MCT Book - Chapter 7

7/30/2019 MCT Book - Chapter 7

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This version: 22/10/2004

Chapter 7

Frequency domain methods for stability

In Chapter 5 we looked at various ways to test various notions of stability of SISO controlsystems. Our stability discussion in that section ended with a discussion in Section 6.2.3 of how interconnecting systems in block diagrams affects stability of the resulting system. Thecriterion developed by Nyquist [1932] deals further with testing stability in such cases, andwe look at this in detail in this chapter. The methods in this chapter rely heavily on somebasic ideas in complex variable theory, and these are reviewed in Appendix D.

Contents

7.1 The Nyquist criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .279

7.1.1 The Principle of the Argument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

7.1.2 The Nyquist criterion for single-loop interconnections . . . . . . . . . . . . . . . . 281

7.2 The relationship between the Nyquist contour and the Bode plot . . . . . . . . . . . . . . 293

7.2.1 Capturing the essential features of the Nyquist contour from the Bode plot . . . . 293

7.2.2 Stability margins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .294

7.3 Robust stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .303

7.3.1 Multiplicative uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304

7.3.2 Additive uncertainty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

7.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .311

7.1 The Nyquist criterion

The Nyquist criterion is a method for testing the closed-loop stability of a system basedon the frequency response of the open-loop transfer function.

7.1.1 The Principle of the Argument In this section we review one of the essentialtools in dealing with closed-loop stability as we shall in this chapter: the so-called Principleof the Argument. This is a result from the theory of complex analytic functions. That suchtechnology should be useful to us has been made clear in the developments of Section 4.4.2concerning Bodes Gain/Phase Theorem. The Principle of the Argument has to do, as weshall use it, with the image of closed contours under analytic functions. However, let usrst provide its form in complex analysis. Let U C be an open set, and let f : U Cbe an analytic function. A pole for f is a point s0 U with the property that the limit
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280 7 Frequency domain methods for stability 22/10/2004

limss 0 f (s ) does not exist, but there exists a k N so that the limit lim ss 0 (s s 0)k f (s )

does exist. Recall that a meromorphic function on an open subset U C is a functionf : U C having the property that it is dened and analytic except at isolated points(i.e., except at poles). Now we can state the Principle of the Argument, which relies on theResidue Theorem stated in Appendix D.

7.1 Theorem (Principle of the Argument) Let U be a simply connected open subset of C and let C be a contour in U . Suppose that f is a function which

(i) is meromorphic in U ,(ii) has no poles or zeros on the contour C , and (iii) has n p poles and n z zeros in the interior of C , counting multiplicities of zeros and

poles.Then

C f (s)f (s ) ds = 2 i (n z n p),provided integration is performed in a counterclockwise direction.

Proof Since f is meromorphic,f f is also meromorphic, and is analytic except at the poles

and zeroes of f . Let s0 be such a pole or zero, and suppose that it has multiplicity k.Then there exists a meromorphic function f , analytic an s 0, with the property that f (s ) =(s s 0)k f (s ). One then readily determines that

f (s )f (s)

=k

s s 0+

f (s )f (s )

for s in a neighbourhood of s 0. Now by the Residue Theorem the result follows since k ispositive if s 0 is a zero and negative if s 0 is a pole.

The use we will make of this theorem is in ascertaining the nature of the image of a closedcontour under an analytic function. Thus we let U C be an open set and c : [0, T ] U a closed curve. The image of c we denote by C , and we let f be a function satisfying thehypotheses of Theorem 7.1. Let us denote by c the curve dened by c(t) = f c(t). Sincef has no zeros on C , the curve c does not pass through the origin and so the functionF : [0, T ] C dened by F (t) = ln( c(t)) is continuous. By the chain rule we have

F (t) =f (c(t ))f (c(t))

c (t), t [0, T ].

Therefore

C

f (s )

f (s)ds =

C

f (c(t))

f (c(t))c (t) d t = ln( f (c(t)))

T

0.

Using the denition of the logarithm we have

ln(f (c(t )))T

0= ln |f (c(t )) |

T

0+ i f (c(t ))

T

0.

Since c is closed, the rst term on the right-hand side is zero. Using Theorem 7.1 we thenhave

2 (n z n p) = f (c(t))T

0. (7.1)

In other words, we have the following.
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22/10/2004 7.1 The Nyquist criterion 281

7.2 Proposition If C and f are as in Theorem 7.1, then the image of C under f encircles the origin n z n p times, with the convention that counterclockwise is positive.

Let us illustrate the principle with an example.

7.3 Example We take C to be the circle of radius 2 in C . This can be parameterised, forexample, by c : t 2eit , t [0, 2 ]. For f we take

f (s) = 1(s + 1) 2 + a2

, a R .

Lets see what happens as we allow a to vary between 0 and 1 .5. The curve c = f c isdened by

t 1

(e2it + 1) 2 + a2, t [0, 2 ].

Proposition 7.2 says that for a < 1 the image of C should encircle the origin in C two timesin the counterclockwise direction, and for a > 1 there should be no encirclements of theorigin. Of course, it is problematic to determine the image of a closed contour under a givenanalytic function. Here we let the computer do the work for us, and the results are shownin Figure 7.1. We see that the encirclements are as we expect.

7.1.2 The Nyquist criterion for single-loop interconnections Now we apply thePrinciple of the Argument to determine the stability of a closed-loop transfer function. Theblock diagram conguration we consider here is shown in Figure 7.2. The key observation isthat if the system is to be IBIBO stable then the poles of the closed-loop transfer function

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

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

must all lie in C

. The idea is that we examine the determinant 1 + R C R P to ascertain

when the poles of T are stable.We denote the loop gain by R L = R C R P . Suppose that R L has poles on the imaginary

axis at i1, . . . , ik where k > > 1 0. Let r > 0 have the property that

r k +r2

. (7.3)

With r and R so chosen we may dene a contour R,r which will be comprised of a collectionof components. For i = 1 , . . . , k we dene

r,k, + = i i + re i 2 2 , r,k, = i i + re

i 2 2 .

Now for i = 1 , . . . , k 1 dene

r,i, + = { i | i + r < < i+1 r } , r,i, = { i | i r > > i+1 + r } ,

and also dene

r,i, + = { i | k + r < < R } , r,i, = { i | k r > > R } .


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-3

-2

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Figure 7.1 Images of closed contours for a = 0 (top left), a = 0 .5(top right), and a = 1 .5 (bottom)

r (s ) R C (s ) R P (s ) y(s )

Figure 7.2 A unity feedback loop

Finally we deneR = Re i 2

2 .


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The union of all these various contours we denote by R,r :

R,r =k

i=1

r,i, + r,i, k

i=1

r,i, + r,i, R .

When R L has no poles on the imaginary axis, for R > 0 we write

R, 0 = { i | R < < R } R .

Note that the orientation of the contour R,r is taken by convention to be positive in theclockwise direction. This is counter to the complex variable convention, and we choose thisconvention because, for reasons will soon see, we wish to move along the positive imaginaryaxis from bottom to top. In any case, the idea is that we have a semicircular contourextending into C + , and we need to make provisions for any poles of R L which lie on theimaginary axis. The situation is sketched in Figure 7.3. With this notion of a contour behind

Re

Im

R

Figure 7.3 The contour R,r

us, we can dene what we will call the Nyquist contour.

7.4 Denition For the unity feedback loop of Figure 7.2, let R L be the rational functionR C R P , and let R and r satisfy the conditions ( 7.3) and (7.2). The (R, r )-Nyquist contour is the contour R L (R,r ) C . We denote the ( R, r )-Nyquist contour by N R,r .

When we are willing to live with the associated imprecision, we shall often simply sayNyquist contour in place of ( R, r )-Nyquist contour.

Let us rst state some general properties of the ( R, r )-Nyquist contour. At the same timewe introduce some useful notation. Since we are interested in using the Nyquist criterionfor determining IBIBO stability, we shall suppose R L to be proper, as in most cases weencounter.

7.5 Proposition Let R L be a proper rational function, and for > 0 let D (0, ) ={ s C | |s | } be the disk of radius centred at the origin in C . The following statements hold.
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(i) If R L has no poles on the imaginary axis then there exists M > 0 so that for any R > 0 the (R, 0)-Nyquist contour is contained in the disk D (0, M ). Furthermore limR N R, 0 is well-dened and we denote the limit by N , 0 .

