c29 - 18.03 Class 29, Apr 24 Laplace Transform IV: The pole...

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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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This note was uploaded on 03/12/2011 for the course MATH 180 taught by Professor David during the Spring '06 term at Mapúa Institute of Technology.

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c29 - 18.03 Class 29, Apr 24 Laplace Transform IV: The pole...

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