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16 linear systems

1.4 T I M E - D O M A I N S O L U T I O N O F L I N E A R D I F F E R E N T I A LE Q U A T I O N S

A n ordinary linear differential equation wi t h constant coefficients ischaracteristic of linear lumped time-invariant systems. In this section, wereview briefly the theory of solving this class of differential equations. Werestrict our discussion here to single input-output systems. The m u l t i d i mensional case is the subject of Chapter 3. Suppose the differential equationis of order n

dnv(t) d^vit)K~dT + K~x + "' + b y { t ) = x ( t ) ( L 2 1 )

It is convenient to write (1.21) in operational form as

(* fm + bn_, ^ + - ' + b0)[y]= x{t) (1.22)o r more briefly as

L[y] = x(t) (1.23)where L is the operator

dn dn~xL = bn+ &n_! ^- + + b0 (1.24)dtn dtn~xIt is easy to verify that L is a linear operator, that is

L\C\y\ + c 22/ 2] = c\ L[Vi\ + c2 L[y2]Thus, if yx{t) and y2(t) are solutions to L[y] 0, then so also is c-^y^t)+c2y2(t) a solution to L[y] = 0.

T he solution to (1.21) is made up of two components: (1) the source-free(transient, natural , complementary, homogeneous) solution, and (2) thecomponent resulting from the source (forced, steady-state, nonhomogeneous,particular solution). The transient solution yc(t) is found from the homogeneous equation corresponding to (1.21). That is, yc(t) must satisfy

L[yc(t)) = 0 (1.25)F o r the special case of constant coefficients, the solution to (1.25) has

been completely solved.2

In this case, the solution has a general formVc (t) = cVl{t) + c2y2(t) + + cnyn{t) (1.26)

2 Wylie, C. R. , Advanced Engineering Mathematics, McGraw-Hill, New York, 1966.


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1.4 Time-Domain Solution of Linear Differential Equations 17The functions yi(t),y2(t), . , depend on the roots of the associatedcharacteristic equation,

f(r) = bnr + bn_xr^ + + b0 = 0 (1.27)where bn, . . . , bQ are the same coefficients as in (1.24). If (1.27) has n distinctroots rlt r2, . . . , rn, the functions yt(t), i = 1, 2 , . . . , n are y t(t) = eTit andthe transient or complementary solution is

yc(t) = cer + c2er* + + cne r^ (1.28)I Spending on the multiplicity of the roots in (1.27), the functions yt(t) takeon various forms. We summarize the results below. After finding the n rootso f (he characteristic equation (1.27), assign the y(0> i = 1,2, . . . ,n asfollows:

(1) fo r each real root r, the function ert(2) for each real roo t r of multiplicity k, the functions e rt, tet,... , tk~xet(3) for each simple complex pair of roots a jb, the functions cos bt

; i n d e a i sin(4) fo r each complex pair of roots a y7> of multiplicity k, the functions


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18 Linear SystemsTherefore the left-hand side of (1.29) is zero and ye(t) o f (1.28) is asolution to the homogeneous equation (1.25).E X A M P L E 1.11. Consider the differential equation

A ( 0 _ A ( 0 M O _ y ( t ) = oThe characteristic equation is

f(r) = r 3 - r 2 + r - l = 0which has roots j, j, and 1. The homogeneous solution function istherefore

yc{t) = cef + c2e~ ji + c3e jt= + c 2 cos t c2j sin t + c 3 cos t + c 3j" sin f= cxe f + c 2 cos t + C g sin f

We c a l l yc(t) a solution function if the constants c i ? i = 1, 2 , . . . , n arenot specified.

The solution resulting from the forced response is somewhat more involved.There are several methods used i n obtaining yv(t) including educatedguessing. The method o f undetermined coefficients can be used i f thederivatives of x(t) result in a finite number of independent functions. W eoutline the method o f undetermined coefficients briefly.

Let D = djdt so that the operator L can be writtenL = bnD + bn_xD + + b0 (1.30)

i.e., as a polynomial operator in the operator D. We wish to f i n d an operatorwhich "annihilates" x{t). That is , given x(t), we wish to f i n d an operatorL A such that

LJx{t)] = 0 (1.31]Such an operator can always be found i f x{t) is a solution o f a homogeneousequation with constant coefficients. F o r example, i f x(t) = eai, then theannihilator operator is LA = D a. I f x(t) = A cos bt + B sin bt, therthe annihilator operator is LA = D2 + b2. Fo r a sum of such terms, theannihilator operator is the product of the operators fo r each term in the sum

Once the annihilator operator has been found, applying it to both sides o:the original nonhomogeneous equation results i n a homogeneous equationThat is , suppose LA is the annihilator fo r x(t) and we wish to solve

