mct book - chapter 8

7/30/2019 MCT Book - Chapter 8

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This version: 22/10/2004

Chapter 8

Performance of control systems

Before we move on to controller design for closed-loop systems, we should carefully in-vestigate to what sort of specifications we should make our design. These appear in variousways, depending on whether we are working with the time-domain, the s-domain, or thefrequency-domain. Also, the interplay of performance and stability can be somewhat subtle,and there is often a tradeoff to be made in this regard. Our objective in this chapter is toget the reader familiar with some of the issues that come up when discussing matters ofperformance. Also, we wish to begin the process of familiarising the reader with the ways

in which various system properties can affect the performance measures. In many cases,intuition is ones best friend when such matters come up. Therefore, we take a somewhatintuitive approach in Section 8.2. The objective is to look at tweaking parameters in a coupleof simple transfer function to see how they affect the various performance measures. It ishoped that this will give a little insight into how one might, say, glean some things aboutthe step response by looking at the Bode plot. The matter of tracking and steady-state errorcan be dealt with in a more structured way. Disturbance rejection follows along similar lines.The final matter touched upon in this chapter is the role of the sensitivity function in theperformance of a unity gain feedback loop. The minimisation of the sensitivity function isoften an objective of a successful control design, and in the last section of this chapter wetry to indicate why this might be the case.

Contents

8.1 Time-domain performance specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318

8.2 Performance for some classes of transfer functions . . . . . . . . . . . . . . . . . . . . . . 320

8.2.1 Simple first-order systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320

8.2.2 Simple second-order systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

8.2.3 The addition of zeros and more poles to second-order systems . . . . . . . . . . . 326

8.2.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3278.3 Steady-state error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328

8.3.1 System type for SISO linear system in input/output form . . . . . . . . . . . . . . 329

8.3.2 System type for unity feedback closed-loop systems . . . . . . . . . . . . . . . . . 333

8.3.3 Error indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

8.3.4 The internal model principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

8.4 Disturbance rejection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

8.5 The sensitivity function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342
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318 8 Performance of control systems 22/10/2004

8.5.1 Basic properties of the sensitivity function . . . . . . . . . . . . . . . . . . . . . . 343

8.5.2 Quantitative performance measures . . . . . . . . . . . . . . . . . . . . . . . . . . 344

8.6 Frequency-domain performance specifications . . . . . . . . . . . . . . . . . . . . . . . . . 346

8.6.1 Natural frequency-domain specifications . . . . . . . . . . . . . . . . . . . . . . . . 346

8.6.2 Turning time-domain specifications into frequency-domain specifications . . . . . 350

8.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350

8.1 Time-domain performance specifications

We begin with describing what is a common manner for specifying the desired behaviourof a control system. The idea is essentially that the system is to be given a step input, andthe response is to have various specified properties. You will recall in Proposition 3.40, itwas shown how to determine the step response by solving an initial value problem. In thischapter we will be producing many step responses, and in doing so we have used this initial

value problem method. In any case, when one gives a BIBO stable input/output system astep input, it will, after some period of transient response, settle down to some steady-statevalue. With this in mind, we make some definitions.

8.1 Definition Let (N, D) be a BIBO stable SISO linear system in input/output form, let1N,D (t) be the step response, and suppose that limt 1N,D (t) = 1ss R. (N, D) is step-pable provided that 1ss = 0. For a steppable system (N, D), let y(t) be the response tothe reference r(t) = 1(t)(1ss)1 so that limt y(t) = 1. We call y(t) the normalised stepresponse.

(i) The rise time is defined by

tr = sup

y(t) t for all t [0, ].

(ii) For (0, 1) the -settling time is defined byts, = inf

|y(t) 1| < for all t [, ).

(iii) The maximum overshoot is defined by yos = supt{y(t) 1} and the maximumpercentage overshoot is defined by Pos = yos 100%.

(iv) The maximum undershoot is defined by yus = supt{y(t)} and the maximumpercentage undershoot is defined by Pus = yus 100%.

For the most part, these definitions are self-explanatory once you parse the symbolism. Apossible exception is the definition of the rise time. The definition we provide is not the usualone. The idea is that rise time measures how quickly the system reaches its steady-statevalue. A more intuitive way to measure rise time would be to record the smallest time atwhich the output reaches a certain percentage (say 90%) of its steady-state value. However,the definition we provide still gives a measure of the time to reach the steady-state value,and has the advantage of allowing us to state some useful results. In any event, a typicalstep response is shown in Figure 8.1 with the relevant quantities labelled.

Notice that in Definition 8.1 we introduce the notion of a steppable input/output system.The following result gives an easy test for when a system is steppable.
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22/10/2004 8.1 Time-domain performance specifications 319

y(

t)

t

yus

yos

tr ts,

2

Figure 8.1 Performance parameters in the time-domain

8.2 Proposition A BIBO stable SISO linear system (N, D) in input/output form is steppableif and only if lims0 TN,D (s) = 0. Furthermore, in this case limt 1N,D (t) = TN,D (0).Proof By the Final Value Theorem, Proposition E.9(ii), the limiting value for the stepresponse is

limt

1N,D (t) = lims0

s1N,D (s) = lims0

TN,D (s).

Therefore limt 1N,D (t) = 0 if and only if lims0 1N,D (s) = 0, as specified. So it is quite simple to identify systems which are not steppable. The idea here is quitesimple. When lims0 TN,D (s) = 0 this means that the input only appears in the differentialequation after being differentiated at least once. For a step input, this means that the right-hand side of the differential equation forming the initial value problem is zero, and since(N, D) is BIBO stable, the response must decay to zero. The following examples illustrateboth sides of the story.

8.3 Examples 1. We first take (N(s), D(s)) = (1, s2 +2s +2). By Proposition 3.40 the initialvalue problem to solve for the step response is

1N,D (t) + 2 1N,D (t) + 2 1N,D (t) = 1, y(0) = 0, y(0) = 0.

The step response here is computed to be

1N,D (t) =12 1

2et(cos t + sin t).

Note that limt 1N,D (t) =1

2 , and so the system is steppable. Therefore, we can computethe normalised step response, and it is

y(t) = 1 et(cos t + sin t).2. Next we take (N(s), D(s)) = (s, s2 + s2+2). Again by Proposition 3.40, the initial value

problem to be solved is

1N,D (t) + 2 1N,D (t) + 2 1N,D (t) = 0, y(0) = 0, y(0) = 1.

The solution is 1N,D (t) = et sin t which decays to zero, so the system is not steppable.

This is consistent with Proposition 8.2 since lims0 TN,D (s) = 0.
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8.2 Performance for some classes of transfer functions

It is quite natural to specify performance criteria in the time-domain. However, in order toperform design, one needs to see how system parameters affect the time-domain performance,and how various performance measures get reflected in the other representations of controlsystemsthe Laplace transform domain and the frequency-domain. To get a feel for this,in this section we look at some concrete classes of transfer functions.

8.2.1 Simple first-order systems The issues surrounding performance of control sys-tems can be difficult to grasp, so our approach will be to begin at the beginning, describingwhat happens with simple systems, then moving on to talk about things more general. Inthis, the first of the three sections devoted to rather specific transfer functions, we will beconsidering the situation when the transfer function

TL(s) =RL(s)

1 + RL(s)

has a certain form, without giving consideration to the precise form of the loop gain RL.

The simplest case one can deal with is one where the system transfer function is first order.If we normalise the transfer function so that it will have a steady-state response of 1 to aunit step input, the transfer functions under consideration look like

T(s) =1

s + 1,

and so essentially depend upon a single parameter . The step response of such a system isdepicted in Figure 8.2. The parameter is exactly the rise time for a first-order system, at

y(t)

t

t =

Figure 8.2 Step-response for typical first-order system

least by our definition of rise time. Thus, by making smaller, we ensure the system willmore quickly reach its steady-state value.

Lets see how this is reflected in the behaviour of the poles of the transfer function, and inthe Bode plot for the transfer function. The behaviour of the poles in this simple first-ordercase is straightforward. There is one pole at s = 1. Thus making smaller moves this
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22/10/2004 8.2 Performance for some classes of transfer functions 321

pole further from the origin, and further into the negative complex plane. Thus the issues indesigning a first-order transfer function are one: move the pole as far to the left as possible.Of course, not many transfer functions that one will obtain are first-order.

