State Transition Matrix, Determining States

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Discussion Overview

The discussion revolves around solving a state-space representation of a linear time-invariant (LTI) system, specifically focusing on determining the state of the system after a given time when subjected to a unit-step input. Participants explore methods to apply the inverse Laplace transform to obtain time-domain solutions from the derived expressions in the s-domain.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the state-space equations and attempts to derive the expressions for X1(s) and X2(s) using the Laplace transform, expressing concern about the -0.5 in the denominator complicating the inverse transform.
  • Another participant questions the initial conditions, suggesting that if x(0) equals zero, certain terms in the equations for X1 and X2 should be removed.
  • Several participants propose forms of inverse Laplace transforms that could be applicable, including expressions involving sine and hyperbolic sine functions.
  • A later reply corrects an earlier misunderstanding regarding the input vector and suggests that due to the initial condition X2(0)=0, the expression for X2(s) may need to be revised.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct form of the expressions for X1(s) and X2(s) or the implications of the initial conditions. Multiple viewpoints regarding the initial conditions and the appropriate forms for the inverse Laplace transform remain evident.

Contextual Notes

There are unresolved assumptions regarding the initial conditions and their impact on the derived expressions. The discussion highlights the complexity of applying the inverse Laplace transform in this context, particularly with the presence of the -0.5 term in the denominator.

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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