choices. Both Zt and tanh y must be
can never be complex. We next turn our atten\302\254
of the conditions under which tanh y can be
sinh
y
\342\200\224L
cosh
=
a + j(3,tanh
sinh
and
denominator
be
can
y
a cos P +
a cos P +
\342\200\224r
cosh
y
numerator
a cos
imaginary\n
the
Since y =
imaginary.
Dividing
y
Ri\n
and Xu,
same for Xu
but
imaginary,
+\n
y\n
real\n
jXi\n
\342\200\224\n
are
These
tanh
\nfactor
are four possiblesign
\342\200\224\n
\342\200\224\n
imaginary.
to
There
+\n
signs are the
\nimaginary and tanh
\ntion
as the
the sign as well
Xu\n
the
or
relates to
quotient
changes,
changes.
+\n
4\n
\nreal
their
although
jXi\n
3\n
\nis
discussed in
networks
LC
the reactance functionchanges
reactances.
Here Xu and Xu are
frequency
Xu\n
1\n
If
of
properties
below.\n
summarized
\nditions
(13-40)\n
i.)
(\302\261jX
('W\n
functions,
function.
transfer
(13-39)\n
y/jjk
driving-point
reactance
\ndriving-point
yjZ^/Zio\n
the sign of
that
know
we
\342\226\240\\/(\302\261jX
i.)
* =
tanh
12,
(13-38)\n
In the
Zi =
\nChapter
y/ZuZu\n
equations to follow, the subscript1
that it can be replaced by a 2 asin
understanding
as long as the network is symmetrical.
above,
in the sinusoidal
steady state, equations for the
transfer
function
become\n
image
the
with
used
be
\nwill
VZuZu
networks.
symmetrical
=
\342\200\224
(13-37)\n
\302\261jB(u>)\n
are\n
y = y/Zu/Zt\\0 =
tanh
=
Y(ju>)
for our analysis
equations
Z%
for
and
\302\261jX(ui)
expanded
j cosh a sin
j sinh a sin
P
as\n
,10\n
(13-42\n
/3\n
of the equation by
the
0 gives\n
tanh
y = ,tonh
1
\nT
+j
a
+J a
tanh
tan 3
tan
(3\n
(13-43)\n
13-4
Art.
Our
y =
Let
(2)
tanh
=
0
t/2
Let
(3)
\nnot
sir;
of
then\n
0,
\302\261t,
(13-45)\n
...)
\302\2612x,
of this angle) suchthat
of a and
there
tanh
tan 0
0 =
\302\261
(13_46)\n
\342\200\242
\342\200\242
\302\261
t\342\200\231
\302\261
t
r
)
real.\n
purely
0, as summarized
Value
(13-44)\n
0)
limit,\n
(for
Hence
permitted.
for
the
In
and again tanh y is
Any other values
\n\302\253
and
= 0;
condition,\n
\342\200\224
real.\n
=
tanh'>'
a
(for
(or odd multiples
infinity.
\napproaches
this
For
0.
(for 0 =
a
tanh
y is purely
tanh
and
=
7
\342\200\224
j tan 0
y is purely imaginary.\n
0 = 0 such that tan 0
tanh
and
or purely
possibilities.\n
Let a = 0 such that tanh a
tanh
319\n
either purely real
expression
several
are
There
\nimaginary.
this
make
to
is
problem
(1)
REACTIVE NETWORKS
TWO-TERMINAL-PAIR
0 will maketanh
only three
below.\n
of a and 0
Conditions
y\n
is
this
and
complex,
y
possibilities for values for
are
\n'
=
a
tanh-
^\n
1 0 = 0, +ir,
I
+2t, ...
(13-47)\n
or\n
Real\n
=
I a
tanh-1\n
0 =
a
Imaginary\n
0
=
0\n
=
tan-1
possible to extendthe table
of a and 0 for the four
It is now
values
\nthe
Case\n
2\n
We
\nterms
jXu,\n
Zi\n
+\n
+\n
\342\200\224\n
\342\200\224\n
jXi\n
\342\200\224\n
+\n
\342\200\224\n
4\n
now
of
^
(13j*8)\n
(13-49)\n
318
page
to include
cases.\n
JX<\n
3\n
\342\200\242
\342\200\242
\342\200\242
\302\261
on
given
jXl8\n
1\n
|
\302\261
+\n
have
sufficient
both
a transfer
Ri\n
Ri\n
information
quantity,
tanh
a\n
7\n
0\n
real\n
a
tA
0\n
0
or
real\n
a
5^
0\n
0
or
imaginary\n
0\n
MO\n
imaginary\n
0\n
0*0\n
ir/2\n
x/2\n
to examine the
differentcasesin
the image transfer
function, and
a\n
REACTIVE
TWO-TERMINAL-PAIR
320\n
the
quantity,
driving-point
\nterminals
1-1
\npresented
to
is
of
fre\302\254
con\302\254
the
at
2 apply. Then the image impedance
This
is the impedance of the network
when terminated
in Z\302\273. We associate
this
imaginary.
the
Xu are such that
of Xu and
signs
13\n
Chap.
At some value
impedance.
image
Case
1 or
Case
of
\nditions
the
that
assume
\nquency,
NETWORKS\n
generator
condition of no powertransfer.
the
a
is
\nfer point
of view,
the attenuation
means
positive and real,
is
than
\nthat the
load current
smaller
the
It
h
(see
generator
is
attenuated
The
current
so
to
13-17).
or,
speak
the
of Case 1 and Case 2, the frequencies
\nUnder
conditions
\nnated
and
the band of stop frequencies is designated
stop
frequencies,
load with the
\nreactive
trans\302\254
From
which
current
\342\200\234stopped.\342\200\235
\nEq.
are
\nthe
desig\302\254
band.\n
stop
the
When
of Xu
signs
and Xu are
opposite,
we
3 and
Case
have
and
the
attenuation
a is
\nCase 4, where the image impedanceis real
is now
the
load
\nzero. From the driving-point impedance point of view,
the
\nresistive and there is power transfer. Withno
mag\302\254
to the magnitude of h (althoughthere will be a
\nnitude
of 72 is equal
attenuation,
\nrent
is
\nand
Case
\nis
identified
\nto
stop
The
phase
Such
\342\200\234passed.\342\200\235
4 are
or vice
band
rules
The
(2)
As
discussed
slope
of the
a consequence
of resonance
\npoints
currents). For thesecases,
the
as
frequencies
versa
function
reactance
(1)
of the two
designated
a pass
band.
as
several
\nto
the
in
\ndifference
give
the
conditions
cur\302\254
of Case 3
pass frequencies. A band of pass frequencies
The frequency of transition from passband
is assigned the name cutofffrequency.\n
Xu and Xu vary with frequency according
in Chapter
12:\n
reactance curve dX/do) is alwayspositive.\n
of the slope property, the polesand zeros
(or
and antiresonance)
alternate as a functionof
\nfrequency.\n
(3)
The
external
\neither
poles
frequencies
(that
is,
u>
=
=
\302\253
0 and
\302\273) are
always
or zeros.\n
function
determine
the
Since the critical frequenciesof the
that
the
\nnature of the reactance versus frequency plot,
suspect
\ncritical frequencies
somehow relate to the pass band,
stop
band,
can
the
the
cutoff
We
relationships by
frequencies.
study
and
have
made
for
the
reactance
curves
Xu
Xu. Both plots
\ning
\nthe general
of the plot shown in Fig. 13-10.In
appearance
\nthe
critical
of Xu and Xu, we recognize that there are
frequencies
the
will coincide but be opposite
critical
\npossibilities:
(1)
frequencies
the
will coincide but be the same
critical
(2)
nature,
frequencies
and
the critical
will not coincide. These three
(3)
frequencies
reactance
we
the
\nand
compar\302\254
will
comparing
three
\nin
\ntype,
are illustrated
\npossibilities
in
Fig.
13-11.\n
Art.
reactance
The
in
\nshown
\nXu,
are
for
the
plots for Xu and Xu,
13-12. At noncritical frequencies,
on page
table
the
plot for
Xu\n
!\n
A\n
i\n
i\n
\302\253i\n
1
\n1
\n1\n
1\n
1\n
1\n
1\n
0)
*
i\n
o)\n
Xu\n
1\n
1\n
|\n
1\n
Xl0\n
13-11.
to
correspond
\nwhere Xu
and
\nfrequencies are
holds
\ncondition
The
\nplots
second
having
Case
T\n
i\n
|\n
u>\n
Xu\n
i\n
\302\245\n
\342\200\242\ni\n
0)\n
l\n
i\n
i\n
i\n
i\n
i\n
T
(6)
Comparison
3 and
of critical
thus
A
possibility
the same
sign
band
0)2\n
frequencies of Xl0 and Xu.\n
a = 0.
in sign, thereis
a pass band.\n
shown
in
no
At
all
frequencies
and
attenuation
of frequencies
for which
Fig. 13-11(b) results
for all values
w\n
(0\n
Case 4 for which
pass frequencies.
w
!
\ni\n
i\n
i\n
i\n
\342\200\242\n
0)1\n
012\n
u>2\n
Xu are opposite
is
i\n
i\n
i\n
i\n
i\n
!\n
(a)
Fig.
Xu,\n
LC network.\n
xu\n
1\n
1\n
1\n
1\n
are
the signs for Xu and
the two functions
for
Reactance
*1.\n
possibility
opposite sign nature.
see that oppositesigns Xu and
319, we
13 -10.
Fig.
frequencies,
first
321\n
to preserve this
time
same
the
at
signs
critical
At
opposite.
always
\nchange
\nFrom
Fig.
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-4\n
the
this
in reactance
of frequency. Sucha
reactance\n
shown
is
plot
\nhave
Case
a stop
last
The
\nexists
same sign for Xu and
critical
being in the other one.
l\n
l\n
1\n
A\n
4\n
frequency
At
\342\226\240\n
\302\273-\342\226\240
i\n
M\t\n
l\n
l\n
and
.Yu
\342\200\235'
T\n
1\n
\n1
\302\253
I\n
\nother
the
not.
This
a stop
band
does
\nband
to
\ncutoff
frequencies.
\ntions
existing
\nstop
band,
sign
An
and
and
(a)
condition
-00\n
bands and cutoff frequency.\n
reactance
function changes but
to changing from a pass
the
corresponds
or vice versa. Such critical frequenciesarethus
of a plot with each of the threecondi\302\254
example
with
cutoff
and pass
of one
Xu.\n
stop\n
cutoff'\n
Stop
crit-\n
\302\253
f\n
l\n
l
\n1\n
pass\n
13-15.
the
1\n
i\n
1
frequency,
to
corresponding
Condition for cutoff frequency from
t\342\200\224r\n
ical
we
Xu,
of page 319,
in Fig. 13-14: a
without
or Xu
Zh\n
Fig.
13\n
Xu
is illustrated
Xu
is-14.
ng.
Chap.
band.\n
possibility
either
in
NETWORKS\n
with the same sign for Xu and
of frequencies
A band
therefore
13-13. With the
2 on the table
in Fig.
1 or Case
\nattenuation.
\nis
REACTIVE
TWO-TERMINAL-PAIR
322\n
corresponding
designation of
is shown in Fig. 13-15.\n
frequency
the
pass band,
\nimpedance
\ncircuit
terms of the
Z, in
13-6
Eq.
by
\ngiven
critical frequencies
of
ZXo and
functions,
impedance
of
open-
the
image
and
short-
The image impedance
is
=
(13-50)\n
VZioZu
poles and zeros
and
Zu
of
Zu,
the
impedance
image
written\n
be
\nmay
Zu.
the
323\n
as\n
Zi
In terms of the
the nature
to study further
a position
in
now
are
We
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-4\n
Art.
7
*
+
(s2
/rr 1\342\200\236
(s2 +
_
\\
W42) .
+
\302\25322)(s2
+
\302\253i2)(s2
o,32)
..
H2 (s2 +
. .\n
W\342\200\2362).
(13-51)\n
...
\302\253
(s2
+
o\302\2732).
. .\n
possible impedance forms have been assumedfor Zi<,
In the pass band, the poles of ZXo are zeros
\nand Zu.
of Zu or vice versa.
=
\nSuch
for
and
cancel
if
factors,
(s2 + o>22)
(s2 + a>62)
\302\253&,
example
\nterm by term
not
critical
In
the
and hence are
stop
frequenciesof Z<.
the
and
removed
and
zeros
of
coincide
be
and
so
Zi0
Zi,
\nband,
poles
may
\nfrom the
critical
radical.
critical fre\302\254
are
thus
frequencies
Stop-band
of the
two
where
A
of
\nquencies
\neither
or
Zi0
\nsinusoidal
Zia
cutoff
(never
In a
both).
s =
where
state
steady
the
\nwithin
\nto
an
This
radical.
number
imaginary
13-16.
Fig.
changes as
term
of the
sign
Image
as a critical
ju),\n
(\342\200\224O)2+
the
frequencyin
typical frequency term (in the
appears
frequency
(13-52)\n
<0l2)\n
\302\253
exceeds
\302\253i,causing
of sign changes Z,
change
or vice versa. One plot
variation
impedance
in the
of sign
a change
from a real
of Zi for a
number
filter
is\n
pass band and stop\n
band.\n
\nance
\nof
\nstop
Zu
at
the
and
band.\n
Note that
13-16.
in Fig.
shown
cutoff
Zu
frequency,
in the
Z<
changes
from
and again, that
stop band are
to a
a resistance
react\302\254
the critical frequencies
the criticalfrequencies Z,
of
in
the
the
illustrate
\nnetwork
consider
T network
13-17.
of Example 1
circuit
We will
determine
\nand
and
Zu,
\ncutoff
The
n
Zl
short-circuit
the
\342\200\236
\nhave
zerosof the
/
.
\\
\342\200\242(*)
2
L*
2
impedance
values
tabulated
LCs1
_J_
+
Cs
function
1
,
functions,
Zeros\n
the
and
2\n
(13-53)\n
is\n
_
Cs + 2/Ls
Zu
stop band,
2 Cs\n
L(LC\302\253\302\273 +
4s)\n
2(LCs*+
the poles and
(13-54)\n
2)\n
zeros are
to
found
below.\n
Zu\n
Zu>\n
Poles\n
short-\n
functions
impedance
the pass band, the
impedance is\n
open-circuit
two
these
the
_~
and
open-circuit
determine
impedance
*uW
with
networks.\n
the poles and
this
from
frequency.
From
13\n
of the concepts of this sectionto specific
first the T section shown in Fig. 13-17.\n
application
configurations,
ng.
and
Chap.
1\n
Example
To
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
324\n
zero\n
infinity
infinity\n
=
\n\302\253
=
\302\253
zero\n
V2/LC\n
CO
=
y/2/LC\n
2\n
7=\n
\\/LC\n
\nfrequencies
\ncutoff
Fig. 13-18, together with the designations the
the
the pass band and in the stop band. The
of
for this network is seen to
are shown in
These
in
frequency
of
value
be\n
-
which is a
zero
of
Zu.
As
a filter,
-
(1W\n
this network
passes the low
frequen\302\254\n
Ait
13-4\n
cies
and
\nname
the
rejects
Another
\nsponding
Filters of this
frequencies.
high
NETWORKS\n
325\n
type are giventhe
filters.\n
low-pass
Example
REACTIVE
TWO-TERMINAL-PAIR
#\n
is shown in
T section
and
open-circuit
Fig. 13-19,together
the
corre\302\254
For this network
networks.
short-circuit
with
we\n
cutoff\n
Fig.
13-18.
filter poles
Low-pass
and
zeros.\n
\342\200\242\342\200\224Hf\342\200\224\n
2 C\n
Zlo\n
Z.\n
,
2C\n
Z\\,
Fig.
13-19.
T network
of Example 2 with
circuit
the
=r=2C\n
L\\\n
and
open-circuit
short-\n
networks.\n
o) \302\256oo\n
Fig.
see
18-20.
filter poles
High-pass
zeros.\n
that\n
^2 Cs\n + L.
+
-
poles
and
\342\200\224
7T?T
2Cs
zeros
+\n
1
2 Cs
1\n
(13-56)\n
2Cs\n
4 LCs2
1\n
Zu
The
and
+
l/Ls
for these two
2Cs(2LCs2
+
1\n
+
(13-57)\n
1)\n
functions are shownin Fig.
13-20,\n
the
with
together
This
\nfrequencies.
stop band, and cutoff frequencydesigna\302\254
rejects low frequencies but passeshigh
a high-pass filter.\n
filter is designated
of
type
the
\nsimplify
rig.
example has element values given in orderto
in computing
The network is a T\n
impedances.
this
for
network
The
algebra
of Example 3 with
T network
13-21.
circuit
\t\n
4\nf1
T
?
I
i
|\n
i\n
i
i\n
i
I\n
:
1
the
o \t\n
\342\200\224T
1
f\n
1
1\n
1\n
\n1
1\n
1\n
1\n
1\n
1\n
1\n
w\n
x\n
1\n
T\n
\n1\n
1\n
1\n
J\342\200\224\n
U
1\n
cutoff\n
13-22.
7 u
u
found
s
_
2
+
l
+
2s
s
and
zeros.\n
From the schematic diagramsthe
branch.
are
expressions
\nimpedance
filter poles
Band-pass
in each
elements
two
stop\n
pass\n
Tig.
-00\n
cutoff\n
stop\n
with
short-\n
w\t\n
1\n
1\n
\342\231\246\n
and
open-circuit
networks.\n
!
T
*
13\n
S\n
Example
and
Chap.
pass band,
this network
a filter,
As
\ntions.
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
326\n
to
be\n
1
_
+ 1/s
+ 4s2 +
2s(s2+
1\n
(13-58)\n
1)\n
similarly,\n
_
\342\200\242
2
+,
1
2s
+,
1
s +
l/\302\253
+
l/(s/2
_
+ l/2s)
\302\2736
+
2*
(s4
+ 7s1
+ 4s2 +
7s4
+
1\n
1)\n
(13-59)\n
This
last
equation
looks
rather
formidable,
being of sixth
order
(or\n
the
\nin
equation
\nmight
also
be
13-22
with
all
\nthe
7s4
s\302\253
+
+ 1 =
all zeros can
\nsummary of values for the
from which
(s2 + l)(s4+
be found using
poles and zeros
the
6s2
co =
are
\nthere
cutoff
two
=
\302\253
1\n
\nare
frequencies.
pass-band
An
filters.
\npass
opposite
and
\nfrequencies
\nelimination
V2
\302\261y/Z\n
a/ 2
+\n
CO
=
1\n
w
=
y/3
2
\302\261
shownin Fig.
\\/2\n
It
13-22.
is
seen
that
both low-frequency and
in a center band
are
bands.
Frequencies
stop
Filters
of this type are designated band\302\254
and
high
type of filter with pass bands at low
a stop band
that
and
at a center band
of
is a
frequencies
band-
filter.\n
and
Short-circuit
an unknown
the
that
Suppose
\nlaboratory.
can be used
measurements
open-circuit
of analyzing
means
\ntical
A
formula.
infinity\n
frequencies
bands
\nhigh-frequency
(13-60)\n
1)
zero\n
poles and zeros is
of these
A plot
to
Zu\n
infinity\n
Zeros\n
Fig.\n
follows.\n
zero\n
co =
+
quadratic
Z\\o\n
Poles\n
1)
of
7s2
-f
appears
(s2
results\n
there
Factoring,
suspicion.
term
of
and
poles
327\n
an s2+ 1
leads one to suspectthat
+
of the pole zero plot
Zu> Inspection
zeros plotted except the zeros Zu adds
of
zero
a
NETWORKS\n
the fact that
in s2). However,
for Z\\a as a pole
order
third
rather
REACTIVE
TWO-TERMINAL-PAIR
13-4\n
Art.
as a
prac\302\254
network in the
network is connected\n
two-terminal-pair
two-terminal-pair
O-T\n
I
Sine wave
or
Open
\nclosed\n
\\)\n
\ngenerator
1\n
Fig.
to
a
sine
wave
\nmeasured by
\nwith
a
cathode
13-23.
apparatus
Experimental
generator
as shown in
an ammeter marked
ray
oscillograph)
to study
filters.\n
Fig. 13-23.
I (or a
as frequency
If
dropping
the
resistor
current
together
is changing with
output\n
is
maintained
voltage
a
and
\nance
of
\nresistance
The
\ninfinity.
be
\ncan
applied
\nwhen
is
lot
in
made
the
but
It
\nWhen
Itc and
now
\noff
the
\npass band,
\nconcerned
to
were
cutoff
a low-pass
filter.\n
in locating pass bands, stop bands, and cut\302\254
to the problem of com\302\254
next
turn
our attention
a in the stop band, and the phaseshift
/3 in the
attenuation
for a number of important filter networks.We
with the variation
of the image impedancewith
and phase shift in symmetricalT
for
the
will
be
also
frequency
Eqs.
the significance
to show
\342\200\224
V^i*/4
=\n
+
that
ZiZ,
of the factor
= V%Z*(1
networks\n
(Zi/4Zi)
net\302\254
be
as\n
+ Zi/4Z*)
(13-61)\n
ZiZ,\n
ZiZ,\n
\342\226\240^Zi*/4
shown
and t
of symmetrical T and r
impedance
13-10 and 13-12. These equations may
image
as
derived
Zu
have
for
recorded
practice
We
Ztr
We
or minimumis
given
have
Expressions
\nrearranged
be
other, that frequencyis a
frequency.
maximum
or
minimum
at
a
are
both
It
frequency,
is in the stop band. The case illustrated in Fig.
Attenuation
\nworks
will
networks.\n
\nfor these
13-5.
in Fig. 13-24
not for the
frequencies.
\nputing
When a maximum
band.
corresponds
We
shown
13-24
frequency
\nevidently
plot
13\n
will have
current
pass
or
\n/\342\200\236
\nthat
a
constant,
Chap.
a maximum value at a zeroof imped\302\254
value
at a pole of impedance. The incidental
minimum
the current from becoming zero or
the
elements
prevents
same
rules
that have been used in our networkanalysis
If /,\302\253is at a minimum
value
results.
to the experimental
and vice versa, measurements are being\n
a maximum
at
The
\nobtained.
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
328\n
+
in the pass
ZiZi\n
+ Zi/4Z,\n
is
band the imageimpedance
(13-62)\n
real
and\n
Art.
in the
that
\ntions
determination
329\n
factor\n
the
When
criterion.
impedance
NETWORKS\n
the image impedance is imaginary.Theseequa\302\254
of pass bands and stop bands, using this
band
stop
permit
\nimage
REACTIVE
TWO-TERMINAL-PAIR
13-5\n
1 + Zi/4Z2
or
\nimaginary
image impedance must change,real to
real. Hence cutoff frequencies occur when\n
of the
nature
the
sign,
changes
(13-63)\n
to
imaginary
Zi\n
1\n
(13-64)\n
4Z2\n
This
of
\nterms
frequency.\n
be derived relating the imagetransferfunction
The derivation begins with Eq. 13-30,which
An equation
may
\nthe factor
(Zi/4Z2).
cosh
This equation
\ntion
a T
to
is\n
1 +
(13-65)\n
but also applies to
section,
both sides of
Dividing
13-2).
we
\nrearranging,
=
y
for
derived
was
Prob.
(see
last section.
used in the
in the analysis of filters,
the nature of the network is
the network passes or stops
known,
whether
knowing
by
additional
one
are
frequencies
established
\nreadily
\nat
(Zi/4Z2)
network
impedances rather than in
for analyzing a
impedances
of importance
is evidently
cutoff
once
\nsince
short-circuit
and
open-
factor
\nThe
the
of
terms
in
\ndirectly
alternate
method
series and shunt
an
offers
equation
ax
the equation
and
2
by
sec\302\254
have\n
cosh
\342\200\224
1
7
z
_
1\n
(13-66)\n
\n2\n
By
an
for hyperbolic
identity
cosh 7
this
Expanding
\n7
functions,\n
- 1_ . ,
a
^
t,
= sinh
Z\\
two
\nare
and
in
possibilities
Z2
the
have
\npossibilities
the
2\n
4Z2\n
(13-67)\n
of
part
imaginary
same
\342\200\224jXi
or
Zi
\342\200\224jXi
\302\261jXi
and
Z2
=
sin
^
^
=
(13-68)\n
yj^
+jXz,
so that there
expression depending on
or opposite signs. We will consider
separately.\n
=
=
cosh
radical
(1) When Zi and Z2 have
=
| + j
cos
^
Now for reactive networks,
\nZ2
Zx\n
the real and
in terms of
equation
2\n 7\n
gives\n
sinh
\nZ\\
4Z2\n
and
oppositesigns,that is,
Z2
=
-\\-jXi,
then
Z\\
=
whether
these
+jXi
(Zi/4Z2) is
negative.\n
and
330\n
last
The
equation
sinh
^
cos
^
cosh
|
Isin
Thus
a =
=
2 sin-1
or\n
0
=
+ir,
and\n
a
=
2 cosh-1
When
sinh
which
\nway
the
\nThus
\nwhen
\nto
Eqs.
a stop
\nfor
equations
is negative
(Zi/4Z2)
13-71
=
^
(13-74)\n
\342\200\224Zi/4Z2\n
Eq. 13-68is
0\n
(13-75)\n
(13-76)\n
.\n
f
and
and\n
real,
Uz2\n
is
only
be 0,
\302\261
2?r,
one
etc.
is\n
=
a =
two
\302\2613t,
(13-73)\n
there
0
These
\302\261tt,
. . .\n
\302\26157r,
can never equal zero, so that
can be satisfied: 0 must
equations
these
=
0
(13-72)\n
cosh (a/2)
solution
or
ZX/4Z2\n
\\/
c\302\260s
5
in
a = 0
the radical of
^
case
<13-70)\n
yji\302\247;
V-
sin
cosh
this
=
+3ir,
is 'positive,
(Zi/4Z2)
(13-69)\n
(13-71)\n
0
In
factor
0\n
and\n
(2)
a positive
= 0
in two ways:
be satisfied
may
either\n
equation
\netc.
13\n
Then\n
\n(\342\200\224Zi/4Z2).
This
in terms of
be interpreted
may
Chap.
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-72
0,
\302\2612x,
\302\2614x,
...
(13-77)\n
2 sinh-1 \\/Zi/4Z2
apply
there
(13-78)\n
when (Zi/4Z2)
is positive. However,
are the two possibilities corresponding
for a pass
band, and to
Eqs. 13-73
13-74
and
band.\n
at different conclusionsfor positive for negative
\nvalues
of (Zi/4Z2),
zero value for (Zi/4Z2) is evidently a
of
\ndivision
for the various
forms of equations for a and 0.
(Zi/4Z2)
there
are two possibilities.
The point of division for these
negative
is given
The
1.
different
equations
by Eq. 13-64 as Zi/4Z2 =
are summarized in the following table. The
for
\npossibilities
values
find the most frequent application, because
of (Zi/4Z2)
\nnegative
of
our
studies
will concern
networks
with opposite signs for
Since we arrive
and
point
When
\nis
\342\200\224
\ntwo
equations
\nmost
\nand
X\\
X2.\n
REACTIVE
TWO-TERMINAL-PAIR
Art. 13-6\n
NETWORKS\n
331\n
Zi/4Z2\n
of
Type
Lower\n
\nband\n
Upper\n
limit\n
-1\n
\n-1
00\n
\nrelated
sinh-1
sin-1
2
\\/Zi/4Z2\n
\\/\342\200\224Zi/4Z2
...\n
+47T,
\302\2612ir,
\n0,
...\n
is designed under the conditionthat Zi
T and standard t section)are
the standard
where
R is a constant both positive
below,
of filters
class
the
by
2
\302\2613tt,
filters\n
for
defined
(as
\302\261ir,
\n0\n
stop\n
important
Z2
equation
real.\n
\nand
=
ZXZ2
This equality
of
and
\ntive
a
is
requires that the
sign.
opposite
\nZobel* used
\n11
\n0\n
-\\/\342\200\224Zi/4Z2
pass\n
Constant-#
\nand
2 cosh-1
stop\n
0\n
\n0\n
An
a\n
limit\n
\342\200\224
00
13-6.
shift
Phase
Attenuation\n
symbol
to be
(13-79)\n
impedances
Zx
In the first
Z2
be
purely
of
of
R
the quantity
because
and
discussion of filters
K in place of the
the letter
preferred
R2
our
equation.
this
reac\302\254
type,
Actually
is dimensionally ohms,
the value of the terminating resistance.Even
R
has
K in the defining equation, filters designedon
\nthough
replaced
\nthe
13-79 are universally designated as constant-K
of Eq.
assumption
of the assumedrelation\302\254
if not the justification
The
\nfilters.
advantages,
between
Zi and
\nship
Z2, will become evident by algebraic simplifica\302\254
and
network structures.\n
later
\ntion,
by simple
we will define
To simplify the equationsderivedin the last section,
R
\nand
\na
out
turns
variable
new
for
the
(Zi/4Z2)
quantity
**
-
since
\nThe
Z\\
and
=
ZXZ2
It
is now
\nvery
*
See
possible
simple
reference
form
in
order
to
make
x2 a positive
opposite signs for constant-#
in Eq. 13-79 are, for reactive networks,\n
Z2 have
expressions
impedance
(13-80>\n
if1
This particular choice of sign is made
\nquantity,
as\n
(\302\261jX1)(+jX2)
=
+M2
=
R2
filters.
(13-81)\n
in
to write the expressions for the imageimpedance
in terms of R and x. Equations 13-61and 13-62\n
at end
of
chapter.\n
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
332\n
Chap. 13\n
become\n
=
ZiT
the
Similarly,
become
\nsection
for
\nvalues
be
\nneed
\nwe
x2,
.
(13-82)\n
(13-83)\n
,
v i
\342\200\224
x\n
time
being.
In
the last
considering only positive
in the table on page331
the stop band, by Eq.
=
2 cosh-1
x
(13-84)\n
and phase of
the attenuation
in form.
Since we are
simple
values for (Zi/4Z2)
only
negative
for
expressions
for
considered
the
13-74,
have\n
a
0 = +?r,
In
X2
R
=
Zi,
-
RVl
the
pass
+3ir, etc.
(13-85)\n
by Eq. 13-72,\n
band,
a = 0
=
0
x
for
x
2 sin-1
a, and 0 against
the
\ngeneralizedplots. In order
Plots of ZiT, Zir,
(13-86)\n
(13-87)\n
in Fig.
shown
are
plots
to
These are
to specific\n
13-25.
be specialized
\t\n
N\n
1\n
^^stop
0
-x\n
i1\n
i!
1
pass
\342\200\236
+i
+x\n
(xj\n
\\;x
I
^'\n
stop
i\n
ib)\n
Tig.
13-26.
Normalized
x as
plots
a function
networks,
only
\nOnce x(o>)
is known, the
in terms
of x =
of frequency
\\/
\302\253
need
\342\200\224Zi/AZ*.\n
be
coordinates may be adjusted
for
determined.
the
special\n
Art.
cases. This
Example
the
in
earlier
\nsidered
for several
will be illustrated
procedure
333\n
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-6\n
con\302\254
examples
chapter.\n
4\n
\nFor
this
\nare
opposite
network,
=
Z\\
in
shown in Fig.
T section
standard
the
Consider
and
jwL
as
sign
Z2 =
The
required).
\342\200\224
13-17,(page324).
that
(note
j/u>C
Z\\
and
Z2
variable x then
normalized
\nbecomes\n
x\n
j\342\200\224Z\\
=
/
w\n
2\n
(13-88)\n
4 /LC\n
sm\n
frequency for thisT
13-55, the cutoff
by Eq.
~
\\4/coC
V4Z7
Now
uL
I
=
section
is\n
2\n
=\n
(13-89)\n
\302\253o
y/LC\n
x becomes\n
that
so
W\n
\342\200\224\n
=
x
(13-90)\n
COO\n
we have the
T section,
this
for
Then
=
0
8 = 2
and
following
information:\n
sin-1
\342\200\224
0
^
w
^
w0\n
(13-91)\n
Wo\n
a
=
2 cosh-1
=
8
\342\200\224;
tt,
^
w
wo\n
(13-92)\n
Wo\n
(13-93)\n
=\n
R
The
for
plots
a
\nuation
=
x replaced
with
\nnetwork,
is
and
ZiT, a,
usually
/3
by
(13-94)\n
in
given
discussed
As
\302\253/a>0.