The Bode plot for the transfer function T is very simple, of course: it has a single breakfrequency at = 1

. A typical such plot is shown in Figure 8.3. Obviously, the greater

dB

log

= 1

Figure 8.3 Bode plot for typical first-order system

values of break frequency correspond to quicker response times. Thus, for simple first-ordersystems, the name of the game can be seen as pushing the break frequency as far as possibleto the right. This is, in fact, a general strategy, and we begin an exploration of this bystating an obvious result.

8.4 Proposition The magnitude of T(s) at s =i

is 12

.

Proof This is a simple calculation since

T

i

=

1

i + 1,

whose magnitude we readily compute to be 12

.

We also compute 20 log 12

3.01dB. Thus the magnitude ofT( i) is approximately 3dBwhich one can easily pick off from a Bode plot. We call 1

the bandwidth for the first-order

transfer function. Note that for first-order systems, larger bandwidth translates to betterperformance, at least in terms of rise time.

8.2.2 Simple second-order systems First-order systems are not capable of sustainingmuch rich behaviour, so lets see how second-order systems look. After again requiring atransfer function which produces a steady-state of 1 to a unit step input, the typical second-order transfer function we look at is

T,0(s) =20

s2 + 20s + 20,
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which depends on the parameters and 0. If we are interested in BIBO stable transferfunctions, the Routh/Hurwitz criterion, and Example 5.35 in particular, then we can withoutloss of generality suppose that both and 0 are positive.

The nature of the closed form expression for the step response depends on the value of. In any case, using Proposition 3.40(ii), we ascertain that the step response is given by

y(t) =

1 + 21 1e(

21)0t

21e(+

21)0t, > 1

1 e0t(1 + 0t), = 11 e0t

cos(

1 20t) +

12sin(

1 20t)

, < 1.

This is plotted in Figure 8.4 for various values of . Note that for large there is no

y(t)

t

= 0.2 = 0.4

= 0.7

= 1

= 2

Figure 8.4 Step response for typical second-order system for vari-ous values of

overshoot, but that as decreases, the response develops an overshoot. Let us make someprecise statements about the character of the step response of a second-order system.

8.5 Proposition Let y(t) be the step response for the SISO linear system (N, D) with transferfunction T,0. The following statements hold:

(i) when 1 there is no overshoot;(ii) when < 1 there is overshoot, and it has magnitude yos = e

/

12 and occurs at

time tos = 120 ;(iii) if < 1 the rise time satisfies

(e0tr

1 2) +

1 2 cos(0

1 2tr) + (0tr + ) sin(0

1 2tr) =e0trtr

1 2.

Proof (i) and (ii) The proof here consists of differentiating the step response and setting itto zero to obtain those times at which it is zero. When 1 the derivative is zero only whent = 0. When < 1 the derivative is zero when 0

1 2t = n, n Z+. The smallest
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positive time will be the time of maximum overshoot, so we take n = 1, and from this thestated formulae follow.

(iii) The rise time tr is the smallest positive time which satisfies

y(t0) =y(t0)

t0,

since the slope at the rise time should equal the slope of the line through the points (0 , 0) and(t0, y(t0)). After some manipulation this relation has the form stated in the proposition.

In Figure 8.5 we plot the maximum overshoot and the time at which it occurs as a function of

0.2 0.4 0.6 0.8 1

0.2

0.4

0.6

0.8

1

| |

yos

0.2 0.4 0.6 0.8 1

5

10

15

20

25

30

35

40

| |

tos

0

Figure 8.5 Maximum overshoot (left) and time of maximum over-

shoot (right) as functions of

. Note that the overshoot decreases as increases, and that the time of overshoot increasesas increases.

The rise time tr is not easily understood via an analytical formula. Nevertheless, we maynumerically determine the rise time for various and 0 by solving the equation in part (iii) ofProposition 8.5 and the results are shown in Figure 8.6. In that same figure we also plot thecurve tr =

950

, and we see that it is a very good approximationindeed it is indistinguishablein our plot from the curve for = 0.9. Thus, for second-order transfer functions T,0(s) with < 1, it is fair to use tr 905 . Of course, for more general transfer functions, even moregeneral second-order transfer functions (say, with nontrivial numerators), this relationshipcan no longer be expected to hold.

From the above discussion we see that overshoot is controlled by the damping factor and rise time is essentially controlled by the natural frequency 0. However, we note that thetime for maximum overshoot, tos, depends only upon , and indeed increases as increases.Thus, there is a possible tradeoff to make when selecting if one chooses a small in the-settling time specification. A commonly used rule of thumb is that one should choose = 1

2which leads to yos = e 0.043 and tos0 =

2 4.44.

Lets look now to see how the poles of T,0 depend upon the parameters and 0. One

readily determines that the poles are 0(1 2). As we have just seen, good response
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0.25 0.5 0.75 1 1.25 1.5 1.75 2

5

10

15

20

25

tr

0

= 0.1

= 0.5

= 0.9 and 950

Figure 8.6 Behaviour of rise time versus 0 for various < 1

dictates that we should have poles with nonzero imaginary part, so we consider the situationwhen < 1. In this case, the poles are located as in Figure 8.7. Thus the poles lie on a circle

0

0

0

1 20

Re

Im

Figure 8.7 Pole locations for T,0 when 0 < < 1

of radius 0, and the angle they make with the imaginary axis is sin1 . Our rule of thumb

of taking = 12

then specifies that the poles lie on the rays emanating from the origin intoC at an angle of 45 from the imaginary axis.

Finally, for second-order systems, lets see how the parameters and 0 affect the Bodeplot for the system. In Exercise E4.6 the essential character of the frequency response forT,0 is investigated, and let us just record the outcome of this.

8.6 Proposition If H,0() = T,0(i), then the following statements hold:
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(i) |H,0()| =20

(20 2)2 + 42202;

(ii) H,0() = arctan2020 2

;

(iii) for < 12

, the function |H,0()| has a maximum of 1212 at the frequencym = 01 22;

(iv) H,0(m) = arctan

1 22

.

In Figure 8.8 we label the typical points on the Bode plot for the transfer function T,0

dB

log

m

|H,0(m)|

Figure 8.8 Typical Bode plot for second-order system with < 12

when < 12

. As we decrease the peak becomes larger, and shifted to the left. The phase,as we decrease , tends to 90 at the peak frequency m.

Based on the discussion with first-order systems, for a second-order system with transferfunction T,0, the bandwidth is that frequency ,0 > 0 for which |T,0(i,0)| = 12 . Thefollowing result gives an explicit expression for bandwidth of second-order transfer functions.Its proof is via direct calculation.

8.7 Proposition When < 12

the bandwidth for T,0 satisfies

,00

= 1 22 + 2(1 22) + 44.Thus we see that the bandwidth is directly proportional to the natural frequency 0. Thedependence on is shown in Figure 8.9. Thus we duplicate our observation for first-ordersystems that one should maximise the bandwidth to minimise the rise time. This is one ofthe general themes in control synthesis, namely that, all other things being constant, oneshould maximise bandwidth.


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0.1 0.2 0.3 0.4 0.5 0.6 0.7

1

1.2

1.4

1.6

1.8

2

2.2

2.4

,0

0

Figure 8.9 Dependence of bandwidth on for second-order sys-tems

8.2.3 The addition of zeros and more poles to second-order systems In our generalbuildup, the next thing we look at is the effect of adding to a second-order transfer functioneither a zero, i.e., making a numerator which has a root, or an additional pole. The idea isthat we will investigate the effect that these have on the nature of the second-order response.This is carried out by Jr. [1949].l

Adding a zero To investigate what happens to the time signal when we add a zero to asecond-order system with a zero placed at 0. If we normalise the transfer function tothat it has unit value at s = 0, we get

T(s) =

20

0

s + 0

s2 + 20s + 20 .

For concreteness we take = 12

and 0 = 1. The step responses and magnitude Bode plotsare shown in Figure 8.10 for various values of . We note a couple of things.

8.8 Remarks 1. The addition of a zero increases the overshoot for < 3, and dramaticallyso for < 1.

2. If the added zero is nonminimum phase, i.e., when < 0, the step response exhibitsundershoot. Thus nonminimum phase systems have this property of reacting in a mannercontrary to what we want, at least initially. This phenomenon will be explored further

in Section 9.1.3. The magnitude Bode plot is the same for = 1

2as it is for = 1

2. Where the Bode

plots will differ is in the phase, as in the former case, the system is nonminimum phase.

