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has one zero eigenvalue and n+12		− 1 positive real eigenvalues. Thus D(P ) must be positive-
definite, and P is then Hurwitz by Lemma 5.40 and Theorem 5.38.

Case 4: We assume (i) and that n is odd. As in Case 2, define P −1(s) = snP (1s ). In this case one can ascertain that C(P −1) is obtained from C(P ) by reversing rows and columns, and that D(P −1) is obtained from D(P ) by reversing rows and columns. The di erence from the situation in Case 2 arises because here we are taking n odd, while in Case 2 it was even. In any event, one may now apply Case 3 to P −1 to show that P −1 is Hurwitz. Then P is itself Hurwitz by Exercise E5.20.

Case 5: We assume (ii) and that n is odd. For > 0 define P R[s] by P (s) = (s + )P (s). Thus the degree of P is now even. Indeed,

P (s) = pnsn+1 + (pn−1 + pn)sn + · · · + (p0 + p1)s + p0.

One may readily determine that

C(P ) = C(P ) + C

for some matrix C which is independent of . In like manner, one may show that

D(P ) =	D12	p02
	D(P ) + D11	D12

where D11 and D12 are independent of . Since D(P ) is positive-definite and a0 > 0, forsu ciently small we must have D(P ) positive-definite. From the argument of Case 2 we may infer that P is Hurwitz, from which it is obvious that P is also Hurwitz.

Case 6: We assume (iv) and that n is odd. We define P −1(s) = snP (1s ) so that C(P −1) is obtained from C(P ) by reversing rows and columns, and that D(P −1) is obtained from D(P ) by reversing rows and columns. One can now use Case 5 to show that P −1 is Hurwitz, and so P is also Hurwitz by Exercise E5.20.

Case 7: We assume (iii) and that n is even. As with Case 5, we define P (s) = (s+ )P (s)

and in this case we compute
C(P ) =	C12	p02
	C(P ) + C11	C12

and

D(P ) = D(P ) + D,

where C11, C12, and D are independent of . By our assumption (iii), for > 0 su ciently small we have C(P ) positive-definite. Thus, invoking the argument of Case 1, we may deduce that D(P ) is also positive-definite. Therefore P is Hurwitz by Lemma 5.40 and Theorem 5.36. Thus P is itself also Hurwitz.

Case 8: We assume (iv) and that n is even. Taking P −1(s) = snP (1s ) we see that C(P −1) is obtained from D(P ) by reversing the rows and columns, and that D(P −1) is obtained from C(P ) by reversing the rows and columns. Now one may apply Case 7 to deduce that

P −1, and therefore P , is Hurwitz.
5.5.4 The Li´enard-Chipart criterion	Although less well-known than the criterion of

Routh and Hurwitz, the test we give due to Li´enard and Chipart [1914]3 has the advantage of

3Perhaps the relative obscurity of the test reflects that of its authors; I was unable to find a biographical reference for either Li´enard or Chipart. I do know that Li´enard did work in di erential equations, with the Li´enard equation being a well-studied second-order linear di erential equation.

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delivering fewer determinantal inequalities to test. This results from their being a dependence on some of the Hurwitz determinants. This is given thorough discussion by Gantmacher [1959b]. Here we state the result, and give a proof due to Anderson [1972] that is more elementary than that of Gantmacher.

5.42 Theorem A polynomial

P (s) = sn + pn−1sn−1 + · · · + p1s + p0 R[s]

is Hurwitz if and only if any one of the following conditions holds:

(i)

p2k > 0, k {0, 1, . . . , bn−2

1 c} and

2k+1 > 0, k {0, 1, . . . , bn−2 1 c};

(ii)

p2k > 0, k {0, 1, . . . , b

n−2

1

c} and

2k > 0, k {1, . . . , bn2 c};

(iii)

p0

> 0, p2k+1

> 0, k {0, 1, . . . , b

n−2

2

c} and

2k+1 > 0, k {0, 1, . . . , b

n−2 1

c};

(iv)

p0

> 0, p2k+1

> 0, k {0, 1, . . . , b

n−2

2

c} and

2k > 0, k {1, . . . , bn2 c}.

