05.Stability of control systems

This version: 22/10/2004

Chapter 5

Stability of control systems

We will be trying to stabilise unstable systems, or to make an already stable system even more stable. Although the essential goals are the same for each class of system we have encountered thus far (i.e., SISO linear systems and SISO linear systems in input/output form), each has its separate issues. We first deal with these. In each case, we will see that stability boils down to examining the roots of a polynomial. In Section 5.5 we give algebraic criteria for determining when the roots of a polynomial all lie in the negative complex plane.

Contents

5.1 Internal stability

Internal stability is a concept special to SISO linear systems i.e., those like

x˙ (t) = Ax(t) + bu(t)

(5.1)

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

Internal stability refers to the stability of the system without our doing anything with the controls. We begin with the definitions.

5.1Definition A SISO linear system Σ = (A, b, ct, D) is

(i)internally stable if

lim sup kx(t)k < ∞

t→∞

for every solution x(t) of x˙ (t) = Ax(t);

(ii) internally asymptotically stable if

lim kx(t)k = 0

t→∞

for every solution x(t) of x˙ (t) = Ax(t);

Of course, internal stability has nothing to do with any part of Σ other than the matrix A. If one has a system that is subject to the problems we discussed in Section 2.3, then one may want to hope the system at hand is one that is internally stable. Indeed, all the bad behaviour we encountered there was a direct result of my intentionally choosing systems that were not internally stable—it served to better illustrate the problems that can arise.

Internal stability can almost be determined from the spectrum of A. The proof of the following result, although simple, relies on the structure of the matrix exponential as we discussed in Section B.2. We also employ the notation

With this we have the following result, recalling notation concerning eigenvalues and eigenvectors from Section A.5.

5.2 Theorem Consider a SISO linear system Σ = (A, b, ct, D). The following statements hold.

(i)Σ is internally unstable if spec(A) ∩ C+ 6= .

(ii)Σ is internally asymptotically stable if spec(A) C−.

(iii)Σ is internally stable if spec(A) ∩ C+ = and if mg(λ) = ma(λ) for λ spec(A) ∩ (iR).

(iv)Σ is internally unstable if mg(λ) < ma(λ) for λ spec(A) ∩ (iR).

22/10/2004 5.1 Internal stability 163

Proof (i) In this case there is an eigenvalue α + iω C+ and a corresponding eigenvector u + iv which gives rise to real solutions

x1(t) = eαt(cos ωtu − sin ωtv), x2(t) = eαt(sin ωtu + cos ωtv).

Clearly these solutions are unbounded as t → ∞ since α > 0.

(ii) If all eigenvalues lie in C−, then any solution of x˙ (t) = Ax(t) will be a linear combination of n linearly independent vector functions of the form

for α > 0. Note that all such functions tend in length to zero as t → ∞. Suppose that we have a collection x1, . . . , xn(t) of such vector functions. Then, for any solution x(t) we have, for some constants c1, . . . , cn,

lim kx(t)k = lim kc1x1(t) + · · · + cnxn(t)k

t→∞ t→∞

≤ |c1| tlim→∞ kx1(t)k + · · · + |cn| tlim→∞ kxn(t)k

= 0,

where we have used the triangle inequality, and the fact that the solutions x1(t), . . . , xn(t) all tend to zero as t → ∞.

(iii) If spec(A) ∩ C+ = and if further spec(A) C−, then we are in case (ii), so Σ is internally asymptotically stable, and so internally stable. Thus we need only concern ourselves with the case when we have eigenvalues on the imaginary axis. In this case, provided all such eigenvalues have equal geometric and algebraic multiplicities, all solutions will be linear combinations of functions like those in (5.2) or functions like

x`+1(t), . . . , xn(t) be linearly independent functions of the form (5.3) so that x1, . . . , xn forms a set of linearly independent solutions for x˙ (t) = Ax(t). Thus we will have, for some constants c1, . . . , cn,

Since each of the terms kx`+1(t)k , . . . , kxn(t)k are bounded, their lim sup’s will exist, which is what we wish to show.