4. When comparing the step response and the magnitude Bode plots for positive s, onesees that the general tendency of larger bandwidths1 to produce shorter rise times ispreserved.

1We have not yet defined bandwidth for general transfer functions, although it is clear how to do so. Thebandwidth, roughly, is the smallest frequency above which the magnitude of the frequency response remainsbelow 1

2times its zero frequency value. This is made precise in Definition 8.25.
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2 4 6 8 10 1 2 14

PSFt

-1

0

1

2

3

PS

F1Sigma(t)

= 0.5

= 1

= 2

= 0

= 0.5

-1.5 -1 -0.5 0 0.5 1 1.5 2

-80

-60

-40

-20

0

dB

log

= 0

= 2

= 1

= 12

Figure 8.10 The effect of an added zero for = 12 , 0, 12 , 1, 2

Adding a pole Now we look at the effect of an additional pole at 0. The normalisedtransfer function isT(s) =

30(s + 0)(s2 + 20s + 20)

.

We once again fix = 12

and 0 = 1, and plot the step response for varying in Figure 8.11.We make the following observations.

8.9 Remarks 1. If a pole is added with < 3, the rise time will be dramatically increased.This is also reflected in the bandwidth of the system increasing with .

2. The larger bandwidths in this case are accompanied by a more pronounced peak in

the Bode plot. As with second-order systems where this is a consequence of a smallerdamping factor, we see that there is more overshoot.

8.2.4 Summary This section has been something of a mixed bag of examples andinformal observations. We do not try to make it more than that at this point. Some of thethings covered here have a more general and rigorous treatment in Chapter 9. However,it is worth summarising the gist of what we have said in an informal way. These are nottheorems. . .

1. Increased bandwidth can mean shorter rise times.


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2 4 6 8 10 1 2 14

PSFt

0.2

0.4

0.6

0.8

1

1.2

P

SF1Sigma(t)

= 10

= 2

= 0.5

= 1

-1.5 -1 -0.5 0 0.5 1 1.5 2

-120

-100

-80

-60

-40

-20

0

dB

log

Figure 8.11 The effect of an added pole for = 12 , 1, 2, 10

2. In terms of poles, larger bandwidth sometimes means closed-loop poles that are farfrom the imaginary axis.

3. Large overshoot can arise when the Bode plot exhibits a large peak at some frequency.This is readily seen for second-order systems, but can happen for other systems.

4. In terms of poles of the closed-loop transfer function, large overshoot can arise whenpoles are close to the imaginary axis, as compared to their distance from the real axis.

5. Zeros of the closed-loop transfer function lying in C+ can lead to undershoot in thestep response, this having a deleterious effect on the systems performance.

These rough guidelines can be useful in predicting the behaviour of a system based uponthe location of its poles, or on the shape of its frequency response. The former connectionforms the basis for root-locus design which is covered in Chapter 11. The frequency responseideas we shall make much use of, as they form the basis for the design methodology ofChapters 12 and 15. It is existence of the rigorous mathematical ideas for control design inChapter 15 that motivate the use of frequency response methods in design.

8.3 Steady-state error

An important consideration is that the difference between the reference signal and theoutput should be as small a possible. When we studied the PID controller in Section 6.5


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22/10/2004 8.3 Steady-state error 329

we noticed that with an integrator it was possible to make at least certain types of transferfunction have no steady-state error. Here we look at this in a slightly more systematicmanner. The first few subsections deal with descriptive matters.

8.3.1 System type for SISO linear system in input/output form For a SISO system(N, D) in input/output form, a controlled output is a pair (r(t), y(t)) defined for t [0, )

with the property that Dddt

y(t) = N

ddt

r(t)

The error for a controlled output (r(t), y(t)) is e(t) = r(t) y(t), and the steady-stateerror is

limt

(r(t) y(t)),and we allow the possibility that this limit may not exist. With this language, we have thefollowing definition of system type.

8.10 Definition Let k 0. A signal r U defined on [0, ) is of type k if r(t) = Ctk forsome C > 0. A BIBO stable SISO linear system (N, D) is oftype k if limt(r(t)

y(t))

exists and is nonzero for every controlled output (r(t), y(t)) with r(t) of type k.

In the definition of the type of a SISO linear system in input/output form, it is essentialthat the system be BIBO stable, i.e., that the roots of D all lie in C. If this is not the case,then one might expect the output to grow exponentially, and so error bounds like those inthe definition are not possible.

The following result attempts to flush out all the implications of system type.

8.11 Proposition Let (N, D) be a BIBO stable SISO linear system in input/output form. Thefollowing statements are equivalent:

(i) (N, D) is of type k;

(ii) limt(r(t) y(t)) exists and is nonzero for some controlled output (r(t), y(t)) withr(t) of type k.

(iii) lims0

1

sk

1 TN,D (s)

exists and is nonzero.

The preceding three equivalent statements imply the following:

(iv) limt(r(t) y(t)) = 0 for every controlled output (r(t), y(t)) with r(t) of type with < k.

Proof (i) (ii) That (i) implies (ii) is clear. To show the converse, suppose thatlimt(r(t) y(t)) = K for some nonzero K and for some controlled output (r(t), y(t))with r(t) of type k. Suppose that deg(D) = n and let y1(t), . . . , yn(t) be the n linearlyindependent solutions to D

ddt

y(t) = 0. If yp(t) is a particular solution to D

ddt

y(t) = r(t)

we must havey(t) = c1y1(t) + + cnyn(t) + yp(t)

for some c1, . . . , cn R. By hypothesis we then have

limt

r(t) c1y1(t) cnyn(t) yp(t)

= lim

t

r(t) yp(t)

= K.

Here we have used the fact that the roots of D are in C so the solutions y1(t), . . . , yn(t) alldecay to zero.
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Now let (r(t), y(t)) be a controlled output with r(t) be an arbitrary signal of type k.Note that we must have r(t) = Ar(t) for some A > 0. We may take yp(t) = Ayp(t) as aparticular solution to D

ddt

y(t) = N

ddt

r(t) by linearity of the differential equation. This

means that we must have

y(t) = c1y1(t) + + cnyn(t) + Ayp(t)

for some c1, . . . , cn R

. Thus we havelim

t

r(t) y(t) = lim

t

Ar(t) c1y1(t) cnyn(t) Ayp(t)

= lim

tA

r(t) yp(t)

= AK.

This completes this part of the proof.(ii) (iii) Let (r(t), y(t)) be a controlled output with r(t) = tkk! , and suppose that

y(0) = 0, y(1)(0) = 0 . . . , y(n1)(0) = 0. Note that r(s) = 1sk+1

. Taking the Laplace trans-form of the differential equation D

ddt

y(t) = N

ddt

r(t) gives D(s)y(s) = N(s)r(s). By

Proposition E.9(ii) we have

limt

r(t) y(t) = lims0 sr(s) y(s)= lim

s0s

r(s) TN,D (s)r(s)

= lims0

s1 TN,D (s)

sk+1

from which we ascertain that

limt

r(t) y(t) = lim

s01 TN,D (s)

sk. (8.1)

From this the result clearly follows.(iii) = (iv) Suppose that1 TN,D (s)

sk= K

for some nonzero constant K and let {0, 1, . . . , k 1}. Let (r(t), y(t)) be a controlledoutput with r(t) a signal of type . Since the roots of D are in C, we can without loss ofgenerality suppose that y(0) = 0, y(1)(0) = 0, . . . , y(n1)(0) = 0. We then have

limt

r(t) y(t) = lim

s01 TN,D (s)

s

= lims0

sk1 TN,D (s)

sk

= Klims0

sk = 0.

This completes the proof.

Let us examine the consequences of this result by making a few observations.

8.12 Remarks 1. Although we state the definition for systems in input/output form, it ob-viously applies to SISO linear systems and to interconnected SISO linear systems sincethese give rise to systems in input/output form after simplification of their transfer func-tions.


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2. The idea is that a system of type k can track up to a constant error a reference signalwhich is a polynomial of degree k. Thus, for example, a system of type 0 can track astep input up to a constant error. A system of type 1 can track a ramp input up to aconstant error, and can exactly track a step input for large time.

Lets see how this plays out for some examples.

8.13 Examples In each of these examples we look at a transfer function, decide what is itstype, and plot its response to inputs of various types.

