Electric Circuits, 9th Edition P53 ppt

Thus Steady-state sinusoidal response computed using a transfer function • y«0 = A\Hja>\ cos [ajt + + 6»ft>], 13.120 which indicates how to use the transfer function to find the stea

496 The Laplace Transform in Circuit Analysis

Substituting Eq 13.114 into Eq 13.96 gives the ,-domain expression for the response:

TT/ x A(s cos <f> - a)sm<}>)

We now visualize the partial fraction expansion of Eq 13.115 The number

of terms in the expansion depends on the number of poles of H(s)

Because H(s) is not specified beyond being the transfer function of a

physically realizable circuit, the expansion of Eq 13.115 is

TO-T^r +

T^r-S — )0) T^r-S + )0)

+ 2 terms generated by the poles of H(s) (13.116)

In Eq 13.116, the first two terms result from the complex conjugate poles

of the driving source; that is, s 2 + w2 = (s — jta)(s + joy) However, the terms generated by the poles of H(s) do not contribute to the steady-state response of y(t), because all these poles lie in the left half of the s plane;

consequently, the corresponding time-domain terms approach zero as t

increases Thus the first two terms on the right-hand side of Eq 13.116 determine the steady-state response The problem is reduced to finding

the partial fraction coefficient K\

H(s)A(s cos 4> — a) sin</>) K\ = ; H(jm)A(j(t>eos<f> - fusing)

H(ja))A(cos(b + / s i n 0 ) 1 , ,

In general, H(jo)) is a complex quantity, which we recognize by writing it

in polar form; thus

H(ja>) = \H(jo))\ej9iM) (13.118) Note from Eq 13.118 that both the magnitude, \H(j<o)\, and phase angle,

#(&>), of the transfer function vary with the frequency co When we substi-tute Eq 13.118 into Eq 13.117, the expression for K x becomes

We obtain the steady-state solution for y(t) by inverse-transforming

Eq 13.116 and, in the process, ignoring the terms generated by the poles of

H(s) Thus

Steady-state sinusoidal response computed

using a transfer function • y«(0 = A\H(ja>)\ cos [ajt + <f> + 6»(ft>)], (13.120)

which indicates how to use the transfer function to find the steady-state sinusoidal response of a circuit The amplitude of the response equals the amplitude of the source, A, times the magnitude of the transfer function,

\H{j(t))\ The phase angle of the response, <£ + 0(co), equals the phase angle of the source, cj>, plus the phase angle of the transfer function, 0(a))

We evaluate both \H(ja>)\ and 9(to) at the frequency of the source, io

Example 13.4 illustrates how to use the transfer function to find the steady-state sinusoidal response of a circuit

13.7 The Transfer Function and the Steady-State Sinusoidal Response 497

The circuit from Example 13.1 is shown in Fig 13.46

The sinusoidal source voltage is 120 cos (5000/

+ 30°) V Find the steady-state expression for v 0

The frequency of the voltage source is 5000 rad/s;

hence we evaluate H(s) at 7/(/5000):

1000 ft

—VYV—

50

mH-IjtF

Figure 13.46 A The circuit for Example 13.4

Solution

From Example 13.1,

H(s) = 1000(.v + 5000)

s2 + 6000s + 25*10*

7/(/5000) 1000(5000 + /5000)

-25 * 106 + /5000(6000) + 25 X 10f

1 + / 1 1 - / 1 V 2

Then, from Eq 13.120,

(120)V2

- — 2 cos(5000f + 30° - 45°)

o„

20V2 cos (5000/ - 15°) V

The ability to use the transfer function to calculate the steady-state sinusoidal

response of a circuit is important Note that if we know 77(/a>), we also know

77(^), at least theoretically In other words, we can reverse the process; instead of

using 7/(5) to find 77(ja>), we use H(jco) to find H(s) Once we know H(s), we

can find the response to other excitation sources In this application, we determine

