Electric Circuits, 9th Edition P51 docx

13.40 to see whether it makes sense in terms of the given initial conditions and the known circuit behavior after the switch has been open for a long time.. 13.40 predicts zero initial c

where co = 40,000, a = 32,000, and (3 = 24,000

We can't test the final value of i L with the final-value theorem because

l L has a pair of poles on the imaginary axis; that is, poles at ±/4 X 104

Thus we must first find i L and then check the validity of the expression from known circuit behavior

When we expand Eq 13.36 into a sum of partial fractions, we generate the equation

/ , = s -/40,000 s + /40,000 s + 32,000 - /24,000 K: + Kl + K,

+ s + 32,000 + /24,000 K\

The numerical values of the coefficients K\ and K 2 are

384 X 105(/40,000)

(13.37)

* 1 = 7T

(/80,000)(32,000 + /16,000)(32,000 + /64,000)

= 7.5 X 1 0 ~3/ - 9 0 ° , (13.38)

K, = 384 X 105( -32,000 + /24,000)

(-32,000 - /16,000)(-32,000 + /64,000)(/48,000)

Substituting the numerical values from Eqs 13.38 and 13.39 into

Eq 13.37 and inverse-transforming the resulting expression yields

i L = [15 cos (40,000r - 90°)

+ 25e -32.000? cos(24,000r + 90°)] mA,

= (15 sin 40,000r - 2Se~ nmt sin 24,000/)tt(0 mA (13.40)

We now test Eq 13.40 to see whether it makes sense in terms of the given initial conditions and the known circuit behavior after the switch has

been open for a long time For t = 0, Eq 13.40 predicts zero initial current,

which agrees with the initial energy of zero in the circuit Equation 13.40 also predicts a steady-state current of

i La = 15sin40,000rmA, which can be verified by the phasor method (Chapter 9)

(13.41)

Figure 13.15 • A multiple-mesh RL circuit

The Step Response of a Multiple Mesh Circuit

Until now, we avoided circuits that required two or more node-voltage or

48 ft mesh-current equations, because the techniques for solving simultaneous

differential equations are beyond the scope of this text However, using Laplace techniques, we can solve a problem like the one posed by the multiple-mesh circuit in Fig 13.15

Here we want to find the branch currents /j and i 2 that arise when the

336 V dc voltage source is applied suddenly to the circuit The initial

energy stored in the circuit is zero Figure 13.16 shows the v-domain

equiv-alent circuit of Fig 13.15 The two mesh-current equations are

336

= (42 + 8.45)/] - 42/2,

0 = - 4 2 / j + (90 + 10s)I 2 Using Cramer's method to solve for l x and I 2 , we obtain

A = 42 + 8.4s - 42

- 42 90 + 10s

(13.42)

(13.43)

48 a

Figure 13.16 • The s-domain equivalent circuit for the

circuit shown in Fig 13.15

= 84(52 + 14.9 + 24)

= 84( s + 2)(5 + 12), (13.44)

A', 336/5 - 42

0 90 + 105

3360(5 + 9)

(13.45)

No = 42 + 8.45 336/5

- 42 0

14,112

(13.46)

Based on Eqs 13.44-13.46,

/> = /Vi _ 40(5 + 9)

A " s(s + 2)(5 + 12)' (13.47)

A 5(5 + 2)(5 + 12)'

Expanding I] and /2 into a sum of partial fractions gives

15

5 5 + 2 5 + 1 2 '

(13.48)

(13.49)

8.4 1.4 +

We obtain the expressions tor i { and i 2 by inverse-transforming Eqs 13.49 and 13.50, respectively:

I, = (15 - Ue~ 2t - e~ ]2t )u(t) A, (13.51)

/2 = (7 - 8.4e"z' + \Ae~ l *)u(t) A (13.52)

Next we test the solutions to see whether they make sense in terms of the circuit Because no energy is stored in the circuit at the instant the switch is closed, both /'i(0~) and /2(0-) must be zero The solutions agree with these initial values After the switch has been closed for a long time, the two

induc-tors appear as short circuits Therefore, the final values of i { and i 2 are

, x 336(90)

/ ( 0 0 ) = *—'- = 15 A ,

15(42) / 2 ( o o ) = ^ = 7 A (13.54)

