control system design astrom

Lecture 1

Introduction1

This first lecture has two parts. The first part gives an introduction and overviewof the course, starting from two examples of modern control, a DVD-reader and cardynamics. The second part of the lecture is a brief review of linear input-outputmodels in continuous time used in the basic course. The concepts of signal normand system gain are introduced.

1.1 First example — a DVD player

The appearance of cheap sensors, actuators and computing devices opens new ap-plication areas for feedback control all the time, even in mass produced consumerproducts. The control technology is mostly hidden to the user, but still critical foroperation and performance. A prime example of this is positioning of the pick-uphead in a storage device as a DVD or CD-rom, where the speed of data recoveryis directly correlated to the control performance.A DVD (Digital Versatile Disk) is a data disk of the same physical size as a

CD. Its use is mostly for video, but also for computer software as a large CD-ROMdisk. The storage technology is in principle the same as for the CD, but improved.A CD holds about 650 megabytes of data whereas a DVD holds 4.7 gigabytes (forsingle layer, single side).

Pit

Track

0.74 µm

Figure 1.1 The right picture shows pits forming tracks on the DVD surface.

The disk surface is reflective, so that laser light is reflected back. Data bitsare represented by pits of different lengths in tracks on the disk. These pits makethe laser beam interfere destructively with itself, and therefore the pits look blackto the laser.The surface velocity is constant (about 3.5 m/s), meaning that the disc should

rotate at different speeds depending on the current reading position. The challengeof the control problem is related to the fact that only 0.022 µm deviations from the

1Written by A. Rantzer with contributions by K.J. Åström, B. Lincoln and B. Wittenmark

1

Lecture 1. Introduction1

bit-track can be accepted. At the same time, a disk is always slightly asymmetric,causing it to oscillate up to 100 µm per rotation, and the rotation speed is up to 23Hz (for single speed). The tracking controller must compensate for this oscillation.

A typical DVD player has a pick-up-head consisting of a laser, an astigmaticlens, and a light detector with four fields – see Figure 1.2. The lens is mountedon springs in the axial (focus) and radial direction, and can be moved by electro-magnets. This way, the laser spot can be moved very fast in a small range (a fewhundred tracks sideways). The lens and laser are mounted on the sledge, whichcan move over the whole disk (in radial direction), but with much less precisionand speed.

Radial electromagnet

Focus electromagnet

Springs

Light detectors

Laser

A B

C D

Tracks

Lens

Pick−up headSledge

Disk

Figure 1.2 The pick-up-head has two electromagnets for fast positioning of the lens (left).Larger radial movements are taken care of by the sledge (right).

Four light detectors are available to estimate the focus error and radial errorof the lens. Measurements are taken with a sampling frequency of 40 kHz andthe DVD standard specifies that the speed of control (cross-over frequency) mustbe at least 2.4 kHz.

It turns out that most of the main topics of this course are relevant for thesolution of the DVD control problem and we have therefore chosen to use it as ademonstrator. The focus control will be treated in a lecture, while a lab exercisetowards the end of the course is devoted to disc track following.

Figure 1.3 The DVD reader used in lab 3.

2

1.2 Second example — Control of car dynamics

1.2 Second example — Control of car dynamics

Amodern car contains numerous micro-processors devoted to feedback control. Forexample, feedback from oxygen sensors in the exhaust gas are needed for properoperation of the engine and catalyzer. This is essential for fuel efficiency and toreduce the emission of polluting exhaust gases.Other feedback loops are used to improve safety, by controlling the brakes to

prevent wheel-locks and to prevent skidding on slippery roads. A simplified modelfor car dynamics is given by the state space description

[V

r

]= A

[V

r

]+

[0

b1

](u1 + u2 − u3 − u4) +

[b2

b3

]δ

where V is lateral speed and r is angular velocity. There are five control signals,the steering angle δ and the brake forces u1, u2, u3 and u4 on the four wheels.

δ

U

V

r

Figure 1.4 A modern car relies on feedback control for comfort, safety and fuel efficiency.The left picture shows a test-car used in a research project together with DaimlerChrysler.

The state is generally not available for direct measurement. Even if the angularvelocity of each wheel can be measured, there is always some discrepancy betweenthe rotational speed and the speed over ground. Hence the velocity of the carmust be estimated based on information from several sources and the remaininguncertainty must be taken into account in the control algorithms.A typical sampling frequency for speed measurements is a few milliseconds.

This may sound fast enough compared to typical car dynamics, but when thepurpose is to prevent wheel-lock or accidents, a delay of a few milliseconds can infact be a severe obstacle for proper control performance.

Vehicle

brake forces

steering angle

lateral velocity

yaw rate

Figure 1.5 Input-output diagram for car dynamics control.

3


1.3 Course overview

The objective of the course is that the students should learn how to state and solveadvanced design problems for computer controlled systems. A schematic pictureof such a system is given below.

Process

u t( )

)

uk

y t(u t( )

yk

SamplerHold

Computer

ukyk

tt

t

y t( )

t

D-A A-D

The control signal u(t) is determined using measurements of y(t) to achievedesired behaviour of the process. For the DVD player, y(t) would be a vector offour variables, representing intensities in the four light detectors in Figure 1.2,while u(t) would correspond to the two electromagnets.It is important to note that the dynamics of a real process is never known

exactly. Neither is it possible to precisely state the “true design objectives”. It istherefore necessary to maintain a broader perspective on the engineering designproblem. See Figure 1.6.

Experiment

Implementation

Synthesis

Analysis

Matematical modeland

specification

Idea/Purpose

Figure 1.6 Schematic overview of the design process

Everything starts with an idea about the purpose of the control task. In simplecases, it is possible to directly come up with a solution proposal that can be testedexperimentally and be accepted, possibly after minor modifications. However, in avast number of applications costs and time can be reduced by analyzing or simu-lating a mathematical model before trying real experiments. The purpose of thediagram is to illustrate this methodology. Note that the arrows point in two direc-tions. Failure in the experimental phase could not only require reimplementation,but also new analysis, more accurate models, or even redefinition of the controlpurpose.Imagine stepping through the diagram in order to design a controller for car

dynamics as in the previous example. Suppose that a controller has been synthe-

4

1.4 Linear input-output maps

sized based on the given two state model. Implementing a controller on a prototypecar is costly, so a second step would typically involve computer simulation. For thispurpose a more complex and accurate car model is needed, a model that is lesstransparent from a synthesis perspective but better suited to reveal the deficien-cies of a proposed controller. If the simulations fail, a reason could be that thetwo state model was too simple and that additional features need to be takeninto account in the synthesis phase. After a sequence of attempts, one could hopeto find a solution ready for experimental tests. Alternatively, persisting failurescould be an indication that the original goal was overly optimistic and impossibleto achieve.The main emphasis of this course is on the analysis/synthesis phase of the

diagram in Figure 1.6. However, to keep the big picture in mind, there will alsobe lectures and laboratory sessions devoted to modelling, implementation andexperiments.The main topics of the course are the following

• Design of scalar controllers

• Sampled systems

• Discrete-time systems

• Fundamental system limitations

• Multi–input-multi–output systems

• Synthesis by optimization

During the first four lectures we keep the continuous time perspective that wasused in the basic course. We build on material from that course, but make a deeperstudy of robustness and performance evaluation in controller design. These fourlectures are supported by hand-out notes, but not by the book. The first lab exerciseis aimed to give practical training in scalar controller design by frequency domainloop shaping.Starting from lecture five, the textbook is used to introduce the discrete time

perspective of computer control. The theory for discrete time linear systems isapplied to sampled data control. This is the topic of lectures 5-9.After this, in the fourth week, there is an intermezzo of two more lectures

supported by handout notes. The topic is fundamental limitations in controllerdesign and we also discuss how the previous ideas can be applied to multivariablesystems (i.e. systems with several inputs and outputs). This is the topic of thesecond lab exercise.The last three weeks of the course, lectures 12-16, continue along the lines

of the textbook and bring in the subject of optimization. The theory of linearquadratic optimal control and Kalman filtering is a cornerstone of modern control.It clarifies fundamental relationships between measurement accuracy, control au-thority and achievable performance. Multivariable systems also fit in very nicely.Finally, the course is concluded by a lab exercise devoted to track following in theDVD reader. Most of the main topics in the course are relevant for a successfulsolution to this problem.