(iv) If A has an eigenvalue λ = iω on the imaginary axis for which mg(λ) < ma(λ) then there will be solutions of x˙ (t) = Ax(t) that are linear combinations of vector functions of

5.3Remarks 1. A matrix A is Hurwitz if spec(A) C−. Thus A is Hurwitz if and only if Σ = (A, b, ct, D) is internally asymptotically stable.

2.We see that internal stability is almost completely determined by the eigenvalues of A. Indeed, one says that Σ is spectrally stable if A has no eigenvalues in C+. It is only in the case where there are repeated eigenvalues on the imaginary axis that one gets to distinguish spectral stability from internal stability.

3.One does not generally want to restrict oneself to systems that are internally stable. Indeed, one often wants to stabilise an unstable system with feedback. In Theorem 6.49 we shall see, in fact, that for controllable systems it is always possible to choose a

The notion of internal stability is in principle an easy one to check, as we see from an example.

5.4 Example We look at a SISO linear system Σ = (A, b, ct, D) where

The form of b, c, and D does not concern us when talking about internal stability. The

eigenvalues of A are the roots of the characteristic polynomial s2 + as + b, and these are

√

−a2 ± 12 a2 − 4b.

The situation with the eigenvalue placement can be broken into cases.

1.a = 0 and b = 0: In this case there is a repeated zero eigenvalue. Thus we have spectral stability, but we need to look at eigenvectors to determine internal stability. One readily verifies that there is only one linearly independent eigenvector for the zero eigenvalue, so the system is unstable.

2.a = 0 and b > 0: In this case the eigenvalues are purely imaginary. Since the roots are also distinct, they will have equal algebraic and geometric multiplicity. Thus the system is internally stable, but not internally asymptotically stable.

3.a = 0 and b < 0: In this case both roots are real, and one will be positive. Thus the system is unstable.

4.a > 0 and b = 0: There will be one zero eigenvalue if b = 0. If a > 0 the other root will be real and negative. In this case then, we have a root on the imaginary axis. Since it is distinct, the system will be stable, but not asymptotically stable.

5.a > 0 and b > 0: One may readily ascertain (in Section 5.5 we’ll see an easy way to do this) that all eigenvalues are in C− if a > 0 and b > 0. Thus when a and b are strictly positive, the system is internally asymptotically stable.

6.a > 0 and b < 0: In this case both eigenvalues are real, one being positive and the other negative. Thus the system is internally unstable.

7.a < 0 and b = 0: We have one zero eigenvalue. The other, however, will be real and positive, and so the system is unstable.

Note that one cannot really talk about internal stability for a SISO linear system (N, D) in input/output form. After all, systems in input/output form do not have built into them a notion of state, and internal stability has to do with states. In principle, one could define the internal stability for a proper system as internal stability for ΣN,D, but this is best handled by talking directly about input/output stability which we now do.

5.2 Input/output stability

We shall primarily be interested in this course in input/output stability. That is, we want nice inputs to produce nice outputs. In this section we demonstrate that this property is intimately related with the properties of the impulse response, and therefore the properties of the transfer function.

5.2.1 BIBO stability of SISO linear systems We begin by talking about input/output stability in the context of SISO linear systems. When we have understood this, it is a simple matter to talk about SISO linear systems in input/output form.

5.5 Definition A SISO linear system Σ = (A, b, ct, D) is bounded input, bounded output stable (BIBO stable) if there exists a constant K > 0 so that the conditions (1) x(0) = 0 and (2) |u(t)| ≤ 1, t ≥ 0 imply that y(t) ≤ K where u(t), x(t), and y(t) satisfy (5.1).

Thus BIBO stability is our way of saying that a bounded input will produce a bounded output. You can show that the formal definition means exactly this in Exercise E5.8.

The following result gives a concise condition for BIBO stability in terms of the impulse response.

that either (1) hΣ˜ (t) blows up exponentially as t → ∞ or that (2) hΣ˜ (t) is a sum of terms, one of which is of the form sin ωt or cos ωt. For the first case we can take the bounded input u(t) = 1(t). Using Proposition 2.32 and Proposition 3.24 we can then see that

Z ∞

y(t) = h˜ (t − τ) dτ + Du(t).