L[y(t)] = x(t) (1.32Then

LA{L[y{t)}} = LA[x(t)] = 0 (1.33


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1.4 Time-Domain Solution of Linear Differential Equations 19W e can solve (1.33) by the method outline d previously. The solution functionis

V(t) = c#x{t) + + cnyn{t) + cPiy9l(t) + + c9#9r(t).The first n solution functions satisfy L[y(t)] = 0. Therefore, i f we substitute;/(/) into (1.32), these terms w i l l sum to zero on the left-hand side. The onlylerms left w i l l be cViyVi(t) + + cVryPr{t) which arise from the forcingfunction. I f we now equate coefficients of l i k e functions on both sides of theequation, we can evaluate the constants cp , . . . , cVr, and thus obtain theforced solution. The following examples demonstrate these ideas.

E X A M P L E 1.12. Consider the differential equationL[y(t)] = (D2 + \)[y(t)] = e*

I n this case, the annihilator fo r e* is {D 1) because {D l)[e f] = 0.Thus we operate on both sides by (D 1) an d obtain the homogeneousequation

C D - i ) ( 2 + i)bf(0] = oW e can solve this equation by means of the characteristic equation

(r - l ) ( r 2 + 1) = 0This equation has roots j. Thus, the solution function is

y(t) = cx co s t + c2 sin t + czelI f we now substitute this solution function into the original equation,the first two terms are zero because they are solutions to L[y] = 0. Weobtain on e equation for c 3 , the undetermined coefficient,

( D 2 + l)[cx cos t + c2 sin / + czex\ = elThus

0 + (D2 + l)[c3el] = elcze* + c3e* = e*

or2c3e* = e1

Hence, c3 = 1/2, an d so cvyv(t) = e1]!.E X A M P L E 1.13. Consider the differential equation

L[y(t)}= {D*+ l)[y(t)] = sin?I n this problem, the annihilator is (D2 + 1). Thus, the homogeneousequation we wish to solve is

(D + \)(D2 + \)[y(t)\ = 0


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20 Linear SystemsTh e corresponding characteristic equation is

(r 2 + 1)(Y2 + 1) = 0which has roots j , j each of multiplicity two. Thus, the solutiofunction is

y(t) = a cos t + c2 sin t + czt cos / + c 4 / sin tSubstituting this solution function into the original equation, we ha>

(D + l)[c x cos t + c 2 sin / + c 3/ cos / + c 4f sin /] = sin tPerforming the indicated differentiation and equating coefficients of l i lfunctions on both sides of the equation, we f i n d that c 4 = 0 and c3 * \. Hence, cvyv{t) = \t cos t. If the i n i t i a l conditions are y(0) =and y'(0) = 0, we can solve for cx and c2 using our knowledge of tlforced solution. We have

2/(0 = ci c o s t + c 2 sin 1A n d so2/(0) = 1 = c x

J/'(0) = 0 = c2 -Thus, we know that

cx= 1c 2 2

and the complete solution isy(t) cos / + i sin t -

Transient and Steady-State ComponentsA t this point, it is useful to view the mathematical solution for (1.21)

terms of the system and the input stimulus x(t). We have decomposed ttotal response of the system y(t) into two components. The source-freetransient solutio n is obtained for a zero input stimulus . Hence, it deperionly on the character of the system and not on any external signals. Becaithis response occurs for x(t) = 0, it is also known as the natural respoio f the system. The terms source-free, transient, natural, homogeneous, acomplementary are a ll used to describe this response that results only fnthe character of the system. The term cos t + \ si n t in Example 1.13 isexample of a transient solution.

In contras t, the forced solu tion is characterist ic of bo th the system athe input stimulus. If we change either the system or the input, we change iforced response. The forced response is also known as the steady-stresponse because this response is a driven response and can exist after

\t cos t

it COS t


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1.5 Initial Energy Storage in Linear Systems 21imnsienl response has died away. The terms steady-state, forced, driven,uotnponent resu lting fro m the source, nonhomoge neous , and par tic ula r aren il used to describe this response. The term \t cos t in Exampl e 1.13 is anexample of a forced solution.

To summarize: The transient response is characteristic of the system and isobiniiied by solving the homogeneous differential equati on that models the*ynlem; the steady-state response depends on both the system and the natureo l I he input stimulus. It is the output response that is forced up on the systemby the input stimulus and must be found by solv ing the nonhomogene ousi l l l l c i e n t i a l equation.

1.5 I N I T I A L E N E R G Y S T O R A G E I N L I N E A R S Y S T E M SThere is another decomposition of the total response that is sometimes

Ui o f u l in linear systems analysis. This decomposit ion involves separating theiespouse resulting from the i n i t i a l energy storage and the response resultingfrom the system input. B y separat ing the i n i t i a l energy storage from theremaining system response, we often ob tai n a simpler way of han dli ng andinterpreting i n i t i a l energy storage in a system.