In this section, we will review some different ways of specifying the input-outputrelationship of a finite-dimensional linear time-invariant system. This is a systemthat can be described by a state space equation

x = Ax + Bu

y = Cx + Du

5


The differential equation has the solution formula

y(t) = CeAtx(0) +∫ t

0CeA(t−τ )Bu(τ )dτ + Du(t)

Note that the formula remains valid for multivariable systems, i.e. when both u(t)and y(t) are vector valued.The map from u to y is linear provided that x(0) = 0. Introducing the impulse

response �(t) as

�(t) =

∫ t

0CeA(t−τ )Bδ (τ )dτ + Dδ (t) = CeAtB + Dδ (t)

the input-output map can be written as a convolution

y(t) =

∫ t

0�(t− τ )u(τ )dτ = [� ∗ u](t)

In frequency domain, the convolution becomes multiplication

Y(s) = G(s)U (s)

and the Laplace transform of the impulse response is equal to the transfer functionG(s) = C(sI − A)−1B + D. For multivariable systems, both �(t) and G(s) arematrices. The term impulse response is of course motivated by the fact that thematrix element �i j(t) is the value of output i obtained when input j is an impulse(Dirac function) at t = 0. This is sometimes used to determine �(t) experimentally.

−1 0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

−1 0 1 2 3 4 5 6 7 8 9 10−0.1

0

0.1

0.2

0.3

0.4

0.5

y i(t)=�ij(t)

uj(t)=

δ(t)

A more common experiment in process industry is the step response. Assumingthat the (possibly vector valued) input is a step

u(t) =

{0 t < 0

u0 t ≥ 0

the output becomes

y(t) =

∫ t

0�(t− s)u0ds =

(∫ t

0�(τ )dτ

)u0

Accordingly, the Laplace transform of the step response is G(s)u0/s.

6


−1 0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

−1 0 1 2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

y i(t)

uj(t)

The main use of the Laplace transform is however to characterize the frequencyresponse. The input u(t) = u0 sinω t gives

y(t) =

∫ t

0�(τ )u(t − τ )dτ = Im

[∫ t

0�(τ )e−iωτ dτ ⋅ eiω tu0

]

The integral approaches G(iω ) as t → ∞, so after a transient, also the outputbecomes sinusoidal and y(t) = Im

(G(iω )eiω t

)u0. To summarize, a linear time-

invariant system always gives a sinusodal response to a sinusodal input. For ascalar system, the gain and phase shifts are determined by the amplitude andphase of the complex number G(iω ).

0 2 4 6 8 10 12 14 16 18 20−0.2

−0.1

0

0.1

0.2

0.3

0 2 4 6 8 10 12 14 16 18 20−1.5

−1

−0.5

0

0.5

1

1.5

y i(t)

uj(t)

There are several ways to graphically illustrate the transfer function G(iω ).One is to plot the amplitude and phase separately versus the frequency. This iscalled the Bode diagram:

7


10−2

10−1

100

100

101

−180

−135

−90

−45

0

Magnitude

Phase

Frequency (rad/sec)

It should be noted that each additional factor in the transfer function contributesadditively to the Bode plots:

log pG1G2G3p = log pG1p + log pG2p + log pG3p

argG1G2G3 = argG1 + argG2 + argG3

The Nyquist diagram is obtained by plotting G(iω ) directly in the complexplane for different values of ω :

−1 −0.5 0 0.5

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

argG(iω )

pG(iω )p

Im

Re

If instead logG(iω ) = log pG(iω )p+ i argG(iω ) is plotted in the complex plane,the Nichols diagram is obtained:

−360 −315 −270 −225 −180 −135 −90 −45 0−40

−30

−20

−10

0

10

20

30

40

6 dB

3 dB

1 dB

0.5 dB

0.25 dB

0 dB

−1 dB

−3 dB

−6 dB

−12 dB

−20 dB

−40 dB

Amplitude

PhaseThe level curves of pG/(1 + G)p and argG/(1 + G) are plotted as dotted lines tosupport use of the diagram in controller design.

8

1.5 Signal norm and system gain

1.5 Signal norm and system gain

In order to efficiently analyze and optimize dynamical systems, it is useful to havemathematical notions that measure the size of a signal and the gain of a system.This is the reason for the following definitions.The size of a signal y(t) ∈ Rn can be measured by the L2-norm, defined as

qyq2 :=

√∫ ∞

0py(t)p2dt

According to a theorem known as Parseval’s formula, the same norm can be de-fined in frequency domain as

qyq2 =

√12π

∫ ∞

−∞

pLy(iω )p2dω

For a system S with input u, output S(u) and zero initial state, the L2-gainis defined as the largest possible fraction between the input norm and the outputnorm

qSq := supu

qS(u)q

quq

The system is called input-output stable (or L2-stable) if its L2-gain is finite. Forexample, a time delay does not change the signal norm, so it has gain one. However,an integrator has infinite gain, since an input u(t) that is identically zero for t ≥ 1,can give an output y(t) that is a nonzero constant for t ≥ 1. Hence, the fractionqyq2/quq2 can be arbitrarily large.More generally, the L2-gain of a system can be obtained as the maximum

amplitude in the Bode diagram:

THEOREM 1.1A stable system with transfer function G(s) has the L2-gain

qGq∞ := supωpG(iω )p

Remark. For multivariable systems the pG(iω )p should be interpreted as the ma-trix norm (the largest singular value) of G(iω ). This case will be studied morecarefully later.

Proof. Let y be the output corresponding to the input u. Then

qyq2 =12π

∫ ∞

−∞

pLy(iω )p2dω ≤12π

∫ ∞

−∞

pG(iω )p2 ⋅ pLu(iω )p2dω ≤ qGq2∞quq2

The inequality is arbitrarily tight when u(t) is a sinusoid near the maximizingfrequency. 2

Example 1

a. For a time delay G(s) = e−sT we have pG(iω )p " 1.b. For an integrator pG(iω )p = p 1

iωp = 1

ωwhich is unbounded ω = 0.

c. The Bode diagram plotted in the previous section has a peak magnitudeabout 0.5 at the frequency 2 rad/sec. Hence, the L2-gain of the correspondingsystem is smaller than one and the highest gain is obtained for an input sinusoidof this frequency. 2

9

Lecture 2

Stability and Robustness1

This lecture discusses the role of stability in feedback design. The emphasis isnot on yes/no tests for stability, but rather on how to measure the distance toinstability. The small gain theorem is introduced as a means to verify stability inpresence of model uncertainty.

2.1 Stability of feedback systems

Feedback and stability are closely connected issues. On one hand, the introductionof feedback may potentially create instability. On the other hand, properly appliedfeedback is often the best way to get rid of instabilities.There are several well known examples in the history. One is the construction

of airplanes. The first airplanes in the early 1900s were depending on the pilot tostabilize the dynamics manually using the control-stick. The modern fighter JAS-Gripen was built unstable for the sake of maneuverability and relies on computercontrol for stabilization.

Figure 2.1 Lawrence Sperry demonstrates a stabilizing gyroscopic controller. He waves hishand in the air, while his mechanic is walking on the wing.

Another striking example was the construction of the first electronic feedbackamplifiers, that were necessary to build long distance telephone connections inthe 1930s. In this case, high gain feedback was needed to reduce the nonlinearsignal distorsion. Stability problems became a major issue, and the developmentof frequency domain stability criteria was critical for successful implementation.A modern example is the maneuver test for Mercesdes A-class, that created

unstable oscillations severe enough to turn the car over. The problem was solvedby introducing electronic feedback control.Recall from the previous lecture that a system is called input-output stable

(or L2-stable) if its L2-gain is bounded. A transfer function is called stable if itcorresponds to an input-output stable system. The following stability criterion isavailable for linear time-invariant systems.

THEOREM 2.1A rational transfer function G(s) is stable if and only if all poles of G have negativereal part. In particular, if G(s) = C(sI − A)−1B + D, it is sufficient that alleigenvalues of A have negative real part.

1Written by A. Rantzer with contributions by K.J. Åström

10

2.1 Stability of feedback systems

Figure 2.2 The stability problems of Mercedes A-class were solved by electronic feedback

More generally, a system with impulse response �(t) is stable provided that G(iω ) =∫∞0 e

−iω t�(t)dt is well defined and there is a c with pG(iω )p ≤ c for all ω .

Example 1 Let us determine input-output stability of the following systems

(I)

x =

[−1 2

−3 −2

]+

[1

0

]u

y= [1 1 ] x + u

(I I) y(t) =

∫ t

0eτ−t

(u(τ ) − y(τ )

)dτ

(I) The first system is input-output stable due to stable eigenvalues of the sys-tem matrix. A two-by-two matrix like this is stable if and only if the trace isnegative (here −3) and the determinant is positive (here 8). This is becausethe trace is the sum of the eigenvalues and the determinant is the product.In general, eigenvalues can be computed by the matlab command eig(A):

>> eig([-1 2; -3 -2])

ans =

-1.5000 + 2.3979i

-1.5000 - 2.3979i

Note that the coefficients of the characteristic polynomial should not be com-puted, at least for high order systems, since this generally leads to numericaldifficulties.

