How These Figures Were Computed
Most level curves of convex functions were plotted using a radial bisection method. Con-
sider the problem of plotting the ( ) = level curve on slice, where is convex. Assume
H
H
that we know some point ( 0 0) inside this level curve, i.e.
;
0 (a) + 0 (b) + (1
0
0) (c)
H
H
;
;
H
<
:
The value of along the radial line segment
= 0 + cos( )
= 0 + sin( )
(11.28)
where
0, is
( ) = ;( 0 + cos( )); (a)
(c) + ( 0 + sin( )); (b)
(c) + (c)
'
H
;
H
H
;
H
H
:
For each ,
is a convex function from + to with (0)
, so there is no more
'
R
R
'
<
than one
0 for which
( ) =
(11.29)
'
:
(11.29) can be solved using a number of standard methods, such as bisection or regula
falsi, with only an evaluation of required at each iteration. As sweeps out the angles
'
0
2 , the solution to (11.29), together with (11.28), sweeps out the desired level
<
curve. This method was used for gures 11.4, 11.5, 11.12, 11.13, 11.17, 11.19, 11.20, 11.21,
and 11.23.
The above method also applies to quasiconvex functionals with the following modication:
in place of (11.29) we need to nd the largest for which ( )
.
'
In certain cases (11.29), or its quasiconvex modication, can be solved directly. For ex-
ample, consider the quasiconvex settling-time functional settle from section 11.1.1. For a
xed , the step response along the line (11.28) is of the form
0( ) + 1( )
s
t
s
t
where 0 is the step response of
s
0 (a)
(b)
(c)
13 + 0 13 + (1
0
0) 13
H
H
;
;
H
and 1 is the step response of
s
cos( ) (a)
(b)
(c)
13 + sin( ) 13 + ( cos( ) sin( )) 13
H
H
;
;
H
:
The largest value of for which ( )
max is given by
'
T
=
max
0 95
:
0( ) + 1( ) 1 05 for
max
:
s
t
s
t
:
t
T
NOTES AND REFERENCES
271
This is a linear program in the scalar variable that can be directly solved.
A similar method was used to produce gures 11.8 and 11.10. A similar method could
also be used to produce level curves of
norms at each frequency a quadratic in
H
!
1
can be solved to nd the positive value of that makes the frequency response magnitude
tight at . Taking the minimum over all gives the desired .
!
!
Figures 11.9 and 11.11 were plotted by directly computing the equations of the level
curves using a state-space method. For example, consider 12 22, which is one term in
kH
k
the functional rms yp. Since
(a)
(b)
(c)
(c)
(a)
(c)
(a)
(c)
12 +
12 + (1
) 12 =
+ ;
+ ;
H
H
;
;
H
H
H
;
H
H
;
H
is ane in and , it has a state-space realization
(
) 1
;
( 0 + 1 + 2)
C
sI
;
A
B
B
B
where is a row vector, and 0, 1 and 2 are column vectors. From section 5.6.1, if
C
B
B
B
obs is the solution to the Lyapunov equation
W
T
obs + obs + T = 0
A
W
W
A
C
C
we have
"
1 #
(a)
(b)
(c) 2
12 +
12 + (1
) 12 = 1
H
H
;
;
H
2
E
where
2
3
T
0
= BT
obs
0
1
2
4
1 5
B
B
B
E
W
B
T
2
B
is a positive denite 3 3 matrix. The level curves of 12 2 on slice are therefore
kH
k
H
ellipses.
The convex sets shown in gures 11.16 and 11.18 were produced using the standard radial
method described above. The exact contour along which the gain margin specication
was tight was also found by a radial method. However, since this specication need not
be convex, a ne grid search was also used to verify that the entire set had been correctly
determined.
The step response sensitivity plot in gure 11.15 was produced by forming an indenite
quadratic in and . The sensitivity, from (11.13), is the step response of
1
(a)
(b)
(c)
(a)
(b)
(c)
13 +
13 + (1
) 13
13 +
13 + (1
) 13
;
H
H
;
;
H
H
H
;
;
H
at = 1. After expansion, the step response of each term, at = 1, gives each of the
t
t
coecients in an indenite quadratic form in and . Figure 11.15 shows the , for
272