(ii) If R L is strictly proper, then for any r > 0 satisfying ( 7.2 ) and for any > 0 there exists R 0 > 0 so that N R,r \ N R 0 ,r D (0, ) for any R > R 0. Thus limR N R,r is well-dened and we denote the limit by N ,r .

(iii) If R L is both strictly proper and has no poles on the imaginary axis, then the conse-quences of ( ii ) hold with r = 0 , and we denote by N , 0 the limit limR N R, 0.

Proof (i) Dene R (s) = R L ( 1s ) for s = 0. Since R L is proper, the limit lim s0 R (s) exists.But, since R L is continuous, this is nothing more than the assertion we are trying to prove.

(ii) If R L is strictly proper then lim sR L (s ) = 0. Therefore, by continuity of R L we canchoose R 0 sufficiently large that, for R > R 0, those points which lie in the ( R, r )-Nyquistcontour but do not lie in the ( R 0, r )-Nyquist contour reside in the disk D (0, ). This isprecisely what we have stated.

(iii) This is a simple consequence of (i) and (ii).

The punchline here is that the Nyquist contour is always bounded for proper loop gains,provided that there are no poles on the imaginary axis. When there are poles on the imag-inary axis, then the Nyquist contour will be unbounded, but for any xed r > 0 sufficientlysmall, we may still consider letting R .

Let us see how the character of the Nyquist contour relates to stability of the closed-loopsystem depicted in Figure 7.2.

7.6 Theorem (Nyquist Criterion) Let R C and R P be rational functions with R L = R C R P proper. Let n p be the number of poles of R L in C + . First suppose that 1+ R L has no zeros on iR . Then the interconnected SISO linear system represented by the block diagram Figure 7.2 is IBIBO stable if and only if

(i) there are no cancellations of poles and zeros in C + between R C and R P ;(ii) limsR L (s ) = 1;(iii) there exists R 0, r 0 > 0 satisfying ( 7.3 ) and ( 7.2 ) with the property that for every

R > R 0 and r < r 0, the (R, r )-Nyquist contour encircles the point 1 + i0 in the complex plane n p times in the counterclockwise direction as the contour R,r is traversed once in the clockwise direction.

Furthermore, if for any R and r satisfying ( 7.3 ) and ( 7.2 ) the (R, r )-Nyquist contour passes through the point 1 + i0, and in particular if 1 + R L has zeros on iR , then the closed-loop system is IBIBO unstable.

Proof We rst note that the condition (ii) is simply the condition that the closed-looptransfer function be proper. If the closed-loop transfer function is not proper, then theresulting interconnection cannot be IBIBO stable.

By Theorem 6.38, the closed-loop system is IBIBO stable if and only if (1) all the closed-loop transfer function is proper, (2) the zeros of the determinant 1 + R L are in C , and(3) there are no cancellations of poles and zeros in C + between R C and R P . Thus the rststatement in the theorem will follow if we can show that, when 1 + R L has no zeros on iR ,the condition (iii) is equivalent to the condition

(iv) all zeros of the determinant 1 + R L are in C .
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Since there are no poles or zeros of 1+ R L on R,r , provided that R > R 0 and r < r 0, we canapply Proposition 7.2 to the contour R,r and the function 1 + R L . The conclusion is thatthe image of R,r under 1 + R L encircles the origin n z n p times, with n z being the numberof zeros of 1 + R L in C + and n p being the number of poles of 1 + R L in C + . Note that thepoles of 1 + R L are the same as the poles of R L , so n p is the same as in the statement of thetheorem. The conclusion in this case is that n z = 0 if and only if the image of R,r under1 + R L encircles the origin n p times, with the opposite orientation of R,r . This, however, isequivalent to the image of R,r encircling 1 + i 0 n p times, with the opposite orientation of R,r .

Finally, if the ( R, r )-Nyquist contour passes through the point 1 + i0, this means thatthe contour 1 + R L (R,r ) passes through the origin. Thus this means that there is a points 0 R,r which is a zero of 1 + R L . However, since all points on R,r are in C + , the resultfollows by Theorem 6.38.

Let us make a few observations before working out a few simple examples.

7.7 Remarks 1. Strictly proper rational functions always satisfy the condition (ii).

2. Of course, the matter of producing the Nyquist contour may not be entirely a straight-forward one. What one can certainly do is produce it with a computer. As we will see,the Nyquist contour provides a graphical representation of some important properties of the closed-loop system.

3. By parts (i) and (ii) of Proposition 4.13 it suffices to plot the Nyquist contour only as wetraverse that half of R,r which sits in the positive imaginary plane, i.e., only for thosevalues of s along R,r which have positive imaginary part. This will be borne out in theexamples below.

4. When R L is proper and when there are no poles for R L on the imaginary axis (so wecan take r = 0), the ( R, 0)-Nyquist contour is bounded as we take the limit R . If we further ask that R L be strictly proper, that portion of the Nyquist contour which isthe image of R under R L will be mapped to the origin as R . Thus in this caseit suffices to determine the image of the imaginary axis under R L , along with the originin C . Given our remark 3, this essentially means that in this case we only determinethe polar plot for the loop gain R L . Thus we see the important relationship between theNyquist criterion and the Bode plot of the loop gain.

5. Heres one way to determine the number of times the ( R, r )-Nyquist contour encirclesthe point 1 + i0. From the point 1 + i0 draw a ray in any direction. Choose thisray so that it is nowhere tangent to the ( R, r )-Nyquist contour, and so that it does notpass through points where the ( R, r )-Nyquist contour intersects itself. The number of times the ( R, r )-Nyquist contour intersects this ray while moving in the counterclockwisedirection is the number of counterclockwise encirclements of 1 + i0. A crossing in theclockwise direction is a negative counterclockwise crossing.

The Nyquist criterion can be readily demonstrated with a couple of examples. In eachof these examples we use Remark 7.7(5).

In the Nyquist plots below, the solid contour is the image of points in the positive imaginary plane under R L , and the dashed contour is the image of the points in the negative imaginary plane.
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7.8 Examples 1. We rst take R C (s ) = 1 and R P (s) = 1s + a for a R . Note that con-ditions (i) and (ii) of Theorem 7.6 are satised for all a , so stability can be check byverifying the condition (iii). We note that for a < 0 there is one pole of R L in C + , andotherwise there are no poles in C + .The loop gain R L (s ) = 1s+ a is strictly proper with no poles on the imaginary axis unlessa = 0. So let us rst consider the situation when a = 0. We need only considerthe image under R L of points on the imaginary axis. The corresponding points on theNyquist contour are given by

1i + a

=a

2 + a2 i

2 + a2

, ( , ).

This is a parametric representation of a circle of radius 12|a | centred at1

2a . This can bechecked by verifying that

a2 + a2

1

2a2

+

2 + a 22

=1

4a2.

The Nyquist contour is shown in Figure 7.4 for various nonzero a . From Figure 7.4 wemake the following observations:(a) for a < 1 there are no encirclements of the point 1 + i0;(b) for a = 1 the Nyquist contour passes through the point 1 + i0;(c) for 1 < a < 0 the Nyquist contour encircles the point 1 + i0 one time in the

counterclockwise direction (to see this, one must observe the sign of the imaginarypart as runs from to + );

(d) for a > 0 there are no encirclements of the point 1 + i0.Now let us look at the case where a = 0. In this case we have a pole for R L at s = 0, sothis must be taken into account. Choose r > 0. The image of { i | > r } is

i 0 < < 1r .

Now we need to look at the image of the contour r given by s = re i , [ 2 ,2 ]. One

readily sees that the image of r is

eir

, [2

,2

]

which is a large semi-circle centred at the origin going from + i to in the clockwisedirection. This is shown in Figure 7.5.

