Lecture 9 - MAT UPC · Lecture 9 Design for stability margin and for robust performance This lecture is based on chapters 11 and 12 of Doyle-Francis-Tannenbaum. 17013 — IOC-UPC,

Lecture 9Design for stability margin and for robust

performance

This lecture is based on chapters 11 and 12 of Doyle-Francis-Tannenbaum.

17013 — IOC-UPC, Lecture 9, December 14th 2005 – p. 1/25

Optimal robust stability (I)• Remember the disk multiplicative uncertainty model. We modify it slightly so that the set

of plants is now given by

Pǫ = {P̃ = (1 + ∆W2)P, ||∆||∞ ≤ ǫ},

whereP is the nominal plant and no unstable pole ofP is cancelled when forming̃P . In

Lecture 3 we tookǫ = 1.

• Let ǫsup be the least upper bound onǫ such that someC stabilizes every plant inPǫ, so

ǫsup is the maximum stability margin for this model of uncertainty.

• The key result in Lecture 3 was that to achieve robust stability for this model

||W2T ||∞ <1

ǫ.

• Define

γinf = infC

||W2T ||∞,

where the infimum is taken over all internally stabilizing controllers. Then

ǫsup = γ−1inf .


Optimal robust stability (II)• Computingγinf reduces to a model-matching problem. Indeed, using the Youla-Kucera

parametrization forC, one gets

γinf = infQ∈Q

||W2N(X + MQ)||∞.

• This is similar to the model-matching problem of Lecture 7:

γopt = minQim stable

||T1 − T2Qim||∞,

with T1 = W2NX, T2 = −W2NM .

• So thatT2 has no zeros on the imaginary axis, we will assume thatP has neither poles

nor zeros on the imaginary axis, and thatW2 has no zeros there.

• The key difference between the two problems is thatQ must be stable and proper.

However, it can be shown thatγinf = γopt.

• The way to solve the problem is very much like the performancedesign problem of the

previous lecture, using theJτ functions to take away the improperness ofQim.


Optimal robust stability (III)Procedure to solve the robust stability problem. Input: P , W2.

Step 1. Do a coprime factorization ofP overQ.

Step 2. Solve the model-matching problem forT1 = W2NX,

T2 = −W2NM . Let Qim denote its solution, achievingγopt.

Thenǫsup = 1/γopt.

Step 3. Letǫ < ǫsup. ChooseJτ such thatQimJτ is proper andτ

is small enough that

||W2N(X + MQimJτ )||∞ <1

ǫ.

Step 4. SetQ = QimJτ , C = (X + MQ)/(Y − NQ).


Optimal robust stability (IV)Example.

• Consider the plant

P (s) =s − 1

(s + 1)(s − p), 0 < p 6= 1,

with an unstable pole ats = p and an unstable zero ats = 1.

• Suppose that the uncertainty weight is the high-pass function

W2(s) =s + 0.1

s + 1,

so that|P̃ /P − 1| ∼ 0.1ǫ at low frequencies and|P̃ /P − 1| ∼ 1ǫ at high frequencies.

• The coprime factorization ofP yields

N(s) =s − 1

(s + 1)2, M(s) =

s − p

s + 1, X(s) =

(p + 1)2

p − 1, Y (s) =

s − p+3p−1

s + 1.

Notice thatX is just a constant.


Optimal robust stability (V)• Now T2 is

T2(s) = −W2(s)N(s)M(s) = − s + 0.1

s + 1

(s − 1)(s − p)

(s + 1)3

which has 2 unstable zeros. In order to make the model-matching part of the procedure

easier, we can decomposeN asN = NapNmp with

Nap(s) =s − 1

s + 1, Nmp(s) =

1

s + 1

and keep onlyNmp, since the all-pass part does not contribute to the relevant∞-norm:

||W2N(X + MQ)||∞ = ||W2Nmp(X + MQ)||∞.

The (modified) model-matching problem has data

T1(s) =(p + 1)2(s + 0.1)

(p − 1)(s + 1)2, T2(s) = − (s + 0.1)(s − p)

(s + 1)3.


