18.03 Class 29, Apr 24
Laplace Transform IV:
The pole diagram
[1] I introduced the weight function = unit impulse response with the
mantra
that you know a system by how it responds, so if you let it respond to
the
simplest possible signal (with the simplest possible initial conditions)
then you should be able to determine the system parameters.
How?
Well, take the equation
p(D) w
=
delta(t)
(with rest initial
conditions)
and apply LT to it (in the original form, so
F[f'(t)] = sF(s)):
p(s) W(s)
=
1
so the Laplace transform of the weight function
w(t)
is
W(s)
=
1/p(s)
(*)
That is, Laplace transform is the device for extracting the system
parameters from the unit impulse response.
If the unit impulse response is
e^{t}sin(2t)
for example, then
W(s)
=
(2)/((s+1)^2+4)
and
1/W(s)
=
(1/2)s^2 + s + (5/2)
so we discover, if you like, that the mass is 1/2, the damping constant
is 1, and the spring constant is 2.
[Of course we knew that, too: the impulse response is (for t > 0) a
homogeneous system response, so the roots of the characteristic
polynomial
are visible and must be
 1 + 2i .
The roots don't quite determine
the polynomial, since you can always multiply through by a constant
and get another polynomial with the same roots. If you normalize to
s^2 + bs + k
then
b
=
 (sum of roots) = 2
k
=
product of roots = 5
so up to a constant you get
s^2 + 2s + 5
The constant is the mass, and this can be derived too, from
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 Spring '06
 david
 Math, Laplace, Complex number, Impulse response, unit impulse response

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