12.Ad hoc methods II.Simple frequency response methods for controller design

1+ατs

1+τs

This version: 22/10/2004

Chapter 12

Ad hoc methods II: Simple frequency response methods for controller design

In Chapter 7 we introduced a frequency domain technique, the Nyquist criterion, for studying closed-loop stability. The stability margins introduced in Section 7.2 suggest that the frequency domain techniques may be useful in control design, as they o er tangible objectives for the properties of the closed-loop system. The design ideas we discuss in this chapter are essentially ad hoc, but are frequently useful in situations where controller design is not essentially di cult—at least as concerns obtaining closed-loop stability—but where one would like guidance in improving the performance of a controller. The methods we use in this chapter fall under the broad heading of loop shaping, as the objective is to shape the Nyquist plot to have desired properties. In this chapter the emphasis is on choosing a priori a form for the controller, then using the design methodology to improve the closedloop response. In Chapter 15, the design method does not assume a form for the controller, but rather the form of the controller is decided upon by the objectives of the problem.

Contents

12.1	Compensation in the frequency domain . . . . . . . . . . . . . . . . . . . . . . . . . . . .		469
	12.1.1	Lead and lag compensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .	470
	12.1.2	PID compensation in the frequency domain . . . . . . . . . . . . . . . . . . . . . .	473
12.2	Design using controllers of predetermined form . . . . . . . . . . . . . . . . . . . . . . . .		475
	12.2.1	Using the Nyquist plot to choose a gain . . . . . . . . . . . . . . . . . . . . . . . .	476
	12.2.2	A design methodology using lead and integrator compensation . . . . . . . . . . .	477
	12.2.3	A design methodology using PID compensation . . . . . . . . . . . . . . . . . . .	483
	12.2.4	A discussion of design methodologies . . . . . . . . . . . . . . . . . . . . . . . . .	486
12.3	Design with open controller form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .		488
12.4	Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .		488

12.1 Compensation in the frequency domain

The discussion of the previous few sections has focused on analysis of closed-loop systems using frequency domain techniques. In this discussion, certain key contributors to stability were identified, principally gain and phase margin. Faced with a plant transfer function RP , one will want to design a controller rational function RC which has certain desirable

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characteristics. The term compensation is used for the process of designing a controller rational function, reflecting the fact that one is “compensating” for deficiencies in the innate properties of the plant. In Section 6.5 we investigated the PID controller, and discussed some of its properties. In this section we introduce another class of controller transfer functions which are useful in shaping the frequency response. We also discuss PID compensation in a manner slightly di erent from our previous discussion.

12.1.1 Lead and lag compensation	The first type of compensation we look at deals
essentially with the manipulation of the phase of the closed-loop system.
12.1 Definition A lead/lag compensator is a rational function of the form
RC(s) =	K(1 + ατs)
		1 + τs

for K, α, τ R with α 6 0{, 1} and K, τ 6= 0. It is a lead compensator when |α| > 1 and

	a lag compensator when |α| < 1.
	Note that any rational function of the form A						s+z	can be put into the form of a lead/lag
							s+p
	compensator by defining
			1		p				A
	τ =			, α =			, K =			.
		p			z				α

The form for the lead/lag compensator as we define it is convenient for analysing the nature of the compensator, as we shall see.

The following result gives the essential properties of a lead/lag compensator.

12.2 Proposition If RC(s) = is a lead/lag compensator then, with HC(ω) = RC(iω), the following statements hold:

(i) limω→0 ]HC(ω) = 0;

(ii) (a) limω→∞ ]HC(ω) = 0 if α > 0;

(b) limω→∞ ]HC(ω) = 180◦ if α < 0 and τ < 0;

(c) limω→∞ ]HC(ω) = −180◦ if α < 0 and τ > 0; (iii) limω→0 |HC(ω)| = 0dB;

(iv) limω→∞ |HC(ω)| = 20 log αdB;

(v) for α > 0, the function ω 7→]|HC(ω)| achieves its maximum at ωm = (|τ| √α)−1, and this maximum value is φm = arcsin αα−+11 . Furthermore,

(a)if α > 1, ]HC(ω) has a maximum at ωm and the maximum value is between 0◦ and 90◦, and

(b)if α < 1, ]HC(ω) has a minimum at ωm and the minimum value is between −90◦ and 0◦;

(vi)|HC(ωm)| = √α.

Proof (i) This is evident since limω→0 HC(ω) = 1.

(ii) In this case, the assertion follows since limω→∞ HC(ω) = αττ . Keeping track of the signs of the real and imaginary parts gives the desired result.

(iii) This follows since limω→0 HC(ω) = 1. (iv) This is evident since limω→∞ HC(ω) = α.

