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Consider an Nthorder discretetime LTI system with the state equation in VS .NET
7.33. Consider an Nthorder discretetime LTI system with the state equation Denso QR Bar Code Recognizer In .NET Using Barcode Control SDK for VS .NET Control to generate, create, read, scan barcode image in VS .NET applications. QR Code Creator In Visual Studio .NET Using Barcode encoder for .NET framework Control to generate, create Quick Response Code image in VS .NET applications. 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 QRCode Decoder In .NET Framework Using Barcode reader for VS .NET Control to read, scan read, scan image in VS .NET applications. Create Barcode In Visual Studio .NET Using Barcode creator for VS .NET Control to generate, create bar code image in .NET applications. STATE SPACE ANALYSIS
Barcode Scanner In .NET Framework Using Barcode decoder for VS .NET Control to read, scan read, scan image in VS .NET applications. Denso QR Bar Code Generation In Visual C# Using Barcode encoder for .NET framework Control to generate, create QR Code image in VS .NET applications. [CHAP. 7
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Drawing Code128 In None Using Barcode maker for Font Control to generate, create Code 128 Code Set C image in Font applications. Data Matrix Encoder In C# Using Barcode maker for Visual Studio .NET Control to generate, create Data Matrix image in Visual Studio .NET applications. ~~'b] Printing GTIN  128 In Java Using Barcode generation for BIRT Control to generate, create EAN 128 image in Eclipse BIRT applications. Decoding EAN / UCC  13 In VB.NET Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET framework applications. We assume that no = 0 and q[O] = 0. Then, by Eq. (7.23) we have
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USS128 Creator In ObjectiveC Using Barcode generator for iPad Control to generate, create USS128 image in iPad applications. Reading UPC Code In Java Using Barcode recognizer for Java Control to read, scan read, scan image in Java applications. Thus, if q[N] is to be an arbitrary Ndimensional 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 Nthorder discretetime 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 polezero cancellation. If H ( z ) has polezero cancellation, then the system cannot be both controllable and observable. SOLUTIONS OF STATE EQUATIONS FOR CONTINUOUSTIME LTI SYSTEMS
7.39. Find eA' for
using the CayleyHamilton 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= (AAII)  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 CayleyHamilton theorem method.
First, we find the characteristic polynomial c(A) of A.
A c(A)=lAlA, =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

