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7.26. Consider the matrix A in Prob. 7.25. Let A be decomposed as
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N2= 0. (b) Show that D and N commute, that is, DN = ND.
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( c ) Using the results from parts ( a ) and ( b ) , find An.
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By simple multiplication we see that
(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
Cayley-Hamilton 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 discrete-time 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
n-l-k
n- l -k
+(f)
n-l-k
-,][,]
n- l-k
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 2-Ti
which is the same result obtained in Prob. 4.32(a).
7.30. Consider the discrete-time 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 right-hand 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+l-k
](i)+&)
which is the same result obtained in Prob. 4.38(b).
732. Consider the discrete-time LTI system shown in Fig. 7-19.
(a) Is the system asymptotically stable ( b ) Find the system function H ( z ) . (c) Is the system BIB0 stable
From Fig. 7-19 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. 7-19
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 pole-zero cancellation in H ( z ) at z = : Thus, the only pole of H ( z ) is . I which lies inside the unit circle of the z-plane. 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.
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