04.Frequency response (the frequency domain)

22/10/2004		4.4	Properties of the frequency response			139
	10				10
	0				0
dB	-10			dB	-10
	-20				-20
	-30				-30
	-1.5 -1 -0.5 0	0.5	1 1.5 2		-1.5 -1 -0.5 0 0.5 1 1.5	2
	log ω				log ω

	150								150
	100								100
deg	50							deg	50
	0								0
	-50								-50
	-100								-100
	-150								-150
	-1.5	-1	-0.5	0	0.5	1	1.5	2

log ω

-1.5

-1

-0.5

0

0.5

1

1.5

2

log ω

Figure 4.15 Bode plots for the two systems of (4.5) and (4.6)—HΣ1 is on the left and HΣ2 is on the right

(iv) ]HN,D(−ω) = −]HN,D(ω) provided ]HN,D(ω) (−π, π).

Proof We prove (i) and (ii) together. It is certainly true that the real parts of D(iω) and N(iω) will involve terms that have even powers of ω and that the imaginary parts of D(iω) and N(iω) will involve terms that have odd powers of ω. Therefore, if we denote by D1(ω) and D2(ω) the real and imaginary parts of D(iω) and N1(ω) and N2(ω) the real and imaginary parts of N(iω), we have

N1(−ω) = N1(ω),	D1(−ω) = D1(ω), N2(−ω) = −N2(ω), D2(−ω) = −D2(ω).								(4.7)
We also have
HN,D(ω) =	N1(ω) + iN2(ω)
	D1(ω) + iD2(ω)
=	N1(ω)D1(ω) + N2(ω)D2(ω)				+ i	N2(ω)D1(ω) − N1(ω)D2(ω)		,
	D12(ω) + D22(ω)					D12(ω) + D22(ω)
so that
	Re(HN,D(ω)) =		N1(ω)D1(ω) + N2(ω)D2(ω)				,
			D12(ω) + D22(ω)
	Im(HN,D(ω)) =		N2(ω)D1(ω) − N1(ω)D2(ω)				.
			D12(ω) + D22(ω)
Using the relations (4.7) we see that
N1(−ω)D1(−ω) + N2(−ω)D2(−ω) = N1(ω)D1(ω) + N2(ω)D2(ω),
N2(−ω)D1(−ω) − N1(−ω)D2(−ω) = N2(ω)D1(ω) − N1(ω)D2(ω),
and from this the assertions (i) and (ii) obviously follow.
(iii) This is a consequence of (i) and (ii) and the definition of |·|.
(iv) This follows from (i) and (ii) and the properties of arctan.

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4 Frequency response (the frequency domain)

22/10/2004

Now we turn our attention to the crux of the material in this section—a look at how the magnitude and phase of the frequency response are related. To relate these quantities, it is necessary to represent the frequency response in the proper manner. To this end, for a SISO linear system (N, D) in input/output form, we define ZPN,D C to be the set of zeros and poles of TN,D. We may then define SN,D : C \ ZPN,D → C by

SN,D(s) = ln(TN,D(s)),

noting that SN,D is analytic on C\ZPN,D. We recall that from the properties of the complex logarithm we have

Re(SN,D(s)) = ln |TN,D(s)| , Im(SN,D(s)) = ]TN,D(s)

for s C \ ZPN,D. Along similar lines we define ZPN,D R by

ZPN,D = {ω R | iω ZPN,D} .

Then we may define GN,D : R \ ZPN,D → C by

GN,D(ω) = SN,D(iω) = ln(HN,D(ω)).

(4.8)

We can employ our previous use of the properties of the complex logarithm to assert that

Re(GN,D(ω)) = ln |HN,D(ω)| , Im(GN,D(ω)) = ]HN,D(ω).

Note that these are almost the quantities one plots in the Bode plot. The phase is precisely what is plotted in the Bode plot (against a logarithmic frequency scale), and Re(GN,D(ω)) is related to the magnitude plotted in the Bode plot by

Re(GN,D(ω)) = ln 10 log |HN,D(ω)| =	ln 10		|HN,D(ω)| dB.
	20

Note that the relationship is a simple one as it only involves scaling by a constant factor. The quantity Re(GN,D(ω)) is measured in the charming units of neppers.

Recall that (N, D) is stable if all roots of D lie in C− and is minimum phase if all roots

of N lie in C−. Here we will require the additional assumption that N have no roots on the imaginary axis. Let us say that a SISO linear system (N, D) for which all roots of N lie in

C− is strictly minimum phase.

4.14 Proposition Let (N, D) be a proper SISO linear system in input/output form that is stable and strictly minimum phase, and let ω0 > 0. We then have

	π	Z0	ω2	− ω02
Im(GN,D(ω0)) =	2ω0		∞ Re(GN,D(ω))	− Re(GN,D(ω0))	dω.

Proof Throughout the proof we denote by G1(ω) the real part of GN,D(ω) and by G2(ω) the imaginary part of GN,D(ω).

Let Uω0 C be the open subset defined by

Uω0 = {s C | Re(s) > 0, Im(s) 6= ±ω0} .

We now define a closed contour whose interior contains points in Uω0 . We do this in parts. First, for R > ω0 define a contour R in Uω0 by

R = {Reiθ | − π2 ≤ θ ≤ π2 }.