This allows us to conclude the following:(a) for a < 1 the system is IBIBO unstable since n p = 1 and the number of counter-

clockwise encirclements is 1;(b) for a = 1 the system is IBIBO unstable since the Nyquist contour passes through

the point 1 + i0;(c) for 1 < a < 0 the system is IBIBO stable since n p = 1 and there is one counter-

clockwise encirclement of the point 1 + i0;(d) for a 0 the system is IBIBO stable since n p = 0 and there are no encirclements

of the point 1 + i0.
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Figure 7.4 The ( , 0)-Nyquist contour for R L (s ) = 1s + a , a = 2(top left), a = 1 (top right), a = 12 (middle left), a =

12

(middle right), a = 1 (bottom left), and a = 2 (bottom right)


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-10

-5

0

5

10

Re

I m

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Figure 7.5 The ( , 0.1)-Nyquist contour for R L (s ) = 1s

We can also check this directly by using Theorem 6.38. Since there are no unstablepole/zero cancellations in the interconnected system, IBIBO stability is determined bythe zeros of the determinant, and the determinant is

1 + R L (s ) =s + a + 1

s + a.

The zero of the determinant is a 1 which is in C exactly when a > 1, and this isexactly the condition we derive above using the Nyquist criterion.

2. The previous example can be regarded as an implementation of a proportional controller.

Lets give an integral controller a try. Thus we takeR

C (s) =

1

s andR

P (s) =

1

s + a . Onceagain, the conditions (i) and (ii) of Theorem 7.6 are satised, so we need only checkcondition (iii).In this example the loop gain R L is strictly proper, and there is a pole of R L at s = 0.Thus we need to form a modied contour to take this into account. Let us expand theloop gain into its real and imaginary parts when evaluated on the imaginary axis awayfrom the origin. We have

R L (i ) = 1

2 + a2 i

a(2 + a2)

.

Let us examine the image of { i | > r } as r becomes increasingly small. For nearzero, the real part of the Nyquist contour is near 12 + a 2 , and as increases, it shrinks tozero. Near = 0 the imaginary part is at sgn(a) , and as increases, it goes to zero.Also note that the imaginary part does not change sign. The image of { i | < r }reects this about the real axis. It only remains to examine the image of the contour raround s = 0 given by s = re i where [ 2 ,

2 ]. The image of this contour under R L

is1

re i (re i + a ), [ 2 ,

2 ].
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For sufficiently small r we have

1re i (re i + a)

=1

re i

1a

1 + ra ei

=1

re i1a

1 ra

ei +

=eiar

1a2 + .

Thus, as r 0, the contour r gets mapped into a semi-circle of innite radius, centredat 1a 2 + i0, which goes clockwise from

1a 2 + isgn(a) to

1a 2 isgn(a) . In Figure 7.6 we

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Figure 7.6 The ( , 0.1)-Nyquist contours for R L (s ) = 1s (s + a ) , a = 2 (top left), a = 1 (top right), a = 1 (bottom left), anda = 2 (bottom right)

plot the Nyquist contours.Now we look at the particular case when a = 0. In this case R L (s) = 1s 2 , and so theNyquist contour is the image of the imaginary axis, except the origin, under R L , along
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with the image of the contour r as r 0. On the imaginary axis we have R L (i ) = 12 .As goes from to 0 the Nyquist contour goes from 0 + i0 to + i0, and as goes from 0+ to + the Nyquist contour goes from + i0 to 0+ i0. On the contourr we have

R L (re i ) =e2i

r 2, [ 2 ,

2 ].

As r 0 this describes an innite radius circle centred at the origin which starts at + i0 and goes around once in the clockwise direction. In particular, when a = 0 theNyquist contour passes through the point 1 + i0. The contour is shown in Figure 7.7.

-75 -50 - 25 0 25 50 75 100

-75

-50

-25

0

25

50

75

100

Re

I m

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Figure 7.7 The ( , 0.1)-Nyquist contour for R L (s ) = 1s 2

We can now make the following conclusions regarding stability:(a) when a < 0 the system is IBIBO unstable since n p = 1 and there is one clockwise

encirclement of the point 1 + i0;(b) when a = 0 the system is IBIBO unstable since the Nyquist contour passes through

the point 1 + i0;(c) when a > 0 the system is IBIBO stable since n p = 0 and there are no encirclements

of the point 1 + i0.We can also check IBIBO stability of the system via Theorem 6.38. Since there are nounstable pole/zero cancellations, we can look at the zeros of the determinant which is

1 + R L (s ) =s 2 + as + 1

s 2 + as .

By the Routh/Hurwitz criterion, the system is IBIBO stable exactly when a > 0, andthis is what we ascertained using the Nyquist criterion.

3. Our nal example combines the above two examples, and implements a PI controllerwhere we take R C (s ) = 1 + 1s and R P (s ) =

1s + a . Here the condition (ii) of Theorem 7.6

holds, but we should examine (i) just a bit carefully before moving on. We have R L (s) =s +1

s1

s+ a , and so there is a pole/zero cancellation here when a = 1. However, it is a stablepole/zero cancellation, so condition (i) still holds. We also ascertain that R L has onepole in C + when a < 0.
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In this example, R L is again strictly proper, and has a pole on the imaginary axis at theorigin. Thus to compute the Nyquist contour we determine the image of the imaginaryaxis minus the origin, and tack on the image of the contour r = re i , [ 2 ,

2 ]. We

rst compute

R L (i ) =a 1

2 + a2 i

2 + a(2 + a2)

.

Lets rst consider the situation when a = 0. In this case, as goes from i to 0, thereal part goes from 0 to a1a 2 and the imaginary part goes from 0 to sgn(a) . While thereal part never changes sign, the imaginary part can change sign, so we must take careabout how it is behaving. As goes from 0+ to + , the resulting part of the Nyquistcontour is the mirror image about the real axis of what we have already computed. Nowlet us look at the image of the contour r . When a = 0 we have

R L (re i ) =re i + 1

re i

1a

1 + ra ei

=re i + 1

are i1

ra

ei +

= 1a

+ ei

ar re

i

a 2 1

a 2+

=eiar

+a 1

a 2+

From this we conclude that for a = 0 the image as r 0 of r under R L is an inniteradius semi-circle centered at a1a 2 and going clockwise from a1a 2 + isgn(a) to a1a 2 isgn(a) . In Figure 7.8 we show the Nyquist contours for various values of a = 0.Now let us consider the image of the imaginary axis when a = 0. In this case we have

R L (i ) = 1

2 i

1

.

Thus the image of those points on the imaginary axis away from the origin describe aparabola, sitting in C , passing through the origin, and symmetric about the real axis.As concerns r when a = 0 we have

R L (re i ) = 1 +ei

rei

r

which, as r 0, describes an innite radius circle centred at the origin and goingclockwise from + i0 to + i0. This Nyquist contour is shown in Figure 7.9.With the above computations and the corresponding Nyquist plots, we can make thefollowing conclusions concerning IBIBO stability of the closed-loop system.(a) For a < 1 the system is IBIBO unstable since n p = 1 and there is one clockwise

encirclement of 1 + i0.(b) For a = 1 the system is IBIBO unstable since the Nyquist contour passes through

the point 1 + i0.(c) For 1 < a < 0 the system is IBIBO stable since n p = 1 and there is one counter-

clockwise encirclement of the point 1 + i0.(d) For a 0 the system is IBIBO stable since n p = 0 and there are no encirclements

of 1 + i0.
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-2 -1 0 1 2 3

-2

-1

0

1

2

3

Re

I m

| |

-4 -2 0 2 4 6

-4

-2

0

2

4

6

Re

I m

| |

-7.5 -5 -2.5 0 2.5 5 7.5 10

-7.5

-5

-2.5

0

2.5

5

7.5

10

Re

I m

| |

-7.5 -5 -2.5 0 2.5 5 7.5 10

-7.5

-5

-2.5

0

2.5

5

7.5

10

Re

I m

| |

-4 -2 0 2 4

-4

-2

0

2

4

Re

I m

| |

-4 -2 0 2 4

-4

-2

0

2

4

Re

I m

| |

Figure 7.8 The ( , 0.2)-Nyquist contours for R L (s ) = s +1s (s + a ) , a = 2 (top left), a = 1 (top right), a = 12 (middle left), a =

12

(middle right), a = 1 (bottom left), and a = 2 (bottom right)


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22/10/2004 7.2 The relationship between the Nyquist contour and the Bode plot 293

-20 -10 0 10 20 30

-20

-10

0

10

20

30

Re

I m

| |

Figure 7.9 The ( , 0.2)-Nyquist contour for R L (s ) = s +1s 2

As always, we can evaluate IBIBO stability with Theorem 6.38. To do this we canstill simply look at the zeros of the determinant, because although there is a pole zerocancellation when a = 1, it is a cancellation of stable factors so it does not hurt us. Wecompute the determinant to be

1 + R L (s ) =s 2 + ( a + 1) s + 1

s 2 + as.