1. We take

TN,D (s) =1

s2 + 3s + 2.

This transfer function is type 0, as may be determine by checking the limit of part (iii)of Proposition 8.11. For a step reference, ramp reference, and parabolic reference,

r1(t) =

1, t 00, otherwise

r2(t) =

t, t 00, otherwise

r3(t) =

t2, t 00, otherwise,

respectively, we may ascertain using Proposition 3.40, that the step, ramp, and parabolicresponses are

y1(t) =12 +

12

e2t et, y2(t) = 14(2t 3) 14e2t + et,y3(t) = 1

4(2t2

6t + 7) + 1

4e2t

2et,

and the errors, ei(t) = ri(t) yi(t), i = 1, 2, 3, are plotted in Figure 8.12. Notice thatthe step error response has a nonzero limit as t , but that the ramp and parabolicresponses grow without limit. This is what we expect from a type 0 system.

2. We take

TN,D (s) =2

s2 + 3s + 2

which has type 1, using Proposition 8.11(iii). The step, ramp, and parabolic responsesare

y1(t) = 1 + e2t

2et

, y2(t) =1

2(2t 3) 1

2e2t

+ 2et

,y3(t) =

12(2t

2 6t + 7) + 12e2t 4et,

and the errors are shown in Figure 8.13. Since the system is type 1, the step responsegives zero steady-state error and the ramp response has constant steady-state error. Theparabolic input gives a linearly growing steady-state error.

3. The last example we look at is that with transfer function

TN,D (s) =3s + 2

s2 + 3s + 2.
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2 4 6 8 10

0.2

0.4

0.6

0.8

1

t

e1

(t)| |

2 4 6 8 10

1

2

3

4

5

6

t

e2

(t)

| |

2 4 6 8 10

10

20

30

40

50

60

70

t

e3

(t)| |

Figure 8.12 Step (top left), ramp (top right), and parabolic (bot-tom) error responses for a system of type 0

2 4 6 8 10

0.2

0.4

0.6

0.8

1

t

e1

(t)| |

2 4 6 8 10

0.25

0.5

0.75

1

1.25

1.5

1.75

2

t

e2

(t)| |

2 4 6 8 10

5

10

15

20

25

30

t

e3

(t)

| |

Figure 8.13 Step (top left), ramp (top right), and parabolic (bot-tom) responses for a system of type 1


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2 4 6 8 10

-0.2

0

0.2

0.4

0.6

0.8

1

t

e1

(t)| |

2 4 6 8 10

0.05

0.1

0.15

0.2

0.25

t

e2

(t)| |

2 4 6 8 10

0.2

0.4

0.6

0.8

1

1.2

1.4

t

e3

(t)| |

Figure 8.14 Step (top left), ramp (top right), and parabolic (bot-tom) responses for a system of type 2

This transfer function is type 2. The step, ramp, and parabolic responses are

y1(t) = 1 2e2t + et, y2(t) = t + e2t et, y3(t) = t2 1 e2t + 2e2t.

The errors are plotted in Figure 8.14. Note that the step and ramp steady-state errorsshrink to zero, but the parabolic response has a constant error.

8.3.2 System type for unity feedback closed-loop systems To see how the steady-state error is reflected in a simple closed-loop setting, let us look at the situation depictedoriginally in Figure 6.25, and reproduced in Figure 8.15. Thus we are not thinking here

r(s) RL(s) y(s)

Figure 8.15 Unity gain feedback loop for investigating steady-stateerror

so much about having a controller and a plant, but as combining these to get the transferfunction RL which is the loop gain in this case. In any event, we may directly give conditions
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on the transfer function RL to determine the type of the closed-loop transfer function

TL(s) =RL(s)

1 + RL(s).

These conditions are as follows.

8.14 Proposition Let RL R(s) be proper and define

TL(s) =RL(s)

1 + RL(s).

If (N, D) denotes the c.f.r. of TL, then (N, D) is of type k > 0 if and only if lims0 skRL(s)exists and is nonzero. (N, D) is of type 0 if and only if lims0 RL(s) exists and is not equalto 1.Proof We compute

1 TL(s) = 11 + RL(s)

.

Thus, by Proposition 8.11(iii), (N, D) is of type k if and only if

lims0

1

sk(1 + RL(s))

exists and is nonzero. For k > 0 we have

lims0

1

sk(1 + RL(s))=

1

lims0 skRL(s),

and the result follows directly in this case. For k = 0 we have

lims0

1sk(1 + RL(s))

= 11 + lims0 RL(s)

.

Thus, provided that RL(0) = 1 as hypothesised, the system is of type 0 if and only iflims0 RL(s) exists.

The situation here, then, is quite simple. If RL(s) is proper, for some k 0 we can write

RL(s) =NL(s)

skDL(s)

withD

L monic,D

L andN

L coprime, andD

L(0) = 0. Thus we factor from the denominatoras many factors of s as we can. Each such factor is an integrator. The situation is depictedin Figure 8.16. One can see, for example, why often the implementation of a PID controllaw (with integration as part of the implementation) will give a type 1 closed-loop system.To be precise, we can state the following.

8.15 Corollary For the unity gain feedback loop of Figure 8.15, the closed-loop system is oftype k > 0 if and only if there exists R R(s) with the properties

(i) R(0) = 0 and(ii) RL(s) =

1sk

R(s).


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r(s)1

s. . . 1

sR(s) y(s)

Figure 8.16 A unity feedback loop with k integrators and R(0) = 0

Furthermore, if RL is the product of a plant transfer function RP with a controllertransfer function RC(s) = K(1 + TDs +

1TIs

), then the closed-loop system will be of type 1provided that lims0 RP(s) exists and is nonzero.

Proof We need only prove the second part of the corollary as the first is a direct consequenceof Proposition 8.14. The closed-loop transfer function TL satisfies

1 TL(s) = 11 + K(1 + TDs +

1TIs

)RP(s),

from which we determine that

lims0

1 TL(s)s

=1

lims0 K(s + TDs2 + 1TI)RP(s),

and from this the result follows, since if lims0 RP(s) exists and is nonzero, thenlims0 sRP(s) = 0 for > 0.

Interestingly, there is more we can say about type k systems when k 1. The followingresult gives more a detailed description of the steady-state error in these cases.

8.16 Proposition Let y(t) be the normalised step response for the closed-loop system depictedin Figure 8.15. The following statements hold:

(i) if the closed-loop system is type 1 with lims0 sRL(s) = C with C a nonzero constant,then

0

e(t) dt = 1C;

(ii) if the closed-loop system is type k with k 2 then0

e(t) dt = 0.

Proof (i) Since limt e(t) = 0, e(t) must be strictly proper. Therefore, by Proposi-tion E.10, taking s0 = 0, we have

0

e(t) dt = lims0

e(s).

Since

e(s) = (1 TL(s))r(s) = ss + sRL(s)

1

s,

we have lims0 e(s) =1C.
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(ii) The idea here is the same as that in the previous part of the result except that wehave

e(s) = (1 TL(s))r(s) = sk

sk + skRL(s)

1

s,

with k 2, and so lims0 e(s) = 0. The essential point is that the proposition will hold for any loop gain RL of type 1 (for

part (i)) or type 2 (for (ii)). An interesting consequence of the second part of the propositionis the following.

8.17 Corollary Let y(t) be the normalised step response for the closed-loop system depictedin Figure 8.15. If the closed-loop transfer system is type k for k 2, then y(t) exhibitsovershoot.

Proof Since the error starts at e(0) = 1, in order that0

e(t) dt = 0,

the error must at some time be negative. However, negative error means overshoot.

This issue of determining the behaviour of a closed-loop system which depends onlyon properties of the loop gain is given further attention in Section 9.1. Here the effect ofunstable poles and nonminimum phase zeros for the plant is flushed out in a general setting.

8.3.3 Error indices From standard texts (see Truxal, page 83).h

8.3.4 The internal model principle To this point, the discussion has centred aroundtracking type k signals, i.e., those that are powers of t. However, one often wishes totrack more exotic signals. The manner for doing so is suggested, upon reflection, by the

discussion to this point, and is loosely called the internal model principle.h

8.4 Disturbance rejection

Our goal in this section is to do something rather similar to what we did in the last section,except that we wish to look at the relationships between disturbances and the output. Sincea disturbance may enter a system in any of the signals, the appropriate setting here is thatof an interconnected SISO linear system (S,G). We will suppose that we are dealing withsuch a system and that there are m nodes with the input being node 1 and the output beingnode m.