(I I) The input-output relationship can be written

y= � ∗ (u − y)

where �(t) = e−t, t ≥ 0. After Laplace transformation, this gives

Y(s) =1s+ 1

[U (s) − Y(s)]

[ Y(s) =1s+ 2

U (s)

The transfer function 1/(s + 2) shows that the system is stable, since theonly pole −2 is negative.

2

Our main objective is to study stability of feedback loops. From the basic course,we recall the Nyquist criterion, which supports understanding by graphical illus-trations.

11

Lecture 2. Stability and Robustness1

n G(s)--

−1

-

�

6Σ

−1 −0.5 0 0.5 1−1

−0.5

0

0.5

1

Im

Re

1

m

0

ϕ

Am

ωc

ω

Figure 2.3 The closed loop system remains stable as long as the Nyquist diagram doesnot encircle −1. The amplitude margin Am and phase margin φm measure the distance frominstability.

THEOREM 2.2—THE NYQUIST CRITERIONSuppose that G(s) is stable and that the Nyquist plot G(iω ), ω ∈ R does notencircle −1. Then (1+ G(s))−1 is also stable.

The following example illustrates the use of the theorem.

Example 2 As motivating example, consider a position control in a mechanicalsystem with damping coefficient c. The controller contains a time delay:

x + cx + x = u u(t) = −k[x(t − T) + x(t− T)]

The feedback loop is illustrated in the figure below. Nominal values of the param-eters are k = 1, c = 1 and T = 0. Let us investigate how much margin there is ineach of the parameters before the system becomes unstable?

h k(s+1)s2+cs+1 e−sT

−1

- - - -

�

6

For this purpose, we plot the Nyquist and Bode digrams of the nominal transferfunction (s+ 1)/(s2+ s+ 1). The Matlab command margin gives numerical valuesfor the amplitude- and phase-margins.

0 0.5 1 1.5−1.5

−1

−0.5

0Nyquist diagram

Re

Im

100

101

−180

−135

−90

−45

00

0.5

1

1.5Gm = Inf, Pm = 109.47 deg (at 1.4142 rad/sec)

Magnitude

Phase

Frequency

Figure 2.4 Nyquist and bode plots for the nominal transfer function (s+ 1)/(s2 + s+ 1)

12

2.2 Sensitivity

For k = c = 1 the open loop transfer function is

s+ 1s2 + s+ 1

e−sT

For small values of T the Nyquist plot will not encircle −1, so the systems remainsstable. To find out exactly how large values of T are needed for instability, note thatthe phase margin 109 degrees (or 109π/180 radians) is obtained at the frequency1.41 rad/sec. Hence a time delay of

109π180 ⋅ 1.41

= 1.35 seconds

can be tolerated for k = c = 1.To investigate the robustness to variations in the parameters c and k, we note

that the closed loop characteristic polynomial for T = 0 is

(s2 + cs+ 1) + k(s+ 1) = s2 + (c+ k)s+ 1+ k

The stability condition for a second degree polynomial is positive coefficients, sostability is maintained as long as c+ k > 0 and 1+ k > 0. 2

Having analyzed the example with respect to parametric uncertainty above,it is natural to ask about robustness to unmodelled dynamics. For example, thiswould be relevant to capture the difference between a rigid body and an elasticone. This is also the subject for the remaining part of the lecture.

2.2 Sensitivity

Two transfer functions are of particular interest in the study of the feedback loopbelow.

C(s) P(s)

−1

ΣΣΣr e u

d

x

n

y

Figure 2.5 A simple control loop

These are

S(s) =1

1+ C(s)P(s)(the sensitivity function)

T(s) =C(s)P(s)

1+ C(s)P(s)(the complementary sensitivity function)

The term complementary refers to the fact that

S(s) + T(s) " 1

Note that T(s) is the transfer function from reference signal r to the the plantoutput x. Another name for T(s) is therefore the closed loop transfer function of

13


R−1 = supω

∣∣∣∣1

1+ C(iω )P(iω )

∣∣∣∣

−1

Re

Im

R

C(iω )P(iω )

Figure 2.6 The L2-gain of the sensitivity function is the inverse of the distance from theNyquist plot to −1.

the system, while C(s)P(s) is called the open loop transfer function or just the“loop transfer function”.The name sensitivity function refers to the fact that S measures how small

relative errors in P are mapped into relative errors in T . This is verified by asimple calculation:

dT

dP=d

dP

(1−

11+ CP

)=

C

(1+ CP)2=TS

P[

dT/T

dP/P= S

Notice that T is a nonlinear function of P. Hence the sensitivity calculationonly says something about the response to small pertubations in P. For largerperturbations, the system could have a drastically different behaviour and evenbecome unstable.Given the Nyquist criterion, it is natural to conjecture that the robustness to

unmodelled dynamics should somehow be related to the distance from the Nyquistplot to the point −1. It is therefore striking to note that the L2-gain of the sensi-tivity function turns out to be exactly the inverse of this distance. Consequencesare investigated in the next section.

2.3 Robustness via the small gain theorem

r1

r2

e1

e2

S1

S2

In this section, we will investigate stability robustness using the followingtheorem, based on the notion of input-output L2-gain. For simplicity, calculationsare done assuming zero initial conditions.

THEOREM 2.3—THE SMALL GAIN THEOREMAssume that S1 and S2 are input-output stable systems with L2-gain qS1q andqS2q. If qS1q ⋅ qS2q < 1, then the L2-gain from (r1, r2) to (e1, e2) in the closed loop

14

2.3 Robustness via the small gain theorem

system

e1 = S2(e2) + r1

e2 = S1(e1) + r2

is finite.

Proof. Define qyqT =√∫ T

0 py(t)p2dt. Then qS(y)qT ≤ qSq ⋅ qyqT .

e1 = r1 + S2(r2 + S1(e1))

qe1qT ≤ qr1qT + qS2q(qr2qT + qS1q ⋅ qe1qT

)

qe1qT ≤qr1qT + qS2q ⋅ qr2qT1− qS1q ⋅ qS2q

Bounded gain from (r1, r2) to e1 follows as T → ∞. The gain to e2 is bounded inthe same way. 2

To demonstrate how the small gain theorem can be used for robustness anal-ysis, consider the feedback loop in Figure 2.3, where the plant P(s) has beenreplaced by [1+ ∆(s)]P(s).

fP(s)

∆(s)

?

�

-

-

C(s)

−1

- -

v w

Figure 2.7 Loop diagram with perturbed plant [1+ ∆(s)]P(s)

The transfer function from w to v is equal to −T(s), the complementary sen-sitivity function. Hence, by the small gain theorem, the feedback system remainsstable as long as

q∆q ⋅ qTq < 1

Note that the small gain theorem does not assume linearity or time-invariance.Hence the closed loop system will remain stable for all plants of the form P(s)[1+∆(s)] where ∆ has L2-gain smaller than [supω pT(iω )p]

−1, even for ∆ that arenonlinear or time-varying.As a second example, let us derive a stability criterion for the case that the

perturbation appears additively, i.e. P(s) is replaced by P(s) + ∆(s).Then the transfer function from w to v is equal to C(s)S(s), so the small gain

theorem shows stability for all perturbations satisfying

q∆q ⋅ qCSq < 1

For linear time-invariant perturbations, this criterion can be nicely illustrated inthe Nyquist diagram. Clearly, the condition

p∆ ⋅ Cp < p1+ PCp

guarantees that the Nyquist plot of (P+ ∆)C does not encircle −1.

15


fC(s) P(s)- -

−1 �

-

∆(s)

?

-wv

Figure 2.8 Loop diagram with perturbed plant P(s) + ∆(s)

−1

1+ PC

∆ ⋅ C

16

Lecture 3

Scalar control synthesis1

The lectures reviews the main aspects in synthesis of scalar feedback systems.Another name for such systems is single-input-single-output (SISO) systems. Thespecifications include ability to follow reference signals, to attenuate load distur-bances and measurement noise and to reduce the effects of process variations. Inthe presentation, we separate the solution into feedback control and feedforwardcontrol.

3.1 Specifications

Recall from Lecture 1 the illustration of the design process shown in Figure 3.1.While Lecture 2 was mainly concerned with analysis, we are now focusing on thethree neighboring blocks: Specification, Analysis and Synthesis.