H(jo)) experimentally and then construct H{s) from the data Practically, this

experimental approach is not always possible; however, in some cases it does

pro-vide a useful method for deriving H(s) In theory, the relationship between H{s)

and H(j(x)) provides a link between the time domain and the frequency domain

The transfer function is also a very useful tool in problems concerning the

fre-quency response of a circuit, a concept we introduce in the next chapter

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

Objective 4—Know how to use a circuit's transfer function to calculate the circuit's impulse response, unit step

response, and steady-state response to sinusoidal input

13.12 The current source in the circuit shown is

deliv-ering 10 cos 4/ A Use the transfer function to

compute the steady-state expression for v 0

Answer: 44.7cos(4/ - 63.43°) V

13.13 a) For the circuit shown, find the steady-state

expression for v 0 when

vg = 10cos50,000r V

NOTE: Also try Chapter Problems 13.77 and 13.80

b) Replace the 50 kfl resistor with a variable resistor and compute the value of resistance

necessary to cause v 0 to lead v g by 120°

10 kO 10 k n

/vw-Answer: (a) 10 cos (50,000/ + 90°) V;

(b) 28,867.51 ft

498 The Laplace Transform in Circuit Analysis

:c,

Figure 13.47 • A circuit showing the creation of an

impulsive current

Figure 13.48 • The s-domain equivalent circuit for the

circuit shown in Fig 13.47

R2<R)

Figure 13.49 A The plot of i (t) versus t for two

different values of /?

13.8 The Impulse Function in

Circuit Analysis

Impulse functions occur in circuit analysis either because of a switching operation or because a circuit is excited by an impulsive source The Laplace transform can be used to predict the impulsive currents and volt-ages created during switching and the response of a circuit to an impulsive source We begin our discussion by showing how to create an impulse function with a switching operation

Switching Operations

We use two different circuits to illustrate how an impulse function can be created with a switching operation: a capacitor circuit, and a series induc-tor circuit

-£- Capacitor Circuit

In the circuit shown in Fig 13.47, the capacitor Ci is charged to an initial

voltage of V () at the time the switch is closed The initial charge on C 2 is

zero The problem is to find the expression for /'(/) as R —» 0 Figure 13.48

shows the s-domain equivalent circuit

From Fig 13.48,

R + (1/sCj) + (l/sC2) Vo/R

where the equivalent capacitance C\C 2 /{C X + C 2 ) is replaced by C e

We inverse-transform Eq 13.121 by inspection to obtain

which indicates that as R decreases, the initial current (Vo/R) increases and the time constant (RC e ) decreases Thus, as R gets smaller, the current

starts from a larger initial value and then drops off more rapidly Figure 13.49 shows these characteristics of/

Apparently i is approaching an impulse function as R approaches zero because the initial value of i is approaching infinity and the duration of i is

approaching zero We still have to determine whether the area under the

current function is independent of R Physically, the total area under the

i versus t curve represents the total charge transferred to C2 after the switch

is closed Thus

which says that the total charge transferred to C2 is independent of R and equals V {)Ce coulombs Thus, as R approaches zero, the current approaches

an impulse strength V ()Ce; that is,

13.8 The Impulse Function in Circuit Analysis 499

The physical interpretation of Eq 13.124 is that when R = 0, a finite

amount of charge is transferred to C2 instantaneously Making R zero in

the circuit shown in Fig 13.47 shows why we get an instantaneous transfer

of charge With R = 0, we create a contradiction when we close the switch;

that is, we apply a voltage across a capacitor that has a zero initial voltage

The only way to have an instantaneous change in capacitor voltage is to

have an instantaneous transfer of charge When the switch is closed, the

voltage across C2 does not jump to V () but to its final value of

v 2 C, + C, (13.125)

We leave the derivation of Eq 13.125 to you (see Problem 13.81)

If we set R equal to zero at the outset, the Laplace transform analysis

will predict the impulsive current response Thus,

( 1 / J d ) + (l/sC 2) C, + C2 C,V0 (13.126)

In writing Eq 13.126, we use the capacitor voltages at t = (T The inverse

transform of a constant is the constant times the impulse function;

there-fore, from Eq 13.126,

The ability of the Laplace transform to predict correctly the occurrence of an

impulsive response is one reason why the transform is widely used to analyze

the transient behavior of linear lumped-parameter time-invariant circuits

Series Inductor Circuit

The circuit shown in Fig 13.50 illustrates a second switching operation

that produces an impulsive response The problem is to find the

time-domain expression for v (, after the switch has been opened Note that

opening the switch forces an instantaneous change in the current of L2,

which causes v 0 to contain an impulsive component

Figure 13.51 shows the s-domain equivalent with the switch open In

deriving this circuit, we recognized that the current in the 3 H inductor at

t = 0~ is 10 A, and the current in the 2 H inductor at t = 0~ is zero Using

the initial conditions at t = 0" is a direct consequence of our using 0~ as

the lower limit on the defining integral of the Laplace transform

We derive the expression for V a from a single node-voltage equation

Summing the currents away from the node between the 15 ft resistor and

the 30 V source gives

v:

