Electric Circuits, 9th Edition P32 doc

8.54, the characteristic equation for the series RLC circuit is 8.55 The roots of the characteristic equation are Neper frequency—series RLC circuit • Resonant radian frequency—series

286 Natural and Step Responses of RLC Circuits

which we can rearrange as

Characteristic equation—series

RLC circuit •

—r + —— + = 0

Comparing Eq 8.54 with Eq 8.3 reveals that they have the same form

Therefore, to find the solution of Eq 8.54, we follow the same process that led us to the solution of Eq 8.3

From Eq 8.54, the characteristic equation for the series RLC circuit is

(8.55)

The roots of the characteristic equation are

Neper frequency—series RLC circuit •

Resonant radian frequency—series

RLC circuit •

*l,2

*1

R 2L

i —

±

-a

U

±vV

r

-

Oil-1

LC

or

The neper frequency (a) for the series RLC circuit is

R A,

a = — rad/s,

and the expression for the resonant radian frequency is

1

(8.56)

(8.57)

(8.58)

Current natural response forms in series

RLC circuits F •

Note that the equation for neper frequency of the series RLC circuit differs from that of the parallel RLC circuit, but the equations for resonant and

damped radian frequencies are the same

The current response will be overdamped, underdamped, or critically

damped according to whether col < a 2 , WQ > a2, or a>o = a 2 , respectively

Thus the three possible solutions for the current are as follows:

i(t) = B x e~ ai cos o) d t + B 2 e~ at sin oj d t (underdamped), (8.61) i(t) = Drfe'" 1 + L\e~ al (critically damped) (8.62)

r = 0

- W W

R

+ v

c:

Figure 8.15 • A circuit used to illustrate the step

response of a series RLC circuit

When you have obtained the natural current response, you can find the natural voltage response across any circuit element

To verify that the procedure for finding the step response of a series

RLC circuit is the same as that for a parallel RLC circuit, we show that the

differential equation that describes the capacitor voltage in Fig 8.15 has the same form as the differential equation that describes the inductor cur-rent in Fig 8.11 For convenience, we assume that zero energy is stored in the circuit at the instant the switch is closed

Applying Kirchhoff s voltage law to the circuit shown in Fig 8.15 gives

V = Ri + L^- + v c

The current (i) is related to the capacitor voltage (%•) by the expression

, dv c

i = C

from which

d[ = d 2 v c

dt dt 1 ' (8.65)

Substitute Eqs 8.64 and 8.65 into Eq 8.63 and write the resulting

expression as

d 2 v c R dv c v c V

dr L dt LC LC (8.66)

Equation 8.66 has the same form as Eq 8.41; therefore the procedure for

finding v c parallels that for finding i L The three possible solutions for v c

are as follows:

v c = V f + A\e Slt + A' 2 e S2 ' (overdamped), (8.67)

v c = V f + B\e~ at cos to d t + B' 2 e~ at sin (o d t (underdamped), (8.68) 4 Capadtor voltage step response forms in

v c = V f + D[te~ at + D' 2 e~ a1 (critically damped), (8.69) series RLC drcuits

where Vf is the final value of v c Hence, from the circuit shown in Fig 8.15,

the final value of v c is the dc source voltage V

Example 8.11 and 8.12 illustrate the mechanics of finding the natural

and step responses of a series RLC circuit

Example 8.11 Finding the Underdamped Natural Response of a Series RLC Circuit

The 0.1 /xF capacitor in the circuit shown in

Fig 8.16 is charged to 100 V At t = 0 the capacitor

is discharged through a series combination of a

100 mH inductor and a 560 fl resistor

a) Find i(t) for t > 0

b) Find v c (t) for t > 0

/ = 0

100 mH

Figure 8.16 A The circuit for Example 8.11

Solution

a) The first step to finding /'(?) is to calculate the roots of the characteristic equation For the given element values,

2

a =

1

LC

(103)(1Q6) (100)(0.1)

A

2L

560

108,

2(100)

= 2800 rad/s

X 103

288 Natural and Step Responses of RLC Circuits

Next, we compare a^ to a 2 and note that o)l > a2,

because

a2 - 7.84 X 106

= 0.0784 X 108

At this point, we know that the response is

under-damped and that the solution for /(f) is of the form

/(f) = B^'** 1 cos wj + B 2 e~ at sincojt,

where a = 2800 rad/s and a) d = 9600 rad/s The

numerical values of B { and B 2 come from the initial

conditions The inductor current is zero before the

switch has been closed, and hence it is zero

immedi-ately after Therefore

/(0) = 0 = B {

To find B 2 , we evaluate di(() + )/dt From the circuit, we

note that, because /(0) = 0 immediately after the

switch has been closed, there will be no voltage drop

across the resistor Thus the initial voltage on the

capacitor appears across the terminals of the inductor,

which leads to the expression,

or

di(0 + ) V 0 100

dt " L " 100 X 1 0

= 1000 A/s

Because B } = 0,

di

— = 40032<r28()0'(24 cos 9600f - 7sin9600f)

