Electric Circuits, 9th Edition P50 pot

498 1 Be able to transform a circuit into the s domain using Laplace transforms; be sure you understand how to represent the initial conditions on energy-storage elements in the 5 d

Hi v _\

13.1 Circuit Elements in the s Domain p 468

13.2 Circuit Analysis in the s Domain p 470

13.3 Applications p 472

13.4 The Transfer Function p 484

13.5 The Transfer Function in Partial Fraction

Expansions p 486

13.6 The Transfer Function and the Convolution

Integral p 489

13.7 The Transfer Function and the

Steady-State Sinusoidal Response p 495

13.8 The Impulse Function in Circuit

Analysis p 498

1 Be able to transform a circuit into the s domain

using Laplace transforms; be sure you

understand how to represent the initial

conditions on energy-storage elements in the

5 domain

2 Know how to analyze a circuit in the s-domain

and be able to transform an s-domain solution

back to the time domain

3 Understand the definition and significance of

the transfer function and be able to calculate

the transfer function for a circuit using

s-domain techniques

4 Know how to use a circuit's transfer function to

calculate the circuit's unit impulse response, its

unit step response, and its steady-state

response to a sinusoidal input

The Laplace Transform

in Circuit Analysis

The Laplace transform has two characteristics that make it an

attractive tool in circuit analysis First, it transforms a set of linear constant-coefficient differential equations into a set of linear polynomial equations, which are easier to manipulate Second, it automatically introduces into the polynomial equations the initial values of the current and voltage variables Thus, initial condi-tions are an inherent part of the transform process (This con-trasts with the classical approach to the solution of differential equations, in which initial conditions are considered when the unknown coefficients are evaluated.)

We begin this chapter by showing how we can skip the step of writing time-domain integrodifferential equations and transform-ing them into the s domain In Section 13.1, we'll develop the s-domain circuit models for resistors, inductors, and capacitors so that we can write s-domain equations for all circuits directly Section 13.2 reviews Ohm's and Kirchhoff's laws in the context of

the s domain After establishing these fundamentals, we apply the

Laplace transform method to a variety of circuit problems in Section 13.3

Analytical and simplification techniques first introduced with resistive circuits—such as mesh-current and node-voltage methods

and source transformations—can be used in the s domain as well

After solving for the circuit response in the s domain, we inverse transform back to the time domain, using partial fraction expansion (as demonstrated in the preceding chapter) As before, checking the final time-domain equations in terms of the initial conditions and final values is an important step in the solution process

The s-domain descriptions of circuit input and output lead us,

in Section 13.4, to the concept of the transfer function The trans-fer function for a particular circuit is the ratio of the Laplace transform of its output to the Laplace transform of its input In Chapters 14 and 15, we'll examine the design uses of the transfer function, but here we focus on its use as an analytical tool We continue this chapter with a look at the role of partial fraction

466

Surge Suppressors

With the advent of home-based personal computers, modems,

fax machines, and other sensitive electronic equipment, i t is

necessary to provide protection from voltage surges that can

occur in a household circuit due to switching A

commer-cially available surge suppressor is shown in the

accompany-ing figure

How can flipping a switch to turn on a light or turn off a hair dryer cause a voltage surge? At the end of this chapter,

we will answer that question using Laplace transform tech-niques to analyze a circuit We will illustrate how a voltage surge can be created by switching off a resistive load in a cir-cuit operating in the sinusoidal steady state

467

468 The Laplace Transform in Circuit Analysis

expansion (Section 13.5) and the convolution integral (Section 13.6) in employing the transfer function in circuit analysis We conclude with a look at the impulse function in circuit analysis

13.1 Circuit Elements in the s Domain

The procedure for developing an ^-domain equivalent circuit for each cir-cuit element is simple First, we write the time-domain equation that relates the terminal voltage to the terminal current Next, we take the Laplace transform of the time-domain equation This step generates an algebraic relationship between the s-domain current and voltage Note that the dimension of a transformed voltage is volt-seconds, and the dimension

of a transformed current is ampere-seconds A voltage-to-current ratio in

the s domain carries the dimension of volts per ampere An impedance

in the s domain is measured in ohms, and an admittance is measured in Siemens Finally, we construct a circuit model that satisfies the relation-ship between the ^-domain current and voltage We use the passive sign convention in all the derivations