An application of the Routh/Hurwitz criterion suggests that we have IBIBO stability fora > 1, just as we have demonstrated with the Nyquist criterion.

Note that the Nyquist criterion as we have shown is applicable only to interconnectionswith a simple structure, namely a single loop. Bode [1945] discusses a version of the Nyquistcriterion for systems with multiple loops, and this is explored further by Zadeh and Desoer[1979]. However, the development is too signicant, and the outcome too modest (what areobtained are sufficient conditions for IBIBO stability under restrictive hypotheses) to makea presentation of these results worthwhile.

7.2 The relationship between the Nyquist contour and the Bode plot

The above examples, although simple, demonstrate that obtaining the Nyquist contourcan be problematic, at least by hand. This is especially well illustrated by the third of the

three examples where the capacity to change sign of the imaginary part of the restrictionof R L to the imaginary axis causes some difficulties which must be accounted for. A usefulobservation here is that the Nyquist contour is in essence the polar plot for the loop gain,taking care of the possibility of poles on the imaginary axis. The matter of constructinga Bode plot is often an easier one than that of building the corresponding polar plot, so aplausible approach for making a Nyquist contour is to rst make a Bode plot, and convertthis to a polar plot as discussed in Section 4.3.3.

7.2.1 Capturing the essential features of the Nyquist contour from the Bode plotLet us illustrate this with the third of our examples from the previous section.


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7.9 Example (Example 7.83 contd) The loop gain, recall, is R L (s ) = s +1s (s + a ) . Let us writethis transfer function in the recommended form for making Bode plots. For a = 0 we have

R L (s ) =1a

1s

1sa + 1

(s + 1) .

Thus the frequency response for R L is a product of four terms:

H 1() =1a

, H 2() = i

, H 3() =1

1 + i a, H 4() = 1 + i.

Each of these is a simple enough function for the purpose of plotting frequency response,and the effect of a is essentially captured in H 3.

Let us see if from Bode plot considerations we can infer when the imaginary part of thetransfer function changes sign as goes from 0+ to + . In Example 7.83 we determinedthat as we vary in this way, the real part of the frequency response goes from a 1a 2 to 0and the imaginary part goes from sgn( a ) to 0. The question is, For which values of adoes the imaginary part of the frequency response change sign as goes from 0+ to + ?

Provideda

1

a 2 < 0 this will happen if and only if the phase is at some nite frequency .Lets take a look at the phase of the frequency response function.1. First we take a < 0. Since H 1() = and H 2() = 2 , in order to have the total

phase equal , it must be the case that

H 3() + H 4() {2 , 32 }.

The phase of H 4 varies from 0 to 2 , and the phase of H 3 varies from 0 to2 (because

a < 0!) Therefore we should aim for conditions on when H 3() + H 4() = 2 forsome nite frequency . But one easily sees that for any a < 0 there will always be anite frequency so that this condition is satised since

lim

H 3() + H 4() = .

Thus as long as a < 0 there will always be a sign change in the imaginary part of thefrequency response as we vary from 0+ to + . The Bode plots for R L are shown inFigure 7.10 for various a < 0.

2. Now we consider when a > 0. In this case we have H 1() = 0 and so we must seek so that

H 3() + H 4() {2 ,32 }.

However, since a > 0 the phase of H 3 will go from 0 to 2. Therefore it will not be

possible for H 3() + H 4() to equal either 2 or 32 , and so we conclude that fora > 0 the imaginary part of R L (i ) will not change sign as we vary from 0+ to + .The Bode plots for R L are shown in Figure 7.11 for various a > 0.

7.2.2 Stability margins The above example illustrates how the Bode plot can beuseful in determining certain aspects of the behaviour of the Nyquist contour. Indeed, if onegives this a short moments thought, one reaches the realisation that one can benet a greatdeal by looking at the Bode plot for the loop gain. Let us provide the proper nomenclaturefor organising such observations.
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-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2-40

-30

-20

-10

0

10

20

30

log

log

d B

d e g

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2-40

-20

0

20

40

log

log

d B

d e g

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2-40

-20

0

20

40

log

log

d B

d e g

| |

Figure 7.10 Bode plots for R L (s ) = s+1s (s + a ) for a = 2 (top left),a = 1 (top right), and a = 12 (bottom)

7.10 Denition Let R L R (s) be a proper rational function forming the loop gain for theinterconnection of Figure 7.12. Assume that there is no frequency > 0 for which R L (i ) =

1 + i0.(i) A phase crossover frequency , pc [0, ), for R L is a frequency for whichR L (ipc ) = 180 .

Let pc,1 , . . . , pc, be the phase crossover frequencies for R L , and assume these are orderedso that

R L (ipc,1 ) < < R L (ipc, ).

Also suppose that for some k {1, . . . , } we have

R L (ipc, k ) < 1 < R L (ipc, k + 1 ).
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-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-40

-20

0

20

40

log

log

d B

d e g

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-40

-20

0

20

40

log

log

d B

d e g

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2-40

-30

-20

-10

0

10

20

30

log

log

d B

d e g

| |

Figure 7.11 Bode plots for R L (s ) = s+1s (s + a ) for a =12 (top left),

a = 1 (top right), and a = 2 (bottom)

r (s ) R L (s ) y(s )

Figure 7.12 Unity gain feedback loop for stability margin discus-sion


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(ii) The lower gain margin for R L dened by

K min = R L (ipc, k )1 (0, 1).

If 1 < R L (ipc,1 ) < < R L (ipc, ),

then K min is undened.(iii) The upper gain margin for R L is dened by

K max = R L (ipc, k + 1 )1 (1, ).

If R L (ipc,1 ) < < R L (ipc, ) < 1,

then K max is undened.(iv) A gain crossover frequency , gc [0, ), for R L is a frequency for which

|R L (igc)| = 1.

Let gc,1 , . . . , gc, be the gain crossover frequencies for R L , and assume these are ordered sothat(R L (igc,1 )) < < (R L (igc, )) .

(v) The lower phase margin for R L is dened by

min =180 (R L (igc, )) , (R L (igc, )) 0undened , (R L (igc, )) < 0.

(vi) The upper phase margin for R L is dened by

max = (R L (igc,1 )) + 180 , (R L (igc,1 )) 0undened , (R L (igc,1 )) > 0.

Let us parse these denitions, as they are in actuality quite simple. First of all, we notethat it is possible to read the gain and phase margins off the Nyquist plot; this saves onehaving to compute them directly using the denitions. Rather than try to state in a preciseway how to procure the margins from the Nyquist plot, let us simply illustrate the process inFigure 7.13. The basic idea is that for the gain margins, one looks for the positive frequencycrossings of the negative real axis closest to 1 + i0 in each direction. The reciprocal of thedistances to the imaginary axis are the gain margins, as indicated in Figure 7.13. For thephase margins, one looks for the positive frequency crossings of the unit circle closest to thepoint 1 + i0. The angles to the negative real axis are then the phase margins, again asindicated in Figure 7.13. We shall adopt the convention that when we simply say phasemargin , we refer to the smaller of the upper and lower phase margins.

The interpretations of gain crossover and phase crossover frequencies are clear. At again crossover frequency, the magnitude on the Bode plot for R L will be 0dB. At a phasecrossover frequency, the graph will cross the upper or lower edge of the phase Bode plot.Note that it is possible that for a given loop gain, some of the margins may not be dened.Let us illustrate this with some examples.
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Re

Im

K 1max

1 + i 0

K 1min

Re

Im

min

max

1

Figure 7.13 Getting gain (left) and phase (right) margins from theNyquist plot

-1 0 1 2 3 4

-2

-1

0

1

2

3

Re

I m

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-50

-40

-30

-20

-10

0

10

log

log

d B

d e g

| |

Figure 7.14 Nyquist and Bode plots for R L (s ) = 10s 2 +4 s+3

7.11 Examples 1. We consider the loop gain

R L (s) =10

s2 + 4 s + 3.

