sol 1034
TRANSCRIPT
-
7/25/2019 Sol 1034
1/2
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).
1
-
7/25/2019 Sol 1034
2/2
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.
2