Thus

di(Q + )

dt = 960052,

The solution for /(f) is /(f) = 0.1042<T2800' sin 9600f A, f > 0

b) To find Vc(t), we can use either of the following

rela-tionships:

1 f l

v c = — / idr + 100 or

w o

r di

v c = iR + L—

dt

Whichever expression is used (the second is recom-mended), the result is

v c (t) = (100cos9600f + 29.17sin9600f)e-280UrV, t > 0

Example 8.12 Finding the Underdamped Step Response of a Series RLC Circuit

No energy is stored in the 100 mH inductor or the

0.4 ^iF capacitor when the switch in the circuit

shown in Fig 8.17 is closed Find v c (t) for t > 0

48 V

- \ 0.1 H

f = 0

-*M,

28011 0.4 M F ; +

Figure 8.17 • The circuit for Example 8.12

Solution

The roots of the characteristic equation are

0.2

10f (0.1)(0.4)

= (-1400 +/4800) rad/s,

s = (-1400 - ;4800) rad/s

The roots are complex, so the voltage response is underdamped Thus

v c (t) = 48 + Bie"14()0'cos4800f

+ £^T1400'sin4800f, f > 0

No energy is stored in the circuit initially, so both i;c(0) and dv c (0 + )/dt are zero Then,

v c (0)

dv c (0 + )

dt

= 0 = 48 + S'b

= 0 = 4800B' 2 - 14005;

Solving for B\ and B' 2 yields

B\ = - 4 8 V,

B 2 = - 1 4 V

Therefore, the solution for v c (t) is

v c {t) = (48 - 48<T14()0'cos4800f

- 14e" 'sin 48000 V, ' * 0

^ / A S S E S S M E N T P R O B L E M S

Objective 2—Be able to determine the natural response and the step response of series RLC circuits

8.7 The switch in the circuit shown has been in

position a for a long time At t = 0, it moves to

position b Find (a) /(0+); (b) v c (0 + );

(c) di(Q + )/dt; (d) 5,, s 2 ; and (e) i(t) for t > 0

Answer: (a) 0;

(b) 50 V;

(c) 10,000 A/s;

(d) (-8000 + /6000) rad/s,

(-8000 - /6000) rad/s;

(e) (1.67<r8000f sin 6000/) A for/ > 0

NOTE: Also try Chapter Problems 8.50-8.52

8.8 Find v c {t) for t > 0 for the circuit in

Assessment Problem 8.7

Answer: [ 100 - <T8000f (50 cos 6000?

+ 66.67 sin 6()00/)] V for t > 0

8.5 A Circuit with Two Integrating

Amplifiers

A circuit containing two integrating amplifiers connected in cascade1 is

also a second-order circuit; that is, the output voltage of the second

inte-grator is related to the input voltage of the first by a second-order

differ-ential equation We begin our analysis of a circuit containing two cascaded

amplifiers with the circuit shown in Fig 8.18

Figure 8.18 • Two integrating amplifiers connected in cascade

We assume that the op amps are ideal The task is to derive the

differ-ential equation that establishes the relationship between v ( , and v g We

begin the derivation by summing the currents at the inverting input

termi-nal of the first integrator Because the op amp is ideal

From Eq 8.70,

— + C , - ( 0 - » ol) = 0

dv o]

dt * C , l v r

(8.70)

(8.71)

' In a cascade connection, the output signal of the first amplifier (v <a in Fig 8.18) is the input

signal for the second amplifier

290 Natural and Step Responses of RLC Circuits

N o w we sum the currents away from the inverting input terminal of the second integrating amplifier:

0 — u„i d

or

dv„

dt

Differentiating Eq 8.73 gives

d 2 v„

R-yCt h>V

1 dvoi

(8.72)

(8.73)

(8.74)

We find the differential e q u a t i o n that governs the relationship between va

a n d vg by substituting E q 8.71 into E q 8.74:

dt 2 R [ C l R 2 C 2 *'

E x a m p l e 8.13 illustrates t h e step response of a circuit containing two cas-caded integrating amplifiers

N o energy is stored in the circuit shown in Fig 8.19

when the input voltage v„ j u m p s instantaneously

from 0 to 25 mV

a) Derive the expression for v 0 (t) for 0 < t < /sat

b) H o w long is it before the circuit saturates?

