10.Stabilisation and state estimation

RO(s) =

RC(s) =

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435

det(sIn − A + `ct)

det(sIn − A + bft + `(ct − Dft)) RP (s) = ct(sIn − A)−1b + D

ftadj(sIn − A)` . det(sIn − A + `ct)

Then the following statements hold:

(i)the closed-loop characteristic polynomial for (10.21) is the product of the characteristic polynomials of A − bft and A − `ct.

(ii)the closed-loop system for (10.21) (i.e., the transfer function from r to y) is the same

transfer function for the interconnection of Figure 10.9, and what’s more, both transfer functions are exactly the transfer function for (A − bf, b, ct − Dft, 01).

Proof (i) The “A-matrix” for the observer/feedback system (10.21) is

Acl = A		−bft t			.
`ct	A − bf		− `ct
Let us define an invertible 2n × 2n matrix
T =		In −In			(10.22)
		0	In
= T −1		I	In
		= 0n	In .
It is a simple computation to show that
T AclT −1 =	A − `ct		0		t .
	`ct		A − bf

Thus we see that the characteristic polynomial of T AclT −1, and therefore the characteristic polynomial of Acl, is indeed the product of the characteristic polynomials of A − `ct and

A − bft, as claimed.

(ii) Let us use the new coordinates corresponding to the change of basis matrix T defined

in (10.22). Thus we take			A − bf
	`ct
A¯ cl =	A −	`ct	0		t
b¯cl = T	b = b
	b	0
c¯clt = ct		Dft	T −1	=	ct	ct		−	Dft
D¯ cl = D, −
		¯	¯	¯		t		¯
and determine the transfer function for Σ = (Acl, bcl, c¯cl							, Dcl). We have
(sI2n − A¯ cl)−1 = (sIn − A + `ct)−1						(sIn − A0+ bft)−1 ,

where “ ” denotes a term whose exact form is not relevant. One then readily computes

¯	t	¯	−1¯	t	t	t	−1	b.
TΣ(s) = c¯cl		(sI2n − Acl)	bcl = (c		− Df	)(sIn − A + bf )

436 10 Stabilisation and state estimation 22/10/2004

From this it follows that the transfer function of the closed-loop system (10.21) is as claimed. Now we determine this same transfer function in a di erent way, proceeding directly from

the closed-loop equations (10.21). First let us define

NC(s) = det(sIn − A + `ct),	DC(s) = det(sIn − A + bft + `(ct − Dft))
NP (s) = D det(sIn − A) + ctadj(sIn − A)b,	DP (s) = det(sIn − A)
NO(s) = ftadj(sIn − A)`,	DO(s) = det(sIn − A + `ct).

Now, the equation governing the observer states is
	˙		t	)xˆ(t) + (b − D`)u(t) + `y(t).
	xˆ	(t) = (A − `c
Taking left causal Laplace transforms gives
Let us define	xˆ(s) = (sIn − A + `ct)−1 (b − D`)ˆu(s) + `yˆ(s) .
T 1(s) = (sIn − A + `ct)−1(b − D`), T 2(s) = (sIn − A + `ct)−1`
so that
		xˆ(s) = T 1(s)ˆu(s) + T 2(s)ˆy(s).
Given that u(t) = r(t) − ftxˆ(t), we have
		uˆ(s) = rˆ(s) − ftT 1(s)ˆu(s) − ftT 1(s)ˆy(s)
	=	uˆ(s) = −	ftT 2(s)			1
					yˆ(s) +		rˆ(s).
			1 + ftT 1(s)			1 + ftT 1(s)

From the proof of Lemma A.3 we have that

1+ ftT 1(s) = 1 + ft(sIn − A + `ct)−1(b − D`)

=det 1 + ft(sIn − A + `ct)−1(b − D`)

=det(In + (sIn − A + `ct)−1(b − D`)ft)

=det (sIn − A + `ct)−1(sIn − A + `(ct − Dft) + bft

=det(sIn − A + `(ct − Dft) + bft)

det(sIn − A + `ct)

=DC(s). NC(s)

We also clearly have

ftT 2(s) = ft(sIn − A + `ct)−1`

=ftadj(In − A + `ct)` det(In − A + `ct)

=ftadj(In − A)` det(In − A + `ct)

=NO(s), DO(s)

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State estimation

437

where we have used Lemma A.5. Thus we have

uˆ(s) =

NO(s)

NC(s)

yˆ(s) +

NC(s)

rˆ(s).

