State Transition Matrix, Determining States

AI Thread Summary
The discussion focuses on solving a state-space representation of a linear time-invariant (LTI) system after applying a unit-step input. The participants work through the state equations and the Laplace transform to derive expressions for X1(s) and X2(s). There is a need to simplify these expressions for easier inverse Laplace transformation, particularly addressing the complications introduced by the -0.5 in the denominator. Clarifications about initial conditions are made, emphasizing that X2(0) = 0 affects the correctness of the derived equations. The conversation highlights the importance of accurately applying the initial conditions to ensure valid results in the time domain.
jegues
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Homework Statement



An LTI system is given in state-space form,

\left( \begin{array}{cc}<br /> \dot{x_{1}} \\<br /> \dot{x_{2}}<br /> \end{array} \right)<br /> =<br /> \left( \begin{array}{cc}<br /> -1 &amp; 0.5 \\<br /> 1 &amp; 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<br />

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.

Homework Equations


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}<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)<br />
Thus,
<br /> X_{1}(s) = \frac{s}{s^{2}+s-0.5} + \frac{0.5}{s^{2}+s-0.5}<br />
<br /> X_{2}(s) = \frac{1}{s^{2}+s-0.5} + \frac{0.5}{s(s^{2}+s-0.5)}<br />

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

Any ideas? Did I make a mistake in my simplification perhaps?
 
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1.) what are the initial conditions for the system? i am assuming x(0) equals zero. If x(0) equals zero you need to remove those second sections from your equations for X1 and X2.

2.)
below is one of the forms you were talking about
e^(-at)*sin(wt) = invlap{ω /( (s+a)^2+ω^2)}

why don't you try this one?
e^(-at)*sinh(wt) = invlap{ω /( (s+a)^2-ω^2)}
 
or thise^(at)*sin(wt) = invlap{ω /( (s-a)^2+ω^2)}
 
or thise^(at)*sinh(wt) = invlap{ω /( (s-a)^2-ω^2)}
 
donpacino said:
1.) what are the initial conditions for the system?

They are provided in the original post.
 
jegues said:
<br /> X_{1}(s) = \frac{s}{s^{2}+s-0.5} + \frac{0.5}{s^{2}+s-0.5}<br />
<br /> X_{2}(s) = \frac{1}{s^{2}+s-0.5} + \frac{0.5}{s(s^{2}+s-0.5)}<br />

My apologies. For some reason I though that was the input vector.

due to the fact that X2(0)=0, your expression for X2(s) is not correct. Eliminate the second part of the statement
 
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