A Resistor in the s Domain

We begin with the resistance element From Ohm's law,

v = RL

Because R is a constant, the Laplace transform of Eq 13.1 is

(13.1)

V = RI, (13.2)

4

+

V <

4

1

»

:K\I

1

a

+ T

vi

•

* ! •

b b

(a) (b)

Figure 13.1 A The resistance element, (a) Time domain,

(b) Frequency domain

where

V = %{v\ and J = £{*}

Equation 13.2 states that the s-domain equivalent circuit of a resistor is

simply a resistance of R ohms that carries a current of I ampere-seconds and has a terminal voltage of V volt-seconds

Figure 13.1 shows the time- and frequency-domain circuits of the resistor Note that going from the time domain to the frequency domain does not change the resistance element

vL iW'

An Inductor in the s Domain

Figure 13.2 shows an inductor carrying an initial current of /Q amperes The time-domain equation that relates the terminal voltage to the termi-nal current is

Figure 13.2 A An inductor of L henrys carrying an

The Laplace transform of Eq 13.3 gives

V = L[sl - /(0 )] = sLl - LI (13.4)

Two different circuit configurations satisfy Eq 13.4 The first consists

of an impedance of sL ohms in series with an independent voltage source

of L/() volt-seconds, as shown in Fig 13.3 Note that the polarity marks on

the voltage source L/() agree with the minus sign in Eq 13.4 Note also

that LI () carries its own algebraic sign; that is, if the initial value of i is

opposite to the reference direction for i, then /(, has a negative value

The second y-domain equivalent circuit that satisfies Eq 13.4 consists

of an impedance of sL ohms in parallel with an independent current

source of I {) /s ampere-seconds, as shown in Fig 13.4 We can derive the

alternative equivalent circuit shown in Fig 13.4 in several ways One way

is simply to solve Eq 13.4 for the current I and then construct the circuit to

satisfy the resulting equation Thus

/ — — T

sL sL s

(13.5)

Two other ways are: (1) find the Norton equivalent of the circuit shown in

Fig 13.3 and (2) start with the inductor current as a function of the

induc-tor voltage and then find the Laplace transform of the resulting integral

equation We leave these two approaches to Problems 13.1 and 13.2

If the initial energy stored in the inductor is zero, that is, if /u = 0, the

y-domain equivalent circuit of the inductor reduces to an inductor with an

impedance of sL ohms Figure 13.5 shows this circuit

A Capacitor in the s Domain

An initially charged capacitor also has two s-domain equivalent circuits

Figure 13.6 shows a capacitor initially charged to V {) volts The terminal

current is

Figure 13.3 • The series equivalent circuit for an

inductor of L henrys carrying an initial current of

/ „ amperes

Figure 13.4 • The parallel equivalent circuit for an

inductor of L henrys carrying an initial current of

/ 0 amperes

/ = C dv

VisLU

Transforming Eq 13.6 yields

or

I = C[sV - v(0~)]

I = sCV - CV (h (13.7)

which indicates that the v-domain current I is the sum of two branch

cur-rents One branch consists of an admittance of sC Siemens, and the second

branch consists of an independent current source of CVQ ampere-seconds

Figure 13.7 shows this parallel equivalent circuit

We derive the series equivalent circuit for the charged capacitor by

solving Eq 13.7 for V:

Figure 13.5 • The s-domain circuit for an inductor

when the initial current is zero

+ 4

vCT-+

Figure 13.6 • A capacitor of C farads initially charged

to V 0 volts

Figure 13.8 shows the circuit that satisfies Eq 13.8

In the equivalent circuits shown in Figs 13.7 and 13.8, V () carries its

own algebraic sign In other words, if the polarity of V 0 is opposite to the

reference polarity for v, V is a negative quantity If the initial voltage on

470 The Laplace Transform in Circuit Analysis

l/.vC

Figure 13.7 A The parallel equivalent circuit for a

capacitor initially charged to V 0 volts

+ fa

1/sC

V 0 /s

Figure 13.8 • The series equivalent circuit for a

capacitor initially charged to V 0 volts

/sC

Figure 13.9 • The s-domain circuit for a capacitor

when the initial voltage is zero

the capacitor is zero, both equivalent circuits reduce to an impedance of

\/sC ohms, as shown in Fig 13.9

In this chapter, an important first problem-solving step will be to choose between the parallel or series equivalents when inductors and capacitors are present With a little forethought and some experience, the correct choice will often be quite evident The equivalent circuits are sum-marized in Table 13.1

