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Unilateral Laplace Transform

The Laplace transform so far discussed is bilateral as it can be applied to left sided signals as well as right sided ones. Now we will consider the unilateral Laplace transform of an arbitrary signal $x(t)$ defined as

\begin{displaymath}{\cal UL}[x(t)]=X(s)\stackrel{\triangle}{=}
\int_{-\infty}^\infty x(t)u(t) e^{-st}dt=\int_{0^-}^\infty x(t) e^{-st}dt
\end{displaymath}

This definition always assumes $x(t)=0$ for $t<0$. When the unilateral Laplace transform is applied to find the transfer function $H(s)={\cal UL}[h(t)]$ of an LTI system, it is always assumed to be causal. And the ROC is always right sided in s-plane.

By definition, the unilateral Laplace transform of any signal $x(t)=x(t)u(t)$ is identical to its bilateral Laplace transform. However, when $x(t) \ne x(t)u(t)$, the two Laplace transforms are different.

Unilateral Laplace Transform shares all the properties of bilateral Laplace transform, except some of the properties are expressed in different forms. Here we only consider the differentiation in time domain.


\begin{displaymath}{\cal UL}[\frac{d}{dt} x(t)]={\cal UL}[x'(t)]=sX(s)-x(0^-) \end{displaymath}

Proof:
$\displaystyle {\cal UL}[\frac{d}{dt} x(t)]$ $\textstyle =$ $\displaystyle \int_{0^-}^\infty x'(t)e^{-st}dt
=\int_{0^-}^\infty e^{-st}d[x(t)]
=x(t)e^{-st}\bigg\vert _{0^-}^\infty-\int_{0^-}^\infty x(t)d(e^{-st})$  
  $\textstyle =$ $\displaystyle -x(0^-)+s \int_{0^-}^\infty x(t)e^{-st}dt=sX(s)-x(0^-)$  

We can further get Laplace transform of higher order derivatives

\begin{displaymath}{\cal UL}[x''(t)]=s{\cal UL}[x'(t)]-x'(0^-)=s^2 X(s)-s x(0^-)-x'(0^-) \end{displaymath}

and in general

\begin{displaymath}{\cal UL}[x^{(n)}(t)]=s^nX(s)-\sum_{k=0}^{n-1}s^k x^{(n-1-k)}(0^-) \end{displaymath}

Example: Consider the following 2nd order LCCDE:

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

with initial conditions:

\begin{displaymath}y(0^-)=y_0=3,\;\;\;\;\;\;y'(0^-)=y_1=-5 \end{displaymath}

Taking unilateral Laplace on both sides we get:

\begin{displaymath}s^2 Y(s)-sy_0-y_1+3s Y(s)-3y_0+2 Y(s)=\frac{2}{s},
\;\;\;\;\;\mbox{or}\;\;\;\;[s^2+3s+2] Y(s)=\frac{2}{s}+y_0(s+3)+y_1 \end{displaymath}

Solving for $Y(s)$ we get:
$\displaystyle Y(s)$ $\textstyle =$ $\displaystyle \frac{y_0(s+3)}{(s+1)(s+2)}+\frac{y_1}{(s+1)(s+2)}+\frac{2}{s(s+1)(s+2)}$  
  $\textstyle =$ $\displaystyle \frac{3(s+3)}{(s+1)(s+2)}-\frac{5}{(s+1)(s+2)}+\frac{2}{s(s+1)(s+2)}$  

The first two terms are the homogeneous solution (transient response) and the last term is the particular solution (steady state response). In time domain the response is:
$\displaystyle y_h(t)$ $\textstyle =$ $\displaystyle {\cal L}^{-1}\left[ \frac{3(s+3)}{(s+1)(s+2)}-\frac{5}{(s+1)(s+2)...
...al L}^{-1}\left[ \frac{6}{s+1}-\frac{3}{s+2}-\frac{5}{s+1}+\frac{5}{s+2}\right]$  
  $\textstyle =$ $\displaystyle {\cal L}^{-1}\left[\frac{1}{s+1}+\frac{2}{s+2}\right]=[e^{-t}+2e^{-2t}] u(t)$  


$\displaystyle y_p(t)$ $\textstyle =$ $\displaystyle {\cal L}^{-1}\left[\frac{2}{s(s+1)(s+2)}\right]=
{\cal L}^{-1}\left[\frac{1}{s}-\frac{2}{s+1}+\frac{1}{s+2}\right]$  
  $\textstyle =$ $\displaystyle [1-2e^{-t}+e^{-2t}]u(t)$  


\begin{displaymath}y(t)=y_h(t)+y_p(t)=[1-e^{-t}+3e^{-2t}] u(t) \end{displaymath}


next up previous
Next: Initial and Final Value Up: Laplace_Transform Previous: System Algebra and Block
Ruye Wang 2012-01-28