13-25
Fig.
apply
directly to this
previously,
the
atten\302\254
using the relationship,
computed
in decibels
consider
the T network
a*\n
8.686anepets.\n
Example
For
\nThe
5\n
this
example,
of
variation
x with
\302\253
is
found
ll/uC
x\n
_
\342\200\234
\\ ~4wZ7
where
\nsection
the
has
cutoff
an
frequency
inverse
wo is
relationship
of Fig. 13-19(page325).
as\n
_
I
\342\200\234
\\4^LC
1
_~
Wo\n
(13-95)\n
^\n
identified from Fig. 13-20.This T
to that of Example 1. Theequations\n
NETWORKS\n
REACTIVE\n
TWO-TERMINAL-PAIR
334\n
Chap.
13\n
become\n
=
a
=
a
0 =
and
0
2 cosh-1
\342\200\224
and
\342\200\2242
sin-1
3\n
\342\200\224>\n
All\n 3\n \302\251\n
(13-96)\n
<0\n
=
0
0
\342\200\224x,\n
w
^
^
0)0\n
(13-97)\n
Cd\n
0)0
1\n
=
(13-98)\n
7=\n
2 VLC\n
R =
These
the
the
\ninverted,
\nThe
becoming
origin
for
plots
resulting
ZiT, a,
13-26.
Fig.
filter. In order to
a high-pass
evidently
identify
normalized
of
plots
equations
of
\nuse
(13-99)\n
VL/C\n
make
Fig. 13-25, only the x axis need
infinity and infinity becoming the origin.
and 0 are shown in Fig.
be
13-26.\n
of high-pass
Characteristics
filter.\n
6\n
Example
of Fig. 13-27,
x section
the
For
x for
\nvariable
this network
=
Z\\
and
j<aL
Z2 =
\342\200\224j/oaC.
The
becomes\n
or\n
o\302\273\n
(13-100)\n
(do\n
4/LC\n
since
cutoff
the
condition,
Zi/4Z2
=
\342\200\224
1
with
identical
lo\n
L\n
\ning
that
of
shift
=r^C/2
^C/2\n
\ncal
13-27.
\n13-25.
x
network
section
3.\n
\nExample
being
\nthe
impedance
\nare
identical.\n
this
those
that
given
in Fig.
characteristics
There
of\n
\nference,
\nance
\nent,
with
the
however.
variation
This
<d0.
equation
13-90,
Eq.
are
x network
of the
are
phase\n
identi\302\254
T network of
in
given
Fig.
is one important
The
with
is\n
indicat\302\254
and
attenuation
\nExample1, and
lo\342\200\224\n
Kg.\n
defines
image
dif\302\254
imped\302\254
o)/o>o is
differ\302\254
13-25(5). For these two different
are quite
different even though a and 0
networks,
335\n
7\n
Example
this
For
was
network
This
13-21.
network
this
for
the network
of Example 3
shown to be a band-pass filter.
consider
example
\nfunctions
=
Z\\
shown in
Fig.\n
The reactance
are\n
j\342\200\224
<j\302\273
and
Z2
O)2
that\n
so
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-6\n
Art.
=
\342\200\234
-j
ur
2
(13-101)\n
T
\342\200\224
1\n
\342\200\224
1\n
X\n
(13-102)\n
2o>\n
the frequency w for the filter to the variable
\nx of the
standard
phase shift, and image impedance char\302\254
attenuation,
we perform a frequency transformation.
\nacteristics.
this
By
equation,
\nThe
same
bandcan
be used for any filter\342\200\224band-pass,
technique
a and
or
combination
of
such
Plots
of
\nelimination,
any
specifications.
the
filter
of this example as shown in Fig. 13-28.\n
\n0 for
band-pass
This
last
13-28.
Attenuation
networks
of our
Fig.
The
\nsame
relates
equation
\naccomplish
multiple-pass
\nconstant-#
type
have been very simple, but the
more
networks
necessary to
complicated
or multiple-stop
All networks of the
bands.
have
must
characteristics for a band-passfilter.\n
four examples
to
apply
concepts
and phase
Z\\
and
obeying
the
inverse
relationship\n
Z, = ~
(13-103)\n
Z2\n
and
\nfamiliar
\nFoster
limits
restriction
this
two
with
forms
of
forms
networks.
possible forms for Z\\
of networks
that obey this
the
The
and
Z%.
We
are
relationship,the
relationship\n
(13-104)\n
is
satisfied
for
if,
_
+
(\302\2562
D2
(S2
_
y 2
and\n
as
will
factors
multiplying
+ Si2) .
.
S22). .
+
s(s2
\342\200\242\n
\342\200\242
\302\256i2)\342\200\242
13\n
.\n
(13-106)\n
.\n
also satisfy
Eq. 104.) Hence if
No. 1 type of network and Y2
a Foster
(13-105)\n
+ s22)...\n
s(s2
\nrealized
Chap.
example,\n
7
Zl~R
(Other
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
336\n
as
realized
is
is
Z\\
a Foster\n
Fig. 13-29. Foster networkforms.\n
2
No.
\ntypical
\nit
admittance
term
in
filter will be
is
expansion of
given
the resulting
network,
admittance
the
(\302\253)
Eq.
12-64;
s/L\342\200\236\n
=\n
(S2
+
(13-107)\n
1/LnCn)\n
for a series LC network. Similarly,typical terms
\nexpansion of Z(s) will have a form givenby Eq.
+
(s2
Since by Eq.
\nabove
typical
terms of the
This
l/LmCm)
equality
similar
(13-108)\n
F(s) and
=
Z(s)
the
containing
R2Y2(s)\n
(13-109)\n
product formsfor F(s) and
_
=
is possible
only if, term
network
\342\200\224
LnCn
configurations,
by
Z(s),
N2(s)\n
D2
R2\n
+
(s2 + l/L
LmCm
for
impedance
as\n
have\n
\nN 1(8)
+
12-60,
the
l/LmCm)\n
13-104 these equationsfor
terms must be equal, we
Z1(s)
or, in
of
s/C\342\200\236\n
=\n
Z(s)
or,
by
Y2
A
is\n
Y
(s2
constant-#.
(13-110)\n
l/LnCn)\n
term,\n
\342\200\224\n
in the two
(13-111)\n
Foster
forms,\n
(13-112)\n
Ait.
13-6
for
the
illustration of these conclusions,
in Fig. 13-30. In order for this network
to
shown
network
the
is necessary
it
constant-#,
337\n
As an
network.
constant-#
\nconsider
\nbe
REACTIVE NETWORKS
TWO-TERMINAL-PAIR
that\n
=
L1C1
L2C2
=
(13-113)\n
wo2\n
This
two
We
=
\nstant-iC
\nTo
illustrate,
satisfying the requirement
Note that Zx is the driving-point
of a ladder element.\n
impedance
of
for
Z\\
and
shown
Z2
need
not
recipro\302\254
imped\302\254
Foster forms.
be of the
the requirement
in Fig. 13-31 satisfy
R2.\n
of a
networks
provided
filters.\n
= R*.
the
is
structures
define
will
\nconnection
\nfilter
Z\\
used
ladder
ZxZt
\nthat
that
and
networks
The
\nThe
ZXZ2
impedances,
\nance
design of constant-#
networks
Ladder
13-31.
Fig.
\ncal
in the
is useful
concept
an
a composite
number
filter as a
of standard T
can be connected in
image
consider
impedance
a standard
match
filter made up
or standard
tandem to
ir
form
of
the
sections.
a
cascade
Concomposite
terminal pair.
13-32
shows
Figure
exists at each
T section.
a\n
terminal
\neach
\nmatch
n such
pair
in
the
that
(such
Chap.
13\n
connected in tandem. Note that at
there exists an image
structure
the composite
we have derived hold). The attenua\302\254
equations
of
representation
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
338\n
sections
sections
n
of
\ntion
=
a
is\n
cosh-1
2n
\342\200\224
(13-114)\n
0>o\n
from
13-33. Attenuation in
Fig.
of
\nness
attenuation
\ning
\nincreasedby increasingthe
\nif
a
is
cutoff
\342\200\234sharp\342\200\235
\nstant-#
composite
the
sections
filter
shown in
effective\302\254
in
provid\302\254
stop band is
in tandem.
However,
a large number of T
the
filter. The limitations of
con\302\254
are:\n
filters
composite
of T sections is required to attainhigh
atten\302\254
the stop
band near the cutoff frequency. This large
elements
make
the cost of the compositefilter
may
number
A large
(1)
in the
T
composite
in
specifications,
by
required
used
be
must
\nsections
of
number
the
n are
Evidently
com\302\254
filters.\n
\nposite
13-33.
\nFig.
of
values
several
\nfor
curves
Attenuation
13-92.
Eq.
in
\nuation
of
\nnumber
\nprohibitive.\n
The
(2)
The
for
cutoff
\nnear
\nO.
m-derived
need
The
filter\n
led
o
The
1923.
in
VW\n
attenuation in the stop
to development
of the m-derived filter
filters considered in the last section
section with high
a filter
frequency
Zobel
J.
R introduces
resistance
\nstant
13-7.
exists.
impedance
\nminating
in
shown
impedance
\nimage
be terminated in the required
because no such ter\302\254
Fig. 13-25(a),
the filter with a con\302\254
Terminating
mismatch
at all but one frequency.\n
cannot
network
composite
were
\342\200\224V\\A/
zy
Z[/2\n
band
\302\260\n
2\n
a\n
AAA/\342\200\224\302\260\n
Zi/2\n
z,T\n
\"*~Zix\n
Z$
by
o\n
la)\n
Fig.
with the same imageimpedances:
(a)
Networks
13-34.
new;
(b)\n
old.\n
restricted
very
\nuct
Z\\Z\\
it seems
\nmitted,
\nwill
be
of filters satisfying the requirement that theprod\302\254
a constant.
If other combinations of elements are per\302\254
class
give
\napproached
the
possible that some arrangement of elements
near cutoff frequency. Zobel
high attenuation
with
one specification
for the new network.\n
intuitively
required
this
problem
REACTIVE
TWO-TERMINAL-PAIR
13-7\n
Art.
the
that
Anticipating
filter
\nconstant-#
he
sections,
might be used in tandemwith
that the image impedance
section
resulting
339\n
NETWORKS\n
specified
of
same as the standard T or standardir section.
in Fig. 13-34. The new impedance
terms
is illustrated
\nThis requirement
terms
constant-#
for the
and
\nare
Z\\
Z2'; the impedance
designated
the
and
We
have
are
\nsections
Z2.
Zi
requirement that the
designated
and
\nZ\\
and
Zx,
Z%,
Z\\,
we need somefurther
impedances;
image
by the
related
were
Zi
relate
to
order
in
\npremise
same
the
have
networks
\ntwo
the
be
network
new
\nthe
Z2'.
Zobel
equation\n
Zi' = mZi
where
should
We
\nmake.
of
\nform
properties
impedance.
image
With
Z\\
let
fixed,
Z\302\2611
4
\nfor
\nin
schematic
Z2.
(13-116)\n
equation
and solving
-
z- +
(tjt)
<mi7>\n
new m-derived network
of the
is shown
may be given
that
by assuming
derivation
same
the
\nthe
standard
the
\nmaking
as
image
impedance
tt network
and further
assumption
Z2'
image
\nequation
=
impedance
and
must
structure
network
new
\nhave
The
this
into
vnZx
representation
tt section
the
\nany
and
13-35.\n
Fig.
A similar
\nfor
^r4\n + Zxz2
\342\200\224
Z\\
z*'
The
Zi
results\n
there
Z2',
=
Z/Z2'
+
condition
the
Substituting
beginning.\n
happens to Z2' in termsof
simple
see what
us
impedances,\n
image
\nEquating
this
from
follow
that
\nresults
to
may seem to be an unusualassumption
that our assumption
might instead relate to
new
the
of
filters, or perhaps the desired
to anticipate the surprising
It is difficult
expect
attenuation
\nthe
(13-115)\n
This
a constant.
m,is
that
assumed
13-118,
that\n
\342\200\224
13-35.
Fig.
(13-118)\n
m-derived
T
section.\n
section is given by Eq. 13-12.
Using
the impedance Z\\ is found to have the value\n
of the ir
Zi'
=
*\342\200\224
\342\200\224:
mZ
(13-119)\n
\t\n
1
4m
i
1
\342\200\236\n
\342\200\224
m2\n
this
z2\n
of a parallelcombination the
is the impedance
which
of
,1
13\n
impedances\n
,Z2
(13-120)\n
m2\n
x section is shown
x section.\n
for the standard
schematic
Chap.
\342\200\224
m-derived
resulting
\nthe
4m
and
mZi
The
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
340
in Fig. 13-36,
with
together
(4m/l-m2)Z2\n
rVWi\n
(a)
13-36.
Fig.
(6)\n
of (a)
Comparison
standard section, and (b) m-derived\n
section.\n
a
\nFirst,
attenuation
the
\ncompute
\nshown
that
Suppose
example.
in
structures, our next task is
we have attained our objective.
of the results thus far can be seenina
we select a low-pass filter of the
m-derived
to see that
interpretation
physical
\nspecific
the
have
we
that
Now
type
13-27 for the T andx
and in Fig.
13-17
Fig.
to
respec-\n
sections,
mL\n
o
rflpr-\n
nm'
mL/2\n
\t\n
mL/2\n
HH\n
_
mC/2^-\n
l-m2 r
^
_
^\n \342\200\224
\nAm
mC/2\n
o \t\n
m-derived
13-37.
Fig.
o\n
(6)\n
(a)\n
low-pass
filter sections: (a)
T section;
(b)
r\n
section.\n
The
tively.
\nFor
\ntion
\nFig.
these
at
13-37
\nseries
LC
\nseries
LC
\nrent
\nof
by-passes
infinite
Fig. 13-37.
can possibly give infinite
sections
are shown in
attenua\302\254
what
The m-derived T section shownin
frequency?
particular
will have
infinite
attenuation
(or no transmission) when the
circuit
is in resonance.
Under this resonance condition,the
is the equivalent
of a short circuit, such that allcur\302\254
circuit
networks,
a
m-derived
equivalent
the
attenuation
let
us ask,
load.
This
designated
resonant
as to..
1
=
\302\260\302\260
\\/[(l
\342\200\224
frequency is thus
It has the value\n
\t\n
m2)/4m]L(mC)\n
a frequency
(13-121)\n
Art.
341\n
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-7\n
top\n
or\n
since
the
cutoff
frequency
At
this
same
frequency,
\nin
resonance
parallel
\nthis
and
\nsponds
to
\nthat
one
\nat
\nin
be
can
the
of
terms
Under
impedance.
the series
pass through
ele\302\254
be no current in the output. Thiscorre\302\254
for the t section. We
now
to see
begin
element
a new
transmission
frequency.\n
equations
\nsections
be
will
T, and Z2 in the t has introduced
as to prevent
in such a way
or in parallel
particular
The
can
current
13-37(b)
the
Z\\ in
series
in
\neither
attenuation
infinite
altering
will
there
hence
infinite
having
no
condition,
by Eq. 13-55.\n
2/\\/LC
circuit of Fig.
parallel
(antiresonance)
open-circuit
\nments,
the
m*\n
value
the
has
o>0
(13-122)\n
~
y/l
and phase shift in the
filter
found by computing the factor x definedby Eq. 13-80
functions
new reactance
Z\\ and ZJ.
for attenuation
m-derived
Thus\n
\342\200\224Z\\
_
~
_
~
4Zi
\342\200\224mZ1
4[Zi(l - m2)/4m+
\t\n
(13-123)\n
Zj/m]\n
\342\200\224
to2
1
in
the
used
\nfactor
is substituted
TO2
-|-
\t\n
(13-124)\n
4Z2/Z1\n
we recognizethat
In this expression,
\nfactor
\342\200\224
into Eq. 13-124,
x\n
there
the
x is
where
this
When
results\n
to2\n
'2\n
\342\200\224 \342\200\224
(1
From this equation,
1/x2,
filter sections.
constant-#
of the
study
\342\200\224
\342\200\224\\Z%/Z\\
we see that
+
wi2)
as
x
(13-125)\n
1/x2\n
approaches
the
value
given
by\n
1\n
x2\n
1
x'2
then
approaches
an
\napproaches
be
will
\ntion
infinity
value.
infinite
such that cosh-1
The value of x
such
designated
(13-126)\n
\342\200\224
TO2\n
this
Eq.
definition,
(13-127)\n
13-125 may be
\342\200\224
1/x.2
In
this
\nfilters,
=
x
a
equation,
= a normalized
x'
attenua\302\254
that\n
2\n
x\n 00\n
With
x', the attenuation,also
causing infinite
+
written\n
(13-128)\n
l/x2\n
normalized frequency relating to constant-#
derived for TO-derivedfilters as
frequency
a\n
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
342\n
Chap.
13\n
value of x causing x' to be infinite,and m = a
\nconstant
for the m-derived filter section. Inspection of this equation
while for x
x is less than x\302\253, then
\nshows
that
when
x'2 is negative,
In order to associate these conditions
than
\nlarger
xM, x'2 is positive.
or stop
band and the equation for attenuation, we will
\nwith
band
pass
= the
of x,
function
on page 331. Comparing this table with
the
different
summarized
for the sign of x2, we reach the conclusions
table
the
to
\nrefer
\npossibilities
\nbelow.\n
of
Limits
of
Sign
band
**
\nx'2
positive
0
g
i\342\200\230
^
positive
1
^
x2 ^
1
0
a = 2 cosh-1x'
a
\302\253ostop
=
2 sinh-1
/3
=
0\n
A
+
together
13-38,
Vi
-
(b)
13-74
\n13-77,
13-78\n
m2\n
is
in
shown
Fig.\n
constant-A
filter\n
of the m-derived filter: (a)
Characteristics
13-38.
Fig.
1/x*)
equations
(for positive values of x)
with the same characteristics for the
these
of
plot
\n13-73,
1\n
\"
V(-l/x\302\253*
\n13-71, 13-72
x'\n
+*-\n
m
,
\nequations
2 sin-1
x'
as
Derived
0\n
a =
pass
stop
x2 ^
^
x\342\200\2362
negative
Phase shift
Attenuation
Type of
m-derived;\n
constant-A.\n
of
the
plot illustrates the high attenuation near the edge
of the m-derived filter. Comparing the plot with
feature
\nstop-band
\nthat
for
the
constant-A
filter, it is seen that the attenuation for the
a minimum
\nm-derived
filter
value
for large values of x,
approaches
section.
The
\nwhereas
the
constant-A
Each
type
\nlarge
x.
\nSince
sections
both
13-38
Figure
section
\nequations,
attenuation
a large value for
sections has advantages and disadvantages.
the same image impedance, there is a pos\302\254
have
of both filter types to get both the high
a comMnation
at
the edge of the stop band and
large
=
shows
that when x
xK,the phaseshift
changes
this
approaches
of filter
near
\nattenuation
\nfilter
of
use
of
\nsibility
filter
abruptly
is caused
from x to
in
from
x.\n
m-derived
the
0 degrees.In
by the sign of x'2 changing
of
values
of
terms
positive
the
to\n
Art.
\nAs
from
\nchanges
no
no
gives
to
\ncorresponds
In
with the samekind
is possible
in a network
the \342\200\234effective\342\200\235 inductive
arms,
This resonant
shift.
phase
all
to inductive).
(or from capacitive
positive
shift
in
\nonance
to
LC circuit reactance
the
resonance,
through
negative
phase
\nreactance
seen
Fig.
increases
frequency
\nSince
in
illustrated
filter
\npass
for this phenomenacanbe
in the
low13-37 (a) in terms of the seriesLC circuit.
A physical reason
negative.
343\n
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-7\n
above
network
of
res\302\254
frequency, incidently,
x\342\200\236.\n
m-derived
designing
will arise,
the question
filters,
not
why
make
m = 0 to make the filterhave
an
extremely
sharp
\ncutoff
in the
to
the
atten\302\254
band?
More
what
stop
generally,
happens
\nuation
characteristics
as m is varied?
To answer this question, we will
the
attenuation
of the m-derived filter at large values of
\ninvestigate
\nx. From
the
for attenuation
equation
given in the table on page342,
=
\ni.
1 by
making
for
attenuation
\nthe
a value
x approaches
large
=
atim
2 sinh-1
-\n
the
from
cosh2
identity
&lim\n
=
cosh2\n
Now\n
1
sinh
(aum/2)
cosh
(aum/2)
\ning
zero,
\naiim
also
\nas
this
in
of
the
\nnitude
\nvalues
of
\nsharp
cutoff
\nall
of
x
frequency.
in
illustrated
\nare
\nvalues
of
\nsmaller
results
m
\342\200\224
&iim
2
=
1
m/V
=
l/y/1
2 tanh-1
-
m2\n
-
m2\n
(13-130)\n
\342\200\224
1
m2
m2\n
(13-131)\n
m\n
(13-132)\n
asm becomes
small,
approach\302\254
from
Also,
paid
price
for
attenuation
being
some
These
conclusions
Fig.
13-39
for
func\302\254
\342\200\224>
0.\n
for avoiding
in
finite
13-39.
Fig.
\nation
for two
\nfilters
\n(x
reason
elements
as
+\n
m*\n
* Another
\nfilter
the
reduced
is
frequencies,
\ntion
1
to become small
small
(for any value of
turn
reduces
the mag\302\254
attenuation
all
for
Thus
x.
h
1,\n
is seen
x'
becomes
and
\nx),
small.
becomes
=
m-\n
=
it is seen that
for large x,
attenuation
the
13-125,
m
an
m\n
aKm
1
last equation,
the
sinh2
sinh:\n
a/\302\273m
From
\342\200\224
&lim\n
+
Hence\n
\nEq.
aKm
(13-129)\n
-
Vi
or,
as\n
given
atim
=
as
of attenu\302\254
m-derived
closer to cutoff
of
a:* moves
1).\n
small values of
m
is that
at
zw,
this
attenuation
Variation
characteristics
finite
finite
dissipation
attenuation
in the
being
this
From
is
\ntion
with
\nm-derived
x.
The
T
and
determined
\nare
determine
to
seen
\nfilter
m also determines element values in the
Since in these structures, elementvalues
it follows
that m
factor
(1 \342\200\224
ms),
multiplying
a
by
in value; that
<
0
(Note: This limitationapplies
only
\nin
of
terms
next
\nour
to
(13-133)\n
ladder
the
for
Values
structures.
of
char\302\254
phase
m
impedance
image
subject
1
m <
structures to give linear
attached to the value of
significance
of the m-derived sections. This will be
is another
There
\nacteristics.)
is,\n
used in lattice
1 are
than
larger
13\n
constant
x sections.
\ncannot exceed unity
\nm
Chap.
value of the constant m of the m-derived
the nature of the variation of attenua\302\254
the
discussion,
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
344\n
study.\n
Image impedance of m-derivedhalf (or L) sections\n
and
the
m-derived
T section is shown in Fig. 13-35,
The m-derived
in Fig.
13-36. These sections were found under the assumption
\nx section
are the same as the constant-^ filter sec\302\254
the
\nthat
impedances
image
an
If the
T and x sections are divided into half (or L) sections,
\ntions.
13-8.
characteristic is found. This result
of any
bonus;
certainly it is not a consequence
the
m-derived
filter. A different image
\nunexpected
image
\nmust
as
regard
\nments
a
of
made
impedance
we
require\302\254
impedance\n
(2m/l-m2)Z2\n
Z,\n
filter half-sections: (a)
m-derived
13-40.
vig.
m-derived
behavior
at
\nsection)
might
the
the
section
L
13-40.
\nignated
The
ZiTm
half
T;
(b)\n
r.\n
T or x section
of the imageimpedance
it
was
found that
314. There
image
terminal
pair was ZiT, and at the
pair
m-derived
for the
filter are shownin
terminals
of a divided
be expected
from the discussion
other
on
page
at one terminal
\nimpedance
\nwas Ziw. The
half-sections
\nof
half
m-derived
image
and
impedances
Ziwm.
They
(the
half
the
other
Fig.\n
at the \342\200\234back door\342\200\235 terminals
may be determined by i\302\253dng
are
Eq.
des\302\254
13-6.\n
TWO-TERMINAL-PAIR REACTIVE
Art. 13-8
we
T section,
the
For
345\n
NETWORKS\n
have\n
\342\200\224
ZiTm
VZloZu\n
\342\200\224
rri1
mZi
+\n
Zi
\342\200\236
i
Zl
\n~~2~ \\~~2m~
\342\200\224
+
Zi\\\n
m
J\n
l_^m>
+
(m
Zi +
2m\n
2
m
2
^
\\
Zi)\n
(13-134)\n
z\n
m\n
L\\2\n
1
+
\nSince IB2
- m>)
(i
= Z1Z2,
1
4Z2J
4]\n
(13-135)\n
4\342\200\234
Zi/4Z2
the image impedance
\342\200\224
\342\200\234
\"
ZiTm
(1
[1
=
same
The
defined for the constant-#
be used to find Z\302\273,m for the filter
may
procedure
(13-136)\n
**\n
factor
the
is
Zi/4Z2
m2)x2]\n
filter.\n
of
Fig.\n
thus\n
13-40(b);
Zi* m
\342\200\224
\342\200\224
-
Vi
where x2
becomes\n
\342\200\224
VZ10Z1.\n
1\n
1\n
\342\226\240V\n
2/mZi
-+- (1
+\n
\342\200\224
2/mZi
m2)/2mZ2\n
+
(1
\342\200\224
m2)/2mZ2
]\n
\n(13-137)\n
(1 -
[1 +
+
Zi2/4\n
(13-138)\n
m2)Z!/4Z2]\n
\342\200\224
so
that\n
z2
-\\/l
\342\200\224
iB\n
Z.-Tm
[1
\342\200\224
(1
\t\n
(13-139)\n
\342\200\224
m2)x2]\n
as a function of x for
plotted
=
0.6 givesan
values
of m. The
impedance
plot for m
m
of
\nconstant
within
4% over 90% of the pass band. Other
this. This image impedance variation is
than
\ngive variations
greater
constant
the constant-#
more
than
image impedance functions
=
=
note
that
for m
0, Z,xm Zir by Eq. 13-83, that for m = 1,
=
In
the
other
13-82 and vice versa for
ZiT by Eq.
words,
with
\nm-derived filter sections reduce to constant-# filter
- 1. The new
m = 0.6
or ZiTm for
image impedance function
a constant,
and so an image impedance
nearly
approximates
with
a constant
resistor is a reasonable
terminating
This
is a very
\ntion.
important
advantage for the m-derived half
In Fig.
13-41, Z,Tm and
Ziwm
are
shown
\nseveral
image
values
\nmuch
and
\nWe
\nZi*m
ZiTm-
sections
\nm
Ziwm
\nmore
\nmatch
approxima\302\254
sec-\n
346\n
With
tion.
Composite
The
the
of the m-derived
use of such
filter
estab\302\254
filters in combination
with
filters\n
summarizes
table
following
the advantages
and disadvantages
filters.\n
m-derived
and
constant-#
\nof
properties
13\n
filters.\n
\nconstant-#
13-9.
to
turn
next
we
\nlished,
remarkable
these
Chap.
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
m-derived\n
Constant-#\n
Attenuation near cutoff (x
at
Attenuation
large
= 1):
small\n
x:
large
\nsmall
large\n
\nmore nearly
Image
impedance
in pass band:
not
constant\n
\nconstant; depends
\non
\nm
m;
=
best
when
0.6.\n
the inverse attenuation characteristics of the two
a combination
of the two types wouldhave
that
and
suggests
\ntypes
either
Such a filter is called a composite
over
alone.
type
\nadvantages
the
filter
which forms the nucleus about which
The
constant-#
\nfilter.
is known as the prototype.Them-derived
filter
is designed
\ncomposite
the same image impedance as the prototype
will
have
\nfilter
sections
values
found in terms of the constant-# section
element
\nand
will
have
illustrates
table
The
\nelement
values.\n
In
designing
\tthere
must
(1)
\nfilter
\nbe
\ndesired.
section,
so
selected
The
must be kept in
impedance match at the terminals
attenuation
properties of each section
a composite
be an image
filter, two factors
mind:\n
of
each
must
(2) the
the composite
that
attenuation
characteristic is that
and attenuation
of the various net-\n
properties
impedance
and
work
are
configurations
\nThese
are
blocks
building
\nSeveral
REACTIVE
TWO-TERMINAL-PAIR
13-9\n
Art.
will
examples
o\342\200\224VW
summarized
basis
the
347\n
NETWORKS\n
13-42 and Fig. 13-43.
of filter design on the imagebasis.
in Fig.
illustrate.\n
AAAr\n
A/VV\342\200\224\302\260\n
ZJ2\n
\nZx/2\n
z1\n
Z\342\200\234\342\226\272<
\342\200\242Zit
%\n
-o
2Za
2Z2
\342\200\242\n
o-\n
(c)\n
AAA/
o\342\200\224vw\n
&,r-\n
0\n
Zx! 2\n
ZJ2\n
\342\200\230
Z i\302\253Zjr
2Z2\n
*
^
2Z2\n
o\342\200\224\n
(6)\n
'if am\n
13-42.
Fig.
filter
\nstant-K
\nfilter
Example
sections;
of networks: (a) conimpedance
properties
half (L) sections; (c) m-derived
(b) constant-if
sections;
half (L) sections.\n
(d) m-derived
Image
8\n
prototype constant-/^ filter and one
filter
will
\nm-derived
be used
in a cascade connection. This is to be a
For
this
filter.
we have shown in Eq. 13-90that x =
\nlow-pass
case,
sec\302\254
The
\nco/o>o.
prototype
(a T section in this case) and the m-derived
in Fig. 13-44. These two sections will make up the com\302\254
\ntion are
shown
The m-derived
of filter.
section
section is first split into half
\nposite
to realize
the best impedance properties. An arrange\302\254
\nsections
in order
sections
such that there is an imagematch
at
each\n
of the three
\nment
For
the
first
example,
one
is shown
input
\nat
the
REACTIVE
TWO-TERMINAL-PAIR
348\n
load,
Pig.
in Fig. 13-44. The
the computed
so that
13-48.
approximately\n
of
function
correct.
\nThe
The
attenuation
is
found
by adding
at
as shown in
the figure.
impedance properties as a\n
x.\n
arrangement of sections.
the separate attenuations,\n
is ZiTm for this
impedance
input
13\n
Chap.
match, however, is only approximate
properties are only
and image
phase
Attenuation,
NETWORKS\n
\342\200\224
aje
+
=
ft
+ ft*
am
(13-140)\n
Similarly,\n
ft
(13-141)\n
filter
are
The input impedance and attenuation the composite
filter or the m-derived
the constant-#
\nrior
to
those
of either
series
The
inductors
of the prototype and the m-derived
\nseparately.
are
when actually constructing the
section
together
lumped
of
supe\302\254
filter
\nhalf
filter.\n
REACTIVE
rWO-TERMINAL-PAIR
13-9\n
Aft.
'
QjflP'
'
1
1 o\n
o-\n
m-derived\n
prototype\n
section\n
-o\n
349\n
/Tnnp\342\200\224o\n
constant-*\n
o\n
NETWORKS\n
o-\n
o\n
-\n
nm^>-4
-o-\n
-o-\n
r\n
JiTm\"\n
-ZL\niT\n
JiT\n
HT\n
-o
-O-\n
R-\n
\342\200\230iTm\n
JiT\n
-o
o\n
m-derived
m-derived
\nhalf
section\n
\nnear
cutoff
\ndropped
would
frequency
too
to
property of the low-passfilter of
either because the attenuation
not be satisfactory,
was
not sufficiently
sharp or because the at curve
a value before beginning to rise again. In thiscase,\n
the
cases,
8
\nExample
low
\302\251\342\226\240\n
attenuation
\342\200\224i
o\n
o\n
o\n
o\n
o
13-45.
(a)
o\n
\t\n
1
\342\226\2400\n
lb)\n
(C)\n
Fig.
Load\n
characteristics.\n
low-pass filter
Composite
section\n
9\n
some
In
\nhalf
Prototype\n
13-44.
Fig.
Example
o-\n
Constant-AT
prototype
and (b) m-derivedlow-pass\n
sections.\n
match
can be used as longas an image
\nis realized
at each
terminal
of the cascade connection of sections.
input
\nTo
that
a
ir section is selected for the prototypeof
illustrate,
suppose
\nthe low-pass
filter
and a decision is made to use two m-derived
sections,
\none
with
m = 0.6 and one with m = 0.3. Theprototype
and
m-derived
\nsection are shown in Fig. 13-45. Sincethe m = 0.6 section
has superior\n
two
or more
m-derived
sections
at
use
\ntions
for
\nhalf
sections
in
an
attenuation
\nthe total
image
The
13-46.
