CATbus
Newbie
I'm trying to get the transfer function of the system described by \[\frac{d^2y}{dt^2} + 5\frac{dy}{dt} + 6y = 2\frac{du}{dt} + 1\], with no initial conditions specified.
The answer is supposed to be \[H(s) = \frac{2s + 1}{s^2 + 5s + 6}\], but the furthest I can get is
\[\begin{align}s^2Y(s) + 5sY(s) + 6Y(s) - sy(0) - 5y(0) - y'(0) &= 2sU(s) - 2u(0) + \frac{1}{s} \rightarrow\\
Y(s)(s^2 + 5s + 6) &= 2sU(s) + \frac{1}{s}\end{align}\]
How did they separate \[U(s)\] from the rest of the RHS? Have I made a mistake applying the Laplace transform table?
The answer is supposed to be \[H(s) = \frac{2s + 1}{s^2 + 5s + 6}\], but the furthest I can get is
\[\begin{align}s^2Y(s) + 5sY(s) + 6Y(s) - sy(0) - 5y(0) - y'(0) &= 2sU(s) - 2u(0) + \frac{1}{s} \rightarrow\\
Y(s)(s^2 + 5s + 6) &= 2sU(s) + \frac{1}{s}\end{align}\]
How did they separate \[U(s)\] from the rest of the RHS? Have I made a mistake applying the Laplace transform table?
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