In Figure 7.14 we note that there is one gain crossover frequency, and it is roughly atgc = 2 .1. The phase at the gain crossover frequency is about 110, which gives theupper phase margin as max 70. The lower phase margin is not dened. Also, neitherof the gain margins are dened.

2. We take as loop gainR L (s ) =

10s 2 + 4 s + 3

.

The Bode and Nyquist plots are shown in Figure 7.15. From the Bode plot we seethat there is one gain crossover frequency and it is approximately at gc = 2 .1. Thephase at the gain crossover frequency it is about 70 . Thus the lower phase margin is
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-4 -3 -2 -1 0 1

-2

-1

0

1

2

3

Re

I m

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

-150

-100

-50

0

50

100

150

-1.5 -1 -0.5 0 0.5 1 1.5 2

-50

-40

-30

-20

-10

0

10

log

log

d B

d e g

| |

Figure 7.15 Nyquist and Bode plots for R L (s ) = 10s 2 +4 s +3

min 110. We note that there the upper phase margin is undened, as are the phasecrossover frequencies.

gain margin

3.

For cases when R L is BIBO stable, we can offer an interpretation of the gain and phasemargins in terms of closed-loop stability.

7.12 Proposition Let R L RH+ be a BIBO stable loop gain, and consider a unity gain feed-back block diagram conguration like that of Figure 7.12 . Either of the following statements implies IBIBO stability of the closed-loop system:

(i) K min is undened and either (a) K max > 1 or (b) K max is undened.

(ii) min is undened and either (a) max > 0.(b) max is undened.

Proof We use the Nyquist criterion for evaluating IBIBO stability. For a stable loop gain

R L , the closed-loop system is IBIBO stable if and only if there are no encirclements of 1 + i0. Note that the assumption that R L RH+ be BIBO stable implies that there areno poles for R L on the imaginary axis, so the ( , 0)-Nyquist contour is well-dened andbounded.

(i) In this case, all crossings of the imaginary axis will occur in the interval ( 1, 0). Thisprecludes any encirclements of 1 + i0.

(ii) If (ii) holds, then all intersections at positive frequency of the ( , 0)-Nyquist contourwith the unit circle in C will occur in the lower complex plane. Similarly, those crossings of the unit disk at negative frequencies will take place in the upper complex plane. This clearlyprecludes any encirclements of 1 + i0 which thus implies IBIBO stability.


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Often when one reads texts on classical control design, one simply sees mentioned gainmargin and phase margin without reference to upper and lower. Typically, a result likeProposition 7.12 is in the back of the minds of the authors, and it is being assumed thatthe lower margins are undened. In these cases, there is a more direct link between thestability margins and actual stabilityat least when the loop gain is itself BIBO stableasevidenced by Proposition 7.12. The following examples demonstrate this, and also show thatone cannot generally expect the converse of the statements in Proposition 7.12 to hold.

7.13 Examples 1. We considerR L (s) =

10s2 + 4 s + 3

,

which we looked at in Example 7.11. In this case, the hypotheses of part (ii) of Proposi-tion 7.12 are satised, and indeed one can see from the Nyquist criterion that the systemis IBIBO stable.

2. Next we considerR L (s ) =

10s 2 + 4 s + 3

,

which we also looked at in Example 7.11. Here, the lower phase margin is dened, sothe hypotheses of part (ii) of Proposition 7.12 are not satised. In this case, the Nyquistcriterion tells us that the system is indeed not BIBO stable.

3. The rst of the preceding examples illustrates how one can use the conditions on gainand phase margin of Proposition 7.12 to determine closed-loop stability when R L isBIBO stable. The second example gives a system which violates the sufficient conditionsof Proposition 7.12, and is indeed not IBIBO stable. The question then arises, Is thecondition (ii) of Proposition 7.12 necessary for IBIBO stability? The answer is, No,and we provide an example which illustrates this.We take

R L (s ) =(1 s )2(1 + 3s

25)3

2(1 + s )2(1 + s100 )(1 +s

50 )3 .

This is a BIBO stable loop gain, and its Bode plot is shown in Figure 7.16. From theBode plot we can see that there is a gain crossover frequency gc satisfying somethinglike loggc = 0 .8. The lower phase margin is dened at this frequency, and it is roughly60. Thus the lower phase margin is dened, and this loop gain is thus contrary to thehypotheses of part (ii) of Proposition 7.12. Now let us examine the Nyquist contour for thesystem. In Figure 7.17 we show the Nyquist contour, with the right plot showing a blowuparound the origin. This is a pretty messy Nyquist contour, but a careful accounting of what is going on will reveal that there is actually a total of zero encirclements of 1 + i0.Thus the closed-loop system with loop gain R L is IBIBO stable. This shows that theconverse of Proposition 7.12 is not generally true.argin

4.

The above examples illustrate that one might wish to make the phase margins large andpositive, and make the gain margins large. However, this is only a rough rule of thumb. InExercise E7.10 the reader can work out an example where look at one stability margin whileignoring the other can be dangerous. The following example from the book of Zhou [1996]indicates that even when looking at both, they do not necessarily form a useful measure of stability margin.
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-1 0 1 2 3 4 5

-150

-100

-50

0

50

100

150

-1 0 1 2 3 4 5-20

-10

0

10

20

30

log

log

d B

d e g

| |

Figure 7.16 Bode plot for BIBO stable transfer function with pos-

itive phase margins

0 10 20 30 40 50

-40

-20

0

20

40

Re

I m

| |

-1.5 -1 - 0.5 0 0.5 1 1.5 2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Re

I m

| |

Figure 7.17 Nyquist contour for BIBO stable transfer functionwith positive phase margins

7.14 Example We take as our plant

R P (s ) =2 s2s 1

,

and we consider two controllers, both of which may be veried to be stabilising:

R C, 1(s) = 1 , R C, 2(s) =s + 33103310 s + 1

s + 11201120 s + 1

1710 s

2 + 32 s + 1s 2 + 32 s +

1710

.


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This second controller is obviously carefully contrived, but let us see what it tells us. InFigure 7.18 are shown the Nyquist plots for the loop gain R C, 1R P and R C, 2R P . One can see

-2 -1.5 -1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

1.5

Re

I m

| |

-2 -1.5 -1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

1.5

Re

I m

| |

-2 -1.5 -1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

1.5

Re

I m

| |

Figure 7.18 Nyquist plots for plant R P (s ) = 2s2s1 with controllerR C, 1 (top left), with controller R C, 2 (top right), and both (bot-tom)

that the gain and phase margins for the loop gain R C, 2R P are at least as good as those forthe loop gain R C, 1R P , but that the Nyquist contour for R C, 2R P passes closer to the criticalpoint 1 + i0. This suggests that gain and phase margin may not be perfect indicators of stability margin.

With all of the above machinations about gain and phase margins out of the way, let usgive perhaps a simpler characterisation of what these notions are trying to capture. If theobjective is to stay far way from the point 1 + i0, then the following result tells us thatthe sensitivity function is crucial in doing this.

7.15 Proposition inf > 0

| 1 R L (i )| = S L 1 .
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22/10/2004 7.3 Robust stability 303

Proof We compute

inf > 0

| 1 R L (i )| = inf > 0

|1 + R L (i )|

= sup> 0

11 + R L (i )

1

= S L 1 ,

as desired.

The results says, simply, that the point on the Nyquist contour which is closest to the point 1 + i0 is a distance S L 1 away. Thus, to increase the stability margin, one may wish tomake the sensitivity function small. This is a reason for minimising the sensitivity function.We shall encounter others in Sections 8.5 and 9.3.

Let us illustrate Proposition 7.15 on the two loop gains of Example 7.14.