22/10/2004 4.4 Properties of the frequency response 141

Now for r > 0 define two contours

r,1 = {iω0 + reiθ | − π2 ≤ θ ≤ π2 }, r,2 = {−iω0 + reiθ | − π2 ≤ θ ≤ π2 }.

Finally define a contours r,j, j = 3, 4, 5, by

ω0,3 = {iω | − ∞ < ω0 ≤ −ω0 − r}ω0,4 = {iω | − ω0 + r ≤ ω0 ≤ ω0 − r}

ω0,5 = {iω | ω + r ≤ ω0 < ∞} .

The closed contour we take is then

5

[

R,r = R r,j. j=1

We show this contour in Figure 4.16. Now define a function Fω0 : Uω0 → C by

Im

r,5									R
							r,1
			iω0
r,4

Re

−iω0 r,2

r,3

Figure 4.16 The contour R,r used in the proof of Proposition 4.14

Fω0 (s) =	2iω0(SN,D(s) − G1(ω0))	.
	s2 + ω2
	0

Since (N, D) is stable and strictly minimum phase, SN,D is analytic on C+, and so Fω0 is analytic on Uω0 , and so we may apply Cauchy’s Integral Theorem to the integral of Fω0 around the closed contour R,r.

Let us evaluate that part of this contour integral corresponding to the contour R as R get increasingly large. We claim that

Z

lim Fω0 (s) ds = 0.

R→∞ R

142

4 Frequency response (the frequency domain)

22/10/2004

Indeed we have

2iω

S

(Reiθ)

2iω

G

(ω )

iθ

N,D

Fω0 (Re

) =

0

−

0

1

0

.

−R2e2iθ + ω02

−R2e2iθ + ω02

Since deg(N) < deg(D) the first term on the right will behave like R−2 as R → ∞ and the second term will also behave like R−2 as R → ∞. Since ds = iReiθ on R, the integrand in

R

Fω0 (s) ds will behave like R−1 as R → ∞, and so our claim follows.

R

Now let us examine what happens as we let r → 0. To evaluate the contributions of r,1

and r,2 we write

F

ω0

(s) =

SN,D(s) − G1(ω0)

−

SN,D(s) − G1(ω0)

.

s − iω0

s + iω0

On r,1 we have s = iω0 + reiθ so that on r,1 we have

F

ω0

(iω

0

+ reiθ) =

SN,D(iω0 + reiθ) − G1(ω0)

−

SN,D(iω0 + reiθ) − G1(ω0)

.

reiθ

2iω0 + reiθ

We parameterise r,1

with the curve c: [−π2 , π2 ] → C defined by c(t) = iω0 + reit so that

c0(t) = ireit. Thus, as r → 0, Fω0 (s) ds behaves like

F

(s) ds

≈

(SN,D(iω0) − G1(ω0))ireit

= i(G

(ω

)

−

G

(ω

)),

ω0

reit

N,D

0

1

0

using the parameterisation specified by c. Integrating gives

Z

Fω0 (s) ds = iπ(GN,D(ω0) − G1(ω0)).

r,1

In similar fashion one obtains

Z

Fω0 (s) ds = −iπ(GN,D(−ω0) − G1(ω0)).

r,2

Now we use Proposition 4.13 to assert that

GN,D(ω0) − GN,D(−ω0) = 2iG2(ω0).

Therefore

ZZ

r,1	Fω0 (s) ds +	r,2	Fω0 (s) ds = iπ(GN,D(ω0) − G1(ω0) − (GN,D(−ω0) − G1(ω0)))
			= − 2πG2(ω0).

Finally, we look at the integrals along the contours r,3, r,4, and r,5 as r → 0 and for fixed R > ω0. These contour integrals in the limit will yield a single integral along the a portion of the imaginary axis:

Z

Fω0 (s) ds.

i[−R,R]

We can parameterise the contour in this integral by the curve c: [−R, R] → C defined by c(t) = it. Thus c0(t) = i, giving

Fω0 (s) ds =	R	2iω0(SN,D(it) − G1(ω0))		i dt
Zi[−R,R]	Z−R	ω02 − t2
=	R	2ω0(SN,D(it) − G1(ω0))	dt.
	Z−R	t2 − ω02

22/10/2004	4.4	Properties of the frequency response				143
Using Proposition 4.13 we can write this as
	Fω0	(s) ds =	R	4ω0(G1(t) − G1(ω0))	dt.
	Zi[−R,R]	Z0		t2 − ω02

Collecting this with our expression for the integrals along r,1 and r,2, as well as noting our claim that the integral along R vanishes as R → ∞, we have shown that

lim lim	Fω0	(s) ds =	∞	4ω0(G1(ω) − G1(ω0))	dω	−	2πG2(ω0).
r→0 R→∞ Z R,r		Z0		ω2 − ω02

By Cauchy’s Integral Theorem, this integral should be zero, and our result now follows from straightforward manipulation.

The following result gives an important property of stable strictly minimum phase systems.

4.15 Theorem (Bode’s Gain/Phase Theorem) Let (N, D) be a proper, stable, strictly minimum phase SISO linear system in input/output form, and let GN,D(ω) be as defined in (4.8). For ω0 > 0 define MN,Dω0 : R → R by MN,Dω0 (u) = Re(GN,D(ω0eU )). Then we have

]HN,D(ω0) =	1	∞ dMN,Dω0		ln coth u2	du.
	π	Z−∞ du

Proof The theorem follows fairly easily from Proposition 4.14. By that result we have

]

H

(ω ) =

2ω0

∞

Re(GN,D(ω)) − Re(GN,D(ω0))

dω.

