Electric Circuits, 9th Edition P31 potx

Example 8.5 illustrates the approach for finding the critically damped response of a parallel RLC circuit.. Finding the Critically Damped Natural Response of a Parallel RLC Circuit a Fo

Therefore, the response is underdamped Now,

w (i = V(4 ~ a 2 = V l O6 - 4 X 104 = 100V96

= 979.80 rad/s,

Sl = -a + joo d = - 2 0 0 + /979.80 rad/s,

s 2 = - a - j(o ti = -200 - /979.80 rad/s

For the underdamped case, we do not ordinarily

solve for S] and s 2 because we do not use them

explicitly However, this example emphasizes

why S] and s 2 are known as complex frequencies

b) Because v is the voltage across the terminals of a

capacitor, we have

v(0) = v(0 + ) = V 0 = 0

Because v(0 + ) = 0, the current in the resistive

branch is zero at t = 0+ Hence the current in

the capacitor at t = 0+ is the negative of the

inductor current:

*c(0+) = - ( - 1 2 2 5 ) = 12.25 mA

Therefore the initial value of the derivative is

dv(0 + ) (12.25)(10""3)

, J = f- = 98,000 V / s

c) From Eqs 8.30 and 8.31,5, = 0 and

98,000

co d

Substituting the numerical values of a, co d , B\ s

and B 2 into the expression for v(t) gives v(t) = 100<rZUUf sin 979.80? V, t > 0

d) Figure 8.9 shows the plot of v(t) versus t for the

first 11 ms after the stored energy is released It clearly indicates the damped oscillatory nature

of the underdamped response The voltage v(t)

approaches its final value, alternating between values that are greater than and less than the final value Furthermore, these swings about the final value decrease exponentially with time

v(V)

t(ms)

Figure 8.9 A The voltage response for Example 8.4

Characteristics of the Underdamped Response

The underdamped response has several important characteristics First, as the dissipative losses in the circuit decrease, the persistence of the oscilla-tions increases, and the frequency of the oscillaoscilla-tions approaches w0 In

other words, as R —> oo, the dissipation in the circuit in Fig 8.8 approaches zero because /; = v 2 /R As R —> oo, a —* 0, which tells us that m d —* w()

When a - 0, the maximum amplitude of the voltage remains constant;

thus the oscillation at a>() is sustained In Example 8.4, if R were increased

to infinity, the solution for v(t) would become

v(t) = 98sinl000rV, t > 0

Thus, in this case the oscillation is sustained, the maximum amplitude of the voltage is 98 V, and the frequency of oscillation is 1000 rad/s

We may now describe qualitatively the difference between an under-damped and an overunder-damped response In an underunder-damped system, the response oscillates, or "bounces," about its final value This oscillation is

also referred to as ringing In an overdamped system, the response

approaches its final value without ringing or in what is sometimes described as a "sluggish" manner When specifying the desired response of

a second order system, you may want to reach the final value in the short-est time possible, and you may not be concerned with small oscillations about that final value If so, you would design the system components to achieve an underdamped response On the other hand, you may be con-cerned that the response not exceed its final value, perhaps to ensure that components are not damaged In such a case, you would design the system components to achieve an overdamped response, and you would have to accept a relatively slow rise to the final value

8.2 The Forms of the Natural Response of a Parallel RLC Circuit 2 7 7

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

Objective 1—Be able to determine the natural and the step response of parallel RLC circuits

8.4 A 10 mH inductor, a 1 /xF capacitor, and a

vari-able resistor are connected in parallel in the

circuit shown The resistor is adjusted so that

the roots of the characteristic equation are

-8000 ± /6000 rad/s The initial voltage on the

capacitor is 10 V, and the initial current in the

inductor is 80 mA Find

a) R;

b) dv(0 + )/dt;

c) B x and B 2 in the solution for v; and

d) idt)

NOTE: Also try Chapter Problems 8.7 and 8.19

Answer: (a) 62.5 ft;

(b) -240,000 V/s;

(c) B x = 10 V, B 2 = - 8 0 / 3 V;

(d)i L (t) = 10e-80()0f[8 cos 6000?

