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Consider the ( scalar ) transfer function T ( s ) = pmsm + pm 1 sm 1 + + p 1 s + p

Consider the (scalar) transfer function
T(s)= pmsm +pm1sm1++p1s+p0,(5)
sn +qn1sn1++q1s+q0
with pm =0 and m < n. The coefficients pi and qi are real (and m denotes the degree of the numerator polynomial rather than the number of inputs). The relative degree r of a transfer function is defined as the integer r = n m.
(a) Let (A, B, C,0) be a realization of T (s) with the matrix pair (A, B) controllable. Show that T (s) has relative degree r if and only if
CAi1B =0 for all i in {1,2,..., r 1} and CAr1B =0 Hint. Exploit the controllability canonical form and recall Theorem 6.9.
In the remainder of this problem, we assume r = n and let (A, B, C,0) be the realization of T (s) in controllability canonical form. Denote, as usual, the corresponding linear system by
where y(k) is the kth-order time derivative of y, i.e., y(k)(t)= dky(t).
dtk
(c) Derive the dynamics (7) directly from the transfer function (5).
x (t)= Ax(t)+ Bu(t), y(t)= Cx(t).
(6)
(b) Show that the dynamics (6) can be written as the higher-order differential equation
y(n)(t)+ qn1y(n1)(t)++ q1y(1)(t)+ q0y(t)= p0u(t),(7)
Hint. You may use the fact that the Laplace transform is injective, i.e., L(f1)= L(f2) implies f1= f2.
2
We would like to solve a tracking control problem in which the output y asymptotically tracks a given reference trajectory yref : [0,\infty )-> R, i.e.,
lim y(t) yref (t)=0.(8) t->\infty
To do so, we define the tracking error e(t)= y(t) yref (t).
Now, assuming that yref is sufficiently differentiable and taking p0=1 for simplicity, consider
a controller of the form
k=0
Then, the tracking problem can more formally be stated as: find a matrix
F=f0 f1 fn1
such that, for each initial condition x0 in Rn, the dynamics (6) with x(0)= x0 and input (9)
satisfy (8).
(d) Define the state z as
n1 n1
u(t)= y(n)(t)+ X qky(k)(t)+ X fke(k)(t).(9)
ref ref k=0
z1 e z2e(1)
z=.=...
zn e(n1)
and show that the dynamics (6) with the controller (9) can be written as
z (t)= A + BF z(t), e(t)= Cz(t).
(e) Using the result from (d), prove that (8) holds for any initial condition x0 if and only if \sigma (A + BF) C.
Hint. For the only if part, first show that the matrix pair (A + BF, C) is observable. Then, take some inspiration from the proof of Theorem 6.16.

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