(4.9)

N,D

0

π

Z0

ω2 − ω02

We make the change of variable ω = ω0eu, upon which the integral in (4.9) becomes

2

∞ MN,Dω0

(u) − MN,Dω0 (0)

1

∞ MN,Dω0

(u) − MN,Dω0 (0)

Z−∞

du =

Z−∞

du.

π

eu − e−u

π

sinh u

We note that

Z

du

= ln coth u2 ,

sinh u

and we may use this formula, combined with integration by parts, to determine that

1

∞ MN,Dω0 (u) − MN,Dω0 (0)

Z0

du =

π

sinh u

1

∞ dMN,Dω0

u

1

ω0

ω0

(0)) ln coth

u

∞

(4.10)

π Z0

du

ln coth

2 −

π

(MN,D(u)

− MN,D

2

0 ,

and

1

0 MN,Dω0 (u) − MN,Dω0

(0)

Z−∞

du =

π

sinh u

1

0 dMN,Dω0

u

+

1

ω0

ω0

u

∞

(4.11)

π Z∞

du

ln coth −2

π

(MN,D(u) − MN,D(0)) ln coth −2

0 .

144 4 Frequency response (the frequency domain) 22/10/2004

Let us look at the first term in each of these integrals. At the limit u = 0 we may

compute

u3

coth u2

2

u

≈

+

−

+ · · ·

u

6

360

that near u = 0, ln coth u

≈ −

ln u . Also, since Mω0 (u) is analytic at u = 0, Mω0

(u)

−

so ω0

2

2

N,D

N,D

MN,D(0)

will behave linearly in u for u near zero. Therefore,

lim Mω0 (u)

−

Mω0

(0)) ln coth u

u ln

u

.

u→0

N,D

N,D

2 ≈ −

2

Recalling that limu→0 u ln u = 0, we see that the lower limit in the above integrated ex-

u	→ ±∞	2	ω0	→	+	∞
pressions is zero. At the other limits as u		, ln coth u	behaves like e−u as u

and like e as u → −∞. This, combined with the fact that MN,D(u) behaves like ln u as u → ∞, implies the vanishing of the upper limits in the integrated expressions in both (4.10)

and (4.11). Thus we have shown that these integrated terms both vanish. From this the result follows easily.

It is not perhaps perfectly clear what is the import of this theorem, so let us examine it for a moment. In this discussion, let us fix ω0 > 0. Bode’s Gain/Phase Theorem is telling us that the phase angle for the frequency response can be determined from the slope of the Bode plot with decibels plotted against the logarithm of frequency. The contribution to

the phase of the slope at some u = ω0 ln ω is determined by the weighting factor ln coth u2 which we plot in Figure 4.17. From the figure we see that the slopes near u = 0, or ω = ω0,

	6
	5
2)	4
|u| /	3
coth(
	2
ln
	1
	-1.5	-1	-0.5	0	0.5	1	1.5	2
			u = ω0 ln ω

Figure 4.17 The weighting factor in Bode’s Gain/Phase Theorem

contribute most to the phase angle. But keep in mind that this only works for stable strictly minimum phase systems. What it tells us is that for such systems, if one wishes to specify a certain magnitude characteristic for the frequency response, the phase characteristic is completely determined.

Let’s see how this works in an example.

4.16 Example Suppose we have a Bode plot with y = |HN,D(ω)| dB versus x = log ω, and that y = 20kx+b for k Z and b R. Thus the magnitude portion of the Bode plot is linear.

22/10/2004 4.5 Uncertainly in system models 145

To employ theωGain/Phase Theorem we should convert this to a relation in y˜ = ln |HN,D(ω)|

versus x˜ = ln

. The coordinates are then readily seen to be related by

ω0

x˜

20

x =

+ log ω0, y =

y˜.

ln 10

ln 10

Therefore the relation y = 20kx + b becomes

b ln 10

y˜ = kx˜ + k ln ω0 +

.

20

In the terminology of Theorem 4.15 we thus have

MN,Dω0 (u) = ku + k ln ω0

+

b ln 10

20

dM

ω0

so that

N,D

= k. The Gain/Phase Theorem tells us that we may obtain the phase at any

du

frequency ω0 as

∞

k

]HN,D(ω0) =

Z−∞ ln coth u2

du.

π

2

The integral is one that can be looked up (Mathematica®

evaluated it for me) to be

π

, so

2

that we have ]HN,D(ω0) = kπ2 .

Let’s see if this agrees with what we have seen before. We know, after all, a transfer function whose magnitude Bode plot is 20k log ω. Indeed, one can check that choosing the transfer function TN,D(s) = sk gives 20 log |HN,D(ω)| = 20k log ω, and so this transfer function is of the type for which we are considering in the case when b = 0. For systems of

this type we can readily determine the phase to be	kπ , which agrees with Bode’s Gain Phase
	2
Theorem.

