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Appendix J: Derivationof the Time Domain Solutionof State Equations

J.1 Derivation

Rather than using the Laplace transformation, we can solve the equations directly inthe time domain using a method closely allied to the classical solution of differentialequations. We will find that the final solution consists of two parts that are differentfrom the forced and natural responses

First, assume a homogeneous state equation of the form

_xt Axt J:1Since we want to solve for x, we assume a series solution, just as we did in elementaryscalar differential equations. Thus,

xt b0 b1t b2t2 bktk bk1tk1 J:2Substituting Eq. (J.2) into (J.1) we get

b1 2b2t kbktk1 k 1bk1tk Ab0 b1t b2t2 bktk bk1tk1 J:3

Equating like coefficients yields

b1 Ab0 J:4a

b2 12Ab1 1

2A2b0 J:4b

..

.

bk 1k!Akb0 J:4c

bk1 1k 1!Ak1b0

..

.

J:4d

1

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Substituting these values into Eq. (J.2) yields

xt b0 Ab0t 12A2b0t

2 1k!Akb0t

k 1k 1!Ak1b0tk1

IAt 12A2t2 1

k!Aktk 1k 1!A

k1tk1

b0

J:5

But, from Eq. (J.2),

x0 b0 J:61

Therefore,

xt IAt 12A2t2 1

k!Aktk 1k 1!A

k1tk1

x0 J:7

Let

eAt IAt 12A2t2 1

k!Aktk 1k 1!A

k1tk1

J:8

where eAt is simply a notation for the matrix formed by the right-hand side of Eq.(J.8). We use this definition because the right-hand side of Eq. (J.8) resembles apower series expansion of eat, or

eat 1 at 12a2t2 1

k!aktk 1k 1! a

k1tk1

J:9

Using Eq. (J.7), we have

xt eAtx0 J:10We give a special name to eAt: it is called the state-transition matrix2, since it performsa transformation on x(0), taking x from the initial state, x(0), to the state x(t) at anytime, t. The symbol, F(t), is used to denote eAt. Thus,

Ft eAt J:11and

xt Ftx0 J:12There are some properties ofF(t) that we will use later when we solve for x(t)

in the text. From Eq. (J.12),

x0 F0x0 J:13Hence, the first property of F(t) is

F0 I J:14

1 In this development we consider the initial time, t0, to be 0. More generally, t0 6 0. After completing thisdevelopment, the interested reader should consult Appendix K on www.wiley.com/college/nise for themore general solution in terms of initial time t0 6 0.2 The state-transition matrix here is for the initial time t0 0. The derivation in Appendix Kon www.wiley.com/college/nise for t0 6 0 yields xt eAtt0xt0.

2 Appendix J: Derivation of the Time Domain Solution of State Equations

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Bibliography 3

where I is the identity matrix. Also, differentiating Eq. (J.12) and setting this equal toEq. (J.1) yields

_xt _Ftx0 Axt J:15which, at t = 0, yields

_F0x0 Ax0 J:16Thus, the second property of F(t) follows from Eq. (J.16):

_F0 A J:17In summary, the solution to the homogeneous, or unforced, system is

xt Ftx0 J:18where

F0 I J:19and

_F0 A J:20Let us now solve the forced, or nonhomogeneous, problem. Given the forced

state equation

_xtAxt But J:21rearrange and multiply both sides by eAt:

eAt _xt Axt eAtBut J:22Realizing that the left-hand side is equal to the derivative of the product eAtx(t), weobtain

d

dteAtxt eAtBut J:23

Integrating both sides yields

eAtxt t0 eAtxt x0 Z t0eAtButdt J:24

since eAt evaluated at t 0 is the identity matrix (from Eq. (J.8)). Solving for x(t) inEq. (J.24) we obtain

xt eAtx0 Z t0eAttButdt

Ftx0 Z t0Ft tButdt

J:25

where F(t) eAt by definition.

BibliographyTimothy, L. K., and Bona, B. E. State Space Analysis: An Introduction. McGraw-Hill,New York, 1968.

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