TABLE 13.1 Summary of the s-Domain Equivalent Circuits

TIME DOMAIN FREQUENCY DOMAIN

»i V*R

b

v = Ri

H f a +

i = C dvjdu

sL] V ( | ) / „ A

- * b

V = sLl - Ll {)

*b

v At

I = +

sL s

i/vc:

/ = sCV - CV {)

13.2 Circuit Analysis in the s Domain

Before illustrating how to use the s-domain equivalent circuits in analysis,

we need to lay some groundwork

First, we know that if no energy is stored in the inductor or capacitor, the relationship between the terminal voltage and current for each passive element takes the form:

Ohm's law in the s-domain • V = ZI, (13.9)

where Z refers to the s-domain impedance of the element Thus a resistor

has an impedance of R ohms, an inductor has an impedance of sL ohms

and a capacitor has an impedance of 1/sC ohms The relationship

con-tained in Eq 13.9 is also concon-tained in Figs 13.1(b), 13.5, and 13.9

Equation 13.9 is sometimes referred to as Ohm's law for the s domain

The reciprocal of the impedance is admittance Therefore, the s domain

admittance of a resistor is \/R Siemens, an inductor has an admittance of

1/sL Siemens, and a capacitor has an admittance of sC Siemens

The rules for combining impedances and admittances in the s domain

are the same as those for frequency-domain circuits.Thus series-parallel

sim-plifications and A-to-Y conversions also are applicable to v-domain analysis

In addition, Kirchhoff s laws apply to s-domain currents and voltages

Their applicability stems from the operational transform stating that the

Laplace transform of a sum of time-domain functions is the sum of the

transforms of the individual functions (see Table 12.2) Because the

braic sum of the currents at a node is zero in the time domain, the

alge-braic sum of the transformed currents is also zero A similar statement

holds for the algebraic sum of the transformed voltages around a closed

path The s-domain version of Kirchhoff s laws is

alg2/ = 0, (13.10)

Because the voltage and current at the terminals of a passive element

are related by an algebraic equation and because Kirchhoff s laws still

hold, all the techniques of circuit analysis developed for pure resistive

networks may be used in s-domain analysis Thus node voltages, mesh

currents, source transformations, and Thevenin-Norton equivalents are

all valid techniques, even when energy is stored initially in the inductors

and capacitors Initially stored energy requires that we modify Eq 13.9 by

simply adding independent sources either in series or parallel with the

element impedances The addition of these sources is governed by

Kirchhoff s laws

/ A S S E S S M E N T PROBLEMS

Objective 1—Be able to transform a circuit into the s domain using Laplace transforms

13.1 A 500 O, resistor, a 16 mH inductor, and a 25 11F

capacitor are connected in parallel

a) Express the admittance of this parallel

com-bination of elements as a rational function

of s

b) Compute the numerical values of the zeros

and poles

Answer: (a) 25 X 1 0 " V + 80,000.? + 25 x 1(f)/s;

(b) -z { = -40,000 - /30,000;

- Z2 = -40,000 + /30,000; p x = 0

13.2 The parallel circuit in Assessment Problem 13.1

is placed in series with a 2000 fl resistor

a) Express the impedance of this series

combi-nation as a rational function of s

b) Compute the numerical values of the zeros and poles

Answer: (a) 2000(5 + 50,000)2/C*2 + 80,0005

+ 25 X 108);

(b) -z x = -z 2 = -50,000;

-p 1 = -40,000 - /30,000,

-p 2 = -40,000 + 730,000

NOTE: Also try Chapter Problems 13.4 and 13.6

472 The Laplace Transform in Circuit Analysis

13.3 Applications

We now illustrate how to use the Laplace transform to determine the tran-sient behavior of several linear lumped-parameter circuits We start by ana-lyzing familiar circuits from Chapters 7 and 8 because they represent a simple starting place and because they show that the Laplace transform approach yields the same results In all the examples, the ease of manipulat-ing algebraic equations instead of differential equations should be apparent