We need a notion of output like that introduced in the previous section when talkingabout transfer function types. We let i {2, . . . , m} and let (Si,Gi) be the ith-appendedsystem. A j-disturbed output with input at node i is a pair (d(t), y(t)) defined on[0, ) where y(t) is the response at node j for the input d(t) at node i for the interconnectedSISO linear system (Si,Gi), where the input at node 1 is taken to be zero.

8.18 Definition Let (S,G) be an IBIBO stable interconnected SISO system as above and letj {2, . . . , m}. The system is of disturbance type k at node j with input at nodei if limt y(t) exists and is nonzero for every j-disturbed output (d(t), y(t)) with input atnode i, where d(t) is of type k.


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22/10/2004 8.4 Disturbance rejection 337

The idea here is very simple. We wish to allow a disturbance at any node which mayenter the system at any other node. Just where the disturbance enters a system and whereit is measured can drastically change the behaviour of the system. Typically, however, onemeasures the disturbance at the output of the interconnected system. That is, typicallyin the above discussion j is taken to be m. Before getting to an illustration by example,let us provide a result which gives some obvious consequences of the notion of disturbancetype. The following result can be proved along the same lines as Proposition 8.11 and byapplication of the Final Value Theorem.

8.19 Proposition Let (S,G) be an IBIBO stable interconnected SISO linear system with inputin node 1 and output in node m. For j {2, . . . , m} the following statements are equivalent:

(i) (S,G) is of j-disturbance type k with input at node i;

(ii) limt y(t) exists and is nonzero for some j-disturbed output (d(t), y(t)) with input atnode i, where d(t) is of type k;

(iii) lims0 1sk Tji (s) exists and is nonzero.

The preceding three equivalent statements imply the following:

(iv) limt y(t) = 0 for every j-disturbed output (d(t), y(t)) with d(t) of type with < k.Now let us consider a simple example.

8.20 Example Let us consider the interconnected SISO linear system whose signal flow graphis shown in Figure 8.17. The system has three places where disturbances ought to naturally

r(s)1

// RC(s)

// RP(s)

// 1

//

1

ff y(s)

Figure 8.17 A system ready to be disturbed

be considered, as shown in Figure 8.18. Note that one should strictly add another node to

d1(s)

1

d3(s)

1

d3(s)

1

RC(s) //

RP(s) //

1 //

1

hh y(s)

Figure 8.18 Disturbances

the output node y, but this does not change anything, so it can be safely omitted. This willgenerally be the case for an interconnected SISO linear system. In any case, the transferfunction for the signal flow graph is

TS,G(s) =y(s)

r(s)=

RC(s)RP(s)

1 + RC(s)(s)RP(s),
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the transfer function from d1 to y is

T1(s) =y(s)

d1(s)=

RC(s)RP(s)

1 + RC(s)RP(s),

the transfer function from d2 to y is

T2(s) = y(s)d2(s)

= RP(s)1 + RC1(s)RC2(s)RP(s)

,

and the transfer function from d3 to y is

T3(s) =y(s)

d3(s)=

1

1 + RC(s)RP(s).

Lets attach some concrete transfer functions at this point. Our transfer functions arepretty artificial, but serve to illustrate the point. We take

RC(s) =

1

s , RP(s) =

s

s + 2 .

We compute the corresponding transfer functions to be

TS,G(s) =s2

s2 + s + 2, T1(s) =

s2

s2 + s + 2, T2(s) =

s

s2 + s + 2, T3(s) =

s + 2

s2 + s + 2.

A straightforward application of Proposition 8.11(iii) indicates that TS,G is of type 0. It isalso easy to see from Proposition 8.19(iii) that the system is of disturbance type 2 withrespect to the disturbance d1, of disturbance type 1 with respect to the disturbance d2, andof disturbance type 0 with respect to d3.

As we see in this example, the ability of a system to reject disturbances may differfrom the ability of a system to represent a reference signal. Furthermore, for disturbancerejection, one typically wants the transfer function from the disturbance to the output tohave the property that the expected disturbance type gives zero steady-state error. Thus,if step disturbances are what is expected, it is sufficient that the system be of disturbancetype 0.

Let us look at our falling mass example thinking of the gravitational force as a distur-bance.

8.21 Example (Example 6.60 contd) The block diagram for the falling mass with the grav-itational force is shown in Figure 8.19. We think of the input

mg as a disturbance input

and so write d(t) = mg1(t). This input is then a step input if we think of holding the massthen letting it go. In fact, in our analysis, lets agree to take y(0) = 0 and y(0) = 0 and seeif the controller can return the mass to y = 0 after we let it fall under gravity. The transferfunction from the disturbance to the output is

Td(s) =y(s)

d(s)=

ms2

ms2 + RC(s).

Let us look at the form of this transfer function for the various controllers we employed inExample 6.60.


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r(s) RC(s)1

ms2y(s)

mg

Figure 8.19 Block diagram for falling mass with gravitational force

1. With proportional control we have RC(s) = K and so

Td(s) =ms2

ms2 + K.

This transfer function has poles on the imaginary axis, and so is not a candidate for havingits type defined. Nonetheless, we can compute the time response using Proposition 3.40

given d(s) = mg

s . We ascertain that the output to this disturbance is

yd(t) = mg cos

K

mt.

One can see that the system does not respond very nicely in this case to the step distur-bance, and to further illustrate the point, we plot this step response in Figure 8.20 for

1 2 3 4 5

-10

-5

0

5

10

t

1

(t

Figure 8.20 Response of falling mass to step disturbance with pro-

portional control

m = 1, g = 9.81, and K = 28.

2. For derivative control we take RC(s) = KTDs which gives

Td(s) =ms

ms + KTD.

This disturbance transfer function is type 1 and so the steady-state disturbance to thestep input d(t) = mg is zero. Using Proposition 3.40 we determine that the response is

yd(t) = mgeKTDt/m
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1 2 3 4 5

-8

-6

-4

-2

0

2

t

1

(t

Figure 8.21 Response of falling mass to step disturbance withderivative control

and we plot this response in Figure 8.21 for m = 1, g = 9.81, K = 28, and TD =9

28 .Note that this response decays to zero which is consistent with our observation that theerror to a step disturbance should decay to zero in the steady-state.

3. If we use an integral controller we take RC(s) =1

TIsfrom which we compute

Td(s) =mTIs

3

mTIs3 + K.

To examine this transfer function for type, let us be concrete and choose m = 1, g = 9.81,K = 28, TD =

928

, and TI =710

. One determines (Mathematica!) that the roots ofmTIs

3 + K are then3

5

3

3

5i,

23

5.

Since the transfer function has poles in C+, the notion of type is not applicable. We plotthe response to the step gravitational disturbance in Figure 8.22, and we see that it is

1 2 3 4 5

-7500

-5000

-2500

0

2500

5000

7500

10000

t

1

(t

Figure 8.22 Response of falling mass to step disturbance with in-tegral control

badly behaved. Thus the given integral controller will magnify the disturbance.
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4. Finally we combine the three controllers into a nice PID control law: RC(s) = K +KTDs +

KTIs

. The disturbance to output transfer function is then

Td(s) =mTIs

3

mTIs3 + KTDTIs2 + KTIs + K.

For our parameters m = 1, g = 9.81, K = 28, TD =928

, and TI =710

, we determine

that the roots of the denominator polynomial are 5, 2 2i (recall that we had chosenthe PID parameters in just this way). Since these roots are all in C, we may make theobservation that the disturbance type is 3. Therefore the steady-state error to a stepdisturbance should be zero. The response of this transfer function to the gravitationalstep disturbance is shown in Figure 8.23 for m = 1, g = 9.81, K = 28, TD =

928 , and

1 2 3 4 5

-8

-6

-4

-2

0

2

t

1

(t

Figure 8.23 Response of falling mass to step disturbance with PIDcontrol

TI =710 . This response decays to zero as it ought to.

Let us address the question of how to interpret our computations for the gravitationaldisturbance to the falling mass. One needs to be careful not to misinterpret the disturbanceresponse for the system response. The input/output response of the system for the variouscontrollers was already investigated in Example 6.60. When examining the system includingthe effects of the disturbance, one must take into account both the input/output responseand the response to the disturbance.