Optimal robust stability (VI)• Since now the only unstable zero ofT2 is s = p, the solution of the model-matching

problem is

Qim(s) =T1(s) − T1(p)

T2(s)

and the optimal error is

γopt = |T1(p)| =

˛

˛

˛

˛

p + 0.1

p − 1

˛

˛

˛

˛

Thus the maximum unstability margin is

ǫsup =

˛

˛

˛

˛

p − 1

p + 0.1

˛

˛

˛

˛

and goes to zero asp aproaches1. Hence less and less uncertainty can be tolerated as the

unstable pole and zero of the plant approach each other. As wealready know from

Lecture 5, this is a general fact.

• To proceed, let’s takep = 0.5, for whichǫsup = 0.8333 and

Qim(s) = −1.2(s + 1)(s − 1.25)

s + 0.1.


Optimal robust stability (VII)• We arbitrarily setǫ = 0.8 and, since the relative degree ofQim is −1,

Jτ (s) =1

τs + 1.

• We try several values ofτ and forτ = 0.01 we get

||W2N(X + MQimJ0.01||∞ = 1.2396 < 1.25 =1

ǫ.

• Finally

Q(s) = −1.2(s + 1)(s − 1.25)

(s + 0.1)(0.01s + 1)

and

C(s) = − (s + 1)(124.5s2 + 240.45s + 120)

s3 + 227.1s2 + 440.7s + 220.


Optimal robust stability (VIII)• We can try the above controller for some plants inPǫ. Let’s consider plants of the form

P̃ = ks − a

(s + 1)(s − 0.5).

• It can be shown that ifk anda satisfy(ka − 1)2 + k2(ω − 1)2 < 0.64(ω2 + 1) for all

ω, thenP̃ belongs toP0.8. This essentially requires|k| < 0.8.

• Let us takek = 0.6 anda = 2. The response to a square periodic pulse of the nominal

plant (upper) and of the perturbed one (lower) is displayed below, and both are clearly

bounded.

0 10 20 30 40 50 60 70 80 90 100−2

−1

0

1

2

3

4

5

6

7

8


Gain margin optimization (I)• We now turn to a different model of plant uncertainty, the gain uncertainty model, which

we have already encountered and which is commonly used in elemental control theory.

• To be precise, the set of plants is given by

P = {P̃ = kP, 1 ≤ k ≤ k1}

for a givenk1. This corresponds toW2(s) = k1 − 1 in the disk multiplicative model,

but more precise results can be given for this particular case.

• Let ksup be the supremum value of thek1 such that a controller exists achieving internal

instability for the set of plants. We will present a formula for ksup, assuming thatP has

neither poles nor zeros on the imaginary axis.

• Define the infimum norm of the (unweighted) complementary sensitivity function

γinf = infC

||T ||∞.

• Lemma. One has thatγinf = 0 if P is stable,γinf = 1 if P is unstable but minimum

phase, andγinf > 1 if P is unstable but non-minimum phase.


Gain margin optimization (II)Theorem 1. If P is stable or minimum phase, thenksup = ∞.

Otherwise

ksup =

(

γinf + 1

γinf − 1

)2

.

The fact thatksup = ∞ or ortherwise does not mean that a single controller will internaly

stabilize all the plants withk < ksup; however, givenk1 < ksup, a controller can be computed

which works for all thek ≤ k1.

The proof of the theorem is fairly involved, and uses some conformal mapping theory. From the

proof, a method can be extracted to compute the controller. For P unstable and non-minimum

phase (and with no imaginary axis poles or zeros) it goes likethis:

Step 1. Do a coprime factorization ofP .

Step 2. Solve the model-matching problem forT1 = NX, T2 = −NM . Let Qim denote its

solution and letγopt denote the minimum model-matching error. Then

ksup =

„

γopt + 1

γopt − 1

«2

.


