02.State-space representations (the time-domain)

Let’s see how this meshes with what we said following Definition 2.20. Suppose we start at x(0) = (0, 0). Since any solution to

which has initial condition x1(0) = 0 will have the property that x1(t) = 0 for all t, the control system is essentially governed by the x2 equation:

x˙2 = −x2 + u.

Therefore we can only move in the x2-direction and all points with x1 6= 0 will not be accessible to us. This is what we mean by controllability. You might note that the set of points reachable are those in the columnspace of the matrix C(A, b).

Based on the above discussion, we say that a triple (A, b, c) Rn×n × Rn × Rn is complete if (A, b) is controllable and if (A, c) is observable.

We have a property of the controllability matrix that is sort of like that for the observability matrix in Theorem 2.17.

2.23 Theorem The columnspace of the matrix C(A, b) is the smallest A-invariant subspace containing b.

Proof Obviously b is in the columnspace of C(A, b). We will show that this columnspace is A-invariant. Let x be in the columnspace of C(A, b), i.e., in the range of the linear map C(A, b). Then there is a vector y Rn with the property that

x = C(A, b)y = b Ab · · · An−1b y.

We then have

Ax = Ab A2b · · · Anb y.

The result will follow if we can show that each of the vectors

Ab, . . . , Anb

is in the columnspace of C(A, b). It is clear that

Ab, . . . , An−1b

are in the columnspace of C(A, b). By the Cayley-Hamilton Theorem we have

Anb = −pn−1An−1b − · · · − p1Ab − p0b,

which shows that Anb is also in the columnspace of C(A, b).

Now we show that if V is an A-invariant subspace with b V then V contains the

There is a somewhat subtle thing happening here that should be understood. If a pair (A, b) is controllable, this implies that one can steer between any two points in Rn with a suitable control. It does not mean that one can follow any curve in Rn that connects the two points. This then raises the question, “What curves in Rn can be followed by solutions of the di erential equation

x˙ (t) = Ax(t) + bu(t)?”

Let us explore the answer to this question, following Basile and Marro [1968]. Because we will deal only with the single-input case, things are somewhat degenerate. Let T (A, b) Rn be the subspace

(

span {b} , b is an eigenvector for A

T (A, b) =

{0}, otherwise.

The following lemma asserts the essential property of the subspace T (A, b).

2.24 Lemma T (A, b) is the largest subspace of Rn with the property that

A(T (A, b)) + T (A, b) span {b} .

the case that Ab = λb for some λ R. It then follows that if x1 = a1b and x2 = a2b for a1, a2 R then

Ax1 + x2 = (a1λ + a2)b span {b} .

Now we need to show that T (A, b) is the largest subspace with this property. Suppose that V is a subspace of Rn with the property that A(V ) + V span {b}. Thus for each x1, x2 V we have

Ax1 + x2 span {b} .

In particular, if we choose x1 = 0 we see that if x2 V then x2 span {b}. Similarly, if x2 = 0 we see that if x1 V then Ax1 span {b}. Thus we have shown that if V is a subspace with the property that A(V ) + V span {b}, this implies that

V = span {b} ∩ A−1(span {b})

where

A−1(span {b}) = {v Rn | Av span {b}}

(note that we are not saying that A is invertible!). It now remains to show that T (A, b) = span {b} ∩ A−1(span {b}). We consider the two cases where (1) b is an eigenvector for A and (2) b is not an eigenvector for A. In the first case, b A−1(span {b}) so we clearly have

span {b} ∩ A−1(span {b}) = span {b} .

In the second case, b 6 A−1(span {b}) so that

span {b} ∩ A−1(span {b}) = {0}.

Now we can use this lemma to describe the set of curves in Rn that can be followed exactly by our control system.

2.25 Proposition Let Σ = (A, b, ct, 01) be a SISO linear system and let T (A, b) Rn be the subspace defined above. If I R is an interval and if r: I → T (A, b) is continuously di erentiable, then there exists a continuous function u: I → R with the property that

r˙ (t) = Ar(t) + bu(t).

Furthermore, T (A, b) is the largest subspace of Rn with this property.

Proof For the first part of the proposition, we note that if r: I → T (A, b) then

r˙ (t) = lim r(t + τ) − r(t) T (A, b)

τ→0 τ

since r(t + τ), r(t) T (A, b). Therefore, by Lemma 2.24,

r˙ (t) − Ar(t) T (A, b), t I.

Therefore, for each t I there exists u(t) R so that

r˙ (t) − Ar(t) = u(t)b.