7.16 Example (Example 7.14 contd) We consider the plant transfer function R P and the twocontroller transfer functions R C, 1 and R C, 2 of Example 7.14. The magnitude Bode plots of the sensitivity function for the two loop gains are shown in Figure 7.19. As expected, the

-1.5 -1 -0.5 0 0.5 1 1.5 2

0

2

4

6

8

log

d B

| |

-1.5 -1 -0.5 0 0.5 1 1.5 2

0

2

4

6

8

log

d B

| |

Figure 7.19 Magnitude bode plot for the sensitivity function withloop gain R C, 1R P (left) and R C, 2R P (right)

peak magnitude for the sensitivity with the loop gain R C, 1R P is lower than that for R C, 2R P ,reecting the fact that the Nyquist contour for the former is further from 1 + i0 than forthe latter.

7.3 Robust stability

We now return to the uncertainty representations of Section 4.5. Let us recall here the

basic setup. Given a nominal plantR P and a rational function W u satisfying W u < ,we denote by P (R P , W u ) the set of plants satisfying

R P = (1 + W u )R P ,

where 1. We also denote byP + (R P , W u ) the set of plants satisfying

R P = R P + W u ,

where 1. This is not quite a complete denition, since in Section 4.5 we additionallyimposed the requirement that the plants in P (R P , W u ) and P + (R P , W u ) have the samenumber of unstable poles as does R P .
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r (s ) R C (s ) R P (s ) y(s )

Figure 7.20 Unity gain feedback loop for robust stability

Now we, as usual, consider the unity gain feedback loop of Figure 7.20. We wish todesign a controller R C which stabilises a whole set of plants. We devote this section to aprecise formulation of this problem for both uncertainty descriptions, and to providing usefulnecessary and sufficient conditions for our problem to have a solution.

Let us be precise about this, as it is important that we know what we are saying.

7.17 Denition Let R P , W u R (s ) be a proper rational functions with W u < . Acontroller R C R (s) provides robust stability for P

(R P , W u ) (resp. P + (R P , W u )) if the feedback interconnection of Figure 7.20 is IBIBO stable for every R P P (R P , W )u)(resp. for every R P P + (R P , W u )).

Now we provide conditions for determining when a controller is robustly stabilising,breaking our discussion into that for multiplicative, and that for additive uncertainty.

7.3.1 Multiplicative uncertainty The following important theorem gives simple con-ditions on when a controller is robustly stabilising. It was rst stated by Doyle and Stein[1981] with the rst complete proof due to Chen and Desoer [1982].

7.18 Theorem Let R P , W u R (s ) be a proper rational functions with W u

< , and suppose that R C R (s ) renders the nominal plant R P IBIBO stable in the interconnection of Figure 7.20 . Then R C provides robust stability for P (R P , W u ) if and only if W u T L

< 1,where T L is the complementary sensitivity function, or closed-loop transfer function, for the loop gain R C R P .Proof Let R C R (s ). For R P P (R P , W u ) denote R L (R P ) = R C R P . By denitionof P (R P , W u ) it follows that R P and R P share the same imaginary axis poles for everyR P P (R P , W u ). For r,R > 0 and for R P P (R P , W u ) let N R,r (R P ) be the ( R, r )-Nyquist contour for R L (R P ).

Now we make a simple computation:

W u T L < 1 W u (i )T L (i ) < 1, R

W u (i )R L (i )

1 + R L (i )< 1, R

W u (i )R L (i ) < 1 + R L (i ) , R

W u (i )R L (i ) < 1 R L (i ) , R .

This gives a simple interpretation of the condition W 2T L < 1. We note that

1 R L (i ) is the distance from the point 1+ i0 to the point R L (i ). Thus the condition
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W 2T L < 1 is equivalent to the condition that the open disk of radius W u (i )R L (i )

centred at R L (i ) not contain the point 1 + i0. This is depicted in Figure 7.21. It is this

1 + i 0

W u (i )R L (i )

R L (i )

Figure 7.21 Interpretation of robust stability condition for multi-plicative uncertainty

interpretation of the condition W 2T L < 1 we shall employ in the proof.

First suppose that W u T L < 1. Note that for R and for R P P (R P , W u ) we

haveR L (R P )( i ) = R C (i )R P (i ) = (1 + ( i )W u (i ))R C (i )R P (i ).

Thus the point R L (R P )( i ) lies in the closed disk of radius W u (i )R L (i ) with centreR L (i ). From Figure 7.21 we infer that the point of the the Nyquist contour N R,r (R P ) thatare the image of points on the imaginary axis will follow the points on the Nyquist contourN R,r (R P ) while remaining on the same side of the point 1 + i0. Since W u RH+ andsince is allowable, R L (R P ) and R L have the same poles on iR and the same number of poles in C + . Thus by choosing r 0 < 0 sufficiently small and for R 0 > 0 sufficiently large, thenumber of clockwise encirclements of 1 + i0 by N R,r (R P ) will equal the same by N R,r (R P )for all R > R 0 and r < r 0. From Theorem 7.6 we conclude IBIBO stability of the closed-loopsystem with loop gain R L (R P ).

Now suppose that W u T L 1. As depicted in Figure 7.21, this implies the existence

of 0 so that the open disk of radius W u (i)R L (i) centred at R L (i) contains thepoint 1 + i0. Denote by D (s 0, r ) be the closed disk of radius r centred at s 0. We claimthat for each > 0, the map from P (R P , W u ) to D R L (i ), W u (i )R L (i ) dened by

R P = (1 + W u )R P R C (i )R P (i ) (7.4)

is surjective. The following lemma helps us to establish this fact.

1 Lemma For any 0 and for any ( , ] there exists a function G RH+ with the properties


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(i) G (i) = and (ii) the map |G (i )| has a maximum at = .

Proof Clearly if = 0 we may simply dene G (s ) = 1. Thus suppose that = 0. Considerthe rational function

T , 0 (s) =20

s2 + 2 0s + 20,

, 0 > 0. In Exercise E4.6 it was shown that for 1 2 < < 1 the function |T (i )|achieves a unique maximum at max = 0 1 2 2. Furthermore, the phase at this maxi-mum is given by atan2( , 1 2 2). Thus, as varies in the interval (

1 2 , 1), the phase atthe frequency max varies between 4 and 0.

Now, given ( , ], dene

=5 , < 0 2 10 , > 0.

Thus is guaranteed to live in the interval ( 4 , 0). Therefore, there exists (1 2 , 1) so

that = atan2( , 1 2

2

). Now dene 0 so that max = . Now dene

G =T 5, 0 , < 0T 10, 0 , > 0.

This function G , it is readily veried, does the job.We now resume showing that the map dened by ( 7.4) is surjective. It clearly suffices

to show that the image of every point on the boundary of D R L (i ), W u (i )R L (i ) liesin the image of the map, since all other points can then be obtained by scaling . For ( , ] let

s = R L (i ) + W u (i )R L (i ) ei

be a point on the boundary of D R L (i ), W u (i )R L (i ) . We wish to write

s = R C (i )R P (i ) = R L (i ) + ( i )W u (i )R L (i )

for an appropriate choice of allowable . Thus RH+ should necessarily satisfy( i )W u (i )R L (i ) = W u (i )R L (i ) ei .

It follows that |( i )| = 1 and that ( i ) + (W u (i )R L (i )) = . Therefore, dene

= (W u (i )R L (i )) .

If G is as dened above, then dening

( s ) = G (s)|G (i )|

does the job.The remainder of the proof is now straightforward. Since the map dened by ( 7.4) is

surjective, we conclude that there exists an allowable so that if R P = (1 + W u )R P wehave

R L (R P )( i) = R C (i)R P (i) = 1 + i0.This implies that the Nyquist contour N R,r (R P ) for R sufficiently large passes through thepoint 1 + i0. By Theorem 7.6 the closed-loop system is not IBIBO stable.


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The proof of the above theorem is long-winded, but the idea is, in fact, very simple.Indeed, the essential observation, repeated here outside the connes of the proof, is thatthe condition W u T L < 1 is equivalent to the condition, depicted in Figure 7.21, that,for each frequency , the open disk of radius W u (i )R L (i ) and centred at R L (i ), notcontain the point 1 + j 0.

One may also reverse engineer this problem as well. The idea here is as follows.Suppose that we have our nominal plant R P and a controller R C which IBIBO stabilisesthe closed-loop system of Figure 7.20. At this point, we have not specied a set of plantsover which we want to stabilise. Now we ask, What is the maximum size of the allowableperturbation to R P if the perturbed plant is to be stabilised by R C ? The following resultmakes precise this vague question, and provides its answer.