Here

1, . . . , n are the Hurwitz determinants.

Proof The theorem follows immediately from Theorems 5.28 and 5.41 and after one checks

that the principal minors of C(P ) are exactly the odd Hurwitz determinants 1,	3, . . ., and
that the principal minors of D(P ) are exactly the even Hurwitz determinants	2, 4, . . ..

This observation is made by a computation which we omit, and appears to be first been noticed by Fujiwara [1915].

The advantage of the Li´enard-Chipart test over the Hurwitz test is that one will generally have fewer determinants to compute. Let us illustrate the criterion in the simplest case, when n = 2.

5.43 Example (Example 5.35 cont’d) We consider the polynomial P (s) = s2 + as + b. Recall that the Hurwitz determinants were computed in Example 5.37:

1 = a, 2 = ab.

Let us write down the four conditions of Theorem 5.42:

1.	p0 = b > 0,		1 = a > 0;
2.	p0	= b > 0,	2 = ab > 0;
3.	p0 = b > 0, p1 = a > 0,			1 = a > 0;
4.	p0 = b > 0, p1 = a > 0,			2 = ab > 0.

We see that all of these conditions are equivalent in this case, and imply that P is Hurwitz if and only if a, b > 0, as expected. This example is really too simple to illustrate the potential advantages of the Li´enard-Chipart criterion, but we refer the reader to Exercise E5.22 to see how the test can be put to good use.

5.5.5 Kharitonov’s test It is sometimes the case that one does not know exactly the coe cients for a given polynomial. In such instances, one may know bounds on the coe cients. That is, for a polynomial

P (s) = pnsn + pn−1sn−1 + · · · + p1s + p0,

(5.19)

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one may know that the coe cients satisfy inequalities of the form
	pimin ≤ pi ≤ pimax, i = 0, 1, . . . , n.	(5.20)

In this case, the following remarkable theorem of Kharitonov [1978] gives a simple test for the stability of the polynomial for all possible values for the coe cients. Since the publication of Kharitonov’s result, or more properly its discovery by the non-Russian speaking world, there have been many simplifications of the proof [e.g., Chapellat and Bhattacharyya 1989, Dasgupta 1988, Mansour and Anderson 1993]. The proof we give essentially follows Minnichelli, Anagnost, and Desoer [1989].

5.44 Theorem Given a polynomial of the form (5.19) with the coe cients satisfying the inequalities (5.20), define four polynomials

Q1(s) = pmin0 + pmin1 s + pmax2 s2 + pmax3 s3 + · · ·

Q2(s) = pmin0 + pmax1 s + pmax2 s2 + pmin3 s3 + · · ·

Q3(s) = pmax0 + pmax1 s + pmin2 s2 + pmin3 s3 + · · ·

Q4(s) = pmax0 + pmin1 s + pmin2 s2 + pmax3 s3 + · · ·

Then P is Hurwitz for all

(p0, p1, . . . , pn) [pmin0 , pmax0 ] × [pmin1 , pmax1 ] × · · · × [pminn , pmaxn ]

if and only if the polynomials Q1, Q2, Q3, and Q4 are Hurwitz.

Proof Let us first assume without loss of generality that pminj > 0, j = 0, . . . , n. Indeed, by Exercise E5.18, for a polynomial to be Hurwitz, its coe cients must have the same sign, and we may as well suppose this sign to be positive. If

p = (p0, p1, . . . , pn) [pmin0 , pmin0 ] × [pmin1 , pmin1 ] × · · · × [pminn , pminn ],

then let us say, for convenience, that p is allowable. For p allowable denote

Pp(s) = pnsn + pn−1sn−1 + · · · + p1s + p0.