The first part of the proposition now follows since the T (A, b)-valued function of t, r˙ (t) − Ar(t) is continuous.

Now suppose that V is a subspace of Rn with the property that for every continuously di erentiable r: I → V , there exists a continuous function u: I → R with the property that

r˙ (t) = Ar(t) + bu(t).

Let x1, x2 V and define r: R → V by −x1+tx2. Then we have r(0) = −x1 and r˙ (0) = x2. Therefore there must exist a continuous u: R → R so that

x2 = −Ax1 + bu(0).

Since this construction can be made for any x1, x2 V , we must have Ax1 + x2 span {b} for every x1, x2 V . By Lemma 2.24 this means that V = T (A, b).

Thus we see for single-input systems, the state trajectories we may exactly follow are actually quite limited. Nevertheless, even though one cannot follow all state trajectories, it is possible for a system to be controllable.

2.26 Remark As was the case for observability in Remark 2.18, it is easy to talk about controllability in the MIMO setting. Indeed, if for a MIMO system Σ = (A, B, C, D) we define

that we must have ourselves covered. Surely if a system is complete then our state-space behaviour will be nice if the output is nice. But this is in fact not true, as the following example shows.

First, let’s see that the system is observable and controllable. The respective matrices are

In this example, we do not use a step input, but rather a violent input:

(

et, t ≥ 0

u(t) =

0, otherwise.

Thus our input blows up as time increases. The usual calculations, using zero initial conditions, give

Thus the output is behaving nicely (see Figure 2.9) while the state is blowing up to infinity

Figure 2.9 The output response of (2.9) to an exponential input

Things are a bit more subtle with this example. The problem is that the large input is not being transmitted to the output. Describing the general scenario here is not altogether easy, but we work through it so that you may know what is going on.

2.29 Example (Example 2.27 cont’d) We shall go through the algorithm step by step.

1.We take V0 = R2 as directed.

2.As per the instructions, we need to compute ker(ct) and we easily see that

ker(ct) = span {(1, 1)} .

Now we compute

since Z0 = R2. Therefore Z1 = ker(ct). To compute Z2 we compute

since ker(ct) and span {b} are complementary subspaces. Therefore Z2 = ker(ct) and so our sequence terminates at Z1.

3. We have

ZΣ = ker(ct) = span {(1, 1)} .

4. Let f = (f1, f2). We compute

In order that this matrix leave ZΣ invariant, it must map the basis vector (1, 1) for ZΣ to a multiple of itself. We compute

In order that this vector be a multiple of (1, 1) we must have f1 +f2 −5 = 1 or f1 +f2 = 6. Let us choose f1 = f2 = 3.

5.We choose the basis {v1 = (1, 1), v2 = (1, −1)} for R2, noting that v1 is a basis for ZΣ.

6.We compute

Ab,f (v1) = (1, 1) = 1v1 + 0v2

Ab,f (v2) = (−1, 1) = 0v1 − 1v2.

Okay, so how is our bad behaviour reflected here? Well, note that the zero dynamics are unstable! This, it turns out, is the problem.

2.30 Remarks 1. Systems with stable zero dynamics (i.e., all eigenvalues for the matrix NΣ have nonpositive real part) are sometimes called minimum phase systems. Note that the response Figure 2.9 shows an output that initially does something opposite from what it ends up eventually doing—the output decreases before it finally increases to its final value. This, it turns out, is behaviour typical of a system that is not minimum phase. We shall be investigating the properties of nonminimum phase systems as we go along (see Theorem 3.15).

2.The zero dynamics as we construct them are not obviously independent of the choices made in the algorithm. That is to say, it is not clear that, had we chosen a di erent vector f, or a di erent basis {v1, . . . , vn}, that we would not arrive at an utterly di erent

matrix NΣ. Nonetheless, it is true that if we were to have chosen a di erent vector f and the same basis {v1, . . . , vn} that the resulting matrix NΣ would be unchanged. A choice of a di erent basis would only change the matrix NΣ by a similarity transformation, and

Proof By the inductive procedure of Algorithm 2.28 it is clear that ZΣ Z . We then need only show that ZΣ is the largest subspace in Z . Let V Z and let x V . This means that x ker(ct) and so x Z1 (since in Algorithm 2.28 we always have Z1 = ker(ct)). We also have

Ax V + span {b} Z1 + span {b} .

ground here with a few simple examples, and some general definitions. The material in this section has touched upon some fairly deep concepts in linear control theory, and a recap is probably a good idea. Let us outline the three things we have found that can go wrong, and just how they go wrong.