α−1

α+1

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(v) We compute
	Re(HC(ω)) =	1 + ατ2ω2	,	Im(HC(ω)) =	τω(α − 1)	,
		1 + τ2ω2			1 + τ2ω2

so that

]HC(ω) = ] τω(α − 1) + i(1 + ατ2ω2) .

In di erentiating this with respect to ω, we need to take care of the various branches of arctan. Let us first consider τ > 0 and α > 1 so that

τω(α − 1)

]HC(ω) = arctan 1 + ατ2ω2 .

Di erentiating this with respect to ω gives
	d]HC(ω)	=	τ(α(1 + τ2ω2 − ατ2ω2) − 1)	,
	dω		(1 + τ2ω2)(1 + α2τ2ω2)

which has a zero at ω = √ατ2−1. If τ > 0 and α < 1, let us define β = α1 so that

τω(1 − β1 ) ]HC(ω) = arctan 1 + β1 τ2ω2 .

	Di erentiating this with respect to ω gives
		d]HC(ω)	=	τ(β(1 + τ2ω2 − β) − τ2ω2)	,
		dω		(1 + τ2ω2)(β2 + τ2ω2)

√β
which has a zero at ω = √τ2	. The computations are the same for τ < 0, except that the

phase angle has the opposite sign. One may also check that the second derivative is not zero at ω = √α1|τ|, and so the function must have a maximum or minimum at this frequency. A straightforward substitution gives

tan φ		=	α − 1			.
	m		2√α

If α > 1 this angle is positive, and if α < 1 it is negative. That the value is bounded in magnitude by 90◦ is a consequence of the maximum argument of 1 + iατω being 90◦ for τ > 0 and the minimum argument of (1 + iτω)−1 being −90◦ for τ > 0. A similar statement holds for τ < 0. To get the final form for φm given in the statement of the proposition, we

recall that if tan θ = x then sin θ = √1+x x2 . Taking x = tan φm gives the result.
(vi) This is a simple matter of substituting the expression for ωm into HC(ω).

12.3 Remarks 1. One sees from part (v) of Proposition 12.2 why the names lead and lag compensator are employed. For a lead compensator with α, τ > 0 the phase angle has a frequency window in which the phase angle is increased, and similarly for a lag compensator with α, τ > 0 there is a frequency window in which the phase is decreased.

2.A plot of the maximum phase lead φm for α > 1 is shown in Figure 12.1. When α < 1 we can define β = α1 and then see that

β − 1

φm = − arcsin β + 1.

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	80
	60
m
φ	40
	20
	2.5	5	7.5	10	12.5	15	17.5	20
				α or α−1

Figure 12.1 Maximum phase shift versus α for lead compensator and versus α1 for lag compensator

Therefore, to determine the maximum phase lag when α < 1, one can reads o the phase angle from Figure 12.2 at α1 , and changes the sign of the resulting angle.

Note that the equation sin φm = can be solved for α given φm, and the result is

α =	1		+ sin φm	.	(12.1)
	1		− sin φm

3.	Although in Proposition 12.2 we have stated the characterisation of lead/lag compen-
	sators in fairly general terms, typically one deals with the case when α, τ > 0.
4.	If one plots the Bode plot for a lead/lag compensator by hand using the methods of
	Section 4.3.2, there are two break frequencies, one at ω1 = 1 and another at ω2 =					1	.
					τ	ατ
	Depending on whether the compensator is lead or lag, one or the other of these frequencies
	will be the smaller. In either case, the geometric mean of these two break frequencies is, by
	definition, √		which is equal to ωm. On a Bode plot for the lead/lag compensator we
		ω1ω2
	therefore have log ωm = 1 (log ω1 + log ω2), showing that on the Bode plot, the maximum
	2
	phase shift occurs exactly between the two break frequencies.
5.	The behaviour of the magnitude portion of the frequency response at ωm is also nice on
	a Bode plot. By (vi) the magnitude at ωm is 20 log √				= 10 log α, and so is half of the
				α
	limiting value of the magnitude as ω → ∞.

Let us look at some simple examples of lead/lag compensators. We illustrate the various things which can happen when choosing parameters in the compensator. But do keep in mind that one normally uses a lead/lag compensator with τ > 0, and either 0 < α < 1 (lag compensator) or α > 1 (lead compensator).