The decomposition shown in Figure 1.10 allows us to resolve the problemWe encountered in Exampl es 1.1 and 1.6, where a seemingly linear systemwas describ ed by an inpu t-ou tput rel ati on whi ch di d not satisfy the superposition property and hence was not linear. In the decomposition shown in

x(t) H

A system with initial energy storage

x(t).

(a)

H

System with no initial energy storage

tit) 0-tern andange thedy-state

after the M G U R E 1.10

H

System with the initial energy storage(b)


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22 Linear SystemsFigure 1.10, each of the systems labeled H is identical to the given system inthe top of the figure. We write the system output y2(t) as the sum of yd(t), theoutput of an initially relaxed system driven by the input x{t), plus yh(t), theoutput of an unforced system with i n i t i a l conditions identical to those of ourgiven system. Then we have

y 2(0 = yd(t) + yh(0whereL[yd(t)} = x(t)

VJLO) = Vffl = = y j ^ (0 ) = 0an d L[yh(t)] = 0

K ( 0 ) = | f i ( 0 ) , y&0) = yl(0), . . . , y^XO) = ^ - " ( O ) .Here yx{0), y'i(0), . . . , 2 / i _ 1 ( 0 ) are the given i n i t i a l conditions for our system.T o demonstrate that y2(t) of the decomposed system equals yx(t) of theoriginal system, we note that the two functions satisfy the same differentialequation

L\y2(t)] = L[yh(t) + yd(t)] = x(t)L[yx{t)\ = x(t)

an d that they have the* same i n i t i a l conditionsy2(0) = yh(0) + yd(0) = yx(0)2/2(0) = y'M + y'd{Q>) = y[(0)

2/ 2 - 1 ) (0) = 2/i w~ 1) (0) 4- ydn-1](0) = y{rl)(0)It follows that yx(t) = y2{t) for all t > 0, because the solution to an thorder differential equation with n i n i t i a l conditions is unique.

E X A M P L E 1.14. The system of Example 1.1, described by thealgebraic equation

y(t) = ax{t) + bc a n be decomposed as shown in Figure 1.11. Here the output resultingfrom the input is yd{t) = ax(t) (note that this relation represents a linearsystem), and yh(t) = b is the output of the original system with x{t) 0.E X A M P L E 1.15. Consider the RC network o f Exampl e 1.6 shown inFigure 1.12 and describ ed by the equat ion

e(t) = Ri(t) + - f Kt') dt'


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1.5 Initial Energy Storage in Linear Systems 23

x(t)- H(-) = a(-)+b ~>~y(t) = ax(t) + b

(a )

x(t)- a(>) a x(t)

@ ^y(0+

(b)

IIGURE 1.11

where i(t) is the system input and e(t) the system output. In our systemdecomposition, we set yd(t) equal to the output of the same system, butwilli no energy storage at t = 0. Thus

yd(t) - Ri(t) + 1 I rffC JoThe term equals the output of the original system with zero input,but with the given i n i t i a l condition e(0) = v0.

Vh(t) = voThus we have


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24 Linear Systemsas before. B y decompo sin g our system in this manne r, we can ncapply superposition to the linear partthat is, the x(t) yd(t) relationand then add the term yh(t) to obtain the entire output.E X A M P L E 1.16. Con sider the system of Exa mpl e 1.13 descr ibedthe differential equation

(D*+ \)\y(t)] = x(t)with the i n i t i a l conditions

y(0) 1 y'(0 ) = 0The term yh{t) in the system deco mpos it ion is independent of 1input x(t). To f i n d this term, we need only solve the homogenecdifferential equation

(D* + 1)[*/,(>)] = 0with i n i t i a l conditions

yh(0) = 1, y;(0) = 0Solving this equation, we have

yh(t) = c cos / + c 2 sin /where

yh(0) = 1 gives c = 1^( 0) = 0 gives c2 = 0

Thus yh{t) = cos t, and we can sketch the equivalent system asFigure 1.13. To f i n d yd{t), we must solve the nonhomogeneous equat

C D * + l)fefc(0] = x ( t )with x{t) = sin t and zero i n i t i a l condit ions. As before, we ob tain

X(t) H

Zero initial conditions

*y(t)M +

cos t

F IGURE 1.13


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1.6 Linear Difference Equationsand substituting in the o r i g i n a l differential equation, we f i n d

Coc 4 = 0

as before. A p p l y i n g the i n i t i a l conditionsyM = oyffl = o

we f i n d thatc x = 0Co = 1/2

Thus2/d = = K s m ' ' c o s 0

Summing the terms and yd(t), we f i n d the solutiony(t) = yh(t) + yd(t)

= cos / + J(si n / / cos /)as before.

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