Is this state space representation controllable (a simple "yes" or "no" answer is not suffisant, justify your answer)?
Response to question 1:
One can compute the controllability matrix ...
However, one can note that this state space representation is a controller canonical form. It is necessary controllable.
Question 2:
Give the simplest possible expression of the transfer function between u and y denoted G(s).
Response to question 2:
The transfer function associated to a controller canonical form. is known
What are the eigenvalues associated to the modes of this system? What are the poles of this system?
Response to question 3:
Two eigenvalues :−d2(s) and−b2b1. But only one pole equals to−d2(s). The mode associated to the eigenvalues−b2b1 should
be unobservable : it is not in the transfer function but is controllable (so it has to be unobservable).
1/7 Correction exam COLMS 17/12/2015
Question 4:
Without computing the observability matrix, say (with justification) if this state space representation is observable.
Give the eigenvalues of the observable modes, give the eigenvalues of the unobservable modes if they exist.
Response to question 4:
The transfer function has only one pole which corresponds to a observable mode with eigenvalue−d2. The other eigenvalue
which is simplified by a zero
In this exercise one study the system given by the state space representation
This system is controllable and observable. You don’t have to prove it.
Note that the system defined by (2) has two states, but that x1 is known since x1 = y. So one does not need to estimate
x1. The only part of the state that needs to be estimated through an observer would be x2.
The aim of this exercise is to show that the following equations define a reduced observer for this system.
˙z = −kz−k2y+u (3)
ˆx2 = z+ky (4)
The system defined by (3) and (4) has two inputs (u and y), one state (z) and one output (ˆx2). The aim is to prove that
under some condition on k, ˆx2 →x2 when t →∞.
Question 7:
From equation (4), one gets z = ˆx2−ky. Using this equation, replace z in equation (3) and show that, if one could apply
derivation, the obtained equation would looks like the equation of a classical observer of x2 with inputs u and ˙y.
Response to question 7:
From equation (4), one gets
˙ˆx2−k˙y = −k(ˆx2−ky)−k2y+u
˙ˆx2 = −kˆx2 +k2y−k2y+k˙y+u
˙ˆx2 = −kˆx2 +k˙y+u
This equation looks like a classical observer with input u and ˙y (the matrix A would be zero, the output matrix C would
be 1, and the observer gain k).
Question 8:
Note that from the equation of the system ˙x2 = u and ˙y = ˙x1 = x2.
Compute ˙˜x2 = ˙x2− ˙ˆx2 and show that ˜x2 it is independent of u, y and ˙y.
Response to question 8:
From the equation of the system, one gets ˙x2 = u and ˙y = x2. So
˙˜x2 =˙x2− ˙ˆx2
˙˜x2 =u−(−kˆx2 +kx2 +u)
˙˜x2 =−k˜x2
The dynamic equation of ˜x2 is an autonomous equation (no input). It is independent of u, y and ˙y.
Question 9:
What is the condition on k to assure that ˜x2 →0 when t →∞.
Response to question 9:
k should be positive.
3/7 Correction exam COLMS 17/12/2015
Exercise 3 :
This exercise is based on the same system than exercise 2.
Question 14:
Let us introduce ˜L(s) = 1+2ss2 . From the previous question one can check that L(itarget)(j ωTc) =−1+2jωω2 = ˜L(jω).
So the Nyquist and the Nichols plot ofL(itarget)(jω) and ˜L(jω) are exactly the same but parametrized by different angular
frequencies. From Matlab, one can obtain the stability margins of ˜L(s) : the gain margin is ˜mg =∞, and the phase margin
is ˜mϕ = 76,34°.
Deduce from this, the gain margin, mg(target), and the phase margin, mϕ(target), of L(itarget)(s).
Response to question 14:
Since the Nyquist and the Nichols plot of L(itarget)(jω) and ˜L(jω) are exactly the same, the gain margin and the phase
margin of ˜L(s) and L(itarget)(jω) are exactly the same.
Note that the angular frequencies associated to the points which defined the phases margin would be different; if ˜ωϕ is
the one linked to ˜L(jω), the one linked to L(itarget)(jω) is ωϕ(target) = ˜ωϕTc .
This comes from the fact that the point which characterizes the phase margin for ˜L(s) is ˜L(j˜ωϕ).
The point which characterizes the phase margin for L(itarget)(s)) is L(itarget)(jωϕ(target)).
These two point are equals ˜L(j˜ωϕ) = L(itarget)(jωϕ(target)). But the fact is that ˜L(j˜ωϕ) = L(itarget)(j˜ωϕTc ).
So ωϕ(target) = ˜ωϕTc .
An other property is that if the delay margin of ˜L(s) is ˜Md = ˜mϕ˜ωϕ , then the delay margin of L(itarget)(s) is
Md(target) = mϕ(target)ω
ϕ(target)
= ˜mϕ˜ωϕ
Tc
= ˜MdTc.
Also note that if one adds that ˜ωϕ = 2.0583 rad/s (obtained from Matlab), then ˜Md = ˜mϕ˜ωϕ = 76,34× pi1802.0583 = 0.65. So
˜Md = 0.65 s ( in second). And finally one deduces that Md(target) = 0.65Tc s ( in second).
Question 15:
Compute the virtual target controller K(target)(s) such that L(itarget)(s) = K(target)(s)G(s).
What is the type of this classical controller?
Response to question 15:
K(target)(s) = L(itarget)(s) 1G(s) = 1T2
c
(1 + 2Tcs). So K(target)(s) =−(f1 +f2s).
The control law associated to this framework would be u(t) =−f1y(t)−f2 ˙y(t)
Step 2 : the recovery step.
For the second step of LTRi, the idea here is to use the reduced observer obtained in exercise 2. The observer is
braceleftbigg˙z =−kz+u−k2y
ˆx2 = z+ky (6)
5/7 Correction exam COLMS 17/12/2015
Question 16:
For this question fill the three empty blocks of the following block diagram.
Response to question 16:
Question 17:
Explain why the expression of ˆx2 can be decomposed in two : ˆx2 = ˆx2u + ˆx2y, with
braceleftbigg˙z
u =−kzu +u
ˆx2u = zu
braceleftbigg˙z
y =−kzy−k2y
ˆx2y = zy +ky (7)
Response to question 17:
This is just the property of linearity. ˆx2 is a function of two input u and y the contribution of each one can be computed
separately.
Question 18:
For this question fill the four empty blocks of the following block diagram which is equivalent to the previous one.
Response to question 18:
Question 20:
Compute the transfer function Ty,ˆx2y of the state space representation
braceleftbigg˙z
y =−kzy−k2y
ˆx2y = zy +ky
Response to question 20:
Ty,ˆx2y = −k2s+k +k = −k2s+k + sk+k2s+k = sks+k.
Question 21:
Compute the limit when k→∞of the transfer function Tu,ˆx2u.
Response to question 21:
when k→∞Tu,ˆx2u = 1s+k →0.
Question 22:
Compute the limit when k→∞of the transfer function Ty,ˆx2y.
Response to question 22:
When k→∞Ty,ˆx2y = sks+k →s.
Question 23:
What is your final analysis about the limit of the global controller.
Response to question 23:
At the limit, when k → ∞, the global control law would be : u = f1x1 + f2sy + v and since x1 = y it would be
u = f1y+f2 ˙y+v which is the control law associated to the target controller computed in question 15 (the proportional
and derivative target controller).
For this example, the idea of LTRi can also be applied with a reduced observer.
Note that the same advantages and drawbacks than with full observer appear. The target loop has good stability margins
but it has also the problem of being inapplicable since a pure derivative controller is not feasible.