MCT Book - Chapter 4

7/30/2019 MCT Book - Chapter 4

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This version: 22/10/2004

Chapter 4

Frequency response (the frequency domain)

The final method we will describe for representing linear systems is the so-called fre-quency domain. In this domain we measure how the system responds in the steady-stateto sinusoidal inputs. This is often a good way to obtain information about how your systemwill handle inputs of various types.

The frequency response that we study in this section contains a wealth of information,often in somewhat subtle ways. This material, that builds on the transfer function discussedin Chapter 3, is fundamental to what we do in this course.

Contents

4.1 The frequency response of SISO linear systems . . . . . . . . . . . . . . . . . . . . . . . . 119

4.2 The frequency response for systems in input/output form . . . . . . . . . . . . . . . . . . 122

4.3 Graphical representations of the frequency response . . . . . . . . . . . . . . . . . . . . . 124

4.3.1 The Bode plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

4.3.2 A quick and dirty plotting method for Bode plots . . . . . . . . . . . . . . . . . . 129

4.3.3 The polar frequency response plot . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

4.4 Properties of the frequency response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

4.4.1 Time-domain behaviour reflected in the frequency response . . . . . . . . . . . . . 135

4.4.2 Bodes Gain/Phase Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

4.5 Uncertainly in system models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

4.5.1 Structured and unstructured uncertainty . . . . . . . . . . . . . . . . . . . . . . . 146

4.5.2 Unstructured uncertainty models . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

4.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

4.1 The frequency response of SISO linear systems

We first look at the state-space representation:

x(t) = Ax(t) + bu(t)

y(t) = ctx(t) + Du(t).(4.1)

Let us first just come right out and define the frequency response, and then we can give itsinterpretation. For a SISO linear control system = (A,b, c,D) we let R be defined
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120 4 Frequency response (the frequency domain) 22/10/2004

by = { R | i is a pole ofT} .

The frequency response for is the function H : R\ C defined by H() = T(i).Note that we do wish to think of the frequency response as a C-valued function, and not arational function, because we will want to graph it. Thus when we write T(i), we intendto evaluate the transfer function at s = i. In order to do this, we suppose that all poles

and zeroes of T have been cancelled.The following result gives a key interpretation of the frequency response.

4.1 Theorem Let = (A,b, ct,01) be an complete SISO control system and let > 0. IfT has no poles on the imaginary axis that integrally divide ,

1 then, given u(t) = u0 sin tthere is a unique periodic output yp(t) with period T =

2 satisfying (4.1) and it is given by

y(t) = u0Re(H())sin t + u0Im(H())cos t.

Proof We first look at the state behaviour of the system. Since (A, c) is complete, thenumerator and denominator polynomials ofT are coprime. Thus the poles ofT are exactlythe eigenvalues ofA. If there are no such eigenvalues that integrally divide , this means that

there are no periodic solutions of period T for x(t) = Ax(t). Therefore the linear equationeATu = u has only the trivial solution u = 0. This means that the matrix eAT In isinvertible. We define

xp(t) = u0eAt(eAT In)1

T0

eAb sin d + u0

t0

eA(t)b sin d, (4.2)

and we claim that x(t) is a solution to the first of equations (4.1) and is periodic with period2 . If T =

2 we first note that

xp(t + T) = u0eA(t+T)(eAT In)1

T0

eAb sin d+

u0

t+T

0

eA(t+T)b sin d

= eAteAT(eAT In)1T0

eAb sin d+

u0eAteAT

T0

eAb sin d + u0

t+TT

eA(t+T)b sin d

= eAt(eAT(eAT In)1 + eAT)T0

eAb sin d+

u0t

0 e

A(t

)b

sin d.

Periodicity ofxp(t) will follow then if we can show that

eAT(eAT In)1 + eAT = (eAT In)1.But we compute

eAT(eAT In)1 + eAT = (eAT+ eAT(eAT In))(eAT In)1= (eAT In)1.

1Thus there are no poles for T of the form i where

Z.
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22/10/2004 4.1 The frequency response of SISO linear systems 121

Thus xp(t) has period T. That xp(t) is a solution to (4.1) with the u(t) = u0 sin t followssince xp(t) is of the form

x(t) = eAtx0 + u0

t0

eA(t)b sin d

provided we take

x0 = (eAT In)1 T

0

eAb sin d.

For uniqueness, suppose that x(t) is a periodic solution of period T. Since it is a solutionit must satisfy

x(t) = eAtx0 + u0

t0

eA(t)b sin d

for some x0 Rn. Ifx(t) has period T then we must have

eAtx0 + u0

t0

eA(t)b sin d = eAteATx0 + u0eAteAT

T0

eA(t)b sin d+

u0t+T

T

eA(t+T)b sin d

= eAteATx0 + u0eAteAT

T0

eA(t)b sin d+

u0

t0

eA(t)b sin d.

But this implies that we must have

x0 = eATx0 + u0e

AT

T

0

eAb sin d,

which means that x(t) = xp(t).This shows that there is a unique periodic solution in state space. This clearly implies

a unique output periodic output yp(t) of period T. It remains to show that yp(t) has theasserted form. We will start by giving a different representation ofxp(t) than that givenin (4.2). We look for constant vectors x1,x2 Rn with the property that

xp(t) = x1 sin t + x2 cos t.

Substitution into (4.1) with u(t) = u0 sin t gives

x1 cos t x2 sin t = Ax1 sin t + Ax2 cos t + u0b sin t= x1 Ax2, Ax1 x2 = u0b= i(x1 + ix2) A(x1 + ix2) = u0b.

Since i is not an eigenvalue for A we have

x1 + ix2 = (iIn A)1b= ctx1 = Re(H()), ctx2 = Im(H()).

The result follows since yp(t) = ctxp(t).


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4.2 Remarks 1. It turns out that any output from (4.1) with u(t) = u0 sin t can be writtenas a sum of the periodic output yp(t) with a function yh(t) where yh(t) can be obtainedwith zero input. This is, of course, reminiscent of the procedure in differential equationswhere you find a homogeneous and particular solution.

2. If the eigenvalues of A all lie in the negative half-plane, then it is easy to see thatlimt |yh(t)| = 0 and so after a long enough time, we will essentially be left with theperiodic solution yp(t). For this reason, one calls yp(t) the steady-state response andyh(t) the transient response. Note that the steady-state response is uniquely defined(under the hypotheses of Theorem 4.1), but that there is no unique transient responseitdepends upon the initial conditions for the state vector.

3. One can generalise this slightly to allow for imaginary eigenvalues i ofA for which integrally divide , provided that b does not lie in the eigenspace of these eigenvalues.

4.2 The frequency response for systems in input/output form

The matter of defining the frequency response for a SISO linear system in input/outputform is now obvious, I hope. Indeed, if (N, D) is a SISO linear system in input/output form,then we define its frequency response by HN,D() = TN,D(i).