(10.23)

DO(s) DC(s)

DC(s)

Also noting that

yˆ(s) =

NP (s)

uˆ(s),

(10.24)

DP (s)

we may eliminate rˆ(s) from equations (10.23) and (10.24) to get

T¯ (s) =

yˆ(s)

=

NC(s)NP (s)

,

(10.25)

Σ

rˆ(s)

DC

(s)DP (s) + NP (s)NO(s)

if we use the fact that DO = NC.

Let us now turn to the transfer function of the interconnection of Figure 10.9. The

transfer function may be computed in the usual manner using Mason’s Rule:
	yˆ(s)	=	RC(s)RP (s)					=	DO(s)NC(s)NP (s)		.
			1 + RC(s)RP (s)RO(s)						DC(s)DP (s)DO(s) + NC(s)NP (s)NO(s)
	rˆ(s)
Since DO = NC this may be simplified to
				yˆ(s)		=			NC(s)NP (s)	.
				rˆ(s)			DC	(s)DP (s) + NP (s)NO(s)

Comparing this to (10.25), we see that indeed the transfer function of Figure 10.9 is the same as that of the closed-loop system (10.21).

10.49 Remarks 1. One of the consequences of part (i) of the theorem is that one can separately choose the observer poles and the controller poles. This is a phenomenon that goes under the name of the separation principle.

2.The change of coordinates represented by the matrix defined in (10.22) can easily be seen as making a change of coordinates from (x, xˆ) to (e, x). It is not surprising, then, that in these coordinates we should see that the characteristic polynomial gets represented as a product of two characteristic polynomials. Indeed, we have designed the observer so that the error should be governed by the matrix A−`ct, and we have designed the state feedback so that the state should be governed by the matrix A − bft.

3.Note that for the interconnection of Figure 10.9 there must be massive cancellation in the transfer function since the system nominally has 3n states, but the denominator of the transfer function has degree n. Indeed, one can see that in the loop gain there is directly a cancellation of a factor det(sIn − A + `ct) from the numerator of RC

and the denominator of RO. What is not so obvious at first glance is that there is an additional cancellation of another factor det(sIn − A + `ct) that happens when forming the closed-loop transfer function. Note that these cancellations are all stable provided one chooses ` so that A − `ct is Hurwitz. Thus they need not be disastrous. That there should be this cancellation in the closed-loop equations (10.21) is not surprising since the control does not a ect the error. Thus the closed-loop system with the observer is not controllable (also see Exercise E10.23). One should also be careful that the characteristic

polynomial for the closed-loop system (10.21) is di erent from that for the interconnection Figure 10.9.

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Let us illustrate Theorem 10.48 via an example.

10.50 Example (Example 10.47 cont’d) Recall that we had

1	0		1
A = 0	−1	,	c = 0 .

Suppose that we want both observer poles and controller poles to be roots of the polynomial P (s) = s2 + 4s + 4. In Example 10.47 we found the required observer gain vector to be ` = (1, 2). To find the static state feedback vector we employ Ackermann’s formula from Proposition 10.13. The controllability matrix is

1	0
C(A, b) = 0	−1 .

Thus we compute

ft = 0 1 (C(A, b))−1P (A) = −1 2 .

This gives the closed-loop system matrix

		0	−1	0	0
Acl =		1	0	1	−2
	0		2	2	−4
		0	1	0	2
					−

for the interconnections of Figure 10.8. For the interconnection of Figure 10.8 we determine that the closed-loop input and output vectors are

00

bcl =	1	,	ccl =	1
	0			0 .

10

In Figure 10.10 we show the state x(t) and the estimated state xˆ(t) for the initial conditions x(0) = (1, 1) and xˆ(0) = (0, 0), and for u(t) = 1(t). Note that the quantities approach the same value as t → ∞, as should be the case as the observer is an asymptotic observer. In Figure 10.11 we show the output for the interconnection of Figure 10.8.