Despite the fact that it might be possible to explicitly apply Bode’s Gain/Phase Theorem in some examples to directly compute the phase characteristic for a certain magnitude characteristic, the value of the theorem lies not in the ability to do such computations, but rather in its capacity to give us approximate information for stable and strictly minimum phase systems. Indeed, it is sometimes useful to make the approximation

	dMN,Dω0	π	,	(4.12)
]HN,D(ω0) = du		u=0 2
and this approximation becomes better when one		is in a region where the slope of the
magnitude characteristic on the Bode plot is large at ω0 compared to the slope at other
frequencies.				Uses of Gain/Phase
				theorem

4.5 Uncertainly in system models

As hinted at in Section 1.2 in the context of the simple DC servo motor example, robustness of a design to uncertainties in the model is a desirable feature. In recent years, say the last twenty years, rigorous mathematical techniques have been developed to handle model uncertainty, these going under the name of “robust control.” These matters will be touched upon in this book, and in this section, we look at the first aspect of this: representing uncertainty in system models. The reader will observe that this uncertainty representation is done in the frequency domain. The reason for this is merely that the tools for controller

146

4 Frequency response (the frequency domain)

22/10/2004

design that have been developed up to this time rely on such a description. In the context of this book, this culminates in Chapter 15 with a systematic design methodology keeping robustness concerns foremost.

In this section it is helpful to introduce the H∞-norm for a rational function. Given R R(s) we denote

kRk∞ = sup{|R(iω)|}.

ω R

This will be investigated rather more systematically in Section 5.3.2, but for now the meaning is rather pedestrian: it is the maximum value of the magnitude Bode plot.

4.5.1 Structured and unstructured uncertainty The reader may wish to recall our general control theoretic terminology from Section 1.1. In particular, recall that a “plant” is that part of a control system that is given to the control designer. What is given to the control designer is a model, hopefully in something vaguely resembling one of the three forms we have thus far discussed: a state-space model, a transfer function, of a frequency response function. Of course, this model cannot be expected to be perfect. If one is uncertain about a plant model, one should make an attempt to come up with a mathematical description of what this means. There are many possible candidates, and they can essentially be dichotomised as structured uncertainty and unstructured uncertainty. The idea with structured uncertainty is that one has a specific type of plant model in mind, and parameters in that plant model are regarded as uncertain. In this approach, one wishes to design a controller that has desired properties for all possible values of the uncertain parameters.

4.17 Example Suppose a mass is moving under a the influence of a control force u. The precise value of the mass could be unknown, say m [m1, m2]. In this case, a control design should be thought of as being successful if it accomplishes stated objectives (the reader does not know what these might be at this point in the book!) for all possible values of the mass parameter m.

Typically, structured uncertainty can be expected to be handled on a case by case basis, depending on the nature of the uncertainty. An approach to structured uncertainty is put forward by Doyle [1982]. For unstructured uncertainty, the situation is typically di erent as

¯

one considers a set of plant transfer functions P that are close to a nominal plant RP in some way. Again the objective is to design a controller that works for every plant in the set of allowed plants.

4.18 Example (Example 4.17 cont’d) Let us look at the mass problem above in a di erent

				¯	1/m
manner. Let us suppose that we choose a nominal plant RP (s) =					s2		and define a set of
candidate plants		¯
P = RP R(s) |				∞ ≤			(4.13)
		RP − RP
have clearly allowed a much larger class of uncertainty that
for some > 0. In this case, we

was allowed in Example 4.17. Indeed, not only is the mass no longer uncertain, even the form of the transfer function is uncertain.

Thus we see that unstructured uncertainty generally forces us to consider a larger class of plants, and so is a more stringent and, therefore, conservative manner for modelling uncertainty. That is to say, by designing a controller that will work for all plants in P, we are designing a controller for plants that are almost certainly not valid models for the plant

22/10/2004 4.5 Uncertainly in system models 147

under consideration. Nevertheless, it turns out that this conservatism of design is made up for by the admission of a consistent design methodology that goes along with unstructured uncertainty. We shall now turn our attention to describing how unstructured uncertainty may arise in examples.

4.5.2 Unstructured uncertainty models We shall consider four unstructured uncertainty models, although others are certainly possible. Of the four types of uncertainty we present, only two will be treated in detail. A general account of uncertainty models is the subject of Chapter 8 in [Dullerud and Paganini 1999]. Consistent with our keep our treatment of robust control to a tolerable level of simplicity, we shall only look at rather straightforward types of uncertainty.

The first type we consider is called multiplicative uncertainty. We start with a

¯

R(s) and let Wu R(s) be a proper rational function with no poles in

nominal plant RP

¯

R(s) with the properties

C+. Denote by P×(RP , Wu) the set of rational functions RP

1.

¯

have the same number of poles in C+,

RP and RP

¯

have the same imaginary axis poles, and

2. RP and RP

RP (iω)

3.

− 1 ≤ |Wu(iω)| for all ω R.

R¯P (iω)

P P×

this set of plants is to note that R

(R¯

P

, W

u

) if and only if

Another

way to write

¯

RP = (1 + Wu)RP

where k

k∞ ≤ 1. Of course,

is not arbitrary, even given k

k∞ ≤ 1. Indeed, this condition

only ensures condition 3 above. If further satisfies the first two conditions, it is said to be allowable. In this representation, it is perhaps more clear where the term multiplicative uncertainty comes up.

The following example is often used as one where multiplicative uncertainty is appropriate.