7.19 Proposition Let R P , R C R (s ) with R P proper. Also, suppose that the interconnection of Figure 7.20 is IBIBO stable with R P = R P . Let T L be the closed-loop transfer function for the loop gain R L = R C R P . The following statements hold:

(i) if W u RH+ has the property that W u < T L1

then R C provides robust stability for P

(R P , W u );

(ii) for any T L 1there exists W u RH+ satisfying W u = so that R C does not

robustly stabilise P (R P , W u ).

Proof (i) This follows directly from Theorem 7.18.(ii) This part of the result too follows from Theorem 7.18. Indeed, let > 0 be a frequency

for which T L (i) = T L . One can readily dene W u RH+ so that |W u (i)| =

W u = . It then follows that W u T L 1, so we may apply Theorem 7.18.

Roughly, the proposition tells us that as long as we choose the uncertainty weight W u sothat its H -norm is bounded by T L

1

, then we are guaranteed that R C will provide robust

stability. If the H

-norm of W u exceeds T L 1

, then it is possible, but not certain, that R C will not robustly stabilise. In this latter case, we must check the condition of Theorem 7.18.

Let us illustrate the concept of robust stability for multiplicative uncertainty with a fairlysimple example.

7.20 Example We take as our nominal plant the model for a unit mass. Thus

R P (s) =1s2

.

The PID control law given byR C (s ) = 1 + 2 s +

1

smay be veried to stabilise the nominal plant; the Nyquist plot is shown in Figure 7.22. Tomodel the plant uncertainty, we choose

W u (s) =as

s + 1, a > 0,

which has the desirable property of not tailing off to zero as s ; plant uncertainty willgenerally increase with frequency. We shall determine for what values of a the controller R C provides robust stability for P (R P , W u ).
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-10 -5 0 5 10 15

-10

-5

0

5

10

15

Re

I m

| |

Figure 7.22 Nyquist plot for PID controller and nominal plant

According to Theorem 7.18 we should determine for which values of a the inequalityW u T L

< 1 is satised. To determine W u T L we merely need to produce magnitude

Bode plots for W u T L for various values of a , and determine for which values of a the maximummagnitude does not exceed 0dB. In Figure 7.23 is shown the magnitude Bode plot for W u T L

-1.5 - 1 -0.5 0 0.5 1 1.5 2-40

-30

-20

-10

0

log

d B

| |

-1.5 - 1 -0.5 0 0.5 1 1.5 2

-40

-30

-20

-10

0

log

d B

| |

Figure 7.23 The magnitude Bode plot for W u T L when a = 1 (left)and when a = 34 (right)

with a = 1. We see that we exceed 0dB by about 2 .5dB. Thus we should reduce a by afactor K having the property that 20 log K = 2 .5 or K 1.33. Thus a 0.75. Thus let ustake a = 3

4for which the magnitude Bode plot is produced in Figure 7.23. The magnitude

is bounded by 0dB, so we are safe, and all plants of the form

R P (s ) = 1 +34 s( s )s + 1

1s 2

will be stabilised by R C if is allowable. For example, the Nyquist plot for R P when = s1s +2 (which, it can be checked, is allowable), is shown in Figure 7.24.

7.3.2 Additive uncertainty Now let us state the analogue of Theorem 7.18 for additiveuncertainty.
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-2 0 2 4 6 8

-4

-2

0

2

4

6

Re

I m

| |

Figure 7.24 Nyquist plot for perturbed plant under multiplicative

uncertainty

7.21 Theorem Let R P , W u R (s ) be a proper rational functions with W u < , and suppose that R C R (s ) renders the nominal plant R P IBIBO stable in the interconnec-tion of Figure 7.20 . Then R C provides robust stability for P + (R P , W u ) if and only if

W u R C S L < 1, where S L is the sensitivity function for the loop gain R C R P .

Proof We adopt the notation of the rst paragraph of the proof of Theorem 7.18. Thefollowing computation, mirroring the similar one in Theorem 7.18, is the key to the proof:

W u R C S L

< 1

W u (i )R C (i )S L (i ) < 1, R

W u (i )R C (i )

1 + R L (i )< 1, R

| W u (i )R C (i )| < 1 + R L (i ) , R

| W u (i )R C (i )| < 1 R L (i ) , R .

The punchline here is thus that the condition W u R C S L < 1 is equivalent to the con-

dition that the point 1 + i0 not be contained, for any R , in the open disk of radius|W u (i )R C (i )| with centre R L (i ). This is depicted in Figure 7.25.

The remainder of the proof now very much follows that of Theorem 7.18, so we cansafely omit some details. First assume that W u R C S L < 1. For R and for R P

P + (R P , W u ) we have

R L (R P )( i ) = R C (i )R P (i ) = R L (i ) + ( i )R C (i )W u (i ).

Thus the point R L (R P )( i ) lies in the closed disk of radius |W u (i )R C (i )| with centreR L (i ). We may now simply repeat the argument of Theorem 7.18, now using Figure 7.25rather than Figure 7.21, to conclude that the closed-loop system with loop gain R L (R P ) isIBIBO stable.
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1 + i 0

|W u (i )R C (i )|

R L (i )

Figure 7.25 Interpretation of robust stability condition for addi-tive uncertainty

Now suppose that W u R C 1. Thus there exists 0 so that |W u (i)R C (i)| 1. Thus by Figure 7.25, it follows that 1 + i0 is contained in the open disk of radius |W u (i), R C (i)| with centre R L (i). As in the proof of Theorem 7.18, wemay employ Lemma 1 of that proof to show that the map from P + (R P , W u ) toD R L (i ), |W u (i )R C (i )| dened by

R P = R P + W u R C (i )R P (i )

is surjective for each > 0. In particular, it follows that there exists an allowable givingR P P + (R P , W u ) so that R L (R P )( i) = 1 + i0. IBIBO instability now follows fromTheorem 7.6.

Again, the details in the proof are far more complicated than is the essential idea. Thisessential idea is that for each R the open disk of radius |W u (i )R C (i )| and centreR L (i ) should not contain the point 1 + i0. This is what is depicted in Figure 7.25.

We also have the following, now hopefully obvious, analogue of Proposition 7.19.

7.22 Proposition Let R P , R C R (s ) with R P proper. Also, suppose that the interconnection

of Figure 7.20 is IBIBO stable with R P = R P . Let S L be the sensitivity function for the loopgain R L = R C R P . The following statements hold:

(i) if W u RH+ has the property that W u < R C S L1

then R C provides robust stability for P + (R P , W u );

(ii) for any R C S L 1there exists W u RH+ satisfying W u = so that R C does

not robustly stabilise P + (R P , W u ).

An example serves to illustrate the ideas for this section.
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22/10/2004 7.4 Summary 311

7.23 Example (Example 7.20 contd) We carry on look at the nominal plant transfer function

R P (s) =1s2

which is stabilised by the PID controller

R C (s ) = 1 + 2 s +1s .

Note that we may no longer use our W u from Example 7.20 as our plant uncertainty model.Indeed, since R C is improper, W u is proper but not strictly proper, and S L is proper but notstrictly proper (one can readily compute that this is so), it follows that W u R C S L is improper.Thus we modify W u to

W u =as

(s + 1) 2

to model the plant uncertainty. Again, our objective will be to determine the maximumvalue of a so that R C provides robust stability for P + (R P , W u ). Thus we should nd themaximum value for a so that W u R C S L

< 1. In Figure 7.26 is shown the magnitude

-1.5 -1 - 0.5 0 0.5 1 1.5 2-120

-100

-80

-60

-40

-20

0

log

d B

| |

-1.5 -1 - 0.5 0 0.5 1 1.5 2

-120

-100

-80

-60

-40

-20

0

log

d B

| |

Figure 7.26 The magnitude Bode plot for W u R C S L when a = 1(left) and when a = 12 (right)

Bode plot for W u R C S L when a = 1. From this plot we see that we ought to reduce a bya factor K having the property that 20 log K = 6 or K 2.00. Thus we take a = 12 , andin Figure 7.26 we see that with this value of a we remain below the 0dB line. Thus we areguaranteed that all plants of the form

R P (s) = R P (s ) +12 s( s)(s + 1) 2

are stabilised by R C provided that is allowable. The Nyquist plot for R P obtained bytaking = s1s +2 is shown in Figure 7.27. This is a pretty sophisticated Nyquist plot, butnonetheless, it is one for an IBIBO stable system.