2s + 15 +

V a - [(100/5) + 30]

10 O

VW-© 100 V

3H

_/-Y-Y"Y-\-Li

/ = (>

sC

i 5 i r

2 H

T

-L 2

Figure 13.50 A A circuit showing the creation of an impulsive Figure 13.51 • The s-domain equivalent circuit for the

voltage circuit shown in Fig 13.50

Solving for V 0 yields

40(5 + 7.5) 12(5 + 7.5)

We anticipate that v 0 will contain an impulse term because the second

term on the right-hand side of Eq 13.129 is an improper rational function

We can express this improper fraction as a constant plus a rational

func-tion by simply dividing the denominator into the numerator; that is,

12(5 + 7.5) 30

— •=-*- = 12 + - (13.130)

Combining Eq 13.130 with the partial fraction expansion of the first term

on the right-hand side of Eq 13.129 gives

x, 6 0 2 0 ,„ 30

V„ = r + 12 +

S 5 + 5 5 + 5

„„ 60 10

= 12 + — + - , (13.131)

from which

Does this solution make sense? Before answering that question, let's

first derive the expression for the current when t > 0~ After the switch has

been opened, the current in Li is the same as the current in L2 If we

refer-ence the current clockwise around the mesh, the 5-domain expression is

55 + 25 s(s + 5) 5 + 5

4

+

5 5 + 5 5 + 5

4 2

= - + r (13.133

5 5 + 5 '

Inverse-transforming Eq 13.133 gives

Before the switch is opened, the current in L 1 is 10 A, and the current

in L 2 is 0 A; from Eq 13.134 we know that at t = Q +, the current in L\ and

in L 2 is 6 A.Then, the current in L\ changes instantaneously from 10 to 6 A,

while the current in L 2 changes instantaneously from 0 to 6 A From this

value of 6 A, the current decreases exponentially to a final value of 4 A

13.8 The Impulse Function in Circuit Analysis 501

This final value is easily verified from the circuit; that is, it should equal

100/25, or 4 A Figure 13.52 shows these characteristics of i { and i 2

How can we verify that these instantaneous jumps in the inductor

cur-rent make sense in terms of the physical behavior of the circuit? First, we

note that the switching operation places the two inductors in series Any

impulsive voltage appearing across the 3 H inductor must be exactly

bal-anced by an impulsive voltage across the 2 H inductor, because the sum of

the impulsive voltages around a closed path must equal zero Faraday's

law states that the induced voltage is proportional to the change in flux

linkage (v = dk/dt) Therefore, the change in flux linkage must sum to

zero In other words, the total flux linkage immediately after switching is

the same as that before switching For the circuit here, the flux linkage

before switching is

A = L x i x + L 2 i 2 = 3(10) + 2(0) - 30 Wb-turns (13.135)

Immediately after switching, it is

Combining Eqs 13.135 and 13.136 gives

Thus the solution for i (Eq [13.134]) agrees with the principle of the

con-servation of flux linkage

We now test the validity of Eq 13.132 First we check the impulsive

term 125(f) The instantaneous jump of i 2 from 0 to 6 A at t = 0 gives rise

to an impulse of strength 66(f) in the derivative of i 2 This impulse gives

rise to the 125(f) in the voltage across the 2 H inductor For f > 0+, di 2/dt

is -10e~5' A/s; therefore, the voltage v 0 is

Vo = 15(4 + 2e~ 5t ) + 2(-10<T5')

/L, i 2 (A)

i 2 = i

Figure 13.52 A The inductor currents versus t for the

circuit shown in Fig 13.50

Equation 13.138 agrees with the last two terms on the right-hand side of

Eq 13.132; thus we have confirmed that Eq 13.132 does make sense in

terms of known circuit behavior

We can also check the instantaneous drop from 10 to 6 A in the

cur-rent I'I This drop gives rise to an impulse of -45(f) in the derivative of i h