4.19 Example Recall from Exercise EE.6 that the transfer function for the time delay of a function g by T is e−T sgˆ(s). Let us suppose that we have a plant transfer function

RP (s) = e−T sR(s)

for some R(s) R(s). We wish to ensure that this plant is modelled by multiplicative uncertainty. To do so, we note that the first two terms in the Taylor series for e−T s are 1 − T s, and thus we suppose that when T is small, a nominal plant of the form

¯	R(s)
RP (s) =	1 − T s

will do the job. Thus we are charged with finding a rational function Wu so that

	RP (iω)	− 1	≤ |Wu(iω)| , ω R.
	R¯(iω)

Let us suppose that T = 101 . In this case, the condition on Wu becomes

|Wu(iω)| ≥		e−0.1iω	− 1 , ω R.
		1 − 10s

148

4 Frequency response (the frequency domain)

22/10/2004

20
0
-20
dB
-40
-60
-0.5	0	0.5	1	1.5	2	2.5	3

log ω

		e−	iω	− 1	(solid) and Wu =		1		(dashed)
Figure 4.18			10			100
		1−	10s			(1−ω152 )

From Figure 4.18 (the solid curve) one can see that the magnitude of

	e−iω10		− 1
1 − 10s
has the rough behaviour of tailing o at	40dB/	decade			at low frequency and having constant
magnitude at high frequency. Thus a model of the form
Wu(s) =			Ks2
			τs + 1

is a likely candidate, although other possibilities will work as well, as long as they capture the essential features. Some fiddling with the Bode plot yields τ = 151 and K = 1001 as acceptable choices. Thus this choice of Wu will include the time delay plant in its set of plants.

The next type of uncertainty we consider is called additive uncertainty. Again we

¯	and a proper Wu R(s) having no poles in C+. Denote by
start with a nominal plant RP

¯
P×(RP , Wu) the set of rational functions RP R(s) with the properties
¯	have the same number of poles in C+,
1. RP and RP
¯	have the same imaginary axis poles, and
2. RP and RP
¯
3. |RP (iω) − RP (iω)| ≤ |Wu(iω)| for all ω R.
¯
A plant in P+(RP , Wu) will have the form
	¯	Wu
	RP = RP +
where k k∞ ≤ 1.	As with multiplicative uncertainty, will be allowable if properties 1
and 2 above are met.
4.20 Example This example has a generic flavour.		Suppose that we make measurements

of our plant to get magnitude information about its frequency response at a finite set of frequencies {ω1, . . . , ωk}. If the test is repeated at each frequency a number of times, we

¯

might try to find a nominal plant transfer function RP with the property that at the measured

22/10/2004

4.5 Uncertainly in system models

149

frequencies its magnitude is roughly the average of the measured magnitudes. One then can determine Wu so that it covers the spread in the data at each of the measured frequencies. Such a Wu should have the property, by definition, that

| − ¯ | ≤ | |

RP (iωj) RP (iωj) Wu(iωj) , j = 1, . . . , k.

One could then hope that at the frequencies where data was not taken, the actual plant data

¯
is in the data of the set P+(RP , Wu).
¯	¯

The above two classes of uncertainty models, P×(RP , Wu) and P+(RP , Wu) are the

two for which analysis will be carried out in this book. The main reason for this choice is convenience of analysis. Fortunately, many interesting cases of plant uncertainty can be modelled as multiplicative or additive uncertainty. However, for completeness, we shall give two other types of uncertainty representations.

The first situation we consider where the set of plants are related to the nominal plant

as

¯

RP

k

k∞ ≤ 1.

RP =

,

(4.14)

1 +

WuRP

This uncertainty representation can arise in practice.

4.21 Example Suppose that we have a plant of the form

RP (s) =

1

s2 + as + 1

where all we know is that a [amin, amax]. By taking

¯

1

1

RP (s) =

s2 + aavgs + 1

,

Wu(s) = 2 δas,

where aavg

= 21 (amax + amin) and δa =

amax − amin, then the set of plant is exactly as

in (4.14), if

is a number between −1 and 1.

If

is allowed to be a rational function

satisfying k

k∞ ≤ 1 then we have embedded our actual set of plants inside the larger

uncertainty set described by (4.14). One can view this example of one where the structured uncertainty is included as part of an unstructured uncertainty model.

The final type of uncertainty model we present allows plants that are related to the nominal plant as

	¯
RP =	RP	,	k k∞ ≤ 1.
	1 + Wu
Let us see how this type of uncertainty can come up in an example.
			finish
4.22 Example

¯¯

4.23Remarks 1. In our definitions of P×(RP , Wu) and P+(RP , Wu) we made some assumptions about the poles of the nominal plant and the poles of the plants in the uncertainty set. The reason for these assumptions is not evident at this time, and indeed they can be relaxed with the admission of additional complexity of the results of Section 7.3, and those results that depend on these results. In practice, however, these assumptions are not inconvenient to account for.

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2.All of our choices for uncertainty modelling share a common defect. They allow plants that will almost definitely not be possible models for our actual plant. That is to say, our sets of plants are very large. This has something of a drawback in our employment of these uncertainty models for controller design—they will lead to a too conservative design. However, this is mitigated by the existence of e ective analysis tools to do robust controller design for such uncertainty models.

3.When deciding on a rational function Wu with which to model plant uncertainty with one of the above schemes, one typically will not want Wu to tend to zero as s → ∞. The

reason for this is that at higher frequencies is typically where model uncertainty is the greatest. This becomes a factor when choosing starting point for Wu.