Experiment

Implementation

Synthesis

Analysis

Matematical modeland

specification

Idea/Purpose

Figure 3.1 Schematic overview of the design process

We will restrict attention to the following structure (Figure 3.2), with a scalartransfer function for the plant. This setup was studied in the basic course and issufficient for many practical situations.The controller consists of two transfer functions, the feedback part C(s) and

the feedforward part F(s). The control objective is to keep the process output xclose to the reference signal r, in spite of load disturbances d. The measurementy is corrupted by noise n.


17

Lecture 3. Scalar control synthesis1

F(s) C(s) P(s)

−1

Σ Σ Σr e u

d

x y

n

v

Controller Process

Figure 3.2 A Controller with two degrees of freedom

Several types of specifications could be relevant for this control loop.

A: Reduce the effects of load disturbances

B: Control the effects of measurement noise

C: Reduce sensitivity to process variations

D: Make the output follow command signals

A useful synthesis approach is to first design C(s) to meet the specificationsA, B, and C, then design F(s), to deal with the response to reference changes, D.However, the two steps are not completely independent: A poor feedback designwill have a negative influence also on the response to reference signals.The following relations hold between the Laplace transforms of the signals in

the closed loop system.

X (s) =PCF

1+ PCR(s) −

PC

1+ PCN(s) +

P

1+ PCD(s)

V (s) =CF

1+ PCR(s) −

C

1+ PCN(s) +

11+ PC

D(s)

Y(s) =PCF

1+ PCR(s) +

11+ PC

N(s) +P

1+ PCD(s)

Several observations can be made:

• The signals in the feedback loop are characterized by four transfer functions(sometimes called The Gang of Four)

11+ P(s)C(s)

P(s)

1+ P(s)C(s)C(s)

1+ P(s)C(s)P(s)C(s)

1+ P(s)C(s)

In particular, we recognize the first one as the sensitivity function and thelast one as the complementary sensitivity.

• The total system with a controller having two degrees of freedom is charac-terized by six transfer functions (The Gang of Six).

To fully understand the properties of the closed loop system, it is necessary to lookat all the transfer functions. It can be strongly misleading to only show propertiesof a few input-output maps, for example a step response from reference signal toprocess output. This is a common mistake in the literature.The properties of the different transfer functions can be illustrated in several

ways, by time- or frequency-responses. For a particular example, we show belowfirst the six frequency response amplitudes, then the corresponding six step re-sponses.It is worthwhile to compare the frequency plots and the step responses and to

relate their shape to the specifications A-D:

18

3.1 Specifications

10−1

100

101

10−1

100

10−1

100

101

10−1

100

10−1

100

101

10−1

100

10−1

100

101

100

101

10−1

100

101

100

101

10−1

100

101

100

101

PCF/(1+ PC) PC/(1+ PC)

C/(1+ PC)

P/(1+ PC)

CF/(1+ PC) 1/(1+ PC)

Figure 3.3 Frequency response amplitudes for P(s) = (s+ 1)−4, C(s) = 0.775(s−1/2.05+ 1)when F(s) is designed to give PCF/(1+ PC) = (0.5s+ 1)−4

0 10 20 30

0

0.5

1

1.5

0 10 20 30

0

0.5

1

1.5

0 10 20 30

0

0.5

1

1.5

0 10 20 30

0

0.5

1

1.5

0 10 20 30

0

0.5

1

1.5

0 10 20 30

0

0.5

1

1.5

PCF/(1+ PC) PC/(1+ PC)

C/(1+ PC)

P/(1+ PC)

CF/(1+ PC) 1/(1+ PC)

Figure 3.4 Step responses for P(s) = (s + 1)−4, C(s) = 0.775(s−1/2.05 + 1) when F(s) isdesigned to give PCF/(1+ PC) = (0.5s+ 1)−4

Disturbance rejection The two upper right plots show the effect of the distur-bance d in process output x and input v respectively. The resulting process errorshould not be too large and should settle to zero quickly enough. The control inputwould cancel the disturbance exactly if the mid upper step response would be anideal step. In a short time-scale this is impossible, since the control input will notchange until the effect of the disturbance has appeared in the process output andbeen available for measurement. However, slow disturbances should normally becancelled by u. Equivalently, the sensitivity function 1/(1 + PC) should be smallfor low frequencies. This specification is usually corresponds to an integrator inthe controller.

Supression of measurement noise The second specification was to limit theeffect of measurement noise, typically a high frequency phenomenon. The mid up-per frequency plot shows good attenuation of measurement noise above the “cutoff” frequency of 1 Hz. In this example, this is mainly an effect of the process dy-namics. A more interesting question is maybe the gain from measurement noiseto control input, since fast oscillations in the control actuator are usually undesir-

19


able. For this aspect, the mid lower frequency plot, showing the Bode amplitudefrom n to v, is of interest.

Robustness to process variations As shown in the previous lecture, the ro-bustness to process variations is determined by the sensitivity functions. In thisexample, the lower right frequency plot has a maximal value of 2, which showsthat a small relative error in the process can give rise to a relative error of doublesize in the closed loop transfer function. The maximal amplitude of the frequencyplot for the complementary sensitivity function is 1.35, so the small gain theoremproves stability of the closed loop system as long as the relative error in the pro-cess model is below 74% = 1/1.35. In fact, most process models are inaccurate athigh frequencies, so the complementary sensitivity function PC/(1 + PC) shouldbe small for high frequencies.

Command response The upper left corner plot shows the map from referencesignal r to process output x. Using the prefilter F, it is possible to get a betterstep response here than in the upper mid plot. The prize to pay is that the corre-sponding response in the control signal gets higher amplitude. This can be seenby comparing the lower left plot, showing the map from r to v, to the lower midplot, which shows the corresponding map when F " 1.

3.2 Loop shaping

The closed loop performance depends critically on the loop transfer function

L(s) = P(s)C(s)

In particular, the sensistivity functions can be written as S = (1 + L)−1 andT = L(1 + L)−1 respectively. A popular approach to control synthesis, knownas loop shaping, is to focus on the shape of the loop transfer function and keepmodifying C(s) until the desired shape is obtained.Recall that proper disturbance rejection requires small sensitivity S (large L)

for for small frequencies, while process uncertainty requires the complementarysensitivity function to be small (small L) for high frequencies. On the other hand,if the amplitude of L decreases very rapidly, the phase tends to become lowerthan −180○ and the system becomes unstable. Loop shaping is therefore a trade-off between different kinds of specifications.Many control problems can be adequately solved by PID controllers, which can

be viewed a combination of one lag compensator and one lead compensator. Formore advanced applications, like resonant systems, higher order controllers aredesirable. An example of such a system is the flexible servo treated in lab 1.Graphical illustrations in Bode- or Nichols- diagrams are typically used to

support the design. These diagrams are convenient because of the logarithmicscale, where the controller contributes additively to the loop transfer function:

log pL(iω )p = log pP(iω )p + log pC(iω )p

arg pL(iω )p = arg P(iω ) + argC(iω )

From the basic course, recall the following essential properties of lead and lagcompensators, illustrated in Figure 3.5:

Lag compensator

• Increases low frequency gain: Can be used to reduce stationary errors

• Decreases phase, which may reduce stability margins

20

3.2 Loop shaping

0

10

20

100

101

102

−60

−30

0

Magnitude

Phase

0

10

20

10−2

10−1

100

101

0

30

60

Figure 3.5 Bode diagrams for lag compensator s+10s+1 (left) and lead compensator 10s+1s+1(right)

Lead compensator

• Increases high frequency gain: Can be used for faster closed loop response

• Increases phase, which may improve stability margins

Loop shaping design of high order controllers will be exercised in lab 1. We willfirst design a controller C1(s) for low frequencies, then keep adding compensatorlinks C2(s),C3(s), . . . to modify the dynamics at higher and higher frequenciesuntil a satisfactory controller C(s) = C1(s) ⋅ ⋅ ⋅Cm(s) is obtained. Lead/lag linksare often sufficient, but occasionally it is useful to also consider controllers withpoles or zeros outside the real axis. The figure below shows the Bode diagram forcases with stable complex zeros (left) and complex poles (right).

−30

−20

−10

0

100

101

−90

−45

0

45

90

Magnitude

Phase

0

10

20

30

100

101

−90

−45

0

45

90

Figure 3.6 Notch compensator s2+0.1s+1(s+1)2

(left) and resonant compensator (s+1)2

s2+0.1s+1(right)

21


3.3 Feedforward synthesis

Let us finally consider the design of F(s) to shape the response to reference signals.For this purpose, it is useful to also consider the second diagram below.

Σr

F C P

−1

u y

ΣΣ

r

My

Mu

C P

−1

um

ym e u y

The two configurations are mathematically equivalent provided that

CF = Mu + CMy

The transfer functions Mu and My can be viewed as generators of the desiredoutput ym and the corresponding input um.