Fig.
it will
be divided into
and
Zirm
=
at*
+ am(0.3) +
with
higher
13-46.
attenuation
at
the
section
filter
resulting
Note the improvement
all frequencies
(13-142)\n
in the
attenuation\n
and sharper cutoff in
the
band.\n
\nstop
In
the
practice,
13-46
combined
we
\tFirst,
\nthe
to
should
Fig.\n
may
and
w-derived
be
us
now
formulate
specifications to be required for
a function
of frequency.
The attenuation
with some combination
met
of constant-A
filters
to the equation\n
according
decide
as
attenuation
\nrequirement
\nfilters
first two examples, let
use on any filter sections.\n
of the
conclusions
procedure
design
parallel capacitors of the schematicof
into equivalent
capacitors.\n
various
10\n
the
From
(1)
be
would
Example
\na
in Fig.
am(0.6)
of
properties
shown
sec\302\254
filter. An arrangement of sections
match at each terminal pair
impedance
and
input impedance in this case is
The impedanceandattenuation
also
two half
13\n
is\n
os*
\nare
Chap.
ends of the
the
giving
shown
\nis
characteristics,
impedance
image
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
350\n
on the
at = Aa
k
+\n
(13-143)\n
/-1\n
Art.
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-10\n
351\n
the number of constant-# sections,Bj is the number
m
a specific
for
\nof sections
m, and n is the numberof different
solution
to
the
and
num\302\254
is usually no unique
\nused. There
type
often
must
be used to
and
\nber
of sections
cut-and-try
required,
A is
where
constant-# prototypeand
(2) Select a
and
beginning
values.
find all element values for
values,
the
(4) The type of
ending
m
properties.\n
prototype selected and the
limited
usually
= 0.6
for
filter section with
of the filter. This gives the
section
impedance
image
\noptimum
\nare
element
one m-derived
least
at
Include
\nthe
the
find
sections.\n
\nm-derived
(3)
element
these
\nFrom
the specifications.\n
meet
to
solution
a
\nfind
of
used
sections
considerations.\n
cost
by
number
13-10. The problem of termination\n
\nbe
such
arranged
that
there
Zi
13-47.
Fig.
\ntion
of
\nThere
filter,
are
no
the
\ncutoff
\nthe
frequency.
generator
\nresistance
function
be?
of
Actual
when
is an
resistive
and half sections can easily
image impedance match
at each
point\n
r\n
of image
termination
we come
designed filters.\n
to the beginning or to the termina\302\254
a problem in approximatingan imagematch.
resistors
with the properties of our image impedances,
to change
from resistance to reactance at the
ability
The
best that can be done is to terminate(andmake
What value should this constant
a constant.
impedance)
for image impedances as
the expressions
Let
us review
x. By Eqs. 13-82 and 13-83,\n
the
\nespecially
\na
But
connection.
of
sections
m-derived
and
constant-#
The
we have
ZiT
=
R \\/l
\\/l
~
\342\200\224
x2\n
x2\n
(13-144)\n
(13-145)\n
Eqs. 13-136 and
From
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
352
Chap.
13-139,\n
1
D
\342\200\224
\342\200\224
(1
m2)x2
ZiTm\n
xi\n
\nx
\342\200\224
0.
This
\nstant-A
impedance
Z\302\273-
con-
as\n
prototype
=
R2
given thus far
All results
network
filter
\nthe
of
\nquences
of
\nance
a
\nin
in the
elements
\npationless
(13-147)\n
m2)x2\n
termination
The
\nused.*
-
- (1
ZiTm
m
with
\nZi\342\200\236m
1
=
R when
expressions reduce to
is
a
constant
value
or
good approximation to either
= 0.6 in the pass band. This is the
commonly
has
a
value
resistor
determined
from
the
terminating
these
all
that
VLzl1-\n
r
Zirm\n
(13-146)\n
\342\200\224
\ny/\\
Note
13\n
using
? What
inductors)
impedance,
nonimage
(13-148)\n
have been
on
found
What
termination.
the
finite
with
of (1)
basis
the
dissi\302\254
image match throughout
and (2) an
network
including
elements
+ZiZ2
dissipation
are the consequences
are the
(primarily the
conse\302\254
resist\302\254
of terminating thefilter
R?\n
the computed values of attenuation
of finite dissipation. This
because
correct
only
approximately
in the pass band and finite attenuation
causes
attenuation
\nsipation
A rule of thumb
the
of infinite
attenuation.
so-called
frequencies
the
that
results
of image basis design will be acceptablein
if the
Q of the inductors is 15 or
applications
\nengineering
answer
An
to the second
question requires that we first definethe
when a network is
loss
as the loss resulting
insertion
\nduced
between
a generator
and a load. The insertion loss is defined
to
answer
In
the
first question,
dis\302\254
\nare
\nat
\nstates
most
higher.\n
intro\302\254
\nquantity
by
\nthe
equation\n
eN
=\n
(13-149)\n
u\n
is the
N
where
insertion
loss in nepers,
I%
load
in the
current
the
is
with the
load
current
\nconnecteddirectly to the generator,and /s is
\nnetwork in place. For numerical computationof insertionloss, let
with
same
\nI\\ =
h. We thus assume that the generatorcurrentis
without
the
network
and so have a basis for comparison.This
a unit value for J2 and then
is
best
made
by assuming
the
to find the corresponding value for I\\.
network
through
the
we
the
\nand
\ncomputation
A\n
\ntracing
\342\200\242
A
\nthan
\nsection
better
slightly
R for
as
approximation
the m-derived
the
termination.\n
results
if the
half t section or larger than
terminating resistor is smaller
R
for
the
m-derived
half T
TWO-TERMINAL-PAIR REACTIVE
Art. 13-11
Even
m = 0.6 is
with
13-44
\nFig.
actual
the
to
useful.
be
close
to
\nbeing
simple
effect
The
13-48.
Fig.
and
routine.
attenuation
13-11. Lattice
filters\n
\nA
to
discussion
Another
\nnetworks.
lattice
symmetrical
Fig.
shown
\nCo.,
For
easily
and
Inc.,
New
structure for
attenuation.\n
filters.\n
Fig. 13-49(b). The advantage of the
that
of the bridge is that sectionsconnected
drawn
as lattice structures.\n
short-circuit
impedance
image
impedances.
+
may
be computed
Since\n
Zb
see Terman, Radio Engineers\342\200\231
York,
1943), pp. 228-236.\n
example,
sufficiently\n
in
Za
Zio =
*
are usually
termination on filter
network
the
open-circuit
is only approximately
point
For the symmetricallattice,
\nfrom
13-48.\n
of
on the image basis has the advantage
with
their
Tables showing various networks
are found in handbooks.*\n
characteristics
Lattice
over
more
are
this
13-49.
representation
cascade
\nin
of
network
has related to the ladderstructureof
used in filter design is the lattice.
common
structure
in Fig. 13-49(a), and the \342\200\234bridge cir-\n
is shown
is
cuit\342\200\235 equivalent
\nlattice
of non-image
Design
\ncorresponding
The
shown in Fig.
353\n
a for the
N and attenuation
basis attenuation
image
insertion
loss, the results
the
though
\nequal
loss
insertion
of
comparison
NETWORKS
(13-150)\n
Handbook
(McGraw-Hill
Book
7
and\n
follows
that\n
The
equation
End
Zjf
=
Zi
=
VZuZu
=
2 tanh-1
it is seenthat
cutoff
the
in
lattice
the
for
filter,
when
summary,
made
analysis
frequency
poles
13-6,
is\n
which
(13-154)\n
y/Zi,/Zi0\n
(13-155)\n
for pass-band,
previously
a
the results we
By Eq. 13-43,\n
, 7 _
\ntann
the
sign
critical
A
band.
stop
cutoff
frequency.
functions
is,
(that
only).\n
have
\nphase shift.
If
versa,
with
coincide
Za
Zu
Zy,\\
Zb}
of
zeros
Zb
also
and
Zb
defines
elements
stop-band,
poles and zeros of Zu
with Za replacing Zu and
replacing
of Za coincide with zeros of
or vice
of the
terms
When poles or
a
\npoles or zeros, respectively, of Zb, there is defined
in Za but not in Zb, or vice versa,
\nfrequency
are
reactance
\nThese rules assume that Za and
We
Z\342\200\236\n
(13-153)\n
y/Za/Zb\n
a pass band.
defined
\nthere is
\nLC
of
function
a
also\n
\\ZZioZi,\n
7 = tanh-1
13-39,\n
Eq.
\nIn
is
\342\200\224
Zi
\nholds
(13-152)\n
\\/ZaZb\n
EhS\n
Comparing these equations with Eq.
\nand
Z6\n
image transfer function
for the
y
and
(13-151)\n
+
z0
it
13\n
2ZaZb
=
\nZu
Chap.
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
354\n
of Za
tan
(fi/2)
tanh (a/2) tan
that of
is opposite to
Zb,
and
attenuation
the
tanh (a/2) + j
l+.
g
need to compute
_
\\Za
(13-156)\n
(j8/2)\n
tanh
then
(y/2)
is imaginary,
\nand\n
(13-157)\n
tan!
~yl
=
a
If
Za
and
Zb have
=
by
Eq.
13-45,
y
\nto
infinite
tanh
(13-158)\n
either\n
and
^
(3
=
0\n
(13-159)\n
or\n
tanh
by Eq.
0\n
then
the same sign,
tanh
z>\n
13-46. Now tanh
a.
It
^\n
=
\302\243
\n2
follows
...
-\342\200\224,
.
tanh
(a/2)\n
and
/3
= ir
(a/2) cannot exceedunit
that the choice of these
(13-160)\n
value
corresponding
two
possibilities\n
REACTIVE
TWO-TERMINAL-PAIR
13-11
Art.
the
on
depends
a
=
a
\342\200\224
2
and Zb.
of Za
magnitude
2 tanh-1
tanh-1
Then\n
=
j8
y/Zb/Za,
/3 = it
a
of
computation
Zb >
when
0
sjZa/Zb,
Theseequationspermit
355\n
NETWORKS\n
Za
(13-161)\n
when Zb < Za
pass and stop
in the
and
(13-162)\n
\nbands.\n
From
\nZa/Zb
approaches
\nZa/Zb
is
\nand
Zb,
\nand
their
the
\ntient
\nA
variation
as
remains
Za/Zb
for
of Za and
zeros
and
poles
procedure
quo\302\254
as possible as frequency varies.
poles and zeros has been given by Bode
unity
nearly
these
locating
the
so that
selected
are
Zb
Dietzold.*\n
\nand
From
\nreverses
13-49(b).
\nthe
The
there
being
For
a perfect
must
components
one
disadvantage
attenuation
\ninfinite
if
\ndissipation,
13-162,
that as
it is seen
the
output
exceeds
Zb
output
voltage
action
in the
filtering
Za
by
changes
of the lattice
the
condition,
polarity.
by
and
vice versa, the phase of the
or
this
\nUnder
\nplace
13-161
Eqs.
\nmagnitude,
\nis
poles
linear
the
Za
is determined
position
\nhowever,
when
selecting
pass-band
example,
\n(for
value.
unit
In
small.
the attenuation becomesinfiniteas
the attenuation is small
Similarly,
positions for the poles and zerosfor
and zeros determine the phase variation,
by the desired form of phase variation
In the stop band,
is often
required).
we see that
13-161,
Eq.
in
180\302\260.
effectively
case of the lattice takes
of the bridge circuit shown in
a balance
to infinite attenuation,
balance,
corresponding
be of high quality and carefully matched. This
filters.
However it is possible to get
of lattice
the
of balance even with finite
at
frequency
Fig.\n
resistances
effective
also balance.\n
1 henry\n
Za\n
pass
lattice
shown
*H.
215
poles\n
and
zeros
W.
Bode
(1935).\n
and
13-50 has element
in Fig.
Za
\n14,
and
oo\n
of Za
and
Zb.\n
11\n
Example
The
Lattice
13-60.
>\342\226\240\n
1
6
Fig.
stop\n
|
\342\226\272r*
R. L.
=
s
Dietzold,
and
values such
that\n
Zb =
\342\200\234Ideal Wave
(13-163)\n
Filters,\342\200\235
Bell
System
Tech.
J.,
The
that this isa
with
filter
~
0 = 2
= 0,
a
by Eqs. 13-157
= 2 tanh-1
tan-1
~
V(\302\2532
by Eq. 13-162,
since
>
Z0
-
0 =
1)M
Zb for w
=
Zi
Another
an
\nby
equation
13-12.
A
Bartlett\342\200\231s
\nThe
This
networks.
the
\ning
w2
(13-166)\n
lattice
of the symmetrical
lattice
theorem
bisection
for a lattice
impedances
Zb
provides
for
a means
network equivalentto a
find\302\254
symmet\302\254
network.\n
ladder
\nrical
the
Bartlett\342\200\231s
lattice
-
\\/l
and
the
impedances Za and
short-circuit
impedances was suggested on page 354.
of these
is given in a theorem originallydue
quantities
theorem
applies
only for symmetrical two-terminal-
equivalence
\npair
the
theorem\n
between
and
Bartlett.
For this particularlattice,
page
11-6,
bisection
relationship
\nopen-circuit
^ 1.
(13-165)\n
263. For that particular case, the entire
was
and the phase characteristic was given
pass band,
of the form of Eq. 13-157.\n
range
\nfrequency
(13-164)\n
lgu^oo
t,
properties
illustrating
example
in Art.
given
^ 1
0 ^
1),
\\/w2/(\302\2532
is\n
impedance
\nwas
=\n
Zb
and\n
13-158
and
\nimage
ju and
low-
1/w),\n
i(\302\2532
a
13\n
Chap.
NETWORKS\n
plot shown in Fig. 13-50 indicates
of o>0 = 1. Since Za =
a cutoff
frequency
pole-zero
\npass
\nto
REACTIVE
TWO-TERMINAL-PAIR
356\n
in the application of this theoremis bisection
of
the
ladder
network.
that
we
By the term bisection,we mean
\nsymmetrical
two
\ndivide the network into identical parts such that the
networks,
end for end, have identical geometricalas well
as elec\302\254
\nwhen reversed
a bisected
Such
network
with only connecting wires\n
\ntrical
properties.
The
first
step
Half
section\342\200\224\342\200\224Half
Bisected
\nFig. 13-51.
without
\nhere
Bartlett,
\nSons,
\ntion
Inc.,
Fig.
A. C.,
of
Theory
York,
Phil.
theorem,\342\200\231\342\200\231
The
a series
has
New
13-51.
states:
proof,
network
\nladder
*
in
appears
showing
arm
Electrical
symmetrical
Bartlett\342\200\231s
lattice
Artificial
14,
network.\n
bisection
equivalent
Za equal to
1931), pp. 53-58;
Mag.,
section
Lines
the impedance
a
of
and Filters,
Brune,Otto,
806 (1932).\n
given
theorem,*
of a symmetrical
\342\200\234Noteon
(John
Wiley
Bartlett\342\200\231s
half\n
A
bisec\302\254
terminals
\nother
\nance
of
\nples
will
the
short-circuited;
network
half
the
the
illustrate
is
the
to
equal
with the bisected terminals
of this theorem.\n
application
Example
357\n
1-1 or 2-2, with
at terminals
shunt
arm Z\\>
measured
network
bisected
the
of
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13-12
Art.
the
imped\302\254
open. Two
exam\302\254
IB\n
\nin
13-52.
Fig.
is shown in
T section
standard
The
bisected
bisection
Bartlett\342\200\231s
Following
original form and also
the
theorem,
open-circuit\n
\342\226\240o\n
(c)\n
(6)
13-62.
Kg.
of
Application
bisected
\n(b)
Bartlett\342\200\231s
theorem:
original
(a)
(c) equivalent
network;
network;
lattice.\n
L/2\n
o\n
0\342\200\224\n
L\n
-\n
=C/2
~d
2\n
\302\273\n
0\342\200\224\n
fa)\n
of
Networks
13-63.
Fig.
\nbisected
short-circuit
and
equivalent
Example
13\n
this
For
\nfilter
The
equivalent
\nresulting
in
avoid
To
\nusual
the
figure.\n
v
the
figure.\n
the complicated
lattice
practice
to
a dashed
\nidentical with the
\narms by
in the
network. The
low-pass
and
network
bisected
also
are
lattice
the
of
for the half
found
is shown
the standard
for
shown
case.
\nshown
\nture
lattice
example,
is
\nsection
are
lattice.\n
equivalent
(c)
network;
impedances
\nresulting
13: (a) original network; (b)\n
Example
struc\302\254
in drawings,
one of
replace
Fig.
Conventional
13-64.
\ntation
of the
it is
the series arms
line. Thus the lattice
lattice of Fig. 13-53.\n
represen\302\254
lattice
and one
of Fig. 13-54is
network.\n
of
defined
the
shunt
to
be
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
358\n
Chap.
13\n
READING\n
FURTHER
on the constant-# and ra-derivedfiltersby
in the Bell System Technical Journal underthe
\nO. J.
Zobel
are found
of uniform
and composite electric wave
\342\200\234Theory and
design
\ntitles,
and
volume
for 1923 and \342\200\234Extensions
to
the
\nfilters\342\200\235 in
the
theory
For additional
discussion of
\ndesign of electric wave filters\342\200\235 in 1931.
The
\nthese
see:
topics
&
\nWiley
\nin
Book
Co.,
Lines,
Networks,
\nRyder,
114-163;
pp.
\n1949),
II
(John
York,
2 vols.
W.
preparation);
Vol.
Networks,
1935), Chaps. 5, 8, 9; D. F. Tuttle,
York,
(John Wiley, & Sons, Inc., New
L. Everitt,
Communication
Engineering
(Mc\302\254
D.
New
J.
Inc.,
York, 1937), pp. 179-240;
and
Fields
(Prentice-Hall,
Inc., New York,
and
LePage
Seely, General Network Analysis
New
Synthesis,
\nGraw-Hill
Communication
Guillemin\342\200\231s
Inc.,
Sons,
Network
\nJr.,
articles
original
Book Co., Inc.,
\n(McGraw-Hill
\nReed, A-C Circuit Theory (Harper
New
&
pp.
1952),
York,
New
Brothers,
and
218-236;
1948),
York,
553-597.\n
\npp.
PROBLEMS\n
The
13-1.
\nalso known
T and t
in electrical
networks shown in the
engineering
o
O\n
figure
accompanying
are
as Y and delta
literature
networks,\n
V\\A^\n
\342\200\224VW
o\n
z2\n
Zi\n
z3\n
O\n
o-\n
o\n
lb)\n
Prob.
a
Show
\n(a)
(a)
x or
can
Networks
respectively.
\nfrom
13-1.
Y
to
that,
delta network; (b)
*
be simplified by converting
or from a delta to an
Y.*
of a Y networkexists, following
equivalent
the
_
Z1Z2-|-
^2 \t\n
Z i
article,
of delta-Y
\342\200\234The
equivalence
\nworks,\342\200\235appeared
in
Electric
\t\n
Z\\Zz -|- ZzZz
Yc\n
notion
ZzZ\\\n
ZiZz
Zz
y
The
network.\n
sometimes
an equivalent
delta
if a delta equivalent
y
\nHis
wye
hold.\n
\nrelationships
*
T or
%iZz
-|\342\200\234
Z%Zz
\342\200\234IZzZ\\\n
\t\n
-p
ZzZ
i\n
in 1899.
equivalence is originally due to A. E. Kennelly
of triangles
and three-point
stars in conductingnet\302\254
World and Engineering.\n
the
Find
(b)
and
Yb,
\nYa,
for the
transformation
corresponding
Give the values
network.
a delta
of
\nalent
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13\n
Chap.
359\n
network
Y
for Zi, Zz, and
equiv\302\254
Fc\n
in the
x network is shown
A two-terminal-pair
figure.
=
=
where
a
for this
coth y
y
\"s/Zu/Zu,
network,
=
=
the impedance at terminal pair with
ZXo
ey, and
13-2.
\nthat
1
\nh/Iz
2
\npair
of
terms
in
Z%
=
Zu
open,
the
at terminal pair 1
impedance
Show
+ jfi and
terminal
terminal-
with
\npair 2 short-circuited.\n
lo-\n
nnnY\342\200\22402\n
lh\n
io\342\200\224\n
lh\n
:2f\n
lo\n
13-2.\n
Prob.
13-3.
network shown abovein
image-impedance
Z, for the
Zi =
\nAnswer.
13-4.
Answer.
the
determine
figure,
range
frequency
co
=
and,
0 to
o>
=
1.
\342\200\224
y/l
w2.\n
for the x
13-3
Prob.
Repeat
=
Zi
13-3.\n
Prob.
For the T
the
\nplot
-o2\n
lo\n
the
network shown in
figure.\n
\342\200\224
a>2.\n
l/\\/l
[0 \t\n
\342\200\224o2\n
2h\n
-
=Uf
-\n
-If\n
o\n
T\n
lo \t\n
K\t\n
o\n
1\n
1\n
1\n
1\n
1\n
K \t\n
1\n
Xl\\\342\200\224\n
u) \342\200\2340\n \302\2531\n
\342\200\224o2\n
\342\200\2249\342\200\224\n
i\n
1\n
i\n
1\n
i\n
\t\n
\342\200\2241
1\n
-1\n
1\n
1\n
|\n
\342\200\2246\342\200\224
\nl\n
\342\200\224<\n
012\n
CO-00\n
013\n
f\n
for Zu and
A pole-zero plot
13-5.
\nfigure.
From
\nbands,
(c)
the
\t\n
i\n
i\n
1\n
0)3\n
014\n
i\n
\302\253i\n012\n
T\n
UUU
L/2\n
L/2\n
\napply
to
the
\nthe attenuation
\nband.
Answer,
T network,
a in the
a
=
figure.\n
u\n
-C\n
\\\n
01 -00\n
0)5
02\n
lo\342\200\224\n
Prob.\n 13-7.\n
network shown in the
with Eqs. 13-30 and/or
derive an equation in terms L,
stop band and for the phase
The symmetrical T
\nvalues as indicated.
Starting
13-7.
the
in
shown
uuu\n
13-6.\n
Prob.
accompanying
bands, (b) the stop
|\n
T
the
in
shown
iv\342\200\224\n
1
oi*0\n
is
Prob. 13-5 for the pole-zero
plot
M\342\200\224\n
\342\200\224K\342\200\224\n
0\n
\342\200\224e\342\200\224\n
i\n
1
i\n
i\n
Xuf\342\200\224\n i\n
i\n
\n1\n
I\n
a\n
i\n
1\n
1\n
i\n
i\n
i\n
i\n
i\n
1\n
I\n
1\n
i
i\n
*lof
Z\\0
(a) the pass
Determine:
plots,
cutoff
frequencies.\n
the
13-6. Repeat
13-5.\n
Prob.
13-4.
Prob.
cosh-1
figure
of
shift
(2
\342\200\224
u2LC)/2,
=
/3
cos-1
(2
element
has
13-32,
which
C,
and
/S in
\342\200\224
the
u2LC)/2.\n
\302\253
for
pass
13-8. For the
cutoff
\nall
schematic shown in
Chap.13\n
the
Analyze
(a)
figure,
<oCi = 0.781,
Answer.
frequencies.
the
and the stop
pass bands
the
determine
to
\nwork
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
360\n
bands, (b)
\302\253C2
=
net\302\254
Determine
1.282.\n
lo\n
lo-\n
13-8.\n
Prob.
13-9.
Repeat
13-10.
Design
0
\nat
\nL
=
and a
second.
per
cycles
31.8
t section
below.\n
shown
cutoff frequency 1000
purely resistive image impedance 100 ohms
Give element values. Answer. C = 3.18
filter having a
a low-pass
second
per
\ncycles
13-8 for the
Prob.
of
of
nf,
mh.\n
Prob.
In
13-11.
that Z, = y/\\
found, for example, in
attenuation
\nband. The
it was found
13-3,
lo
GTOID-\n
\342\200\224
\302\2532
in
the
pass
Prob. 13-7
only\n
applies
\342\226\240GHHD-\n
lh\n
lh\n
^2f\n
10-\n
13-11.\n
Prob.
the
when
\ntion,
\nan
let
is:
approximation
\nabove
by
On
\n(b)
\n1
T section is
Zi = R = 1
<
a?
the
< 2.
13-12.
\nas a
for
solving
function
For
same
terminated in this
ohm (constant). To
Z\302\273. As
\\I%/Ii\\
eN
the
filter
\302\253.Make
defined
plot
given
by
loss of the
Eq.
13-149
a
such
good
circuit
shown
for 0 <
\342\200\224
2
cosh-1
w
<
c*>
2.
for
result.\n
in Prob. 13-8,
a careful
how
the function,
approximate
approxima\302\254
investigate
the insertion
\342\200\224
coordinates,
This is the
of
Plot
(a)
a practical
sketch
of:
determine the value
(a) the attenuation a as
of
x
a\n
of
function
a
13-25
in
working
13-13.
Repeat
13-14.
Show
ft
as
of
the
(d) What is
problem,
the value
in
R
of
13-12 for the network givenin Prob.13-9.\n
a single
if F is the number of cutofffrequencies,
filter requires 3F elements.\n
that
13-15. Designa composite
\nspecifications:(a) the termination is a
m-derived
0.127
\n)
to the
filter
low-pass
600-ohm
(b)
resistor,
o
-i\342\200\224'7RRT'-
0.085
(c) the image
normalized
plots of
of co,
a function
use
\302\253.Make
361\n
Prob.
a constant-#
of
\nsection
of
this
\nEq. 13-81?\n
shift
phase
function
as
\nimpedance
\nFig.
the
(b)
w,
NETWORKS\n
REACTIVE
TWO-TERMINAL-PAIR
13\n
Chap.
following
the
cutoff\n
o-\n
h\n
h\n
\342\200\230'6000\n
infinite
of
cycles per second, (c) the frequencies
are
1500 and 1700 cycles per second. Draw the sche\302\254
for the filter with all possible series and parallelelements
is 1200
frequency
\nattenuation,
\nmatic
Solution.\n
13-16.
Prob.
diagram
13-16. Design a
filter to the
0 to
band:
Pass
Cutoff:
values.\n
element
all
Indicate
\ncombined.
Output
specifications:\n
following
2000 cycles per second.\n
must
be no more than
2100
at
input
cycles
per
5% of
second and all higher
the\n
\nfrequencies.\n
Termination:
End
section:
Load
resistor
End
sections
m
\nwith
(a) Draw
values.
\nment
to
\nsolution
\nare
meet
13-17. A network
\nfour
Draw
\nsections can
\nat
each
of
be m-derived half
sections
0.6.\n
specifications?\n
is to be composed
half
constant-#
\nsections.
should
the schematic diagram of the filterandindicateall ele\302\254
As in most design problems, there is no unique
(Note:
this
(b) How many sections of constant-# filter
problem.)
to
needed
=
a value of 600 ohms.\n
will have
of
sections
schematic
be combined
the
\nm-derived half
cascaded
sections.\n
(or L
diagrams
sections)
and
of all the
possible
such that there is an
terminal
pairs.
two
Consider
half
m-derived
ways
image
of
connection
cascade
the
match
impedance
both the
half
these
ir
and
T
362
the
in
1
Zi
band,
-
all
A
\nmine
a
\nMark
\nand
element
Zb
ladder
element
values
be
image impedance.
=
=
(c)
2 tanh\"1
(a)
1 radians/sec,\n
- 1)M
vV
(f)\n
equal.\n
equivalent
of the network
of Prob.
values.\n
shown in the figureis
equivalent of the network
as
known
is shown without
of this lattice
equivalent
structure
possible
must
lattice
the
lattice
(e)
phase
Answer,
radians/sec,
a
the
(f)
\302\2532-\n
13-20. The network
\nDetermine the lattice
13-21.
\342\200\224
\302\273
yV/(o2 - 1),
i/Vi
Show
\n13-8.
1
(b)
Determine
13-19.
in the stop band,
and (g) the
band
stop
radians/sec,
=
shown below, determine (a) the pass
(c) the cutoff frequency, (d) the phaseshift\n
(e) the attenuation
= 2 tan-1
(d)
(g)
the
in
\342\200\224
stop
band,
pass
\nshift
\n0
the
(b)
\nband,
13\n
Chap.
filter
lattice
the
For
13-18.
NETWORKS
REACTIVE
TWO-TERMINAL-PAIR
(Lx, Lt,
if
L\\
=
a
bridged
T.
L2.\n
element values.
by studying and
Deter\302\254
Za
Zb.
Ci, etc.) noting which elementsin Z,
CHAPTER
AMPLIFIER
Shunt peaked
14-1.
\nsome
\nthe
\na
shown
the
be
the
vacuum
The
resistor
of
plate
pentode.
\ncapacitor C may
the
of
\nance
R
represent the wiring
tube.
If the
vacuum
tube
the
of
\ntance
usually
tube,
plate resist\302\254
is very high (as in
the
F2
where
gm
14-1
\nFig.
=
The
denominator
\nand
may
also
\nand
the
natural
+Ls
undamped
function
The
pole-zero
there
\nequation:
analysis
\n(Prentice-Hall,
(14-1)\n
the impedanceof the
impedance for the network of
and Z is
The
has
=
configuration
are
a pair
14-1
Inc.,
C
frequently
+
Rs/L
+
l/LC\\
1
r-.\n
K
}\n
many times before
is\n
_ 0m
CsH
s + 2fa>n
2f<OnS+
,14
U
o\\\n
;\n
COn2
transfer function is evident fromthis
of conjugate poles and one real zero.How\302\254
made in terms of the circuit Q discussed\n
of the
in Ryder, Electronic EngineeringPrinciples,
2d
220.\n
York,
1952), p.
is derived
New
|_s2
been encountered
equation
resulting
is
s + R/L
lT
the dimensionlessdampingratiof
con. From Eq. 14-1, the voltage
frequency
in terms of the impedance
be written
may
V1(s)
Equation
the
pentodes),
in terms of
be written
V*(s)
*
of
i
-gmZV
+ R)
+ R
polynomial
transfer
\nexpression.
\never,
=
plate.
(Ls
(V^s)
l/Cs
The
the case
is\n
Z(s)
\nratio
the
to
connected
\nnetwork
amplifier
\nnetwork.\n
equation*\n
transconductance
tube
the
is
peaked
capaci\302\254
is given by
\noutput voltage
Shunt
14-1.
Fig.
interelectrode
and
\ncapacitance
to
connected
represents
and the
inductor,
the
of
resistance
\nthe
inductor
the
element
only
filter-amplifier. Such a
give
Fig.
networks,
practical
\nmay
in
in conjunction with
used
often
are
a combination
14-1. In
to
amplifiers
is
\nnetwork
network\n
networks
tube
\nvacuum
NETWORKS\n
amplifier
Frequency-sensitive
14\n
363\n
ed.
364\n
AMPLIFIER
and so this equation
Circuit
Q is defined as\n
Art.
in
11-4,
\nquantity.
The
\nnew frequency
this
of
terms
14\n
1\n
con
(14-4)\n
2 R/2L
R
2r\n
s will be normalized
will be designated by
frequency
variable
complex
will be rearrangedin
1
WnL
_
q
^
Chap.
NETWORKS\n
by division
the letter p and
by
This
&>n.
defined
\nas\n
P
these
With
=
=
s/w\342\200\236
+
ff/w\342\200\236
the
substitutions,
jctf/oon
equation
== <Tp +
jo) p
(14-5)\n
for the transfer
function
\nbecomes\n
Vt(p)
gm
_
^i(p)
The
of this
poles
co\342\200\236CLp2
function
are
-
Pa*
P\302\253>
P + 1/Q
[
2Q
+
+
p/Q
1\n
(14-6)\n
lJ\n
evidently\n
~
^(20)
1
Q
(14\342\200\2347)\n
(u-8)\n
\302\253>*
In
of
terms
\nin
considered
be
will
\ntion
Q is much larger than
case. The
the typical
networks,
practical
Q is
circuit
p-plane\n
shown in Fig.
and so
the
second
equa\302\254
pole-zero configuration
14-2. The locus
of
the
poles
and\n
y\302\253/w\302\273\n
D
v.\n
T\n
A-U/2Q)2\n
-1/2Q\n
1\n
*
a
<r/\302\253\302\273\n
l\n
\342\200\2341/Q\n
\t\n
\302\253\342\200\242
i\n
t
1\n
Pa
V\n
Fig.
14-2.
\nof
zero
location
Pole-zero
is shown
in terms
Q.\n
Q.\n
in Fig.
14-3 for various
values of
Q
greater
than
the
crit\302\254
As Q decreases toward -y, the zero approaches
the point
and
the
the point \342\200\2241.\n
poles
\n\342\200\2242,
approach
of the shunt peaked amplifier is found by letting
response
Frequency
\342\200\224
the gain and phase for a number of frequen\302\254
\nV
j\302\260>pand
computing
\ncies.
