Consider an Nth-order discrete-time LTI system with the state equation in VS .NET

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7.33. Consider an Nth-order discrete-time LTI system with the state equation
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The system is said to be controllable if it is possible to find a sequence of N input samples x [ n , ] , x [ n , + 11, . . . ,x [ n , f N - 1 ] such that it will drive the system from q [ n , ] = q , to q [ n , + N ] = q , and q , and q , are any finite states. Show that the system
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STATE SPACE ANALYSIS
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[CHAP. 7
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is controllable if the controllability matrix defined by
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M , = [b A b
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has rank N.
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We assume that no = 0 and q[O] = 0. Then, by Eq. (7.23) we have
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which can be rewritten as
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Thus, if q[N] is to be an arbitrary N-dimensional vector and also to have a nonzero input sequence, as required for controllability, the coefficient matrix in Eq. (7.122) must be nonsingular, that is, the matrix
must have rank N.
734. Consider a n Nth-order discrete-time LTI system with state space representation
The system is said to be obseruable if, starting at an arbitrary time index nu, it is possible to determine the state q[n,] = q , from the output sequence y[n,], y[nO+ I ] , . . ., y[n, + N - 11. Show that the system is observable if the obseruability matrix defined by
has rank N .
We assume that n, = 0 and x [ n ] = 0. Then, by Eq. (7.25) the output y[n] for n = 0,1,. ..,N - 1, with x[n] = 0, is given by
CHAP. 71
STATE SPACE ANALYSIS
Rewriting Eq. (7.125) as a matrix equation, we get
Thus, to find a unique solution for q[O], the coefficient matrix of Eq. (7.126) must be nonsingular; that is, the matrix
M, =
must have rank N.
7.35. Consider the system in Prob. 7.7
( a ) Is the system controllable
( b ) Is the system observable ( c ) Find the system function H ( z ) .
( a ) From the result from Prob. 7.7 we have
Now and by Eq. (7.120) the controllability matrix is
and IM,l ( b ) Similarly,
- 1 # 0. Thus, its rank is 2 and hence the system is controllable.
and by Eq. (7.123) the observability matrix is
and (MoI = - & # 0. Thus, its rank is 2 and hence the system is observable.
STATE SPACE ANALYSIS
[CHAP. 7
( c ) By Eq. (7.44) the system function H ( z ) is given by
7.36. Consider the system in Prob. 7.7. Assume that
Find x[O] and x[l] such that q[2] = 0.
From Eq. (7.23)we have
Thus,
from which we obtain x[O] = -
and x[l] =
7.37. Consider the system in Prob. 7.7. We observe y[O] = 1 and y[l] = 0 with x[O] = x[l] = 0. Find the initial state q[O].
Using Eq. (7.1251,we have
Thus,
Solving for ql[Ol and q,[Ol, we obtain
[:[m j
x[ol
+ :x[ol]
CHAP. 71
STATE SPACE ANALYSIS
7.38. Consider the system in Prob. 7.32.
( a ) Is the system controllable
(6) Is the system observable
( a ) From the result from Prob. 7.32 we have
Now and by Eq. (7.120) the controllability matrix is
and IM,I = ( b ) Similarly,
- 4 # 0. Thus, its rank is 2 and hence the system is controllable.
and by Eq. (7.123) the observability matrix is
and IMol = 0. Thus, its rank is less than 2 and hence the system is not observable. Note from the result from Prob. 7.32(b)that the system function H ( z ) has pole-zero cancellation. If H ( z ) has pole-zero cancellation, then the system cannot be both controllable and observable.
SOLUTIONS OF STATE EQUATIONS FOR CONTINUOUS-TIME LTI SYSTEMS
7.39. Find eA' for
using the Cayley-Hamilton theorem method.
First, we find the characteristic polynomial c(A) of A.
= A ' + 5A + 6
Thus, the eigenvalues of A are A , =
+ 2)(A + 3)
- 2 and A ,
- 3. Hence, by Eqs. (7.66)and (7.67)we have
STATE SPACE ANALYSIS
[CHAP. 7
and b, and b , are the solutions of
from which we get Hence,
7.40. Repeat Prob. 7.39 using the diagonalization method.
Let x be an eigenvector of A associated with A . Then
[ A 1 - A]x
For A = A ,
- 2 we have
The solutions of this system are given by x , are those vectors of the form
- 2 x 1 . Thus, the eigenvectors associated with A ,
with a # 0
For A = A 2 = -3 we have
The solutions of this system are given by x , are those vectors of the form
- 3 x , . Thus, the eigenvectors associated with A ,
Let a = p
1 in the above expressions and let
Then
CHAP. 71
STATE SPACE ANALYSIS
7.41. Repeat Prob. 7.39 using the spectral decomposition method.
Since all eigenvalues of A are distinct, by Eq. (7.33) we have 1 3 El = -A - A 2 1 ) = A + 3 1 = ( AI -A2 [-6 E2= (A-AII)
- A1
-(A+21)
-:] [-:-:I
Then by Eq. (7.70) we obtain
7.42. Repeat Prob. 7.39 using the Laplace transform method.
First, we must find (SI -A)-'
Then, by Eq. (7.71) we obtain
Again we note that when the eigenvalues of A are all distinct, the spectral decomposition method is computationally the most efficient method of evaluating eA'.
7.43. Find eA' for
The characteristic polynomial c(A) of A is
= h 2 + 4 ~ + 3 = ( ~ + +3) 1)(A Thus, the eigenvalues of A are A ,
= -1
and A, = -3. Since all eigenvalues of A are distinct, by
STATE SPACE ANALYSIS
[CHAP. 7
Eq. (7.33) we have
Then, by Eq. (7.70) we obtain
7.44. Given matrix
( a ) Show that A is nilpotent of index 3.
( 6 ) Using the result from part ( a ) find eA'.
By direct multiplication we have
Thus, A is nilpotent of index 3. ( b ) By definition (7.53) and the result from part ( a )
7.45. Find eA' for matrix A in Prob. 7.44 using the Cayley-Hamilton theorem method.
First, we find the characteristic polynomial c(A) of A.
A c(A)=lAl-A, =I0 0 2 A -il=A3
CHAP. 71
STATE SPACE ANALYSIS
Thus, A
is the eigenvalues of A with multiplicity 3. By Eq. (7.66) we have
eA' = b,I
+ b,A + b2A2
where b,, b,, and b2 are determined by setting A (A. 5 9 ) and ( A . 60 )I:
in the following equations [App. A, Eqs.
+ b,A + b2h2 = eA'
Thus,
Hence,
eA' = I
+ tA + -A' 2
which is the same result obtained in Prob. 7.44(b).
7.46. Show that
eA+B
= eAeB
provided A and B commute, that is, AB = BA.
By Eq. (7.53)
and Thus, if AB = BA, then
1 1 1 1 =I+A+B+-A~+-AB+-BA+-B~+ 2! 2 2 2! eAeB - e A + B = L (AB - BA) + .
eA+B
= eAeB
7.47. Consider the matrix
STATE SPACE ANALYSIS
[CHAP. 7
Now we decompose A as A=A+N where
A = [ O 2 00 ] 2 0 0 0 2
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