The transfer function from r to the error signal e = ym − y is (My − PMu)S.The error is zero provided that

Mu = My/P

Notice that this condition does not depend on C! Since Mu = My/P should bestable, causal and not include derivatives we find that

• Unstable process zeros must be zeros of My

• Time delays of the process must be time delays of My

• The pole excess of My must be greater than the pole excess of P

Apparently all three conditions put limitations on the achievable command re-sponse.

Example 1 If

P(s) =1

(s+ 1)4My(s) =

1(sT + 1)4

then

Mu(s) =My(s)

P(s)=

(s+ 1)4

(sT + 1)4Mu(∞)

Mu(0)=1T4

Fast response (T small) requires high gain of Mu. Bounds on the control signaltherefore limit how fast response we can obtain. 2

22

Lecture 4

DVD focus control1

Imagine the following: You are traveling at half the speed of light, along a line

from which you may only deviate 1 m. The line is not straight but oscillates up to

4.5 km sideways, 23 times per second.

This is a scaled version of the control task in a DVD player, where the pick-uphead needs to follow the bit-track. The real numbers are 3.5 m/s, with maximally0.022 µm deviations from the track. A disk is always slightly asymmetric, causingit to oscillate up to 100 µm per rotation, and the rotation speed is up to 23 Hz(for single speed).

Pit

Track

trackpitch

Pick−up headSledge

Disk

Figure 4.1 Pits forming tracks on DVD surface (left). Larger radial movements are takencare of by the sledge (right).

The surface velocity is constant (about 3.5 m/s), meaning that the disc shouldrotate at different speeds depending on the current reading position.

4.1 The DVD player

A typical DVD player has a pick-up-head consisting of a laser, an astigmatic lens,and a light detector with four fields. See Figure 4.1.The lens is mounted on springs in the axial (focus) and radial direction, and

can be moved by electromagnets. This way, the laser spot can be moved very fastin a small range (a few hundred tracks sideways). The lens and laser are mountedon the sledge, which can move over the whole disk (in radial direction), but withmuch less precision and speed. As the disk rotates, the track moves both radiallyand axially because of asymmetries, so feedback control is needed for both theradial and the axial position of the lens.Light is emitted by the laser, focused through the lens onto the disk surface.

The disk surface is reflective, so that laser light is reflected back. Data bits are1This text is based on an original manuscript by Bo Lincoln

23

Lecture 4. DVD focus control

represented by pits of different lengths in tracks on the disk. The “image” of thesurface is reflected back through the lens and read by a set of four photo detectors.These pits make the laser beam interfere destructively with itself, and thereforethe pits look black to the detector.

Radial electromagnet

Focus electromagnet

Springs

Light detectors

Laser

A B

C D

Tracks

Lens A B

C D

Figure 4.2 The pick-up-head has two electromagnets for fast positioning of the lens (left).The four photo detectors A− D (right).

The photo detectors

The photo detectors sense the reflected laser light from the disk. It can be thoughtof as a very simple camera with only four pixels (see Figure 4.2). This informationis enough to sense position radially (are we to the left or right of the track?),axially (thanks to the astigmatic lens) as well as the current bit. There are noother sensors in the pick-up head to help keep the laser in the track.

Focus error

The pick-up lens is astigmatic diagonally, so bad focus results in brighter light ineither detectors A + D (too high) or B + C (too low). Therefore, the focus error(FE) can be calculated as

FE = A+ D − (B + C)

which gives a surprisingly good result. Sweeping the focus lens from low to highusually results in an FE as in Figure 4.3. There is a linear range in the centerwhere the FE signal is useful to control the lens around the correct focus. Theslope of the curve depends on the reflectivity of the disc.

Too low Too high

Correct focus

Possible to see tracks

Focus Error

Lens height

Figure 4.3 Sweeping the lens axially from low to high results in a curve like this.

24

4.1 The DVD player

Radial error

The radial error (RE), meaning sideways deviation from the track center, can bemeasured in two ways using the photo detectors. The simple way is to let

RE = A+ C − (B + D),

i.e., use the difference in light from the left and right pair of detectors. For example,if the reflected light is brighter to the left, the radial error is positive, and weshould move right. This measurement method is called radial push-pull (PP). SeeFigure 4.4.

A B

C D

Pit

Figure 4.4 Push-pull: Calculating radial error as RE = A+ C − (B + D)..

There is a second measurement method (DPD), which usually gives better result,but requires the disk to have pits. This is not true for a non-written DVD-R(writable), for example. The signals f1 = A + D and f2 = B + C are created,and phase compared (see Figure 4.5). For example, if f1 comes before f2 the lensis too far to the right. The time difference forms the error signal RE.

A B

C D

Pit

Radial error

f1f2

Figure 4.5 DPD: Radial error from phase difference between f1 and f2.

According to the DVD specification, DPD should be used whenever possible. Sweep-ing the lens radially over the disk creates an RE as in Figure 4.6 for PP and DPD.

DPD

PP

Radial position

1 track

Correct radial position

Figure 4.6 DPD- and PP-signals as the disk rotates and the lens is swept over the tracksradially. As can be seen, DPD is linear in a larger range.

25


4.2 Dynamics

The DVD player system can be viewed as a two-input, two-output dynamical sys-tem. The main dynamics are due to the springs and masses in the lens system.The inputs are voltages to the electromagnets moving the lens, and the outputsare voltages corresponding to FE and RE. See Figure 4.7.

ufocus

uradial

FE

RE

Pick-up & Disc

Figure 4.7 The pick-up head and disk seen as a two-input, two-output dynamical system.

The department has a “raw” DVD player without any controller. Using systemidentification techniques, the transfer function P f (s) from ufocus to FE and thetransfer function Pr(s) from uradial to RE have been estimated. The cross-couplingbetween inputs and outputs have been ignored for simplicity. The resulting Bodediagrams can be seen in Figure 4.8.

Bode Diagram

Frequency (rad/sec)

Pha

se (

deg)

Mag

nitu

de (

dB)

−50

0

50

100

102

103

104

105

106

−270

−180

−90

0

Bode Diagram

Frequency (rad/sec)

Pha

se (

deg)

Mag

nitu

de (

dB)

−50

0

50

100

101

102

103

104

105

−270

−180

−90

0

Figure 4.8 Left: Transfer function estimate for the focus servo. The model is of secondorder. Right: Transfer function estimate for the radial servo.

4.3 Specifications and control structure

The control structure is simple: the focus and radial parts are separated (seeFigure 4.9). The task of the focus controller Cf is to keep the disk sufficiently infocus for the system to see the tracks, and the task of the radial controller Cr isto keep the laser spot in the track.In the DVD specification, bounds on the open-loop transfer functions for both

the focus and radial loops are given. The open-loop transfer function is (as youknow) the product of all elements in the control loop: Lr = CrPr, L f = Cf P f . Thecontrol specifications in the DVD standard are as follows:

pC(iω )P(iω )p ≥ 1000 for ω ≤ 23.1 Hz

pC(iω )P(iω )p ≤ 1 for ω > 2 kHz

The first specification is chosen to remove enough of the disk oscillation distur-bances to stay in track. The second specification is chosen to reduce the effects ofmeasurement noise and small disturbances (dirt and scratches).

26

4.4 Design of the Focus Controller

PUH & Disk

Focus controller

Radial controller

ufocus

uradial

FE

RE

Figure 4.9 The control structure.

Meeting these specifications is not a trivial task. In fact, most DVD readersprobably don’t! Instead, the manufacturers modify the specifications accordingto the circumstances. For example, a CD player for a car would need very gooddisturbance rejection (high gain at low frequencies) and this is obtained at theexpense of high gain also at high frequncies, which gives less robustness to discscratches.The goal of this lecture is to design the focus controller according to the spec-

ifications and try it on the experimental setup. In Lab 3 in this course, you willdesign your own radial controller to keep the DVD player in track!

4.4 Design of the Focus Controller

The specifications require that the magnitude of the loop transfer function de-creases by a factor 1000 in the frequency interval between 23.1 Hz and 2.0 kHz.Without compensation, the magnitude decreases only by a factor 200. See leftplot in Figure 4.10. A proportional controller with gain 0.4 satisfies the amplitudespecification at 2.0 kHz (right plot). However, the phase at 2.0 kHz is −184○, sothe gain has to be even smaller in order to keep the closed loop stable. Hence, adynamic compensator is needed.

Figure 4.10 Uncompensated system (left) and with proportional gain 0.4 (right).