The
of the voltage ratio or the amplifiergain found\n
magnitude
\nical
value
of 7.
is
the
from
NETWORKS\n
AMPLIFIER
14-2\n
Art.
365\n
relationship\n
Vl(jUp)
=
0)nC
Vl(jup)
~
| jtJp
9m
\\joip
-
Pll\n
(14-9)\n
Pa\\\\jo)p
V*\\\n
\342\200\224
0 to \302\273, the length
changes
|ja>p
pa|
a
when
nearest
minimum
value
is
through
\302\253p
pa. The
\nrapidly,
going
reaches a maximum value at a frequencynear
\nresponse magnitude
As frequency increases, the magnitude
minimum.
not
this
at)
\n(but
the
rate
of 6 db
\nfunction
falls
at
the
As
for
octave
\nper
\nFig.
14-4.
\nRLC
circuit
\nthe
angle
be
\ncan
\ntion
the
by
res\302\254
fact
inspec\302\254
of visualizing the
of the
terms
\nsults
in
\nbraic
investigation
Pig.
configuration.\n
pole-zero
This method
14-2.
11-4,
at
This
established
readily
of
zero
network.
this
for
series
Art.
in
not
is
in
shown
to the
contrast
considered
phase
\nonance
is
curve
In
A
frequencies.
high
response
\ntypical
from
varies
frequency
peaked
response curvefor
amplifier
network.\n
re\302\254
involves
amplifier
14-4. Typical
shunt
configuration
pole-zero
which
Sta3ser-tuned
\na
is simpler than
manipu\342\200\231ation
of
complex
an
alge\302\254
numbers.\n
networks\n
of the shunt peaked amplifier
in
\nshown
was not discussed in the previous section.This
14-1,
Fig.
follows
fact that the input to the grid of the
from
the
\ntube
draws
current
and thus has high internal impedance.
negligible
the
is that
this amplifier network may be
property
\nnected
to another
without loading; that is, without causing
network
flow
such that the output voltage of the
current
to
significant
network
would
be altered
by connecting the amplifier network.
this
successive
of amplifier networks may be
reason,
stages
\nnected
in
cascade
and each network will be
together
(or tandem),
An important
property
network,
vacuum
\nproperty
con\302\254
\nSpecifically,
\nany
\nother
\nFor
con\302\254
of
\n-independent
others.\n
all
Stage 2
Stage 1
14-6.
Fig.
A
\neach
\nis
cascade
connection
stage
is represented
to
connected
\namplifier
is
the
connected
Cascade
connection
of amplifier
Stage3\n
networks.\n
stages is shown in Fig. 14-5,where
a
block.
The output of the first amplifier
by
of the second, the output of the second
input
to the input of the third, and so on. Any
num\302\254\n
of amplifier
Chap.
NETWORKS\n
AMPLIFIER
366\n
14\n
If the amplifier input has a
and
the
has a low impedance, the
impedance
\nhigh
amplifier
output
of
and
will be independent
in the sense
amplification
filtering
the
not
second
will not affect the first, the third
amplifier
other
\naffect
the
and so on. Each stage is isolated
all
second,
it
each
the
voltage
\nstages;
stage lives in a world of its
accepting
and
\nreceives without influencing the
in turn
without
being
\ninfluenced by the stage that receivesits
connection of stages of amplifiers
The
reason
for cascade
practical
that
one
does not provide enough voltage gain. The ordinary
stage
uses
at least two stages of amplification (the
receiver
it is common for
intermediate
or IF amplifiers);
frequency,
to
of amplification
to be used in radar receivers.
eight
stages
in a network made up of resistors,
is ordinarily
terminated
\nstage
the output voltage is a
and
For such networks,
\ntors,
capacitors.
we will investigate desirable forms
In this section,
\ntion
of frequency.
of the output
the
variation
radian
voltage to the input voltage
of
ber
be so
may
stages
connected.
very
\nstages
will
\nthat
from
own,
\342\200\234giver/\342\200\231
output.\n
\nis
\nsuperheterodyne
\nso-called
Each
\nsix
induc\302\254
func\302\254
with
\nof
\nfrequency.\n
the
Under
\ntions
assumption
the voltage ratio
transfer
func\302\254
written\n
be
may
of no loading,
r\302\273(\302\253)\n
vm\342\200\231\n
where
G2,
Gi,
\ntions
for
\nmay
be
the
found
Gz are successively the voltage ratio transferfunc\302\254
and third stages. The output voltage F4(s)
first,
second,
in terms
of the input voltage V i(s) by the following
and
\nmanipulation.\n
F2(s) Fj(s) F4(s)
Vi(s)
Vi(*)
where Gt(s) is the
had
\npossible
not
voltage
This
cascade.
in
\nnected
Vz(s)
the
output
F4(s)
Fi(s)
transfer
mathematical
/nr
/
Gt^
\\
function
/\342\200\236\\/nr
/\342\200\236\\/i
(\342\200\236\\/nr
Gi(s)G2(s)Gz(s)
for
the
three
a\n
(14-11)\n
stages
con\302\254
operation would not have been
been a function of the inputvoltage
voltage
each stage isolated
from the
In
the
sinusoidal
state, two properties of the total transfer
steady
in design.
The first is the maximum
are
important
G(j<a)
or the maximum gain. The otheris
\nnitude
transfer
of the
function,
with
radian
\nvariation
of magnitude
It has been found that
frequency.
and gain variation
combinations
of gain
with frequency cannot
attained
identical
by
cascading
simply
amplifier network stages.
can
be realized
if each stage is made slightly
performance
tuned.
\nent.
The
network
is then said to be
composite
amplifier
\nThe
of stagger-tuned
networks is easily
design
amplifier
\nonly
for
each
stage
(i.e.,
others).\n
\nfunction
mag\302\254
the
\ndesired
\nbe
differ\302\254
\nBetter
stagger
accomplished\n
of
the
will
be
terms
in
\nproblem
\nfor
Our approach to this design
configuration.
first the desirable pole-zefo configurations
consider
We will then show how such pole-zero
amplifiers.
be realized
by the design of the amplifiernetworks
to
can
\nconfigurations
each
stage.\n
\na
filter;
low-pass
\nThe
techniques
filters,
\nhigh-pass
time
\nthe
being,
functions
\ntransfer
high
the
having
Vn+l(s)
=
form\n
1\t\n
K
Fi(s)
+
bosn
...
bis\342\200\235-1+
Fn+l(ico)\n
1
=
K'\n
Fi(jco)\n
This
\nis
of
raised
\nfrequency
co
\nexponent
when
For
squared.
a
the
=
0.
of
\nrate
w
for
of
the
as
magnitude
(14-14)\n
+
ju>\n
+
\\A>2
contains
0 up
curve
The
co
to
a2\n
even
as
and
a
j
exponents
only.\n
then
approaches
octave
rate
the
\nflat
can
\nflat
characteristic
the
|
v\n
Ns\n
!\n
^
!
~\n
asymptoti/J u_6 IdealflatcharacterW\"
the
n X 6 db per\n
Justhow
We now face a number of problems:
of decrease.
this
How do we go about accomplishing
be made?
curve
\342\200\224
\ncally
\342\200\242
is flat from
called the
for
_______\n
to have
is shown
to a frequency
frequency
CUH
curve
frequencies
14-6.
Fig.
be
may
function
1\n
magnitude
frequency
intermediate
=
ikf(oo2),
of a complex
by Eq. 14-13 has the value K'/y/~An
represented
As frequency
becomes larger, the magnitude decreasesat
\342\200\224n
X 6 db
per octave. An
function
in
(14-13)\n
An\n
example,\n
function
magnitude
form
\nideal
for
+
function
The
\nco
...
+
designated
1\n
and
form\n
is equal to the squarerootof the real
magnitude
squared,
plus the imaginary part squared. The
either
an even or an odd power has an even
to
function
the
way
the
has
\t\n
AiO)2n~2
<o2,
the
sinusoidal
In the
is, the
that
formed;
\npart
the
reviewing
by
of
a function
is
magnitude
+
vVn
(14-12)\n
+bn\n
where n is the number of stages
of amplification.
the
of
this
transferfunction
\nsteady-state,
magnitude
\nseen
of
frequencies
by means of a low gain.
will
to the case of band-passfiltersand
apply
directly
as
will
later
be illustrated
by several examples. For
we will also restrict
our discussion to voltageratio
attenuates
and
gain
\nhigh
to follow, we will restrict ourselvesto the case
that
is, a filter which passes low frequencieswith
discussion
the
In
367\n
pole-zero
stagger-tuned
\nof
NETWORKS\n
AMPLIFIER
14-2\n
Ait
in
terms
of Eq.
14-13?\n
NETWORKS\n
AMPLIFIER
368\n
Chap.
14\n
ratio function with frequency can be made
to
that the derivative of M with
to zero
respect
\nequal
If the
M is a function of <o2) be equal
to zero.
than
<a since
\no)2 (rather
to
of
of the slope of the Jlf(to2) curve is also made
\nrate
equal
change
of the curve will be improved.The pattern becomes
\nzero, the flatness
The
of M(u>2) zero as we can.
as many
derivatives
make
\nclear:
just
the curve will be
that are zero at w = 0, the flatter
\nmore derivatives
above w = 0.\n
a band
of frequencies
\nat
with respect to to2, there
results\n
is differentiated
If Eq.
14-13
The
of the
voltage
by requiring
slope
Fn+lO'to)
_d_\n
dto2\n
This
(to2\"
\nVl(jto)
zero.
to
make
all
\nEq.
14-13
has
A
successively
zero
A-coefficients
the
+
=
to
V
A
0
An-i
setting
by
equal to zero
these
Under
i.
'\n
AnY'2
derivatives
up to
conditions,
K'\n
(14-16)\n
l(i\302\253)
+
Vto2\"
An\n
let
of this equation,
form
the
...
An-1
\342\226\240
\342\200\242
+
form\n
y
simplify
.
+
at
higher
Fn+i(ja>)
To
4
2^<02w
+ Axto2\"-2+
Setting
the
\nwill
+
equal to zero
be made
can
expression
\nequal
A /CO2\" 2
_
=
An
ft2n
and
to//3 =
<op.
\nThen\n
Vn+i(j<*p)
a
1\n
=
(14-17)\n
K'
Vi(j<*p)
are designed to satisfy the relationship
this
they are said to be maximallyflat. Suchstag\302\254
by
equation,
Butterworth
also called
are
amplifiers after an author
amplifiers
in 1930. (Networks with no
such
first
described
design
amplifier
and
no isolation
may also have the maximally flat character\302\254
When
amplifiers
staggered
\ngiven
\ngered
\nwho
\namplifiers
last
\nthe
\nThe
\nmade,
\nto
the
Maximally
flat
character\302\254
\nistic.\n
\nthe
in
The
\nratio
0.707
point
\nlater
the
(the
larger
the
the
better
value
the
curve.
ideal
given
by
fil\302\254
by
not
exactly
in Fig. 14-6.
of n can be
approximation
A plot of
Eq.
14-17
the
for
\nseveral values of n is shown
in Fig.\n
14-7.
All the curves pass through
=
at
1. This
will be shown
point)
wp
chapter.\n
next problem
transfer
half-power
shown
ideal
\nfunction
14-7.
does
equation
the
\nfit
realized
curve
actual
The
\nters.)
Butterworth
called
are
and
\nistic
Fig.
1\n
+
V\342\200\234P2n
function
is to find the
that
will
positions
of
the
poles
of the
give an absolute magnitude
voltage
function\n
of
The
\np/j.
14-17. To simplify notation, let
root factor of Eq. 14-17 then becomes\n
square
\342\200\224p2\"_|_
+
p2n
We
The
in
proceed
route.
indirect
of
poles
=
H(p)
=
the
of H (p)
poles
\342\200\224
the
find
2wth
of
roots
=
(14-19)\n
1
+ 1 =
_1
=
+1
=
Setting
these
\nsides of
the equation
pm
pk
=
=
p2n =
or
we
\302\261
1,
=
this
write
(14-21)\n
of the first
these
of
\342\200\224
1
(14-23)\n
number
=
...
=
e\302\261*2\302\273
=
...
e*iT
in generalized
to p2n
(14-22)\n
condition\n
=
e\302\261\302\2738ir
e\302\261)'5T
=
1, 2,
3,
k
=
0, 1,
2,
in polar form
as\n
(14-24)\n
(14-25)\n
form
m
e\302\261>-2*',
equal
(14-20)\n
p2n = +1
or
c\302\261f(jm-i>*
equations
an\n
is,\n
occur under the
e\302\2610r
even
n is odd
+ t.
written
be
may
equations
*s
be
what appears to
taking
the denominator
e\302\2613> =
\342\200\224
These
is even
1=0
+
p2n
To
if n
1=0
+
p2n
Similarly,
1
_pif
to zero; that
co\342\200\236
(14-18)\n
w
when
=
or
is odd
p2n\\^.
occur
F(p)
j<ap
if n
investigation
by
the functions\n
Consider
F(p)
p=
i
our
is equal
\nequations
369\n
of Eq.
form
the
NETWORKS
AMPLIFIER
14-2
Art.
as\n
...
...
and taking the
(14-26)\n
(14-27)\n
2nth root
of
both
gives\n
**<\302\253\342\200\224\302\273*/\302\273*,m
=
k =
e***'*,
3,
..., n
0, 1, 2,
..., n
1, 2,
(14-28)\n
(14-29)\n
H(p) given above. The
of each
root
of the last two equations is unity, and the
\nare
radians.
The location of the roots for an odd
separated
ir/n
by
=
=
and
an
even
n (n
in Fig.
14-8.
Similar plots
(n
3)
4) are
other
values
of n are readily made by following these
These
expressions
locate
the
poles
of F(p) and
roots
\nmagnitude
\nn
\nfor
shown
rules:\n
370\n
NETWORKS\n
AMPLIFIER
\342\200\224
\nthe
<rp
For even values
\nRoots are located
(2)
(3) There is a
symmetry
are
the
and
on
+o>
roots
from these real
displaced
are
negative
occur
roots
of roots
Location
14-8.
real axis.
of n, no roots
located
on
the
r/2n radians from the positiveand
are displaced
by t/u radians.\n
with
to both
the real axis and the
respect
No
axis.
\nimaginary
Kg.
roots
All
axis.
\nreal
on
roots
Other
axis.
root is alwayslocated
14\n
radians.\n
ir/n
\nby
of n, a
values
odd
For
(1)
Chap.
of
on the jup
\302\2611:
(a)
=
n
3
axis.\n
(odd);
(b) n =
4\n
(even).\n
\nand
in Fig. 14-8 are really polesof the functions
F(p)
and
14-21. We know that impedance
14-20
defined
by Eqs.
cannot
functions
have poles in the right half
transfer
and
These
have.
and H(p)
functions, however, are not neces\302\254
F(p)
roots
The
H(p)
\nfunctions
\nplane
as
\nsarily
network
invented
\nbeen
transfer
\nthe
Because
functions
\nthe
\nnated
functions that have
only arbitrary
that they might somehow relate to
in the
expectation
of the maximally flat form.\n
functions
magnitudes
having
of rule
above, there are always as many polesof
(3) stated
half
in the right half plane as in the
left
and H(p)
F(p)
functions.
the
If
\nplane.
plotted
fr(p),
we
\nfi(p),
poles are grouped together and
poles are similarly grouped as
half-plane
half-plane
right
left
the
and
are
They
desig\302\254
write\n
can
= fr(p)fi(p)
Fip)
When p =
equal
\nalways
in
ju>p
to
axis
\ncerned.
the
sinusoidal
can
Because
be
matched
of
this
steady
of fi(p).
magnitude
of F(joi)
|F0't0p)|
of fr{p) is
pole
the
con\302\254
equality,\n
|/r0'*p)|
the
the magnitude
state,
This can be seenfrom the
each of the poles to a point on
from
drawn
in identical pairs as far as magnitudeis
magnitude
phasors
\nconfiguration:
\nj(ap
the
(14-30)\n
may be
=
=
(14-31)\n
I
written\n
\\friM\\\\fl(jo>p)\\
=
|/lO\342\200\230\"p)|*\n
(14-32)\n
Art.
NETWORKS\n
AMPLIFIER
14-2\n
the
Extracting
of this equation
root
square
-
l/<(i\302\273,)l
371\n
gives\n
(U-33)\n
VWGZil.
for
even
Now F(p) is definedby Eq. 14-20
\nof n,
is a real number having the
F(jo>)
n
even values
these
For
only.
value\n
=
\302\273\302\253*>\n
rrsjs
this
Substituting
into
magnitude
/iO\342\200\231u)
This
equation
\nmaximally
\nfi(p)
\nistic.
the
is
These
by
\nnately
\nplane
are
the
ne-
+
W(1
yields\n
(14-35)\n
<\302\273p2n)
of Eq. 14-17
defined
the
flat function.
Thus the pole configuration described
one required
to give a maximally flat magnitude
are
found by Eqs. 14-28 and
poles
readily
of the form
is precisely
which
by
character\302\254
14-29\342\200\224alter\302\254
the
of page
rules
retained.
=
same
14-9.
V H
Polo
(jo)p) = \\/l/(l
that
requirement
locations
that
in
poles
only
370\342\200\224provided
it follows
Similarly,
\\hi(jo)p)\\
under
=
Eq. 14-33
for odd values
+
of n,
half
that\n
(14-36)\n
\"P2n)
only left half-plane
for maximally flat frequency
the negative real axis).\n
the left
poles be
response
con-\n
($
\nmeasured from
sidered.
\ncharacteristics
The
pole
for
configurations
several
values
to give maximally flat
of n are shown in Fig. 14-9.\n
magnitude
NETWORKS\n
AMPLIFIER
372\n
Chap.
the two network functions we have found, Eq.
is normalized
radian frequency. If this term is rewritten
Eq.
14-36,
then
in each
case we are considering terms of the form\n
\302\253/\302\253\342\200\236,
\nand
\nas
b), of
the
Now
14\n
14-35
1\n
(14-37)\n
+
\\/i
(\302\253A>\342\200\236)2n\n
=
=
when
1 or
value 1 at <*>/\302\253\342\200\236
0. And
co/co\342\200\236
\nto =
function
has the value (l/\\/2) independent of the value
<on, the
the half-power frequency as was done
\nof n!
This
is designated
frequency
and there is another half-power frequency at negative
\nin Eq.
11-79,
\noon.
The
\nfor
larger
14-3.
the
has
function
This
of
values
Overstaggered
function
\nnitude
\nbut
\nillustrated
for the
as
amplifiers
in
14-10.
Fig.
as a dashed
disadvantage
of
flat (or Butterworth)
maximally
a constant
becomes
Fig.
polynomials)\n
(Chebyshev
points to a
approximates
frequency
a constant value
approximates
nearly
n.\n
This last discussion
\namplifer
more
curve
response
larger
the
14-10.
approximation
Comparison
line. The Butterworth
condition. The
for a range
The ideal characteristic
response
an
stagger-tuning
mag\302\254
of frequencies,
is poor. This is
is a constant
shown\n
of responses.\n
closely
the
approximates
but the differencebetween
and
actual
characteristic
becomes
\nideal
large as frequency increases.
between the ideal and the actual) is
the
error
(the difference
total
characteristic
would seem to be better
\nat high
The
frequency.
low
this
error
could
be spread out over the entire band
frequencies
a
is
in
Such
this
filter
case).
frequency response
\n(for
low-pass
=
=
as
an
is spread out from
0 to
The
error
\nFig.
14-10(b).
a series
of hills and dales. The maximum error is the
a frequency
Such
for
several
response appears to be
points.
We face two problems: Can
than
flat response.
the
maximally
form for this response? Can
write
an expression
in mathematical
that gives this response?\n
find
the
pole configuration
\nideal characteristic
for low frequencies,
the
\nAll
lumped
of
\nif
shown
\302\253
w
w\342\200\236
\n\342\200\234equal ripple,\342\200\235
\nsame
\nbetter
\nwe
\nwe
in Fig.
functions
illustrated
The equal ripple
P.
\noriginally studied by the Russian mathematician
type
14-10(b)
of
L.
Chebyshev\n
were
AMPLIFIER
14-3\n
Art.
100
some
in
\nuseful
\nThese
the
The
by
polynomials,
Chebyshev*
equal ripple form we have
polynomial of order n is defined\n
Chebyshev
general
equation\n
=
C\302\253(a>)
(n cos-1
cos
as follows.\n
Ci = cos
(cos-1
C2 =
2\302\2532
-
1
(14-40)\n
C3 =
4o>3
-
3co
(14-41)\n
8co4
-
8<o2 +
C4 =
C6
C6
Cn+1
may
equation
are
n
of
last
(14-38)\n
to)
Chebyshev polynomials for severalvalues
The
engines.
call
in the characteristic
will
we
which
a constant
\nillustrated.
of polynomial
in steam
polynomials,
\napproximate
373\n
found a certain type
Chebyshev
of the action of linkages used
ago.
studies
years
his
NETWORKS\n
=
16co6
=
32(o8
=
2(0Cn
o>)
=
(14-39)\n
1
- 20(o3+
- 48w4
+
-
a)
(14-42)\n
5co
(14-43)\n
-
18(o2
1
(14-44)\n
Cn\342\200\224i
(14-45)\n
calculate higher-orderedChebyshev
be used to
\npolynomials.\n
The
\ntion
(for
the
defined
case
\nflat
of the voltage ratio transferfunc\302\254
magnitude
sinusoidal
state) corresponding to the maximally
steady
14-17 has the form, for n stages,\n
by Eq.
the
for
equation
V n+lQ\342\200\230(Op)\n
Vi
where
be
\n\302\273
will
of
number
\nthe
(to
to the
related
G\n
(14-46)\n
0\302\253,)\n
constant
\342\202\254
is a
1\n
=
+
\\/l
be defined).
and
number
eC\342\200\2362((op)\n
the maximally
flat case,
location of the poles(andhenceto
Just as in
stages).\n
amplifier
this equal ripple function, must start with a
n and square it. The squaredfunctionis
\nshev
of order
polynomial
e and added to unity. The reciprocal
constant
the
\ntiplied
by
root
of the resulting function is the transfer function
\nsquare
can be duplicated
This
\ntude.
by performing each step
process
To construct
we
Cheby\302\254
mul\302\254
of
the
magni\302\254
graphi\302\254
The
\ncally.
polynomial\n
Chebyshev
= cos (n
Cn(io)
is
defined
for
\342\200\242Chebyshev
\nforms
\nman
apparently
to
French,
\ntranslation.\n
a range
of
also
variously
from
is
resulted
etc.
u>
from
The spelling
+1
spelled
repeated
to
cos-1 a)
\342\200\2241
(that
(14-47)\n
is
\342\200\2241
^
w
^
+1).\n
as Tchebycheff, Tchebichef, etc. These
translation:
Russian to German, Ger\302\254
used is consideredthe bestRussian
to
English
374
AMPLIFIER
When
\\a\\
>
1, cos-1
*=
<a
cosh-1
j
Cn(\302\253)
=
This function does not have a
\nat
a
rate
\nof
n
are
\nthe
Chap.
NETWORKS\n
14\n
and\n
u>
cosh
cosh-'
(n
\342\200\234rippling\342\200\235
(14-48)\n
w)
hut
nature
with
increases
o>
value of n. Plots of C\342\200\236(w)for several
values
in Fig. 14-11. The functions vary from +1 to \342\200\2241
over
of the Chebyshev\n
0 ^ w 5* 1. Several
features
range
by the
determined
shown
frequency
Fig. 14-11.
(a)
polynomials:
Chebyshev
for
>
\302\253
plot
for 0
<
<
\302\253
1\n
0\n
n -
Fig.
14-12.
Squared
1; (b)
plot\n
1.\n
5
Cl>\n
(odd)\n
polynomials.\n
Chebyshev
all
from the plots of Fig. 14-11. For
odd
values
=
\nof
the initial slope is alternately
has zero value at \302\253 0 and
n,
Cn(o>)
For even values
of n, C\342\200\236(o>)alternately
has
the
and
\npositive
negative.
=
\nvalue +1 and \342\200\2241at <o
0. When the Chebyshev polynomials are
values
of the Cn(w) plot will be \342\200\234reflected\342\200\235
the
all
negative
\nsquared,
from
zero and have
\nas positive
For odd n all plots of Cn2(<o)
start
values.
can
polynomials
be seen
\ninitially increasing values;
\nand
\nfunction
have
an
are
in
Fig.
even
n
all
values.
14-12.\n
decreasing
initially
shown
for
plots
of
Typical
C\302\273J(w)
start
from
plots of the
+1
squared
Ait
our
construct
To
\nFig.
14-12
\nscale)
and
frequency response, the curves
by e (which will merely reducethe
equal-ripple
are
above
substituted
375\n
NETWORKS\n
AMPLIFIER
14-3\n
multiplied
the
into
of
equation\n
1\n
=\n
I G(M\\
(14-49)\n
+
Vl
When C'n(ttp) has zero value, G(je>p)
the value
of unity (the maximum
\nhas
will
\nfunction
will
eC'n2(cop)\n
have
unit
when
value;
value it canhave),
the
C\302\273(<op)
magnitude
be\n
1\n
(14-50)\n
Vl+~*\n
The
14-13
in Fig.
will vary between these two limitsas shown
to Fig. 14-12). The ripple\n
values
of n (corresponding
typical
ripples
two
\nfor
Fig.
width is often
14-13.
frequency
Equal-ripple
specified in decibels.This
width
response.\n
and
related
t are
by the
\nequation\n
width
ripple
=
*
8 =
+20
1\342\200\22420
log\342\204\242
log\342\204\242
.
V1
8
or
8 is
where
a
\n(in
design
At this
\na
in
\ntified
order
later
in decibels.
=
+
(14-51)\n
\302\253\n
(1 + e)
10 logio
From this equation c can be foundif
(14-52)\n
8 is
specified
problem).\n
in terms of a new factor,
point, the quantity e will be defined
the computation.
This relationship will be jus\302\254
to simplify
and will appear as Eq. 14-71.Then,by
in this
chapter
\ndefinition,\n
1\n
sinh2
o
or\n
where
n is
the
order of the
=
(no)\n
- sinh-1
Chebyshev
\342\200\224;=\n
polynomial\n
(14-53)\n
(14-54)\n
the
In
flat
maximally
to
\ndropped
at
0.707
\ncosh
Now the factor
in
value
another
a)
C\302\273(cosh
\na
Chebyshev
equal-ripple frequency
+ c at the same frequency. In
1/-\\A
Let coPl =
has significance.
frequency
to a particular frequency
for the Chebyshev
polynomial,\n
equation
14-56
Eq.
a is
defined
14\n
response curve had
frequency
corresponding
a,
the
\nBy
Chap.
the
For
1.
\302\253p
case,
Chebyshev
the
case,
=
\nresponse,the curve has a
\nthe
NETWORKS\n
AMPLIFIER
376\n
=
cosh
[n cosh-1
\n=
cosh
na\n
(cosh a)]
(14-55)\n
(14-56)\n
14-54.
Eq.
by
= 1.
larger than
this value for
Substituting
gives\n
(14-57)\n
The hyperbolic sine
and hyperboliccosine
related
are
the
by
identity\n
(14-58)\n
(14-59)\n
cosh a; = -\\/l
x
or\n
Then
14-57
Eq.
=
+
cosh-1
y/\\
(14-61)\n
1/e\n
+ 1/e)
cosh (cosh-1 \\/l
If this equation is squaredand
to
added
is
unity
=
+
y/l
both
(14-62)\n
1/e
of the
sides
equa\302\254
results\n
there
1
This
+
be written\n
may
<7n(cosh a) =
\ntion,
(14-60)\n
l/c\n
is in
equation
+ eCB2(cosh a)
the form of
= 2+
cosh
\\G(j
(14-63)\n
the square root factor
the
of
given by Eq. 14-49. Makingthis
response
e\n
. ^
a) | =
substitution
frequency\n
gives\n
\302\253
0.707\n
(14-64)\n
>/2Tl\n
if
\302\243
is
\n(the
much
usual
\npower\342\200\235
\nFig.
14-14
\nfar).
The
smaller
case),
frequency
(along
approximate
1. Thus for
than
the frequency
in
with
\302\253Pl
=
the approximationthat
cosh
flat
maximally
other
information
the
half-power
frequency,
is
to the
a corresponds
case. This is
that we have
cosh a, is
e
small
\342\200\234half\302\254
illustrated in
deduced thus
determined
by\n
Art.
14-3\n
the
value
\nof
a large
value
of a, which is determinedby bothn ande. For
than
a
value
is small and cosh a has
onlyslightlylarger
unity.
the
faster
the
value
of
the
the
frequency
n,
larger
words,
=
1
better
above
falls
off with
(giving
filtering
<op
frequency
a
n,
other
\nIn
377\n
NETWORKS\n
AMPLIFIER
\nresponse
is also
characteristic
dependent
\naction). This steepness of the response
size
of
the
is in turn
which
ripplesin the
upon the
dependent
\nupon \342\202\254,
= 1. The summary of our knowledge
= 0 to
range
\302\253p
o>p
\nfrequency
shown in Fig. 14-14,\n
\nof equal-ripple
characteristics,
response
frequency
Fig. 14-14.
point
Half-power
half
that
indicates
the
\ngive
next
We
n.
\nand
of
frequency
response
of ripple).\n
cycles
(n =
number of\n
of
specifications are complete and givenin terms
e, a,
turn our attention to the pole configuration
that will
5,
characteristics.\n
equal-ripple
for determining the locationsfor the
parallels
flat (or Butterworth) case. In this procedure,
a
\nthat for the
maximally
form
14-49
with
some
other
of
the
of Eq.
replaced
by
is written,
and the poles of this function are determined.As in
\niable
flat
the poles in the right half plane are
maximally
case,
the discussion leading to
function.
\nto give
the
Paralleling
magnitude
=
=
and examine the
and
we let p
or
14-18
p/j
14-19,
cop
14-49
which
under
the
radical
in
Eq.
appearing
The procedure
poles
\nfunction
var\302\254
cop
\nthe
rejected
\nEqs.
joyp
is\n
\nfunction
1
Now
\n|x|
= 0
*Cn*(p/j)
x is
where
Cn(x),
^
+
Cn2(p/j) = -1/e
or
any variable, is
defined as
\ndefine
the
cosine
= a
inverse
cos-1
(p/j)
of a complex
\342\200\224
ja
cos
Expansion
of the
cos
n cos-1
cos
x,
the last equation becomes\n
1, so that
cos n cos-1 (p/j) =
Since
=
Cn(x)
(14-65)\n
(na
such
na cosh
na + j
number is complexin
general,
we
that\n
\342\200\224
cosine of the
(14-66)\n
\302\261j/s/e
jna)
=
\302\261j/
differenceof
\\/e
(14-67)\n
two
sin na sinhna
angles
=
+j/\\/e
gives\n
(14-68)\n
NETWORKS\n
AMPLIFIER
378\n
and
real
The
Since
\ngiven
the
by
=
2AT -I- 1
r\n
ft
z\n
sin na =
of a,
-
any value
-
sinh-'
\nequation,p/j
or
\nstep
\nthe
=
\342\200\224
cos
a\n
(a
\342\200\236
=
W
~ja)f
2
, .
2AT
+ 1x
\342\200\224\342\200\224
.
,
4
\302\253
811.
+ J
2
the form of
to modify
the
of
=
p
~
=
b
where
a
cos
1
N
*>>
on
\nlocated
found
with
a circle
p
where
\nEq.
b'
14-29
are
values
for
odd
given
values
e*v
=
...,
2N + l
0\n
cos b'
n\n
(14-72)\n
n\n
(14-73)\n
r\\\n
<\342\200\242<\302\253
=
0, 1,
2.\n
next
Our
14-73
= 0,
+ j
by Eq. 14-28
of n as\n
with
comparison
identities, cos
be
written\n
b)
1, 2,
-\n
x
may
+ i sin
6
that the poles for
locations
given by
as
.
0, 1, 2,
for
\342\200\224^
already
a in the
a and
14-65 as required.
Eq.
a ( \342\200\224tanh
n
this equation
x\n
cosh
to
of
case. From the
flat
maximally
sin
14-54in order
2)'\n
roots of Eq.
the
(14-70)\n
(14-71)\n
values
. \\
lr
+
n
write\n
we
ja),
defines
equation
will be
have
...,
required
I\342\200\224-\342\200\224
(
and
We
0, 1, 2,
values
have
point.\n
\342\200\224
(2N
3 cos
p = cosh
results
must
proof as Eq.