It is clear that if all polynomials Pp are allowable then the polynomials Q1, Q2, Q3, and Q4 are Hurwitz. Thus suppose for the remainder of the proof that Q1, Q2, Q3, and Q4 are Hurwitz, and we shall deduce that Pp is also Hurwitz for every allowable p.

For ω R define

R(ω) = {Pp(iω) | p allowable} .

The following property of R(ω) lies at the heart of our proof. It is first noticed by Dasgupta [1988].

1 Lemma For each ω R, R(ω) is a rectangle in C whose sides are parallel to the real and imaginary axes, and whose corners are Q1(iω), Q2(iω), Q3(iω), and Q4(iω).

Proof We note that for ω R we have

Re(Q1	(iω)) = Re(Q2	(iω)) = p0min − pmaxω2			+ p4minω4 + · · ·
Re(Q3	(iω)) = Re(Q4	(iω)) = p0max − pminω2			+ p4maxω4 + · · ·			.
Im(Q1	(iω)) = Im(Q4	(iω)) = ω pmin − pmaxω2				+ p4minω4 + · · ·
Im(Q2	(iω)) = Im(Q3	(iω)) = ω pmax	−	pminω2		+ p4maxω4 +	· · ·

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From this we deduce that for any allowable p we have

Re(Q1(iω)) = Re(Q2(iω)) ≤ Re(Pp(iω)) ≤ Re(Q3(iω)) = Re(Q4(iω))

Im(Q1(iω)) = Im(Q4(iω)) ≤ Im(Pp(iω)) ≤ Im(Q2(iω)) = Im(Q3(iω)).

This leads to the picture shown in Figure 5.4 for R(ω). The lemma follows immediately

Q2(iω)

Q3(iω)

Q1(iω)

Q4(iω)

Figure 5.4 R(ω)

from this.

H

Using the lemma, we now claim that if p is allowable, then Pp has no imaginary axis roots. To do this, we record the following useful property of Hurwitz polynomials.

2 Lemma If P R[s] is monic and Hurwitz with deg(P ) ≥ 1, then ]P (iω) is a continuous and strictly increasing function of ω.

Proof Write

				n
				Yj
			P (s) =		(s − zj)
				=1
where zj = σj + iωj with σj		< 0. Thus
]		n	| j| −		j	n		|σj|
		j=1 ]				j=1
	P (iω) =	X	(iω + σ		iω ) =	X	arctan	ω − ωj	.

Since |σj| > 0, each term in the sum is continuous and strictly increasing, and thus so too is

]P (iω). H

To show that 0 6 R(ω) for ω R, first note that 0 6 R(0). Now, since the corners of R(ω) are continuous functions of ω, if 0 R(ω) for some ω > 0, then it must be the case that for some ω0 [0, ω] the point 0 C lies on the boundary of R(ω0). Suppose that 0 lies on the lower boundary of the rectangle R(ω0). This means that Q1(iω0) < 0 and Q4(iω0) > 0 since the corners of R(ω) cannot pass through 0. Since Q1 is Hurwitz, by Lemma 2 we must have Q1(i(ω0 + δ)) in the (−, −) quadrant in C and Q4(i(ω0 + δ)) in the (+, +) quadrant in C for δ > 0 su ciently small. However, since Im(Q1(iω)) = Im(Q4(iω)) for all ω R, this cannot be. Therefore 0 cannot lie on the lower boundary of R(ω0) for any ω0 > 0. Similar arguments establish that 0 cannot lie on either of the other three boundaries either. This then prohibits 0 from lying in R(ω) for any ω > 0.

Now suppose that Pp0 is not Hurwitz for some allowable p0. For λ [0, 1] each of the polynomials

λQ1 + (1 − λ)Pp0

(5.21)

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is of the form Ppλ for some allowable pλ. Indeed, the equation (5.21) defines a straight line from Q1 to Pp0 , and since the set of allowable p’s is convex (it is a cube), this line remains in the set of allowable polynomial coe cients. Now, since Q1 is Hurwitz and Pp0 is not, by continuity of the roots of a polynomial with respect to the coe cients, we deduce that for some λ [0, 1), the polynomial Ppλ must have an imaginary axis root. However, we showed above that 0 6R(ω) for all ω R, denying the possibility of such imaginary axis roots. Thus all polynomials Pp are Hurwitz for allowable p.