One final test involves the numerical values of the exponents and calcu-lating the voltage drop across the 42 11 resistor by three different methods

From the circuit, the voltage across the 42 Q resistor (positive at the top) is

dii d'h

v = 42(/, - h) = 336 - 8.4—- = 48/, + 1 0 -2

/ ' A S S E S S M E N T P R O B L E M

You should verify that regardless of which form of Eq 13.55 is used, the voltage is

v = (336 - 235.2e~2r - 100.80e"12r)«(0 V

We are thus confident that the solutions for i x and i 2 are correct

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

13.5 The dc current and voltage sources are

applied simultaneously to the circuit shown

No energy is stored in the circuit at the instant

of application

a) Derive the 5-domain expressions for V\

and V 2

b) For t > 0, derive the time-domain

expres-sions for V\ and v 2

c) Calculate ^(0+) and v 2 (0 + )

d) Compute the steady-state values of V\

and Vi

Answer: (a) V l = [5(s + 3)]/[s(s + 0.5)(5 + 2)],

V 2 = [2.5(s 2 + 6)]/[s(s + 0.5)(5 + 2)]; (b) v, = (15 50 -0.5/ + f<r2>(0 v,

v2 = 05 - l-fe-°-51 + ?e*)«(0 V;

(c) u,(0+) = 0, v 2 (0 + ) = 2.5 V;

(d) vi = v 2 = 15 V

1 H

_ / - W Y - \ _

+

v\ 3 n f

v-15a

15VI

NOTE: Also try Chapter Problems 13.22 and 13.29

The Use of Thevenin's Equivalent

In this section we show how to use Thevenin's equivalent in the s domain

Figure 13.17 shows the circuit to be analyzed The problem is to find the

capacitor current that results from closing the switch.The energy stored in

the circuit prior to closing is zero

To find i c, we first construct the 5-domain equivalent circuit and then

find the Thevenin equivalent of this circuit with respect to the terminals of

the capacitor Figure 13.18 shows the 5-domain circuit

The Thevenin voltage is the open-circuit voltage across terminals a, b

Under open-circuit conditions, there is no voltage across the 60 Q

resis-tor Hence

V T I , =

{480/s)(0.002s)

20 + 0.002*

480

5 + 104' (13.56) The Thevenin impedance seen from terminals a and b equals the 60 ft

resistor in series with the parallel combination of the 20 fl resistor and the

2 mH inductor Thus

ZT h = 60 + 0.0025(20) 80(5 + 7500)

20 + 0.0025 .v + KV (13.57)

Using the Thevenin equivalent, we reduce the circuit shown in Fig 13.18

to the one shown in Fig 13.19 It indicates that the capacitor current I c

equals the Thevenin voltage divided by the total series impedance Thus,

4) [80(5 + 7500)/(5 + 104)] + [(2 X 105)/5]

We simplify Eq 13.58 to

52 + 10,0005 + 25 X 106 (s + 5000)2

A partial fraction expansion of Eq 13.59 generates

Ir = -30,000 +

(5 + 5000)' the inverse transform of which is

5 + 5000'

i c = (-30,000f«T50,M)' + 6e- 5m ')u(t) A

(13.58)

(13.59)

(13.60)

(13.61)

We now test Eq 13.61 to see whether it makes sense in terms of

known circuit behavior From Eq 13.61,

This result agrees with the initial current in the capacitor, as calculated from

the circuit in Fig 13.17 The initial inductor current is zero and the initial

capacitor voltage is zero, so the initial capacitor current is 480/80, or 6 A

The final value of the current is zero, which also agrees with Eq 13.61 Note

also from this equation that the current reverses sign when t exceeds

6/30.000, or 200 /AS The fact that ic reverses sign makes sense because,

when the switch first closes, the capacitor begins to charge Eventually this

charge is reduced to zero because the inductor is a short circuit at t = co

The sign reversal of i c reflects the charging and discharging of the capacitor

Let's assume that the voltage drop across the capacitor v c is also of

inter-est Once we know i c we find v c by integration in the time domain; that is,