Let us see how one may recover the transfer function from the frequency response. Notethat it is not obvious that one should be able to do this. After all, the frequency responsefunction only gives us data on the imaginary axis. However, because the transfer functionis analytic, if we know its value on the imaginary axis (as is the case when we know thefrequency response), we may assert its value off the imaginary axis. To be perfectly preciseon these matters requires some effort, but we can sketch how things go.

The first thing we do is indicate a direct correspondence between the frequency responseand the impulse response. For this we refer to Section E.2 for a definition of the Fouriertransform. With the notion of the Fourier transform in hand, we establish the correspondence

between the frequency response and the impulse response as follows.

4.3 Proposition Let (N, D) be a strictly proper SISO linear control system in input/outputform, and suppose that the poles of TN,D are in the negative half-plane. Then HN,D() =hN,D().

Proof We have HN,D() = TN,D(i) and so

HN,D() =

0+

hN,D(t)eit dt.

By Exercise EE.2, min(hN,D) < 0 since we are assuming all poles are in the negative half-plane. Therefore this integral exists. Furthermore, since hN,D(t) = 0 for t < 0 we have

HN,D() =

hN,D(t)eit dt = hN,D().

This completes the proof.

Now we recover the transfer function TN,D from the frequency response HN,D . In thefollowing result we are thinking of the transfer function not as a rational function, but as aC-valued function.
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22/10/2004 4.2 The frequency response for systems in input/output form 123

4.4 Proposition Let (N, D) be a strictly proper SISO linear control system in input/outputform, and suppose that the poles of TN,D are in the negative half-plane. Then, providedRe(s) > min(hN,D), we have

TN,D(s) =1

2

HN,D()

s i d.

Proof By Proposition 4.3 we know that hN,D is the inverse Fourier transform of HN,D :

hN,D(t) =1

2

HN,D()eit d.

On the other hand, by Theorem 3.22 the transfer function TN,D is the Laplace transform ofhN,D so we have, for Re(s) > min(hN,D).

TN,D(s) =

0+

hN,D(t)est dt

=1

2

0+

HN,D()eitest dt d

=1

2

HN,D()

s i d.

This completes the proof.

4.5 Remarks 1. I hope you can see the importance of the results in this section. What wehave done is establish the perfect correspondence between the three domains in whichwe work: (1) the time-domain, (2) the s-plane, and (3) the frequency domain. In eachdomain, one object captures the essence of the system behaviour: (1) the impulse re-sponse, (2) the transfer function, and (3) the frequency response. The relationships aresummarised in Figure 4.1. Note that anything you say about one of the three objects in

impulse response(time domain)

L

//

F

%%JJJ

JJJJ

JJJJ

JJJJ

JJJJ

J

transfer function(Laplace domain)

L1

oo

restrict to i

yyrrrr

rrrr

rrrr

rrrr

rrrr

r

frequency response(frequency domain)

F1

eeJJJJJJJJJJJJJJJJJJJJanalytic continuation

99rrrrrrrrrrrrrrrrrrrrr

Figure 4.1 The connection between impulse response, transferfunction, and frequency response

Figure 4.1 must be somehow reflected in the others. We will see that this is true, andwill form the centre of discussion of much of the rest of the course.

2. Of course, the results in this section may be made to apply to SISO linear systems in theform (4.1) provided that D = 01 and that the polynomials c

tadj(sIn A)b and PA(s)are coprime.
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4.3 Graphical representations of the frequency response

One of the reasons why frequency response is a powerful tool is that it is possible tosuccinctly understand its primary features by means of plotting functions or parameterisedcurves. In this section we see how this is done. Some of this may seem a bit pointless atpresent. However, as matters develop, and we get closer to design methodologies, the powerof these graphical representations will become clear. The first obvious application of these

ideas that we will encounter is the Nyquist criterion for stability in Chapter 7.

4.3.1 The Bode plot What one normally does with the frequency response is plot it.But one plots it in a very particular manner. First write H in polar form:

H() = |H()| eiH()

where |H()| is the absolute value of the complex number H() and H() is theargument of the complex number H(). We take 180 < H() 180. One thenconstructs two plots, one of 20 log |H()| as a function of log , and the other ofH()as a function of log . (All logarithms we talk about here are base 10.) Together these

two plots comprise the Bode plot for the frequency response H. The units of the plot of20log |H()| are decibels.2 One might think we are losing information here by plottingthe magnitude and phase for positive values of (which we are restricted to doing by usinglog as the independent variable). However, as we shall see in Proposition 4.13, we donot lose any information since the magnitude is symmetric about = 0, and the phase isanti-symmetric about = 0.

Lets look at the Bode plots for our mass-spring-damper system. I used Mathematica

to generate all Bode plots in this book. We will also be touching on a method for roughlydetermining Bode plots by hand.

4.6 Examples In all cases we have

A =

0 1

km

dm

, b =

01m

.

We take m = 1 and consider the various cases of d and k as employed in Example 2.33. Herewe can also consider the case when D = 01.1. We take d = 3 and k = 2.

(a) With c = (1, 0) and D = 01 we compute

H() =1

2

+ 3i + 2

.

The corresponding Bode plot is the first plot in Figure 4.2.

(b) Next, with c = (0, 1) and D = 01 we compute

H() =i

2 + 3i + 2 ,

and the corresponding Bode plot is the second plot in Figure 4.2.

2Decibels are so named after Alexander Graham Bell. The unit of bell was initially proposed, but whenit was found too coarse a unit, the decibel was proposed.
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22/10/2004 4.3 Graphical representations of the frequency response 125

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Figure 4.2 The displacement (top left), velocity (top right), andacceleration (bottom) frequency response for the mass-springdamper system when d = 3 and k = 2

(c) If we have c = ( km

, dm

) and D = [1] we compute

H() = 2

2 + 3i + 2 ,The Bode plot for this frequency response function is the third plot in Figure 4.2.

2. We take d = 2 and k = 1.


H() =1

2 + 2i + 1 .



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H() =

i

2 + 2i + 1 ,and the corresponding Bode plot is the second plot in Figure 4.3.

(c) If we have c = ( km , dm) and D = [1] we compute

H() =2

2 + 2i + 1 ,

The Bode plot for this frequency response function is the third plot in Figure 4.3.

3. We take d = 2 and k = 10.
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H() =1

2 + 2i + 10 .


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H() =i

2 + 2i + 10 ,



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, dm


H() =2

2 + 2i + 10 ,

The Bode plot for this frequency response function is the third plot in Figure 4.4.

4. We take d = 0 and k = 1.


H() =1

2 + 1 .


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H() =i

2 + 1 ,



, dm


H() = 2

2 + 1 ,The Bode plot for this frequency response function is the third plot in Figure 4.5.