12.4 Examples 1. We look at a lead compensator with transfer function

1 + 10s RC(s) = 1 + s ,

meaning that τ = 1 and α = 10. Based on Proposition 12.2 and Remark 12.3–4, it is actually easy to guess what the Bode plot for this transfer function will look like. The

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473

two break frequencies are ω1 = 1 ad ω2 = 101 . Thus the maximum phase lead will occur at log ωm = 12 (log 1 + log 101 ) = −12 . The magnitude part of the Bode plot will start at 0dB and turn up at about log ω = −1 until it reaches log ω = 0 where it will level o at 20dB. Given that α = 10, from Figure 12.1 we determine that the maximum phase lead is about 55◦ (the calculation gives φm ≈ 54.9◦). The actual Bode plot can be found in Figure 12.2. One should note that the maximum phase shift does indeed occur between

20
15
dB10
5
0	-1	-0.5	0	0.5	1	1.5	2
-1.5

log ω

0
-5
dB-10
-15
-20	-1	-0.5	0	0.5	1	1.5	2
-1.5

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 12.2 Bode plots for a lead compensator RC(s) =									1+10s
					1				1+s
(left) and lag compensator RC(s) =		1+					s	(right)
					10
			1+s
the two break frequencies, for example.
2. Here we take the lag compensator
1 +		1		s
		10
RC(s) =						,
	1 + s

so that τ = 1 and α = 101 . Here on determines that the two break frequencies are ω1 = 1 and ω2 = 10. Thus log ωm = 12 (log 1 + log 10) = 12 . The magnitude portion of the Bode plot will start at 0dB and turn down at log ω = 0 until it reaches log ω = 1 where it levels o at −20dB. The maximum phase lag can be determined from Figure 12.1. Since α < 1 we need to work with α1 which is 10 in this case. Thus, from the graph, we determine that the maximum phase lag is about −55◦. The actual Bode plot is shown in Figure 12.2.

12.1.2 PID compensation in the frequency domain We now look at how a PID controller looks in the frequency domain. First we need to modify just a little the type of PID controller which we introduced in Section 6.5.

12.5 Definition A PID compensator is a rational function of the form

	K		1
RC(s) =		(1 + TDs) s +			(12.2)
	s		TI
for K, TI , TD > 0.

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If we expand this form of the PID compensator we get
RC(s) = K 1 +	TD	+	1		+ TDs ,
	TI		TI s

and this expression is genuinely di erent from the PID compensator we investigated in Section 6.5. However, the alteration is not one of any substance (the di erence is merely

a constant factor K TD ), and it turns out that the form (12.2) is well-suited to frequency

TI

response methods.

The following result gives some of the essential features of the transfer function RC in (12.2).

12.6 Proposition Let RC be a PID compensator of the form (12.2) and define HC(ω) = RC(iω) for ω > 0. The following statements hold:

(i) limω→0				K
		ω |HC(ω)| = TI ;
(ii) limω→∞			1	|HC(ω)| = KTD;
			ω
(iii) limω→0		]HC(ω) = −90◦;
(iv) limω	→∞		]HC(ω) = 90◦;			√
					TD+TI			−1
(v) the function ω 7→H|C(ω)| has a minimum of K
					TI	at ωm =	TDTI		;
(vi) ]HC(ωm) = 0;

Proof The first four assertions are readily ascertained from the expression

RC(s) = K 1 +	TD	+	1		+ TDs
	TI		TI s

for PID compensator. The final two assertions are easily derived from the decomposition

	K	(TD + TI ) + i	T	D	T ω2	−	1
HC(ω) =					I
	TI				ω

of HC(ω) into its real and imaginary parts, and then di erentiating the magnitude to find extrema.

12.7 Remarks 1. The frequency ωm at which the magnitude of the frequency response

achieves its minimum is the geometric mean of the frequencies	1	and	1	. Thus on
	TD
		TI
a Bode plot, this minimum will occur at the midpoint of these two frequencies.
				˜
2. If one redefines a normalised frequency ω˜ = TI ω and a normalised derivative time TD =

TD , the frequency response for a PID compensator, as a function of the normalised

TI

frequency, is given by

	2
	HC(ω) = K T˜D + 1) + i	T˜Dω˜	− 1	.
		ω˜

This identifies the normalised derivative time and the gain K as the essential parameters in describing the behaviour of the frequency response of a PID compensator. Of these, of course the dependence on the gain is straightforward, so it is really the normalised

˜	TD
derivative time TD =	TI	which is most relevant.

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We have a good idea of what the Bode plot will look like for a PID compensator based on the structure outline in Proposition 12.6. Let’s look at two “typical” examples to see how it looks in practice.

12.8 Example In each of these examples, we take K = 1 since the e ect of changing K is merely reflected in a shift of the magnitude Bode plot by log K.