1.Unseen unstable dynamics due to lack of observability: This was illustrated in Example 2.13. The idea was that any input we gave the system leads to a nice output. However, some inputs cause the states to blow up. The problem here is that lack of observability causes the output to not recognise the nasty state-space behaviour.

2.Lurking unstable dynamics caused by lack of controllability: It is possible, as we saw in Example 2.19, for the dynamics to be unstable, even though they are fine for some initial conditions. And these unstable dynamics are not something we can get a handle on with our inputs; this being the case because of the lack of controllability.

3.Very large inputs can cause no output due to the existence of unstable zero dynamics:

This is the situation illustrated in Example 2.27. The problem here is that all the input energy can be soaked by the unstable modes of the zero dynamics, provided the input is of the right type.

It is important to note that if we have any of the badness of the type listed above, there ain’t nothing we can do about it. It is a limitation of the physical system, and so one has to be aware of it, and cope as best one can.

We shall see these ideas arise in various ways when we discuss transfer functions in Chapter 3. As we say, the connection here is a little deep, so if you really want to see what is going on here, be prepared to invest some e ort—it is really a very neat story, however.

2.4 The impulse response

Exercise E3.1.

2.4.1 The impulse response for causal systems Typically, we will use the impulse response in situations where we are interested in positive times. Thus we consider everything before t = 0 to be zero, and then at t = 0 the action starts. It is this “standard” situation we deal with in this section.

We wish to determine the output when we start with zero initial condition, and at t0 = 0 give the system a sharp “jolt.” Let us argue intuitively for a moment. Our input will be zero except for a short time near t = 0 where it will be large. One expects, therefore, that the integration over τ in (2.10) will only take place over a very small interval near zero. Outside this interval, u will vanish. With this feeble intuition in mind, we define the causal

More succinctly, we may write h+Σ(t) = 1(t)cteAtb, where 1(t) is the unit step function. In the next section we will define h−Σ. However, since we shall almost always be using h+Σ, let us agree to simply write hΣ for h+Σ, resorting to the more precise notation only in those special circumstances where we need to be clear on which version of the impulse response we need. The idea is that the only contribution from u in the integral is at τ = 0. A good question is “Does there exist u U so that the resulting output is hΣ(t) with zero state initial condition?” The answer is, “No there is not.” So you will never see the impulse response if you only allow yourself piecewise continuous inputs. In fact, you can allow inputs that are a whole lot more general than piecewise continuous, and you will still not ever see the impulse response. However, the impulse response is still an important ingredient in looking at the input/output behaviour of the system. The following trivial result hints at why this is so.

2.32 Proposition For any u U the output of the system

x˙ (t) = Ax(t) + bu(t)

y(t) = ctx(t)

with the initial condition x = x0 is

Z t

y(t) = cteAtx0 + hΣ(t − τ)u(τ) dτ.

0

That is to say, from the impulse response one can construct the solution associated with any input by performing a convolution of the input with the impulse response. This despite the fact that no input in U will ever produce the impulse response itself!

We compute the impulse response for the mass-spring-damper system.

Since the nature of eAt changes character depending on the choice of m, d, and k, let’s choose specific numbers to compute the impulse response. In all cases we take m = 1. We also have the two cases of output to consider (we do not in this section consider the case when D 6= 01).

1. We first take d = 3 and k = 2. The matrix exponential is then

(a)We first consider the case when c = (1, 0), i.e., when the output is displacement. The impulse response is then

hΣ(t) = 1(t)(e−t − e−2t)

which we show in the left plot in Figure 2.11.

2.4.2 The impulse response for anticausal systems Now we turn our attention to a situation that we will only have need to resort to in Section 15.3; the situation is one where we deal with functions of time that end at t = 0. Thus functions are defined on the interval (−∞, 0]. The definition of the impulse response in these cases has the same motivation as in the causal case. We shall use Theorem 2.34 for our motivation. For > 0, let us define

We then define

Z 0

y (t) = cteA(t−τ)bu (τ) dτ, t ≤ 0,

t

and then h−Σ = lim →0 y . Let us determine the expression for h−Σ be performing the computations carried out in the proof of Theorem 2.34, but now for t ≤ 0:

Z 0

y (t) = cteA(t−τ)bu (τ) dτ

t

This may be written as h−Σ(t) = −1(−t)cteAtb, which we call the anticausal impulse response.

The anticausal impulse response is useful for solving a final value problem, as the following result states.