Solution

a) Figure 8.19 indicates that the amplifier scaling

factors are

0,1 /*F

1 1000

= 40,

Now, because vg = 25 m V for t > 0, E q 8.75

b e c o m e s

d 2 v (>

dt 2

To solve for v() , we let

= (40)(2)(25 X 10" 3 ) = 2

t h e n ,

dg(t)

dt

,,v dv ()

g(0 = -^-

2, a n d dg(t) = 2dt

Figure 8.19 • The circuit for Example 8.13

H e n c e

from which

However,

dy = 2 I dx,

g(0) /()

g(t) ~ g(0) = 2t

dvJQ)

m - - ^ i - o

b e c a u s e the e n e r g y s t o r e d in t h e circuit ini-tially is z e r o , a n d the o p a m p s a r e ideal (See

P r o b l e m 8.57.) Then,

^ = 2t, and v a = t 2 + v o (0)

But vo (0) = 0, so the experssion for v a b e c o m e s

v„ = t 2 , 0 < t < ;Sill

b) The second integrating amplifier saturates when

v a reaches 9 V or t = 3 s But it is possible that

the first integrating amplifier saturates before

t = 3 s To explore this possibility, use Eq 8.71 to

find dv„i/dt:

dVoi

dt -40(25) X 10"3 = - 1

Solving for v nl yields

%] = -t

Thus, at t = 3 s, v ()] = —3 V, and, because the power supply voltage on the first integrating amplifier is

±5 V, the circuit reaches saturation when the second amplifier saturates When one of the op amps satu-rates, we no longer can use the linear model to predict the behavior of the circuit

NOTE: Assess your understanding of this material by trying Chapter Problem 8.63

Two Integrating Amplifiers with Feedback Resistors

Figure 8.20 depicts a variation of the circuit shown in Fig 8.18 Recall from

Section 7.7 that the reason the op amp in the integrating amplifier

satu-rates is the feedback capacitor's accumulation of charge Here, a resistor is

placed in parallel with each feedback capacitor (C { and C2) to overcome

this problem We rederive the equation for the output voltage, v tr and

determine the impact of these feedback resistors on the integrating

ampli-fiers from Example 8.13

We begin the derivation of the second-order differentia] equation that

relates v a] to v g by summing the currents at the inverting input node of the

first integrator:

() - Vg () - v „i d ,

(8.76)

We simplify Eq 8.76 to read

dV a i , 1 _ - ¾

For convenience, we let T\ = R\C\ and write Eq 8.77 as

dt T]

•Vo

R n C ]

(8.78)

The next step is to sum the currents at the inverting input terminal of the

second integrator:

(8.79)

Figure 8.20 • Cascaded integrating amplifiers with feedback resistors

We rewrite Eq 8.79 as

dv p v a -v oi

where r2 = R 2 C 2 Differentiating Eq 8.80 yields

d 2 v a J_dv (> _ 1 dv ol

It? + V2~d7 ~ ~W2~^' (8-81)

From Eq 8.78,

dv ol -v ol v g

(8.82)

and from Eq 8.80,

dv.y R b Ci

We use Eqs 8.82 and 8.83 to eliminate dv a Jdt from Eq 8.81 and obtain

the desired relationship:

From Eq 8.84, the characteristic equation is

Ji T 2 J T\T 2

The roots of the characteristic equation are real, namely,

Si = —, (8.86)

r i

- 1

ft = (8.87)

Example 8.14 illustrates the analysis of the step response of two cascaded

integrating amplifiers when the feedback capacitors are shunted with

feedback resistors

Example 8.14 Analyzing Two Cascaded Integrating Amplifiers with Feedback Resistors

The parameters for the circuit shown in Fig 8.20

are R a = 100 kfi, R Y = 500 kO, C x = 0.1 ^ F ,

R h = 25 kH, R 2 = 100 kO, and C2 = 1 /xF The

power supply voltage for each op amp is ±6 V The

signal voltage (v g ) for the cascaded integrating

amplifiers jumps from 0 to 250 mV at t = 0 No

energy is stored in the feedback capacitors at the

instant the signal is applied

a) Find the numerical expression of the differential

equation for v Q

b) Find v () (t) for t > 0

c) Find the numerical expression of the differential

equation for v a]

d) Find v (A (t) for/ > 0

The solution for v» thus takes the form:

Solution

a) From the numerical values of the circuit

parame-ters, we have TJ = R\C] = 0.05 s; r2 = R 2 C 2

= 0.10 s, and v g /R A CiR h C 2 = 1000 V/s2

Substi-tuting these values into Eq 8.84 gives

10f

v a = 5 + A[e~ m + A' 2 e -2i.tr

With v o (0) = 0 and dv o (0)/dt = 0, the numeri-cal values of A\ and A' 2 are A\ — —10 V and