For example, with the PID controller we had chosen, our analysis of both these responsessuggests that the system with the chosen parameters should behave nicely with the gravita-

tional force. To verify this, let us look at the situation where we hold the mass still at y = 0and at time t = 0 let it go. Our PID controller is then charged with bringing the mass backto y = 0. The initial value problem governing this situation is

my + KTDy(t) + Ky(t) +K

TI

t0

y() d = mg, y(0) = 0, y(0) = 0.

To solve this equation we differentiate once to get the initial value problem

m...y + KTDy(t) + Ky(t) +

K

TIy(t) = 0, y(0) = 0, y(0) = 0, y(0) = g.


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1 2 3 4 5

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

t

y(t

Figure 8.24 Response of falling mass with PID controller

The solution is plotted in Figure 8.24 for the chosen PID parameters. As should be the case,

the response is nice with the PID controller bringing the mass back to its initial height inrelatively short order.

This example is one which contains a lot of information, so you would be well served tospend some time thinking about it.

8.5 The sensitivity function

In this section we introduce and study a transfer function which is, at least for certainblock diagram configurations, complementary to the transfer function. The block diagramconfiguration we look at is the one we first looked at in Section 6.3.2, and the configuration

is reproduced in Figure 8.25. One may extend the discussion here to more general block

r(s) RL(s) y(s)

e(s)

n(s)

d(s)

Figure 8.25 Block diagram for investigating the sensitivity func-tion

diagram configurations, but this unity gain feedback setup is one which can be handled nicely.In this scenario, we are typically supposing that the loop gain RL is the product of a planttransfer function RP and a controller transfer function RC. But unless we say otherwise, we

just take RL as a transfer function in its own right, and do not concern ourselves with fromwhere it comes.

h this
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22/10/2004 8.5 The sensitivity function 343

8.5.1 Basic properties of the sensitivity function Let us first make some elementaryobservations concerning the closed-loop transfer function and the sensitivity function. Thefollowing result follows immediately from the definition of TL and SL.

8.22 Proposition For the feedback interconnection of Figure 9.3, the following statementshold:

(i)p

C

is a pole forR

L if and only ifS

L(p

) = 0 andT

L(p

) = 1;(ii) z C is a zero for RL if and only if SL(z) = 1 and TL(z) = 0.The import of this simple result is that it holds for any loop gain RL. Thus the zeros and

poles for RL are immediately reflected in the closed-loop transfer function and the sensitivityfunction. Let us introduce the following notation:

Z(SL) = {s C | SL(s) = 0}Z(TL) = {s C | TL(s) = 0} .

(8.2)

This notation will be picked up again in Section 9.2.1.Let us illustrate this with an example some of the simpler features of the sensitivity

function.

8.23 Example (Example 6.60 contd) Let us carry on with our falling mass example. Whatwe shall do is plot the frequency response of the closed-loop transfer function and the sen-sitivity function on the same Bode plot. We have RL(s) = RC(s)

1ms2 , and we shall use the

four controller transfer functions of Example 6.60. In all cases, we take m = 1.

1. We take RC(s) = K with K = 28. In Figure 8.26 we give the Bode plots for the closed-

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-20

-10

0

10

20

30

log

log

dB

deg

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-25

-20

-15

-10

-5

0

5

10

log

log

dB

deg

| |

Figure 8.26 Bode plots for the closed-loop transfer function andsensitivity function for the falling mass with proportional con-troller (left) and derivative controller (right). In each case, thesolid line represents the closed-loop transfer function and thedashed line the sensitivity function.

loop transfer function (the solid line) and the sensitivity function (the dashed line).


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2. Next we take RC(s) = KTDs with K = 28 and TD =928

. In Figure 8.26 we give the Bodeplots for the closed-loop transfer function and the sensitivity function.

3. Now we consider pure integral control with RC(s) =1

TIswith TI =

710

. In Figure 8.26 we

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-25

-20

-15

-10

-5

0

5

10

log

log

dB

deg

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-25

-20

-15

-10

-5

0

5

10

log

log

dB

deg

| |

Figure 8.27 Bode plots for the closed-loop transfer function andsensitivity function for the falling mass with integral controller(left) and PID controller (right). In each case, the solid linerepresents the closed-loop transfer function and the dashed linethe sensitivity function.

give the Bode plots for the transfer function (the solid line) and the sensitivity function

(the dashed line).4. Finally, we look at a PID controller so that RC(s) = K + KTDs +

1TIs

, and we use thesame numerical values as above. As expected, in Figure 8.27 you will find the Bode plotsfor the closed-loop transfer function and the sensitivity function.

In all cases, note that the gain attenuation at high frequencies leads to high sensitivities atthese frequencies, and all the associated disadvantages.

8.5.2 Quantitative performance measures Our emphasis to this point has been es-sentially on descriptive measures of performance. However, it is very helpful to have onhand quantitative measures of performance. Indeed, such performance measures form the

backbone of modern control theory, as from these precise performance characteristics springuseful design methodologies. The reader will wish to recall from Section 5.3.1 the definitionsof the L2 and L-norms.

The quantitative measures we will provide are those on the error of a system to a stepinput. Thus we work with the block diagram of Figure 6.25 which we reproduce in Fig-ure 8.28. We shall assume that the closed-loop system is type 1 so that the steady-stateerror to a step input is zero. In order to make meaningful comparisons, we deal with theunit step response to the reference 1(t).

For design methodologies with a mathematical basis, the following two measures of per-


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22/10/2004 8.5 The sensitivity function 345

r(s) RL(s) y(s)

Figure 8.28 Block diagram for quantitative performance measures

formance are often the most useful:

e2 =

0

e2(t) dt1/2

e = sup0

{|e(t)| for almost every t}.

The first is of course the L2-norm of the error, and the second the L-norm of the error.The error measure

e1 =0

|e(t)| dt

is also used. For example, you will recall from Section 13.1 that the criterion used by Zieglerand Nicols in their PID tuning was the minimisation of the L1-norm. In practice, one willsometimes wish to consider alternative performance measures. For example, the followingtwo, related to the L2 and L1-norms, will serve to penalise errors which occur for large times,but deemphasise large initial errors:

et,2 =

0te

2

(t) dt1/2

et,1 =0

t |e(t)| dt.

One can imagine choosing other weighting functions of time by which to multiply the errorto suit the particular nature of the problem with which one is dealing.

One of the reasons for the usefulness of the L2 and L norms is that they make it possibleto formulate statements in terms of transfer functions. For example, we have the followingresult which was derived in the course of the proof of Theorem 5.22, if we recall that thetransfer function from the input to the error in Figure 8.28 is the sensitivity function.

8.24 Proposition Consider the block diagram of Figure 8.28 and suppose that the closed-loopsystem is type k for k 1. If e(t) is the error to an input u L2[0, ), then

e2 SL u2 .

where SL is the sensitivity function of the closed-loop system.

This proposition tells us that if a design objective is to minimise the transfer of energyfrom the input signal to the error signal, then one should aim to minimise the H-norm ofSL.


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8.6 Frequency-domain performance specifications

The time-domain performance specifications of Section 8.1 are traditionally what oneencounters as these are the easiest to achieve a feeling for. However, there are powerfulcontroller design methods which rely on specifying the performance requirements in thefrequency-domain, not the time-domain. In this section we address this in two ways. Firstwe look at some specifications that are actually most naturally given in the frequency-

domain. These are followed by an explanation of how time-domain specifications can beapproximately turned into frequency-domain specifications.

8.6.1 Natural frequency-domain specifications There are a significant number ofperformance specifications that are very easily presented in the frequency-domain. In thissection we list some of these. In a given design problem, one may wish to invoke only someof these constraints and not others. Also, the exact way in which the specifications are bestpresented can vary in detail, depending on the particular application. However, it can behoped that what we say here is a helpful guide in getting started in a given problem.

Before we begin, we remark that all of the specifications we give will be are what Helton

and Merrino [1998] call disk constraints. This means that the closed-loop transfer functionTL will be specified to satisfy a constraint of the form

|M(i) TL(i)| R(i), [1, 2]

for functions M: i[1, 2] C M: i[1, 2] R+, and for 1 < 2 > 0. Thus at eachfined?frequency in the interval [1, 2], the value ofTL at i must lie in the disk of radius R(i)with centre M(i).