2.35 Proposition Let Σ = (A, b, ct, 01) be a SISO linear system. If u: (−∞, 0] → R is admissible then the output of the final value problem

x˙ (t) = Ax(t) + Bu(t) x(0) = x0

y(t) = ctx(t),

is given by

Z 0

y(t) = cteAtx0 + h−Σ(t − τ)u(τ) dτ, t ≤ 0.

t

Proof The result will follow if we can show that the solution to the final value problem

x˙ (t) = Ax(t) + bu(t) x(0) = x0

54 2 State-space representations (the time-domain) 22/10/2004

is given by

Z 0

x(t) = eAtx0 − eA(t−τ)bu(τ) dτ, t ≤ 0.

t

Clearly the final condition x(0) = x0 is satisfied. With x(t) so defined, we compute

Z 0

x˙ (t) = AeAtx0 + bu(t) − AeA(t−τ)bu(τ) dτ

t

Z0

=AeAtx0 − A eA(t−τ)bu(τ) dτ + bu(t)

t

= Ax(t) + bu(t).

Thus the anticausal impulse response acts for anticausal inputs in much the same way as the causal impulse response acts for causal inputs.

Note again that we shall only rarely require h−Σ, so, again, whenever you see hΣ, it implicitly refers to h+Σ.

2.5 Canonical forms for SISO systems

In this section we look at the appearance of a “typical” SISO linear system of the form (2.2). To do so, we shall take an arbitrary system of that form and make a linear change of coordinates. So let us first make sure we understand what is a linear change of coordinates, and how it manifests itself in the multi-input, multi-output system equations (2.1). We take as our state coordinates x, and define new state coordinates ξ = T x where T is an invertible n × n matrix.2 We can easily derive the equations that govern the behaviour of the state variables ξ. The following result holds in the MIMO case.

One may consider more general changes of variable where one defines η = Q−1y and µ = R−1u, but since our interest is mainly in the SISO case, such transformations simply boil down to scaling of the variables, and so constitute nothing of profound interest.

then ξ = T x are exactly the components of x Rn in the basis {f1, . . . , fn}.

2.5.1 Controller canonical form We now revert to the SISO setting, and prove a “normal form” result for controllable SISO linear systems. Recall that a pair (A, b) Rn×n × Rn is controllable if the vectors {b, Ab, A2b, . . . , An−1b} form a basis for Rn.

2.37 Theorem If (A, b) Rn×n × Rn is controllable then there exists an invertible n × n matrix T with the property that

Proof We begin with some seemingly unrelated polynomial constructions. Let the characteristic polynomial of A be

PA(λ) = λn + pn−1λn−1 + · · · + p1λ + p0.

Define n + 1 polynomials in indeterminant λ by

n−i

X

Pi(λ) = pk+iλk, i = 0, . . . , n.

k=0

Note that P0 = PA and Pn(λ) = 1 if we declare that pn = 1. These polynomials satisfy the relation

k=−1

n−(i−1)

=X pk0 +(i−1)λk0 − pi−1Pn(λ)

k0 =0

= Pi−1(λ) − pi−1Pn(λ),

as asserted.

Now we define n + 1 vectors by

fi = Pi(A)b, i = 0, . . . , n.

Note that Pi(A) is simply given by

n−i

X

Pi(A) = pk+iAk, i = 0, . . . , n.

k=0

56 2 State-space representations (the time-domain) 22/10/2004

By the Cayley-Hamilton Theorem, f0 = 0, and we claim that the vectors {f1, . . . , fn} are linearly independent. To see this latter assertion, note that since (A, b) is controllable the

which is clearly invertible. Therefore {f1, . . . , fn} are themselves linearly independent and so form a basis.

We define the matrix T by asking that its inverse be given by

from which we get the representation of A in the basis {f1, . . . , fn} as in the theorem statement. It is trivially true that the coordinates of b in this basis are (0, 0, . . . , 0, 1) since fn = b.

The pair (T AT −1, T b) of the theorem are sometimes called the controller canonical form for the pair (A, b). This is also sometimes known as the second LuenbergerBrunovsky canonical form for (A, b).

What is the import of this theorem? Well, let us suppose that we are handed an n × n matrix A and an n-vector b in the form of that in the theorem statement:

What does the system look like, really? Well, define a scalar variable x by x = x1. We then note that if x˙ (t) = Ax(t) + bu(t) then

x˙ = x˙1 = x2

x¨ = x¨1 = x˙2 = x3

.

.

