(from (12.43), h does not contain any impulse at t = 0). We can express the
interpolation conditions in terms of h as
Z
1
h(t)e t dt = 0
(12.44)
;
0
Z
1
h(t)e 2t dt = (1 + 2T
1
(12.45)
;
trk);
0
h(0) = T 1
;
trk :
(12.46)
We will solve the feasibility problem by solving the optimization problem
Z
min
1
h(t) dt:
(12.47)
subject to (12.44{12.46) 0 j j
In chapters 13{15 we will describe general numerical methods for solving an in nite-
dimensional convex optimization problem such as (12.47) here we will use some
speci c features to analytically determine the solution. We will rst guess a solution,
based on some informal reasoning, and then prove, using only simple arguments,
that our guess is correct.
We rst note that the third constraint, on h(0), should not a ect the minimum,
since we can always adjust h very near t = 0 to satisfy this constraint, without
a ecting the other two constraints, and only slightly changing the objective. It
can be shown that the value of the minimum does not change if we ignore this
constraint, so henceforth we will.
Now we consider the two integral constraints. From the second, we see that h(t)
will need to be positive over some time interval, and from the rst we see that h(t)
will also have to be negative over some other time interval. Since the integrand
e 2t falls o more rapidly than e t, it seems that the optimal h(t) should rst be
;
;
positive, and later negative, to take advantage of these di ering decay rates. Similar
reasoning nally leads us to guess that a nearly optimal h should satisfy
h(t)
(t)
(t T)
(12.48)
;
;
where and are positive, and T is some appropriate time lapse. The objective is
then approximately + .
Given this form, we readily determine that the optimal , , and T are given
by
=
1
p
1 + 2T
1 + 1 2
(12.49)
trk
2
=
1
p
1 + 2T
2 + 3 2
(12.50)
trk
2
T = log(1 + p2)
(12.51)
12.5 A WEIGHTED PEAK TRACKING ERROR EXAMPLE
289
which corresponds to an objective of
1
1 + 2T
3 + 2p2 :
(12.52)
trk
Our guess that the value of (12.47) is given by (12.52) is correct. To verify this,
we consider : +
given by
R
!
R
(t) = (2 + 2p2)e t
2t
;
+ (3 + 2p2)e;
(12.53)
;
and plotted in gure 12.5. This function has a maximum magnitude of one, i.e.,
= 1.
k
k
1
1:5
1
0:5
(t)
0
;0:5
;1
;1:5
0
0:5
1
1:5
t2
2:5
3
3:5
4
The function ( ) from (12.53).
Figure
12.5
t
Now suppose that h satis es the two integral equality constraints in (12.47).
Then by linearity we must have
Z
1
h(t) (t)dt =
1
3 + 2p2 :
(12.54)
0
1 + 2Ttrk
Since (t) 1 for all t, we have
j
j
Z
1
h(t) (t)dt Z 1 h(t) dt = h 1:
(12.55)
0
0 j
j
k
k
Combining (12.54) and (12.55), we see that for any h,
Z
Z
1
h(t)e t dt
1
h
2t dt
1
;
= 0 and
(t)e;
= (1 + 2Ttrk);
0
0
290