10.6Summary

1.Stabilisability and detectability extend the notions of controllability and observability. The extensions are along the lines of asking that the state behaviour that you cannot control or observe is stable.

2.Ackermann’s formula is available as a means to place the poles for a SISO linear system in a desired location.

				Exercises for Chapter 10							439
	2						2
	1.5						1.5
	1						1
x	0.5					ˆx	0.5
2						2
	0						0
	-0.5						-0.5
							-1 -0.5	0	0.5	1	1.5
	-1	-0.5	0	0.5	1	1.5		xˆ1
			x1

Figure 10.10 The state (left) and the estimated state (right) for the closed-loop system

	1.5
	1.25
	1
)	0.75
y(t
	0.5
	0.25
	0
	-0.25
	1	2	3	4	5	6	7	8
				t

Figure 10.11 The output for the interconnection of Figure 10.8

Exercises

E10.1 Show that (A, v) is stabilisable if and only if (At, v) is detectable.

E10.2 Show that Σ = (A, b, ct, D) is stabilisable if and only if the matrix

	has rank n for all s		sIn − A		b
			+.
		C
E10.3	Show that Σ = (A, b, ct, D) is detectable if and only if the matrix
			sInc−t	A
	has rank n for all s		+.
		C
E10.4	If (A, c) is detectable and if P Rn×n is positive-semidefinite and satisfies
			AtP + P A = −cct,			(E10.1)

440 10 Stabilisation and state estimation 22/10/2004

show that A is Hurwitz.

Hint: Show that (E10.1) implies that

Z t

P = eAttPeAt + eAtτ ccteAτ dτ.

0

In the next exercise, we will introduce the notion of a linear matrix inequality (LMI ). Such a relation is, in general, a matrix equation, invariant under transposition (i.e., if one takes the matrix transpose of the equation, it remains the same), for an unknown matrix. Since the equation is invariant under transposition, it makes sense to demand that the unknown matrix render the equation positive or negative-definite or semidefinite. In recent years, LMI’s have become increasingly important in control theory. A survey is [El Ghaoui and Niculescu 2000]. The reason for the importance of LMI’s is one can often determine their solvability using “convexity methods.” These are often numerically tractable. This idea forms the backbone, for example, of the approach to robust control taken by Dullerud and Paganini [1999].

E10.5 Consider a SISO linear system Σ = (A, b, ct, D).

(a)Use the Liapunov methods of Section 5.4 to show that Σ is stabilisable if and only if there exists f Rn so that the linear matrix inequality

(A − bft)P + P (At − fbt) < 0

has a solution P > 0.

(b)Use your result from part (a) to prove the following theorem.

Theorem For a stabilisable SISO linear system Σ = (A, b, ct, D), a state feedback vector f Rn stabilises the closed-loop system if and only if there exists P > 0 and g Rn so that

(i)f = P −1g and

(ii)so that the LMI

A b		P		+	P g		At		< 0
		gt					bt

is satisfied.

The point of the exercise is that the LMI gives a way of parameterising all stabilising state feedback vectors.

E10.6 Consider the two polynomials that have a common factor in C+:

P1(s) = s2 − 1, P2(s) = s3 + s2 − s − 1.

Construct three SISO linear systems Σi = (Ai, bi, cti, 01), i = 1, 2, 3, with the following properties:

1.TΣ = P1 , i = 1, 2, 3;

i P2

2.Σ1 is stabilisable but not detectable;

3.Σ2 is detectable but not stabilisable;

4.Σ3 is neither stabilisable nor detectable.

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441

E10.7 Let Σ = (A, b, ct, D) be a SISO linear system.

(a)If Σ is stabilisable show how, using Proposition 10.13, to construct a stabilising state feedback vector f in the case when Σ is not controllable.

(b)Show that if Σ is not stabilisable then it is not possible to construct a stabilising state feedback vector.

E10.8 Let (A, b) Rn×n × Rn be controllable and let P R[s] be a monic polynomial of degree n. Show that there exists a unique static state feedback vector f Rn with the property that P (s) = det(sIn − (A − bft)).

Hint: Refer to Exercise E2.38.