Σ

0

Since hΣ˜ (t) blows up exponentially, so too will y(t) if it is so defined. Thus the bounded input u(t) = 1(t) produces an unbounded output. For case (2) we choose u(t) = sin ωt and compute

Therefore, y(t) will be unbounded for the bounded input u(t) = sin ωt. We may then conclude that Σ is not BIBO stable.

Now suppose that limt→∞ |hΣ˜ (t)| = 0. By Proposition 3.24 this means that hΣ˜ (t) dies o exponentially fast as t → ∞, and therefore we have a bound like

Z ∞

|h˜ (t − τ)| dτ ≤ M

Σ

≤ |h˜ (t − τ)u(τ)| dτ + |D|

Σ

0

Z t

≤ |h˜ (t − τ)| |u(τ)| dτ + |D|

Σ

0

Z t

≤ |h˜ (t − τ)| dτ + |D|

Σ

0

≤ M + |D| .

This result gives rise to two easy corollaries, the first following from Proposition 3.24, and the second following from the fact that if the real part of all eigenvalues of A are negative then limt→∞ |hΣ(t)| = 0.

5.7 Corollary Let Σ = (A, b, ct, D) be a SISO linear system and write

N(s)

TΣ(s) = D(s)

where (N, D) is the c.f.r. Then Σ is BIBO stable if and only if D has roots only in the negative half-plane.

5.8 Corollary Σ = (A, b, ct, D) is BIBO stable if spec(A) C−.

The matter of testing for BIBO stability of a SISO linear system is straightforward, so let’s do it for a simple example.

5.9 Example (Example 5.4 cont’d) We continue with the case where

We compute

1

TΣ(s) = s2 + as + b.

From Example 5.4 we know that we have BIBO stability if and only if a > 0 and b > 0. Let’s probe the issue a bit further by investigating what actually happens when we do

not have a, b > 0. The cases when Σ is internally unstable are not altogether interesting since the system is “obviously” not BIBO stable in these cases. So let us examine the cases when we have no eigenvalues in C+, but at least one eigenvalue on the imaginary axis.

√

1. a = 0 and b > 0: Here the eigenvalues are ±i b, and we compute

√

sin bt hΣ(t) = √ .

b

Thus the impulse response is bounded, but does not tend to zero as t → ∞. Theorem 5.6

predicts that there will be a bounded input signal that produces an unbounded output

√

signal. In fact, if we choose u(t) = sin bt and zero initial condition, then one verifies that the output is

Thus a bounded input gives an unbounded output.

2.a > 0 and b = 0: The eigenvalues here are {0, −a}. One may determine the impulse response to be

hΣ(t) = 1 − e−at . a

This impulse response is bounded, but again does not go to zero as t → ∞. Thus there ought to be a bounded input that gives an unbounded output. We have a zero eigenvalue, so this means we should choose a constant input. We take u(t) = 1 and zero initial condition and determine the output as

As usual, when dealing with input/output issues for systems having states, one needs to exercise caution for the very reasons explored in Section 2.3. This can be demonstrated with an example.

5.10 Example Let us choose here a specific example (i.e., one without parameters) that will illustrate problems that can arise with fixating on BIBO stability while ignoring other considerations. We consider the system Σ = (A, b, ct, 01) with

We determine that hΣ(t) = −e−2t. From Theorem 5.6 we determine that Σ is BIBO stable. But is everything really okay? Well, no, because this system is actually not observable.

We compute

and since this matrix has rank 1 the system is not observable. How is this manifested in the system behaviour? In exactly the way one would predict. Thus let us look at the state behaviour for the system with a bounded input. We take u(t) = 1(t) as the unit step input, and take the zero initial condition. The resulting state behaviour is defined by

168 5 Stability of control systems 22/10/2004

Let us state a result that provides a situation where one can make a precise relationship between internal and BIBO stability.

5.11 Proposition If a SISO linear system Σ = (A, b, c0, D) is controllable and observable, then the following two statements are equivalent:

(i)Σ is internally asymptotically stable;

(ii)Σ is BIBO stable.