+ (82/3) sin 6000f] mA when t

The Critically Damped Voltage Response

The second-order circuit in Fig 8.8 is critically damped when coj, = a 2 , or

tu() = a When a circuit is critically damped, the response is on the verge of

oscillating In addition, the two roots of the characteristic equation are

real and equal; that is

1

Si = $ 2

When this occurs, the solution for the voltage no longer takes the form

of Eq 8.18 This equation breaks down because if s% = s^ = - a , it

pre-dicts that

{A x + A 2 )e~ at = A Q e (8.33)

where A {) is an arbitrary constant Equation 8.33 cannot satisfy two

inde-pendent initial conditions (VQ, /0) with only one arbitrary constant, A {)

Recall that the circuit parameters R and C fix a

We can trace this dilemma back to the assumption that the solution

takes the form of Eq 8.18 When the roots of the characteristic equation

are equal, the solution for the differential equation takes a different

form, namely

v(t) = D { te~ a ' + D 2 e (8.34)

Thus in the case of a repeated root, the solution involves a simple

expo-nential term plus the product of a linear and an expoexpo-nential term The

jus-tification of Eq 8.34 is left for an introductory course in differential

equations Finding the solution involves obtaining D x and D 2 by following

Ihe same pattern set in the overdamped and underdamped cases: We use

the initial values of the voltage and the derivative of the voltage with

respect to time to write two equations containing D\ and/or D

< Voltage natural response—critically damped parallel RLC circuit

From Eq 8.34, the two simultaneous equations needed to determine

D x and D 2 are

,,(0+) = V () = D2,

dv(Q + ) /c(0+)

dt c = Di - aDj

(8.35)

(8.36)

Example 8.5

As we can see, in the case of a critically damped response, both the

equation for v(t) and the simultaneous equations for the constants D\ and

D 2 differ from those for over- and underdamped responses, but the general approach is the same You will rarely encounter critically damped systems

in practice, largely because co 0 must equal a exactly Both of these quanti-ties depend on circuit parameters, and in a real circuit it is very difficult to choose component values that satisfy an exact equality relationship Example 8.5 illustrates the approach for finding the critically damped

response of a parallel RLC circuit

Finding the Critically Damped Natural Response of a Parallel RLC Circuit

a) For the circuit in Example 8.4 (Fig 8.8), find the

value of R that results in a critically damped

volt-age response

b) Calculate v(t) for t > 0

c) Plot v(t) versus t for 0 < t =s 7 ms

Solution

a) From Example 8.4, we know that o>o = 106

Therefore for critical damping

or

R

a = 1()-1 =

106

1 2i?C

= 4000 a

(2000)(0.125) b) From the solution of Example 8.4, we know that

v(0 + ) = 0 and dv{0 + )/dt = 98,000 V/s From

Eqs 8.35 and 8.36, D2 = 0 and D x = 98,000 V/s

Substituting these values for a, D l% and D 2 into

Eq 8.34 gives

v(t) = 98,000*e_1000f V, t s> 0

c) Figure 8.10 shows a plot of v(t) versus t in the

interval 0 < r ^ 7 ms

1 2 3 4 5 6 7 Figure 8.10 • The voltage response for Example 8.5

L t (ms)

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

Objective 1—Be able to determine the natural and the step response of parallel RLC circuits

8.5 The resistor in the circuit in Assessment

Problem 8.4 is adjusted for critical damping

The inductance and capacitance values are

0.4 H and 10 fiF, respectively The initial energy

stored in the circuit is 25 mJ and is distributed

equally between the inductor and capacitor

Find (a) R\ (b) V 0 ; (c) /(); (d) D x and D 2 in the

solution for v; and (e) i R , t S 0+

Answer: (a) 100 O;

(b)50V;

(c) 250 mA;

(d) -50,000 V/s, 50 V;

(e) i R (t) = (-500te~ 5m + 0.50e-500') A,

t £= 0+

NOTE: Also try Chapter Problems 8.9 and 8.20

8.2 The Forms of the Natural Response of a Parallel RLC Circuit 2 7 9

A Summary of the Results

We conclude our discussion of the parallel RLC circuit's natural response

with a brief summary of the results The first step in finding the natural

response is to calculate the roots of the characteristic equation You then

know immediately whether the response is overdamped, underdamped, or

critically damped

If the roots are real and distinct (c_2, < a2), the response is

over-damped and the voltage is

v(t) = A { e Sit + A 2 e s *,

where

S\ = —a + voT

s 2 = —a — Va 2

1

2RC

2 _ 1

"° "

LC w {2 ,,

- <4h

The values of A { and A 2 are determined by solving the following

simulta-neous equations:

v(0 + ) = Ai + A 2 , dv{0 + ) /C(0+)