Stability margin lower bound Recall from Proposition 7.15 that the point on the Nyquistcontour closest to

1 + i0 is a distance

SL

away. Thus, to improve stability margins,

one should provide a lower bound for this distance. This means specifying an upper boundon the H-norm of the sensitivity function. In terms of the transfer function this gives

|1 TL(i)| sm, > 0. (8.3)

It is possible that one may wish to enforce this constraint more or less at various frequencies.For example, if one knows that a system will be operating in a certain frequency range, thenstability margins in this frequency range will be more important that in others. Thus onemay relax (8.3) by introducing a weighting function W and asking that

|W(i)(1 TL(i))| sm, > 0, > 0.

As a disk constraint this reads.

|1 TL(i)| sm|W(i)| , > 0. (8.4)

Tracking specifications Recall that the tracking error is minimised as in Proposition 8.24by minimising SL. However, the kind of tracking error minimisation demanded byProposition 8.24 is extremely stringent. Indeed, it asks that for any sort of input, weminimise the L2-norm of the error. However, in practice one will wish to minimise the
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22/10/2004 8.6 Frequency-domain performance specifications 347

tracking error for inputs having a certain frequency response. To this end, we may specifyvarious sorts of specifications that are less restrictive than minimising SL.

A first case we consider is a constraint

|1 TL(i)| < Rtr, [1, 2]. (8.5)This corresponds to tracking well signals whose frequency response is predominantly sup-ported in the given range.

Another approach that may be taken occurs when one knows, or approximately knows,the transfer function for the reference one wishes to track. That is, suppose that we wishto track the reference r(t) whose Laplace transform is r(s). Generalising the analysis ofSection 8.3 in a straightforward way from type k reference signals to reference signals witha general Laplace transform, we see that to track r well we should require

|(1 TL(i))r(i)| tr, [1, 2].This is then made into the disk constraint

|1 TL(i)| tr|r(i)| , [1, 2]. (8.6)

Bandwidth constraints One of the more important features of a closed-loop transfer func-tion is its bandwidth. As we saw in Section 8.2, larger bandwidth generally means quickerresponse. However, one wishes to limit the bandwidth since it is often destructive to a sys-tems physical components to have the response to high-frequency signals not be adequatelyattenuated.

Let us take this opportunity to define bandwidth for fairly general systems. Motivatedby our observations for first and second-order transfer functions, we make the followingdefinition.

8.25 Definition Let (N, D) be a proper SISO linear system in input/output form and let

TN,D = sup

{|HN,D ()|},

as usual. When lim0 |HN,D ()| < , we consider the following five cases:(i) (N, D) is strictly proper and steppable: the bandwidth of (N, D) is defined by

N,D = inf

|HN,D()||HN,D(0)| 12 for all > ;

(ii) (N, D) is not strictly proper and not steppable: the bandwidth of (N, D) is definedby

N,D = sup

|HN,D()|

|HN,D()

| 1

2for all < ;

(iii) (N, D) is strictly proper, not steppable, and TN,D < : the lower cutoff fre-quency of (N, D) is defined by

lowerN,D = sup

|HN,D()|TN,D

12

for all <

and the upper cutoff frequency of (N, D) is defined by

upperN,D = inf

|HN,D()|TN,D

12

for all >

,

and the bandwidth is given by N,D = upperN,D lowerN,D ;
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(iv) (N, D) is strictly proper, not steppable, and TN,D = : the bandwidth of (N, D)is not defined;

(v) (N, D) is not strictly proper and steppable: the bandwidth of (N, D) is not defined.

When lim0 |HN,D ()| = , bandwidth is undefined. Thus N,D is, for typical systems, simply the frequency at which the magnitude part

of the Bode plot drops below, and stays below, the DC value minus

3dB. For most othertransfer functions, definition of bandwidth is still possible, but must be modified to suitthe characteristics of the system. Note that the bandwidth is not defined for systems withimaginary axis poles. However, this is not an issue since such systems are not BIBO stable,and we are typically interested in bandwidth for the closed-loop transfer function, whereBIBO stability is essential.width and

crossover

ency With is general notion of bandwidth, the typical bandwidth constraint one may posemight take the form

|TL(i)| < bw, > bw. (8.7)

High-frequency roll-off In practise one does not want the closed-loop transfer function to

be proper, but strictly proper. This will ensure that the system will not be overly susceptibleto high-frequency disturbances in the reference. Furthermore, one will often want the closed-loop transfer function to not only be proper, but to have a certain relative degree, so that itfalls off at a prescribed rate as . Note that

TL(i) =RC(i)RP(i)

1 + RC(i)RP(i).

Therefore, assuming that RCRP is proper, for large frequencies TL(i) behaves likeRC(i)RP(i). To render the desired behaviour as a disk inequality, we first assume thatwe insist on controllers that satisfy an inequality of the form

|RC(i)| C||nC

for some C > 0 and nC N. This is tantamount to making a relative degree constraint onthe controller. Since we are primarily concerned with high-frequency behaviour, the choiceof C is not critical. Assuming that such a constraint has been enforced, the high-frequencyapproximation

TL(i) RC(i)RP(i)gives rise to the inequality

|TL(i)|

C |R

P(i

)|||nC , [1, 2]. (8.8)

Controller output constraints The transfer function

RCSL =RC

1 + RCRP

is sometimes called the closed-loop controller. It is the transfer function from the errorto the output from the plant. One would wish to minimise this transfer function in order toensure that the controller is not excessively aggressive. In practise, an excessively aggressive


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22/10/2004 8.6 Frequency-domain performance specifications 349

controller might cause saturation, i.e., the controller may not physically be able to supplythe output needed. Saturation is a nonlinear effect, and should be avoided, unless its effectsare explicitly accounted for in the modelling.

In any event, the constraint we consider to limit the output of the closed-loop controlleris

|RC(i)(1 TL(i))| clc,

where we leave the frequency unspecified for the moment. Using the relation

TL =RCRP

1 + RCRP= RCRP = TL

1 TL ,

this gives the disk constraint|TL(i)| clc |RP(i)| .

Now let us turn our attention to the reasonable ranges of frequency for the invocation ofsuch a constraint. Clearly the constraint will have the greatest effect for frequencies that arezeros for RP that are on the imaginary axis. This we let iz1, . . . , i z k be the imaginary axiszeros for RP, and take as our disk constraint

|TL(i)| clc |RP(i)| , [z1 r1, z1 + r1] [zk rk, zk + rk], (8.9)

for some r1, . . . , rk > 0.

Plant output constraints The idea here is quite similar to that for the closed-loop con-troller. To wit, the name closed-loop plant is often given to the transfer function

RPSL =RP

1 + RCRP.

This is the transfer function from the output of the controller to the output of the plant.That this should be kept from being too large is a reflection of the desire to not have theplant overreact to the controller.

The details of the computation of the ensuing disk constraint go very much like that forthe closed-loop controller. We begin by considering the constraint

|RP(i)(1 TL(i))| clp,

immediately giving rise to the disk constraint

|1 TL(i)| clp

|RP(i)

|.

To determine when one should invoke this constraint, note that it most restrictive for fre-quencies near imaginary axis poles for RP. Thus we let p1, . . . , pk be the imaginary axispoles for RP and take the constraint

|1 TL(i)| clp|RP(i)| , [p1 r1, p1 + r1] [pk rk, pk + rk], (8.10)

for given r1, . . . , rk > 0.


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Overshoot attenuation In Section 8.2.2 we saw that there is a relationship, at least forsecond-order transfer functions, between a peak in the frequency response for the closed-looptransfer function and a large overshoot. While no theorem to this effect is known to theauthor, it may be desirable to enforce a constraint on TL of the type

|TL(i)| os, 0. (8.11)

Summary We have given a list of typical frequency-domain constraints. In practise onerarely enforce all of these simultaneously. Nevertheless, in a given application, any of theconstraints (8.3), (8.4), (8.5), (8.6), (8.8), (8.7), (8.9), (8.10), or (8.11) may prove useful.Note that specifying the constants and various weighting functions will be a process of trialand error in order to ensure that one specifies a problem that is solvable, and still hassatisfactory performance. In Chapter 9 we discuss in detail the idea that not all types offrequency-domain performance specification should be expected to be achievable, particu-larly for unstable and/or nonminimum phase plants. The methods for obtaining controllerssatisfying the type of frequency-domain constraints we have discussed in this section may befound in the book [Helton and Merrino 1998], where a software is demonstrated for doing

such design. This is given further discussion in Chapter 15 in terms of H methods.