7.4 Summary

In this chapter we have provided a graphical method for analysing the closed-loop stabilityof a single loop feedback interconnection. You should understand the following things.1. You need to understand what the Nyquist criterion, in the form of Theorem 7.6, is saying.
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-5 0 5 10 15 20

-10

-5

0

5

10

15

Re

I m

| |

Figure 7.27 Nyquist plot for perturbed plant under additive un-certainty

2. Drawing Nyquist plots can be a bit tricky. The essential idea is that one should focuson the image of the various parts of the Nyquist contour R,r . In particular, one shouldfocus on the image of the imaginary axis for very large and very small frequencies. Fromthere one can try to understand of something more subtle is happening.

3. The gain and phase margins are sometimes useful measures of how close a system is tobeing unstable.


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Exercises for Chapter 7 313

Exercises

E7.1 For the following rational functions R and contours C , explicitly verify the Principleof the Argument by determining the number of encirclements of the origin by theimage of C under R .(a) R (s ) = 1s 2 and C = e

i < .(b) R (s ) = s and C = ei < .(c) R (s ) = ss 2 +1 and C = 2e

i < .E7.2 Let F (s ) = ln s , and consider the contour C = ei < . Does the

Principle of the Argument hold for the image of C under F ? Why or why not?E7.3 For the following loop gains,

(a) R L (s) = s 2 (s +1) , > 0,(b) R L (s) = s (s1) ,

> 0, and

(c) R L (s) = (s +1)s (s1) , > 0,

do the following.

1. Determine the Nyquist contour for the following loop gains which depend on aparameter satisfying the given conditions. Although the plots you produce maybe computer generated, you should make sure you provide analytical explanationsfor the essential features of the plots as they vary with .

2. Draw the unity gain feedback block diagram which has R L as the loop gain.3. For the three loop gains, use the Nyquist criterion to determine conditions on

for which the closed-loop system is IBIBO stable.4. Determine IBIBO stability of the three closed-loop systems using the

Routh/Hurwitz criterion, and demonstrate that it agrees with your conclusionsusing the Nyquist criterion.

In this next exercise you will investigate some simple ways of determining how to close aNyquist contour for loop gains with poles on the imaginary axis.

E7.4 Let R L R (s) be a proper loop gain with a pole at i0 of multiplicity k. Forconcreteness, suppose that 0 0. If 0 = 0, this implies that i0 is also a pole of multiplicity k. Thus we write

R L (s) =1

s k R (s ), 0 = 01

(s 2 + 20 )kR (s ), 0 = 0 ,

where i0 is neither a pole nor a zero for R .(a) Show that

limr 0

R L (i0 + re i ) =ar k

eik + O(r 1k ).

where a is either purely real or purely imaginary. Give the expression for a .(b) For 0 = 0 show that

lim0 ,

R L (i ) 2 , 0,2 , .

Denote 0 = lim 0 , R L (i ).


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(c) Determine a relationship between 0, and k and a .(d) Determine the relationship between

lim0 , +

R L (i )

and 0. This relationship will depend on k .(e) Conclude that for a real rational loop gain, the closing of the Nyquist contour

is always in the clockwise direction and the closing arc subtends an angle of k .E7.5 Let (N, D ) be a proper SISO linear system in input/output form, and let n = deg( D )

and m = deg( N ).(a) Determine lim T N,D (i ) and lim T N,D (i ).(b) Comment on part (a) as it bears on the Nyquist plot for the system.

E7.6 Consider the SISO linear system ( N (s), D (s )) = (1 , s 2 + 1) in input/output form.(a) Sketch the Nyquist contour for the system, noting that the presence of imaginary

axis poles means that the ( , r )-Nyquist contour is not bounded as r 0.Consider the SISO linear system ( N (s), D (s )) = (1 , s 2 + s + 1) which now has a

bounded Nyquist contour for > 0.(b) Show, using the computer, that the Nyquist contour for ( N , D ) approaches that

for the system of part (a) as 0.E7.7 Let (N, D ) be a SISO linear system in input/output form with deg( N ) = 0 and let

= max { Im( p) | p is a root of D } .

Answer the following questions.(a) Show that |T D,N (i )| is a strictly decreasing function of for || > .(b) Comment on part (a) as it bears on the Nyquist plot for the system.

E7.8 Formulate and prove a condition for IBIBO stability for each of the two interconnectedsystems in Figure E7.1, using the Principle of the Argument along the lines of the

r (s ) R C (s ) R P (s ) y(s )

R S (s )

r (s ) R I (s ) R C (s ) R P (s ) y(s )

Figure E7.1 Alternate congurations for Nyquist criterion

Nyquist criterion of Theorem 7.6.
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Exercises for Chapter 7 315

E7.9 LetR P (s ) =

20s 2 + 2 0s

, R C (s ) = 1 .

Plot the upper phase margin of the closed-loop system as a function of .

In the next exercise you will investigate gain and phase margins for a simple plant andcontroller. You will see that it is possible to have one stability margin be large while the

other is small. This exercise is taken from the text of [Zhou 1996].E7.10 Take

R P (s) =a s

as 1, R C (s) =

s + bbs + 1

,

where a > 1 and b > 0. Consider these in the standard unity gain feedback loop.(a) Use the Routh/Hurwitz criterion to show that the closed-loop system is IBIBO

stable if b( 1a , a ).Take b = 1.(b) Show that K min = 1a , K max = a , min is undened, and max = arcsin

a 2 1a 2 +1 .(c) Comment on the nature of the stability margins in this case.Fix a and take b( 1a , a ).(d) Show that

limba

K min =1a2

, limba

K max = a2, limba

max = 0 .

(e) Comment on the stability margins in the previous case.(f) Show that

limb1a

K min = 1 , limb1a

K max = 1 , limb1a

max = 2 arcsina 2 1a2 + 1

.

(g) Comment on the stability margins in the previous case.In this chapter we have used the Nyquist criterion to assess IBIBO stability of input/outputfeedback systems. It is also possible to use the Nyquist criterion to assess stability of staticstate feedback, and the following exercise indicates how this is done.

E7.11 Let = ( A , b, c t , D ) be a SISO linear system.(a) Let f R n . Show that if is controllable then the closed-loop system f is

internally asymptotically stable if and only if

f t (sI n A )1b1 + f t (s I n A )1b

RH+.

Thus, we may consider closed-loop stability under static state feedback to be equiva-lent to IBIBO stability of the interconnection in Figure E7.2. The Nyquist criterioncan then be applied, as indicated in the following problem.(b) Using part (a), prove the following result.

Proposition Let = ( A , b, c t , 01) be a controllable SISO linear system and let n pbe the number of eigenvalues of A in C + . For a state feedback vector f R ndene the loop gain R f (s ) = f t (sI n A )1b. Assuming that (A , f ) is observable,the following statements are equivalent:
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r (s ) f t (s I n A ) 1 b y (s )

Figure E7.2 A feedback interconnection for static state feedback

(i) the matrix A bf t is Hurwitz;(ii) there exists r 0, R 0 > 0 so that the image of R,r under R f encircles the point

1 + i0 n p times in the clockwise direction for every r < r 0 and R > R 0.

To be concrete, consider the situation where

A = 0 1 1 1 , b =01 .

For this system, answer the following questions.(c) Show that if we take f = (0 , 2) then A bf t is Hurwitz. Verify that the result

of part (a) of the problem holds.(d) How many eigenvalues are there for A in C + ?(e) Plot the Nyquist contour for Figure E7.2 and verify that the Nyquist criterion of

part (b) holds.E7.12 For the plant uncertainty set

R P =R P

1 + W u R P , 1,

use the idea demonstrated in the proofs of Theorems 7.18 and 7.21 to state and provea theorem providing conditions for robust stability.

E7.13 For the plant uncertainty set

R P =R P

1 + W u, 1,

use the idea demonstrated in the proofs of Theorems 7.18 and 7.21 to state and provea theorem providing conditions for robust stability.

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