5.45 Remarks 1. Note the pattern of the coe cients in the polynomials Q1, Q2, Q3, and Q4 has the form (. . . , max, max, min, min, . . . ) This is charmingly referred to as the

Kharitonov melody.

2.One would anticipate that to check the stability for P one should look at all possible extremes for the coe cients, giving 2n polynomials to check. That this can be reduced to four polynomial checks is an unobvious simplification.

3.Anderson, Jury, and Mansour [1987] observe that for polynomials of degree 3, 4, or 5, it su ces to check not four, but one, two, or three polynomials, respectively, as being Hurwitz.

4.A proof of Kharitonov’s theorem, using Liapunov methods (see Section 5.4), is given by

Mansour and Anderson [1993].

Let us apply the Kharitonov test in the simplest case when n = 2.

5.46 Example We consider

P (s) = s2 + as + b

with the coe cients satisfying

(a, b) [amin, amax] × [bmin, bmax].

The polynomials required by Theorem 5.44 are

Q1(s) = s2 + amins + bmin

Q2(s) = s2 + amaxs + bmin

Q3(s) = s2 + amaxs + bmax

Q4(s) = s2 + amins + bmax.

We now apply the Routh/Hurwitz criterion to each of these polynomials. This indicates that all coe cients of the four polynomials Q1, Q2, Q3, and Q4 should be positive. This reduces to requiring that

amin, amax, bmin, bmax > 0.

That is, amin, bmin > 0. In this simple case, we could have guessed the result ourselves since the Routh/Hurwitz criterion are so simple to apply for degree two polynomials. Nonetheless, the simple example illustrates how to apply Theorem 5.44.

5.6 Summary

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1.It should be understood that internal stability is a notion relevant only to SISO linear systems. The di erence between stability and asymptotic stability should be understood.

2.The conditions for internal stability are generally simple. The only subtleties occur when there are repeated eigenvalues on the imaginary axis. All of this needs to be understood.

3.BIBO stability is really the stability type of most importance in this book. One should understand when it happens. One should also know how, when it does not happen, to produce an unbounded output with a bounded input.

4.Norm characterisations if BIBO stability provide additional insight, and o er a clarifying language with which to organise BIBO stability. Furthermore, some of the technical results concerning such matters will be useful in discussions of performance in Section 9.3 and of robustness in Chapter 15.

5.One should be able to apply the Hurwitz and Routh criteria freely.

6.The Liapunov method o er a di erent sort of characterisation of internal stability. One should be able to apply the theorems presented.

The matter of stability is, of course, of essential importance. What we have done in this chapter is quite simple, so let us outline the major facts.

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205

Exercises

E5.1 Consider the SISO linear system Σ = (A, b, c, D) of Example 5.4, i.e., with

A =	−b	−a .
	0	1

By explicitly computing a basis of solutions to x˙ (t) = Ax(t) in each case, verify by direct calculation the conclusions of Example 5.4. If you wish, you may choose specific values of the parameters a and b for each of the eight cases of Example 6.4. Make sure that you cover sub-possibilities in each case that might arise from eigenvalues being real or complex.

E5.2 Determine the internal stability of the linearised pendulum/cart system of Exercise E1.5 for each of the following cases:

(a)the equilibrium point (0, 0);

(b)the equilibrium point (0, π).

E5.3 For the double pendulum system of Exercise E1.6, determine the internal stability for the linearised system about the following equilibria:

(a)the equilibrium point (0, 0, 0, 0);

(b)the equilibrium point (0, π, 0, 0);

(c)the equilibrium point (π, 0, 0, 0);

(d)the equilibrium point (π, π, 0, 0).

E5.4 For the coupled tank system of Exercise E1.11 determine the internal stability of the linearisation.