4.3.2 A quick and dirty plotting method for Bode plots It is possible, with varyinglevels of difficulty, to plot Bode plots by hand. The first thing we do is rearrange thefrequency response in a particular way suitable to our purposes. The form desired is

H() =

Kk1j1=1

(1 + ij1)k2j2=1

1 + 2ij2

j2

( j2

)2

(i)k3k4j4=1

(1 + ij4)k5j5=1

1 + 2ij5

j5

( j5 )2 (4.3)

where the s, s, and s are real, the s are all further between 1 and 1, and the s areall positive. The frequency response for any stable, minimum phase system can always beput in this form. For nonminimum phase systems, or for unstable systems, the variationsto what we describe here are straightforward. The form given reflects our transfer functionhaving

1. k1 real zeros at the points 1j1 , j1 = 1, . . . , k1,

2. k2

pairs of complex zeros at j2(

j2 1

2

j2), j

2= 1, . . . , k

2,

3. k3 poles at the origin,

4. k4 real poles at the points 1j4 , j4 = 1, . . . , k4, and

5. k5 pairs of complex poles at j5(j5

1 2j5), j5 = 1, . . . , k5.Although we exclude the possibility of having zeros or poles on the imaginary axis, one cansee how to handle such functions by allowing to become zero in one of the order two terms.

Let us see how to perform this in practice.

4.7 Example We consider the transfer function

T(s) =s + 1

10

s(s2 + 4s + 8).

To put this in the desired form we write

s + 110 =110

1 + 10s

s2 + 4s + 8 = 8

1 + 12s +

18s

2

= 8

1 + 2 14

8 s

8+ ( s

8)2)

= 8

1 + 2 12s8

+ ( s8

)2

.


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Thus we have a real zero at 110

, a pole at 0, and a pair of complex poles at 2 2i. Thuswe write

T(s) =1

80

1 + 10s

s(1 + 2 12s8

+ ( s8

)2)

and so

H() =1

80

1 + i10

i(1 + 2i 12

8 ( 8

)2).

I find it easier to work with transfer functions first to avoid imaginary numbers as long aspossible. You may do as you please, of course.

One can easily imagine that one of the big weaknesses of our computer-absentee plots is thatwe have to find roots by hand. . .

Let us see what a Bode plot looks like for each of the basic elements. The idea is to seewhat the magnitude and phase looks like for small and large , and to fill in the gaps inbetween these asymptotes.

1. H() = K: The Bode plot here is simple. It takes the magnitude 20 log K for all valuesof log . The phase is 0 for all log if K is positive, and 180 otherwise.

2. H() = 1 + i: For near zero the frequency response looks like 1 and so has thevalue of 0dB. For large the frequency response looks like i, and so the log of themagnitude will look like log . Thus the magnitude plot for large frequencies we have|H()| 20log dB. These asymptotes meet when 20 log = 0 or when = 1. Thispoint is called the break frequency. Note that the slope of the frequency response forlarge is independent of (since log = log + log ), but its break frequency doesdepend on . The phase plot starts at 0 for small and for large , since the frequencyresponse is predominantly imaginary, becomes 90. This Bode plot is shown in Figure 4.6for = 1.

3. H() = 1 + 2i0 ( 0 )2: For small the magnitude is 1 or 0dB. For large thefrequency response looks like ( 0 )2 and so the magnitude looks like 40 log 0 . The twoasymptotes meet when 40 log

0= 0 or when = 0. One has to be a bit more careful

with what is happening around the frequency 0. The behaviour here depends on thevalue of , and various plots are shown in Figure 4.6 for 0 = 1. As decreases, theundershoot increases. The phase starts out at 0 and goes to 180 as increases.

4. H() = (i)1: The magnitude is 1

over the entire frequency range which gives |H()| =20log dB. The phase is 90 over the entire frequency range, and the simple Bodeplot is shown in Figure 4.7.

5. H() = (1 + i)1: The analysis here is just like that for a real zero except for signs.The Bode plot is shown in for = 1.

6. H() = (1 + 2i0 ( 0 )2)1: The situation here is much like that for a complex zerowith sign reversal. The Bode plots are shown in Figure 4.8 for 0 = 1.

4.8 Remark We note that one often sees the language 20dB/decade. With what we havedone above for the typical elements in a frequency response function. A first-order element inthe numerator increases like 20 log for large frequencies. Thus as increases by a factorof 10, the magnitude will increase by 20dB. This is where 20dB/decade comes from. Ifthe first-order element is in the denominator, then the magnitude decreases at 20dB/decade.Second-order elements in the numerator increase at 40dB/decade, and second-order elementsin the denominator decrease at 40dB/decade. In this way, one can ascertain the relative


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Figure 4.6 Bode plot for H() = 1 + i (left) and for H() =1 + 2i 2 for = 0.2, 0.4, 0.6, 0.8 (right)

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Figure 4.7 Bode plot for H() = (i)1 (left) and for H() =(1 + i)1 (right)


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0

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150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-30

-20

-10

0

10

20

30

40

log

log

dB

deg

Figure 4.8 Bode plot for H() = (1 + 2i 2)1 for =0.2, 0.4, 0.6, 0.8

degree of the numerator and denominator of a frequency response function by looking at itsslope for large frequencies. Indeed, we have the following rule:

The slope of the magnitude Bode plot for HN,D at large frequencies is20(deg(N)deg(D))dB/decade.

To get a rough idea of how to sketch a Bode plot, the above arguments illustrate that theasymptotes are the most essential feature. Thus we illustrate these asymptotes in Figure 4.19(see the end of the chapter) for the essential Bode plots in the above list. From these onecan determine the character of most any Bode plot. The reason for this is that in (4.3) wehave ensured that any frequency response is a product of the factors we have individuallyexamined. Thus when we take logarithms as we do when generating a Bode plot, the graphssimply add! And the same goes for phase plots. So by plotting each term individually bythe above rules, we end up with a pretty good rough approximation by adding the Bodeplots.

Let us illustrate how this is done in an example.

4.9 Example (Example 4.7 contd) We take the frequency response

H() =1

80

1 + i10

i

1 + 2i 128 (

8)2 .

Four essential elements will comprise the frequency response:

1. H1() =180

;

2. H2() = 1 + i10;
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3. H3() = (i)1;

4. H4() =

1 + 2i 128 (

8)21

.

Lets look first at the magnitudes.

1. H1 will contribute 20 log180

38.1dB. The asymptotes for H1 are shown in Fig-ure 4.9.

-3 -2 -1 0 1 2 3 4

-150

-100

-50

0

50

100

150

-3 -2 -1 0 1 2 3 4

-100

-50

0

50

log

log

dB

deg

-3 -2 -1 0 1 2 3 4

-150

-100

-50

0

50

100

150

-3 -2 -1 0 1 2 3 4

-100

-50

0

50

log

log

dB

deg

Figure 4.9 Asymptotes for Example 4.7: H1 (left) and H2 (right)

2. H2 has a break frequency of =110 or log = 1. The asymptotes for H2 are shown

in Figure 4.9.

3. H3 gives 20log across the board. The asymptotes for H3 are shown in Figure 4.10.4. H4 has a break frequency of =

8 or log 0.45. The asymptotes for H4 are shown

in Figure 4.10. Note that here we have = 12

, which is a largish value. Thus we donot need to adjust the magnitude peak too much around the break frequency when weuse the asymptotes to approximate the actual Bode plot.

Now the phase angles.