.

x(n) = x˙n = −p0x1 − p1x2 − · · · − pn−1xn = − p0x − p1x˙ − · · · − pn−1x(n−1) + u.

22/10/2004 2.5 Canonical forms for SISO systems 57

Thus the vector equation x˙ (t) = Ax(t) + bu(t) reduces to the scalar equation x(n)(t) + pn−1x(n−1)(t) + · · · + p1x(1)(t) + p0x(t) = u(t).

Therefore, when we study the controllable SISO linear system

x˙ (t) = Ax(t) + bu(t)

(2.12)

y(t) = ctx(t) + Du(t),

we can always make a change of coordinates that will render the system an nth order one whose state variable is a scalar. This is important. It is also clear that if one conversely starts with a scalar system

1cn−1

by rescaling variables (for example, if d = 0 we may choose u˜(t) = bu(t)).

We shall see that for controllable SISO systems, we may move easily from the linear systems formulation, i.e., equation (2.2), to the scalar equation formulation in all settings we examine, and here we have provided the time-domain setting for making this change. We look at alternative canonical forms for controllable pairs (A, b) in Exercises E2.31 E2.32, and E2.33.

2.5.2 Observer canonical form Now let us focus on the situation when the system is observable. The proof here is simpler than for the controllable case, since we use a “duality” between controllability and observability.

2.38 Theorem If (A, c) Rn×n ×Rn is observable then there exists an invertible n×n matrix T so that

58 2 State-space representations (the time-domain) 22/10/2004

Proof We make the simple observation that (A, c) is observable if and only if (At, c) is controllable. This follows from the easily seen fact that C(A, c) = O(At, c)t. Therefore, by

The pair (T AT −1, T −tc) in the theorem statement are said to be in observer canonical form or in second Luenberger-Brunovsky canonical form

Let us look at the value of this canonical form by expression the system equations for a system that has this form. The di erential equation x˙(t) = Ax(t) + bu(t) reads

x˙1 = − p0xn = b0u

x˙2 = x1 − p1xn = −p0x−p1xn = b1u

.

.

.

x˙n = xn−1 − pn−1xn = bn−1u,

where we write b = (b0, b1, . . . , bn−1). Now we di erentiate the expression for x˙n with respect to t n − 1 times, and use the equations for x˙1, . . . , x˙n to get the equation

x(nn) + pn−1x(nn−1) + · · · + p1x(1)n + p0xn = bn−1u(n−1) + · · · + b1u(1) + b0u.

The equation y = ctx + Du simply reads y = xn + Du. The upshot, therefore, is that for an observable system one can always make a change of coordinates so that the system is e ectively described by the equation

y(n) + pn−1y(n−1) + · · · + p1y(1) + p0y = bnu(n) + bn−1u(n−1) + · · · + b1u(1) + b0u,

where bn is defined by D = [bn]. Thus an observable system can be immediately put into the form of a di erential equation for the output in terms of the output and its derivatives. This is an essential observation for the discussion of input/output systems that is initiated in Section 3.4. As with controllable pairs, in the exercises (See E2.34, E2.35, and E2.36) we provide alternate canonical forms for observable pairs.

seen, for systems that are either controllable or observable, it is possible to find a set of coordinates in which the system looks simple, in some sense. Now let us address the situation when we know that the system is not both controllable and observable.

First we consider the situation when (A, b) is not controllable. The following result expresses the “simplest” form such a pair may take.

2.39 Theorem If (A, b) Rn×n × Rn is not controllable then there exist an invertible matrix T and a positive integer ` < n with the property that

Proof Let V be the smallest A-invariant subspace containing b, and suppose that dim(V ) = `. Since (A, b) is not controllable, by Theorem 2.23, ` < n. Choose a basis {v1, . . . , vn} for Rn with the property that {v1, . . . , v`} is a basis for V . We define T so that

T −1 = v1 · · · vn .

Since V is A-invariant and since b V , the relations in (2.13) must hold. Note that A11 is simply the representation in the basis {v1, . . . , v`} of the restriction of A to V , and that b1 is the representation of b V in this same basis. Now we look at the final assertion. By the very definition of V , the pair (A11, b1) is controllable. Therefore, by Theorem 2.37 there exists an ` × ` invertible matrix T t so that

Now let us do the same thing, except now we look at the situation when (A, c) is not observable.

2.40 Theorem If (A, c) Rn×n × Rn is not observable then there exist an invertible matrix T and a positive integer k < n with the property that

Furthermore, T may be chosen so that (A11, c1) Rk×k ×Rk are in observer canonical form.