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Solving LCCDEs by Unilateral Laplace Transform

Due to its differentiation property, the unilateral Laplace transform is a powerful tool for solving LCCDEs with arbitrary initial conditions.

Example: A system is described by this LCCDE

\begin{displaymath}\frac{d^2}{dt^2}y(t)+3\frac{d}{dt}y(t)+2y(t)=x(t)=\alpha u(t) \end{displaymath}

with initial conditions

\begin{displaymath}y(0^-)=\beta,\;\;\;\; y'(0^-)=\gamma \end{displaymath}

To find $y(t)$, we first apply unilateral Laplace transform to the differential equation to get

\begin{displaymath}s^2Y(s)-\beta s-\gamma+3 sY(s)-3\beta+2Y(s)
=(s^2+3s+2)Y(s)-\beta s-\gamma-3\beta=\alpha/s \end{displaymath}

which can be then solved for $Y(s)$ algebraically to get

\begin{displaymath}Y(s)=\frac{\alpha}{s(s+1)(s+2)}+\frac{\beta(s+3)}{(s+1)(s+2)}
+\frac{\gamma}{(s+1)(s+2)}=Y_p(s)+Y_h(s) \end{displaymath}

This is the general solution of the LCCDE which is composed of two parts: Given specific values for $\alpha$, $\beta$ and $\gamma$ such as $\alpha=2$, $\beta=3$ and $\gamma=-5$, $Y(s)$ can be partial fraction expanded to be
$\displaystyle Y(s)$ $\textstyle =$ $\displaystyle Y_p(s)+Y_h(s)=\frac{2}{s(s+1)(s+2)}+\frac{3(s+3)}{(s+1)(s+2)}-\frac{5}{(s+1)(s+2)}$  
  $\textstyle =$ $\displaystyle \left[\frac{1}{s}-\frac{2}{s+1}+\frac{1}{s+2}\right]
+\left[\frac{6}{s+1}-\frac{3}{s+2}\right]+\left[\frac{5}{s+2}-\frac{5}{s+1}\right]$  
  $\textstyle =$ $\displaystyle \frac{1}{s}-\frac{1}{s+1}+\frac{3}{s+2}$  

Finally, the inverse Laplace transform of $Y(s)$ gives the time domain solution

\begin{displaymath}y(t)={\cal UL}^{-1}[Y(s)]=[1-e^{-t}+3e^{-2t}]u(t) \end{displaymath}

The particular solution under zero initial condition (caused by the input $x(t)=2u(t)$ only) is

\begin{displaymath}y(t)={\cal UL}^{-1}\left[\frac{1}{s}-\frac{2}{s+1}+\frac{1}{s+2}\right]
=[1-2e^{-t}+e^{-2t}]u(t)
\end{displaymath}

which is quite different from the output with non-zero initial conditions.

If bilateral Laplace transform is applied to the same DE, we get

\begin{displaymath}s^2Y(s)+3 sY(s)+2Y(s)=(s^2+3s+2)Y(s)=\alpha/s \end{displaymath}

which can be solved for $Y(s)$

\begin{displaymath}Y(s)=\frac{\alpha}{s(s+1)(s+2)} \end{displaymath}

This is the particular solution above with zero initial conditions. From this we see that bilateral Laplace transform can only solve systems of zero initial conditions. When the initial conditions of the system are not all zero, unilateral Laplace transform has to be used.
next up previous
Next: About this document ... Up: Laplace_Transform Previous: Initial and Final Value
Ruye Wang 2012-01-28