1.We take TD = 1 and TI = 10. In this case, Proposition 12.6 tells us that the magnitude Bode plot will have its minimum at ωm = √10−1 or log ωm = −12 . The value of the

minimum will be TD+TI = 11 . At the frequency ωm, the phase will be zero. Below this

TI 10

frequency the phase is negative and above it the phase is positive. Putting this together gives the frequency response shown in Figure 12.3.

60
50
40
dB30
20
10
0	-2	-1	0	1	2	3

log ω

80
70
60
dB50
40
30
20	-2	-1	0	1	2	3

log ω

	150						150
	100						100
deg	50					deg	50
	0						0
	-50						-50
	-100						-100
	-150						-150
	-2	-1	0	1	2	3

log ω

-2

-1

0

1

2

3

log ω

Figure 12.3 The Bode plots for a PID compensator with K = 1,

TD = 1, and TI = 10 (left) and with K = 1, TD = 10, and

TI = 1 (right)

2. Now we take TD = 10 and TI	= 1. Again we have ωm = √				−1. The magnitude Bode
				10
plot has its minimum at ωm =	√		−1 and the value of the minimum is TD+TI
		10				= 11. At
					TI
the frequency ωm the phase is 0◦, and it is negative below this frequency and positive
above it. The Bode plot is shown in Figure 12.3.

12.2 Design using controllers of predetermined form

In this section we indicate how the procedures of the preceding sections can be helpful in controller design. The basic idea of the frequency response methods we discuss here are that one designs controllers which do various things to the system’s frequency response, and so alter both the Bode plot for the loop gain as well as the Nyquist plot. The technique of using the frequency response to design a controller is called loop shaping since, as we shall see, the idea is to shape the Nyquist plot to have a shape we like.

In Sections 12.2.2 and 12.2.3 we discuss ways to design controller rational functions which shape the Nyquist contour, or equivalently the Bode plot, to have desired characteristics.

476 12 Ad hoc methods II: Simple frequency response methods for controller design 22/10/2004

Let us begin, however, with a simple design situation where one wants to determine the e ects of adjusting a gain.

12.2.1 Using the Nyquist plot to choose a gain We work with the unity gain feedback setup of Figure 6.21 which we reproduce here in Figure 12.4. In this scenario,

rˆ(s)	RC (s)	RP (s)	yˆ(s)
	−

Figure 12.4 Closed-loop system

we have chosen RC (perhaps it is a PID controller) and we wish to tune the gain K. To determine whether a given gain K produces an IBIBO stable closed-loop system, we could, for example, plot the Nyquist contour for the loop gain RL = KRCRP . However, if we have to do this for very many gains, it might get a bit tedious. The following result relieves us of this tedium.

12.9 Proposition For the closed-loop interconnection of Figure 12.4, define RL = RCRP , and let R, r > 0 be selected so that the (R, r)-Nyquist contour is defined. Then, for K 6= 0, the number of encirclements of −1 + i0 by the (R, r)-Nyquist contour for KRL is equal to the number of encirclements of −K1 + i0 by the (R, r)-Nyquist contour for RL.

Proof A point s C is on the (R, r)-Nyquist contour for KRL if and only if K1 s is on the (R, r)-Nyquist contour for RL. That is, the (R, r)-Nyquist contour for KRL is the (R, r)- Nyquist contour for RL scaled by a factor of K. From this the result follows. If K < 0 the result still holds as the (R, r)-Nyquist contour for KRL is reflected about the imaginary axis.

This result allows one to simply plot the Nyquist contour for RCRP , and then determine stability for the closed-loop system with gain K merely by counting the encirclements of −K1 + i0. The following example illustrates this.

12.10 Example Let us look at the open-loop unstable system with RP (s) = s−1 1 , and we wish to stabilise this using a proportional controller, i.e., RC(s) = 1. Let us use the Nyquist criterion to determine for which values of the gain the closed-loop system is stable. The Nyquist contour for RL(s) = RC(s) = RP (s) = s−1 1 is shown in Figure 12.5. Since the loop gain has 1 pole in C+, by Proposition 12.9, for IBIBO stability of the closed-loop system we must have one counterclockwise encirclement of −K1 +i0, provided K 6= 0. From Figure 12.5 we see that this will happen if and only if K > 1.

Of course, we can determine this directly also. The closed-loop transfer function is

	KRL(s)			K
TL(s) =			=		.
	1 + KRL(s)			s + K − 1

The Routh/Hurwitz criterion (if you wish to hit this with an oversized hammer) says that we have closed-loop IBIBO stability if and only if K > 1.