If the roots are complex OJ2} > a" the response is underdamped and

the voltage is

v(t) = B { e~ n ' cos o>/ + B 2 e~ a! sin ctjt,

where

to (l = vo){) — or

The values of B\ and B 2 are found by solving the following simultaneous

equations:

v(Q + ) - % = B h

dv(0 + ) ic(0 + ) _ , _ _ = _ _ aB] + ^

If the roots of the characteristic equation are real and equal (OJQ = a"),

the voltage response is

v(t) = D^e-o" + D 2 e~ a \ where a is as in the other solution forms To determine values for the

con-stants D) and D 2 , solve the following simultaneous equations:

v(0 + ) = V 0 = D 2 , dv(0 + ) i c (0 + ) _

- _ - _ — A - aD,

RLC Circuit

• Finding the step response of a parallel RLC circuit involves finding the

voltage across the parallel branches or the current in the individual

branches as a result of the sudden application of a dc current source

There may or may not be energy stored in the circuit when the current

source is applied The task is represented by the circuit shown in Fig 8.11

To develop a general approach to finding the step response of a

second-order circuit, we focus on finding the current in the inductive branch (//)

This current is of particular interest because it does not approach zero as

t increases Rather, after the switch has been open for a long time, the

inductor current equals the dc source current I Because we want to focus

on the technique for finding the step response, we assume that the initial

energy stored in the circuit is zero This assumption simplifies the

calcula-tions and doesn't alter the basic process involved In Example 8.10

we will see how the presence of initially stored energy enters into the

general procedure

To find the inductor current i L , we must solve a second-order

differ-ential equation equated to the forcing function /, which we derive as

fol-lows From Kirchhoffs current law, we have

k + hi + k = U

or

Because

v _ dv

dir

v = L—-, (8.38)

we get

dt '

dv T d 2 i L

Substituting Eqs 8.38 and 8.39 into Eq 8.37 gives

L dir afit

R dt dt For convenience, we divide through by LC and rearrange terms:

— £ + + —^ = (8.41)

Comparing Eq 8.41 with Eq 8.3 reveals that the presence of a nonzero

term on the right-hand side of the equation alters the task Before

show-ing how to solve Eq 8.41 directly, we obtain the solution indirectly

When we know the solution of Eq 8.41, explaining the direct approach

will be easier

The Indirect Approach

We can solve for i L indirectly by first finding the voltage v We do this with

the techniques introduced in Section 8.2, because the differential equation

that v must satisfy is identical to Eq 8.3 To see this, we simply return to

Eq 8.37 and express i L as a function of v; thus

I f , v „dv

Differentiating Eq 8.42 once with respect to t reduces the right-hand side

to zero because I is a constant Thus

— + — — + C—r = 0,

or

^ + 7 ^ + ^ = 0 - (8'43)

As discussed in Section 8.2, the solution for v depends on the roots of the

characteristic equation Thus the three possible solutions are

v = B x e~ a ' cos o) d t + B 2 e~ al sin a ) / , (8.45)

A word of caution: Because there is a source in the circuit for t > 0, you

must take into account the value of the source current at t = 0+ when you

evaluate the coefficients in Eqs 8.44-8.46

To find the three possible solutions for /L, we substitute Eqs 8.44-8.46

into Eq 8.37 You should be able to verify, when this has been done, that

the three solutions for i L will be

//, = I + B\e~ at cos o) d t + B' 2 e~ at sin wj, (8.48)

where A\, A 2 , B[, B 2 , D\, and D' 2 , are arbitrary constants

In each case, the primed constants can be found indirectly in terms of

the arbitrary constants associated with the voltage solution However, this

approach is cumbersome

The Direct Approach

It is much easier to find the primed constants directly in terms of the

ini-tial values of the response function For the circuit being discussed, we

would find the primed constants from /L(0) and dii(0)/dt

The solution for a second-order differential equation with a constant

forcing function equals the forced response plus a response function

identical in form to the natural response Thus we can always write the solution for the step response in the form

_ [function of the same form!

' \ as the natural response J '

or

v = V f + {function of the same form 1

as the natural response J '

(8.50)

(8.51)

where If and Vf represent the final value of the response function The final value may be zero, as was, for example, the case with the voltage v in

the circuit in Fig 8.8

Examples 8.6-8.10 illustrate the technique of finding the step

response of a parallel RLC circuit using the direct approach

Example 8.6 Finding the Overdamped Step Response of a Parallel RLC Circuit

The initial energy stored in the circuit in Fig 8.12 is

zero At t = 0, a dc current source of 24 mA is

applied to the circuit The value of the resistor is

400 ft

a) What is the initial value of i L l

b) What is the initial value of dijdtl

c) What are the roots of the characteristic equation?

d) What is the numerical expression for //,(/1) when

t > 0?

