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(7.45)
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then the system is said to be asymptotically stable; that is, if, undriven, its state tends to zero from any finite initial state q,. It can be shown that if all eigenvalues of A are distinct and satisfy the condition (7.45), then the system is also BIB0 stable.
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STATE SPACE ANALYSIS
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[CHAP. 7
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76 SOLUTIONS OF STATE EQUATIONS FOR CONTINUOUS-TIME LTI SYSTEMS .
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Laplace Transform Method: Consider an N-dimensional state space representation
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where A, b, c, and d are N x N, N X 1, 1 X N, and 1 X 1 matrices, respectively. In the following we solve Eqs. ( 7 . 4 6 ~ )and (7.46b) with some initial state q(0) by using the unilateral Laplace transform. Taking the unilateral Laplace transform of Eqs. ( 7 . 4 6 ~ and ) (7.466) and using Eq. (3.441, we get
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Rearranging Eq. (7.47~1, have we ( s I - A)Q(s)
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= q(0)
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+ bX(s)
(7.49)
Premultiplying both sides of Eq. (7.48) by (sI - A)-' yields
Q(S)
= (SI-
A)-'~(o+ ( S I - A)-'~x(s) )
Substituting Eq. (7.49) into Eq. (7.47b1, we get
Taking the inverse Laplace transform of Eq. (7.501, we obtain the output y(t). Note that c(sI - A)-'q(0) corresponds to the zero-input response and that the second term corresponds to the zero-state response.
System Function H(s): As in the discrete-time case, the system function H(s) a continuous-time LTI system of is defined by H(s) = Y(s)/X(s) with zero initial conditions. Thus, setting q(0) = 0 in Eq. (7.501, we have
Thus,
CHAP. 71
STATE SPACE ANALYSIS
C. Solution in the Time Domain:
Following
we define
where k!= k(k
- 1) . . -2.1.If
t = 0,then
Eq. (7.53)reduces to
e0 = I
where 0 is an N x N zero matrix whose entries are all zeros. As in e - a r at e , we can show that
eA(t-r) ea('-')
(7.54) = ea'e-a' = (7.55) (7.56)
=eA~e-Ar =e-AreA~
Setting Thus,
in Eq. (7.55),we have
eAte-Ar
= e-AteAt = e O = I
which indicates that e-A' is the inverse of eA'. The differentiation of Eq. (7.53)with respect to
yields
which implies
-eAt
= AeA' = e A t ~
Now using the relationship [App. A, Eq. (A.70)]
and Eq.(7.581,we have
STATE SPACE ANALYSIS
[CHAP. 7
Now premultiplying both sides of Eq. (7.46a) by e--A', obtain we
ePA'q(t = L - A ' ~ q ( +)e A ' b x ( t ) ) t
e P A ' q ( t - e P A ' A q ( t= e - A ' b x ( t ) ) )
From Eq. (7.59)Eq. (7.60) can be rewritten as
-[ e - A ' q ( t ) = C A ' b x ( t ) ]
Integrating both sides of Eq. (7.61)from 0 to I , we get
Hence
e - * ' q ( t ) = q ( 0 ) + /'e-*'bx(r) d i
(7.62)
Premultiplying both sides of Eq. (7.62) by eA' and using Eqs. (7.55)and (7.561, we obtain
If the initial state is q(t,,)and we have x( t ) for t
2 I,,
then
which is obtained easily by integrating both sides of Eq. (7.61) from t , to t . The matrix function eA' is known as the state-transition matrix of the continuous-time system. Substituting Eq. (7.63) into Eq. (7.466),we obtain
D. Evaluation of eA':
Method 1: As in the evaluation of An, by the Cayley-Hamilton theorem we have
When the eigenvalues A , of A are all distinct, the coefficients b,, b , , . . ., b N - , can be found from the conditions
For the case of repeated eigenvalues see Prob. 7.45.
CHAP. 71
STATE SPACE ANALYSIS
Method 2: Again, as in the evaluation of An we can also evaluate eA' based on the diagonalization of A. If all eigenvalues A, of A are distinct, we have
eA' = P
where P is given by Eq. (7.30).
Method 3: We could also evaluate eA' using the spectral decomposition of A, that is, find constituent matrices E, (k = 1,2,. . ., N ) for which
A=A,El
+ A2E2+ . . . +ANEN
( 7.69)
where A, ( k = 1,2,. . ., N ) are the distinct eigenvalues of A. Then, when eigenvalues A, of A are all distinct, we have
e A t= e A ~ ' E l e A ~ + ~ ., . +eAN'E, + ' .
(7.70)
Method 4: Using the Laplace transform, we can calculate eA'. Comparing Eqs. (7.63)and (7.49),we see that
E. Stability: From Eqs. (7.63) and (7.68) or (7.70), we see that if all eigenvalues A, of the system matrix A have negative real parts, that is, Re{A,) < 0 all k
(7.72)
then the system is said to be asymptotically stable. As in the discrete-time case, if all eigenvalues of A are distinct and satisfy the condition (7.721, then the system is also B I B 0 stable.
Solved Problems
STATE SPACE REPRESENTATION 7.1. Consider the discrete-time LTI system shown in Fig. 7-1. Find the state space representation of the system by choosing the outputs of unit-delay elements 1 and 2 as state variables q,[n] and q2[n], respectively.
From Fig. 7-1 we have 4 1 b + 11 = 4 2 M 42[n + 11 = 2 q , [ n l + 3 q 2 b I + x b I y b I = 2q,[nl+ 3 q h I + x b I
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