Therefore the voltage across Lt contains an impulse of —125(f) at the

ori-gin This impulse exactly balances the impulse across L2; that is, the sum of

the impulsive voltages around a closed path equals zero

Impulsive Sources

Impulse functions can occur in sources as well as responses; such sources

are called impulsive sources An impulsive source driving a circuit imparts

a finite amount of energy into the system instantaneously A mechanical

analogy is striking a bell with an impulsive clapper blow After the energy

has been transferred to the bell, the natural response of the bell

deter-mines the tone emitted (that is, the frequency of the resulting sound

waves) and the tone's duration

502 The Laplace Transform in Circuit Analysis

V Q 8(t)

Figure 13.53 A An RL circuit excited by an impulsive

voltage source

In the circuit shown in Fig 13.53, an impulsive voltage source having a strength of V0 volt-seconds is applied to a series connection of a resistor and an inductor When the voltage source is applied, the initial energy in the inductor is zero; therefore the initial current is zero There is no voltage

drop across R, so the impulsive voltage source appears directly across L

An impulsive voltage at the terminals of an inductor establishes an instan-taneous current The current is

Given that the integral of 8(t) over any interval that includes zero is 1, we

find that Eq 13.139 yields

.•«n = v~l A (13.140)

Thus, in an infinitesimal moment, the impulsive voltage source has stored in the inductor

in

2 L

(13.141)

The current V {)/L now decays to zero in accordance with the natural

response of the circuit; that is,

l = Te

Figure 13.54 • The 5-domain equivalent circuit for the

circuit shown in Fig 13.53

where T = L/R Remember from Chapter 7 that the natural response is

attributable only to passive elements releasing or storing energy, and not

to the effects of sources When a circuit is driven by only an impulsive source, the total response is completely defined by the natural response; the duration of the impulsive source is so infinitesimal that it does not contribute to any forced response

We may also obtain Eq 13.142 by direct application of the Laplace transform method Figure 13.54 shows the 5-domain equivalent of the cir-cuit in Fig 13.53

Hence

ion

>vw- _/-Y"VY>_ 3H

505(/)

100 V

12 H

Figure 13.55 A The circuit shown in Fig 13.50 with

an impulsive voltage source added in series with the

100 V source

L

-(R/L)t

L

Thus the Laplace transform method gives the correct solution for i £: 0+ Finally, we consider the case in which internally generated impulses and externally applied impulses occur simultaneously The Laplace

trans-form approach automatically ensures the correct solution for t > 0+ if

inductor currents and capacitor voltages at t = 0~ are used in constructing

the j'-domain equivalent circuit and if externally applied impulses are rep-resented by their transforms To illustrate, we add an impulsive voltage

13.8 The Impulse Function in Circuit Analysis 503

source of 505(f) in series with the 100 V source to the circuit shown in

Fig 13.50 Figure 13.55 shows the new arrangement

At t = 0", ii(Q~) = 10 A and /2(0~) = 0 A The Laplace transform of

505(0 = 50 If we use these values, the s-domain equivalent circuit is as

shown in Fig 13.56

The expression for I is

I = 50 + (100/5) + 30

25 + 5s circuit shown in Fig 13.55 Figure 13.56 • The s-domain equivalent circuit for the

16

+

20

s + 5 s(s + 5)

16 4 4 _

5 + 5 s 5 + 5

12 4 +

-5 + -5 -5

(13.145)

from which

/(f) = (12e_5/ + 4 ) M ( 0 A (13.146)

The expression for V 0 is

V0 = (15 + 2s) J =

-5 + -5 5(5 + 5)

2.5 \ 60 20

= 32 1 + r ) +

5 + 5 ) 5 5 + 5

+ — ,

5 + 5 5

(13.147)

from which

v„ = 325(f) + (60e"5' + 60)»(f) V (13.148)