1. H1 has phase exactly 0 for all frequencies.

2. For H2, the phase is approximately 0 for log 10 < 1 or log < 2. For log 10 > 1

(or log > 0) the phase is approximately 90. Between the frequencies log = 2and log = 0 we interpolate linearly between the two asymptotic phase angles.

3. The phase for H3 is 90 for all frequencies.4. For H4, the phase is 0

for log 8

< 1 or log < log 8 1 0.55. For log 8

> 1,

(log > log

8 + 1 1.45), the phase is approximately 180.
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-150

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0

50

100

150

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-100

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0

50

log

log

dB

deg

-3 -2 -1 0 1 2 3 4

-150

-100

-50

0

50

100

150

-3 -2 -1 0 1 2 3 4

-100

-50

0

50

log

log

dB

deg

Figure 4.10 Asymptotes for Example 4.7: H3 (left) and H4 (right)

-3 -2 -1 0 1 2 3 4

-150

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0

50

100

150

-3 -2 -1 0 1 2 3 4

-100

-50

0

50

log

log

d

B

deg

-3 -2 -1 0 1 2 3 4

-150

-100

-50

0

50

100

150

-3 -2 -1 0 1 2 3 4

-100

-50

0

50

log

log

d

B

deg

Figure 4.11 The sum of the asymptotes (left) and the actual Bodeplot (right) for Example 4.7


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22/10/2004 4.4 Properties of the frequency response 135

The sum of the asymptotes are plotted in Figure 4.11 along with the actual Bode plot so youcan see how the Bode plot is essentially the sum of the individual Bode plots. You shouldtake care that you always account for , however. In our example, the value of in H4 isquite large, so not much of an adjustment had to be made. If were small, we have to adda little bit of a peak in the magnitude around the break frequency log = log

8, and also

make the change in the phase a bit steeper.

4.3.3 The polar frequency response plot We will encounter in Chapter 12 anotherrepresentation of the frequency response H. The idea here is that rather than plottingmagnitude and phase as one does in a Bode plot, one plots the real and imaginary part ofthe frequency response as a curve in the complex plane parameterised by (0, ). Doingthis yields the polar plot for the frequency response. One could do this, for example, bytaking the Bode plot, and for each point on the independent variable axis, put a point ata distance |H()| from the origin in the direction H(). Indeed, given the Bode plot, onecan typically make a pretty good approximation of the polar plot by noting (1) the maximaand minima of the magnitude response, and the phase at these maxima and minima, and(2) the magnitude when the phase is 0,

90, or

180.

In Figure 4.12 are shown the polar plots for the basic frequency response functions.Recall that in (4.3) we indicated that any frequency response will be a product of thesebasic elements, and so one can determine the polar plot for a frequency response formed bythe product of such elements by performing complex multiplication that, you will recall, isdone in polar coordinates merely by multiplying radii, and adding angles.

For a lark, lets look at the minimum/nonminimum phase example in polar form.

4.10 Example (Example 4.12) Recall that we contrasted the two transfer functions

H1() =1 + i

2 + i + 1

, H2() =1 i

2 + i + 1

.

We contrast the polar plots for these frequency responses in Figure 4.13. Note that, asexpected, the minimum phase system undergoes a smaller phase change if we follow it alongits parameterised polar curve. We shall see the potential dangers of this in Chapter 12.

Note that when making a polar plot, the thing one looses is frequency information. Thatis, one can no longer read from the plot the frequency at which, say, the magnitude of thefrequency response is maximum. For this reason, it is not uncommon to place at intervalsalong a polar plot the frequencies corresponding to various points.

4.4 Properties of the frequency response

It turns out that in the frequency response can be seen some of the behaviour we haveencountered in the time-domain and in the transfer function. We set out in this section toscratch the surface behind interpreting the frequency response. However, this is almost anart as much as a science, so plain experience counts for a lot here.

4.4.1 Time-domain behaviour reflected in the frequency response We have seenin Section 3.2 we saw that some of the time-domain properties discussed in Section 2.3were reflected in the transfer function. We anticipate being able to see these same features
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-2 -1 0 1 2 3

-2

-1

0

1

2

3

Re

Im

| |

-2 -1 0 1 2 3

-2

-1

0

1

2

3

Re

Im

| |

-2 -1 0 1 2 3

-2

-1

0

1

2

3

Re

Im

| |

-2 -1 0 1 2 3

-2

-1

0

1

2

3

Re

Im

| |

Figure 4.12 Polar plots for H() = 1 + i (top left), H() =1 + 2i 2, = 0.2, 0.4, 0.6, 0.8 (top right), H() =(1 + i)1 (bottom left), and H() = (1 + 2i 2)1,= 0.2, 0.4, 0.6, 0.8

reflected in the frequency response, and ergo in the Bode plot. In this section we explorethese expected relationships. We do this by looking at some examples.

4.11 Example The first example we look at is one where we have a pole/zero cancellation.As per Theorem 3.5 this indicates a lack of observability in the system. It is most beneficialto look at what happens when the pole and zero do not actually cancel, and compare it towhat happens when the pole and zero really do cancel. We take

A =

0 11

, b =

01

, c =

11

, (4.4)

and we compute

H() =1 + i

2 + i 1 .
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-1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

Re

Im

| |

-1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

Re

Im

| |

Figure 4.13 Polar plots for minimum phase (left) and nonmini-mum phase (right) systems

The Bode plots for three values of are shown in Figure 4.14. What are the essential featureshere? Well, by choosing the values of = 0 to deviate significantly from zero, we can seeaccentuated two essential points. Firstly, when = 0 the magnitude plot has two regionswhere the magnitude drops off at different slopes. This is a consequence of there being twodifferent exponents in the characteristic polynomial. When = 0 the plot tails off at oneslope, indicating that the system is first-order and has only one characteristic exponent. Thisis a consequence of the pole/zero cancellation that occurs when = 0. Note, however, thatwe cannot look at one Bode plot and ascertain whether or not the system is observable.

There is also an effect that can be observed in the Bode plot for a system that is notminimum phase. An example illustrates this well.

4.12 Example We consider two SISO linear systems, both with

A =

0 1

1 1

, b =

01

. (4.5)

The two output vectors we look at are

c1 =

11

, c2 =

1

1

. (4.6)

Let us then denote 1 = (A, b, c1,01) and 2 = (A,b, c2,01). The two frequency responsefunctions are

H1() =1 + i

2 + i + 1 , H2() =1 i

2 + i + 1 .The Bode plots are shown in Figure 4.15. What should one observe here? Note that themagnitude plots are the same, and this can be verified by looking at the expressions for H1and H2. The differences occur in the phase plots. Note that the phase angle varies onlyslightly for 1 across the frequency range, but it varies more radically for 2. It is from thisbehaviour that the term minimum phase is derived.


