The problem statement, all variables and given/known data

An LTI system is given in state-space form,

\(\left( \begin{array}{cc} \dot{x_{1}} \\ \dot{x_{2}} \end{array} \right) = \left( \begin{array}{cc} -1 & 0.5 \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} x_{1} \\ x_{2} \end{array} \right) + \left( \begin{array}{cc} 0.5 \\ 0 \end{array} \right) u
\)

A unit-step signal is applied to the input of the system. If,

\(x_{1}(0) = 1, \quad x_{2}(0) = 0\)

determine the state of the system after t = 0.1 sec.

The attempt at a solution

\(\underline{\dot{x}} = \underline{A} \underline{x} + \underline{B}u\)

\(\mathcal{L} \Rightarrow s \underline{X(s)} - \underline{x(0)} = \underline{A}\underline{X(s)} + \underline{B} U(s)\)

\(\Rightarrow \underline{X(s)} = (s\underline{I} - \underline{A})^{-1} \underline{x(0)} + (s\underline{I} - \underline{A})^{-1}\underline{B}U(s)\)

Working through the simplification I obtain,

\(\left( \begin{array}{cc} X_{1}(s) \\ X_{2}(s) \end{array} \right) = \left( \begin{array}{cc} \frac{s}{s^{2}+s-0.5}\\ \frac{1}{s^{2}+s-0.5} \end{array} \right) + \left( \begin{array}{cc} \frac{0.5}{s^{2}+s-0.5}\\ \frac{0.5}{s(s^{2}+s-0.5)} \end{array} \right)
\)
Thus,
\(
X_{1}(s) = \frac{s}{s^{2}+s-0.5} + \frac{0.5}{s^{2}+s-0.5}
\)
\(
X_{2}(s) = \frac{1}{s^{2}+s-0.5} + \frac{0.5}{s(s^{2}+s-0.5)}
\)

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,
\(
\frac{\omega}{(s+a)^{2} + \omega^{2}} \quad \text{ or } \quad \frac{s+a}{(s+a)^{2} + \omega^{2}}
\)

Any ideas? Did I make a mistake in my simplification perhaps?

 

I didn't really go through all the math but couldn't you just complete the square in the denominators to get it in the form you listed below:

\(

s^2 + s - \frac{1}{2} = s^2 + s + \frac{1}{4} - \frac{3}{4} = (s + \frac{1}{2})^2 - \frac{3}{4} = (s + \frac{1}{2})^2 - (\frac{\sqrt{3}}{2})^2

\)

Then for that one term in X2(s) with the s term multiplied by the quadratic term you could expand it using partial fractions then get to something you could invert.

Hope that helped.

 

I didn't really go through all the math but couldn't you just complete the square in the denominators to get it in the form you listed below:

\(

s^2 + s - \frac{1}{2} = s^2 + s + \frac{1}{4} - \frac{3}{4} = (s + \frac{1}{2})^2 - \frac{3}{4} = (s + \frac{1}{2})^2 - (\frac{\sqrt{3}}{2})^2

\)

The denominator in the forms I've listed has \(+\omega^{2}\) in the denominator, so that \(- (\frac{\sqrt{3}}{2})^2
\) term won't cut it.