4.6 Summary

In this section we have introduced a nice piece of equipment—the frequency response function—and a pair of slick representations of it—the Bode plot and the polar plot. Here are the pertinent things you should know from this chapter.

1.For a given SISO linear system Σ, you should be able to compute ΩΣ and HΣ.

2.The interpretation of the frequency response given in Theorem 4.1 is fundamental in understanding just what it is that the frequency response means.

3.The complete equivalence of the impulse response, the transfer function, and the frequency response is important to realise. One should understand that something that one observes in one of these must also be reflected somehow in the other two.

4.The Bode plot as a representation of the frequency response is extremely important. Being able to look at it and understand things about the system in question is part of the “art” of control engineering, although there is certainly a lot of science in this too.

5.You should be able to draw by hand the Bode plot for any system whose numerator and denominator polynomials you can factor. Really the only subtle thing here is the dependence on ζ for the second-order components of the frequency response. If you think of ζ as damping, the interpretation here becomes straightforward since one expects larger magnitudes for lower damping when the second-order term is in the denominator.

6.The polar plot as a representation of the frequency response will be useful to us in Chapter 12. You should at least be able to sketch a polar plot given the corresponding Bode plot.

7.You should be aware of why minimum phase system have the name they do, and be able to identity these in “obvious” cases.

8.The developments of Section 4.4.2 are somehow essential, and at the same time somewhat hard. If one is to engage in controller design using frequency response, clearly the fact that there are essential restrictions on how the frequency response may behave is important.

9.It might be helpful on occasion to apply the approximation (4.12).

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4.6 Summary

151

	40							40
	30							30
	20							20
dB	10						dB	10
	0							0
	-10							-10
	-20							-20
	-30							-30
	-1.5 -1 -0.5	0	0.5	1	1.5	2		-1.5 -1 -0.5	0	0.5	1	1.5	2

log ω

log ω

	150							150
	100							100
deg	50						deg	50
	0							0
	-50							-50
	-100							-100
	-150							-150
	-1.5 -1 -0.5	0	0.5	1	1.5	2		-1.5 -1 -0.5	0	0.5	1	1.5	2

log ω

log ω

	40							40
	30							30
	20							20
dB	10						dB	10
	0							0
	-10							-10
	-20							-20
	-30							-30
	-1.5 -1 -0.5	0	0.5	1	1.5	2		-1.5 -1 -0.5	0	0.5	1	1.5	2

log ω

log ω

	150							150
	100							100
deg	50						deg	50
	0							0
	-50							-50
	-100							-100
	-150							-150
	-1.5 -1 -0.5	0	0.5	1	1.5	2		-1.5 -1 -0.5	0	0.5	1	1.5	2

log ω	log ω
Figure 4.19 Asymptotes for magnitude	and phase plots for

H(ω) = 1 + iω (top left), H(ω) = 1 + 2iζω − ω2 (top right), H(ω) = (1+iω)−1 (bottom left), and H(ω) = (1+2iζω−ω2)−1 (bottom right)

152 4 Frequency response (the frequency domain) 22/10/2004

Exercises

E4.1 Consider the vector initial value problem

x˙ (t) = Ax(t) + u0 sin ωtb, x(0) = x0

where there are no eigenvalues for A of the form iω˜ where ω˜ integrally divides ω. Let xp(t) be the unique periodic solution constructed in the proof of Theorem 4.1, and write the solution to the initial value problem as x(t) = xp(t) + xh(t). Determine an expression for xh(t). Check that xh(t) is a solution to the homogeneous equation.

E4.2 Consider the SISO linear system Σ = (A, b, ct, 01) with

A =	−ω0	σ0	, b =	1	, c =	0
	σ0	ω0		0		1

for σ0 R and ω0 > 0.

(a)Determine ΩΣ and compute HΣ.

(b)Take u(t) = u0 sin ωt, and determine when the system satisfies the hypotheses of Theorem 4.1, and determine the unique periodic output yp(t) guaranteed by that theorem.

(c)Do periodic solutions exist when the hypotheses of Theorem 4.1 are not satisfied?

(d)Plot the Bode plot for HΣ for various values of σ0 ≤ 0 and ω0 > 0. Make sure you get all cases where the Bode plot assumes a di erent “character.”

We have two essentially di ering characterisations of the frequency response, one as the way in which sinusoidal outputs appear under sinusoidal inputs (cf. Theorem 4.1) and one involving Laplace and Fourier transforms (cf. Proposition 4.3). In the next exercise you will explore the di ering domains of validity for the two interpretations.

E4.3 Let Σ = (A, b, ct, D) be complete. Show that the Fourier transform of h+Σ is defined if and only if all eigenvalues of A lie in C−. Does the characterisation of the frequency response provided in Theorem 4.1 share this restriction?

E4.4 For the SISO linear system (N(s), D(s)) = (1, s + 1) in input/output form, verify explicitly the following statements, stating hypotheses on arguments of the functions where necessary.

(a) TN,D is the Laplace transform of hN,D.

(b) hN,D is the inverse Laplace transform of TN,D.

(c) HN,D is the Fourier transform of hN,D.

(d) hN,D is the inverse Fourier transform of HN,D.

(e) TN,D(s) = 1 Z ∞ HN,D(ω) dω. 2π −∞ s − iω

(f) HN,D(ω) = TN,D(iω).

