Linear Controller Design: Limits of Performance by Stephen Boyd and Craig Barratt - HTML preview

CHAPTER 13 ELEMENTS OF CONVEX ANALYSIS

mag !0 at H0 is a subgradient of at H0. But since H0(j!0) = 0, this functional

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is di erentiable at H0, with derivative

sg(H) = 1(H H

0)

0(j!0)H(j!0) :

<

This linear functional is a subgradient of at H0. The reader can directly verify

that the subgradient inequality (13.3) holds.

13.4.5

Norm of a Transfer Matrix

H

1

Now suppose that H is an m p transfer matrix, and is the

norm:

H

1

(H) = H :

k

k

1

We will express directly as the maximum of a set of linear functionals, as follows.

For each !

, u

m, and v

p, we de ne the linear functional

2

R

2

C

2

C

u v !(H) = (u H(j!)v):

<

Then we have

(H) = sup u v !(H) !

u = v = 1

f

j

2

R

k

k

k

k

g

using the fact that for any matrix A

m p,

2

C

max(A) = sup (u Av)

u = v = 1 :

f<

j

k

k

k

k

g

Now we can determine a subgradient of at the transfer matrix H0. We pick

any frequency !0

at which the

norm of H0 is achieved, i.e.

2

R

H

1

max(H0(j!0)) = H0 :

k

k

1

(Again, we ignore the case where there is no such !0, commenting that for rational

H0, there always is such a frequency, if we allow !0 = .) We now compute a

1

singular value decomposition of H0(j!0):

H0(j!0) = U V :

Let u0 be the rst column of U, and let v0 be the rst column of V . A subgradient

of at H0 is given by the linear functional

sg = u0 v0 !0:

13.4 COMPUTING SUBGRADIENTS

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13.4.6

Peak Gain

We consider the peak gain functional

Z

(H) = H

1

pk gn =

h(t) dt:

k

k

0 j

j

In this case our functional is an integral of a family of convex functionals. We will

guess a subgradient of at the transfer function H0, reasoning by analogy with

the sum rule above, and then verify that our guess is indeed a subgradient. The

technique of the next section shows an alternate method by which we could derive

a subgradient of the peak gain functional.

Let h0 denote the impulse response of H0. For each t 0 we de ne the functional

that gives the absolute value of the impulse response of the argument at time t:

abs h t(H) = h(t) :

j

j

These functionals are convex, and we can express as

Z

(H) = 1 abs h t(H)dt:

0

If we think of this integral as a generalized sum, then from our sum rule we might

suspect that the linear functional

sg

Z

(H) = 1 sg t(H)dt

0

is a subgradient for , where for each t, sg t is a subgradient of abs h t at H0.

Now, these functionals are di erentiable for those t such that h0(t) = 0, and 0 is a

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subgradient of abs h t at H0 for those t such that h0(t) = 0. Hence a speci c choice

for our guess is

sg

Z

(H) = 1 sgn(h0(t))h(t)dt:

0

We will verify that this is a subgradient of at H0.

For each t and any h we have h(t) sgn(h0(t))h(t) hence

j

j

Z

(H) = 1 h(t) dt Z 1 sgn(h0(t))h(t)dt:

0 j

j

0

This can be rewritten as

Z

(H)

1

( h0(t) + sgn(h0(t))(h(t) h0(t))) dt = (H0) + sg(H H0):

0

j

j

;

;

This veri es that sg is a subgradient of at H0.

306