Now we test the results to see whether they make sense From

Eq 13.146, we see that the current in L x and L 2 is 16 A at f = 0+ As in the

previous case, the switch operation causes /, to decrease instantaneously

from 10 to 6 A and, at the same time, causes /2 to increase from 0 to 6 A

Superimposed on these changes is the establishment of 10 A in L\ and L 2

by the impulsive voltage source; that is,

1

Therefore i { increases suddenly from 10 to 16 A, while /2 increases

sud-denly from 0 to 16 A The final value of i is 4 A Figure 13.57 shows i h /2,

circuit shown in Fig 13.55

We may also find the abrupt changes in i x and /2 without using

super-position The sum of the impulsive voltages across L x (3 H) and L 2 (2 H)

equals 505(f) Thus the change in flux linkage must sum to 50; that is,

Because A = Li, we express Eq 13.150 as

But because /5 and /2 must be equal after the switching takes place,

Then,

Solving Eqs 13.151 and 13.153 for A/j and A/2 yields

A/j - 6 A, (13.154)

These expressions agree with the previous check

Figure 13.57 also indicates that the derivatives of ij and i 2 will contain

an impulse at t = 0 Specifically, the derivative of i x will have an impulse

of 65(0, and the derivative of /2 will have an impulse of 165(f)

Figure 13.58(a), (b), respectively, illustrate the derivatives of/^ and i 2

Now let's turn to Eq 13.148 The impulsive component 325(f) agrees

with the impulse 165(f) that characterizes di 2/dt at the origin The term

(60<T5' + 60) agrees with the fact that for t > 0 +,

di

v (> = 15/ + 2 —

dt

We test the impulsive component of di^jdt by noting that it produces

an impulsive voltage of (3)65(f), or 185(f), across L\ This voltage, along

with 325(f) across L2, adds to 505(f) Thus the algebraic sum of the

impul-sive voltages around the mesh adds to zero

To summarize, the Laplace transform will correctly predict the creation

of impulsive currents and voltages that arise from switching However, the

.y-domain equivalent circuits must be based on initial conditions at t — 0~,

that is, on the initial conditions that exist prior to the disturbance caused by

the switching The Laplace transform will correctly predict the response to

impulsive driving sources by simply representing these sources in the

,v domain bv their correct transforms

NOTE: Assess your understanding of the impulse function in circuit

analysis by trying Chapter Problems 13.87 and 13.88

Practical Perspective

Practical Perspective

Surge Suppressors

As mentioned at the beginning of this chapter, voltage surges can occur in

a circuit that is operating in the sinusoidal steady state Our purpose is to

show how the Laplace transform is used to determine the creation of a surge

in voltage between the line and neutral conductors of a household circuit

when a load is switched off during sinusoidal steady-state operation

Consider the circuit shown in Fig 13.59, which models a household

cir-cuit with three loads, one of which is switched off at time t = 0 To simplify

the analysis, we assume that the line-to-neutral voltage, \ ()r is

120 / 0 ° V (rms), a standard household voltage, and that when the load

is switched off at t = 0, the value of V^ does not change After the switch

is opened, we can construct the s-domain circuit, as shown in Fig 13.60

Note that because the phase angle of the voltage across the inductive load

is 0°, the initial current through the inductive load is 0 Therefore, only the

inductance in the line has a non-zero initial condition, which is modeled in

the s-domain circuit as a voltage source with the value L// 0 , as seen in

Fig 13.60

Just before the switch is opened at t = 0, each of the loads has a

steady-state sinusoidal voltage with a peak magnitude of 120V2 = 169.7 V

All of the current flowing through the line from the voltage source y g is

divided among the three loads When the switch is opened at t =? 0, all of

the current in the line will flow through the remaining resistive load This is

because the current in the inductive load is 0 at t = 0 and the current in an

inductor cannot change instantaneously Therefore, the voltage drop across

the remaining loads can experience a surge as the line current is directed

through the resistive load For example, if the initial current in the line is

25 A (rms) and the impedance of the resistive load is 12 0 , the voltage

drop across the resistor surges from 169.7 V to (25)( V 2 )(12) = 424.3 V

when the switch is opened If the resistive load cannot handle this amount

of voltage, it needs to be protected with a surge suppressor such as those

shown at the beginning of the chapter

A / = 0

V* ( T ) h\k*a.Voh\\iX a fcjiife

Figure 13.59 • Circuit used to introduce a switching surge voltage Figure 13.60 A Symbolic s-domain circuit

NOTE: Assess your understanding of this Practical Perspective by trying Chapter

Problems 13.92 and 13.93