E5.5 Determine the internal stability of the coupled mass system of Exercise E1.4, both with and without damping. You may suppose that the mass and spring constant are positive and that the damping factor is nonnegative.

E5.6 Consider the SISO linear system Σ = (A, b, ct, 01) defined by

A =	−ω σ	, b =	1	, c =	0
	σ ω		0		1

for σ R and ω > 0.

(a)For which values of the parameters σ and ω is Σ spectrally stable?

(b)For which values of the parameters σ and ω is Σ internally stable? Internally asymptotically stable?

(c)For zero input, describe the qualitative behaviour of the states of the system when the parameters σ and ω are chosen so that the system is internally stable but not internally asymptotically stable.

(d)For zero input, describe the qualitative behaviour of the states of the system when the parameters σ and ω are chosen so that the system is internally unstable.

(e)For which values of the parameters σ and ω is Σ BIBO stable?

(f)When the system is not BIBO stable, determine a bounded input that produces an unbounded output.

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E5.7 Consider the SISO linear system Σ = (A, b, ct, 01) defined by												0
	A =	−ω σ			0		1		, b =	0	, c =		,
			σ ω		1		0			0		1
			0	0		ω σ				1		0
			0	0	σ		ω
					−

where σ R and ω > 0.

(a)For which values of the parameters σ and ω is Σ spectrally stable?

(b)For which values of the parameters σ and ω is Σ internally stable? Internally asymptotically stable?

(c)For zero input, describe the qualitative behaviour of the states of the system when the parameters σ and ω are chosen so that the system is internally stable but not internally asymptotically stable.

(d)For zero input, describe the qualitative behaviour of the states of the system when the parameters σ and ω are chosen so that the system is internally unstable.

(e)For which values of the parameters σ and ω is Σ BIBO stable?

(f)When the system is not BIBO stable, determine a bounded input that produces an unbounded output.

E5.8

Show that (N,

D) is BIBO stable if and only if for every u

L

∞

[0,

∞

), the function

d

d

u(t) also lies in L∞[0, ∞).

y satisfying D

y(t) = N

E5.9

dt

dt

Determine

whether the pendulum/cart system of Exercises E1.5 and E2.4 is BIBO

stable in each of the following linearisations:

(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;

(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.

E5.10 Determine whether the double pendulum system of Exercises E1.6 and E2.5 is BIBO stable in each of the following cases:

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

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

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

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

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

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

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

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

In each case, use the angle of the second link as output.

E5.11 Consider the coupled tank system of Exercises E1.11 and E2.6. Determine the BIBO stability of the linearisations in the following cases:

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

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207

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

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

E5.12 Consider the coupled mass system of Exercise E1.4 and with inputs as described in Exercise E2.19. For this problem, leave α as an arbitrary parameter. We consider a damping force of a very special form. We ask that the damping force on each mass be given by −d(x˙1 + x˙2). The mass and spring constant may be supposed positive, and the damping constant is nonnegative.

(a)Represent the system as a SISO linear system Σ = (A, b, c, D)—note that in Exercises E1.4 and E2.19, everything except the matrix A has already been determined.

(b)Determine for which values of α, mass, spring constant, and damping constant the system is BIBO stable.

(c)Are there any parameter values for which the system is BIBO stable, but for which you might not be confident with the system’s state behaviour? Explain your answer.

E5.13 Let	(N, D) be a SISO linear system		in input/output form. In this exercise, if
u: [0, ∞) → R is an input, yu will be the output defined so that yˆu(s) = TN,D(s)ˆu(s).
(a)	For > 0 define an input u by
	u	(t) =	1 ,	t [0, ]
			(0,	otherwise.

Determine

(i)lim →0 kyu k2;

(ii)lim →0 kyu k∞;

(iii)lim →0 pow(yu ).

(b)If u(t) = sin(ωt) determine

(i)kyuk2;

(ii)kyuk∞;

(iii)pow(yu).