'(p'X

Figure 8.12 • The circuit for Example 8.6

Solution

a) No energy is stored in the circuit prior to the

application of the dc current source, so the initial

current in the inductor is zero The inductor

pro-hibits an instantaneous change in inductor

cur-rent; therefore i L (0) = 0 immediately after the

switch has been opened

b) The initial voltage on the capacitor is zero

before the switch has been opened; therefore it

will be zero immediately after Now, because

v = Ldifjdt,

di

dt - ( 0

+ ) = 0

c) From the circuit elements, we obtain

(-0() =

a =

LC (25)(25)

2RC (2)(400)(25)

16 x 10*,

= 5 x 104rad/s,

or

a 2 = 25 X 108

Because a»o < a 2 , the roots of the characteristic

equation are real and distinct Thus

Sl = - 5 X 104 + 3 X 104 = -20,000 rad/s,

s 2 = - 5 X 104 - 3 X 104 = -80,000 rad/s d) Because the roots of the characteristic equation are real and distinct, the inductor current response

will be overdamped Thus i L (t) takes the form of

Eq 8.47, namely,

i L = I f + A[e Sit + A! 2 e s -'

• Inductor current in overdamped parallel

RLC circuit step response

Hence, from this solution, the two simultaneous

equations that determine A\ and A^are

*t(0) = If + A\ + A' 2 = 0,

di L

di (0) = s x A\ + s 2 A ,2 = 0

Solving for A\ and A' 2 gives

A[ = - 3 2 mA and A' 2 = 8 mA

The numerical solution for i L (t) is

i L {t) = (24 - 32e-20-000' + 8e-" )mk ) mA, t > 0

8.3 The Step Response of a Parallel RLC Circuit 2 8 3

Finding the Underdamped Step Response of a Parallel RLC Circuit

The resistor in the circuit in Example 8.6 (Fig 8.12)

is increased to 625 ft Find i L (t) for t 2s 0

Solution

Because L and C remain fixed, col has the same

value as in Example 8.6; that is, col = 16 X 108

Increasing R to 625 ft decreases a to

3.2 X 104rad/s With col > oc\ the roots of the

characteristic equation are complex Hence

s x = - 3 2 X 104 + /2.4 X 104rad/s,

s 2 = - 3 2 X 104 - /2.4 X 104 rad/s

The current response is now underdamped and

given by Eq 8.48:

kit) = I/ + B[e~°" cos co d t + B' 2 e~ at sin to d t

• Inductor current in underdamped parallel

RLC circuit step response

Here, a is 32,000 rad/s, co d is 24,000 rad/s, and

If is 24 mA

As in Example 8.6, B[ and B' 2 are determined from the initial conditions Thus the two simultane-ous equations are

I'ZXO) = I f + B[ = 0,

^(0)=co (l B 2 -aB[ = Q

Then,

and

B\ = - 2 4 mA

54 = - 3 2 mA

The numerical solution for i L (t) is

i L (t) = (24 - 24e-3Z000'cos24,000r

- 32e"32'000' sin 24,000?) mA, t > 0

Example 8.8 Finding the Critically Damped Step Response of a Parallel RLC Circuit

The resistor in the circuit in Example 8.6 (Fig 8.12)

is set at 500 ft Find i L for t > 0

Solution

We know that col remains at 16 X 108 With R set at

500 ft, a becomes 4 X 104s~\ which corresponds

to critical damping Therefore the solution for i L (t)

takes the form of Eq 8.49:

kit) = I f + D\te~ at + D' 2 e~°"

• Inductor current in critically damped parallel

RLC circuit step response

Again, D\ and D 2 are computed from initial conditions, or

i L (0) = If + D' 2 = 0,

~ ( 0 ) = D\ - aD'o = 0

at

Thus

D\ = -960,000 mA/s and D' 2 = - 2 4 mA

The numerical expression for i L {i) is

i L (t) = (24 - 960,000re-4(X()0{)' - 24e_40'000') mA, t > 0

Example 8.9 Comparing the Three-Step Response Forms

a) Plot on a single graph, over a range from 0 to

220 jus, the overdamped, underdamped, and

critically damped responses derived in

Examples 8.6-8.8

b) Use the plots of (a) to find the time required for

i L to reach 90% of its final value

c) On the basis of the results obtained in (b), which

response would you specify in a design that puts

a premium on reaching 90% of the final value of

the output in the shortest time?

d) Which response would you specify in a design

that must ensure that the final value of the

cur-rent is never exceeded?

i L (mA)

Underdamped {R = 625 il)

jOverdamped (R = 400 SI)

I Critically damped {R = 500 Q)

tQa)

