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from which we get QR Code 2d Barcode Recognizer In .NET Framework Using Barcode Control SDK for VS .NET Control to generate, create, read, scan barcode image in VS .NET applications. Making QR In .NET Using Barcode encoder for Visual Studio .NET Control to generate, create Denso QR Bar Code image in Visual Studio .NET applications. b,= ( 1 n)2" Decoding QR Code ISO/IEC18004 In VS .NET Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET framework applications. Barcode Drawer In .NET Framework Using Barcode generator for VS .NET Control to generate, create barcode image in .NET applications. b, = n2"' Barcode Decoder In VS .NET Using Barcode decoder for Visual Studio .NET Control to read, scan read, scan image in .NET applications. Create Quick Response Code In Visual C# Using Barcode printer for VS .NET Control to generate, create QR Code 2d barcode image in .NET applications. 7.26. Consider the matrix A in Prob. 7.25. Let A be decomposed as
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(b) Since the diagonal matrix D can be expressed as 21, we have DN = 21N = 2N = 2NI = N(2I) = ND
that is, D and N commute.
CHAP. 7 1
STATE SPACE ANALYSIS
(c) Using the binomial expansion and the result from part (b), we can write
Since N*= 0, then Nk = 0 for k 2 2, and we have Thus [see App. A, Eq. (A.431, which is the same result obtained in Prob..7.25. Note that a square matrix N is called nilpotent of index r if Nr'# 0 and Nr= 0. 7.27. The minimal polynomial m(A) of A is the polynomial of lowest order having 1 as its leading coefficient such that m(A) = 0. Consider the matrix 0 0 ( a ) Find the minimal polynomial m(A) of A. ( b ) Using the result from part ( a ) , find An. (a) The characteristic polynomial c(A) of A is Thus, the eigenvalues of A are A , Now
 3 and
= A, = 2. Consider
Thus, the minimal polynomial of A is
From the result from part (a) we see that An can be expressed as a linear combination of I and A only, even though the order of A is 3. Thus, similar to the result from the STATE SPACE ANALYSIS
[CHAP. 7
CayleyHamilton theorem, we have
where 6, and 6 , are determined by setting A =  3 and A
in the equation
b0+b,A =An
Thus, b, 36, = (  3 ) " b,+ 26, = 2" from which we get
b O = $(3)" + $(2)" =  f(3)"+
f(2)" 7.28. Using the spectral decomposition method, evaluate An for matrix A in Prob. 7.27.
Since the minimal polynomial of A is
which contains only simple factors, we can apply the spectral decomposition method to evaluate A". Thus, by Eq. (7.33) we have CHAP. 71
STATE SPACE ANALYSIS
Thus, by Eq. (7.34) we get
which is the same result obtained in Prob. 7.27(6). 7.29. Consider the discretetime system in Prob. 7.7. Assume that the system is initially relaxed.
(a) Using the state space representation, find the unit step response of the system. ( b ) Find the system function H(z). From the result from Prob. 7.7 we have q [ n + 11 = A q [ n ] + b x [ n ] Y [ ~ = cq[nl+dub1 I
Setting q[O] = 0 and x [ n ] = u [ n ] in Eq. ( 7 . 2 9 , the unit step response s [ n ] is given by
Now, from Prob. 7.20 we have
c ~ n  '  k= [ b
nlk
n l k
+(f) nlk
,][,] n lk
n k
STATE SPACE ANALYSIS
[CHAP. 7
Thus, which is the same result obtained in Prob. 4.32(c). (6) By Eq. (7.44) the system function H ( z ) is given by Now Thus, ( IA
z112] ( [ i
3 2Ti
which is the same result obtained in Prob. 4.32(a). 7.30. Consider the discretetime LTI system described by
+ 11 = A q [ n ] + b x [ n ] Y [ n ] = c q [ n ]+ h [ n ] Show that the unit impulse response h [ n ]of the system is given by
(6) Using Eq. (7.] I T ) , find the unit impulse response hlnl of the system in Prob. 7.29.
(a> By setting q[O] = 0, x[k] = 6[k], and x[n] = 6[n] in Eq. (7.25), we obtain
CHAP. 71
STATE SPACE ANALYSIS
Note that the sum in Eq. (7.118) has no terms for n = 0 and that the first term is cA"'b for n > 0. The second term on the righthand side of Eq. (7.118) is equal to d for n = 0 and zero otherwise. Thus, we conclude that (6) From the result from Prob. 7.29 we have
&I,, = ( + ) "  '  '(')"I 4 4
Thus, by Eq. (7.117) h [ n ] is
which is the same result obtained in Prob. 4.32(b). 7.31. Use the state space method to solve the difference equation [Prob. 4.38(6)] 3 y [ n ]  4 y [ n  11 + y [ n
 21 =x[n] (7.119) with x [ n ] = ( ; ) " u [ n ] and y [  11 = 1 , y [  2 ] = 2 . Rewriting Eq. (7.119), we have
y [ n ]  $ y [ n 11 d n + 1 1 =+I s z [ n+ 11 =
In matrix form
+ + y [ n  21 = i x [ n ] Let q , [ n ]= y[n  2 and q,[n] = y [ n  11. Then 1
 f q l [ n ]+ ; q 2 [ n ] + i x [ n ] ~ [ n l =f q l [ n l + ; q 2 [ n ] + + [ n l  and Then, by Eq. (7.25) STATE SPACE ANALYSIS
[CHAP. 7
Now from the result from Prob. 7.24 we have
c A n q [ 0 ] =[  4 =f +
'(9)" ,.An I k
b = [) Thus, n+lk
](i)+&) which is the same result obtained in Prob. 4.38(b). 732. Consider the discretetime LTI system shown in Fig. 719. (a) Is the system asymptotically stable ( b ) Find the system function H ( z ) . (c) Is the system BIB0 stable From Fig. 719 and choosing the state variables q , [ n ] and q , [ n ] as shown, we obtain
9 , [ n+ 11
= tclz[n1 + x [ n l
92[n + 11 =  $ l [ n ] + 2 9 2 [ n ] Y ~ = qI , [ n l  9 * [ n l
In matrix form
q[n+ 1] = Aq[n] + bx[n] Y [ ~ = cq[nl I
CHAP. 71
STATE SPACE ANALYSIS
Fig. 719 where
Thus, the eigenvalues of A are A , = $ and A, = ; Since IAzl > 1, the system is not . asymptotically stable. B Eq. ( 7 . 4 4 ) the system function H ( z ) is given by y Note that there is polezero cancellation in H ( z ) at z = : Thus, the only pole of H ( z ) is . I which lies inside the unit circle of the zplane. Hence, the system is BIBO stable. Note that even though the system is BIBO stable, it is essentially unstable if it is not initially relaxed.