Figure 13.10 • The capacitor discharge circuit

+

/

RkV

Figure 13.11 A An s-domain equivalent circuit for the

circuit shown in Fig 13.10

The Natural Response of an RC Circuit

We first revisit the natural response of an RC circuit (Fig 13.10) via

Laplace transform techniques (You may want to review the classical analysis of this same circuit in Section 7.2)

The capacitor is initially charged to V0 volts, and we are interested in

the time-domain expressions for i and v We start by finding i In transfer-ring the circuit in Fig 13.10 to the s domain, we have a choice of two

equiv-alent circuits for the charged capacitor Because we are interested in the current, the series-equivalent circuit is more attractive; it results in a single-mesh circuit in the frequency domain Thus we construct the s-domain cir-cuit shown in Fig 13.11

Summing the voltages around the mesh generates the expression

s sC (13.12)

Solving Eq 13.12 for /yields

RCs + 1 s + (1/RC)' (13.13)

Figure 13.12 A An s-domain equivalent circuit for the

circuit shown in Fig 13.10

Note that the expression for I is a proper rational function of s and can be

inverse-transformed by inspection:

(13.14)

which is equivalent to the expression for the current derived by the classi-cal methods discussed in Chapter 7 In that chapter, the current is given by

Eq 7.26, where T is used in place of RC

After we have found /, the easiest way to determine v is simply to

apply Ohm's law; that is, from the circuit,

V = Ri = V {) e~' /RC u(t) (13.15)

We now illustrate a way to find v from the circuit without first finding /'

In this alternative approach, we return to the original circuit of Fig 13.10 and transfer it to the 5 domain using the parallel equivalent circuit for the charged capacitor Using the parallel equivalent circuit is attractive now because we can describe the resulting circuit in terms of a single node volt-age Figure 13.12 shows the new s-domain equivalent circuit

The node-voltage equation that describes the new circuit is

Solving Eq 13.16 for V gives

V Vn (13.17)

s + {i/Rcy

Inverse-transforming Eq 13.17 leads to the same expression for v given by

Eq 13.15, namely,

v = V {) er ! l RC = V 0 e^ T u(t) (13.18)

Our purpose in deriving by direct use of the transform method is to

show that the choice of which 5-domain equivalent circuit to use is

influ-enced by which response signal is of interest

/ A S S E S S M E N T PROBLEM

Objective 2—Know how t o analyze a circuit in the s domain and be able to transform an s domain solution to the

time domain

13.3 The switch in the circuit shown has been in

position a for a long time At t = 0, the switch

is thrown to position b

a) Find I, V h and V 2 as rational functions of s

b) Find the time-domain expressions for i, v h

and v 2

Answer: (a) I = 0.02/(5 + 1250),

V x = 80/(5 + 1250),

V 2 = 20/(5 + 1250);

NOTE: Also try Chapter Problems 13.9 and 13.12

(b) i = 20<T1250fM(0 m A ,

v 2 = 20e- u50t u(t) V

io kn \

100V

A

(j\ 0.2 MF;

0.8 JUF;

f = 0

+ +

| | 5 k a

The Step Response of a Parallel Circuit

Next we analyze the parallel RLC circuit, shown in Fig 13.13, that we first

analyzed in Example 8.7 The problem is to find the expression for i L after

the constant current source is switched across the parallel elements The

initial energy stored in the circuit is zero

As before, we begin by constructing the 5-domain equivalent circuit

shown in Fig 13.14 Note how easily an independent source can be

trans-formed from the time domain to the frequency domain We transform the

source to the s domain simply by determining the Laplace transform of its

time-domain function Here, opening the switch results in a step change in

the current applied to the circuit Therefore the 5-domain current source is

!£{I\} C u(t)}, or lfe/$ To find I L , we first solve for V and then use

I, =

to establish the 5-domain expression for 1 L Summing the currents away

from the top node generates the expression

sCV + — H — - = —

R sL 5

Solving Eq 13.20 for V gives

5 + {i/RC)s + (1/LC)

(13.20)

(13.21)

R h\L

625 Ct

2 5 m H |

Figure 13.13 • The step response of a parallel

RLC circuit

Figure 13.14 • The s-domain equivalent circuit for the

circuit shown in Fig 13.13

Substituting Eq 13.21 into Eq 13.19 gives

hJLC

L s[s 2 + (1/RQs + (1/LC)]'