When Σ is not both controllable and observable, the situation is more delicate. The diagram in Figure 5.1 provides a summary of the various types of stability, and which types

internally

asymptotically +3internally stable stable

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G

G Â'

BIBO stable

Figure 5.1 Summary of various stability types for SISO linear systems

imply others. Note that there are not many arrows in this picture. Indeed, the only type of stability which implies all others is internal asymptotic stability. This does not mean that if a system is only internally stable or BIBO stable that it is not internally asymptotically stable. It only means that one cannot generally infer internal asymptotic stability from internal stability or BIBO stability. What’s more, when a system is internally stable but not internally asymptotically stable, then one can make some negative implications, as shown in Figure 5.2. Again, one should be careful when interpreting the absence of arrows from this diagram. The best approach here is to understand that there are principles that underline when one can infer one type of stability from another. If these principles are understood, then matters are quite straightforward. A clearer resolution of the connection between BIBO stability and internal stability is obtained in Section 10.1 when we introduce the concepts of “stabilisability” and “detectability.” The complete version of Figures 5.1 and 5.2 is given by Figure 10.1. Note that we have some water to put under the bridge to get there. . .

5.2.2 BIBO stability of SISO linear systems in input/output form It is now clear how we may extend the above discussion of BIBO stability to systems in input/output form, at least when they are proper.

M

M

M

M

M

MM

M

M

M

M

M

M

M

M

M

M

MM

if controllable and observable MM M

M

M

M

M

M

M

M

MM

M

M

M

M

M

M

M

M

M

M

M "*

not BIBO stable

Figure 5.2 Negative implication when a system is internally stable, but not internally asymptotically stable

5.12 Definition A proper SISO linear system (N, D) in input/output form is bounded input, bounded output stable (BIBO stable) if the SISO linear system ΣN,D is BIBO stable.

From Corollary 5.7 follows the next result giving necessary and su cient conditions for BIBO stability of strictly proper SISO systems in input/output form.

5.13 Proposition A proper SISO linear system (N, D) in input/output form is BIBO stable if and only if TN,D has no poles in C+.

The question then arises, “What about SISO linear systems in input/output form that are not proper?” Well, such systems can readily be shown to be not BIBO stable, no matter what the character of the denominator polynomial D. The following result shows why this is the case.

5.14 Proposition If (N, D) is an improper SISO linear system in input/output form, then there exists an input u satisfying the properties

(i)|u(t)| ≤ 1 for all t ≥ 0 and

(ii)if y satisfies D ddt y(t) = N ddt u(t), then for any M > 0 there exists t > 0 so that

|y(t)| > M.

Proof From Theorem C.6 we may write

N(s) = R(s) + P (s)

D(s)

where R is a strictly proper rational function and P is a polynomial of degree at least 1. Therefore, for any input u, the output y will be a sum y = y1 + y2 where

If R has poles in C+, then the result follows in usual manner of the proof of Theorem 5.6.

So we may as well suppose that R has no poles in C+, so that the solution y1 is bounded. We will show, then, that y2 is not bounded. Let us choose u(t) = sin(t2). Any derivative of u will involve terms polynomial in t and such terms will not be bounded as t → ∞. But y2, by (5.4), is a linear combination of derivatives of u, so the result follows.

5.3 Norm interpretations of BIBO stability

In this section, we o er interpretations of the stability characterisations of the previous section in terms of various norms for transfer functions and for signals. The material in this section will be familiar to those who have had a good course in signals and systems. However, it is rare that the subject be treated in the manner we do here, although its value for understanding control problems is now well established.

5.3.1 Signal norms We begin our discussion by talking about ways to define the “size” of a signal. The development in this section often is made in a more advanced setting where the student is assumed to have some background in measure theory. However, it is possible to get across the basic ideas in the absence of this machinery, and we try to do this here.

For p ≥ 1 and for a function f : (−∞, ∞) → R denote

separately by defining

kfk∞ = sup{|f(t)| ≤ α for almost every t}

α≥0

as the L∞-norm of y. The L∞-norm is sometimes referred to as the sup norm. Here “almost every” means except on a set T (−∞, ∞) having the property that

Z

dt = 0.