When discussing the impulse response, it was declared that to obtain the output for an arbitrary input, one can use a convolution with the impulse response (plus a bit that depends upon initial conditions). In the next exercise, you will come to an understanding of this in terms of Fourier transforms.

|HΣ(ω)| = p

Exercises for Chapter 4

153

E4.5 Suppose that fˇ and gˇ are functions of ω, and that they are the Fourier transforms of functions f, g : (−∞, ∞) → R. You know from your course on transforms that the inverse Fourier transform of the product fˇgˇ is the convolution

Z ∞

(f g)(t) = f(t − τ)g(τ) dτ.

−∞

Now consider a SISO linear system Σ = (A, b, ct, 01) with causal impulse response h+Σ. Recall that the output for zero state initial condition corresponding to the input u: [0, ∞) → R is

Z t

y(t) = h+Σ(t − τ)u(τ) dτ.

0

Make sense of the statement, “The output y is equal to the convolution h+Σ u.” Note that you are essentially being asked to resolve the conflict in the limits of integration between convolution in Fourier and Laplace transforms.

E4.6 We will consider in detail here the mass-spring-damper system whose Bode plots are produced in Example 4.6. We begin by scaling the input u by the spring constant k so that, upon division by m, the governing equations are

						x¨ + 2ζω0x˙ + ω02x = ω02u,
where ω0	=		k	and ζ =		2√dkm	. As output we shall take y = x. We allow d to be
			m
zero, but	neither m nor k can be zero.
	q

(a)Write this system as a SISO linear system Σ = (A, b, ct, D) (that is, determine A, b, c, and D), and determine the transfer function TΣ.

(b)Determine ΩΣ for the various values of the parameters ω0 and ζ, and then determine the frequency response HΣ.

(c)Show that

ω2

0 . (ω02 − ω2)2 + 4ζ2ω02ω2

(d)How does |HΣ(ω)| behave for ω ω0? for ω ω0?

(e)Show that ddω |HΣ(ω)| = 0 if and only if ddω |HΣ(ω)|2 = 0.

1
(f) Using the previous simplification, show that for ζ < √2	there is a maximum for

HΣ(ω). The maximum you determine should occur at the frequency

p

ωmax = ω0 1 − 2ζ2,

and should take the value

1

|HΣ(ωmax)| = p .

2ζ 1 − ζ2

(g) Show that

]HΣ(ω) = atan2(ω02 − ω2, −2ζω0ω),

where atan2 is the smart inverse tangent function that knows in which quadrant you are.

(h) How does ]HΣ(ω) behave for ω ω0? for ω ω0?

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(i)Determine an expression for ]HΣ(ωmax).

(j)Use your work above to give an accurate sketch of the Bode plot for Σ in cases

when ζ ≤ 1 , making sure to mark on your plot all the features you determined

√

2

in the previous parts of the question. What happens to the Bode plot as ζ is decreased? What would the Bode plot look like when ζ = 0?

E4.7 Construct Bode plots by hand for the following first-order SISO linear systems in input/output form:

(a)(N(s), D(s)) = (1, s + 1);

(b)(N(s), D(s)) = (s, s + 1);

(c)(N(s), D(s)) = (s − 1, s + 1);

(d)(N(s), D(s)) = (2, s + 1).

E4.8 Construct Bode plots by hand for the following second-order SISO linear systems in input/output form:

(a)(N(s), D(s)) = (1, s2 + 2s + 2);

(b)(N(s), D(s)) = (s, s2 + 2s + 2);

(c)(N(s), D(s)) = (s − 1, s2 + 2s + 2);

(d)(N(s), D(s)) = (s + 2, s2 + 2s + 2).

E4.9 Construct Bode plots by hand for the following third-order SISO linear systems in input/output form:

(a)(N(s), D(s)) = (1, s3 + 3s2 + 4s + 2);

(b)(N(s), D(s)) = (s, s3 + 3s2 + 4s + 2);

(c)(N(s), D(s)) = (s2 − 4, s3 + 3s2 + 4s + 2);

(d)(N(s), D(s)) = (s2 + 1, s3 + 3s2 + 4s + 2).

E4.10 For each of the SISO linear systems given below do the following:

1.calculate the eigenvalues of A;

2.calculate the transfer function;

3.sketch the Bode plots of the magnitude and phase of the frequency response;

4.by trial and error playing with the parameters, on your sketch, indicate the rˆole played by the parameters of the system (e.g., ω0 in parts (a) and (b));

5.try to justify the name of the system by looking at the shapes of the Bode plots. Here are the systems (assume all parameters are positive):

(a)low-pass filter:

(b)	high-pass filter:	−	ω	0			0
	A =			, b = 1 , c = ω				, D = 0 ;
(c)	notch filter: A = −ω0			,	b = 1 ,	c = −ω0		,	D = 1 ;
	A = −ω02	−δω0			, b = 1	, c =	−δω0 , D = 1 ;
	0		1		0			0
(d)	bandpass filter:
	A = −ω02 −δω0 , b = 1 , c = δω0 , D = 0 .
	0			1	0			0

Exercises for Chapter 4

155

E4.11 Consider the coupled mass system of Exercises E1.4, E2.19, and E3.14. Assume no damping and take the input from Exercise E2.19 in the case when α = 0. In Exercise E3.14, you constructed an output vector c0 for which the pair (A, c0) was unobservable.