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150

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0

log

log

dB

deg

| |

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0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

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0

log

log

dB

deg

| |

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-150

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-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-40

-30

-20

-10

0

log

log

dB

deg

| |

Figure 4.14 Bode plots for (4.4) for = 5, = 0, and = 5

4.4.2 Bodes Gain/Phase Theorem In Bodes book [1945] one can find a few chapterson some properties of the frequency response. In this section we begin this development, andfrom it derive Bodes famous Gain/Phase Theorem. The material in this section relies onsome ideas from complex function theory

that we review in Appendix D. We start by examining some basic properties of frequencyresponse functions. Here we begin to see that the real and imaginary parts of a frequencyresponse function are not arbitrary functions.

4.13 Proposition Let (N, D) be a SISO linear system in input/output form with HN,D thefrequency response. The following statements hold:

(i) Re(HN,D()) = Re(HN,D());(ii) Im(HN,D()) = Im(HN,D());(iii) |HN,D()| = |HN,D()|;


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150

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-10

0

10

log

log

dB

deg

| |

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-150

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-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-30

-20

-10

0

10

log

log

dB

deg

| |

Figure 4.15 Bode plots for the two systems of (4.5) and (4.6)H1is on the left and H2 is on the right

(iv) HN,D() = HN,D() provided HN,D() (, ).Proof We prove (i) and (ii) together. It is certainly true that the real parts of D(i)and N(i) will involve terms that have even powers of and that the imaginary parts ofD(i) and N(i) will involve terms that have odd powers of . Therefore, if we denote byD1() and D2() the real and imaginary parts ofD(i) and N1() and N2() the real andimaginary parts of N(i), we have

N1(

) = N1(), D1(

) = D1(), N2(

) =

N2(), D2(

) =

D2(). (4.7)

We also have

HN,D() =N1() + iN2()

D1() + iD2()

=N1()D1() + N2()D2()

D21() + D22()

+ iN2()D1() N1()D2()

D21() + D22()

,

so that

Re(HN,D()) =N1()D1() + N2()D2()

D21() + D22()

,

Im(HN,D()) = N2()D1() N1()D2()

D21() + D22()

.

Using the relations (4.7) we see that

N1()D1() + N2()D2() = N1()D1() + N2()D2(),N2()D1() N1()D2() = N2()D1() N1()D2(),

and from this the assertions (i) and (ii) obviously follow.(iii) This is a consequence of (i) and (ii) and the definition of ||.(iv) This follows from (i) and (ii) and the properties of arctan.
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Now we turn our attention to the crux of the material in this sectiona look at how themagnitude and phase of the frequency response are related. To relate these quantities, it isnecessary to represent the frequency response in the proper manner. To this end, for a SISOlinear system (N, D) in input/output form, we define ZPN,D C to be the set of zeros andpoles of TN,D . We may then define SN,D : C \ ZPN,D C by

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

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

Re(SN,D(s)) = ln |TN,D(s)| , Im(SN,D(s)) = TN,D(s)for s C \ ZPN,D . Along similar lines we define ZPN,D R by

ZPN,D = { R | i ZPN,D} .Then we may define GN,D : R \ ZPN,D C by

GN,D() = SN,D(i) = ln(HN,D()). (4.8)

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

Re(GN,D()) = ln |HN,D()| , Im(GN,D()) = HN,D().Note that these are almost the quantities one plots in the Bode plot. The phase is preciselywhat is plotted in the Bode plot (against a logarithmic frequency scale), and Re( GN,D())is related to the magnitude plotted in the Bode plot by

Re(GN,D()) = ln 10 log |HN,D()| = ln1020

|HN,D()| dB.

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

Recall that (N, D) is stable if all roots of D lie in C and is minimum phase if all rootsof N lie in C. Here we will require the additional assumption that N have no roots on theimaginary axis. Let us say that a SISO linear system (N, D) for which all roots of N lie inC is strictly minimum phase.

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

Im(GN,D(0)) =20

0

Re(GN,D()) Re(GN,D(0))

2

2

0

d.

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

Let U0 C be the open subset defined byU0 = {s C | Re(s) > 0, Im(s) = 0} .

We now define a closed contour whose interior contains points in U0. We do this in parts.First, for R > 0 define a contour R in U0 by

R = {Rei | 2 2}.


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Now for r > 0 define two contours

r,1 = {i0 + rei | 2 2}, r,2 = {i0 + rei | 2 2}.

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

0,3 = {i | < 0 0 r}0,4 = {i | 0 + r 0 0 r}

0,5 = {i | + r 0 < } .

The closed contour we take is then

R,r = R

5j=1

r,j.

We show this contour in Figure 4.16. Now define a function F0 : U0 C by

Re

Im

i0

i0

r,1

r,2

r,3

r,5

r,4

R

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

F0(s) = 2i0(SN,D(s) G1(0))s2 + 20 .

Since (N, D) is stable and strictly minimum phase, SN,D is analytic on C+, and so F0 isanalytic on U0, and so we may apply Cauchys Integral Theorem to the integral of F0around the closed contour R,r.

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

limR

R

F0(s) ds = 0.
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Indeed we have

F0(Rei) =

2i0SN,D(Rei)

R2e2i + 20 2i0G1(0)R2e2i + 20

.

Since deg(N) < deg(D) the first term on the right will behave like R2 as R and thesecond term will also behave like R2 as R . Since ds = iRei on R, the integrand in

RF0(s) ds will behave like R

1 as R , and so our claim follows.Now let us examine what happens as we let r

0. To evaluate the contributions of

r,1and r,2 we write

F0(s) =SN,D(s) G1(0)

s i0 SN,D(s) G1(0)

s + i0.

On r,1 we have s = i0 + rei so that on r,1 we have

F0(i0 + rei) =

SN,D(i0 + rei) G1(0)

rei SN,D(i0 + re

i) G1(0)2i0 + rei

.

We parameterise r,1 with the curve c : [2 , 2 ] C defined by c(t) = i0 + reit so thatc(t) = ireit. Thus, as r 0, F0(s) ds behaves like

F0(s) ds (SN,D(i0) G1(0))ireit

reit= i(GN,D(0) G1(0)),

using the parameterisation specified by c. Integrating givesr,1

F0(s) ds = i(GN,D(0) G1(0)).

In similar fashion one obtainsr,2

F0(s) ds = i(GN,D(0) G1(0)).

Now we use Proposition 4.13 to assert thatGN,D(0) GN,D(0) = 2iG2(0).

Thereforer,1

F0(s) ds +

r,2

F0(s) ds = i(GN,D(0) G1(0) (GN,D(0) G1(0)))

= 2G2(0).Finally, we look at the integrals along the contours r,3, r,4, and r,5 as r 0 and for

fixed R > 0. These contour integrals in the limit will yield a single integral along the aportion of the imaginary axis:

i[R,R]F0(s) ds.

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

i[R,R]F0(s) ds =

RR

2i0(SN,D(it) G1(0))20 t2

i dt

=

RR

20(SN,D(it) G1(0))t2 20

dt.
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Using Proposition 4.13 we can write this as

i[R,R]

F0(s) ds =

R0

40(G1(t) G1(0))t2 20

dt.

Collecting this with our expression for the integrals along r,1 and r,2, as well as notingour claim that the integral along R vanishes as R

, we have shown that

limr0

limR

R,r

F0(s) ds =

0

40(G1() G1(0))2 20

d 2G2(0).