T

We denote

L∞(−∞, ∞) = {f : (−∞, ∞) → R | kfk∞ < ∞}

as the set of functions that we declare to be L∞-integrable. Note that we are dealing here with functions defined on (−∞, ∞), whereas with control systems, one most often has functions that are defined to be zero for t < 0. This is still covered by what we do, and the extra generality is convenient.

Most interesting to us will be the Lp spaces L2(−∞, ∞) and L∞(−∞, ∞). The two sets of functions certainly do not coincide, as the following collection of examples indicate.

5.15 Examples 1. The function cos t is in L∞(−∞, ∞), but is in none of the spaces Lp(−∞, ∞) for 1 ≤ p < ∞. In particular, it is not L2-integrable.

2.The function f(t) = 1+1 t is not L1-integrable, although it is L2-integrable; one computes kfk2 = 1.

3.Define

q

(

1t , t (0, 1]

0, otherwise.

One then checks that kfk1 = 2, but that f is not L2-integrable. Also, since limt→1− f(t) = ∞, the function is not L∞-integrable.

generally one has kfkp = (1 + p)1/p where the -function generalises the factorial to non-integer values.

There is another measure of signal size we shall employ that di ers somewhat from the above measures in that it is not a norm. We let f : (−∞, ∞) → R be a function and say that f is a power signal if the limit

that pow(f) = 0 even though f is nonzero. Thus pow is certainly not a norm. Nevertheless, it is a useful, and often used, measure of a signal’s size.

The following result gives some relationships between the various Lp-norms and the pow operation.

5.16Proposition The following statements hold:

(i)if f L2(−∞, ∞) then pow(f) = 0;

(ii)if f L∞(−∞, ∞) is a power signal then pow(f) ≤ kfk∞;

The result follows since as T → ∞, the right-hand side goes to zero. (ii) We compute

= kfk2∞ .

172 5 Stability of control systems 22/10/2004

(iii) We have

Z ∞

kfk22 = f2(t) dt

−∞

Z∞

=|f(t)| |f(t)| dt

The relationships between the various Lp-spaces we shall care about and the pow operation are shown in Venn diagrammatic form in Figure 5.3.

pow

L2

L∞

L1

Figure 5.3 Venn diagram for relationships between Lp-spaces and pow

5.3.2 Hardy spaces and transfer function norms For a meromorphic complex-valued function f we will be interested in measuring the “size” by f by evaluating its restriction to the imaginary axis. To this end, given a meromorphic function f, we follow the analogue of our time-domain norms and define, for p ≥ 1, the Hp-norm of f by

In like manner we define the H∞-norm of f by

kfk∞ = sup |f(iω)| .

ω

While these definitions make sense for any meromorphic function f, we are interested in particular such functions. In particular, we denote

RLp = {R R(s) | kRkp < ∞}

H+p = {f | f is a Hardy function with kfkp < ∞},

for p [0, ∞) {∞}. We also have RH+p = R(s) ∩ Hp as the Hardy functions with bounded Hp-norm that are real rational. In actuality, we shall only be interested in the case when p {1, 2, ∞}, but the definition of the Hp-norm holds generally. One must be careful when one sees the symbol k·kp that one understands what the argument is. In one case we mean it to measure the norm of a function of t defined on (−∞, ∞), and in another case we use it to define the norm of a complex function measured by looking at its values on the imaginary axis.

Note that with the above notation, we have the following characterisation of BIBO stability.

5.17 Proposition A proper SISO linear system (N, D) in input/output form is BIBO stable if and only if TN,D RH+∞.

The following result gives straightforward characterisations of the various rational function spaces we have been talking about.

5.18Proposition The following statements hold:

(i)RL∞ consists of those functions in R(s) that

(a)have no poles on iR and

(b)are proper;

(ii)RH+∞ consists of those functions in R(s) that

(a)have no poles in C+ and

(b)are proper;

(iii)RL2 consists of those functions in R(s) that

(a)have no poles on iR and

(b)are strictly proper.

(iv)RH+2 consists of those functions in R(s) that

(a)have no poles in C+ and

(b)are strictly proper.

Proof Clearly we may prove the first and second, and then the third and fourth assertions together.

(i) and (ii): This part of the proposition follows since a rational Hardy function is proper if and only if lims→∞ |R(s)| < ∞, and since |R(iω)| is bounded for all ω R if and only if R has no poles on iR. The same applies for RL∞.