E10.9 Suppose that (A, b) Rn×n × Rn is controllable, and let P R[s] be a monic polynomial of degree n − 1.

(a)Show that there exists a static state feedback vector f Rn with the following properties:

1.A − bft possesses an invariant (n − 1)-dimensional subspace V Rn with

the property that for each basis {v1, . . . , vn−1} for V , {v1, . . . , vn−1, b} is a basis for Rn;

2.The characteristic polynomial of (A − bft)|V is P .

(b)For V as in part (a), define a linear map AV : V → V by AV (v) = prV ◦A(v), where prV : Rn → V is defined by

prV (a1v1 + · · · + an−1vn−1 + anb) = a1v1 + · · · + an−1vn−1,

for {v1, . . . , vn−1} a basis for V .

E10.10 Consider the pendulum/cart system of Exercises E1.5. In this problem, we shall change the input for the system from a force applied to the cart, to a force applied to the mass on the end of the pendulum that is tangential to the pendulum motion; see Figure E10.1. For the following cases,

x

θ

F

Figure E10.1 The pendulum/cart with an alternate input

(a)the equilibrium point (0, 0) with cart position as output;

(b)the equilibrium point (0, 0) with cart velocity as output;

(c)the equilibrium point (0, 0) with pendulum angle as output;

(d)the equilibrium point (0, 0) with pendulum angular velocity as output;

(e)the equilibrium point (0, π) with cart position as output;

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(f)the equilibrium point (0, π) with cart velocity as output;

(g)the equilibrium point (0, π) with pendulum angle as output;

(h)the equilibrium point (0, π) with pendulum angular velocity as output,

do the following:

1.obtain the linearisation Σ = (A, b, ct, D) of the system;

2.obtain the transfer function for Σ;

3.determine whether the system is observable and/or controllable;

4.determine whether the system is detectable and/or stabilisable.

E10.11 Consider the pendulum/cart system of Exercises E1.5 and E2.4. Construct a state feedback vector f that makes the linearisation of pendulum/cart system stable about the equilibrium point with the pendulum pointing “up.” Verify that the closed-loop system has the eigenvalues you asked for.

E10.12 Consider the double pendulum system of Exercises E1.6 and E2.5. For the following equilibrium points and input configurations, construct a state feedback vector f that makes the linearisation of double pendulum about that equilibrium point stable:

(a)the equilibrium point (0, π, 0, 0) with the pendubot input;

(b)the equilibrium point (π, 0, 0, 0) with the pendubot input;

(c)the equilibrium point (π, π, 0, 0) with the pendubot input;

(d)the equilibrium point (0, π, 0, 0) with the acrobot input;

(e)the equilibrium point (π, 0, 0, 0) with the acrobot input;

(f)the equilibrium point (π, π, 0, 0) with the acrobot input.

In each case, take the output to be the angle of the second link. Verify that the closed-loop system has the eigenvalues you asked for.

E10.13 Consider the coupled tank system of Exercises E1.11, E2.6. For the following equilibrium points and input configurations, construct a state feedback vector f that makes the linearisation of double pendulum about that equilibrium point stable:

(a)the output is the level in tank 1;

(b)the output is the level in tank 2;

(c)the output is the di erence in the levels.

In each case, take the output to be the angle of the second link. Verify that the closed-loop system has the eigenvalues you asked for.

E10.14 Let Σ = (A, b, ct, 01) with

0	1	0	0	· · ·	0
0	0	1	0	· · ·	0

A =		0 0 0 1 · · ·	0 .
	.. .. .. .. ...		..
		. . . .	.

0	0	0	0	·· ·· ··	0
0	0	0	0		1

Answer the following questions.

(a)Show that Σ is stabilisable by static output feedback only if all components of c are nonzero and of the same sign.

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443

(b)If A is as given, and if all components of c are nonzero and of the same sign, is it true that Σ can be stabilised using static output feedback? If so, prove it. If not, find a counterexample.

E10.15 (a) Show that the closed-loop transfer function for any one of the stabilising controllers from Theorem 10.37 is an a ne function of θ. That is to say, show that the closed-loop transfer function has the form R1θ + R2 for rational functions R1 and R2.