Figure 8.13 A The current plots for Example 8.9

Solution

a) See Fig 8.13

b) The final value of i L is 24 mA, so we can read the

times off the plots corresponding to i L = 21.6 mA

Thus t od = 130 ^is, t cd = 97 /xs, and t ud = 74 fis

c) The underdamped response reaches 90% of the

final value in the fastest time, so it is the desired

response type when speed is the most important

design specification

d) From the plot, you can see that the under-damped response overshoots the final value of current, whereas neither the critically damped nor the overdamped response produces currents

in excess of 24 mA Although specifying either of the latter two responses would meet the design specification, it is best to use the overdamped response It would be impractical to require a design to achieve the exact component values that ensure a critically damped response

Energy is stored in the circuit in Example 8.8

(Fig 8.12, with R = 500 n) at the instant the dc

cur-rent source is applied The initial curcur-rent in the

inductor is 29 mA, and the initial voltage across the

capacitor is 50 V Find (a) /L(0); (b) di L {0)/dt;

(c) i L {t) for t > 0; (d) v(t) for t > 0

Solution

a) There cannot be an instantaneous change of

cur-rent in an inductor, so the initial value of i L in the

first instant after the dc current source has been

applied must be 29 mA

b) The capacitor holds the initial voltage across the

inductor to 50 V Therefore

c) From the solution of Example 8.8, we know that the current response is critically damped Thus

idO = if + D\te~ at + D 2 e~ at ,

where

2RC = 40,000 rad/s and I f = 24 mA

Notice that the effect of the nonzero initial stored energy is on the calculations for the

con-stants D[ and D' 2 , which we obtain from the

ini-tial conditions First we use the iniini-tial value of the inductor current:

Lf«n = 50,

^ ( 0 + ) = § X 103 = 2000 A/s

at 25

iL(0) = I f + D' 2 = 29 mA,

from which we get

D' 2 = 29 - 24 = 5 mA

8.4 The Natural and Step Response of a Series RLC Circuit 285

The solution for D[ is

~(0 + ) = D{ - aD' 2 = 2000,

or

D\ = 2000 + aD' 2

= 2000 + (40,000)(5 X 10~3)

= 2200 A/s = 2.2 X 106 m A/s

Thus the numerical expression for i L (t) is

i L (t) = (24 + 2.2 X lO'Ve"40-000'

+ 5e~ 4{UMH ) mA, t > 0

d) We can get the expression for v(t), t > 0 by

using the relationship between the voltage and current in an inductor:

dij

v(t) = L ^

= (25 X 10~3)[(2.2 X 106)(-40,000)te-4ao,1()'

+ 2.2 X ioV4 0-0 0 0'

+ (5)(-40,000)<r4(M)00'] X 10~3

= - 2 2 x lO6^4 0-0 0 0' + 5 0 e ^w u* V, / s> 0

To check this result, let's verify that the initial voltage across the inductor is 50 V:

v(0) = - 2 2 X 106(0)(1) + 50(1) = 50 V

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

Objective 1—Be able to determine the natural response and the step response of parallel RLC circuits

8.6 In the circuit shown, R = 500 O, L = 0.64 H,

C = 1 fxF, and I = —1 A The initial voltage

drop across the capacitor is 40 V and the initial

inductor current is 0.5 A Find (a) //?(0+);

(b) ic(0+); (c) di L (0 + )/dt; (d) st, s 2 ; (e) i L (t) for

t > 0; and (f) v(t) for t > 0+

Answer: (a) 80 mA;

(b)-1.58 A;

(c) 62.5 A/s;

(d) (-1000 + ;750) rad/s, (-1000 - /750) rad/s;

( e ) [ - l + <r1000'[1.5cos750f

+ 2.0833 sin 750f] A, for t > 0;

(f) £r1000'(40cos750; - 2053.33 sin 750r) V,

for t > 0+

NOTE: Also try Chapter Problems 8.29-8.31

8.4 The Natural and Step Response

of a Series RLC Circuit

The procedures for finding the natural or step responses of a series RLC

circuit are the same as those used to find the natural or step responses of a

parallel RLC circuit, because both circuits are described by differential

equations that have the same form We begin by summing the voltages

around the closed path in the circuit shown in Fig 8.14 Thus

/* 1 p

Ri + L~ + - idr + V 0 = 0

dt CJ ()

We now differentiate Eq 8.52 once with respect to t to get

n di , dri i R— + L— + — = 0,

dt dt 2 C

(8.52)

(8.53)

R

•"•"N,

iy

L

X ) d

+

^V a

Figure 8.14 A A circuit used to illustrate the natural

response of a series RLC circuit