It is natural to introduce lag compensation to increase the gain at low fre-quencies. However, the break points need to be at frequencies well below 2 kHzin order to avoid additional phase lag at the cut-off frequency. Using a lag filterC1(s) = 0.4 s+600s gives the modified plots in Figure 4.11 (left).

27


However, the closed loop system is still unstable, so further compensation isneeded. Hence, we add a lead filter to increase the phase near 2 kHz; C2(s) =1+s/50001+s/50000C1(s) (right plot).

Figure 4.11 A lag filter has been used to increase the gain at low frequencies (left). A leadfilter improves stability by increasing the phase near 2 kHz

The gain needs to be adjusted to keep the cross-over frequency unchanged.See Figure 4.12 (left). Now the closed loop system is stable with good margins,but the gain at 23.1 Hz is still too low, just 100 instead of 1000. This can becorrected by modifying the break point of the lag filter to get the final controllerC(s) = 0.15 1

s(s+1600)1+s/50001+s/50000 . See Figure 4.12 (right).

Figure 4.12 Bode and Nichols plots with three lag filters and one lead filter (dashed) andwithout compensation (solid).

The controller is verified by experiments done in the lecture. Notice that the fi-

nal design is very similar to a PID controller of the form C(s) = K(1sTi+ sTd1+sTd/N

).

28

Lecture 10

Fundamental limitations1

10.1 Introduction

Maybe the cardinal mistake in control engineering would be to consider the processas fixed once and for all. In fact, the control system specifications could very wellbe impossible to meet once the process is constructed and fixed. A striking exampleof this is given in the following citation from F. R. Whitt and D. G. Wilson (MITPress, 1974), Bicycling Science - Ergonomics and Mechanics:

“Many people have seen theoretical advantages in the fact that front-drive,

rear-steered recumbent bicycles would have simpler transmissions than

rear-driven recumbents and could have the center of mass nearer the front

wheel than the rear. The U.S. Department of Transportation commissioned

the construction of a safe motorcycle with this configuration. It turned out

to be safe in an unexpected way: No one could ride it.”

This lecture is devoted to the fundamental limitations that are inherited fromproperties of the controlled plant and will address questions like the followingtwo:

• Why are some bicycles impossible to ride?

• How short inverted pendulums can be balanced by hand?

One of the practically most important restrictions to control performance is thefact that actuators have limited capacity and may saturate. However, saturationis a non-linear effect and will not be studied further here. Instead, the focus willbe on limitations caused by unstable zeros, unstable poles and time-delays.An unstable pole p means that the response to a disturbance grows exponen-

tially as ept. It is intuitively clear that in order to stabilize such a system, thefeedback loop must be “faster” than the time constant 1/p. A formal argument forthis will be given in this lecture.In case the unstable system also includes a time-delay, the control problem

could become “impossible”. A time-delay T means that control action at time t doesnot have any effect until time t+T . Hence, it is intuitively clear that an unstablepole can not be stabilized unless T is small compared to 1/p. The argument appliesto the question about pendulum balancing above, since the human feedback systemthrough the eyes always involves a time delay.A more intricate performance limitation is imposed by unstable zeros. It is

well known that an unstable zero results in a step response that initially goes inthe “wrong” direction. In fact, this can be seen directly in the expression for theLaplace transform Y(s) of the step response y(t), where the zero at z means that

0 = Y(z) =∫ ∞

0y(t)e−ztdt


29

Lecture 10. Fundamental limitations1

Figure 10.1 Schematic picture of a bicycle. The top view is shown on the left and the rearview on the right.

Clearly the integral cannot be zero unless y(t) takes both positive and negativevalues. The time duration of such dynamics is approximately 1/z and limits theachievable rate of control. Hence, while an unstable pole requires a fast feedbackloop, an unstable zero gives an upper bound on how fast it can be. A combinationof the two phenomena can make the system impossible to control.Example 1 A tourque balance for a bicycle can be written as

Jd2θ

dt2= m�{θ +

mV0{

b

(V0β + a

dβ

dt

)

where the physical parameters have typical values as follows:

Mass: m = 70 kg

Distance rear-to-center: a = 0.3m

Height over ground: { = 1.2 m

Distance center-to-front: b = 0.7 m

Moment of inertia: J = 120 kgm2

Speed: V0 = 5 ms−1

Acceleration of gravity: � = 9.81 ms−2

The transfer function from β to θ is

P(s) =mV0{

b

as+ V0Js2 −m�{

The system has an unstable pole p with time-constant

p−1 =

√J

m�{( 0.4 s

The closed loop system must be at least as fast as this. Moreover, the transferfunction has a zero z with

z−1 = −a

V0( 0.06s

30

10.2 The Maximum Modulus Theorem

Riding the bicycle at this speed, the zero is not really an obstacle for control.However, with a rear-wheel-steered bicycle, the speed gets a negative sign andthe zero becomes unstable. In particular, for slow speed (( 0.7m/s) there is anunstable pole-zero cancellation, which is impossible to stabilize. 2

10.2 The Maximum Modulus Theorem

More formal arguments about fundamental limitations can be obtained using the-ory for analytic functions. It is natural that analytic functions appear, since wehave a seen that a controller is stabilizing if and only if the closed loop transferfunction is analytic in the right half plane (all poles in the left half plane). Themain mathematical theorem to be used is the following:

THEOREM 10.1—THE MAXIMUM MODULUS THEOREMSuppose that the function f is analytic in a set containing the unit disc. Then

maxpzp≤1

p f (z)p = maxpzp=1

p f (z)p

In Laplace transform applications, the stability boundary will be the imaginaryaxis. It is therefore convenient to note that for every stable rational transfer func-tion G(s), analytic in the right half plane, the function

f (z) = G

(1+ z1− z

)

is analytic in the unit disc. Hence the Maximum Modulus Theorem can be appliedto give the following corollary:

COROLLARY 10.1Suppose that all poles of the rational function G(s) have negative real part. Then

maxRe s≥0

pG(s)p = maxω∈R

pG(iω )p

10.3 Sensitivity bounds from unstable zeros and poles

It is easy to see that the sensitivity function must be equal to one at an unstablezero of the transfer function:

P(z) = 0 [ S(z) :=1

1+ C(z)P(z)= 1

Notice that the unstable zero in the plant can not be cancelled by an unstable polein the controller, since this would give an unstable transfer function C/(1+ CP)from measurement noise to control input.Similarly, the complimentary sensitivity must be one at an unstable pole:

P(p) = ∞ [ T(p) :=C(p)P(p)

1+ C(p)P(p)= 1

31


log pWap

a logω

0

log pWbp

b logω

0

Figure 10.2 Amplitude plots for weighting functions. The left weighting function is usedto bound the sensitivity at small frequencies, while the right function is used to boud thecomplementary sensitivity at high frequencies

In this case, cancellation by an unstable zero in the controller would give anunstable transfer function P/(1+ CP) from input disturbance to plant output.Combining the first constraint S(z) = 1 with Corollary 10.1 immediately gives

a lower bound on the norm of the sensitivity function:

maxω∈R

pS(iω )p = maxRe s≥0

pS(s)p ≥ pS(z)p = 1 [ qSq∞ ≥ 1

This bound is however not particularly interesting, since usually S(iω ) ( 1 for highfrequencies anyway. A much more interesting conclusion will next be obtained byusing a weighting function.Recall that disturbance rejection requires small sensitivity for small frequen-

cies. One way to formalize this condition is to define

Wa(s) =s+ a

2s

and require that

supωpWa(iω )S(iω )p ≤ 1 (10.1)

for some value of a. See Figure 10.2, left. Satisfying (10.1) with a high value of ameans fast disturbance rejection.The specification requires that S(s) has a zero in the origin. This is often

obtained by an integrator in the controller. Moreover, Corollary 10.1 implies that

supωpWa(iω )S(iω )p = sup

Re s≥0pWa(s)S(s)p ≥ pWa(zi)p

for every unstable zero zi of the plant P. In particular, the specification (10.1)is impossible to satisfy unless pWa(zi)p ≤ 1, or in other words a ≤ zi, for everyunstable zero zi. Hence the unstable zeros give an upper bound on the achievablebandwidth. In the following theorem, this discussion is summarized together witha corresponding argument for unstable poles:

THEOREM 10.2Suppose that the plant P(s) has unstable zeros zi and unstable poles pj . Definethe weighting functions Wa = (s + a)/(2s) and Wb(s) = (s + b)/(2b). Then thespecifications

supωpWa(iω )S(iω )p ≤ 1 sup

ω

∣∣Wb(iω )T(iω )∣∣ ≤ 1

32

10.4 Bode’s integral formula

are impossible to meet with a stabilizing controller unless

a ≤ minizi b ≥ max

jpj

Proof. The statement about the sensitivity function was proved above, and thestatment about the complementary sensitivity function is analogous. 2

Example 2 Let us see what Theorem 10.2 has to say about the bicycle example.The unstable pole gives a bound qWbTq ≤ 1 for b ≥

√m�{/J. This shows that the

closed loop transfer function from measurement noise to process output can notbe forced small for frequencies below

√m�{/J. A loose interpretation is that it

is impossible to ride the bicycle and keep the eyes shut except for a sample everysecond. This applies for the bicycle with normal steering regardless of speed.For a rear wheel steering bike, there is the second complication of an unstable

zero at V0/a, which gives a bound on how fast disturbances one can reject. Forlow speed, only slow disturbances can be rejected.The special difficulties corresponding to a combination of an unstable pole and

an unstable zero nearby are however not apparent in Theorem 10.2. Such problemswill be treated next. 2

The following theorem gives simple expressions for the limitations caused byan unstable pole/zero pair.