Gain margin optimization (III)Step 3. Letk1 be arbitrary with1 < k1 < ksup. SetJτ (s) = 1

(τs+1)nwith n large enough so

thatQimJτ is proper andτ small enough so that

||N(X + MQimJτ )||∞ <

√k1 + 1√k1 − 1

.

Step 4. Set

K = N(X + MQimJτ ),

G =1 −

√k1

1 +√

k1K,

T =1

k1 − 1

„

1 − G

1 + G

«2

− 1

!

,

Q =T − NX

NM.

Step 5. GetC using Youla-Kucera and the aboveQ.


Gain margin optimization (IV)As an example, let us return to the plant

P (s) =s − 1

(s + 1)(s − p), 0 < p 6= 1

and study now the gain margin problem,P̃ = kP .

The coprime decomposition yields

N(s) =s − 1

(s + 1)2, M(s) =

s − p

s + 1, X(s) =

(p + 1)2

p − 1, Y (s) =

s − (p + 3)/(p − 1)

s + 1.

Let us factorN asN = NapNmp with

Nap(s) =s − 1

s + 1, Nmp(s) =

1

s + 1

and consider the equivalent model-matching problem with

T1 = NmpX =(p + 1)2

(p − 1)(s + 1), T2 = −NmpM = − s − p

(s + 1)2.


Gain margin optimization (V)We get

γopt = |T1(p)| =

˛

˛

˛

˛

p + 1

p − 1

˛

˛

˛

˛

and

ksup =

„

p + 1 + |p − 1|p + 1 − |p − 1|

«2

=

8

<

:

p2 p ≥ 1

1/p2 p < 1

which, as in the case ofǫsup for the disk multiplicative model, as a minimum atp = 1.

Let us take arbitrarilyp = 2, for whichksup = 4. The model-matching problem yields, sinceT2

has a single unstable zero,

Qim =T1 − T1(p)

T2= 3(s + 1).

Again arbitrarily, let us setk1 = 3.5. With Jτ (s) = 1/(τs + 1), the valueτ = 0.01 yields

||N(X + MQimJ0.01||∞ = 3.0827 <

√k1 + 1√k1 − 1

= 3.2967.

Finally K(s), G(s), T (s) andQ(s) could be computed, and thenC(s) from Youla-Kucera.


Phase margin optimization• Now the set of plants to be stabilized is

P ={

P̃ = e−jθP,−θ1 ≤ θ ≤ θ1

}

,

whereP is the nominal plant andθ1 ∈ (0, π].• Let θsup denote the supremumθ1 such that a stabilizing controller for the set does exist.

As in the preceeding section, letγinf = infC ||T ||∞.

• Under the assumption thatP has neither poles nor zeros on the imaginary axis, we have

the

Theorem 2. If P is stable or minimum phase, thenθsup = π.

Otherwise

θsup = 2 arcsin1

γinf.


The modified problem (I)• Remember that the robust performance problem is to design

a (proper) controller so that the feedback system for the

nominal plant is internally stable and the inequality (RPT)

|| |W1S| + |W2T | ||∞ < 1

holds.

• As stated, the problem has not been solved in general, so

we look for a nearby problem that is solvable.

• Fix a frequency and letx = |W1S|, y = |W2T |. Then

x2 + y2 <1

2⇒ x + y < 1.


The modified problem (II)• Thus, a sufficient condition for the RPT to hold is the

modified robust performance test(MRPT)

|| |W1S|2 + |W2T |2 ||∞ <1

2.

• Notice that it is entirely possible for the MRPT to have no

solution and yet the RPT be solvable.

• We will try to solve the MTPT under the following

simplifying assumptions

• P is strictly proper and has neither poles nor zeros on

the imaginary axis.

• W1 andW2 are stable and proper, and have no common

zeros on the imaginary axis.


Spectral factorization (I)• For a rational functionF (s) with real coefficients, let

F (s) = F (−s). This is the complex conjugate value when

s = jω. Thus

F (jω) = F (−jω) = F (jω).