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	0.4
	0.2
Im	0
	-0.2
	-0.4
	-1	-0.8	-0.6	-0.4	-0.2	0
				Re

Figure 12.5 Stabilising an open-loop unstable system using the

Nyquist criterion

12.2.2 A design methodology using lead and integrator compensation In the previous section we assumed that we had in hand the nature of the controller we would be using, and that the only matter to account for was the gain. In doing this, we glossed over how one can choose the controller rational function RC. There are, in actuality, many ways in which one can use PID control elements, in conjunction with lead/lag compensators, to make a frequency response look like what we need it to look like. We shall explore just a few design methodologies, and these can be seen as representative of how one can approach the problem of controller design.

We look at a methodology which first employs a lead compensator to obtain a desired phase margin for stability, and then combines this with an integrator for low frequency performance and disturbance rejection. The rough idea is described in the following “algorithm.”

12.11 A design methodology Consider the closed-loop system of Figure 12.6. To design a

rˆ(s)	RC (s)	RP (s)	yˆ(s)
	−

Figure 12.6 Closed-loop system for loop shaping

controller transfer function RC, proceed as follows.

(i)Select a gain crossover frequency for your system. This choice of frequency is based upon requirements for transient response since larger gain crossover frequencies are related to shorter rise times. One cannot expect to specify an arbitrary gain crossover frequency, however. The bandwidth of a system will be limited by the bandwidth of the system components.

478 12 Ad hoc methods II: Simple frequency response methods for controller design 22/10/2004

(ii)Design a phase lead controller rational function RC,1 so that the closed-loop system will meet the phase and gain margin requirements.

(iii)Use phase lag compensation or integral control to boost the low frequency gain of the controller transfer function. These terms will assist in the tracking of step inputs, and the rejection of disturbances. Take care here not to degrade the stability of the system.

(iv)Should the plant not have su cient attenuation at high frequencies, add a term which “rolls o ” at high frequencies to alleviate the systems susceptibility to high frequency

noise. Most plants will have su cient attenuation at high frequencies, however.

We can illustrate this design methodology with an example. Although this example is illustrative, it is simply not possible to come up with a design methodology, or an example, which will work in all cases.

12.12 Example We consider the problem of controlling a mass in space in the absence of gravity. We again suppose that the motive force is applied through a propeller, as in Example 6.60. The governing di erential equations are

my¨(t) = f(t) + d(t),

where f is the input force from the propeller, and d(t) is a disturbance force on the system. In the Laplace transform domain, the block diagram is as depicted in Figure 12.7. In this

		ˆ
		d(s)
	ˆ	1
rˆ(s)	f (s)		yˆ(s)
	RC (s)	ms2
	−

Figure 12.7 Block diagram for controller design for mass

example we fix m = 1. Let us suppose that we have determined that a respectable value for the gain crossover frequency is ωgc (in stating this, we are implicitly assuming that we will have just one gain crossover frequency). For suitable stability we require that the phase margin for the system be at least 40◦. We also ask that the system have zero steady-state error to a step disturbance.

Now we can go about ascertaining how to employ our design methodology to obtain the specifications. First we look at the Bode plot for the plant transfer function RP (s) = ms1 2 . This is shown in Figure 12.8. At the gain crossover frequency ωgc = 10rad/ sec (i.e., log ωgc = 1), the phase of the plant transfer function is 180◦. Indeed, the phase of the plant transfer function is 180◦ for all frequencies. To get the desired phase margin, we employ a lead compensator. Indeed, we should choose a lead compensator which gives us the phase margin we desire, and hopefully with some to spare. We need a minimum phase lead of 40◦, and from (12.1) we can see that this means that α ≥ 1+sin1−sin 4040 ≈ 4.60. Let us be really conservative and choose α = 20. Now, we not only need for the phase shift to be at least 40◦, we need it to exceed this value at a specific frequency. But Proposition 12.2 tells us how to manage this: we need to specify ωm. We have ωm = (|τ| √α)−1, and solving this for τ with α = 20 and

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	75
	50
	25
dB	0
	-25
	-50
	-75
	-1.5	-1	-0.5	0	0.5	1	1.5	2

log ω

		150
		100
	deg	50
		0
		-50

-100 -150

-1.5 -1 -0.5 0 0.5 1 1.5 2 log ω

Figure 12.8 Plant Bode plot for mass system

	ωm = 10 gives τ =	1	≈ 0.022. We have been su ciently conservative with our choice of
		20√5
	α that we can a ord to make τ a nice number, so let’s take τ =				1	. With this specification,
				50

the lead compensator is

1 + 2s RC,1(s) = 1 + 501 s.

