# Determine States, State Transition Matrix

Discussion in 'Homework Help' started by jegues, Mar 8, 2014.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
732
45
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?

2. ### blah2222 Distinguished Member

May 3, 2010
582
38
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.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
732
45
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.