N
This
a
a,
of
(14-69)\n
(l/\\/\302\253)
Since we have now determinedthe
V
\302\261
l/\\/\302\253
and\n
without
at that
discussion
the
N =
\302\2611
\302\261
ft\n
introduced
was
equation
\nsimplify
=
radians,
a
This
give\n
equation\n
values
these
and
equal zero for
na cannot
cosh
a
For
= 0
na
cosh
na
cos
equation may he equatedto
sin na sinh na
parts of this
imaginary
14\n
Chap.
(14-74)\n
..., n
(14-75)\n
the maximallyflat case
are
the
equation\n
sin b'
(14-76)\n
for even values
of
n
and
by
1
V
\342\200\234
V =
We
next
compare
these
kr/n,
angles
\302\273even
x,
n odd
with
those of Eq.
(14-77)\n
(14-78)\n
14-73.These
two\n
NETWORKS
AMPLIFIER
14-3
Art.
are
angles
14-75
Eq.
\nequating
following integer values
14-77 and 14-78\n
to Eqs.
the
for
equal
379\n
=
m
\342\200\224
(n
2N)/2
for m and k, found
by
for
even
for
odd
n\n
(14-79)\n
\342\200\224
k
and\n
\nare
= 0,
N
where
the
in
\nroots
be
can
\nflat
case
\nfor
the
\nin
The
14-15.
Fig.
are
multiplied
=
<fp
+
two
\nthese
can
This
\nan ellipse.
jo)p,
\nlength
roots
the
then
equations
of
by tanh
for
poles
pt
= cosh
for the equal
for the maximally
flat case
equation
a
ripple
the
by
the circle the maximally
real part the poles located
a. This constructionis illustrated
in Fig. 14-9 will be foundusefulin
of
of
for the equal
configurations
pole
and
6'
sin
ripple
for
Fig. 14-16. Pole locations
shev
case.\n
Cheby'
case,\n
of
equal ripple case are located on the periphery
be demonstrated
by noting from Eq. 14-74that if
\342\200\224
=
a
sinh
cos
b and wp = cosh a sin 6. From
av
it follows
that\n
a
+\n
=
cosh2
1\n
(14-80)\n
a\n
of an ellipse with its majoraxisalongthe j<>>p
a major
semiaxis of length cosh a and a minor
semiaxis
These
features
sinh a.
are illustrated in Fig. 14-16.\n
is the
\nhaving
that
roots
tabulated
sinha
This
= cos6' + j
parts of poles on
by (tanh a).\n
the
for
roots
The
Real
case.
\nChebyshev
\np
Location
14-15.
Fig.
case
angles
new
the
\nconstructing
circle
flat
maximally
the angles b and
of n,
value
(1) Change the radius of
to cosh a. (2) Multiply the
1
from
given
see
we
the
from
found
steps:
\nfollowing
case.\n
j sin b),
\n(-tanh a cosb +
\ncase
a
For
the equations, pi
Comparing
n\n
the
ripple
equal
2N)/2
max\302\254
equality
is the
cases
flat
\nimally
1
b'
are specified
in different orders according to
of these angles for the equal rippleand
basis for a simplified method for locating
they
The
14-79.
\342\200\224 \342\200\224
..., n.
1, 2,
although
equal
\nEq.
(n
axis
of
NETWORKS\n
AMPLIFIER
380\n
All
cutoff
unit
and
\nnitude
be
may
\nfrequency
\nmined
by
the
\nscaled
by
multiplying
the
\nas
of
\nin
the
of the
under
system
the
leaving
scaling
required.
network
a
\nor
\nbeen
found.
\nArt.
14-1.
the
\nfor
of
Q
the
case,
poles
in
\nbe
varied).
\nand
that
plate
high
\ntion
the
for
resistance
voltage
in
mind,
have
another
to
attention
might
a zero,
be ignored.
Keep
and let us turn
still
we
cannot
zero
problem
\nour
network.\n
But
adjusting
a tuning
resistance
the
(alternately,
\nthis
RC amplifier
practice
by
means
of
by
inductance
\nslug
\na
plane
two poles (conjugate pair) and one zero.The
in the complex plane can be controlledby control\302\254
This is ac\302\254
the
Q of the network.
\nling
\nthe
14-17.
complex
that
\ncomplished
Fig.
the
of
examination
only poles are present; no zeroshave
with this transfer characteristic,
networks
There
are
can
be constructed
by using results that have already
the shunt
Consider
amplifier network studied in
peaked
function for this network has,
transfer
ratio
The
voltage
high
\nlocation
of
these stagger-tuned characteristics.\n
to realize
shows
case
Chebyshev
\nbeen
there will be advantage
the
An
the
maximum
design, because the
actual
small numbers. We turn next to
Low-Pass Filter Amplifier.
\nfor
taken
(usually
ton
which is the
the magnitude
and
with
of working
used
networks
\namplifier
by
frequencies
In many cases,
study.
to the last step in the
ease
\nrelative
cutoff
be
normalized
frequency)
mag\302\254
deter\302\254
tube),
all
half-power
\ngain
frequency.
kilocycles/sec
gain
14\n
transfer function
case where the
and the magnitude
is large (as
the magnitude and frequencycan
been for unit
In a practical
far have
thus
computations
Chap.
network
\nshown in Fig. 14-17. If the tube
has
(and acts as a current source), the transferfunc\302\254
ratio
is\n
(14-81)\n
where
\nthe
Z(s)
network
is
the
shown,
of the plate
has the
impedance
impedance
the
circuit (termination).For
value\n
=
s
C
and the voltagetransfer
function
YM
ViCp)
The
network
has a pole
=
in
is,
^
(14-82)\n
+ l/RC\n
1
(
0>\302\253C Vj>
=
of p
terms
s/w\302\273\n
\\\n
(14-83)\n
+
1/cOnRC)\n
on the negativerealaxis
\nposition can be adjusted by adjustingeither R
of
or
C.
the
Return
s plane;
now to
its
our\n
\nlated
(and
\nfiers),
the
Here
zero:
we have a
iso\302\254
of
=
vm
=
is\n
\t\n
~
Vout(p)
p
the
function
transfer
(14-84)\n
G1(p)G,(p)
transfer functions are Eqs. 14-6and 14-83;
The two appropriate
where
ampli\302\254
function\n
G(p)
\nover-all
Might
two
pole
connected
\nbe
381\n
network with a singlepole.
can
be used
to cancel the unwanted zero? If the
stages
in cascade
such that the two networksare
together
the
this
is the case because of the isolating action
has
a
transfer
answer
is yes. The cascade connection
the
of
problem
\nthis
NETWORKS\n
AMPLIFIER
14-3\n
Art.
<o\342\200\2362CiC2\n
+
(p2
normalized
is
\302\253/&>\342\200\236
+ 1)(P
P/Q
If we
frequency.
=
JL
+ 1
(14-85)\n
/anRC)_\n
set\n
1\n
(14-86)\n
u>nRC\n
Q
the transferfunctionhas
two
cancellationis illustratedin Fig.
the pole and zero cancel, and
\ndefined in terms of Q. This
14-18.*\n
\\
1
Stage
poles
only
1\n
*\n
2\n
Stage
ju>\n
jw\n
a\n
a\n
<*\n
x\n
Composite
poles
x\n
<r\n
x\n
Fig.
14-18.
Use
of two
stages of amplification to give
* Another
\n14-6.
For
\nProc. IRE,
amplifier network with
a discussion
43, 923
of stagger-tuned
(1955).\n
conjugate
poles\n
zeros).\n
(no
the
same
characteristics
is shown
amplifier design, seeMcWhorter
and
in Prob.
Pettit,
NETWORKS\n
AMPLIFIER
382\n
a
\nbuild
This
response.
\nfrequency
are
\nworks
in Fig. 14-18 may now be usedto
to give either maximally flat or Chebyshev
in Fig. 14-19, where the net\302\254
is illustrated
to give maximally flat
varying
parameters
configuration
pole
by
adjusted
adjustedto
\nresponse.These poles might be
of
networks
\nthe
basic
same
the
using
response,
\nquency
14\n
shown
unit
network
basic
This
Chap.
a
give
fre\302\254
Chebyshev
blocks in the
building
form
of
14-18.\n
Fig.
Pole pair\n
Filter
\nto
made
\nadjusted
as
\nple,
we
\nhalf
of
\nwork
as
where,
was
in the
Vi(p)
9m
Vl(p)
0>nC
L(p
pi)\n
- Pa)(p
filter case, p
low-pass
-
(p
r
=
s/\302\253n.
-
(14-87)\n
Pa*)\n
\nfour
in Fig.
and the polesare adjusted
14-20. This time we
The
zeros.
is
stages
VUP)
V inip)
have
The over-all transfer
numbers
to designate
account.
into
taken
be
for
subscript
(using
four
that
Suppose
\nnetworks are cascaded
\nmust
moved
of Fig. 14-18. The transfer function of this net\302\254
zero
as\n
derived
as Eq. 14-6 and has two polesand one
network
the
amplifier
\nshown
zeros can be
the poles and
s plane by merely adjusting parameters(of
the
within
of practical adjustment:
the network Q can
range
with practical
so large
only
elements), the poles can be
filter characteristics,
to
give
band-pass
using the same basic
in the low-pass
filter case. As the building block in thisexam\302\254
will
make
use of the shunt peaked network which is the
first
\ncourse,
\nideas
4).\n
the
in
position
any
\nbe
Since
Amplifier.
=
(n
response
Band-Pass
to give maximallyflat\n
of networks
connection
Cascade
14-19.
Fiff.
the
such
configuration
of the zeros
effect
function for the
the
stage)\n
QmiQmiQmjQmj
\t\n
_
2C
1C
WftjO7n1b)n|C0n4Cr
zC
(p
w
~
(P
4\n
- Pn)(p
\342\200\234
P\302\253,)(P
Pa*)iP
(p
w
~
(P
Pat)
(.P
~
- pn)(p
- Pa*)(p
\t\n
Pat)(p
P12)
~
Pat*)\n
~ Pu)
\t\n
- Pat)(P~
(14-88)\n
Pa*)\n
Art.
NETWORKS\n
AMPLIFIER
14-3\n
383\n
thus has eight poles and four zeros.The
position
and
zeros
can
be
fixed
the
\nof the
circuit
for
each
by
adjusting
poles
Q
the poles also fixes the zeros, however).Fora maximally
\nstage (fixing
\nflat response,
that
the poles are assigned positions as shownin
suppose
one pole in the upper half plane,
14-20
contributes
One
\nFig.
(b).
stage
\none pole
in the
lower half plane, and one zero. The poleshavepositions\n
The cascaded
system
14-20.
Fig.
of
\nstages
for
\nand
the
are
\nthe
sinusoidal
\nEq.
14-88
of a
frequencies
\nthe
greatest
\nfrom
these
\nfrom
the
many
state,
steady
be
represented
from
poles
are
in the
practical
to
\302\253pi
on
effect
poles
circle having
frequencies
half-power
can
\nFor
\nIn
network);
periphery
copi
maximally
peaking
\n(shunt
on
networks):
amplifier
\nconfiguration
band-pass filter characteristics (four
(a) four cascaded stages; (b) pole-zero
flat response;
(c) basic unit of each stage
flat response curve.\n
(d) maximally
flat
Maximally
the
changing
when
a diameter
for
p =
the
jo>p,
(\302\253pi
\342\200\224
\302\253P2),
maximally
each
where
wP2
flat case. For
term
frequency
in
by a phasor as shown in Fig.
o>P2, the poles in the upper half plane have
function magnitude. The phasors
transfer
14-20(b).
rapidly
in magnitude, while the
lower half plane and the zerosarechanging
the mid-band frequency coq is high
designs,
phasors
slowly.
and
the\n
384\n
of
band
last
\nthe
frequencies
pass
Voutjp)
_\n
K\n
\342\200\234
is only
K
~
(P
~
(p
=\302\273\n
~
(p
<*nl0>nt0>ni0>ntClC2CzCi
~
as\n
Pu)(P ~
Pa*){p ~
~ Pli)(P ~
~ Pa*)(p ~
Pa*)(P
(14-89)\n
P\302\253\302\253)
P\302\253*)(P
and is given
Pll)(p
as\n
~
~
?>\302\253,)(?
constant
approximately
gmigmtgm&ml
K
in approximate form
Pai)(P
%4\n
these conditions
Under
is small.
wpi)
be written
Vin(p)
\nwhere
to
(\302\253p2
can
equation
Chap.
NETWORKS\n
AMPLIFIER
Pu)\n
Pa*)\n
(14-90)\n
This
\ntypes
\ncan
\nresponse
\nThe
\nThe
the
for
equation
the form required
have been studied. The poles this
that
of responses
the
for
of
two
equation
flat (Butterworth) frequency
be
to give either maximally
adjusted
as illustrated
or equal
frequency response.
ripple (Chebyshev)
in Fig. 14-20(d).
case
is
shown
flat
for
the
maximally
response
are evident. Such response
features
of this
response
band-pass
are
\ncharacteristics
over-all
the
outside
\nresponse
\nthese
the
n
specifications,
\ntions.
The
to
\nadjusted
For
the
in
\nspecified
case
of stages)
is determined.\n
the
(Butterworth),
frequency
over-all gain. From
pole
configuration
is
of Fig. 14-9 or from correspondingequa\302\254
of the actual
network are used and then
parameters
the required
real and imaginary part for each pole.\n
give
(Chebyshev)
equal
case, the ripple width is usually
ripple
addition
to the list given above. From these specifications,\n
(1) Determine
from
8
8
=
14-52
Eq.
from
cosh
a and
decibels\n
Eq.
(14-91)\n
14-54.\n
-- sinh-1\n
a =
Compute
as\n
(1 + e)
10 logio
(2) Calculate the factora
(3)\n
frequency
chart
the
from
selected
\nthen
(a) the mid-band
of decrease of the
(c) the rate
band, and (d) the
pass
number
(the
For the maximally
flat
fre\302\254
receivers.\n
for design are:
bandwidth,
of
as intermediate
applications
superheterodyne
The usual specifications
(b)
such
in
required
in
amplifiers
\nquency
\no)0,
ratio is of
voltage
tanh a. Draw
(14-92)\n
a circle
of
radius
(cosh
a),
the bandwidth, with a center at the mid-band
frequency.
the
on this circle as in the maximallyflatcase(given
poles
the
real part of each pole by tanh o. This
Multiply
\nequal to
\nLocate
\nabove).
\ngives
(4)
\tAdjust
\npole
the
pole
the
locations.\n
locations
parameters
for
of
the
the
Chebyshev case.\n
network
being used to
give these
AMPLIFIER
14-3\n
Ait
NETWORKS\n
385\n
procedure just outlined. Three
of
with
shunt
are to be stagger-tuned
\nstages
amplification
peaking
\nwith
an
The mid-band
characteristic.
equal-ripple
frequency is to be
\n5.0
and
the bandwidth
to the half-power frequencies
megacycles/sec,
be 500 kilocycles/sec.
The ripple width is specifiedas 1.0 db.
\nis to
\nWe
are
the mid-band
to find
required
frequency and the Q for eachof
\nthe three
The
R, L, and C of the shunt peaking
stages.
parameters
it often
is
\nnetwork
can
in turn
be found if one of the three is fixed
(as
the
interelectrode
and wiring capacitance).\n
\nin practice\342\200\224for
example,
An
the
illustrate
will
example
Following the steps just
8 = 10
Since
8
=\n
1
given
logio (1 +
1 =
10 logio (1
next
of the
value
this
e
=
-
1001
factor a, the
the
(14-94)\n
14-92
as\n
0.475\n
(14-96)\n
hyperbolic tangenthas
a =
our attention on
1\n
(14-95)\n
\\ sinh-1 1.963 =
tanh
We next turn
(14-93)\n
0.259\n
a from Eq.
- sinh\"1
=
a
For
factor
the
compute
equation\n
decibels\n
\302\253)
+ e) or
e =
and\n
We
the
e from
find
first
we
have\n
we
db,
design
value\n
the
0.442
(14-97)\n
configuration\342\200\224in
pole-zero
partic\302\254
of the poles. Figure 14-9 showsthe pole
location
\nular
to the
with
to the negative
and
at 6 =
3 as occurring
\nfor n =
respect
of 0.500 megacycleare
for a bandwidth
locations
These
axis.
pole
the
14-21. With respect to the midbandfrequency,
in Fig.
locations
0\302\260
60\302\260
\302\261
\nreal
\nshown
poles\n
3 poles\n
3
JU\n
zeros\n
3 poles\n
<\n
(a)\n
yig.
14-21.
\n(a)
Pole
full
location
scale
for maximally
(approximately);
flat
(b) region
frequency
of
interest.\n
response:
AMPLIFIER
386\n
=
Sl
+ >0.866)0.25,\n
(-0.5
s2 = (-1.0
=
\302\2533
+
>0.866)0.25\n
(\342\200\2240.5
plane are shownin Fig.
For the equal ripple
is multiplied
\nlocation
by (tanh a =
\npoles then become\n
\nscale even
so).
si' = (-0.221
s2'
=
Eqs.
Q is
and
defined
for
11-87
by Eq. 11-77
the
(14-99)\n
->0.866)0.25\n
these
to
are
equations
f <K
1), the
made
with
bandwidth
B is
as\n
=
B
pole
+ >0.866)0.25,\n
(corresponding
and
11-86
the
of
new
by
from
to
(hardly
response,the real part
locations
0.442). The
indicated
where all measurements
\nrespect to 5.0 megacycles/sec.\n
\nfound
14-21
+>0)0.25,
(-0.442
S3' = (-0.221
For the high-Qcase
(14-98)\n
>0)0.25,
\342\200\224
in the s
locations
actual
The
14\n
Chap.
locations:\n
the
have
NETWORKS\n
(14-100)\n
2{*b>n\n
as\n
Q
=
(14-101)\n
2T\n
these
Combining
two
we
equations,
r\\
*
where
\nBf
\nIn
is
/\302\273
is
the
the
bandwidth
14-22(b)
Fig.
natural
Wn
B
undamped
in cycles
the
distance
_
have\n
/\302\273\n
(1+102)\n
Bf\n
frequency
per second (the
from the ><o
in cycles per
second,and
common 2irterm
cancels).
axis
to
a
location
pole
jw\n
la)\n
Tig.
14-22.
(6)\n
pole location for the equalripplecase;(b)
to show distance to jo> axis.\n
\nenlarged
(a) New
Si
pole
is\n
d.
marked
NETWORKS\n
AMPLIFIER
14-3\n
Ait
\n(fa,,).
But
\nthis it
is seen
real part of the complex
polehasthe
11-70 this
Eq.
By
by Eq.
387\n
14-100, the bandwidth
is givenasB
=
value
From
2fo>,.
that\n
=
i
(14-103)\n
\302\247
d is the \342\200\234half bandwidth.\342\200\235\n
the
We
next
make
use of these last two identities to design
stagger\ntuned amplifier
to the computed pole configuration,
corresponding
\nunder
the
the
that
zeros have negligible effect. Assume
assumption
\nthat
1 will be made to correspond to the poles/ and
stage
conjugate,
or
1
stage
/,
\nBf
\nQ
\nFor
+ (0.866 X 0.25) = 5.217 megacycles/sec
X 0.221 X 0.25 = 0.110 megacycle/sec
=
5.0
=
2
=
fn/B,
=
\nQ
=
2 X
=
22.6
megacycles/sec
0.442
X 0.25 =
0.221 megacycle/sec
3 (s3'),\n
stage
fn =
\nBf
\nQ
5.0
=
2
=
43.2\n
\342\200\224
0.866
X 0.221
amplifier-network
\nacteristic
47.2
2 (s2'),\n
stage
\nB/
The
conjugate.\n
(\302\253/),\n
/, = 5.0
\nFor
its
stage 3 to poless'andits
and
its conjugate,
and
2 to
\nstage
For
distance
the
that
shown
is
in
X 0.25 =
\nBf,
\nQ\n
mc/sec
mc/sec
0.110 megacycle/sec
1
Stage
\n5.217
5.00\n
\n0.110
0.221\n
\n47.2\n
megacycles/sec
of the required equal-ripplechar\302\254
If required, the circuit parameters\n
realization
Fig. 14-23.
Stage
fn,
= 4.783
0.25
X
22.6\n
2\n
Stage
4.783\n
0.110\n
43.2\n
3\n
AMPLIFIER
388\n
of the
characteristics
the
\nhow
found, if one is
C can be
and
L,
R,
to
connection
\ntandem
the
give
The frequenciesmarked
fa
the
of
\nend
\ncies
was
fb are
fre\302\254
25\n
a
cogh
the
of
the
that
is
gain
the
because
\nfollows
5.224
megacycles/sec,
One advantage
to
\ncase.
Referring
\n(5.00
megacycles/sec)
\nratio
of
\ngiven
as\n
of
gains
(14-104)\n
megacycles/sec
\302\261
= 4.776
higher
poles
Fig.
the
ratio of gains
the maximally
over
case
equal-ripple
megacycles/sec (14-105)\n
gain at the mid-band frequency
be found
in terms of phasor lengths. The
may
case to the maximally flat case is
equal-ripple
the
14-22(a),
pole
same
\nzero
\nconverting
For
\nform.
and
(14-106)\n
distances)\n
the
zeros
by
may
the
this
maximally
megacycles/sec.\n
polar
problem\n
particular
equal
5.00
pole
the other polesand
is
approximately
cases. The three significant distances
be
found
numbers of Eqs. 14-98 and 14-99to
complex
to each of
both
\nsame for
zero
and
\ndistances)\n
=\n
(the
distances
flat case
for the equal-ripple configuration. This
are closer to the ja> axis in the equal-ripple
(other
at
to the
corresponding
frequencies
the
fb
ft
= 500
fb, fa
The
shows
individual
stages are combined by
frequency response.\n
equal-ripple
as\n
\ngiven
\nis
14-24
Figure
frequen\302\254
and
(4.75
fixed.
assumed
the preceding discussionthis
normalized
to unity.
By the specifications of this
more
convenient
to work with the 3-db point
are
5.25 megacycles/sec).
The frequencies fa and
been
it
\nproblem,
and
14\n
Chap.
In most of
band.
ripple
has
\nquency
NETWORKS\n
ripple gain
flat
gain
_
~~
(0.25)8\n
(0.223)2
X 0.1105\n
(14-107)\n
389\n
NETWORKS\n
AMPLIFIER
14*3\n
Art.
phase
disadvantagethat
a
as
frequency
input
in the maximally
this gain advantage is the
Offsetting
with respect to the
\nangle of the output
\nis not
so linear
in the equal-ripple
case as
such
in
amplifiers
\ndesigned
applications
Series
\ntory
(McGraw-Hill
is
account
\nThis
Wallman
\nHenry
\nare
based
on
issued
in
536
7,
\nneer,
\nmaximal
\nCircuits
Network
Jour.
\namplifiers,\"
on
\nbibliography
CT-2,
IRE,
found
New
F.
R.
the
to
with
(1941). Further discussions
in T. L. Martin, Jr., Electronic
of
Book
Baum,
and Seely,
Inc., New York,
LePage
1955),
York,
Co.,
\342\200\234Design
of
IF
broad-band
721 (1946). extensive
by H. A. Wheeler,
is given
subject
Engi\302\254
amplifiers
17, 519 and
Phys.,
Appl.
the
applied
analog
\npotential
and
Labora\302\254
Wireless
\342\200\234Cascade
(McGraw-Hill
Analysis
238-256,
pp.
\n1952),
to be
Inc.,
(Prentice-Hall,
\nGeneral
\nProc.
Rev., 5, 347
are
amplifiers
\nstagger-tuned
in Valley and
is that
amplifiers/'
V. D. Landon,
RCA
flatness,\"
of filter
theory
and
(1930)
tuning
stagger
Chap. 4.
an MIT Radiation Laboratory report by
1944. Two earlier papers on stagger tuning
\342\200\234Onthe
Butterworth,
of equal-ripple
Vol. 18 of the Radiation
Amplifiers,
Book
Co., Inc., New York, 1948)
Tube
Vacuum
\nWallman,
of
treatment
known
case.
flat
READING\n
FURTHER
The best
of
function
the application
as television.\n
restricts
characteristic
nonlinear
\nThis
the
An
\342\200\234The
of stagger-tuned
synthesis
filters,\"
86 (1955).\n
PROBLEMS\n
14-1. For the
the
\nfind
voltage
\nspond
\nin
to
the
\naccording
which
\namplifier
The
transfer
can
be
oscillatory
case.
components
Networks
figure,
accompanying
a current
considered
network shown in
has
high\n
source. Show a
have values to
of this type
amplifiers.\n
amplifier
the
The vacuum tube
function.
if the
configuration
cascade-connected
14-2.
\nfor
and
pole-zero
\ntypical
in
shown
ratio
resistance
plate
amplifiernetwork
corre\302\254
find application
Fig. 14-3is to
be
designed
The tube used in a 6AK5,
following specifications.
be
assumed. The capacitance
of
the
a gm of 4500 /xmho may
the
is
the
tube,\n
interstage capacitance only (including
stage
to
the
AMPLIFIER
390\n
NETWORKS\n
Chap.
14\n
has a value of 12 wxf.
If the
circuit
Q has a
\nvalue
of
the frequency
75
and
at resonance is 12.00 megacycles/sec,
\ndetermine:(a) the values
of
R
and
L, (b) the maximum
gain of the
and
bandwidth.
Answer, (a) 14.7 ohms,
\namplifier
stage,
(c) the
\n14.7
Mh,
(b)
(c) 160 kc.\n
375,
is to be used as a voltage couplingnetwork,
14-3.
The
network
For
the
ratio transfer function to have a maximally
flat\n
voltage
\n(a)
and
socket,
which
wiring)
Prob.
the
must
what
\nradians/sec,
\nDetermine
\n(of
of the
characteristic
frequency
a value for
flat
maximally
14-4. Plot the
14-3.\n
form,
l/-\\/l +
characteristic)
functionl/\\/l +
the
co
where
between
be the relationship
R and for L suchthat
\302\2532n>
and
R
L?
in
(b)
frequency
half-power
is 10 radians/sec.\n
0 ^ co ^ 4
o>2n
for
is
for n =
1, 2,
3.\n
\nand
\nButterworth
filter.
in the figure
shown
network
The
14-6.
Find
the
is known as a
of
magnitude
the
transfer
second-order
impedance
as\n
\ndefined
F2(io,)\n
hU\\n")
and
show
\nthe
frequency
it has maximally
that
response
Sketch
etc.\n
frequencies,
\nhalf-power
and
curve
flat frequencycharacteristics.
identify significant points such as the
i
In
14-6.
14-5
\nProb.
\nthat
the
two
\nwork
\n50
ohms
this
is
problem,
to be used
frequency
changes
(purely
the
second-order
-
-gmVg\n
Butterworth
network of
in the amplifier network shown. It is desired
characteristic
be maximally flat, but for this net\302\254
must be made: (1) the load resistance R must
be
and (2) the half-power frequency must be\n
resistive)
250
network,
|Zi20\342\200\231\302\253)l
transfer
the
\342\200\234
=
sx
=
\342\200\224
s2
1,
1
\342\200\224|
<\n
If-\n
'\n -If
18^
a
of
\nform
a
of
1
O\n
14-8.\n
defined
8th
where
eCV^co)
C\342\200\236(w) is
14-38 if c =
by Eq.
nth
the
0.25 for 0 ^
3.\n
order
Chebyshev
polynomial,
C\302\273(u>),
in
the
in
the
polynomial.\n
the 9th order
Write
14-11.
\nform
the
Write
14-10.
<10\n
the transfer impedance, Zi2(s)= U2(s)//i(s),
in the figure differs from that given in Prob.
the constant.\n
Determine
multiplier.
1, 2, and
n =
4 for
0\n
1
0\342\200\224\342\200\224\n
Prob.
polynomial
Chebyshev
^
O
if-\n ^
-
if-\n
v2
functionl/\\/l +
14-9. Plot the
\norder
\342\200\224|
\n4h\n
14-7
a constant
by
s2*.\n
of
shown
network
and
g3)\n
II\n
14-7.
that
Show
14-8.
transfer
characteristic
0 \t\n
0
1
1
Prob.
\n(j>
and s3 =
' <<10
*\t\n
>
\nonly
S2)(S _
_
\n2h\n
jl
\nthe
Sl)(s
+ j
*
last
the
form\n
\342\200\224
\t\n
>
_
Butterworth
flat frequency
the
has
(s
where
nf.\n
is a third-order
in a pentode circuit as shownin
show that the magnitude of the
in the figure
impedance
Zi2(s)
C to meet these
C = 0.113
X 10~4 henry,
1.414
a maximally
has
\nimpedance,
\nthat
this
For
used
be
might
(that
\nproblem).
shown
network
The
14-7.
=
L
Answer.
\nrequirements.
391\n
of L and
values
the
Determine
kilocycles/sec.
\nJUter
NETWORKS
AMPLIFIER
14
Chap.
Chebyshev polynomial,
Cg(w),
polynomial.\n
An equal-ripple frequency response of the form
given
by
14-49 has the following fixed parameters: n = 5 and e = 0.1. For
value of |G(ico)|, (b) the
determine:
(a) the maximum
response,
14-12.
\nEq.
\nthis
(d)
\ndecibels,
\n0.414
the
half-power
Repeat
0.913,
|G(j\302\253)|
0.79
(c)
14-14. It is
the
ripple
the
pass
frequency.
Prob. 14-12 for n
db, (d) 1.076.\n
requiredthat a
(c) the ripple width 8 in
Answers, (a) 1, (b) 0.953, (c)
band,
system
= 4 and
having
by Eq. 14-49 have a half-power
width
is limited to 0.5 db, what
\nequation given
\nIf
in
1.07.\n
(d)
db,
14-13.
\n(b)
of
value
\nminimum
\nmay have?\n
The constant
14-15.
\nUnder this condition,
of n.\n
\nand a function
e
=
0.2.
the
Answers,
(a)
1,
frequency-response
frequency
at
=
\302\253
1.10.
is the minimum
value
is
in
factor a in Eq. 14-54
fixed
derive a relationship betweenn and8.
assumed
Sketch
n
value.
8
NETWORKS
AMPLIFIER
392
to
Refer
14-16.
of the
\nthe locations
\nhalf-power
to
be
\nof
n,
and
plotted
14\n
give
case,
=\n
poles
a
with n = 3 has
low-pass
filter-amplifier
An equal ripple (Chebyshev) filter
of o> = 1.00.
frequency
with the same half-power frequency, the samevalue
designed
a ripple
fre\302\254
width of 0.5 db. Determine the upper
lower
limits
of the pass band
(the frequency at which the equalripple
A
14-17.
\nis
for n =
5. For this
for the Chebyshev caseif a
14-16
Fig.
Chap.
flat
maximally
and
\nquency
\nends).\n
14-18.
is
\nIt
specified
is to
A filter-amplifier
that
the magnitude
be designed on the Chebyshevbasis.
of the voltage transfer ratio must
be\n
1\n
G \342\226\240
1\n
0.5
V//////A\n
db\n
t\n
G-0.2\n
777777777777/\n
0
l.(\n 1
Prob.
=
\302\253
\nrange
\nflat
frequency
\nfier
is
to have
the
\nto
1.0
The
14-19.
half-power
\nrequirements.
\ntem.
ripple
made
(gain vs.
to)
for
composite
sys\302\254
for
response
frequency
first
the
to
\nDetermine
the
mid-band
network is to
stage.
the
frequency
be designedto
com\302\254
the
fol\302\254
is 10.0 megacycles/sec,
the
half-power
to 0.5 db.
frequencies
is to be
of
\nnetwork
mid-band
14-21.
\nfrequency
\nflat
gain for the
500 kilocycles/sec
is limited
Based on gain requirements, a decision
to use 3 stages.
Stagger tune these three stages
a Chebyshev
give
equal ripple characteristic, using the basic
shown
in Fig.
14-23.
Determine
Bf, and Q for each stage.
ratio
of the maximally
the
flat gain to the equal ripplegain
to
bandwidth
\ntion
amplifier
the
specifications:
\nlowing
\nthe
response
frequency
A band-pass
14-20.
\nthe
maximum
system.\n
\nposite
\nat
the
the
Compute
the second stage. Plot the
\nRepeat for
\nis
(b)
Plot
(c)
14-18.\n
sections. (No specifications are given for the
to 1.1.) What is the required value of n?\n
a maximally
network
of Fig. 14-18 is to be used to give
for a low-pass
characteristic
filter amplifier. The ampli\302\254
a bandwidth
of 100 kilocycles/sec (measuredfromw = 0
an amplifier to meet these
(a) Design
frequency),
a vacuum
tube for each stage. Specify all com\302\254
Select
values,
\nponent
it)\n
crosshatch
the
inside
1.1
frequency.\n
amplifier network having
given in Prob. 14-20 but for the
Design a band-pass
and
amplifica\302\254
bandwidth
the
mid-band
maximally
case.\n
14-22.