v c = 2 x 10 5 / ( 6 - 30, -5(K«h

20 fi 60 a a

'VW-Figure 13.17 A A circuit to be analyzed using

Thevenin's equivalent in the s domain

-^vw

•-N 4| ^ 10.002 s V c / ct -±z2x 105

Figure 13.18 • The 5-domain model of the circuit

shown in Fig 13.17

v c i c ; ^ 2 x i o5

Figure 13.19 • A simplified version of the circuit

shown in Fig 13.18, using a Thevenin equivalent

Although the integration called for in Eq 13.63 is not difficult, we may

avoid it altogether by first finding the s-domain expression for V c and then

finding v c by an inverse transform Thus

V c = -pic

sC

2 X 105 6.v

s (s + 5000)2

from which

12 X 105

(s + 5000)^

vc = 12 x io5rtr50f)0

'u(0-(13.64)

(13.65)

You should verify that Eq 13.65 is consistent with Eq 13.63 and that it

also supports the observations made with regard to the behavior of i c (see Problem 13.33)

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

13.6 The initial charge on the capacitor in the circuit

shown is zero

a) Find the 5-domain Thevenin equivalent

cir-cuit with respect to terminals a and b

b) Find the s-domain expression for the current

that the circuit delivers to a load consisting of

a 1 H inductor in series with a 2 ft resistor

Answer: (a) V Th = V &h = [20(5 + 2A))/[s(s + 2)),

ZT t l = 5(5 + 2.8)/(s + 2);

NOTE: Also try Chapter Problem 13.34

(b) Ja b = [20(5 + 2A)]/[s(s + 3)(5 + 6)]

0.5 F

A Circuit with Mutual Inductance

The next example illustrates how to use the Laplace transform to analyze the transient response of a circuit that contains mutual inductance Figure 13.20 shows the circuit The make-before-break switch has been in

position a for a long time At t = 0, the switch moves instantaneously to position b The problem is to derive the time-domain expression for i 2

We begin by redrawing the circuit in Fig 13.20, with the switch in position b and the magnetically coupled coils replaced with a T-equivalent circuit.1 Figure 13.21 shows the new circuit

9 f i

©

3 a

-'VW-/ = 0

2 H ^vw-2D,

60 V

Figure 13.20 • A circuit containing magnetically coupled coils

h ion

Sec Appendix C

We now transform this circuit to the s domain In so doing, we note that

f l ( °" ) = f = 5A '

WO = 0

(13.66)

(13.67)

Because we plan to use mesh analysis in the s domain, we use the series

equivalent circuit for an inductor carrying an initial current Figure 13.22

shows the s-domain circuit Note that there is only one independent

volt-age source This source appears in the vertical leg of the tee to account for

the initial value of the current in the 2 H inductor of /,(0-) + /2(0~), or 5 A

The branch carrying /, has no voltage source because Lt - M = 0

The two i'-domain mesh equations that describe the circuit in

Fig 13.22 are

(3 + 2s)I } + 2sl 2 = 10

2sT t + (12 + 85)/2 = 10

Solving for /2 yields

/ 2 =

2.5

(s + l ) ( j + 3)

Expanding Eq 13.70 into a sum of partial fractions generates

1.25 1.25

/ 7 =

s + 1 s + 3

Then,

i 2 = (1.25e"' - 1.25e"3/)w(0 A

(13.68) (13.69)

(13.70)

(13.71)

(13.72)

Equation 13.72 reveals that i 2 increases from zero to a peak value of

481.13 mA in 549.31 ms after the switch is moved to position b Thereafter,

/2 decreases exponentially toward zero Figure 13.23 shows a plot of i 2

ver-sus t This response makes sense in terms of the known physical behavior

of the magnetically coupled coils A current can exist in the L2 inductor

only if there is a time-varying current in the L, inductor As i\ decreases

from its initial value of 5 A, /2 increases from zero and then approaches

zero as i, approaches zero

(L, - M) (L 2 - M) 3H OH 6 H 2ti

( M H 2 H 10 a

Figure 13.21 A The circuit shown in Fig 13.20, with

the magnetically coupled coils replaced by a T-equivalent circuit

Figure 13.22 A The s-domain equivalent circuit for the

circuit shown in Fig 13.21

i 2 (mA) 481.13

Figure 13.23 A The plot of i 2 versus t for the circuit

shown in Fig 13.20

/ 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

13.7 a) Verify from Eq 13.72 that /2 reaches a peak

value of 481.13 mA at t = 549.31 ms

b) Find i h for t > 0, for the circuit shown in

Fig 13.20

c) Compute di x fdt when i 2 is at its peak value

d) Express i 2 as a function of dii/dt when i 2 is

at its peak value

e) Use the results obtained in (c) and (d) to

calculate the peak value of i 2

Answer: (a) di 2 /dt = 0 when t = | l n 3 (s);