A 2 = 5 V Therefore, the solution for v () is

2QC\

v 0 (t) = (5 - 10e~lu' + 5e"iU0 V, f > 0

The solution assumes that neither op amp saturates We have already noted that the final

value of v a is 5 V, which is less than 6 V; hence the second op amp does not saturate The final value

of v ol is (250 X 10"3)(-500/100), or -1.25 V Therefore, the first op amp does not saturate, and our assumption and solution are correct

c) Substituting the numerical values of the parame-ters into Eq 8.78 generates the desired differen-tial equation:

—£ + 30—^ + 200vo = 1000

dv 0\

dt + 20v ol = - 2 5

b) The roots of the characteristic equation are

S] = - 2 0 r a d / s and s 2 = - l O r a d / s The final

value of v 0 is the input voltage times the gain of

each stage, because the capacitors behave as

open circuits as t —» oo Thus,

Vo(°°) (250 X 1 0- 3) --500) (-100)

100 25 5 V

d) We have already noted the initial and final

val-ues of v 0 \, along with the time constant T\ Thus

we write the solution in accordance with the technique developed in Section 7.4:

>o\ -1.25 + [0 - (-1.25)]e -20/

m

= -1.25 + 1.25<T/l"V, t > 0

NOTE: Assess your understanding of this material by trying Chapter Problem 8.64

294 Natural and Step Responses of RLC Circuits

Practical Perspective

Figure 8.21 A The circuit diagram of the

conven-tional automobile ignition system

An Ignition Circuit

Now let us return to the conventional ignition system introduced at the beginning of the chapter A circuit diagram of the system is shown in Fig 8.21 Consider the circuit characteristics that provide the energy to ignite the fuel-air mixture in the cylinder First, the maximum voltage avail-able at the spark plug, v sp , must be high enough to ignite the fuel Second, the voltage across the capacitor must be limited to prevent arcing across the switch or distributor points Third, the current in the primary winding of the autotransformer must cause sufficient energy to be stored in the system to ignite the fuel-air mixture in the cylinder Remember that the energy stored in the circuit at the instant of switching is proportional to the

primary current squared, that is, a> 0 = | L / 2 ( 0 )

EXAMPLE

a) Find the maximum voltage at the spark plug, assuming the following

val-ues in the circuit of Fig 8.21: V dc = 12 V , R = 4 ft, L = 3 m H ,

C - OAfiF, and a = 100

b) What distance must separate the switch contacts to prevent arcing at the time the voltage at the spark plug is maximum?

Solution

a) We analyze the circuit in Fig 8.21 to find an expression for the spark

plug voltage v sp We limit our analysis to a study of the voltages in the

circuit prior to the firing of the spark plug We assume that the current

in the primary winding at the time of switching has its maximum

possi-ble value V&JR, where R is the total resistance in the primary circuit

We also assume that the ratio of the secondary voltage (v 2 ) to the

pri-mary voltage (v{) is the same as the turns ratio N 2 /Ni We can justify

this assumption as follows With the secondary circuit open, the voltage induced in the secondary winding is

di

dt

and the voltage induced in the primary winding is

(8.88)

v l — L di

It

I t follows from Eqs 8.88 and 8.89 that

« i

M_

L"

(8.89)

(8.90)

I t is reasonable to assume that the permeance is the same for the fluxes

4>u and <f> 2 \ in the iron-core autotransformer; hence Eq 8.90 reduces to

V2

«1

N t N 2 ^ N 2 Nty Ni — a (8.91)

We are now ready to analyze the voltages in the ignition circuit

The values of R, L, and C are such that when the switch is opened, the

primary coil current response is underdamped Using the techniques

I = Y*£ e -at

R cosw d t + I — ] sm o) d t (8.92)

where

R_

2 V

a =

" * " ^ L C- °

(See Problem 8.66(a).) The voltage induced in the primary winding

of the autotransformer is

V\ = L— = di -V Ac _ at

e ut sin (o d t

dt co d RC

(See Problem 8.66(b).) I t follows from Eq 8.91 that

(8.93)

v 2 = -«Kic _, io

The voltage across the capacitor can be derived either by using the

relationship

or by summing the voltages around the mesh containing the primary

winding:

at

In either case, we find

*V = Kfc[l _ e~ a ' cos co d t + Ke~ at sin <a d t], (8.97)

where

cod\RC J

(See Problem 8.66(c).) As can be seen from Fig 8.21, the voltage

across the spark plug is

T/ ttV ^ -at •

= Vr dc 1

-io RC e sin a) d t (8.98)