Let’s evaluate where we are for the moment. In Figure 12.9 we give the Nyquist and Bode

							80
							60
	20						40
					dB		20
							0
	10						-20
							-40
Im	0
							150
							100
	-10				deg		50
							0
							-50
	-20					-100
						-150
	-20	-10	0	10	20
			Re

-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 12.9 The (100, 0.25)-Nyquist contour and the Bode plot for mass and lead compensator

plots for RC1 RP . From the Nyquist plot we see that there are no encirclements of −1 + i0,

480 12 Ad hoc methods II: Simple frequency response methods for controller design 22/10/2004

and so the system is IBIBO stable. From the Bode plot, we see that the phase is about −105◦ (about −104.2◦ to be more precise) at ωgc, which gives a phase margin of about 75◦, well in excess of what we need. There is a problem at the moment, however, in that the desired gain crossover frequency is, in fact, not a gain crossover frequency since the transfer function has less magnitude than we desire at ωgc. We can correct this by boosting the gain of the lead compensator. The matter of just how much to boost the gain is easily resolved. At ω = ωgc, from the Bode plot of Figure 12.9 we see that the magnitude is about −15dB (about −14.1dB to be more precise). So we need to adjust K so that 20 log K = 15 or K ≈ 5.62. Let us take K = 512 . The Nyquist and Bode plots for the gain boosted led compensator are provided in Figure 12.10. Also note from the Bode plot that the magnitude is falling o at

	60								80
									60
	40						dB		40
									20
									0
	20								-20
									-1.5	-1	-0.5	0	0.5	1	1.5	2
Im											log ω
	0
									150
									100
	-20						deg		50
									0
	-40								-50
								-100
								-150
	-40	-20	0	20	40	60			-1.5	-1	-0.5	0	0.5	1	1.5	2
											log ω
			Re

Figure 12.10 The (100, 0.25)-Nyquist contour and the Bode plot for mass and lead compensator with increased gain

40dB/decade, which means that the system is type 2 by Proposition 8.14. Therefore this system meets our objective to track step inputs. In Figure 12.11 we plot the response of the closed-loop system to a unit step input which we obtained by using Proposition 3.40. Note that the error decays to zero as predicted.

Let’s now look at how the system handles step disturbances since we have asked that these be handled gracefully (specifically, we want no error to step disturbances). In Figure 12.12 we display the response of the system to a unit step disturbance (no input). Clearly this is not satisfactory, given our design specifications. To see what is going on here, we determine the transfer function from the disturbance to the output to be

RP (s)

Td(s) = 1 + RC,1(s)RP (s).

We compute

lim Td(s) =			1	=	1	= 1,
	1
					lims→0 RC,1(s)
s→0	lims→0	RP (s)	+ lims→0 RC,1(s)

and so the system has disturbance type 0 with respect to this disturbance. This is not satisfactory for the purpose of eliminating error resulting from a step input, so we need to

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481

	1
	0.8
)	0.6
(t
Σ
1	0.4
	0.2
	0
	2	4	6	8	10	12	14
				t

Figure 12.11 Step response of mass with lead compensator

	0.175
	0.15
	0.125
)	0.1
(t
Σ
1	0.075
	0.05
	0.025
	0
	2	4	6	8	10	12	14
				t

Figure 12.12 Step disturbance response of mass with lead compensator

repair this in some way. As we have seen in the past, a good way to do this is to add an integrator to the controller transfer function. Let’s see how this works. We work with the new controller rational function

	KI		1 + 2s
RC,2(s) =		+				.
	s		1 +	1	s
			10

First note that lims→0 RC,2(s) = ∞ and so this immediately means that the system type with respect to the disturbance is at least 1 (and is in fact exactly 1). We look to ascertain how changing the integrator gain KI a ects system stability. In Figure 12.13 the Bode plots are shown for various values of KI . We can see that, as expected since an integrator will decrease the phase, the phase margin becomes worse as KI increases. What’s more, one can check the Nyquist criterion to show that the system is in fact unstable for K su ciently large. Thus we choose a not too large integrator gain of KI = 101 . Let’s see how this choice of controller performs. First, in Figure 12.14 we display the step response (no disturbance) of the system with the integrator added to the controller. Note that the overshoot and the settling time

482 12 Ad hoc methods II: Simple frequency response methods for controller design 22/10/2004

	100
	80
	60
dB	40
	20
	0
	-20
	-1.5	-1	-0.5	0	0.5	1	1.5	2

log ω

	125
	100
dB	75
	50
	25
	0
	-25
	-1.5	-1	-0.5	0	0.5	1	1.5	2

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

	150
	125
	100
dB	75
	50
	25
	0
	-25
	-1.5	-1	-0.5	0	0.5	1	1.5	2

log ω

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

log ω

Figure 12.13 Bode plots for mass with lead and integrator compensation: KI = 101 (top left), KI = 10 (top right), and KI = 100 (bottom)

have increased from the situation when we simply employed the lead compensator, but may be considered acceptable. If they are not, then one should iterate the design methodology until satisfactory performance is achieved. Also in Figure 12.14 we display the response of the system to a step disturbance (no input). Note that this response ends up at zero as we have ensured by designing the controller as we have done. The decay of the e ect of the error is quite slow, however. One may wish to improve this by increasing the integrator gain.