(b) /, = 2.5(6'"' + e "3> ( 0 A;

(c) -2.89 A/s;

(d) /2 = -{MdiJdt)/\2\

(e) 481.13 mA

NOTE: Also try Chapter Problems 13.39 and 13.40

+ y

Vj < Ri

Figure 13.24 • A circuit showing the use of

super-position in s-domain analysis

Figure 13.25 • The s-domain equivalent for the circuit

of Fig 13.24

Figure 13.26 • The circuit shown in Fig 13.25 with V^

acting alone

The Use of Superposition

Because we are analyzing linear lumped-parameter circuits, we can use superposition to divide the response into components that can be identi-fied with particular sources and initial conditions Distinguishing these components is critical to being able to use the transfer function, which we introduce in the next section

Figure 13.24 shows our illustrative circuit We assume that at the instant when the two sources are applied to the circuit, the inductor is car-rying an initial current of p amperes and that the capacitor is carcar-rying an initial voltage of y volts The desired response of the circuit is the voltage

across the resistor R 2 , labeled v 2

Figure 13.25 shows the s-domain equivalent circuit We opted for the

parallel equivalents for L and C because we anticipated solving for V 2

using the node-voltage method

To find V 2 by superposition, we calculate the component of V 2 result-ing from each source actresult-ing alone, and then we sum the components We

begin with V g acting alone Opening each of the three current sources deactivates them Figure 13.26 shows the resulting circuit We added the

node voltage V{ to aid the analysis The primes on V t and V 2 indicate that

they are the components of Vj and V 2 attributable to V g acting alone The two equations that describe the circuit in Fig 13.26 are

J_

Ri

l

sL

V*

+ — + sC \V\ - sCV' 2 = -P-,

fli

\-jf + sC

(13.73)

(13.74)

For convenience, we introduce the notation

K 22 = 4 - + ^

K 2

(13.77)

Substituting Eqs 13.75-13.77 into Eqs 13.73 and 13.74 gives

Y n V\ + Y n V 2 = V g /R h (13.78)

Y„V\ + YnV'r, = 0 12" 1 (13.79)

Solving Eqs 13.78 and 13.79 for V' 2 gives

V'o

^11¾ _ *12

(13.80)

With the current source I g acting alone, the circuit shown in Fig 13.25

reduces to the one shown in Fig 13.27 Here, V'{ and V 2 are the

compo-nents of V x and V 2 resulting from L If we use the notation introduced in

Eqs 13.75-13.77, t h e two node-voltage equations that describe t h e circuit

in Fig 13.27 a r e

and

1/sC

-1(-sLAV'[ V'^R,

Y l2 Vl + Y 22 V'i = I r

Solving Eqs 13.81 and 13.82 for V'i yields

V'i = ^ - j - /

M l ' 2 2 — ' 1 2

Figure 13.27 • The circuit shown in Fig 13.25, with

(13.82) j g acting alone

(13.83)

To find the component of V 2 resulting from the initial energy stored in

the inductor {V%), we must solve the circuit shown in Fig 13.28, where Fi9U"-e 13.28 • The circuit shown in Fig 13.25, with

the energized inductor acting alone

YnV'{' + Y l2 V 2 " = -p/s, (13.84)

Y 12 Vf + Y 22 V'{ = 0

Til us

V'i' = Yjs

Ml*22 -yl

V-* 12

(13.85)

(13.86)

From the circuit shown in Fig 13.29, we find the component of

V 2 iy'i') resulting from the initial energy stored in the capacitor The

node-voltage equations describing this circuit are

YuVT + Y l2 V? = yC,

K „ V r + Yy>V7 = -yC 22^ 2 Solving for Vf yields

Y\\Y 2 2 Yyi

The expression for V 2 is

V 2 =V' 2 + V'i + V'i' + V'i"

(13.87)

(13.88)

(13.89)

Figure 13.29 A The circuit shown in Fig 13.25, with

the energized capacitor acting alone

-(Yn/Ri) Y\\Y 22 - Y\\ *i + — — 7 ?

