Biomedical Engineering Reference
In-Depth Information
Impulse Response
The complete response of the system,
‥
, to an impulse with zero
initial conditions is given in Table 13.6. Let us consider estimating z and o
n
þ
a
y
þ
a
y
¼
d
ð
t
Þ
a
y
2
1
0
for the over-
damped case. Here, the solution is
1
e
s
1
t
e
s
2
t
y
ð
t
Þ¼
p
z
2
ð
13
:
77
Þ
2o
n
1
p
z
2
where
s
1
,
2
¼
zo
n
o
n
1
:
As before, we use the time to peak overshoot,
T
p
, to find
the parameter values. The time to peak overshoot is found by first calculating
@
y
@
t
¼
1
e
s
1
t
s
e
s
2
t
p
z
2
ð
13
:
78
Þ
s
1
2
2o
n
1
from
@
y
@
t
and then determining
T
p
t
¼
T
p
¼
0
. This gives
s
2
s
1
ln
T
p
¼
ð
13
:
79
Þ
s
1
s
2
In this situation, the time to peak overshoot is easy to estimate from the data.
If one of the time constants is considerably larger than the other, then the larger or
dominant time constant can be determined from the data by locating when the response
is less than 1 percent of the initial value. For example, consider the impulse response
y
ð
t
Þ¼
e
4
t
) goes
to zero by approximately 1 s, well before the response of the dominant time constant (
e
t
e
4
t
as shown in Figure 13.68. The response of the smaller time constant (-
e
t
).
So the impulse response is essentially equal to the response of the dominant time constant
(
e
t
) for
1 s. To determine the dominant time constant, locate the time at which the
response for the dominant time is at 1 percent of the value at
t
>
0 s; that time is 5t.To
estimate the initial value of the response for the dominant time constant at
t
¼
0, project
the exponential back, as shown in Figure 13.68, to give a value of 1, then take 1 percent
of that value to find 5t. The dominant time constant is then easily determined. Thus, given
T
p
and one of the time constants, the smaller time constant can then be determined from
Eq. 13.79 and using MATLAB's “solve” command, from which z and o
n
are calculated
(i.e., o
2
t
¼
¼
s
þ
s
1
2
n
¼
s
s
2
and z
).
1
2o
n
TABLE 13.6
Impulse Response for a Second-Order System
Damping
Natural Response Equation
1
e
s
1
t
e
s
t
2
Overdamped
y
ð
t
Þ¼
p
z
2
2o
n
1
o
n
e
o
n
z
t
1
Underdamped
y
ð
t
Þ¼
p
sin o
d
ðÞ
z
2
y
ð
t
Þ¼
te
o
n
t
Critically damped