E5.14 Let Σ = (A, b, ct, D) be a SISO linear system.

(a)Show that if A + At is negative-semidefinite then Σ is internally stable.

(b)Show that if A+At is negative-definite then Σ is internally asymptotically stable.

E5.15 Let (A, P, Q) be a Liapunov triple for which P and Q are positive-definite. Show that A is Hurwitz.

E5.16 Let

0 1 A = 0 a

for a ≥ 0. Show that if (A, P, Q) is a Liapunov triple for which Q is positivesemidefinite, then (A, Q) is not observable.

E5.17 Consider the polynomial P (s) = s3 + as2 + bs + c.

(a)Use the Routh criteria to determine conditions on the coe cients a, b, and c that ensure that the polynomial P is Hurwitz.

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(b)Use the Hurwitz criteria to determine conditions on the coe cients a, b, and c that ensure that the polynomial P is Hurwitz.

(c)Verify that the conditions on the coe cients from parts (a) and (b) are equivalent.

(d)Give an example of a polynomial of the form of P that is Hurwitz.

(e)Give an example of a polynomial of the form of P for which all coe cients are strictly positive, but that is not Hurwitz.

E5.18 A useful necessary condition for a polynomial to have all roots in C− is given by the following theorem.

Theorem If the polynomial

P (s) = sn + pn−1sn−1 + · · · + p1s + p0 R[s]

is Hurwitz, then the coe cients p0, p1, . . . , pn−1 are all positive.

(a)Prove this theorem.

(b)Is the converse of the theorem true? If so, prove it, if not, give a counterexample.

The Routh/Hurwitz method gives a means of determining whether the roots of a polynomial are stable, but gives no indication of “how stable” they are. In the following exercise, you will examine conditions for a polynomial to be stable, and with some margin for error.

E5.19 Let P (s) = s2 + as + b, and for δ > 0 denote

Rδ = {s C | Re(s) < −δ} .

Thus Rδ consists of those points lying a distance at least δ to the left of the imaginary axis.

(a) Using the Routh criterion as a basis, derive necessary and su cient conditions for all roots of P to lie in Rδ.

˜

Hint: The polynomial P(s) = P(s + δ) must be Hurwitz.

(b)Again using the Routh criterion as a basis, state and prove necessary and su cient conditions for the roots of a general polynomial to lie in Rδ.

Note that one can do this for any of the methods we have provided for characterising Hurwitz polynomials.

E5.20 Consider a polynomial

				P (s) = pnsn + pn−1sn−1 + · · · + p1s + p0 R[s]
with p		, p	= 0, and define P −1		R	[s] by P −1(s) = snP (1 ).
	0		n 6			s
(a)	Show that the roots for P −1 are the reciprocals of the roots for P .
(b)	Show that P is Hurwitz if and only if P −1 is Hurwitz.

E5.21 For the following two polynomials,

(a)P (s) = s3 + as2 + bs + c,

(b)P (s) = s4 + as3 + bs2 + cs + d,

do the following:

1.Using principal minors to test positive-definiteness, write the conditions of the Hermite criterion, Theorem 5.38, for P to be Hurwitz.

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209

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2.Again using principal minors to test positive-definiteness, write the four conditions of the reduced Hermite criterion, Theorem 5.41, for P to be Hurwitz, and ascertain which is the least restrictive.

E5.22 For the following two polynomials,

(a)P (s) = s3 + as2 + bs + c,

(b)P (s) = s4 + as3 + bs2 + cs + d,

write down the four conditions of the Li´enard-Chipart criterion, Theorem 5.42, and determine which is the least restrictive.

E5.23 Consider a general degree three polynomial

P (s) = s3 + as2 + bs + c,

where the coe cients satisfy

(a, b, c) [amin, amax] × [bmin, bmax] × [cmin, cmax].

(E5.1)

Use Kharitonov’s test, Theorem 5.44, to give conditions on the bounds for the intervals for a, b, and c so that P is Hurwitz for all coe cients satisfying (E5.1).