By Cauchys Integral Theorem, this integral should be zero, and our result now follows fromstraightforward manipulation.

The following result gives an important property of stable strictly minimum phase sys-tems.

4.15 Theorem (Bodes Gain/Phase Theorem) Let (N, D) be a proper, stable, strictly mini-

mum phase SISO linear system in input/output form, and let GN,D() be as defined in (4.8).For 0 > 0 define M0N,D : R R by M0N,D(u) = Re(GN,D(0eU)). Then we have

HN,D(0) =1

dM0N,Ddu

lncothu2

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

HN,D(0) =20

0

Re(GN,D()) Re(GN,D(0))2 20

d. (4.9)

We make the change of variable = 0eu, upon which the integral in (4.9) becomes

2

M0N,D(u) M0N,D(0)eu eu du =

1

M0N,D(u) M0N,D(0)sinh u

du.

We note that du

sinh u= ln coth u2 ,

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

1

0

M0N,D(u) M0N,D(0)

sinh u

du =

1

0

dM0N,Ddu

ln coth u2 1

(M0N,D(u) M0N,D(0)) ln coth u2

0

, (4.10)

and

1

0

M0N,D(u) M0N,D(0)sinh u

du =

1

0

dM0N,Ddu

lncoth u2

+1

(M0N,D(u) M0N,D(0)) ln coth u2

0

. (4.11)
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Let us look at the first term in each of these integrals. At the limit u = 0 we maycompute

coth u2 2

u+

u

6 u

3

360+

so that near u = 0, ln coth u2

ln u2

. Also, since M0N,D(u) is analytic at u = 0, M0N,D(u)

M0N,D(0) will behave linearly in u for u near zero. Therefore,

limu0

M0N,D(u) M0N,D(0)) ln coth u2 u ln u2 .

Recalling that limu0 u ln u = 0, we see that the lower limit in the above integrated ex-pressions is zero. At the other limits as u , ln coth u2 behaves like eu as u +and like eu as u . This, combined with the fact that M0N,D(u) behaves like ln u asu , implies the vanishing of the upper limits in the integrated expressions in both (4.10)and (4.11). Thus we have shown that these integrated terms both vanish. From this theresult follows easily.

It is not perhaps perfectly clear what is the import of this theorem, so let us examine it

for a moment. In this discussion, let us fix 0 > 0. Bodes Gain/Phase Theorem is tellingus that the phase angle for the frequency response can be determined from the slope of theBode plot with decibels plotted against the logarithm of frequency. The contribution tothe phase of the slope at some u = 0 ln is determined by the weighting factor ln coth

u2

which we plot in Figure 4.17. From the figure we see that the slopes near u = 0, or = 0,

-1.5 -1 -0.5 0 0.5 1 1.5 2

1

2

3

4

5

6

u = 0 ln

ln

coth

u

/2

Figure 4.17 The weighting factor in Bodes Gain/Phase Theorem

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

Lets see how this works in an example.

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


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22/10/2004 4.5 Uncertainly in system models 145

To employ the Gain/Phase Theorem we should convert this to a relation in y = ln |HN,D()|versus x = ln 0 . The coordinates are then readily seen to be related by

x =x

ln10+ log 0, y =

20

ln10y.

Therefore the relation y = 20kx + b becomes

y = kx + k ln 0 +b ln10

20.

In the terminology of Theorem 4.15 we thus have

M0N,D(u) = ku + k ln 0 +b ln10

20

so thatdM

0N,D

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

frequency 0 as

HN,D(0) =k

lncoth

u2 du.

The integral is one that can be looked up (Mathematica evaluated it for me) to be 2

2, so

that we have HN,D(0) =k2

.Lets see if this agrees with what we have seen before. We know, after all, a transfer

function whose magnitude Bode plot is 20k log . Indeed, one can check that choosingthe transfer function TN,D(s) = s

k gives 20 log |HN,D()| = 20k log , and so this transferfunction is of the type for which we are considering in the case when b = 0. For systems ofthis type we can readily determine the phase to be k2 , which agrees with Bodes Gain PhaseTheorem.

Despite the fact that it might be possible to explicitly apply Bodes Gain/Phase Theo-

rem in some examples to directly compute the phase characteristic for a certain magnitudecharacteristic, the value of the theorem lies not in the ability to do such computations, butrather in its capacity to give us approximate information for stable and strictly minimumphase systems. Indeed, it is sometimes useful to make the approximation

HN,D(0) =dM0N,D

du

u=0

2, (4.12)

and this approximation becomes better when one is in a region where the slope of themagnitude characteristic on the Bode plot is large at 0 compared to the slope at otherfrequencies. Uses of G

theorem

4.5 Uncertainly in system models

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


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

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

R = supR

{|R(i)|}.

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

4.5.1 Structured and unstructured uncertainty The reader may wish to recall ourgeneral control theoretic terminology from Section 1.1. In particular, recall that a plant isthat part of a control system that is given to the control designer. What is given to the controldesigner is a model, hopefully in something vaguely resembling one of the three forms we havethus far discussed: a state-space model, a transfer function, of a frequency response function.Of course, this model cannot be expected to be perfect. If one is uncertain about a plantmodel, one should make an attempt to come up with a mathematical description of what

this means. There are many possible candidates, and they can essentially be dichotomisedas structured uncertainty and unstructured uncertainty. The idea with structureduncertainty is that one has a specific type of plant model in mind, and parameters in thatplant model are regarded as uncertain. In this approach, one wishes to design a controllerthat has desired properties for all possible values of the uncertain parameters.

4.17 Example Suppose a mass is moving under a the influence of a control force u. Theprecise value of the mass could be unknown, say m [m1, m2]. In this case, a control designshould be thought of as being successful if it accomplishes stated objectives (the reader doesnot know what these might be at this point in the book!) for all possible values of the massparameter m.

Typically, structured uncertainty can be expected to be handled on a case by case basis,depending on the nature of the uncertainty. An approach to structured uncertainty is putforward by Doyle [1982]. For unstructured uncertainty, the situation is typically different asone considers a set of plant transfer functions P that are close to a nominal plant RP insome way. Again the objective is to design a controller that works for every plant in the setof allowed plants.

4.18 Example (Example 4.17 contd) Let us look at the mass problem above in a different

manner. Let us suppose that we choose a nominal plant RP(s) =1/ms2 and define a set of

candidate plantsP=

RP R(s) |

RP RP (4.13)for some > 0. In this case, we have clearly allowed a much larger class of uncertainty thatwas allowed in Example 4.17. Indeed, not only is the mass no longer uncertain, even theform of the transfer function is uncertain.

Thus we see that unstructured uncertainty generally forces us to consider a larger classof plants, and so is a more stringent and, therefore, conservative manner for modellinguncertainty. That is to say, by designing a controller that will work for all plants in P, weare designing a controller for plants that are almost certainly not valid models for the plant
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under consideration. Nevertheless, it turns out that this conservatism of design is made upfor by the admission of a consistent design methodology that goes along with unstructureduncertainty. We shall now turn our attention to describing how unstructured uncertaintymay arise in examples.