\namplifier
In this
networks
problem,
designed
we will investigate the transient
on the equal ripple and maximally
response
of
flat\n
this
For
basis.
NETWORKS
AMPLIFIER
14
Chap.
the
of
\ns plane
* 2 corresponding
let n
problem,
and
\nand
maximally
rise
time
(s
\nthe
rise
of
\ninition
the
times
value.
\nstate)
Two
overshoot
response,
The
overshoot
\ndetermine
\namplifier
the
these
\namplifier
seems
overshoot
for
two cases (equal ripple
of interest are the\n
quantities
for the
input. One
def\302\254
as\n
\342\200\224
value)
(final
value)
X 100%\n
value)\n
current is a
step
function;
that
is,
i(t)
= u(t),
for (a) a maximally flat designed
(b) an equal-ripple
design amplifier network.
for
two
the
quantities
amplifier networks. Which
to
have
transient
response? (Check point:
superior
the maximally
flat amplifier is about 4%.)\n
time
and
network,
\nCompare
\nthe
rise
s2)\n
from a step-function
is defined
\n(final
If the driving-force
\302\253i)(s
time interval, measured in secondsbetween
that has 10% and 90% of the final (steady-
transient
(maximum
-
time-domain
resulting
is the
time
values
different
have
\302\2532
flat).
and
the
1\n
Vt(8)
/l(s)
si
to two polesin
form,\n
Z12 (s)\n
where
393\n
and
overshoot
15\n
CHAPTER
DIAGRAMS\n
BLOCK
widely used
Block diagrams are
\nentirely,
in
\nindicate
the
\ndiagrams,
Severalblocks
\ncomponents.
block
one
or
blocks
the
Indeed,
\ncomputer.
a single
represent
compo\302\254
a complex mechanism such as a digital
the solution of a math\302\254
may represent
represent
may
to
used
be
may
\nnent
sym\302\254
block represents components:
describing
system.
or in combination.
Lines given direction by arrows
part,
of operations
that take place in the system. Block
sequence
as
we
shall
use them,
are not pictorial representations of
The
a
in
\nbolism
a shorthand
as
engineers
by
significance. The block
a system in such a
for representing
as
a method
provides
between the input and the
\nto express
a cause-and-effect
relationship
are also referred
to by the descriptive name
Block
diagrams
the
where
we define
signal as any causal quantity
flow
diagrams,
introduced
into
a system
(in contrast to
As
we study
we should keep in mind the
block diagrams,
\nematical
no
with
equation
direct
physical
\ndiagram
way
\noutput.
\nsignal
noise).\n
\nintentionally
underlying
\nobjectives
(1) Block
diagrams
The
\ndiagrams.
Block
(2)
\nthe
(3)
use:\n
their
in
it
\nonly
The
construction
Once
\nysis.
of a
in
reduction
\nThis
system
the
Let
\n(where
does.\342\200\235\n
one step in
system
anal\302\254
be manipulated by a
a simplified
equivalent block diagram.
is equivalent
to algebraic reduction
blocks
may
equations.\n
block
diasrams\n
as
system be designated as tq and the output
such as voltage, current, etc.). The relation\302\254
any variable
input
v is
find
complexity
Basic operations for
15-1.
it
block diagram is
the
to
algebra
function, indicate
They tell the engineer not
system.
\342\200\234what
constructed,
of
\nsystem
is\342\200\235
but
notation.\n
the transfer
with
of the
behavior
dynamic
schematic
detailed
draw
together
diagrams,
\342\200\234what
\nof
than
are easier to
block diagram is shorthand
to a
the
between
\nship
be
\nmay
expressed
input
and
in
terms
output
of an
\noperator G, such that
\nFig.
15-1.
We define the
Block
y2 =
diagram.
blockdiagram
in
shown
394\n
Fig.
15-1
Qvl
(15-1)\n
to be equivalent
to\n
Ait
395\n
DIAGRAMS\n
BLOCK
15-1\n
G may be any operator any linear
of operators.
For example,
\nbination
suppose that G represents
is
This means that the input
\ntiation
with
to time.
respect
The
block
as dvi(t)/dt
to be equal to the output,
\nentiated
diagram
is shown in Fig. 15-2(a).If
operation
\nequivalent to this statement
of a transform function Fi(s) by a
be
is multiplication
performed
the block diagram and algebraic equivalent
then
function
Kis,
as
A linear combination of
shown
in
15-2(b).
Fig.
this equation.
The function
com\302\254
or
differen\302\254
\302\273i(f)
differ\302\254
v%(t).
the
\nto
\ntransfer
\nare
operations,\n
doM\n
v2(t)w'\n
dt\n
V2(8)-KlsVl(s)\n
V2M-.K1\302\2532ViW+*2ViM\n
(0\n
and equation equivalents.\n
Block diagram
15-2.
Fig.
-Vl
v3-vx\302\261v2
+>Q\342\200\224=>
\n\302\261T
\nv2\n
15-3.
Fig.
Summing
V2\n
point
symbol.\n
V4\n
\342\226\240*\302\273\n
Vi-Vx
+ V2-V3\n
n\n
\342\200\224
v3\n
16-4.
Fig.
as
expressed
\nin
the
G(s)
=
Kis2
operates
expression
Summing
+ K3, is
on
Another basic symbolis
shown
Fi(s)
in
point.\n
shown in Fig. 15-2(c).
Each
term
as illustrated.\n
independently
The circle with arrows
15-3.
Fig.
point. Quantities entering the circleor
are
either
added
or subtracted according to directions
\nsumming
point
\342\200\224
If the sign on the
arrow.
\nin the
form
of a + sign or a
on the
sign
to be positive.
in a diagram, it is presumed
\narrow
marked
V i is omitted
a block
\nIn constructing
however, it is wise to use a signfor
diagram,
be doubt. This is very
in
whenever
there
\neach
might
necessary
input
when
more
two inputs enter a summing point as illus\302\254
\nthe
case
than
into
\npointing
\ntrated
in
Fig.
it
is a
15-4.\n
summing
BLOCK DIAGRAMS\n
396\n
a
When
line
single
diagram separates into
on a block
all lines carry the
a
is called
\nSuch a point
pickoff point. pickoff
\nillustrated in Fig.
the same point,
\nlines leaving
Chap. 15\n
A
in
point
more
variable.
unaltered
same
or
two
is
a system
15-5.\n
Cj Vi\n
\342\200\242\n
Vi
>\n
=\t\n
G\n
Vi\n
Vi\n
V!\n
Vl
HVi\n
H\n
>\n
\342\226\272
V.\n
(6)\n
(a)\n
Fig.
Pickoff
15-5.
Tandem
15-6.
Fig.
points.\n
system.\n
that several blocks are connectedin cascade
as
(or
tandem)
\nshown in Fig. 15-6. The transfer function for each blockis given
on
and is the ratio of the output to the
\nthe figure,
for
each
of the
input
Suppose
Since\n
cases.
\nthree
total
the
(or over-all)
Vt
Vz
Vj
V*\n
Vi
V,
Vz
Vi\n
function
transfer
Gt
This
\nas
equality
be
will
in a
being
(15-3)\n
\342\200\234no
loading\342\200\235
a
between
Thus blocks connected in
Expansion
blocks,
tandem\n
of blocks.\n
a single equivalentblock multiplying
the
functions
for the equivalent transfer function.
together
be expanded
into several parts as illustrated in
system
may
be combined into
\ntransfer
\nwise
15-7.
is\n
GiGiGz\n
later section.
Fig.
may
=
depends on there
discussed
(15-2)\n
by
Like\302\254
15-7.\n
\nFig.
15-2.
Block
diagrams
for electrical elements\n
Instantaneous value of voltageand
ments
are
related
by the
following
current
equations.\n
for
the
electrical
ele\302\254\n
DIAGRAMS\n
BLOCK
15-2\n
Ait
resistance:
\302\273*(<)
or
* Qvn(t),
G
ii{t) = 1/L j
dt or
\342\204\242
1/R\n
dt
v^t)
(15-4)\n
\342\200\224
C
dvc(t)/dt\n
ic(t)
to the block diagrams
are equivalent
These equations
ia(t)
L dtL(t)/dt
= 1/C
J ic(t)
Vc(0
capacitance:
or
Riii(t)
=
i>l(0
inductance:
*
397\n
in
shown
Fig.\n
15-8.\n
Current
output
Voltage
output\n
Resistance\n
Inductance\n
Capacitance\n
To
\nthe
\nwith
chart
a similar
construct
of
transformation
Laplace
conditions
initial
resistance:
ignored.
=
Fr(s)
inductance:
Vl(s)
Vc(s)
capacitance:
The
Block diagrams for electricalelements.\n
15-8.
Tig.
same
or
RIr(s)
=
LsIl(s)
=
(1 /Cs)Ic(s)
Ir(s)
GVr(s),
sinusoidal
the
for
\nFig.
15-9
\nand
the
for
transform
for
expressions
impedance
or admittance.
impedance
and
admittance
Impedance
chart
diagrams
and
elements.\n
Admittance\n
transfer
of
The block diagrams
shown in Fig. 15-9\n
Capacitance\n
Block
the
admit\302\254
Inductance\n
15-9.
of
the transform
Resistance\n
Tig.
8
with
voltage transform),a
may be constructed as shownin
elements
electrical
(15-5)\n
steady state,
impedance is the ratio
Since
transform
by
(jw).
to
the
transform
current
\nvoltage
(and, similarly,
is the
\ntance
ratio
of the current to the
\nblock diagrams
1/R\n
Cslc(s)\n
the
\nreplaced
G =
\342\200\224
Il(s) = (1/Ls)Vl(s)
or Ic(s) =
or
for
apply
equations
for the transform voltages and currents,
each expression
in Eq. 15-4 will be taken
The resulting
equations are:\n
functions for
electrical\n
398\n
\nfor
sinusoidal
the
to
reduce
Chap.
DIAGRAMS\n
BLOCK
with the substitution
state
steady
of
15\n
(jo>)
s.\n
\nillustrated
\nThe
this
of
use
The
an
by
of block diagrams for the elementswill
be
the electric circuit of Fig. 15-10.
Consider
example.
chart
and
current
relating
equation
voltage
is\n
=
v(t)
in
or,
L
^
of voltage
terms
+ Ri(t)
(15-6)\n
and current
trans\302\254
\nforms,\n
if
RL
15-10.
Fig.
circuit.\n
=
15-7 and 15-8
Equations
be
may
=
1l(s)
There remains
these
this
\n/($), corresponding to
Block
16-11.
Fig.
for
\nblock
\nmay
\nis,
at
arrive
be
current
the block diagram equivalent
the input and output quantities.
15-1 and the
as
the
V
(s)
the
each
found
Now,
following
equivalent block of Fig.
15-1,\n
Fig. 15-12. Block
of net-
as
output
diagram
for
of Fig. 15-11. Instead of this diagramwith
in the network, a single blockrepresentation
solving
Eq. 15-7 for the ratio of outputto input,that
Thus\n
voltage.
1
I(s)
V(s)
RL\n
network.\n
system
element
by
to
the
and
response.
of Fig.
work
we
diagram
15-10.
Eq.
(15-9)\n
(15-11)\n
the excitationand
by
suggested
pattern
form and
(15-10)\n
identify
the input
let us identify
problem,
following
draw
must
We
equations.
the
in
Fl(s)
we
before
task
one
drop
(15-8)\n
= RI(s)
7*(s)
is
voltage
7*(s)
written
(15-7)\n
voltage
VR(s) = VL(s)
-
V(s)
+
VL(s)
\norder:\n
\nthe
+ RI(s)
This last equation tells us that the
applied
across
the
inductor
addedto
the
to the voltage drop
that is,\n
the resistor;
V(s)
\nFor
Lsl(s)
= 0.
t(0+)
\nequal
\nacross
\nof
=
7(s)
Ls
+
_
R
1/R\n
Ls/R
+
(15-12)\n
1\n
a
Thus
an
\nform,
is
block diagram of Fig. 15-11,amplifiedin
The two diagrams for one networksug\302\254
in Fig.
15-12.
shown
can
be
found
for reducing one to the other.
a method
Rules
a
will
in
later
be
section.\n
given
manipulations
such
\nfor
15-3.
An
Open-loop
and
open-loop
system
block
This
\nput.
block
closed-loop
in
15-13.
Fig.
in Fig. 15-13
=
G(s)E(s)
loop is closedaround
the
input identified as
by two equations:\n
\ndescribed
E(s) = Fi(s)
=
y2(s)
The quantity E(s)
E(s)
\ning
equations
F2(s)\n
=
this
\nfunction
\nrelates
equation
and
the
to
Vx(8)
new
system
is
(15-15)\n
error
the
transform.
Eliminat\302\254
gives\n
G(s)\n
L(s)\n
1 +
(15-16)\n
G(s)\n
spoken of as the open-loop
closed-loop transfer function. The last equation
transfer function.\n
the closed-loop
as the
open-loop
shown\n
(15-14)\n
is sometimes
G(s)
L(s)
Fi(s). The
G(s)E(s)
Vi(s)\n
In
15-13 as
Closed-loopsystem.\n
\302\261V2(s)
is now identifiedas
two
the
from
of Fig.
Fig. 15-14.
a new
with
out\302\254
(15-13)\n
system
system.
Open-loop
15-14,
with E(s) as the
the error input) and 72(s)as the
to the algebraic equation\n
is equivalent
7*(s)
Fig.
equivalents\n
be called
diagram
Suppose that a
diagram
is represented
later
will
(which
\ninput
399\n
of the
equivalent
that
\ngest
DIAGRAMS\n
BLOCK
Art. 15-3\n
}
Bis)
Q
transfer
1\n
V2{s)\n
y\n
G(s)\n
A(s)\n
H{8)\n
Fig.
In a
\nelement
\nhaving
16-15.
Closed-loop
system.\n
an
more generalized representation of a closed-loop
system,
of elements is included in the feedback
or combination
loop
in
a
is
shown
15-15.\n
function
Such
a transfer
Fig.
H(s).
system
400
The
for
equations
this
system
=
and
B(s)
_~ L(s)
-
from these
are eliminated
\302\253\302\253.
.
1
\nF.(\302\253)
Thus, the two blocks
\nsubstituted for the other.\n
16-16.
Pig.
Block
15-4.
(15-19)\n
equa\302\254
results\n
there
\ntions,
(15-18)\n
G(s)B(s)\n
--- H(s)V,(s)\n
A(t)
(s)
(15-17)\n
\302\261A(s)\n
r,(\302\273)
=
Vi(s)
A
15\n
are\n
B(s)
If the two quantities
Chap.
DIAGRAMS\n
BLOCK
of Fig. 15-16
are
one
and
equivalent,
open- and
Equivalent
(15-20)\n
G(s)H(s)\n
+
closed-loop
may
be
system.\n
transformations\n
diagram
be
sections
can
of block diagrams in the last two
Manipulations
in
into a system of block diagram algebra. The objective
\ngeneralized
num\302\254
\nthe use
of this algebra is simplification; that is, reduction the
of
\nber
of blocks
and
number
the number
of loops of a system. A large
in
\nrules for
manipulations
are tabulated
\nimportant
Rule 1.
\nthe
Vt
\n(by
the
=
Vi
law
associative
be
may
in the
illustrated
As
Fa = Vi +
+ Vt +
2.
F, +
Va
\342\200\224P\342\200\224*9\342\200\224\n
|v3
|v3
in tandem
Blocks
|v2\n
1\n
are multiplied.\n
V2\n
V2\n
Ox\n
G2\n
0\\G2\n
Rule
*
See
references
at the
the
most
altering
of algebra).\n
Rule
Rule
without
interchanged
figure,\n
\342\200\224*<?\342\200\224[v2
of
few
below.\n
Summationpoints
system.
the literature;* a
are given in
end of the
chapter.\n
2\n
3.
Rule
DIAGRAMS\n
BLOCK
15-4\n
Aft.
Blocks
in parallel
are added.\n
3\n
Rule
The
following
four
or
summing-points
\npast
Rule
rules
the procedure for
indicate
take-off
4.\n
\nRule
5\n
6.\n
Rule
The
shifting blockc
points.\n
Rule
Rule
401\n
following
8.
Removing
rules
are applications
a block
7\n
of the rules
in the feedback
loop.\n
given above.
DIAGRAMS\n
BLOCK
402\n
from
Transformation
9.
Rule
open-loop
Rule
15-5. Limitations the
in
\nsented
block
by
to
\nence
be
must
Care
Fig.
G
\nfunctions
from
Transformation
10.
Rule
exercised
diagrams.
15-17.
and
Fig.
block
Chap.
to closed-loop.\n
9\n
closed-loop
to open-loop.\n
representation
diagram
15\n
of physical
in dividing a system into parts to
This limitation
may be illustrated
systems\n
be
by
repre\302\254
refer\302\254
Two systems are shown, characterized by transfer
two systems can be connected together in\n
H. The
16-17.
Fig.
16-18.
Two-terminal-pair
net\302\254\n
work.\n
F3, only if in making this connectionV2 is
This
is the assumption
of negligible loading of one system
\nunaffected.
of the electrical
In
terms
network shown in Fig. 15-18,
another.
\nby
means that the output current
\nthe
of negligible
loading
assumption
so
If
this
is
zero
or be
small that it may be
\n12 must
equal
must be considered in writingthe dynamic
\nnot the case, the interaction
the system. In other words, a sectionof a
to describe
\nequations
making
tandem,
V2 =
neglected.
system
\ndynamic
\nwithout
\nthe
cannot
the
considering
rest
of
the
be
separated
interaction
from the system
of this section of the
for analysis
system
with
system.\n
shown
To illustrate, consider the double RC networkof Fig.
15-19,
for
into two separate RC networks. The transferfunction
\nseparated
be found by multiplying the transfer func\302\254
\nthe entire
cannot
system
\ntions
of the
of the system, since the second network loadsthe
parts
first
RC
that
\nfirst;
is, it causes a current to flow in the output of the
in
5
of
this
section.
\nnetwork.
This will be discussed
Example
However,
an
is
connected
between
the
as
such
\nif some
amplifier,
isolating
device,
\ntwo
networks,
then\n
Gt
=
GiGi&cf\n
(15-21)\n
Art.
BLOCK
15>;5\n
the over-all transfer function, G\\ is the transfer
function
RC
Gt is the transfer function of the secondRC
section,
function of the cathode followeramplifier\n
Get is the transfer
Gt is
where
first
\nthe
and
\ntion,
403\n
DIAGRAMS\n
sec\302\254
O\n
\342\200\224VV\\r\n
Cathode\n
of
R2\n
follower\n
C2:\n
(isolation)\n
o\n
(b)\n
15-19.
Fig.
a
(considered
\nhas
Several
\nfer
the
\nof
block
of
function,
restrictions
draws
and
impedance
will
examples
input
high
follows
This
constant).
(b) modificationof (a).\n
RC network;
Double
(a)
the cathode follower
because
negligible current at
its
input.\n
to illustrate the conceptof the trans\302\254
of physical systems, and
representation
be given
diagram
on tandem
connection\n
o\342\200\224
of blocks.\n
\302\260\n
'WV-\n
T
^pC\n
1\n
Example
\"\342\226\240
15-20
as
\napplication
<Rz\n
V\n
Fig.\n
lag network. It finds
as a
is known
in
shown
network
electrical
The
network
a compensating
in
=
V* =
these
=
are
Lag
network.\n
by
given
Kirch-
as\n
Vi
From
15-20.
Fig.
\nservomechanisms. The relationships be-\n
tween the instantaneous
voltages and currents
\nhoff\342\200\231s
law
\302\2562\n
<
rT\\\n
equations,
(5T+
+
q J
%
dt
(15-22)\n
idt + R2i
^ J
the transfer
R2Cs
_
R2)i
(Ri
1
r2)Cs~+1
(15-23)\n
function is found
-
be\n
(s 4- 1/R2C)
#2
Rl'+Tt2
to
[S
+
i/{Ri
+
__\n
k2)U\\\n
(15-24)\n
block
The
DIAGRAMS\n
BLOCK
404\n
*
of the lag
equivalent
diagram
r,\302\253+i\n
in
Fig.
r2
=
Chap.
network of Fig. 15-20is
where
15-21
(\302\253x
+
shown\n
\342\200\224
T\\
15\n
and\n
R2C
r2)C.\n
,\n
r*+i\n
2\n
Example
16-21.
Fig.
Lag
block
network
jn
circuit, the transfer functionis
to
found
tube
Vacuum
be\n
hRl\n
Vi\n
16-22.
the equivalent\n
From
15-22.
Fig.
h\n
Fig.
amplifier is shown
An electronic
\ndiagram.
Tp
(15-25)\n
+
Ri\n
(a) schematic;
amplifier:
equivalent\n
(b)
circuit.\n
S\n
Example
\nin
15-23.
Fig.
Fig.
equivalent
schematic
follower
cathode
A
16-23.
circuit
transfer
The
Cathode
to
follower:
Vi
value
electronic
of the transfer
circuit.\n
(a) schematic; (b)
are shown
from
the\n
equivalent
circuit.\n
be\n
Yj
The
and the equivalent circuit
function
can be determined
+
(m/1
_
r9/{
1
+
p)Rk\n
m)
+
(15-26)\n
Rk\n
function is a constant
equal
to
the
gain
of
the\n
Art.
15-5\n
Example
4\n
for
network
The
a very
has
\nRg
is
high
16-25.
Tig.
the
\ngrid
draws
is actually
cathode
a
these assumptions, it is seen that
of a lag network, shownin Fig.
shown
15-20,
in
Fig.\n
0\n
diagram
representa\302\254
is
in Fig-\n
shown
system
the
of
\ntion
the
that
assumed
With
composed
follower,
be
may
block
The
15-23.
Also, it
network).
current.
negligible
network
Block diagram equivalent of Fig. 15-24.\n
of the
remainder
load
\nand
example is shown in Fig. 15-24.Theresistance
value such that the current flowingthroughit
with
that
Ri (in other words, R0 does not\n
through
this
compared
\nnegligible
\nthe
405\n
DIAGRAMS\n
BLOCK
r2\n
Ri\n
Vl
-C2
\302\260l~\n
V2\n
15-25.\n
n\n
5\n
Example
this
For
network
\nRC
\nt2.
By
shown
Kirchhoff\342\200\231s
V\\
in Fig.
the
law,
=
Riii
ii)
+\n
(*2
\342\200\224
4*1)
dt
+
C2\n
\ninitially
corresponding
uncharged,\n
4*1
and
are\n
+
4*2\n dt\n
-q~
(15-27)\n
J
4*2 dt\n
v2\n
The
network.\n
dt\n
Riiz
I\n
RC
currents are marked
equations
system
Double
16-26.
Tig.
the double
15-26. The two
consider
example,
I\n
transform
equations
are, if the capacitors
are
BLOCK
406\n
=
F,(\302\253)
=
\nThe
15-27
block
-
15\n
/\342\200\242(\302\253)]\n
RMs) + els
/iW
(15\342\200\23028)
/.(.)\n
input and V* as the output, the blockdiagram
from the algebraic equations of Eq. 15-28.
is constructed
Fi
Considering
Fig.
\302\261
l/i(\302\253)
~ /l(s)1 +
chlh(s)
\nr,(s)
\nof
+
*./\342\200\242(\302\273)
=
0
Chap.
DIAGRAMS\n
diagram
Fig.
as the
reduction
15-27.
following
procedure,
Double
BC network
block
the rules stated
in\n
diagram.\n
Combined\n
(6)\n
Fig.
15-28.
Block
diagram
reduction.\n
in Fig.
The
of
blocks
Art. 15-4, is outlined by the sequence
15-28(a).
are
shown
in
\nsimplified block diagram and compositetransferfunction
result
could be found by algebraic manipula\302\254
The
same
15-28(b).
\nFig.
of Eq. 15-28 instead of the manipulationof blocks.\n
\ntion
of the
equations
READING\n
FURTHER
discussions
Excellent
literature
by
of the use
T. M. Stout,
of block diagrams are
\342\200\234A
block
diagram
approach
found
to
in
the\n
network\n
Chap.15\n
AIEE
Trans.
analysis,\342\200\235
(Applications
for
solutions
\342\200\234
\nand
Block-diagram
\nAIEE
\342\200\234Block
\nGraybeal,
\n985
and
approach
3, 255 (1952),
Industry),
vacuum-tube
transformation,\342\200\235Elec.
to the same
Feedback
Mason,
\ngraphs,\342\200\235
Proc.
1144 (1953) is
especially
Eng.,
of signal
theory\342\200\224some properties
41,
IRE,
by T. D.
70,
subject, the paper
\342\200\234
J.
Trans.
circuits,\342\200\235
9, 561 (1953), and
Electronics),
network
diagram
a different
For
(1951).
S.
\nby
and
(Communication
407\n
DIAGRAMS\n
BLOCK
flow
recommended.\n
PROBLEMS\n
For
the network
diagram
with
15-1.
\na
block
excitation
the
\nbe
15-2.
15-3.
\na
15-
\t4.
Repeat
16-
\t5.
The
\ntation
shows
\nidentification.
\nthe
quantity
amplifier.
number
Give
F*(\302\253)
the
associated
Rl
as
the
draw
response.\n
shown
in
the
figure.\n
shown
in
the
figure.\n
Prob.
schematic
a
but
the network
the network
16-4.\n
the network shown in the figure.\n
shown
in the figure is the equivalent circuitof
block diagram represen\302\254
The accompanying
of blocks and arrows labeled a, b, ..., g for
function for each block and identify
transfer
all marked arrows.\n
with
15-1 for
Prob.
tube
vacuum
as the excitation
16-3.
Prob.
(a)
figure,
for each element, considering Vi(s) to
and I(s) the response (or output), (b)
Prob. 15-1 for
Prob. 15-1 for
Repeat
Repeat
accompanying
block
one
(or input)
with
Fi($)
(a)
part
\nRepeat
shown in the
v\342\200\236\n
)-ltv\n
o\n
Prob.
16-6.\n
16-6.
Simplify
value
the
Give
the block diagram shown as
of Gt in terms of Gi, G%, G%, and
(a) to
form
the
of
(b).\n
GY\n
00\n
Prob.
Reduce
16-7.
the
giving
\n(b),
16-6.\n
shown
the block diagram shown in (a) to the form
value
for A and B in the feed-forwardand feedback
in
\nloops.\n
(a)
(6)\n
Prob.
16-8.
Repeat
Prob.
15-7 for the
16-7.\n
block diagram
shown
Prob.
T. M. Stout* has shown that many
\nnonlinear
element
can be represented by
shown
\nform
diode
\nequivalent
T.
\nReport
accom\302\254
16-8.\n
16-9.
*
the
figure.\n
\npanying
\nis a
in
M.
No.
networks
one
containing
a blockdiagram
The nonlinear element in the circuit
having
in (a).
of
\342\200\224
=
described
by a conductance
function, i2
g(t>i v2).
block
for the system is shown in (c). Donotattempt\n
diagram
below
\342\200\234Block
Stout,
E.
E.
9,
Dept.,
diagram
simplification
of
some
University of Washington,
1952.\n
nonlinear
the
(b)
The
circuits,\342\200\235
409\n
DIAGRAMS\n
BLOCK
15\n
Chap.
the nonlinear block with other blocks(thisisnot permitted
does not hold). By manipulation and combination
superposition
reduce
the block diagram of (c) to the standardform
blocks,
in (a).
Give the values for Gi(s), Gi(s), and H(s).\n
to combine
\nsince
the
\nof
\nshown
Input
Forward
block
block
Output
block\n
(a)\n
(c)\n
Prob.
The
16-10.
\nwith
\nShow
\nrepresented
series
and
and
discussion
by
figure shows a general ladder network
shunt elements or combinations of elements.
that
the ladder network can be
equations
of the figure.\n
diagram
accompanying
alternating
by
15-9.\n
the
block
-X/3-/5\n
/i-/3\n
1/Z3\n
Prob.
15-10.\n
to\n
CHAPTER
16-1. Feedback
\nfeedback
\nat
by
For
input.
electromechanical
and
path
the
systems\n
electric
Many
SYSTEMS\n
FEEDBACK
IN
STABILITY
16\n
Fig.
example,
incorporate
a so-called
the output is
16-1 (a) shows a plate-tuned
a part of
of which
means
systems
reintroduced
oscillator.\n
(a)\n
(0\n
Fig.
16-1.
(b)
temperature
Examples
of feedback
regulating
system;
Feedback, which is essentialto
\nplate
means
by
\nplished
the
to
\nthe
Figure
system.
\nture-regulating
\nmeans
grid.
of the
of
temperature
the
the
systems: (a) plate-tunedoscillator;\n
(c) feedback system representation.\n
operation
of
an
oscillator,
is
accom\302\254
coupled coils returning the output from the
16-1 (b) shows an electromechanicaltempera\302\254
is accomplished
in this system by
Feedback
thermocouple,
in the vat.
which
produces
a voltage proportional
This feedback is compared with the
410\n
stand-\n
to
16-2\n
ard
to produce
\nof
an
\342\200\234error
into
introduced
steam
FEEDBACK
IN
STABILITY
Art.
voltage\342\200\235
the
in
which
411\n
SYSTEMS\n
system.\n
in last chapter,
block diagram algebra,discussed
as the oscillator and the temperature controlsystemcan
form of a feedback system shownin
to a standard
By means of
\nsuch systems
reduced
\nbe
the amount
controls
turn
the
Fig.\n
16-1(c).\n
Feedback
\nand
such
A
capable
to be stable
being
systems
is said
system
or does not
\nremains small
such as accurate control
advantages
These advantages are partially offset
of instability.\n
many
of response.
time
improved
\nby
afford
systems
if, for small values
increase
of
time.
with
We
the
input,
do
not
output
ordinarily
capable of instability by this
system
\ntion.
If
the
is made up entirely of passive linear elements,
system
time
is no
source
to supply an output which increases
energy
\nand thus
without
limit. There will be a distinct relationship
and
as expressed
\ninput
by the transfer function. Linear
output
\nments
can
form
or the power level the input.\n
at most
the
modify
If the output
is to increase without limit,energymust supplied
to
This
must be within the system closedby the
system.
supply
\nback
for
the output
to increase with the input remainingsmall.
loop
is required
for an unstable system to introduce
feedback
path
new
into
the system from the output to overridethe initial
input
a
of
\nthink
as being
linear
defini\302\254
\nthere
with
between
ele\302\254
of
be
\nthe
feed\302\254
\nThe
\na
\nsmall
input.\n
essential features of a system capableof being
within
the system,
must
be a sourceof energy
and (2)
(1) there
\nthere
must
be at least one feedback path.\n
These
but not sufficient conditions for instability.
are
necessary
\nThat
is
feedback
to say,
are, with proper design, not only
systems
in many
\nstable
but
respects to systems without feedback. An
superior
is to meet specifications and yet
for
the engineer
\nimportant
problem
This
can be done from transfer functions by a math\302\254
\navoid
instability.
It is these procedures we will con\302\254
\nematical
or a graphical
procedure.
are
there
Thus
two
\nunstable:
\nsider
The
\nand
is, by the definition just given,fundamen\302\254
in nature,
and is related to the transientresponse
of
the
The transfer
function of the closed loop written
system.
transfer function was studied in Chapter15,
the open-loop
question
transient
\nclosed-loop
\nin
chapter.\n
equation\n
System stability in terms of the characteristic
16-2.
\ntally
this
in
terms
of
of stability
is\n
V2(5)\n
VAs)\n
=
G(s)\n
L(s)\n
1
+
G(s)H(s)\n
(16-1)\n
STABILITYIN
412\n
where
and
\noutput
a
\nof
and
G(s)
quotient
V*
or
=
(a0sn
form
polyno\302\254
aoSn
+
OiSn 1 +
+ aisn_1
+
... +
when
\nG(s)H(s)],
the
of
\nequation
na\302\253x\n
...
a\342\200\236-is +
a\342\200\236)Fj(s)
=
an\n
(16-3)\n
G(s)Vi(s)
form, the denominator polynomial, that is -f
set equal
to zero is recognized as the
If the characteristic
equation is factoredinto
system.*
this
in
Written
n
the
Vi(s)
(\302\253)
V i(s)
\nits
are
Fig. 15-15; Fj(s) and
This transfer function has the
s. Expanding the denominator
write\n
we
\nmial,
respectively.
in
of polynomials
16\n
Chap.
are defined in
H(s)
input,
SYSTEMS\n
FEEDBACK
16-2 can
Eq.
roots,
[1
characteristic
F\302\273(\302\253)
=
Vi(s)
be written\n
\t\n
G(s)
o0(s -
-
sb)...
s\302\253)(s
(s
-
(164)\n
sn)\n
n\n
y-o\n
In
\nspecified,
\nfractions,
\nportion
initial
\nall
\nIf
solution of this equation
vi(t)
this
is
to
the
usual
expand
equation by partial
procedure
of the transient
the
constants
thus
arbitrary
evaluating
of
the
solution.