8.6.2 Turning time-domain specifications into frequency-domain specifications Itis generally not possible to make a given time-domain performance specification and pro-duce an exact, tractable frequency-domain specification. Nonetheless, one can make someprogress in producing reasonably effective and manageable frequency-domain specificationsfrom many types of time-domain specifications. The idea is to produce a transfer func-tion having the desired time-domain behaviour, then using this as the basis for forming thefrequency-domain specifications.h

8.7 SummaryWhen designing a controller the first step is typically to determine acceptable perfor-

mance criterion. In this chapter, we have come up with a variety of performance measures.Let us review what we have said.

1. Based upon a systems step response, we defined various performance features (rise time,overshoot, etc.). These features should be understood at least in that one should be ableto compare two signals and determine which is the better with respect to certain of theseperformance features.

2. Some of the character of system response are exhibited by a simple second-order transferfunction. In particular, the tradeoffs one has to make in controller design begin to showup for such systems in that one cannot perfectly satisfy all performance measures.

3. For simple systems, often one can obtain an adequate understanding of the problems tobe encountered in system performance by using the observations seen when additionalpoles and zeros are added to a second-order transfer function.

4. The effects of the existence of unstable poles and nonminimum phase zeros on the stepresponse should be understood.

5. System type is an easily understood concept. Particularly, one should be able to readilydetermine the system type of a unity gain feedback loop with ease.
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22/10/2004 8.7 Summary 351

6. For disturbance rejection, one should understand that a disturbance may affect a systemsdynamics in a variety of ways, and that the way these are quantified are via appendedsystems, and systems types for these appended systems.

7. For design, the sensitivity function is an important consideration as concerns designingcontrollers which are in some sense robust. The reader should be aware of some of thetradeoffs which are necessitated by the need to have a good closed-loop response along

with desirable robustness and disturbance rejection characteristics.


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Exercises

E8.1 Consider the SISO linear system = (A, b, ct,D) with

A =

, b =

01

, c =

10

, D = 01,

where 0 and > 0. For = 1 and {0, 0.1, 0.7, 1}, do the following.(a) Determine the steady-state value of the output for a unit step input.

(b) For a unit step input and zero initial conditions, plot the output response (youdo not have to provide an analytical expression for it).

(c) From your plot, determine the rise time, tr.

(d) From your plot, determine the -settling time, ts,, for = 0.1.

(e) Determine the percentage overshoot Pos.

(f) Plot the location of the poles of the transfer function T in the complex plane.

(g) Produce the magnitude Bode plot for H(), and from it determine the band-width of the system.

E8.2 Denote by 1 the SISO linear system of Exercise E8.1. Add in series with the system1 the first-order SISO system 2 = (A2, b2, c

t2,D2) with

A2 = , b2 = 1 , c2 = 1 , D2 = 01.

That is, consider a SISO linear system having as its input the input to 2 and as itsoutput the output from 1 (see Exercise E2.1).

(a) Determine the transfer function for the interconnected system, and from thisascertain the steady-state output arising from a unit step input.

Fix = 1 and = 1, and for {0, 0.1, 1, 5}, do the following.(b) For a unit step input and zero initial conditions, plot the output response (you

do not have to provide an analytical expression for it).



(e) From your plot, determine the percentage overshoot Pos.

(f) Plot the location of the poles of the transfer function in the complex plane.

(g) Produce the magnitude Bode plot for the interconnected system, and from itdetermine the bandwidth of the system.

E8.3 Denote by 1 the SISO linear system of Exercise E8.1. Interconnect 1 with blockscontaining s and

R as shown in the block diagram Figure E8.1. Thus the input

gets fed into the parallel block, whose output becomes the input into 1.(a) Determine the transfer function for the interconnected system, and from this

ascertain the steady-state output arising from a unit step input.

Fix = 1 and = 1, and for {1, 0, 1, 5}, do the following.(b) For a unit step input and zero initial conditions, plot the output response (you

do not have to provide an analytical expression for it).



(e) From your plot, determine the percentage overshoot Pos.
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Exercises for Chapter 8 353

u(s)

1 y(s)

s

Figure E8.1 Block diagram for system interconnection

(f) Plot the location of the poles and zeros of the transfer function in the complexplane.

(g) Produce the magnitude Bode plot for the interconnected system, and from itdetermine the bandwidth of the system.

(h) Produce the phase Bode plot for the interconnected system and make some com-ments on what you observe.

E8.4 Consider the closed-loop interconnection in Figure E8.2, and assume that it is IBIBO

r(s) RC(s) RP(s) y(s)

Figure E8.2 A closed-loop system

stable. Suppose that the loop gain RCRP satisfies lims0 RP(s)RC(s) = K whereK = 1.(a) Show that the closed-loop transfer function is type 0. If one desires zero steady-

state error to a step input, will the closed-loop system be satisfactory?

(b) Design a rational function RC which has the property that, when put into theblock diagram of Figure E8.3, the resulting closed-loop transfer function will be

r(s)

RC(s)

RP(s) y(s)

RC(s)

Figure E8.3 A modified closed-loop system

of type 1. For the closed-loop system you have just designed, what will be theerror of the system to a step input?


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Now suppose that you wish to consider the effect of a disturbance which enters thesystem as in Figure E8.4.

r(s) RC(s) RP(s) y(s)

d(s)

Figure E8.4 Adding a disturbance

(c) What is the system type with respect to this disturbance for the controller de-picted in the block diagram Figure E8.2?

(d) What is the system type with respect to this disturbance for the modified con-

troller depicted in the block diagram Figure E8.3 (i.e., that obtained by replacingthe RC block with the parallel blocks containing your RC and RC)?

(e) Comment on the effectiveness of the modified controller as concerns rejection ofstep disturbances in this case.

E8.5 Let RP R(s) be a strictly proper plant with RC S(RP) a proper controller.(a) Show that if the Nyquist plot for RL = RCRP lies in C+ then SL < 1. What

can you say about the performance of the resulting closed-loop system?

Now consider the two plants

RP,1 =1

s + 1

, RP,2 =s 1

s2

+ 2s + 1

.

(b) Produce the Bode plots for both plants with the controller RC(s) =12

S(RP,1)S(RP,2). How do they differ?

(c) Produce the Nyquist plots for both loop gains RL,1 = RCRP,1 and RL,2 = RCRP,2.

(d) Comment on the comparative performance of the closed-loop systems in light ofyour work done in part (a).

(e) Produce the step response for both closed-loop systems and comment on theircomparative behaviour.

E8.6 In problems (a)(d) there are two SISO linear systems (N1, D1) and (N2, D2) in in-

put/output form, but you are not told what they are. Instead, you are given a pairof Bode plots, and a pair of step responses, one each for the pair of transfer functionsTN1,D1 and TN2,D2 . You are not told which Bode plot and which step response comefrom the same transfer function.

In each case, do the following:

1. Indicate which Bode plot goes with which step response. correctly.

2. Indicate clearly (but not necessarily at great length) the features of the plotsthat justify your choice in step 1.

(a)


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The material in this chapter has focused upon the unity gain feedback loop and its relationto the solution of Problem 6.41 concerning design for input/output systems. In the nexttwo problems, you will investigate a few aspects of performance for Problem 6.48 wherestatic state feedback is considered, and for Problem 6.53 where static output feedback isconsidered. Recall that the block diagram representation for static state feedback is as inFigure 8.5, and that the block diagram representation for static output feedback is as in

r(s) b (sIn A)1

ct y(s)

D

ft

x0

Figure 8.5 Block diagram for static state feedback


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Figure 8.6. You may also wish to refer to Theorems 6.49 and 6.54 concerning the form of

r(s)b

(sInA

)1

c

t

y(s)

D

x0

F

Figure 8.6 Block diagram for static output feedback

the transfer function under static state feedback and static output feedback.

E8.7 In this exercise we consider a controllable SISO linear system = (A, b, ct, 01) and astate feedback vector f. You may suppose that (A, b) is in controller canonical form,and that f Ss().(a) Compute the transfer function from the reference r(s) to the error r(s) y(s) for

the closed-loop system of Figure 8.5.

(b) Determine the system

mct book - chapter 8

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