ure 5.13 shows unit-energy excitations for
5 that produce square root output
t
energies for
5 close to the Hankel norms of (a)
(b)
13 and 13 .
t
H
H
3
3
2
2
( )
z
t
( )
@
@
R
1
( )
z
t
1
w
t
;
;
@
@
R
0
0
1
1
;
;
;
;
( )
w
t
2
2
;
0
2
4
6
8
10 ; 0
2
4
6
8
10
t
t
(a)
(b)
(a) shows a unit-energy input signal that is zero for
5,
Figure
5.13
w
t
together with the output when (a)
13 is driven by . The square root
z
H
w
energy in for
5 is close to
(a)
13 hankel = 1 07. (b) shows a unit-
z
t
kH
k
:
energy input signal that is zero for
5, together with the output
w
t
z
when (b)
13 is driven by . The square root energy in for
5 is close to
H
w
z
t
(b)
13 hankel = 2 04.
kH
k
:
5.2.10
Example 2: the Gain of an Amplifier Circuit
Consider the band pass lter circuit shown in gure 5.14. We will assume that the
opamp saturates at 14V. The input of the circuit (which is produced by another
opamp) is no larger than 14V (i.e.,
14). We ask the question: can this
kw
k
1
lter saturate?
Assuming the opamp does not saturate, the transfer function from to is
w
z
( ) =
2 104
;
s=
H
s
( 104 + 1)2
(5.19)
s=
The maximum magnitude of this transfer function is 1.0 (
= 1), so, provided
kH
k
1
the opamp does not saturate, the RMS value of the lter output does not exceed
the RMS value of the lter input. It is tempting to conclude that the opamp in the
lter will not saturate.
This conclusion is wrong, however. The peak gain of the transfer function H
is
pk gn = 1 47, so there are inputs bounded by, say, 10V that will drive the
kH
k
:
opamp into saturation. Figure 5.15 gives an example of such an input signal, and
the corresponding output that would be produced if the opamp did not saturate.
Since it exceeds 14V, the real lter will saturate with this input signal.
5.2 NORMS OF SISO LTI SYSTEMS
109
5nF
q
q
10nF 10K
20K
q
w
;
q
z
+
A bandpass lter circuit. The amplier has very large open-
Figure
5.14
loop gain. The output clips at 14 , and the input lies between 14 .
V
V
When the circuit is operating linearly the transfer function from to is
w
z
given by (5.19).
15
10
@
I
@
( )
w
t
5
V
0
@
I
;5
@
( )
z
t
;10
;15
0
0:2
0:4
0:6
0:8
1
1:2
1:4
1:6
(ms)
t
If the circuit shown in gure 5.14 did not saturate, the input
Figure
5.15
shown would produce the output . Even though
= 10V, we have
w
z
kw
k
1
= 14 7V. Thus the input will drive the real circuit in gure 5.14
kz
k
:
w
1
into saturation. Of course,
rms
rms, since
rms gn = 1.
kz
k
kw
k
kH
k
110