• We saw that, ifF ∈ Q, it has a factorization of the form

F = FapFmp. The all-pass factor has the property

F ap(s)Fap(s) = 1.

• WhenF = F andF has no zeros or poles on the imaginary

axis there is a related factorization, calledspectralfactorization.


Spectral factorization (II)• If F = F and no poles or zeros on the imaginary axis, we

can write

F (s) = cF1(s), F1(s) =

∏

(zi − s)(zi + s)∏

(pi − s)(pi + s),

where{zi} and{pi} are the right half-plane zeros and

poles. Note thatF1(0) > 0, since theF is real-rational and

complex zeros or poles must appear in conjugate pairs.

• FromF1 form a functionG by selecting the poles and zeros

in ℜ(s) < 0:

G(s) =

∏

(zi + s)∏

(pi + s).


Spectral factorization (III)• With thisG we have

F (s) = G(s)cG(s), with G, G−1 stable.

• Finally, if c > 0, we define anspectral factor of F , Fsf, as

Fsf(s) =√

c

∏

(zi + s)∏

(pi + s).

Note thatc > 0 iff F (0) > 0.

• Hence we have aspectral factorization (it is not unique)

F = F sfFsf, with Fsf andF−1sf stable.


Solution of the MRPT (I)• The modified RPT can be transformed into a

model-matching problem using an spectral factorization.

• In terms of the Youla-Kucera parametrization, the MRPT is

|| |W1M(Y − NQ)|2 + |W2N(X + MQ)|2 ||∞ <1

2. (1)

• SettingR1 = W1MY , R2 = W1MN , S1 = W2NX,

S2 = −W2MN , (1) becomes

|| |R1 − R2Q|2 + |S1 − S2Q|2 ||∞ <1

2. (2)


Solution of the MRPT (II)• The first key step is to findU1, U2 ∈ Q, andU3 real rational

and satisfyingU3 = U3 such that (2) becomes

|| |U1 − U2Q|2 + U3 ||∞ <1

2. (3)

• The second key step is to introduceU4, a spectral factor of12− U3. Then (2) can be written as

||U−14 U1 − U−1

4 U2Q||∞ < 1. (4)

This is already an standard model-matching problem, and

the, by now, well known machinery can be started.


Solution of the MRPT (III)The whole procedure can be partitioned as follows. First goes a

routine to computeU1 andU2.

Procedure A.GivenR1, R2, S1, S2,

Step A1. SetF = R2R2 + S2S2.

Step A2. Compute a spectral factorFsf of F .

Step A3. Choose an all-pass funtionV such that

R2R1 + S2S1

FsfV ∈ Q.

Step A4. Set

U1 =R2R1 + S2S1

FsfV, U2 = FsfV.


Solution of the MRPT (IV)The main procedure is as follows.

Procedure.GivenP , W1, W2,

Step 1. Compute

U3 =W1W1W2W2

W1W1 + W2W2

.

Check if||U3||∞ < 1/2. If not, the problem is not solvable;

exit.

Step 2. Do a coprime factorization ofP . GetN , M , X andY .

Step 3. SetR1 = W1MY , R2 = W1MN , S1 = W2NX,

S2 = −W2MN .

Step 4. Apply Procedure A to getU1 andU2.


Solution of the MRPT (V)Step 5. Compute a spectral factor,U4, of 1

2− U3 (this is

guaranteed to exist by Step 1).

Step 6. SetT1 = U−14 U1, T2 = U−1

4 U2.

Step 7. Computeγopt for the model-matching problem of Step

6. If γopt < 1 continue; otherwise the MRPT is not

solvable; exit.

Step 8. ComputeQ, the solution to the above model-matching

problem. IfQ is not proper, roll it off at high frequency

while maintaining||T1 − T2Q||∞ < 1.

Step 9. GetC = (X + MQ)/(Y − NQ).


Documents

Lecture 9 - MAT UPC · Lecture 9 Design for stability margin and for robust performance This lecture is based on chapters 11 and 12 of Doyle-Francis-Tannenbaum. 17013 — IOC-UPC,