(a)Construct a family of output vectors c by defining c = c0 + c1 for R and c1 R4. Make sure you choose c1 so that c is observable for 6= 0.

(b)Determine ΩΣ, and the frequency response HΣ using the output vector c (allowto be an arbitrary real number).

(c)Choose the parameters m = 1 and k = 1 and determine the Bode plot for values of around zero. Do you notice anything di erent in the character of the Bode plot when = 0?

E4.12 Consider the pendulum/cart system of Exercises E1.5, E2.4, and E3.15. For 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,

do the following:

1.determine ΩΣ, and the frequency response HΣ for the system;

2.for parameters M = 112 , m = 1, g = 9.81, and ` = 12 , produce the Bode plot for the pendulum/cart system;

3.can you see reflected in your Bode plot the character of the spectrum of the zero dynamics as you determined in Exercise E3.15?

E4.13 Consider the double pendulum system of Exercises E1.6, E2.5, E1.6 and E3.16. For each of the following linearisations:

(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,

do the following:

1.determine ΩΣ, and the frequency response HΣ for the system;

2.for parameters m1 = 1, m2 = 2, `1 = 12 , and `2 = 13 , produce the Bode plot for the double pendulum;

3.can you see reflected in your Bode plot the character of the spectrum of the zero dynamics as you determined in Exercise E3.16?

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

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E4.14 Consider the coupled tank system of Exercises E1.11, E2.6, and E3.17. For the linearisations in the following cases:

(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,

do the following:

1.determine ΩΣ, and the frequency response HΣ for the system;

2.for parameters α = 13 , δ1 = 1, A1 = 1, A2 = 12 , a1 = 101 , a2 = 201 , and g = 9.81, produce the Bode plot for the tank system;

3.can you see reflected in your Bode plot the character of the spectrum of the zero dynamics as you determined in Exercise E3.17?

E4.15 Suppose you are shown a Bode plot for a stable SISO linear system Σ. Can you tell from the character of the plot whether Σ is controllable? observable? minimum phase?

E4.16 Construct two transfer functions, one minimum phase and the other not, whose frequency response magnitudes are the same (you cannot use the one in the book). Make Bode plots for each system, and make the relevant observations.

E4.17 Let (N, D) be a proper, stable SISO linear system of relative degree m. Characterise the total phase shift,

]HN,D(∞) − ]HN,D(0),

in terms of the roots of N, assuming that N has no roots on iR.

In the next exercise, you will provide a rigorous justification for the term “minimum phase.”

E4.18 Let (N, D) be a SISO linear system in input/output form, and suppose that it is minimum phase (thus N has no roots in C+). Let M(N, D) denote the collection of

˜ ˜

SISO linear systems (N, D) in input/output form for which

		|HN,D(ω)| =		HN,D˜ ˜ (ω) , ω R.
			systems in M(N, D) are exactly the magnitude
Thus the magnitude Bode plots for all
Bode plot for (N, D).
(a) How are systems in M(N, D) related to (N, D)? This boils down, of course, to
	identifying SISO linear systems that have a plot magnitude of 1 at all frequencies.
For	˜ ˜		˜	˜	=	˜ ˜
	(N, D) M(N, D) denote		φ0(N, D)			limω→0 ]HN,D˜ ˜ (ω) and φ∞(N, D) =
limω→∞ ]HN,D˜ ˜ (ω). Now let
		˜ ˜		˜	˜	˜ ˜
		φ(N, D) = φ∞(N, D)				− φ0(N, D).
With this notation prove the following statement.
(b)	φ(N, D) ≤	˜ ˜	˜	˜
		φ(N, D) for any (N, D) M(N, D).

E4.19 Construct polar plots corresponding to the Bode plots you made in Exercise E4.7.

E4.20 Construct polar plots corresponding to the Bode plots you made in Exercise E4.8.

E4.21 Construct polar plots corresponding to the Bode plots you made in Exercise E4.9.

E4.22 For the SISO linear system Σ of Exercise E4.2, plot the polar plots for various σ0 ≤ 0 and ω0 > 0.

T (s) =

Exercises for Chapter 4

157

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It is not uncommon to encounter a scheme for control design that relies on a plant being stable, i.e., having all poles in C−. This is in conflict with many plants—for example, the simple mass— that have a pole at s = 0. In the next exercise, you will investigate a commonly employed hack to get around plants having poles at s = 0.

E4.23 The transfer function

K τs + 1

is put forward as providing a “stable approximation to 1s .”

(a)Comment on how fiddling K and τ, particularly τ, render the approximation better or worse.

(b)Explain in what sense this approximation is valid, and also how it is invalid.

E4.24 For the following SISO linear systems in input/output form, determine whether they satisfy the hypotheses of Bode’s Gain/Phase Theorem. If the system does satisfy the hypotheses, verify explicitly that the theorem holds, and determine how good is the approximation (4.12) at various frequencies. If the system does not satisfy the hypotheses, determine explicitly whether the theorem does in fact hold.

(a)(N(s), D(s)) = (1, s).

(b)(N(s), D(s)) = (s, 1).

(c)(N(s), D(s)) = (1, s + 1).

(d)(N(s), D(s)) = (1, s − 1).

(e)(N(s), D(s)) = (s + 1, 1).

(f)(N(s), D(s)) = (s − 1, 1).

E4.25 Exercise on the approximate version of Bode’s Gain/Phase Theorem.

complete