(b)Show that in the expression R1θ + R2 from part (a) that R1 RH+∞, and is strictly proper if and only if RP is strictly proper.

(c)Conclude that if

RC = ρ1 + θP2 ρ2 − θP1

for an admissible θ then the unity gain feedback loop with RL = RCRP is well-posed.

E10.16	Let RP RH∞+ be a stable plant transfer function. Show that the set of controllers
	RC for which the closed-loop system in Figure 10.4 is given by
		θ		θ RH∞+ .
	1 − θRP
				1
E10.17	Consider the plant transfer function RP (s) =					, and consider the interconnection
					s2
	of Figure 10.4. Show that there is a rational function RC(s), not of the form given

in Theorem 10.37, for which the closed-loop system is IBIBO stable.
Hint: Consider PD control.
E10.18 Exercise on restricted domain for poles.	Finish

In the next exercise you will demonstrate what is a useful feature of RH+∞. Recall that a ring is a set R with the operations of addition and multiplication, and these operations should satisfy the “natural” properties associated with addition and multiplication (see [Lang 1984]). A ring is commutative when the operation of multiplication is commutative (as for example, with the integers). A commutative ring is an integral domain when the product of nonzero elements is nonzero. An ideal in a ring R is a subset I for which

1.0 I,

2.if x, y I then x + y I, and

3.if x I then ax I and xa I for all a R. An ideal I is principal if it has the form

I = {xa | a R} .

The ideal above is said to be generated by x. An integral domain R is a principal ideal domain when every ideal I is principal.

E10.19 Answer the following questions.

(a)Show that RH+∞ is an integral domain.

(b)Show that RH+∞ is a principal ideal domain.

(c)Show that H+∞ is not a principal ideal domain.

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Normally, only condition 1 is given in the definition of admissibility for a function θ RH+∞ to parameterise Spr(RP ). This is a genuine omission of hypotheses. In the next exercise, you will show that, in general, the condition 2 also must be included.

E10.20 Consider the plant transfer function

s RP (s) = s + 1.

For this plant, do the following.

(a) Show that (P1(s), P2(s)) = (s+1s , 1) is a coprime fractional representative for RP .

(b) Show that

(ρ1(s), ρ2(s)) = −	s + 1		,		(s + 1)(s + 4)
	(s + 2)2				(s + 2)2
is a coprime factorisation of (P1, P2).
(c) Define θ RH∞+ by		s
θ(s) =				.
	s + 1
Is θ admissible?

(d) For the function θ RH+∞ from part (c), show that the controller

RC = ρ1 + θP2 ρ2 − θP1

is improper. Thus for the conclusions of Theorem 10.37 to hold, in particular, for the controller parameterisation to be one for proper controller transfer functions, the condition 2 for admissibility is necessary.

(e)Show that despite our conclusions of part (d), the unity gain interconnection with loop gain RL = RCRP , using RC from part 2, is IBIBO stable.

E10.21 Consider the plant

s

RP (s) = (s + 1)(s − 1).

For this plant, do the following.

(a)Find a coprime fractional representative (P1, P2) for RP (choose the obvious one, if you want to make life easy).

(b)Show that

2(s + 2)

RC(s) = s − 12

stabilises RP .

(c)Use the stabilising controller from the previous step to construct a coprime factorisation (ρ1, ρ2) for (P1, P2).

(d)Write the expression for the set of proper stabilising controllers for RP depending on the parameter θ RH+∞.

(e)For two di erent values of θ, produce the Nyquist plots for the loop gain RCRP and comment on the properties of the controller you have chosen.

E10.22 Let Σ = (A, b, ct, 01) be a SISO linear system.

Exercises for Chapter 10

445

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(a)Show that for a Luenberger observer with observer gain vector ` the state and the estimated state together satisfy the vector di erential equation

x˙ (t)	=	A
˙		`c	t
xˆ(t)

A − `ct	xˆ(t)	+	b u(t).
0	x(t)		b

(b)What is the di erential equation governing the behaviour of x(t) and the error e(t) = x(t) − xˆ(t).

E10.23 Verify that the system (10.21) is not controllable.

Hint: Consider the system in the basis defined by the change of basis matrix of (10.21).