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N = O O
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( a ) Show that the matrix N is nilpotent of index 3.
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( b ) Show that A and N commute, that is, AN = N h ( c ) Using the results from parts ( a ) and ( b ) ,find eA'.
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By direct multiplication we have
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0 N~=NN= o
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0 0 1 = o 01 ( 0 0 0 1 = 0
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0 0 N3=N2N=0 0 [o 0
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Thus, N is nilpotent of index 3. ( b ) Since the diagonal matrix A can be expressed as 21, we have
that is, A and N commute. Since A and N commute, then, by the result from Prob. 7.46
Now [see App. A, Eq. (A.4911
and using similar justification as in Prob. 7.44(b),we have
Thus,
CHAP. 71
STATE SPACE ANALYSIS
7.48. Using the state variables method, solve the second-order linear differential equation y"(t) + 5y'(t)+ 6 y ( t )= x ( t )
(7.127)
with the initial conditions y(0) = 2, yl(0) = 1 , and x ( t ) = e - ' u ( t ) (Prob. 3.38). Let the state variables q J t ) and q,(t) be
qdt)=Y(I)
~ 2f ) (
=~'(t)
Then the state space representation of Eq. (7.127) is given by [Eq. (7.19)l
q ( t ) = Aq(t) + b x ( t )
with
A=[-:
~ t t =)c q ( t )
b=[y]
c = [ l O]
q'ol
[ q d o 0 ) ]= 2( )
Thus, by Eq.(7.65)
with d = 0. Now, from the result from Prob. 7.39
c ~ q ( 0= [ I )
~](e-~'[-:
+e-''[-:
]:[)I:-
Thus,
7.49. Consider the network shown in Fig. 7-20. The initial voltages across the capacitors C, and C, are f V and 1 V, respectively. Using the state variable method, find the voltages across these capacitors for t > 0 . Assume that R , = R , = R, = 1 0 and C ,= C 2 =1 F.
Let the state variables q , ( t ) and q2(t ) be
STATE SPACE ANALYSIS
[CHAP. 7
Fig. 7-20
Applying Kirchhoffs current law at nodes 1 and 2, we get
Substituting the values of R , , R 2 , R,, C , , and C 2 and rearranging, we obtain
41(t) = - 2 9 1 ( t ) + q 2 ( t )
42(t) = 9 l ( t )-292(t)
In matrix form
il(t) =Aq(t)
with Then, by Eq. (7.63) with x ( t ) = 0 and using the result from Prob. 7.43, we get
7.50. Consider the continuous-time LTI system shown in Fig. 7-21.
(a) Is the system asymptotically stable ( b ) Find the system function H(s). ( c ) IS the system B I B 0 stable
From Fig. 7-21 and choosing the state variables q , ( t ) and q 2 ( t )as shown, we obtain
41(t) = 9 2 ( t ) + x ( t )
42(t) =
29l(t) +92(t)- x ( t )
~ ( t =) 9 L t ) - % ( t )
CHAP. 71
STATE SPACE ANALYSIS
In matrix form
where Now Thus, the eigenvalues of A are A , = - 1 and A t = 2. Since Re{A,) > 0, the system is not asymptotically stable. ( b ) By Eq. (7.52) the system function H(s) is given by
Note that there is pole-zero cancellation in H(s) at s = 2. Thus, the onIy pole of H(s) is - 1 which is located in the left-hand side of the s-plane. Hence, the system is B I B 0 stable. Again it is noted that the system is essentially unstable if the system is not initially relaxed.
7.51. Consider an Nth-order continuous-time LTI system with state equation
The system is said to be controllable if it is possible to find an input x ( t ) which will drive the system from q(t,) = q , to q(t ,) = q , in a specified finite time and q , and q , are any finite state vectors. Show that the system is controllable if the controllability matrix defined by
has rank N.
STATE SPACE ANALYSIS
[CHAP. 7
We assume that r,
and q[Ol = 0. Then, by Eq. (7.63) we have
Now, by the Cayley-Hamilton theorem we can express e-A' as
Substituting Eq. (7.130) into Eq. (7.129) and rearranging, we get
Let Then Eq. (7.131) can be rewritten as
For any given state q , we can determine from Eq. (7.132) unique Pk's (k = 0,1,. . . , N - I), and hence x(t), if the coefficients matrix of Eq. (7.132) is nonsingular, that is, the matrix
has rank N.
7.52. Consider an Nth-order continuous-time LTI system with state space representation
q ( t ) = Aq(t) + b x ( t )
~ ( t= c 4 t ) )
T h e system is said to be observable if any initial state q(t,) can be determined by examining the system output y ( t ) over some finite period of time from to to t , . Show that the system is observable if the observability matrix defined by
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