Y\]Y 22 - Yyi

+ Y]\Y Y 22 ]2 Y\ /s 2 p ^ T~ -C(Y U + Y n )

y-Y\\Y 22 — Y i2

(13.90)

We can find V 2 without using superposition by solving the two

node-voltage equations that describe the circuit shown in Fig 13.25 Thus

(13.91)

Y X2 V X + Y 22 V 2 = I - yC (13.92)

You should verify in Problem 13.43 that the solution of Eqs 13.91 and

13.92 for V 2 gives the same result as Eq 13.90

• 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

13.8 The energy stored in the circuit shown is zero

at the instant the two sources are turned on

a) Find the component of v for t > 0 owing to

the voltage source

b) Find the component of v for t > 0 owing to

the current source

c) Find the expression for v when t > 0

NOTE: Also try Chapter Problem 13.42

Answer: (a) [(100/3)e"2 ' - (100/3)<T 8r jw(0 V ;

(b) [(50/3)<T2' - (50/3)e-*')u(t) V;

(c) [50<T2' - 50e"8']«(0 V

2 f l

20u(t)f +

1.25IH » : 5 0 m F ( t ^ 5u{t)

13.4 The Transfer Function

The transfer function is defined as the s-domain ratio of the Laplace

trans-form of the output (response) to the Laplace transtrans-form of the input (source) In computing the transfer function, we restrict our attention to circuits where all initial conditions are zero If a circuit has multiple inde-pendent sources, we can find the transfer function for each source and use superposition to find the response to all sources

The transfer function is

Definition of a transfer function • H(s) = Y(s)

sL

\/sC

Figure 13.30 • A series RLC circuit

where Y(s) is the Laplace transform of the output signal, and X(s) is the

Laplace transform of the input signal Note that the transfer function depends on what is defined as the output signal Consider, for example, the series circuit shown in Fig 13.30 If the current is defined as the response signal of the circuit,

V~R + sL+ 1/sC ~ S2 LC + RCs + 1 (13.94)

In deriving Eq 13.94, we recognized that /corresponds to the output Y(s) and V g corresponds to the input X(s)

If the voltage across the capacitor is defined as the output signal of the

circuit shown in Fig 13.30, the transfer function is

R + sL + 1/sC s 2 LC + RCs + 1 (13.95)

Thus, because circuits may have multiple sources and because the definition

of the output signal of interest can vary, a single circuit can generate many

transfer functions Remember that when multiple sources are involved, no

single transfer function can represent the total output—transfer functions

associated with each source must be combined using superposition to yield

the total response Example 13.1 illustrates the computation of a transfer

function for known numerical values of R, L, and C

Example 13.1 Deriving the Transfer Function of a Circuit

The voltage source v„ drives the circuit shown in

Fig 13.31 The response signal is the voltage across

the capacitor, v (>

a) Calculate the numerical expression for the

trans-fer function

b) Calculate the numerical values for the poles and

zeros of the transfer function

1000 a

AM,

250fi

50 mH

+

1 ju.F v t>

Figure 13.31 A The circuit for Example 13.1

Solution

a) The first step in finding the transfer function is to

construct the 5-domain equivalent circuit, as

shown in Fig 13.32 By definition, the transfer

function is the ratio of V 0 /V s , which can be

com-puted from a single node-voltage equation

Summing the currents away from the upper

node generates

V - V

<> ?

1000 + 250 + 0.055 10V' V„s 6 = 0

ooo n

AM-—

s

Figure 13.32 • The s-domain equivalent circuit for the circuit shown in Fig 13.31

Solving for V () yields

1000(5 + 5()00)¾

Vo =

52 + 60005 + 25 X 106' Hence the transfer function is

s

1000(5 + 5000)

" 52 + 60005 + 25 X 106 "

b) The poles of H{s) are the roots of the

denomina-tor polynomial Therefore

-p ] = -3000 - y'4000, -p 2 = -3000 + /4000

The zeros of H(s) are the roots of the numera-tor polynomial; thus H(s) has a zero at

-Zi = -5000