Re: Laplacle transform
- From: Badger <badger@xxxxxxxxxxxxxxx>
- Date: Tue, 26 Sep 2006 20:58:57 -0400
On Tue, 26 Sep 2006 13:11:02 EDT, Ruffe <ante86@xxxxxxxxx> wrote:
Hello!
I´ve been stucked on a math equation and i can´t seem to solve it. I would be very happy if someone here could help me, here we go!
The linear system is described by the differential equation:
y''(t) + y'(t) + 6,50 * y(t) = x(t)
a) Decide the systems transaction function and the systems stepanswer.
Have a nice day
/Rolf
Taking y'(0) = y(0) = 0, the Laplace transform of the differential
equation is
s^2 Y(s) + s Y(s) + 6.5 Y(s) = X(s)
Y(s) (s^2 + s + 6.5) = X(s)
The transfer function H(s) is "output / input", that is, Y(s) / X(s)
H(s) = 1 / (s^2 + s + 6.5)
The Laplace transform of the unit step function is 1/s. So, taking
X(s) = 1/s,
Y(s) = X(s) H(s) = (1/s) [ 1 / (s^2 + s + 6.5) ]
Finally, for the unit step function, y(t) is the inverse Laplace
transform of Y(s).
.
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