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Solutions
1 Elementary concepts
1.b) The Nyquist frequency is the half of the sampling rate (see
chap. 1 of CCS and lecture 1).
2.c) The transport delay of2hresults into two extra states (see
chap. 2 of CCS and lecture 2).
3.a) Asnx(0) = 0, the state of the system will go to zero for
zero input. According toDefinition 3.7 on page 94 in CCS, the
system is controllable. Nothing is said about the
rank of the controllability matrix Wc, the system is thus not
necessarily reachable (see
also lecture 3).
4.b) For the eigenvalues of and A, it holds that i() = ei(A)h
(see page 61 in CCS and
lecture 3).
5.a) The Lyapunov function is a function of states (see page 88
in CCS and lecture 3).
6.a) All poles are strictly in the unit circle (see page 78 in
CCS and lecture 3). The system is
even asymptotically stable.
7.b) Controllability is related to the ability to control the
system. Therefore, even without
remembering the formulas, it is clear that a) and c) cannot be
true as they do not contain
(see page 94 in CCS and lecture 3).
8.b) State feedback is a static control law. It cannot change
the order of the system.
9.b) To eliminate steady-state errors, the loop must contain an
integrator. If there is one (or
more) in the system, there is no need to insert another one
(this may, in fact, deteriorate
the performance or destabilize the closed-loop system).
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2 Open questions
1. a) The pulse-transfer function is:
H(z) = C(zI )1 = (1 0)
z 1 1.5
0 z 0.4
1
21
= 2z 2.3z2 1.4z+ 0.4
The system has one pole in 1 and thus the stationary gain is
.
b) The poles are 1 and 0.4. The system is not asymptotically
stable as|i| < 1 doesnot hold for the first pole.
c) The controllability matrix is
Wc = ( ) =
2 0.51 0.4
Asdet(Wc)= 0, the system is reachable.
d) The observability matrix is
Wo=
C
C
=
1 01 1.5
Asdet(Wo)= 0, the system is reachable.
2. With the state-feedback control lawu(k) = (l1 l2)x(k), the
closed loop characteristicpolynomial is
det(zI ( L)) = det
z 1.7 + l1 0.72 + l2
1 z
= z2 (1.7 l1)z+ 0.72 + l2
This polynomial must be equal to z2 1.2z+ 0.5. By comparing the
coefficients we getl1 = 0.5and l2=0.22.
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