(In using the transfer function, we assumethat
time domain solution.)
conditions
are zero in this particular
are no repeated
roots in the characteristic equation, the
for
solving
there
with
expres\302\254
for
\nsion
time-domain
the
will
v2(<)
be\n
n\n
=
Vi(t)
K,*'*
^
(16-6)\n
y-1\n
other
In
in
\ndetermined
\nthe
form
the
case,
general
is
portion of the time-domainsolution
of the characteristic equation. For
by the roots
roots of the equation are complex and
the transient
words,
written
s
form of the
The
the
on
\ndepends
\342\200\224
<r
+
jco
(16-7)\n
time domain solutioncorrespondingto
magnitude
and
for
a finite
as\n
signs of
<r
and
o>. Several
root
each
are
cases
of
\ninterest:\n
Case
\n(for
the
1.
<r
negative
combined
terms
of the conjugate
Ke-'
*
Compare
this
equation
case, the solution
pair a j<a)}\n
o>. For this
sin
+
(\302\253<
will
be
\302\261
<f>)
with Eq. 6-88 and Eqs. 7-37
(16-8)\n
and
7-38.\n
FEEDBACKSYSTEMS
STABILITY IN
Art. 16-2
413\n
a damped sinusoid; the
as time
\nto zero
t becomes
large.\n
Case 2. <r = 0 and finite a. For the conjugate
\nthe solution
will
be similar
in form to Eq. 16-8with
This
sin
K
and
16-8
\nEq.
a finite
and
tr positive
S.
Case
of
Case
4.
for
<r =
and
\nconstant
large
0 and
does
16-2.
Fig.
5.
\nthis
case
<r
(for
=
0; that
as a function of
not
=
either
+
time.\n
to
or
root)
(16-10)\n
increases
values
negative
\ntime. For positive
case, the solution
limiting
illustrated
<r,
the
in
term
Fig.
negative,
but
of a system(Vo)
with
plane.\n
<a
=
0.
The
for
solution
form\n
(16-11)\n
magnitude of the
increases
16-2.
is a
time.\n
has the
of a, the
and
exponentially
t.\n
this
with
0)
of the time response
in the complex s
position
positive
a simple
(<at
oscillations
For
0.
change
Comparison
For
are
is,\n
is similar
solution
the
Ke^
\ncases
\302\261jco,
(16-9)\n
amplitude
of time
values
<a
\npole
Case
<r
<f>)
\302\253.Again,
sin
In this case, the magnitude
limit
pair of roots
is\n
Ke+vl
\nwithout
+
{(at
at constant
oscillates
function
This
magnitude reduces
is termed
expression
From
term
diminishes
exponentially.
the figure, the
These
with
several
effect of
the\n
For
of
values
negative
Chap.
the
<r,
\302\253
response
increases
time, but for positive values of <r, the response
\nwithout limit with time. Our concept of stability evidently
ties
with
the sign of the real part of the roots of the characteristic
\ndirectly
\ndecreases
16\n
value of
nonzero
A finite
seen.
be
\302\253
can
oscillation.
to
\ncorresponds
role of
the
<r and
of
sign
SYSTEMS\n
FEEDBACK
IN
ST ABILITY
414\n
with
in
\nequation.\n
small
of stability, that the output should
of
16-2
\nor not
increase
with time, the responses of parts (e) and
Fig.
These both correspond to positive
identified
with
unstable
systems.
of
case shown as part (e) of Fig. 16-2is,
transition
<r. The
The output does not increase without
a stable
case.
speaking,
It
to sustained
oscillations
as, for example, in an
corresponds
values
of a, there is damping, and
oscillator.
With
negative
as stability
is related directly to the
is stable.
Just
system
real part of the roots of the characteristic equation,so
basic
\nof the
is finding an answer to this question:
in determining
stability
\nproblem
= 0,
1 + G(s)II(s)
the
characteristic
with
roots
any
equation,
\npositive real partst This is the basis of all stability
The
of determining
is thus one of findingthe
problem
stability
our definition
By
remain
(f)
\nare
\nvalues
\nstrictly
\nlimit.
\nelectronic
\nthe
sign
the
have
\nDoes
studies.\n
roots
the
\nof
1
equation\n
=
+ G(s)H(s)
do8n
+
ai8\"_l
+
...
+
at\302\253\"-5
+
an-i\302\253
+
=
a\302\273
0\n
(16-12)\n
some
or
\nequations,
\nequation
\nincreases,
\nconsuming
equivalent
1
n =
Suppose
For first- and
the roots.
finding
second-order
factor
the
2, this is a simplematter:
and so find the roots. But as the
of
the
equation
and
time
the task of finding the roots becomes
tedious
machines are available), and alternate
(unless computing
and n =
merely
order
very
used.\n
are
\nmethods
to
that
the characteristic
do*
equation
+
oi
=
-
is of first
order,\n
= 0
(16-13)\n
The root of this equation
is\n
s
The
\nand
\na
real
part
real;
this
of this root
requirement
is met
have
two
roots
given
longas
Ui
for all physical
and
a0
are
positive
systems. Likewise
equation,\n
do\302\253*
will
(16-14)\n
is negative as
characteristic
second-order
\342\200\224
+
di\302\253
+
dj
=
0\n
(16-15)\n
by the equation\n
dj. ,
X
2d0
/oi*\n
dj\n
\\4d02\n
do\n
(16-16)\n
STABILITY IN
Art. 16-3
Si*
both
\nsign
of
\nreal
part
of
the
roots
\nthis
rule
is
not
sufficient.
+ 4)(s2
(s
this
In
even
There
\nparts.
the
to
\ntion
of
form
\nthe
\npolynomial
\nthis
criterion,
the
be
polynomial
one
least
\ncoefficient
16-3. The
of a
\nroots
is
polynomial
first
the
alternate
\nof
etc.
\nfifth,
coefficients
\netc.
made
+
(16-17)\n
or
the
negative,
has
polynomial
only
the
way
negative).\n
Routh
rule*\n
gives a procedure for determiningthe number
with positive real parts without actuallyfinding
polynomial
the
...
d,n\342\200\224is
-}\342\200\242
-f-
in the application of the rule,form
that is,
coefficients
of the equation;
coefficient
form
one row, and from the
form
a second
row as follows:\n
step
disn_1
of
polynomial\n
row
doSn
40
+ 2s +
(since that is the
root
dosn -|- ais\"-1 -(- a2s71\342\200\2242-|As
2s2
rule
Consider
roots.
\nthe
+
positive.\n
real
positive
could
be
criterion
Routh
Routh
The
s8
with
be no roots of a polynomial
positive
it is necessary but not sufficientthat the coefficients
(2) If a coefficient of a
\nat
=
there
that
\nreal parts,
\nof
-2s + 10)
of the third-order polynomial are al
there
are clearly two roots with positive real
though
are
further
that must be satisfied in addi\302\254
requirements
coefficient
requirement. The requirements take
positive
of the magnitudes of the coefficientsof the
relationships
Routh\342\200\231s
criterion.
Before
to a study of
turning
given
by
let us summarize
our findings thus far:\n
order
In
(1)
poly\302\254
coefficients
the
example,
\npositive
will
follows.\n
as
formed
\nnomial
as
long
system, a requirementthat the
is sufficient to guarantee that the
positive
be negative. But for higher-orderedsystems,
consider the third-order
To illustrate,
be
coefficients
all
real parts onlyas
will have negative
,
do, ai, and 02 are positive.\n
the first- and second-order
\nthe coefficients
For
Si and
roots
two
The
415\n
SYSTEMS
FEEDBACK
+ a2sn_2 +
a3s\"-3+
\342\200\224
0
two
rows
(10\342\200\22418)\n
made
up
from the first, third,
second, fourth, sixth,
1\n
a4sn_4
row
dn
+
a6sn_6
+ a6sn~6
+ ...\n
2\n
(16-19)\n
*
\n(6th
E. <J. Routh,
ed.),
(Macmillan
Part of Dynamics of a
Co. Ltd., London, 1930).\n
Advanced
&
System
of
Rigid
Bodies,
Vol. II
next
the
As
in the
rows
two
the
Write
\na
IN FEEDBACK
ST ABILITY
416
sixth-order
form\n
do\n
d 2
a4\n
do\n
d8\n
dl\n
\n#3\n
dg\n
d7\n
dg\n
(16-20)\n
the following array
complete
step,
16\n
Chap.
SYSTEMS\n
of numbers
for
(shown
system):\n
use\n
a0\n
CL 2
d4\n
ai\n
d3
dg\n
bi\n
bz
&6\n
Ci\n
Cz\n
di\n
dz\n
a6\n
C\\\n
fi\n
and g coefficientsare
\ncoefficients by the following
the
where
6, c,
d, e, /,
in
defined
of the
terms
a
pattern:\n
\342\200\224
ctiUz
_
a0a3.
*\n
a\\'*at\n
\n3
a<Kv
-'CU _
ai'^as
\nax
aiQ4
~~ <*<#6
a\\\n
\342\200\224
bzdi
biQz
Cl\n
;\n
etc.\n
\nbi
new element is found from the two elementsabove
element
in the same column and the two elementsabovebut
in
\nthe
\nthe column
to the right. These elements form a determinant-like
elements
The
\nstructure.
by a line with positive slope have a
joined
In
any
general,
a
\nhave
negative
subtract
\nWe
on
\nelement
joined
the opposite
and
two
these
products
left-hand
corner
the
lower
(just
sign
to
\ncontinued
elements
the
while
sign,
\npositive
give
the
Routh
with negativeslope
of the rule for determinants).
divide this difference by the
of the array. This processis
by a line
array.\n
The number of changes in sign in elements
of
the
first
column
with
\n(marked use) indicates the number of rootsof the equation
posi\302\254
\ntive
real
For
there
to
be
no
roots
with
real
parts.
positive
parts,all
\nelements
of the
first column
must have the same sign.\n
For
the
rule as given to hold, it is necessarythat no
of s in
powers
\nthe
\nterms
\nroots
at
has
\nequation
\ntest
be
equation
missing.
least
one
An
inspection.
by
which
are
are
purely
all
However,
if any such
terms are missing,the
fails
root with a positive real part andso
the
occurs when the equation contains
exception
even
imaginary.\n
powers
or all odd
powers, indicatingthat all
Art.
Example
417\n
1\n
the
Consider
identity\n
l)(s + 2)(s
(s +
SYSTEMS\n
FEEDBACK
IN
STABILITY
16-3\n
+ 3)(s +
-
4)
+ 10s3 +
s<
35s2 + 50s +
24\n
(16-21)\n
Routh
The
to give the
is formed
array
following:\n
35
1
24\n
10
50\n
30
24\n
42
\n24\n
\nreal
first
the
From
(agreeing
parts
Example
As
positive
Eq. 16-17.\n
consider
example,
2s +
s8 + 2s2 +
which
with
2\n
second
a
it is seen that there are no roots
the known roots).\n
with
column,
is known to
have two roots
with
40
=
positive
0
(16-22)\n
real
parts.\n
to
\342\200\22418
40)
2
1
40\n
\n2
-18
\n40\n
There
\nExample
changes
of sign (2
a third-order
equation,\n
two
are
to
as required.
8\n
Consider
OoS3-f- Ois2 +
The
and
\342\200\22418
Routh
array
for this
equation
did2
a2s -J-a3 = 0
(16-23)\n
is\n
do
d2\n
dl
#3\n
\342\200\224
dods
\ndl\n
d3\n
From
the
\nthe equation
\nthat
there
that it is necessarythat all coefficients
in
be positive
and, in addition, that Oi\302\2532 > Ooa8 in order
no roots with positive real parts in a third-order
equation.\n
array
be
we conclude
polynomial is a
A Hurwitz
stable
\nwith
even
\ntient
of
parts
a
polynomial.\n
(16-24)\n
+\n
aiS
Q(\302\253)\n
aaS
+\n
ass
-f\n
+\n
en$
a*s
+\n
+
<*6S
to
polynomial
*
be positive
the
\na-coefficients
\nto
+
(s*
\302\253!*)
+
odd
and
even
\nthe
+
\302\253i
*)[P(\302\253)
(\302\253*
=
Q(s)
is
\nQi(s)
is
\nnomial
the
of
\ntors
either
an
\nby
type
it
\nexample.
comes
+
(s\342\200\231
A
step
to
Consider
\nchapter.\n
derivation
au
on*)
\342\200\231)Q(s)
cancel
=
+
Px(s)
when
(16-25)\n
Qi(s)
the quotient
the test. The test does assurethat the
or a Hurwitz polynomial multiplied
fraction
Pi(s)/
poly\302\254
by
fac\302\254
only.
an
end
the
of this
is most easily
accomplished
of step (1)
After
completion
of
the
other
lower degree into
part
part and
Invert
the remainder and continue the process
(as it must).
This is best illustrated
by
an
polynomial\n
4s4 + 2s* +
*
Thus\n
procedure.
the
divide
one
\ncomplete
+
(s\342\200\231
continued
the
of
above,
\nuntil
+
\342\200\234invert-and-divide\342\200\235
\nlisted
in both
+
(s\342\200\231
\302\253i\342\200\231).\n
form,
Formation
correspond
are incorporated
which
1\n
in applying
Hurwitz
formed
under test. Suchroots
polynomial
+
(s\342\200\231
Wl*)P(\302\253)
Hence, terms of the
of
of the polynomial.
parts
.\n
the
the
in
.
the
it is necessary that all
However, the test we have describeddoes
of roots without real parts (i.e., on
j<a
in the polynomial
factors
.
be Hurwitz,
possibility
s plane)
of the
\naxis
the
out
rule
\nnot
frac-
thus\n
\nion;
For
steps:\n
continued
the quotient of polynomials as a Stieltjes
Expand
(2)
following
polynomial
numerator
the
\nas
equations
into even and odd parts (that is,
in s and with odd powers in s). Form quo\302\254
powers
two
with the part of higher degree
these
polynomials
the
Separate
(1)
negative
To apply the
polynomials.
carry out the
a polynomial,
to
criterion
\nHurwitz
with
the characteristic
Hurwitz
therefore
are
systems
polynomialhavingroots
representing
Polynomials
only.
\n\342\200\242eal
parts
\n)f
criterion\n
Hurwitz
The
16-4.
8s\342\200\231
+
3s
criterion is given by
+ 1.5
Guillemin;
- 0
see
(16-26)\n
reference
at end
of
The
IN FEEDBACK
STABILITY
16-5
Ait.
is formed as
of polynomials
quotient
\n2s8
one
Dividing
step
(16-27)\n
3s\n
+
gives\n
2s8
+
6s2\n
and divide
as\n
2s8 + 3s
1.5)
+
2s2
1.5\n
+
2s2
we invert
1.5 (2s
4s4 + 8s2 +
+ 3s)
\n4s4
Again,
follows.\n
1.5
+ 8s2 +
4s4
419\n
SYSTEMS\n
(s
1.5s
+
\n2s8
\n1.5s\n
and
+ 1.5
2s2
1.5s)
again,\n
(|s\n
2s2\n
1.5\n
and
fraction
continued
the
Hence
4s4
+
8s2
+
\n2s8
\nThe
2s8
a continued
+
expansion
+ 1-5
2s
is\n
+\n
(16-28)\n
3s\n
are positive, the polynomialis Hurwitz.
in four steps above are conveniently carried
shown
as illustrated below.\n
operation
operations
as
\nout
(Is\n
a-coefficients
the
all
Since
1.5s
1.5)
finally,
4s4
+ 8s2 +
\n4s4
+
3s)
1.5 (2s
6s2\n
2s2
+
1.5)
2s8 + 3s
2s8
+
(s\n
1.5s\n
L5s)
2s2 + 1.5
(|s
\n2s2\n
1.5)
1.5s
(s\n
1.5s\n
0\n
The
16-5.
The
\nof
the
stability
Bell
is
\ncriterion
\nthe
Nyquist
criterion\n
Telephone
the
same
approach
* H. Nyquist,
study next was developedby Nyquist*
Laboratories
in 1932. While the objectiveof this
as the Routh criterion and the Hurwitzcriterion,
criterion
differs
we will
in several
\342\200\234Regeneration
respects.\n
theory,\342\200\235
Bell
System
Tech.
J., 11,
126 (1932).\n
the
than
\nrather
the
of
of
\nanswer
characteristic
closed-loop
(3) Analysis is madein
of
terms
\nconceptsof phase and
sinusoidal
the
be
will
shown,
the
are
inspec\302\254
\342\200\234yes
or
no\342\200\235
where the
state
steady
ratio
magnitude
transfer function
equation.\n
more than
criteria.\n
plot
gives
and Hurwitz
graphical
Routh
the
Chap. 16\n
of the open-loop
is partially graphicaland,as
(2) The method
\ntion
in terms
made
is
Analysis
(1)
SYSTEMS
IN FEEDBACK
STABILITY
420
to
related
readily
\nexperiments.\n
The basic operation in
s plane
the
\nfrom
of values
a set
\nthat
to the
applying
the
These
\nF(sa)].
\nin
Here
16-3.
Fig.
infinite
an
three\342\200\224and
and
s2,
16-3.
of
number
for a
have,
F(si),
F(s2), and
are
other\342\200\224points
s planeis
given
\342\200\234
shown
\342\200\235
into\n
mapped
illustration.\n
Mapping
Flsl-plane\n
6-plane\n
s3)
[namely,
F(s),
contour in the
an arbitrary
Fig.
of
values
mean
we
of s (for example,Si,
of
mapping
the term mapping,
F(s) plane. By
\nF(s),a corresponding set
is a
criterion
Nyquist
jImF{8)\n
ju>2\n
7a)i\n
a\n
Re
00\n
F{s)\n
0*f\n
\302\253l\\\n
16-4.
Fig.
a
\nin
\ns
As
\none
The
16-4.
Fig.
=
j<a for
w
another
given
example
above,
suppose
\342\200\224
l/s(sT
F(s) plane. A specific
is made for imaginary
mapping
^ 0. The
for F(s)
in the
contour
corresponding
example
Mapping
specific function
mapping operation,
that two function are
-
P(s)
1).\n
example
is shown
values of s; that is,
is\n
of a
F(s)
+
+
1\n
not so difficultas
related by the
the
equation\n
(16-30)\n
Art.
16-5\n
that
is,
\ntion
F(s).
IN FEEDBACK
STABILITY
421\n
plus a constant (unity) is equalto the func\302\254
plot in the two planes is shownin Fig.16-5.The
typical
moves
the plot one unit to the left.\n
evidently
function
the
SYSTEMS\n
A
\ntransformation
P(s)
F-plane\n
jlmF\n
jlmP\n
P-plane\n
ReF\n
ReP\n
PA\n
U\n
suppose
Next,
\nare
in
given
the
Tig.
16-6.
Mapping
that
F(s)
is factored
=
is
\302\253i,
\nin Eq.
16-31.
\342\200\224
\n(sa
are
in
isolated
$i)
-
Si)(s
-
Mi
\nFig.
\nfor
Fig.
may
be
is the
This
and
and
other
are so
=
<f>i
M
the
These
poles.
plane
in
shown
is
phase
the
sb) =
phase
Fig.\n
zero,\n
single
\342\200\224
Si)
value
as\n
le\342\200\231*1
(16-32)\n
angle
are shown on
in Eq. 16-31
term
-
terms
are
from the term (s
in polar form
Si)
magnitude
magnitude
Any
(16-31)\n
Sm)\n
frequency sa, this termhas a
expressed
example,\n
all
8n)
\342\200\224
. (s
This zero comes
16-6(b).
(s\302\253
When
-
\342\200\242.
(s
A
At some particular
which
16-6(b).
which
array for purposes of illustration).
\342\200\224
\342\200\224
\n(s\302\253 Si).
s2).
\342\200\224
\302\253fc)..
Sa)(s
(sa
where
polesand zeros
the zeros and s0,sb, ..., sm
on a plot of the s
displayed
(an arbitrary
16-6(a)
1.\n
+
sn are
...,
zeros
and
\npoles
\342\200\224
P(\302\253)
to find its
\342\200\224
(S
K\n
\n(s
\302\253i,
s2,
P(\302\253)
equation\n
F(s)
where
of
of the phasor
the s plane in
can be similarlyexpressed;
Mbe\342\200\231+>
(16-33)\n
expressed, Eq. 16-31 takes the form\n
KMiMiMiMi...
|F(s)|e,An*y(*)\n
MaMbMcMd...
e
,.A<\n
\342\200\230\n
(16-34)\n
IN FEEDBACK
STABILITY
422\n
where\n
<l>t
last
This
the
\nfor
and
^1
02 +
+
..
~
~
\342\200\242 <t>a
\342\200\242
\342\200\242
\342\200\242\n
<f>b\n
the total phase at some
may be found by adding the phase
F(s)
the
(16-35)\n
sa
frequency
the
of
subtracting
16\n
Chap.
us that
tells
equation
function
\nphasors\342\200\235
=
SYSTEMS\n
of the
phase
\342\200\234pole
\342\200\234zero
other
in
phasors\342\200\235;
\nwords,\n
F(s)
Ang
=
\342\200\224
(s
Ang
si)
+
Ang
\342\200\224
(s
\342\200\224
-
\342\200\224
(s
Ang
sa)
...\n
(16-36)\n
for any value
s.\n
the
shows
16-7
Figure
of
two zeros, Si and
s plane with
\nthe
\nSi
\nin
C
contour
\nping
s2)
(ignoring
a
(sa is a
clockwise
sa moves along C
After one
direction.
\nterm
(s
Si)
\nradians.
\ns2
on
the
C,
\342\200\224
of the closed
has
factor
the
of the
phase
Next,
of
effect
as
\ncomplete traversing
\ncontour
map\302\254
point on
the
Consider
contour).
a
and
s2,
phasor
increased
\342\200\224
s2),
\342\200\2242r
of
effect
the
consider
(s
by
time
this
same
same closed contourC is
in the
\nwise
There
is no net gain in phase of the phasor
direction.
s2).
(s
\nIn summary,
if the closed contour encircles a zero in traversinga
2r
in the
clockwise
the function changes in phase
\npath
direction,
if no zero is encircled, there is no changein
\nradians;
a pole
the
same
conclusion
Exactly
may be reached in the case
the phase is changed by +2tt
\nexcept that
next
a contour
is selected in the s plane Fig. 16-6(a)
that
Suppose
is traversed
\nsuch that
P poles and Z zeros are encircledas
contour
a clockwise
direction.
The net change in the phase of the
\nignoring
si, as the
clock\302\254
traversed
\342\200\224
term
closed
\342\200\224
by
phase.\n
of
radians.\n
of
the
function
\nin
will
\nF(s)
be
by the
given
A
F(s)
=
2t(P
\342\200\224
Z)
radians
(16-37)\n
Let
the
of the s plane
F(s)
plane.
examine
of the F(s) plot in the complexplaneas
the
behavior
contour
in the s plane is traversed. An increase in the phase
manifests
itself
in the F(s) plane by an encirclement of the
2?r radian
increase.
Further, every zero encircled will cause
every
of the origin just as every pole will
counterclockwise
encirclement
next
Return
\nus
Ang
equation\n
to the mapping
into
the
of
\nclosed
\nF(s)
\nfor
\none
origin
\ncause
a
\npoles
or
\ncontour
\nthe
closed
encirclement.
clockwise
zeros\342\200\224or
in
the
contour
F(s)
if
it
encircles
Should
equal
the contour not encircle any
of poles and zeros\342\200\224the
numbers
plane will not encircle the origin. In summary,
C in the s plane encircles in a clockwise
(or
negative)\n
if
Art.
P
direction
\nplane
in
times
Z)
examples
are
to a
closed-loop
system\n
reviewed
in the
F(s)
a counterclockwise
(or\n
16-8 to illustrate
given in Fig.
conclusion.\n
16-6.
Application
The
concepts
the closed-loop
in terms
written
of
related
are
\noutput
\ntransfer
by
function,
and
H(s)
His)\n
G(s)\n
The poles and zeros
of
\nNyquist
1
+
(16'38)\n
\302\251(.)\302\273(.)
=
two
must be consideredin the
=
(s
two
functions
Si).
Si)(s
(s
Sq)
\342\200\224
have the same
s0)(s
(s
.
\342\200\224
Sfi).
.(s
-
. .(s
of
derivation
the
\342\200\224
sn)\n
(16-39)\n
Sb) . . . (s
Sa) (s
K'\n
\n(s
The
\342\200\224
\342\200\224
K\n
\342\200\224
=
system.\n
Closed-loop
func\302\254
(8
G(s)H(s)
16-9.
Tig.
the
GH) and (GH)
Let\n
criterion.
+
(1
V.\n
G[s)\n
equation\n
1 + G(s)H(s)
Vl(s)
v;\n
O\n
the
by
V2(s)
\ntions
a
last section will next be appliedto a
feed-forward
transfer function G(s) and
shown in Fig. 16-9. The input and
H(s),
function
transfer
feedback
\nG(s)
having
system
\nclosed-loop
\na
\342\200\224
(P
origin
Two
direction.
positive)
\nthis
the
encircles
423\n
corresponding contour in the
Z zeros, the
and
poles
SYSTEMS\n
FEEDBACK
IN
STABILITY
16-6\n
Sm)\n
\342\200\224
s\342\200\236)
\342\200\224
(16-40)\n
Sm)\n
*)...(\302\253
poles. In Eq. 16-39,
the
order
of
the\n
is
polynomial
P(s)
\nNyquist
criterion,
FEEDBACK
IN
STABILITY
424\n
SYSTEMS\n
and the order of Q(s)
orders are restricted to
n,
the
Chap.
is m. In
the case n <
16\n
the
deriving
such
m,
\nthat\n
lim
or a constant
=0
G(s)H(s)
(16-41)\n
oo\n
8\342\200\224>
is
important
\ntabulated
as
It
follows:\n
8i,
are
\302\2532.. .s\342\200\236
\nsa,
sb..
.8m
are
\ns0,
sb..
.8m
are
8a, Sp..
The si,
the
for
\nof
1
equation
+
(1
In
GH)
are, by Eq.
+
(1
to us because
0, whichis
characteristic
the
must
roots
interest
specific
equation
not have positive
to be stable.
16-38, the polesof
these zerosare
Note that the
the
s
real
zeros
(V2/V1).\n
is the zeros of the
polyno\302\254
choose
positive real parts. This suggeststhat
plane to include the entire right half planeas
in Fig.
16-10.
This contour will enclose
the
zeros
of interest.
The contour is
with
GH)
in
contour
of G(s)H(s).\n
concern
represent
our
stability,
studying
\nmial
[1 + G(s)H(s)].
poles of G(s)H(s).\n
zeros
the
These
they
system
poles of
also the
+ GH =
system.
closed-loop
\nparts
\na
the
+ G(s)H(s)].
of [1
zeros
of vital
are
roots
the
\nof
are
.8\342\200\236
s\302\253
\302\2532,\342\200\242..,
of
\nzeros
and zeros. They are
the various poles
to distinguish
we
shown\n
\nall
\ning
\n(s
s =
at
=
to
\ntinuing
\nof
for
0)
\342\200\224jo0,
s =
to
start\302\254
the
avoiding
+jo0,
contour
This
1-2-3-4-1,
time
the
radius
infinite
\nning.
direction
the
in
\ntraced
being,
origin
and
con\302\254
thence on a circle
the
is
point of
traversed
begin\302\254
in a
\nclockwise(or negative) direction.
\ncontour
in
the
s plane can be mapped
plane.\n
either
the
(1 + GH) plane or the
between
these mappings was
(the
simple
relationship
If any poles or zeros of (1 +
are
Eq.
16-30).
half
s
in
of
the
then
locus
the
right
plane,
(1)
(1 + GH)
encircle
the
or
locus
in
the
will
(2) the
origin,
plane
The
\nin
\nGH
con\302\254
plane
in
\nsidered
\ncled
\nencircle
Let
\npoles
\nplane.
GH
the
point (\342\200\2241+ j0).\n
Z = the zeros of (1 + GH)
real
with
positive
parts,
of
with
real
the
(1 +
GH)
positive
polesof
parts (also
\npositive
\nthe
the
will
\nplane
point
encir\302\254
GH)
the
in
real
parts),
(\342\200\2241+
jO)
P = the
(GH)
with
R = the
in
the
net counterclockwiseencirclements
(GH) plane or the origin in the (1 +
of
GH)
Then\n
R
= P
-
Z\n
(16-42)\n
Since
be equal
must
parts,
characteristic
\nwith the
of
the
\342\200\224
=
R
In most cases
\342\200\224
0 for
\nR
P = 0, and
the Nyquist criterion,
\342\200\224
\nof
with
\nof
G(s)H(s)
\nthe
system
pole
\nthere
the
\nat
transfer
a
\nthrough
is
the
the
\nthe
point).
\nsee,
vital
joined
If the
and
together
(as
is, as we
point.\n
small, only the
function
\ntransfer
This
unstable.
becomes
As s
\nbe
is
system
a
(+0)
points
the
\342\200\224
\302\273<
range
\302\253
locus closes
the
half
the sys\302\254
right
plane,
=
since
R
0; however,
stable,
locus
closes
the other direction
left
half
then R = 1 and
plane,
same
\ntem
The
be
\n(\342\200\2240)should
\nthe
function,\n
detail.
one
\nin
the right half plane),
in
in Fig. 16-11 for frequencies
plot is complete except
The
co.
+
\nif
and
(and so in
parts
range
net counterclockwise
P is the number of poles
the
is
is plotted
which
\nfor
jO)
the
that
if and
is stable
\nconsider
\n<
for
(\342\200\2241+
real
positive
requirement
P.\n
only if R \342\200\224
have
thus
far avoided
of
any problems that might arisebecause
of G(s)H(s)
at the origin or several poles at the origin.Actually,
is a practical
matter
involved in taking into account thesepoles
To illustrate the problem,
of special attention.
origin
deserving
We
\na
point
to the
plot the G(s)H(s)locus
R
if\n
(16-43)\n
reduces
If
\302\273<\302\253<\302\253.
frequencies,
\nencirclements*of the
P
criterion
the
stability.\n
To apply
425\n
OH) and the poles V%/V\\ with positive
to zero for the system to be stable,
system
0 is stable
and
equation
(1 + OH)
only
if
of (1 +
zeros
the
Z,
\nreal
IN FEEDBACK SYSTEMS\n
STABILITY
16-6\n
Ait
G(s)H(s).
pole at the
Thus
for small
Plot
16-11.
Tig.
origin
has
an
s, the transfer
of\n
effect
on the
function can
written\n
G(s)H(s)
n
where
is the
order
(or multiplicity)
=
(16-45)\n
\302\247
of the polesat
the
origin.
For\n
the value of R, imagine
a phasor with one end securelytied to the point
away from this point. Let the end of this phasor trace the
\342\200\224
\342\200\224
\302\273
at
0 and
\nlocus
+0 finally ending at + 00. Count the net
through
starting
of
counterclockwise
rotations
of this phasor. This is the value
of
\nnumber
R. A
is
for R.\n
number
\nclockwise rotation
designated by a negative
* To find
\342\200\224
1
\n(
+ jO)
pointing
STABILITY
426
semicircular
the
locus
\nphasor
shown
path
IN FEEDBACK
SYSTEMS
in Fig. 16-12,
the equation of the s plane
is\n
from
directed
0)
\n(s
function
\nthe transfer
lim
s->o
as
Hence
\nof
G(s)H(s)
\nthe
origin
at
\ntions
=
G(s)H(s)
^on\n e~in9 =
lim
s->o
=
s
the
odd,
\npart
= +0.
follows
Im
\nfor
the
\nthis
\nbe
\nthe
\n(w)
function
Routh
or Hurwitz
Nyquist
such
of
advantage.
plot.
nearest
\nunstable; if on
n = 1
see that
locus of G(j<a)H(jui) ingoing
phasor
=
values of
transfer
to
we
16-44,
Eq.
plot completed.\n
16-13.\n
be
con\302\254
imaginary
+Re
G(+ju)H(+ja>)\n
made
be
\302\253
can
(16-48)\n
by
the plot
reflecting
real axis of
plane.\n
making
rule
Trace
from
(\342\200\234walk\342\200\235)
If the point
approach
=
-ImO(+i\302\253)H(+i\302\253)
the GH
has no poles in the righthalfplane
G(s)H(s)
criteria can be used to advantagein
of
thumb
that P = 0 in Eq. 16-43,a
may
the
upon
determination)
used
Nyquist
co
\342\200\224ju))H(\342\200\224jo>)
frequency
the
\n(and
(16-47)\n
Figure 16-11 is completedin Fig.
plot, only positive values for need
part of G(jo))H(ju>) is even and the
G{-ju)H(-jv)
plot for negative
positive
If
5 \342\200\224>
0,
As
that\n
Re G(
The
of the
Nyquist
the real
Because
it
of
rotation
\342\200\2240
to
making
\nsidered.
phasor
the circle.
ooe-*nfl
16-13.
the example
this rule to
clockwise
\ncauses
In
the
in Fig. 16-12 varies from \342\200\224
the phase
to +t/2,
t/2
from
to \342\200\224nir/2. In summary,
the n poles at
ranges
nx/2
in the
transfer
function G(s)H(s) cause (n/2) clockwiserota\302\254
infinite
radius
of the phasor locus of G(j<a)H(j<a).\n
Applying
s
of
angle
6 shown
Fig.