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

The first type we consider is called multiplicative uncertainty. We start with anominal plant RP R(s) and let Wu R(s) be a proper rational function with no poles inC+. Denote by P(RP, Wu) the set of rational functions RP R(s) with the properties

1. RP and RP have the same number of poles in C+,

2. RP and RP have the same imaginary axis poles, and

3.RP(i)

RP(i) 1 |Wu(i)| for all R.

Another way to write this set of plants is to note that RP P(RP, Wu) if and only ifRP = (1 + Wu)RP

where 1. Of course, is not arbitrary, even given 1. Indeed, this conditiononly ensures condition 3 above. If further satisfies the first two conditions, it is said to beallowable. In this representation, it is perhaps more clear where the term multiplicativeuncertainty comes up.

The following example is often used as one where multiplicative uncertainty is appropri-

ate.

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

RP(s) = eTsR(s)

for some R(s) R(s). We wish to ensure that this plant is modelled by multiplicativeuncertainty. To do so, we note that the first two terms in the Taylor series for eTs are1 T s, and thus we suppose that when T is small, a nominal plant of the form

RP(s) =R(s)

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

RP(i)R(i)

1 |Wu(i)| , R.

Let us suppose that T = 110

. In this case, the condition on Wu becomes

|Wu(i)|

e0.1i

1 s10

1, R.


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-0.5 0 0.5 1 1.5 2 2.5 3

-60

-40

-20

0

20

log

dB

Figure 4.18 e i101 s

10

1 (solid) and Wu = 1100(12

15)

(dashed)

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

ei

10

1 s10 1

has the rough behaviour of tailing off at 40dB/decade at low frequency and having constantmagnitude at high frequency. Thus a model of the form

Wu(s) =Ks2

s + 1

is a likely candidate, although other possibilities will work as well, as long as they capturethe essential features. Some fiddling with the Bode plot yields = 1

15and K = 1

100as

acceptable choices. Thus this choice of Wu will include the time delay plant in its set ofplants.

The next type of uncertainty we consider is called additive uncertainty. Again westart with a nominal plant RP and a proper Wu R(s) having no poles in C+. Denote byP(RP, Wu) the set of rational functions RP R(s) with the properties

1. RP and RP have the same number of poles in C+,

2. RP and RP have the same imaginary axis poles, and

3. |RP(i) RP(i)| |Wu(i)| for all R.A plant in P+(RP, Wu) will have the form

RP = RP+ Wu

where 1. As with multiplicative uncertainty, will be allowable if properties 1and 2 above are met.

4.20 Example This example has a generic flavour. Suppose that we make measurementsof our plant to get magnitude information about its frequency response at a finite set offrequencies {1, . . . , k}. If the test is repeated at each frequency a number of times, wemight try to find a nominal plant transfer function RP with the property that at the measured


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

|RP(ij) RP(ij)| |Wu(ij)| , j = 1, . . . , k .One could then hope that at the frequencies where data was not taken, the actual plant data

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

The above two classes of uncertainty models, P(RP, Wu) and P+(RP, Wu) are thetwo for which analysis will be carried out in this book. The main reason for this choiceis convenience of analysis. Fortunately, many interesting cases of plant uncertainty can bemodelled as multiplicative or additive uncertainty. However, for completeness, we shall givetwo other types of uncertainty representations.

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

RP =RP

1 + WuRP, 1. (4.14)

This uncertainty representation can arise in practice.

4.21 Example Suppose that we have a plant of the form

RP(s) =1

s2 + as + 1

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

RP(s) =1

s2 + aavgs + 1, Wu(s) =

12as,

where aavg =12(amax + amin) and a = amax amin, then the set of plant is exactly as

in (4.14), if is a number between 1 and 1. If is allowed to be a rational functionsatisfying 1 then we have embedded our actual set of plants inside the largeruncertainty set described by (4.14). One can view this example of one where the structureduncertainty is included as part of an unstructured uncertainty model.

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

RP =RP

1 + Wu, 1.

Let us see how this type of uncertainty can come up in an example.

finish

4.22 Example

4.23 Remarks 1. In our definitions ofP(RP, Wu) and P+(RP, Wu) we made some assump-tions about the poles of the nominal plant and the poles of the plants in the uncertaintyset. The reason for these assumptions is not evident at this time, and indeed they canbe relaxed with the admission of additional complexity of the results of Section 7.3, andthose results that depend on these results. In practice, however, these assumptions arenot inconvenient to account for.
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2. All of our choices for uncertainty modelling share a common defect. They allow plantsthat will almost definitely not be possible models for our actual plant. That is to say,our sets of plants are very large. This has something of a drawback in our employmentof these uncertainty models for controller designthey will lead to a too conservativedesign. However, this is mitigated by the existence of effective analysis tools to do robustcontroller design for such uncertainty models.

3. When deciding on a rational function Wu with which to model plant uncertainty withone of the above schemes, one typically will not want Wu to tend to zero as s . Thereason for this is that at higher frequencies is typically where model uncertainty is thegreatest. This becomes a factor when choosing starting point for Wu.

4.6 Summary

In this section we have introduced a nice piece of equipmentthe frequency responsefunctionand a pair of slick representations of itthe Bode plot and the polar plot. Hereare the pertinent things you should know from this chapter.

1. For a given SISO linear system , you should be able to compute

and H

.

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

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

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

5. You should be able to draw by hand the Bode plot for any system whose numerator

and denominator polynomials you can factor. Really the only subtle thing here is thedependence on for the second-order components of the frequency response. If you thinkof as damping, the interpretation here becomes straightforward since one expects largermagnitudes for lower damping when the second-order term is in the denominator.

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

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

8. The developments of Section 4.4.2 are somehow essential, and at the same time somewhat

hard. If one is to engage in controller design using frequency response, clearly the fact thatthere are essential restrictions on how the frequency response may behave is important.

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


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22/10/2004 4.6 Summary 151

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dB

deg

Figure 4.19 Asymptotes for magnitude and phase plots forH() = 1 + i (top left), H() = 1 + 2i 2 (top right),H() = (1+i)1 (bottom left), and H() = (1+2i2)1(bottom right)


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Exercises

E4.1 Consider the vector initial value problem

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

where there are no eigenvalues for A of the form i where integrally divides . Let

xp(t) be the unique periodic solution constructed in the proof of Theorem 4.1, andwrite the solution to the initial value problem as x(t) = xp(t) + xh(t). Determine anexpression for xh(t). Check that xh(t) is a solution to the homogeneous equation.

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

A =

0 0

0 0

, b =

01

, c =

10

for 0 R and 0 > 0.(a) Determine and compute H.

(b) Take u(t) = u0 sin t, and determine when the system satisfies the hypotheses ofTheorem 4.1, and determine the unique periodic output yp(t) guaranteed by thattheorem.

