jegues
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Homework Statement
An LTI system is given in state-space form,
[itex]\left( \begin{array}{cc}<br /> \dot{x_{1}} \\<br /> \dot{x_{2}}<br /> \end{array} \right)<br /> =<br /> \left( \begin{array}{cc}<br /> -1 & 0.5 \\<br /> 1 & 0<br /> \end{array} \right)<br /> \left( \begin{array}{cc}<br /> x_{1} \\<br /> x_{2}<br /> \end{array} \right)<br /> +<br /> \left( \begin{array}{cc}<br /> 0.5 \\<br /> 0<br /> \end{array} \right)<br /> u[/itex]
A unit-step signal is applied to the input of the system. If,
[itex]x_{1}(0) = 1, \quad x_{2}(0) = 0[/itex]
determine the state of the system after t = 0.1 sec.
Homework Equations
The Attempt at a Solution
[itex]\underline{\dot{x}} = \underline{A} \underline{x} + \underline{B}u[/itex]
[itex]\mathcal{L} \Rightarrow s \underline{X(s)} - \underline{x(0)} = \underline{A}\underline{X(s)} + \underline{B} U(s)[/itex]
[itex]\Rightarrow \underline{X(s)} = (s\underline{I} - \underline{A})^{-1} \underline{x(0)} + (s\underline{I} - \underline{A})^{-1}\underline{B}U(s)[/itex]
Working through the simplification I obtain,
[itex]\left( \begin{array}{cc}<br /> X_{1}(s) \\<br /> X_{2}(s)<br /> \end{array} \right)<br /> =<br /> \left( \begin{array}{cc}<br /> \frac{s}{s^{2}+s-0.5}\\<br /> \frac{1}{s^{2}+s-0.5}<br /> \end{array} \right)<br /> +<br /> \left( \begin{array}{cc}<br /> \frac{0.5}{s^{2}+s-0.5}\\<br /> \frac{0.5}{s(s^{2}+s-0.5)}<br /> \end{array} \right)[/itex]
Thus,
[itex] X_{1}(s) = \frac{s}{s^{2}+s-0.5} + \frac{0.5}{s^{2}+s-0.5}[/itex]
[itex] X_{2}(s) = \frac{1}{s^{2}+s-0.5} + \frac{0.5}{s(s^{2}+s-0.5)}[/itex]
How can I put this in a form that I can easily pull out of the s-domain back into time domain using inverse Laplace transform? If weren't for the -0.5 in the denominator I think I could work something out by reworking it into one of the following two forms,
[itex] \frac{\omega}{(s+a)^{2} + \omega^{2}} \quad \text{ or } \quad \frac{s+a}{(s+a)^{2} + \omega^{2}} [/itex]
Any ideas? Did I make a mistake in my simplification perhaps?