Now that we have added the integrator to reject disturbances, we need to ensure that the other performance specifications are still met. A look at Figure 12.13 shows that our gain crossover frequency has decreased. We may wish to repair this by further boosting the gain on the lead compensator.

This example, even though simple, shows some of the tradeo s which must take place in

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483

	1
	0.8
t)	0.6
(
Σ
1	0.4
	0.2
	0
	2	4	6	8	10	12	14
				t

	0.15
	0.125
)	0.1
(t
Σ	0.075
1
	0.05
	0.025
	0
	20	40	60	80	100
			t

Figure 12.14 Step response of mass (top) and response to step disturbance (bottom) with lead and integrator compensation

designing a controller to meet sometimes conflicting performance requirements.

12.2.3 A design methodology using PID compensation Let’s see in this section how one can employ a PID controller in a systematic way in the frequency domain to obtain a desired response. The rough idea of PID design is outlined as follows.

12.13 A design methodology Consider the closed-loop system of Figure 12.6. To design a controller transfer function RC, proceed as follows.

(i)Select a gain crossover frequency consistent with the demands for quick response and the limitations of the system components.

(ii)Commit to TI being quite a bit larger than TD.

(iii)Adjust TD so that the phase margin requirements are met.

(iv) Adjust K so that the gain crossover frequency is as desired.

Let us illustrate this in an example.

12.14 Example (Example 12.12 cont’d) We again take the plant transfer function

1

RP (s) = ms2 .

484 12 Ad hoc methods II: Simple frequency response methods for controller design 22/10/2004

The Bode plot for this transfer function is reproduced in Figure 12.15. We suppose that we

	75
	50
	25
dB	0
	-25
	-50
	-75
	-1.5	-1	-0.5	0	0.5	1	1.5	2

log ω

		150
		100
	deg	50
		0
		-50

-100 -150

-1.5	-1 -0.5 0 0.5 1 1.5 2
	log ω
Figure 12.15	Plant Bode plot for mass system

are given the criterion of having a phase margin of at least 65◦ at as high a gain crossover frequency as possible.

Let us decide to go with TD = 1 . In order to achieve the prescribed phase margin,

TI 5

we must use the PID compensator to produce a phase for the loop gain RCRP which is

somewhere negative and greater than		−	Note that given the relative degree of the
		115◦.
plant and the PID controller, the phase for large frequencies will approach						−		1
								90◦. The
contribution to the phase made by the controller changes sign at ωm = √							− . Given
					TDTI
TD	1	1	−1	. Thus choosing a large derivative time
our choice of TI	= 5 , this means that ωm = √5 TD

will decrease the frequency at which the phase becomes negative. In Figure 12.16 we show the situation for TD = 15 and TD = 2. For the smaller value of TD, we see that the phase margin requirements are met at a quite high frequency (around 20rad/ sec). However, a peek at the Nyquist contour for TD = 15 shows that there are 2 clockwise encirclements of −1 + i0. Since RCRP has no poles in C+, this means the system is not IBIBO stable for the given PID parameters. When the derivative time is TD = 2, the phase requirement is met at roughly 2rad/ sec, but now the Nyquist contour is predicting IBIBO stability. Thus we see that we should expect there to be a tradeo between stability and performance in this design methodology. Let us fix TD = 1 (and hence TI = 5), for which we plot the Bode plot and Nyquist contour in Figure 12.17.

We now choose the gain K in order to make the gain crossover frequency large. From Figure 12.17 we see that when the phase is −115◦ the frequency is roughly given by log ω = 0.5, so we take ωgc = 100.5 ≈ 3.16. At this frequency the magnitude of the frequency response is about −10dB. Thus we need to choose K so that 20 log K = 10, or K ≈ 3.16. Let’s make

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12.2 Design using controllers of predetermined form

485

	100
	80
	60
dB	40
	20
	0
	-20
	-1.5	-1	-0.5	0	0.5	1	1.5	2

log ω

	100
	75
dB	50
	25
	0
	-25
	-50
	-1.5	-1	-0.5	0	0.5	1	1.5	2

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
	75								30
	50								20
	25								10
Im	0							Im	0
	-25								-10
	-50								-20
	-75								-30
	-50	-25	0	25	50	75	100		-30	-20	-10	0	10	20	30
			Re									Re
		Figure 12.16		E ect of varying PID derivative time for RP									= 12 :
			= 1										s
		TD		(left) and TD = 2 (right). In both cases K = 1 and
			5
		TI = 5TD, and Nyquist plots are on top and Bode plots are on

the bottom

this K = 314 so that our final controller is given by

	31	(1 + TDs) s +	1	.
RC(s) =	4
	s		TI

The Bode and Nyquist plots for the corresponding loop gain are shown in Figure 12.18. From the Nyquist plot we see that the system is IBIBO stable, and we also see that out phase margin requirements are satisfied.