THEOREM 10.3If P(s) has an unstable pole p and an unstable zero z, then

∥∥∥∥1

1+ CP

∥∥∥∥∞

≥

∣∣∣∣z+ p

z− p

∣∣∣∣

for every stabilizing C(s).

Note that if S is very large, then the same is true for T , since S+T " 1. Hence, ifp(z+ p)/(z− p)p is significantly larger than one, the system is impossible to controlbecause of poor robustness to model errors and amplification of measurementnoise.

Proof. Assume that P(s) = (s− z)(s− p)−1 P(s), with P proper and P(p) ,= 0. Thenthe sensitivity function satisfies

qSq∞ = supω

∣∣∣∣1

1+ CP

∣∣∣∣ = supω

∣∣∣∣∣1

1+ CP(iω − z)(iω − p)−1

∣∣∣∣∣

= supω

∣∣∣∣∣iω − p

iω − p+ CP(iω − z)

∣∣∣∣∣ = supω

∣∣∣∣∣iω + p

iω − p+ CP(iω − z)

∣∣∣∣∣

= supRe s≥0

∣∣∣∣∣s+ p

s− p+ CP(s− z)

∣∣∣∣∣ ≥∣∣∣∣z+ p

z− p

∣∣∣∣

The fourth inequality uses that piω − pp =√

ω 2 + p2 = piω + pp and the fifthinequality is Corollary 10.1. 2

A similar argument can be applied to a system involving a time delay butapplication of the maximum modulus theorem is less straightforward in this case.

10.4 Bode’s integral formula

Another striking performance limitation, known as Bode’s integral formula, showsthat the effort to make the sensitivity function small is always a trade-off betweendifferent frequency regions:

33


0 1 2 3 4 510

−0.3

10−0.2

10−0.1

100

100.1

frequency

magnitude

Figure 10.3 The amplitude curve of the sensitivity function always enclose the same areabelow the level pSp = 1 as above.

THEOREM 10.4If P(s), C(s) and S(s) = [1 + C(s)P(s)]−1 are stable and s2C(s)P(s) is bounded,then

∫ ∞

0log pS(iω )p dω = 0

Proof. Proof sketch. From the theory of analytic functions, recall that Cauchy’sformula states that

∫

γ

f (z)dz = 0

for every closed path γ in the region where the function f is analytic. Bode’sintegral formula follows by application of Cauchy’s formula to

f (z) = log S(z)

The stability of C and P guarantee that f is well-defined and analytic in thewhole right half plane. Integration along the imaginary axis can be extended tointegration along a closed path by adding a large half-circle in the right halfplane. The condition that s2C(s)P(s) is bounded is needed to make sure that thecontribution from the half-circle vanishes as the radius tends to infinity. 2

The invariance of Bode’s integral is sometimes referred to as the “water-bed”effect: If the designer tries to push the magnitude of the sensitivity function downat some point, it will inevitably pop up somewhere else!The assumptions behind Bode’s integral formula deserve some discussion. The

expression s2C(s)P(s) is always bounded whenever C(s) and P(s) correspond toreal sensor/actuator interconnections, since direct terms are not physically imple-mentable. With unstable poles in C(s)P(s) the integral formula changes into

∫ ∞

0log pS(iω )p dω = π

∑

i

Re pi

which makes it even harder to push down the sensitivty magnitude! The fasterunstable modes, the harder it is. In fact, this can be used as an argument whyunstable controllers should in general be avoided.

34

Lecture 11

Multivariable control1

There is a clear trend in modern engineering toward systems of higher and highercomplexity. One reason is that demands for efficiency give tighter interconnectionsbetween subsystems. For example, to get minimally pollutive emissions in theexhaust gas of a car, it is necessary that engine, carburettor, catalyst, gearbox, etc.all cooperate in an optimal manner. Similarly, the demand for efficient productionand distribution of electrical power has led to tighter coupling between productionunits in different geographical regions and more complex large scale dynamics.A system with several inputs and several outputs is sometimes called a MIMO

(Multiple-Input-Multiple-Output) system. Control theory for such systems is ahighly active research area and a review of the available methods is outside thescope of this course. However, many of the ideas that were developed for scalarsystems can be easily adapted also to a multivariable setting. This lecture willpresent a few such items:

• Multivariable performance specifications

• Limitations due to unstable multivariable zeros

• Decentralized control by pairing of signals

Figure 11.1 A modern car, a power plant and an oil refinery all make extensive use ofmultivariable control systems

11.1 Multivariable specifications

Consider again the feedback loop illustrated in Figure 11.2 but assume that allsignals are vector valued. Then P(s), F(s) and C(s) are matrices, so the closedloop transfer functions need to be derived with some more care:

X (s) = PCF ⋅ R(s) + P ⋅ D(s) − PC ⋅ [N + X ](s)

[I + PC]X (s) = PCF ⋅ R(s) + P ⋅ D(s) − PC ⋅ N(s)

X (s) = [I + PC]−1PCF ⋅ R(s) + [I + PC]−1P ⋅ D(s) − [I + PC]−1PC ⋅ N(s)

1Written by A. Rantzer. The examples come from the book by Glad and Ljung, Control Theory —Multivariable and Nonlinear Methods, Taylor & Francis, 2000

35

Lecture 11. Multivariable control1

F(s) C(s) P(s)

−I

Σ Σ Σr e u

d

x y

n

v

Controller Process

Figure 11.2

where a dot denotes matrix-vector multiplication.Similarly

V (s) = [I + CP]−1CF ⋅ R(s) − [I + CP]−1C ⋅ N(s) + [I + CP]−1 ⋅ D(s)

Notice that S = [I + PC]−1 is generally not the same as [I + CP]−1. The first iscalled the sensitivity function and has a matrix size determined by the numberof outputs. The second is called input sensitivity function and its size correspondsto the number of inputs. The convention is to call T = [I + PC]−1PC the comple-mentary sensitivity function. Notice the following identities:

[I + PC]−1P = P[I + CP]−1

C[I + PC]−1 = [I + CP]−1C

T = P[I + CP]−1C = PC[I + PC]−1

S + T = I

The first equality follows by multiplication with I+CP from the right and I+PCfrom the left. The second one is analogous. Using the first two equalities, weimmediately get the third. The last one is straight from definitions as well.Also for multivariable systems, it is common to require S to be small at low

frequencies and T to be small at high frequencies. The first specification meansthat y follows Fr well at small frequencies, while the second means that highfrequency measurement noise n does not influence x significantly. Another wayto state these requirements is to say that the loop transfer matrix

P(iω )C(iω )

should have small norm qP(iω )C(iω )q at high frequencies, while at low the fre-quencies instead q[P(iω )C(iω )]−1q should be small. See Figure 11.3.

11.2 Limitations due to unstable zeros

Just like for scalar systems, there are fundamental limitations on the achievableperformance in a multivariable system. For a multivariable system with squaretranfer matrix P(s), i.e. the same number of inputs and outputs, the zeros can bedefined as the poles of P(s)−1. The following theorem captures the influence of anunstable zero:

THEOREM 11.1Let WS(s) be stable and let S(s) = [I + P(s)C(s)]−1 be the sensitivity function ofa stable closed loop system. Then, the specification

qWSSq∞ ≤ 1

is impossible to satisfy unless qWS(z)q ≤ 1 for every unstable zero z of P(s).

36

11.2 Limitations due to unstable zeros

100

101

−70

−60

−50

−40

−30

−20

−10

0

10

20

w0 w1

P(iω )C(iω )

Disturbance rejection

RobustnessMagnitude

Frequency

Figure 11.3 Specifications on the singular values of the loop transfer function often havethis form. A lower bound on the singular values for low frequencies is needed for disturbancerejection. This means that q[P(iω )C(iω )]−1q should be small. An upper bound on the singularvalues for high frequencies, making qP(iω )C(iω )q small enough, is needed for robustness tomodel errors and measurement noise.