Substituting the numerical values of R, L, C, and /d c into Eq 13.22 yields

384 X 105

h = — > 5T« (13.23)

i'(.?2 + 64,000.v + 16 X 108) Before expanding Eq 13.23 into a sum of partial fractions, we factor the

quadratic term in the denominator:

384 X 105

L " s(s + 32,000 - /24,000)(5 + 32,000 + /24,000) * ( 1 3'2 4 )

Now, we can test the ,v-domain expression for I L by checking to see

whether the final-value theorem predicts the correct value for i L at

t = co All the poles of I L , except for the first-order pole at the origin, lie

in the left half of the s plane, so the theorem is applicable We know from

the behavior of the circuit that after the switch has been open for a long

time, the inductor will short-circuit the current source.Therefore, the final

value of i L must be 24 m A The limit of si L a s s —> 0 is

,. T 3 8 4 X 1 05 n, A

hm sir = r = 24 mA (13.25)

*-o 16 X 108 (Currents in the v domain carry the dimension of ampere-seconds, so the

dimension of sI L will be amperes.) Thus our s-domain expression checks out

We now proceed with the partial fraction expansion of Eq 13.24:

s s + 32,000 - /24,000

Kl

+ s + 32,000 + /24,000' ( 1 3'2 6 ) The partial fraction coefficients are

384 X 103 ,„ ,

K ] = jT- = 2 4 X 10"3, (13.27)

16 x 1 08

3 8 4 X 1 05

A.'

(-32,000 + /24,000)(/48,000)

Substituting the numerical values of K x and K 2 into Eq 13.26 and

inverse-transforming the resulting expression yields

i L = [24 + 40e"320(M)'cos(24,000f + 126.87° )]«(f)mA (13.29)

The answer given by Eq 13.29 is equivalent to the answer given for

Example 8.7 because

40cos(24,000f + 126.87°) = - 2 4 cos 24,000/ - 32 sin 24,000/

If we weren't using a previous solution as a check, we would test

Eq 13.29 to make sure that /L(0) satisfied the given initial conditions and

i L (oo) satisfied the known behavior of the circuit

/ A S S E S S M E N T PROBLEM

Objective 2—Know how to analyze a circuit in the s domain and be able to transform an s domain solution to the

time domain

13.4 The energy stored in the circuit shown is zero Answer: (a) I = 40/(52 + 1.2s + 1);

at the time when the switch is closed. ( b ) , = (5Qe-o.6r s i n Q 8 t ) m A ;

a) Find the 5-domain expression for I (c) v = 160^/(^2 + 1.2s + 1);

(d) v = [200£fa6'cos(0.8r + 36.87°)]u(0 V

b) Find the time-domain expression for i when

t > 0 V 4.8 II 4 H

c) Find the s-domain expression for V ^J?\ ' = ()

d) Find the time-domain expression for v when

t > 0

NOTE: Also try Chapter Problems 13.10 and 13.21

The Transient Response of a Parallel RLC Circuit

Another example of using the Laplace transform to find the transient

behav-ior of a circuit arises from replacing the dc current source in the circuit shown

in Fig 13.13 with a sinusoidal current source The new current source is

if> = /mcostof A, (13.30) where /„, = 24 mA and o> = 40,000 rad/s As before, we assume that the

initial energy stored in the circuit is zero

The v-domain expression for the source current is

Sim

h r + w2

The voltage across the parallel elements is

(I s /C)s

s 2 + (l/RC)s + (1/L

Substituting Eq 13.31 into Eq 13.32 results in

(IJC)s 2

L + (1/RQs

from which

V (IJLC)s

sL (s 2 + a> 2 )[ s2 + (1/RQs + {1/LC)]

Substituting the numerical values of /m, w, R, L, and C into Eq 13.34 gives

384 X 105s

(s z + 16 X 108)(s2 + 64,000^ +

We now write the denominator in factored form:

-vw-+ v —

V = 2 , - / rJ ^ -777777- (1 3-3 2)

y - 7 X 7 2u.2 • 7717777, 777777;, (13.33)

J L = — == 7 T T - ^ 2 ,+ ,— 7777- (13.34)

h = ~h T7^ 5T- (13.35)

:0.25 F

384 X 1055

'" ~ (5 - j(o)(s + j(o)(s + a - //3)(5 + a+Wy