In Figure 12.19 is the response of the system to a unit step input. We may also compute the e ect of a disturbance entering between the controller and the plant. The corresponding

486 12 Ad hoc methods II: Simple frequency response methods for controller design 22/10/2004

		100								30
		80
		60
dB		40								20
		20
		0
		-20								10
		-40	-1	-0.5	0	0.5	1	1.5	2
		-1.5
				log ω					Im	0
		150
		100								-10
deg		50
		0								-20
		-50
	-100
	-150									-30
		-1.5	-1	-0.5	0	0.5	1	1.5	2	-20	-10	0	10	20	30
				log ω
													Re

Figure 12.17 Nyquist plot (left) and Bode plot (right) for mass with PID controller and K = 1, TD = 1, and TI = 5

	100								100
	80
dB	60
	40
	20								50
	0
	-20
	-1.5	-1	-0.5	0	0.5	1	1.5	2
			log ω					Im	0
	150
	100
deg	50								-50
	0
	-50
-100
-150									-100
	-1.5	-1	-0.5	0	0.5	1	1.5	2	-50	0	50	100
			log ω
											Re

Figure 12.18 Nyquist plot (left) and Bode plot (right) for mass with PID controller and K = 314 , TD = 1, and TI = 5

transfer function is
	RP (s)				s
Td(s) =		=						,
	1 + RC(s)RP (s)		s3 + 3	1 s2	+ 3	9	s + 13
						10
				4			20
and the response to a unit step is shown in Figure 12.20.
12.2.4 A discussion of design methodologies				The above examples give an idea of

how one can use ideas of frequency response in designing controller rational functions. In-

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12.2 Design using controllers of predetermined form

487

	1.2
	1
	0.8
(t)	0.6
Σ
1
	0.4
	0.2
	0
	2	4	6	8	10	12	14
				t

Figure 12.19 Step response of mass with PID controller

	0.2
(t)	0.15
Σ
1	0.1
	0.05
	0
	2	4	6	8	10	12	14
				t

Figure 12.20 Unit step disturbance of mass with PID controller

deed, it is interesting to compare the examples since they use the same plant with di erent control design strategies. For example, the one thing we notice straight away is the better disturbance response for the PID compensator. This should come as no surprise since the reset time for the integrator is significantly larger for the PID compensator.

We should emphasise that the process is rarely as straightforward as the examples suggest, and that in practice one will usually have to iterate the design process to account for the various tradeo s which exist between stability and performance. Also, the methodologies we discuss above can only be expected to have a reasonable chance at success when the plant transfer function RP has no poles or zeros in C+. If RP does have poles in C+, then one must design the controller RC so that the point −1 + i0 is encircled. If RP has zeros in C+, then it typically turns out to be more di cult to design a stabilising controller that meets goals for stability margins. In such cases, mundane considerations of gain and phase margin become less satisfactory measures of a good design unless weighed against other factors.

488 12 Ad hoc methods II: Simple frequency response methods for controller design 22/10/2004

12.3 Design with open controller form

In the previous section, the emphasis was on tuning controllers of a specified type. This can be a di cult exercise for plants which are unstable and/or nonminimum phase (see Exercise E12.6). The di culty is that be fixing the form on the controller, one also limits what can be done to the sensitivity function and the closed-loop transfer function. It may be possible that the plant will not allow stabilisation by a controller of a certain form.

12.4 Summary

In this section, we have been able to assimilate our knowledge gained to this point to generate some design methods for controller rational functions. One should note that our methods are not guaranteed to work, but for “simple” applications, they provide a starting point for serious controller design. Let us review some of the basic ideas.

1.The basis for the frequency response design methodology is the Nyquist criterion. This can be a somewhat subtle notion, so it would be best to understand it. A good way to do this is to study the proof of the Nyquist criterion since it is quite simple, given the Principle of the Argument.

2.We have presented two design methodologies using frequency response: one for lead compensation and the other for PID compensation. One should make sure to understand the idea behind these design methodologies.