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

(d) Plot the Bode plot for H for various values of 0 0 and 0 > 0. Make sureyou get all cases where the Bode plot assumes a different character.

We have two essentially differing characterisations of the frequency response, one as theway in which sinusoidal outputs appear under sinusoidal inputs (cf. Theorem 4.1) and oneinvolving Laplace and Fourier transforms (cf. Proposition 4.3). In the next exercise you willexplore the differing domains of validity for the two interpretations.

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

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

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

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

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

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

(e) TN,D(s) =1

2

HN,D()s i d.

(f) HN,D() = TN,D(i).

When discussing the impulse response, it was declared that to obtain the output for anarbitrary input, one can use a convolution with the impulse response (plus a bit that dependsupon initial conditions). In the next exercise, you will come to an understanding of this interms of Fourier transforms.
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Exercises for Chapter 4 153

E4.5 Suppose that f and g are functions of , and that they are the Fourier transformsof functions f, g : (, ) R. You know from your course on transforms that theinverse Fourier transform of the product fg is the convolution

(f g)(t) =

f(t )g() d.

Now consider a SISO linear system = (A, b, ct,01) with causal impulse responseh+. Recall that the output for zero state initial condition corresponding to the inputu : [0, ) R is

y(t) =

t0

h+(t )u() d.

Make sense of the statement, The output y is equal to the convolution h+ u. Notethat you are essentially being asked to resolve the conflict in the limits of integrationbetween convolution in Fourier and Laplace transforms.

E4.6 We will consider in detail here the mass-spring-damper system whose Bode plots areproduced in Example 4.6. We begin by scaling the input u by the spring constant k

so that, upon division by m, the governing equations are

x + 20x + 20x =

20u,

where 0 =km and =

d2km

. As output we shall take y = x. We allow d to be

zero, but neither m nor k can be zero.

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

(b) Determine for the various values of the parameters 0 and , and then deter-mine the frequency response H.

(c) Show that

|H()| = 20

(20 2)2 + 42202.

(d) How does |H()| behave for 0? for 0?(e) Show that d

d|H()| = 0 if and only if dd |H()|2 = 0.

(f) Using the previous simplification, show that for < 12

there is a maximum for

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

max = 0

1 22,

and should take the value

|H(max)| = 12

1 2 .

(g) Show thatH() = atan2(

20 2, 20),

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

(h) How does H() behave for 0? for 0?
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(i) Determine an expression for H(max).

(j) Use your work above to give an accurate sketch of the Bode plot for in caseswhen 1

2, making sure to mark on your plot all the features you determined

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

E4.7 Construct Bode plots by hand for the following first-order SISO linear systems in

input/output form:(a) (N(s), D(s)) = (1, s + 1);

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

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

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

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

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

(c) (N(s), D(s)) = (s

1, s2 + 2s + 2);

(d) (N(s), D(s)) = (s + 2, s2 + 2s + 2).E4.9 Construct Bode plots by hand for the following third-order SISO linear systems in

input/output form:

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

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

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

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

1. calculate the eigenvalues ofA;

2. calculate the transfer function;3. sketch the Bode plots of the magnitude and phase of the frequency response;

4. by trial and error playing with the parameters, on your sketch, indicate the roleplayed by the parameters of the system (e.g., 0 in parts (a) and (b));

5. try to justify the name of the system by looking at the shapes of the Bode plots.

Here are the systems (assume all parameters are positive):

(a) low-pass filter:

A =0 , b = 1 , c = 0 , D = 0 ;

(b) high-pass filter:

A =0 , b = 1 , c = 0 , D = 1 ;

(c) notch filter:

A =

0 1

20 0

, b =

01

, c =

0

0

, D =

1

;

(d) bandpass filter:

A =

0 1

20 0

, b =

01

, c =

0

0

, D =

0

.


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E4.11 Consider the coupled mass system of Exercises E1.4, E2.19, and E3.14. Assume nodamping and take the input from Exercise E2.19 in the case when = 0. In Ex-ercise E3.14, you constructed an output vector c0 for which the pair (A, c0) wasunobservable.

(a) Construct a family of output vectors c by defining c = c0 + c1 for R andc1 R4. Make sure you choose c1 so that c is observable for = 0.

(b) Determine , and the frequency response H using the output vectorc (allow to be an arbitrary real number).

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

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

(a) the equilibrium point (0, 0) with cart position as output;

(b) the equilibrium point (0, 0) with cart velocity as output;

(c) the equilibrium point (0, 0) with pendulum angle as output;

(d) the equilibrium point (0, 0) with pendulum angular velocity as output;(e) the equilibrium point (0, ) with cart position as output;

(f) the equilibrium point (0, ) with cart velocity as output;

(g) the equilibrium point (0, ) with pendulum angle as output;

(h) the equilibrium point (0, ) with pendulum angular velocity as output,

do the following:

1. determine , and the frequency response H for the system;

2. for parameters M = 112 , m = 1, g = 9.81, and =12 , produce the Bode plot for

the pendulum/cart system;

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

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

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

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

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

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

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

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

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

do the following:

1. determine , and the frequency response H for the system;

2. for parameters m1 = 1, m2 = 2, 1 =12 , and 2 =

13 , produce the Bode plot for

the double pendulum;


In each case, use the angle of the second link as output.
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E4.14 Consider the coupled tank system of Exercises E1.11, E2.6, and E3.17. For the lin-earisations in the following cases:

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

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

(c) the output is the difference in the levels,

do the following:

1. determine , and the frequency response H for the system;2. for parameters = 1

3, 1 = 1, A1 = 1, A2 =

12

, a1 =110

, a2 =120

, and g = 9.81,produce the Bode plot for the tank system;


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

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

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

HN,D() HN,D(0),in terms of the roots of N, assuming that N has no roots on iR.

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

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

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

|HN,D()| =HN,D() , R.

Thus the magnitude Bode plots for all systems in M(N, D) are exactly the magnitudeBode plot for (N, D).

(a) How are systems in M(N, D) related to (N, D)? This boils down, of course, toidentifying SISO linear systems that have a plot magnitude of 1 at all frequencies.

For (N, D) M(N, D) denote 0(N, D) = l i m0HN,D() and (N, D) =limHN,D(). Now let

(N, D) = (N, D) 0(N, D).With this notation prove the following statement.

(b) (N, D) (N, D) for any (N, D) M(N, D).E4.19 Construct polar plots corresponding to the Bode plots you made in Exercise E4.7.

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

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

E4.22 For the SISO linear system of Exercise E4.2, plot the polar plots for various 0 0and 0 > 0.
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It is not uncommon to encounter a scheme for control design that relies on a plant beingstable, i.e., having all poles in C. This is in conflict with many plantsfor example, thesimple mass that have a pole at s = 0. In the next exercise, you will investigate acommonly employed hack to get around plants having poles at s = 0.

E4.23 The transfer function

T(s) =K

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

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

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

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

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

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

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

(f) (N(s), D(s)) = (s 1, 1).E4.25 Exercise on the approximate version of Bodes Gain/Phase Theorem. complete
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MCT Book - Chapter 4

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