\nfrom
(16-46)\n
of the semicircle and 6 is the
the origin to a point on
has the limiting value\n
radius
the
8 is
\342\200\224
0) = he*
-
(s
where
Chap. 16\n
of
the left the
+
(\342\200\2241
system
is
is on
jO)
G(ju>)H(ja))
stable
=
\302\253
to
(P
0to\302\253
=
the right at
(\342\200\2241+
= 0
jO),
only).\n
the
+\302\273
on
the point
system
is
Several
the
\nof
IN FEEDBACK
SYSTEMS
be considered
to illustrate the
STABILITY
16-6
Art.
will
examples
next
studies
to the
criterion
Nyquist
this
For
the transfer
consider
example,
=
The
Nyquist
plot
\ncircle. No
matter
is shown
stability.\n
\nstable
jaTK+ i
(16-49)\n
of the
diameter
becomes, the locus
the
encircle
cannot
function
transfer
the
function\n
in Fig. 16-14, where K is the
how large K
\342\200\2241.Hence
\npoint
an unconditionally
represents
system.\n
locus
impossible
\nvalue
Example
The
+ I) plotted.
Fig. 16-16.
of
plot
Nyquist
16-50.\n
Eq.
5\n
the
Let
is
= K/(sT
GH
16-14.
Example
\nit
application
4\n
Example
Fig.
of system
427\n
of gain
plotted
for
the
as Fig. 16-13 serve as a secondexample.
for
locus to encircle the point \342\200\2241
except infinite
Again,
any
positive
gain.\n
6\n
transfer
function\n
-
awuw
O\342\200\235\302\256\n
jSgar+oo.:r,+T)
shown
in Fig.
a value
of
16-15 for two values of the constantK.
=
such
that
B results,
curve
the system is stable, since
0.
then
if the
value
of K is increased to give the curve
A,
=
=
and
P
is
since
0 by inspection of Eq. 16-50,
2,
system
\nunstable.
Such a system is described as a conditionally
stable
is
For
\nK
How\302\254
R
marked
\never,
\342\200\224
\nR
the
system.\n
Example
The
16-16
7\n
exact
is not
nature
of the
transfer
given, but it is
function for the
known that P
=
0
plots
for
both
in
shown
cases.
Fig.\n
The\n
of
locus
\nof
in
shown
\nlocus
the
locus
net
(a)
Fig.
the
about
\none
point
the
\342\200\2241,
decreases
a
\nrepresents
Example
for
Loci
two
of the
lb)\n
16-16.
or
\nincreases
16\n
Chap.
a conditionally stable system. The
is similar to that of (a) except
shape
16-16(b)
no
are
rotations\n
there
altered.
Since
has been
Fig.
the
of
part
SYSTEMS\n
represents
16-16(a)
Fig.
FEEDBACK
IN
STABILITY
428\n
conditionally stable systems:
p = R = Z = 0.\n
is stable.
system
corresponding
other loops, the
if the gain
However,
of the
a shift
(b)\n
either
into
\342\200\2241
point
system becomesunstable.This
locus
stable.\n
is conditionally
that
system
to
and
(a)
8\n
a system
shows
16-17(a)
Figure
having
a transfer
function\n
de-si)\n
For
this
locus,
P = 1,
R =
\342\200\224
1
clockwise
(one
(a)
Wff.
\nreal
are
two
\nbe
unstable
\nunconditionally
Loci
parts:
(a) P
of (1
zeros
for
any
so
there\n
(6)\n
transfer
for
16-17.
and
rotation)
=*
1,
R
=
functions
\342\200\224
1,
Z
=
having poles
2; (b) P -
1,
R
with
=
positive
1, Z
=
0.\n
will
+ GH) with positive real parts andthe system
value of gain. Such a system can be designated
as
unstable.\n
16-6\n
Ait.
Example
transferfunction
particularsystem,
plotted
with
pole
one
this
\nclosed,
that Z = 0, and
This
\nfeedback
the
in
once
1
a counterclockwise
With the loop
is stable.
with the loop open, the system
is, of course, caused by another
system
However,
\342\200\234open-loop\342\200\235
the
at
discussed
(as
beginning
chapter).\n
READING\n
FURTHER
comparison of the variousstability
For an interesting
Both
E.
\nF.
\342\200\224
point
instability
open-loop
the
within
path
the
is stable.
system
unstable.
the
encircles
locus
the
\ndirection such
\nof
429\n
in Fig. 16-17(b) comes from a
a positive
real part. For this
locus
\nhowever,
\nis
SYSTEMS\n
9\n
The
\nwith
FEEDBACK
IN
STABILITY
\nstability
\nthat
their
in
\nprocedures
1345
(1950).
\n\342\200\234Regeneration
Routh-Hurwitz
Bothwell points out
all employed essentially the
The articles of these
writings.
original
Bell
theory,\342\200\235
the
Hurwitz
Dynamics
Routh,
Co., Ltd., London, Part
&
\n(Macmillan
diagrams
and
J.
E.
are:
38,
IRE,
and
Routh,
Nyquist,
\nauthors
\342\200\234Nyquist
Proc.
criterion,\342\200\235
see
criteria,
well\342\200\231sarticle
of a System of
System
Bodies
Rigid
II, 1905),Chap.6; H.
Tech.
J.,
same
three
Nyquist,
11, 126 (1932); and
welchen
eine
Gleichung
Math.
Teilen
46,
besitzt,\342\200\235
Ann.,
negativen
\n273
For
additional
on the Routh-Hurwitz criterion,
(1895).
reading
\nsee
The
Mathematics
of Circuit Analysis (John Wiley &
Guillemin,
New
Syn\302\254
\nSons,
Inc.,
York,
1949), pp. 395-409, or Tuttle, Network
2
in
New
vols.
&
\nthesis,
Wiley
York, preparation).
(John
Sons, Inc.,
\nA.
Hurwitz,
\nIn
reelen
Wiley
(John
Design
\nSystem
Mayer, Servomechanisms and
& Sons, Inc., New York,
and
Chesnut
see
addition,
unter
Bedingungen
mit
Wurzeln
\nnur
die
\342\200\234Ueber
Regulating
1951), pp.
\n124-156.\n
an
For
\nin
terms
\nHomer
Jacobson,
\nAmerican
explanation
interesting
of feedback
Scientist,
system
\342\200\234Information,
43,
of the process of organic evolution
concepts and language, see p. 126 of
and
the origin of life,\342\200\235
reproduction
119-127
(1955).\n
PROBLEMS\n
16-1. Determine by
\nthe
\nnot.
+
(d)
having
systems
\tnot
of
Routh\342\200\231s
+
2s3
(b)
6s
s2
+
whether
criterion
equations are stable
s2 - 5s + 2 = 0. (c)
+ 2 = 0. Answers, (a)
or
s3\n
stable,\n
stable.\n
\n6s3
+
Repeat Prob. 16-1 for the
4s2 + 2s + 3 = 0. (b) s4 +
3s3
+
6s2
16-2.
stability
characteristic
following
4s3 + 7s2 + 7s + 2 = 0.
4s + 1 =0. (d) 5s8+
+
(a)
3s2
the
means
+ 7s +
2 = 0.
Answer,
characteristicequations:
3s3 + 2s2 + s + 1 = 0. (c)
(a)
(a)
not
stable.\n
5s4
+
2s4
+\n
STABILITY IN
430\n
16-3.
144s4
\n+
+ 1 = 0. Answer,
20s
+
\n120s2
A
16-4.
10s + 1
+ 38s2 +
214s3
+
s3 + 5s2+
=
K
\nwhen
25s5
stable.\n
equation,\n
Ks +
1
=
0\n
determine the range of the valuesof
criterion,
the
make
system stable, (b) Investigate system stability
Discuss
your results.\n
will
that
(b)
equations: (a) 720s6
+ 105s4 + 120s3 +
Routh
the
Using
(a)
\nK
0.
the characteristic
has
system
(a)
=
16\n
Chap.
for the characteristic
16-1
Prob.
Repeat
SYSTEMS\n
FEEDBACK
16-6. The feedback system shown in
accompanying
\nbeen analyzed
by J. F. Koenig in his paper
the
if\n
L\n
Ri
for\n
diagrams
\342\200\234Stability
i\342\200\224w\\\342\200\224\n
has
figure
Amplifier\n
Vi\n
gain-X\n
Prob.
feedback
Routh\342\200\231s
Using
\n(a)
\nbetween
Ri,
Ri,
Ri,
same
\nOn the
that
\ncriterion,
\nhave
negative
\nsystem.
Assume
equation,
characteristic
fourth-order
must
exist
+ ais3 +
plot K as
of
will
insure
real
parts
that
all
system
between
Ri/Ri.
equation,\n
aiS2 + a3s +
=
\302\2534
0\n
to those given in Example3,
that all roots of the characteristicequation
such that the equation will represent a stable
must be positive.\n
coefficients
similar
of rules,
set
a
that
relationship
1950.
a function
regions of stability and instability.\n
this
From
aos4
find
the
Oct.,
the system to be stable, (b) For the
what must be the relationship
damping,
plot, show
16-6.For a
find
Baltimore,
K for
and
K?
and
Paper,
Conference
criterion,
without
\nto oscillate
\nRi,
AIEE
systems,\342\200\235
16-6.\n
by
Routh\342\200\231s
characteristic
16-7. Repeat Prob. 16-6 a fifth-order
equation.\n
16-8. Classify the polynomials given in Prob. 16-1as
all
roots
in the left half plane) or
\n(having
test to the polynomials given in Prob.
16-9.
the Hurwitz
Apply
16-10.
Test
the polynomials
given in Prob. 16-3 using the
for
Hurwitz
not.\n
16-2.\n
Hurwitz
\ncriterion.\n
16-11.
Rework
Prob.
16-4 making
use of the
Hurwitz
criterion.\n
16\n
Chap.
\noscillator
43
figure
first
described
\nis
=
<*>o
The
\nmay
by
represented
Show
schematic.
\ning
the
that
equivalent
the smallest
when
Fig. 16-1
of
G(.)ff
(b)
G(\302\273)ff(*)
(C)
0(.)*(.)
each
For
values
\nof
\nNyquist
16-16.
\ntem
with
K
=
K\n
text
the
is
gm
=
and
LP/MR
=
that
the
1 /y/LpC.\n
~\n
-\n
of
these
=
\302\253
from
\n(///-plane
=
(\302\253)
29
shown in the accompany\302\254
value that the tube constant
is o>0
\nfrequency of oscillation under this condition
16-14. Consider the following transfer functions:\n
(a)
=
gmRL
circuit
are to start
if oscillations
have
can
\ngm
of oscillation
oscillator shown in
tuned-plate
431\n
SYSTEMS\n
shows an equivalent circuit of a phase-shift
in Proc. IRE,
and Hollingsworth
Ginzton
the necessary condition for oscillation is
that the frequency
(b) Show
RC.\n
1/y/Q
29.
16-13.
be
by
that
Show
(1941).
^
\n9mRL
below
The
16-12.
\n29,
IN FEEDBACK
STABILITY
of
K
that
(a) plot G(j<a)H(ju>)in the complex
K = 1. (b) Determine
the range
functions:
0 to
w
=
\302\253>
with
will result in
a stable system
by
means
of
the
criterion.\n
in the
is for a sys\302\254
(a) The locus of G(ja)H(ju) shown
figure
two
of G(s)//(s)
with positive real parts. Apply the\n
poles
Prob.
16-16.\n
\nclosed,
The
(b)
G(s)H(s) in the
the loop closed?\n
with
stable
be
\nsystem
right half plane. Will
the
16-16. The following transfer functions
servomech\302\254
two
to
relate
\nanisms
16\n
Chap.
if the system is stable with the loop
locus shown in (b) is known to represent
G(j<t))H(ju))
no poles
of
with
system
SYSTEMS\n
to determine
criterion
Nyquist
\na
IN FEEDBACK
STABILITY
432\n
:\n
3000\n
=\n
G(s)H(s)
(a)
+
s2(l
=\n
G(s)H(s)
(b)
+
In
16-17.
\nfarad
(these
gmRL
\n(a)
the
\nrepresent
16-18.
=
stability
closed-loop
by
unstable.\n
(7
for
two
a stable
A
0.004s)2\n
of system 1 and system 2
criterion.
Answer.
Nyquist
System (a) is
of Prob. 16-12, let R = 1 ohm and = 1
the
network
are normalized
values). Plot the Nyquist diagram
conditions
will
10 and
(b) gmRi. = 40. Which of the
the
of
\nmeans
+ 0.04s)
1500(1
\ns2(l
Investigate
0.004s)\n
system?\n
(nonoscillating)
certain
by the transfer
is described
system
closed-loop
\nfunctions\n
K
=\n
GOO
Determine
\nthe
system
16-19.
the
\t\n
s(TiS + 1)(T*2S+
value
maximum
and
making
unstable.\n
(a) Consider two
functions:\n
H<s>
H(s)
these
\nthe
s plane
1\n
may be used without
of K that
ew -
Plot
=
H(s)
1)\n
functions
two
in (a) of
for values
-1\n
=
1\n
of s along the
the figure. Discuss
how
contour
shown
results
your
\nNyquist criterion.\n
6-plane\n
s-plane\n
Ju
ju)\n
3\n
\\*\n
rC
r-1^\n
-45-y
r)
\nJ
3\n
/!'\n
(6)\n
(a)\n
Prob.
16-19.\n
relate
for
to the
(b)
IN FEEDBACK
STABILITY
16\n
Chap.
two
Consider
G^
G(S)
functions:\n
~
a2
+
+
5\342\200\231\n
4.
+
5\342\200\231\n
~
\nthe
Does
figure.
\neralized
\nDiscuss.\n
as
a
this
criterion
//(\302\253)
=\302\273
1\n
=
1\n
of s along the contourshown
in (b)
of
how the Nyquist criterion might be gen\302\254
suggest
for other than poles in the right half plane?
for values
functions
these
H(s)
2s
\302\253\342\200\231
+
Plot
433\n
SYSTEMS\n
INDEX\n
Active
element,
Admittance
27\n
(also
197
215
200
\nCathode
217
363
\nAmplifier networks,
of
\ncomparison
responses,
frequency
\nCharge, 2
382
270
\nChart,
peaked,
363
365
\nCircuit
280
\nAntiresonant frequencies,
of magnitude,
\nAuxiliary equation,
\nComplete solution, 75, 113
Fourier
\nComplex
328
315,
\nAttenuation,
260
inversion
\nComplex
179
series,
194
\nComplex frequency,
98\n
75, lib
function,
Complementary
change
Asymptotic
97
59\n
\nCofactors,
18\n
\nApproximation,
30
elements,
\nClassical solution,
228
difference,
phase
249, 251
30
\nCircuit,
\nstagger-tuned,
372
polynomials,
\nCircle diagram,
372
118
circuit,
55\n
Chebyshev
\noverstaggered,
\nAngular
2\n
series
RLC
380
411
98,
equation,
\nelectron,
\ndouble-tuned,
296
of networks,
\nCharacteristic
\n388\n
\nlow-pass,
365
403
follower,
\nCauer form
11
band-pass,
91
divider,
\nvoltage
\nCascade connection,
\nAmpere\342\200\231s
law,
\nshunt
24
\nnonlinear,
199\n
\nresistor, 198
\ntransfer,
22
in,
\nCapacitor, 22
combinations,
parallel
stored
\nenergy
\ndriving-point,
\ninductor,
7
\ndefinition,
200
\ncapacitor,
5
Capacitance,
see Immittance),
integral,
127
\nComplex plane, 220
Band-pass
335
filter,
to
\nrelated
scheme, 1
\nConceptual
386
Q,
18
\nConductance,
353
\nBartlett\342\200\231sbisection
theorem,
\nBilateral
21
\nConductors,
353
\nConjugate, 135
elements,
\nBisection theorem,
\nBlock
diagrams:
systems,
331
of, 14
296, 418
principle
linkages,
232,
fractions,
\nContinuous
\n402\n
spectrum, 183,
frequency
\n185\n
399
open-loop,
\nreduction
pulse,
406
procedure,
pulse,
\nConvolution
30
network,
186
\nrecurrent
\ntransformations (table), 400
\nBridge
filters,
flux
\nContinued
in representing
15
laws,
\nConstant
397
elements,
\nlimitations
\nBranch,
3
\nConstant-if
\ndefinitions, 394
\nelectrical
17
\nConductivity,
\nConservation
399
\nclosed-loop,
346
337,
filter,
\nComposite
384
\nBandwidth, 259,
integral,
\nCoulomb\342\200\231s
263
\nButterworth
amplifiers,
\nButterworth
filters,
\nCoupled
368
\nCoupling,
\n435\n
167
6
law,
coils, 44
magnetic,
\nCramer\342\200\231srule,
391\n
183
60
15,
33, 44
INDEX\n
436\n
Critical
220
frequencies,
287
\ninternal,
287
\nCritical resistance,
\ncurrent
2
\nCurrent,
28
Thfrvenin\342\200\231s
\nvacuum
41\n
42
direction,
\nError
in networks,
\nCurrent
ratio
\nCurrent
source,
\nCurrent
source
315
407
tubes,
399
transform,
\nEven
48
equivalent,
function,
320,
frequency,
\nEven
277
174
symmetry,
of phasors,
representation
\nExponential
367\n
179
107,
101,
174
\nEven polynomial,
27
205
theorem,
\nEuler\342\200\231s
equation,
17
density,
\nCutoff
200\n
\nseries,
\nelectron,28
positive
207
202
\nparallel,
2
\ndefinition,
\nCurrent
358
\nNorton\342\200\231s theorem,
\nconventional,
\nloop,
48
source,
\ndelta-Y,
41
\nbranch,
275
circuits,
\nEquivalent
108
solution,
damped
40
equations,
\nEquilibrium
103
372
characteristics,
ripple
Equal
\nexternal,
\nCritically
110\n
Envelope,
\n241\n
Damped
196
sinusoid,
\nDamping
factor,
\nDamping
ratio,
\nDatum
Exterior
103
Faraday\342\200\231s
\nDecade, 262
\nFilters
per
\nDelta
octave,
261
358
transformation,
\nDependent variable, 72
\nresistive
62
network,
60\n
Differential
equation,
\nDirichlet
72
definitions,
173
conditions,
\nDriving-point,
impedances,
194, 274
\nresistrictions on poles and
\nDuality,
zeros,222
52
quantities,
impedance
\nimage
transfer
\nlattice,
353
\nFlux
52
\nFoster
6
27
51
\nextraneous,
\npassive,
Elements
\nEnergy,
\nEnergy
131
14
constant,
79
222
oscillations,
72
function,
of
form
145
14,
11,
of
340
288, 336
networks,
288
theorem,
\nFourier analysis,
171
\nFourier
integral,
183
\nFourier
series,
170
\ncoefficients,173
30\n
of
314
339
section,
\nFoster\342\200\231s
reactance
\nElements:
\nactive,
T
\nForced response,
\nForcing
7
\nElectric field,
of,
324
234,
linkages,
\nForced
51\n
Edastance,
of, 312
function
\nm-derived *- section,
\nFinal value theorem,
\nprinciple
construction,
graphical
346
325
\nimage
\nTO-derived
214
\nDriving-point
\nDuals,
337,
\nm-derived, 338
\nDoublet, 159
\nDual
327
\ncomposite,
\nlow-pass,
331
\nDot convention,
\nband-pass,
\nhigh-pass,
\nsymmetry, 62
\nsystem,
filters),
Amplifier
\nconstant-K, 331
\nexperimental study, 327
58
\nDeterminants,
also
(see
\nattenuation band,
358
networks,
\nDelta-Y
410
systems,
319
3
\nband-elimination, 27
\nDecibel,261, 315
change
12
law,
\nFeedback
31
29,
node,
\nDecibel
70\n
branch,
108
determinant,
\ncomplex
exponential
\ngraphical evaluation,
22
sources,
58
3\n
\nsquare
wave,
175\n
form,
177
179
310
437\n
INDEX\n
sweep voltagefunction,
Fourier series,
\n180\n
\nHomogeneous differential equations, 72,
173
symmetry,
\n111\n
\nsymmetry
rules,
175
Hurwitz
\ntriangular
wave,
177
\nHurwitz
187
\nFourier transformation,
transform
\nFourier
to
\nrelated
185
pairs,
Image
186
transforms,
Laplace
\nL
367
320,
\ncutoff,
\nx
368, 376
258,
infinite
383
\nmid-band,
242
181,
\nnegative,
\nFrequency
spectra,
181
\nFrequency
scaling,
\nFrequency
plane,
\nFrequency
response,
\nImpulse function,
interpre\302\254
30
loops,
\nIndependent variable,
\nInductance
75, 113
of
interpretation
\nself,
deriva\302\254
\nInitial
29\n
of duals,
of Fourier
construction
evaluation
\ncoefficients,
Half-power
impedance
sections,
13
series
258
properties,
\nInitial
171,
\nHeaviside, Oliver,
182
352
\nInstability
conditions,
\nInsulators,
3
factor,
\nIntegrating
\nInverse
roots,
repeated
139\n
(multiple)
J,
roots,
140\n
74,
81
178
43
equations,
transformation,
101
defined,
\nJunction
411
numerical,
\nHeaviside's
expansiontheorem,138,142
\nreal
145
theorem,
\nInsertion loss,
\nIntegrodifferential
125
87
procedure,
value
\nIntegration,
344\n
84
conditions,
\nevaluation
52
177\n
frequencies,
10
parameter,
\nInductor, 22
29
of networks,
72
13
\nmutual,
86\n
Harmonics,
157
158\n
\nunit,
Independent
solution,
\n314,
391
232,
217,
\ntransfer,
graphical
200
combinations,
304
364\n
\nGraphical
198\n
\nresistor,
series
256\n
\nGeometrical
\nHalf
177
220
of, 222
zeros
and
\npoles
126,
200
combinations,
\nparallel
222
230
215,
198
\ninductor,
79
domain,
Graphical
200
\ndriving-point,
225
\nFrequency
Geometry
232, 391
217,
\nImpedance, 197
natural,
\ntation,
230
215,
\ncapacitor,
\nFree oscillations,
246
275
\nequivalent,
320\n
response,
of function,
\ndriving-point,
257
undamped
101
215
\ntransfer,
\nresonant,
314
function,
part
323
338
312,
match,
\nImmittance,
241
194,
\ntives,
with frequency,
numbers,
320
General
313\n
section,
\nImaginary
\nnormalization, 304
\nradian,
313
section,
\nImaginary
195
\nneper,
344
L sections,
\nImage transfer
340
attenuation,
312
314
section,
\nImage
224\n
\ninfinite,
\nGraph,
106\n
variation
211\n
Gain,
functions,
impedance,
\nT
half-power,
\nFree
418
\nm-derived
\nforced, 211
\nstop,
polynomials,
\nHyperbolio
\nFrequency:
\npass,
418
criterion,
188
\nunilateral,
\nfree,
333
325,
filters,
High-pass
(aee
Node)\n
126\n
INDEX\n
438\n
equations, 40
Kirchhoff-law
laws:
\nKirchhoff\342\200\231s
\ncurrent
\nvoltage
58\n
Minors,
of phasors,
Multiplication
law,
57
law,
55\n
\nMutual
Natural
L Section
\nLadder
of
analysis
\nLagging
phase
\nLaplace
transform
\nopen-
\nLaplacetransformation,
Fourier
to
related
\nNeper
127
125\n
186
transformation,
\nLaplacetransformation theorems, 129
\ninitial
value,
filter,
network,
183
41
of nodes,
separate
parts,
31
30
29
datum,
\nNode,
30\n
differential
Nonhomogeneous
324, 333
22\n
Nonlinear
338
filter,
\nMagnetic
field,
\nMagnetic
flux,
10
wave analysis,
\nloop
analysis
of
\nMagnitude
\nMapping,
Maximally
\nMesh
(also
see
205
integration,
\npoles at
178
419
criteria,
\nNyquist
33
origin,
426\n
of, 44
phasor,
228
420\n
flat
frequency,
\nNumerical
170
304
\nNorton\342\200\231s theorem,
11
convention,
21
elements,
\nNormalized
\nMagnetically coupled systems, 15
\ndot
equations,
\n113\n
\nNonsinusoidal
m-DERivED
205
47
\nNode analysis,
\nNode pair,
352
system,
\nLumped
48
30
\nNode,
filter,
\nLow-pass
205
interchange,
\nTh&venin\342\200\231s,
number
insertion,
288
\nsuperposition, 79
44
and
358
transformation,
\nNorton\342\200\231s,
\nLoops, independent,
\nLoss,
353\n
\nsource
30\n
\nbranches
248
\nFoster\342\200\231sreactance,
43
to the
220
theorems:
delta-Y
systems,
247
245
\nbisection,
21
\nLoop currents,
related
fre\302\254
230
procedure,
locus,
\nphasor
125
\nLoops,
at high
245
\nNetwork
complex frequencies, 105
\ncoupled
214\n
asymptotes,
\nphase,
angle, 250
analysis,
\nLoop
53
\nmagnitude interpretation,
35
\nLogarithm,
general,
260\n
\nmagnitude,
elements,
\nLocus of
61
magnitude
\nlow-frequency
353
66, 231,
\nLine spectrum,
\nLinear
356
219
\nLenz\342\200\231s
law,
of
\nquency,
of networks,
phase
resistive,
\nhigh-frequency asymptotes, 246
network,
\nLeading
310
computation
353
\nLattice
\nLattice
214, 374
pair,
equations,
change
equivalent
\nLead
terminal-pair,
\nNetwork
129\n
Lattice
terminal
\ntwo
analysis,
130
\nlinearity,
30\n
one
145
\nintegration,
195
315
\nNetwork
145
\nfinal value,
181, 242
\nNetwork functions,
129
\ndifferentiation,
108, 225
321
short-circuit,
unit,
\nNetwork,
164
\ntables, 129,
undamped,
frequency,
\nNeper
angle, 250
pairs,
and
\nNegative frequency,
403
network,
frequencies:
\ndamped and
232, 310
for, 230
65,
network,
\nmethod
\nLag
311
network,
66\n
45,
inductance,
228, 421
368
characteristics,
Loop),
30
\nMinimum-phase functions,
Odd
function,
\nOdd
polynomial,
\nOhm\342\200\231s
law,
Open-circuit
225\n
174
277
17,
84\n
function,
impedance
\nOperational calculus,
125\n
313
INDEX\n
Oscillator:
439\n
\nResistance, 18
\ntuned-plate, 431\n
108
response,
\nOverdamped response, 108
18
\ndefinition,
22
\nResistor,
255
\nResonance,
of admittances,
combination
Parallel
\n200\n
\nfor
75,
354
332,
320,
band,
of phasor,
328
shift,
421
254,
28
on
locations,
222
direction,
42
current
\nPotential,
5,
\nPower,
415\n
220\n
177
waveform,
219, 287
\nSchedule,
55
\nSelectors,
258
13
22
\nPrototype, 346
resonance,
\nSignal
255
network,
function,
impedance
flow
394
diagrams,
159
\nSingular functions,
Q, 255,
to
related
\nQuad,
\ntransforms
363\n
386
bandwidth,
266\n
112, 414
equation,
Quadratic
\nQuasi-stationary state,
21\n
\nSinusoid,
damped,
\nSolution,
general,
\nSolution,
particular,
\nRamp
loss, 20
72
\nRational
\nline,
215
function,
\nReactive networks,
\nCauer
form,
\ncomparison
\nFoster
form,
\ncases,
288
\nspecifications,
plots,
287
284\n
48
274
\nStability,
302
183
155, 164
wave,
Hurwitz,
\nNyquist,
\nRouth,
410
criteria:\n
418
419
415\n
\nStaircase waveform,
\nStandard
365
amplifiers,
Stagger-tuned
284
\nReactance
of,
181\n
\nStability
functions
\nReactance
Square
296
of features,
72
\nSources, 27
\ncontinuous,
157
function,
196
\nSpectrum:
194
frequency,
\nRadiation
166
(table),
\ntransformation
Radian
200
255
\nShort-circuit
162\n
154,
of impedances,
combination
\nSeries RLC
58
and zeros,
of poles
295\n
\n279,
\nSeries
30\n
property
Separation
Series
226
coptours,
\nSawtooth
3
diagonal,
\nPrincipal
\nPulse,
phasor,
\nSeparateparts,
219\n
restriction
\nPositive
242
\nSelf-inductance,
44
markings,
possible forms, 112
of equations,
\nScale factor,
396
point,
\nPolarity
time,
constant
311
231,
network,
34
rule,
393\n
Right-hand
s-plane,
228,
locations,
pole-zero
\n225\n
\nRotating
241
addition,
\nPolarity,
142
315
function,
\nunit rotating,
\nPoles,
228
195
\nPickoff
from
\nRouth criterion,
\nimage transfer
\nPhasor,
252\n
transient,
Roots
\nPhase
\nPhasor
72
\nsinusoidal,
\nRise
171
function,
\nPhase angle
\nit
roots,
113
conjugate
integral,
\nPeriodic
134
expansion,
complex
\nParticular
\nPass
20
effects,
fraction
\nPartial
280
\nResonant frequencies,
\nResponse,
Parasitic
17
\nparameter,
393\n
\nOvershoot,
103
\ncritical,
transient
Oscillatory
100
roots,
(multiple)
Repeated
\nphase-shift, 431
form
191
of solution,
102\n
313
INDEX\n
440\n
75
Steady-state,
320, 332, 354
\nStop band,
\nSubscript
\nTransient
mutual
for
convention
\nSuperposition, principle of, 79
\nSusceptance,279
\nSustained
\nSymmetrical
lattice
\nSymmetrical
ir
\nSymmetrical
T network,
\nSymmetry
network,
series,
175
\nTandem
\nTerminal
351
\nTermination problem,
\nTheorems
(see
\nThfrvenin\342\200\231s
functions,
phase,
Fourier,
188
\nLaplace,
127\n
161\n
transform,
Vector
29
216
zeros,224
\nVoltage
conventions,
\nVoltage
divider,
\nVoltage
source,
Y
195
(phasor),
\nvon Helmholtz
225
\nrestrictions on poles and
\nTransform:\n
128
153
205
of networks,
\nminimum
165
\ndefinition,
77
networks,
28
218
27
rule,
219\n
31\n
358
\nY-delta transformation,
Zeros,
159
inductor,
function,
step
\nshifted
theorems)
126
domain,
\nTransfer
Network
theorem,
\nTime constants,
\nTopology
\nresponse,
\nUnit
160
capacitor,
168
\ntransform,
214
\nTerminals,
\nTime
373
214
pair,
to
\napplied to
(see Chebyshev),
\nTchebycheff
158
impulse,
\napplied
365
connection,
21
166
\ntransform,
231, 311
method,
21
systems,
\nUnit doublet,
\nUnit
T network,
coefficients
\nUnilateral
60\n
determinant,
\nSystem
231
Fourier
for
frequency,
angular
\n103\n
Undetermined
231
network,
natural
Undamped
231, 263
termsof
241\n
\nexponentials,
110
oscillations,
rules
177
waveform,
\nTrigonometric functions in
395
point,
Summing
225
locations,
pole-zero
\nTriangular
108
oscillatory,
response,
\nfrom
induc\302\254
16\n
\ntance,
129, 164
{table),
pairs
\nperiodic function, 165
170
128,
function,
\nStep
Transform
358\n
113