(iii) and (iv) Clearly if R RH+2 then lims→∞ |R(s)| = 0, meaning that R must be strictly proper. We also need to show that R RH+2 implies that R has no poles on iR. We shall do this by showing that if R has poles on iR then R 6 RH+2 . Indeed, if R has a pole at ±iω0 then near iω0, R will essentially look like

C

R(s) ≈ (s − iω0)k

1After George Harold Hardy (1877-1947).

Thus the contribution to kRk2 of a pole on the imaginary axis will always be unbounded. Conversely, if R is strictly proper with no poles on the imaginary axis, then one can find

these signal norms, we consider a SISO linear system (N, D) in input/output form. We wish to flush out the input/output properties of a transfer function relative to the Lp signal norms and the pow operation. For notational convenience, let us adopt the notation k·kpow = pow and let Lpow(−∞, ∞) denote those functions f for which pow(f) is defined. This is an abuse of notation since pow is not a norm. However, the abuse is useful for making the following definition.

5.19 Definition Let R R(s) be a proper rational function, and for u L2(−∞, ∞) let yu : (−∞, ∞) → R be the function satisfying yˆu(s) = R(s)ˆu(s). For p1, p2 [1, ∞) {∞} {pow}, the Lp1 → Lp2 -gain of R is defined by

If (N, D) is SISO linear system in input/output form, then (N, D) is Lp1 → Lp2 -stable if

kTN,Dkp1→p2 < ∞.

This definition of Lp1 → Lp2 -stability is motivated by the following obvious result.

5.20 Proposition Let (N, D) be an Lp1 → Lp2 stable SISO linear system in input/output form and let u: (−∞, ∞) → R be an input with yu : (−∞, ∞) → R the function satisfying yˆu(s) = TN,D(s)ˆu(s). If u Lp1 (−∞, ∞) then

kyukp2 ≤ kTN,Dkp1→p2 kukp1 .

In particular, u Lp1 (−∞, ∞) implies that yu Lp2 (−∞, ∞).

22/10/2004 5.3 Norm interpretations of BIBO stability 175

Although our definitions have been made in the general context of Lp-spaces, we are primarily interested in the cases where p1, p2 {1, 2, ∞}. In particular, we would like to be able to relate the various gains for transfer functions to the Hardy space norms of the previous section. The following result gives these characterisations. The proofs, as you will see, is somewhat long and involved.

5.21 Theorem For a proper BIBO stable SISO linear system (N, D) in input/output form,

(vii)kTN,Dkpow→2 = ∞;

(viii)kTN,Dkpow→∞ = ∞;

(ix)kTN,Dkpow→pow = kTN,Dk∞.

If (N, D) is strictly proper, then part (v) can be improved to kTN,Dk∞→∞ = khN,Dk1.

Proof (i) By Parseval’s Theorem we have kfk2 = kfˆk2 for any function f L2(−∞, ∞). Therefore

kyuk22 = kyˆuk22

=1 Z ∞ |yˆu(iω)|2 dω

2π −∞

=1 Z ∞ |TN,D(iω)|2 |uˆ(iω)|2 dω

2π −∞

This shows that kTN,Dk2→2 ≤ kTN,Dk∞. We shall show that this is the least upper bound. Let ω0 R+ be a frequency at which kTN,Dk∞ is attained. First let us suppose that ω0 is

If kTN,Dk∞ is not attained at a finite frequency, then we define u so that

In either case we have shown that kTN,Dk∞ is a least upper bound for kTN,Dk2→2. (ii) Here we employ the Cauchy-Schwartz inequality to determine

=khN,Dk2 kuk2

=kTN,Dk2 kuk2 ,

where the last equality follows from Parseval’s Theorem. Thus we have shown that kTN,Dk2→∞ ≤ kTN,Dk2. This is also the least upper bound since if we take

u(t) = hN,D(−t),

kTN,Dk2

we determine by Parseval’s Theorem that kuk2 = 1 and from our above computations that |y(0)| = kTN,Dk2 which means that kyuk∞ ≥ kTN,Dk2, as desired.

