- #1
jegues
- 1,097
- 3
Homework Statement
An LTI system is given in state-space form,
[itex]\left( \begin{array}{cc}
\dot{x_{1}} \\
\dot{x_{2}}
\end{array} \right)
=
\left( \begin{array}{cc}
-1 & 0.5 \\
1 & 0
\end{array} \right)
\left( \begin{array}{cc}
x_{1} \\
x_{2}
\end{array} \right)
+
\left( \begin{array}{cc}
0.5 \\
0
\end{array} \right)
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}
X_{1}(s) \\
X_{2}(s)
\end{array} \right)
=
\left( \begin{array}{cc}
\frac{s}{s^{2}+s-0.5}\\
\frac{1}{s^{2}+s-0.5}
\end{array} \right)
+
\left( \begin{array}{cc}
\frac{0.5}{s^{2}+s-0.5}\\
\frac{0.5}{s(s^{2}+s-0.5)}
\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?