The proof is analogous to the result for scalar systems in Lecture 4. Instead ofgiving the details, we turn our attention to an example.

Example 1 (Non-minimum phase MIMO System) Consider a feedback systemy= (I + PC)−1r with the multivariable process

P(s) =

2s+ 1

3s+ 2

1s+ 1

1s+ 1

Computing the determinant by hand

det P(s) =2

(s+ 1)2−

3(s+ 2)(s+ 1)

=−s+ 1

(s+ 1)2(s+ 2)

shows that the process has an unstable zero at s = 1, which will limit the achiev-able performance. For further understanding of the limitation, consider the fol-lowing three different control structures:

Controller 1 The controller

C1(s) =

K1(s+ 1)s

−3K2(s+ 0.5)s(s+ 2)

−K1(s+ 1)s

2K2(s+ 0.5)s(s+ 1)

gives the diagonal loop transfer matrix

P(s)C1(s) =

K1(−s+ 1)s(s+ 2)

0

0K2(s+ 0.5)(−s+ 1)s(s+ 1)(s+ 2)

Hence the system is decoupled into to scalar loops, each with an unstable zeroat s = 1 that limits the bandwidth. The closed loop step responses are shown inFigure 11.4.

37


Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1.5

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1.5

Figure 11.4 Closed loop step responses with decoupling controller C1(s) for the two outputsy1 (solid) and y2 (dashed). The upper plot is for a reference step for y1. The lower plot is fora reference step for y2.


C2(s) =

K1(s+ 1)s

K2

−K1(s+ 1)s

K2

gives the upper triangular loop transfer matrix

P(s)C2(s) =

K1(−s+ 1)s(s+ 2)

K2(5s+ 7)(s+ 2)(s+ 1)

02K2s+ 1

Now the decoupling is only partial: Output y2 is not affected by r1. Moreover,there is no unstable zero that limits the rate of response in y2! The closed loopstep responses for K1 = 1, K2 = 10 are shown in Figure 11.5.


C3(s) =

K1−K2(s+ 0.5)s(s+ 2)

K12K2(s+ 0.5)s(s+ 1)

gives the lower triangular loop transfer matrix

P(s)C3(s) =

K1(5s+ 7)(s+ 1)(s+ 2)

0

2K1s+ 1

K2(−1+ s)(s+ 0.5)s(s+ 1)2(s+ 2)

In this case y1 is decoupled from r2 and can respond arbitrarily fast for high valuesof K1, at the expense of bad behavior in y2. Step responses for K1 = 10, K2 = −1are shown in Figure 11.6.To summarize, the example shows that even though a multivariable unstable

zero always gives a performance limitation, it is possible to influence where theeffects should show up. 2

38

11.3 Pairing of signals

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1.5

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1.5

2

2.5

3

Figure 11.5 Closed loop step responses with controller C2(s) for the two outputs y1 (solid)and y2 (dashed). The right half plane zero does not prevent a fast y2-response to r2 but at theprice of a simultaneous undesired response in y1.

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

1

1.5

Figure 11.6 Closed loop step responses with controller C3(s) for the two outputs y1 (solid)and y2 (dashed). The right half plane zero does not prevent a fast y1-response to r1 but at theprice of a simultaneous undesired response in y2.


The simplest way to deal with a multivariable control problem is to select an equalnumber of inputs and outputs and make a pairing, so that each input is responsiblefor control of one particular output. The corresponding transfer function from inputto output of the process is determined for each pair, and the scalar feedback loopsare designed ignoring the coupling between the loops inside the plant.When connecting all the feedback loops simultaneously, the cross-coupling may

potentially lead to performance degradation or even instability. The approachtherefore works much better if there is a good way to select input and outputvariables for the scalar loops. One way to do this is to use the Relative Gain Array

39


(RGA) that will be studied next.For an full rank matrix A ∈ Cm$n, define the Relative Gain Array as

RGA(A) := A. ∗ (A−1)T

where “.*” denotes element-by-element multiplication. For non-square matrices,the inverse A−1 is replaced by the pseudo-inverse A† (in Matlab pinv(A)). Ifm > n,then A† = (A∗A)−1A∗ and if m < n then A† = A∗(AA∗)−1. The Relative Gain Arrayhas several interesting properties:

• The sum of all elements in a column or row is one.

• Permutations of rows or columns in A give the same permutations in RGA(A)

• RGA(A) = RGA(D1AD2) if D1 and D2 are diagonal, i.e. RGA(A) is indepen-dent of scaling

• If A is triangular, then RGA(A) is the unit matrix I.

Furthermore, the Relative Gain Array has an interpretation related to controltheory: Let P(s) be the transfer matrix from u to y. Then

• The (k, j) element of P is the transfer function u j → yk when ui = 0 for i ,= j(open loop control)

• The ( j, k) element of P−1 is the inverse of the transfer function u j → yk whenthe other inputs are such that yi = 0 for i ,= k (closed loop control)

Hence, if the (k, j) element of RGA(P(s)) is equal to one, then the value of transferfunction from u j to yk does not depend on whether the remaining inputs operatein open loop or closed loop. This indicates that the cross-coupling with other loopsis weak.The following rules of thumb use RGA to identify input-output pairings that

have small cross-coupling. However, the outcome must always be evaluated usingother tools.

1. Find a permutation of inputs and outputs that brings RGA(P(iω c)) as closeas possible to the identity matrix. Here ω c is the closed loop bandwidth ofthe system.

2. Avoid pairings that give negative diagonal elements of RGA(P(0))

The second rule is motivated by the desire to have the same sign of the staticgain in one loop regardless if the other loops are closed or not. We illustrate therules with an example.

Example: A Distillation Column

A distillation column is used in chemical process industry to separate differentcomponents in a chemical product. This example comes from Shell and describes acolumn where raw oil is separated into various petro-chemical derivatives. Heatedraw oil is inserted at the bottom of the column and evaporates. Subcomponentsthen condense and get extracted at different levels of the column.

Outputs: Inputs:

y1 = top draw composition u1 = top draw flowrate

y2 = side draw composition u2 = side draw flowrate

u3 = bottom temperature control input

[Y1(s)

Y2(s)

]=

450s+ 1

e−27s1.8

60s+ 1e−28s

5.950s+ 1

e−27s

5.450s+ 1

e−18s5.7

60s+ 1e−14s

6.940s+ 1

e−15s

︸︷︷︸P(s)

U1(s)

U2(s)

U3(s)

40


Figure 11.7 Schematic picture of a distillation column

Computing the RGA for the distillation column model gives

P(0) =[4.0 1.8 5.9

5.4 5.7 6.9

]

P(i/50) =[0.6871− 2.7437i 0.1548− 1.1419i 1.0135− 4.0469i

1.5758− 3.4781i 1.4704− 3.3397i 3.0247− 4.4589i

]

RGA(P(0)) =[0.2827 −0.6111 1.3285

0.0134 1.5827 −0.5962

]

RGA(P(i/50)) =[0.4355− 0.3667i −0.6536− 0.0171i 1.2181+ 0.3839i

−0.0906+ 0.3667i 1.5933+ 0.0171i −0.5027− 0.3839i

]

To choose control signal for y1, we apply the rules of thumb to the top row. Thissuggests the bottom temperature u3 for control of the top draw composition y1,since the third column has values slightly closer to 1 than the first column.Based on the bottom row, we choose the side draw flowrate u2 to control the

side draw composition y2. The top draw flow rate u1 is left unused. The matrixtransfer function P(s) from (u3,u2) to (y1, y2) is now to be controlled by a diagonalcontroller C(s), say a PI controller:

P(s) =

5.9e−27s

50s+ 11.8e−28s

60s+ 16.9e−15s

40s+ 15.7e−14s

60s+ 1

C(s) =

60s+ 150s

0

060s+ 150s

Without feedforward, the closed loop transfer matrix from reference to outputbecomes

PC(I + PC)−1

41


Bode Magnitude Diagram

Frequency (rad/sec)

Mag

nitu

de (

dB)

−80

−60

−40

−20

0

20From: U(1)

To:

Y(1

)

10−2

100

102

−60

−40

−20

0

20

To:

Y(2

)

From: U(2)

10−2

100

102

Figure 11.8 Magnitude plots for the closed loop transfer function of the distillation column

and the Bode magnitude plots for the four transfer functions are given in Fig-ure 11.8. As seen in the plots, the resulting cross-coupling is generally small andthere is no static error.

42

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