(iii) Since yu is L2-integrable if u is L2-integrable by part (i), this follows from Proposition 5.16(i).

(iv) Let ω R+ have the property that TN,D(iω) 6= 0. Take u(t) = sin ωt. By Theorem 4.1 we have

yu(t) = Re(TΣ(iω)) sin ωt + Im(TΣ(iω)) cos ωt + y˜h(t) + C

where limt→∞ yh(t) = 0. In this case we have kuk∞ = 1 and kyuk2 = ∞. (v) We compute

= kh ˜ ˜ k1 + |C| kuk∞ .

N,D

This shows that kTN,Dk ≤ kh ˜ ˜ k1 + |C| as stated. To see that this is the least upper

∞→∞ N,D

bound when (N, D) is strictly proper (cf. the final statement in the theorem), fix t > 0 and define u so that

(

+1, hN,D(τ) ≥ 0

u(t − τ) =

−1, hN,D(τ) < 0.

Then we have kuk∞ = 1 and

Z ∞

yu(t) = hN,D(τ)u(t − τ) dτ

−∞

Z∞

=|hN,D(τ)| dτ

−∞

= khN,Dk1 .

Thus kyuk∞ ≥ khN,Dk1.

(vii) To carry out this part of the proof, we need a little diversion. For a power signal f define

and note that ρ(f)(0) = pow(f). The limit in the definition of ρ(f) may not exist for all τ, but it will exist for certain power signals. Let f be a nonzero power signal for which the limit does exist. Denote by σ(f) the Fourier transform of ρ(f):

Z ∞

σ(f)(ω) = ρ(f)(t)e−iωt dt.

−∞

Therefore, since ρ(f) is the inverse Fourier transform of σ(f) we have

Now we claim that if yu is related to u by yˆu(s) = TN,D(s)ˆu(s) where u is a power signal for which ρ(u) exists, then we have

Indeed note that

Z ∞

yu(t)yu(t + τ) = hN,D(α)y(t)u(t + τ − α) dα,

−∞

so that Z ∞

ρ(yu)(t) = hN,D(τ)ρ(yu, u)(t − τ) dτ,

−∞

where

In like manner we verify that

Z ∞

ρ(yu, u)(t) = h−N,D(t − τ)ρ(u)(τ) dτ,

−∞

where h−N,D(t) = hN,D(−t). Therefore we have ρ(yu) = hN,D h−N,D ρ(u), where signifies convolution. One readily verifies that the Fourier transform of h−N,D is the complex conjugate

of the Fourier transform of hN,D. Therefore

¯

σ(yu)(ω) = TN,D(iω)TN,D(iω)σ(u)(ω),

(ix) By (5.7) we have pow(yu) ≤ kTN,Dk∞ pow(u). Therefore kTN,Dkpow→pow ≤ kTN,Dk∞.

To show that this is the least upper bound, let ω0 R+ be a frequency at which kTN,Dk is

√ ∞ realised, and first suppose that ω0 is finite. Now let u(t) = 2 sin ω0t. One readily computes

ρ(u)(t) = cos ω0t, implying by (5.5) that pow(u) = 1. Also we clearly have

σ(u)(ω) = π δ(ω − ω0) + δ(ω + ω0) ,

An application of (5.7) then gives

pow(yu)2 = 12 |TN,D(iω0)|2 + |TN,D(−iω0)|2

=|TN,D(iω0)|2

=kTN,Dk2∞ .

If kTN,Dk∞ is attained only in the limit as frequency goes to infinity, then the above argument is readily modified to show that one can find a signal u so that pow(yu) is arbitrarily close

to kTN,Dk∞.

(vi) Let u L∞(−∞, ∞) be a power signal. By Proposition 5.16(ii) we have pow(u) ≤ kuk∞. It therefore follows that

u Lpow(−∞,∞) u not zero

During the course of the proof of part (ix) we showed that there exists a power signal u with pow(u) = 1 with the property that pow(yu) = kTN,Dk∞. Therefore, this part of the theorem follows.

(viii) For k ≥ 1 define

(

uk(t) = k, t (k, k + k13 ) 0, otherwise,

and define an input u by

∞

X

u(t) = uk(t).

k=1