mag !0 at H0 is a subgradient of at H0. But since H0(j!0) = 0, this functional
6
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
305
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
6
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