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Representation of LTI Systems by Laplace Transform

Due to its convolution property, the Laplace transform is a powerful tool for analyzing LTI systems:

\begin{displaymath}y(t)=h(t)*x(t) \stackrel{{\cal L}}{\longleftrightarrow} Y(s)=H(s)X(s) \end{displaymath}

Also, if an LTI system can be described by a linear constant coefficient differential equation (LCCDE), the Laplace transform can convert the differential equation to an algebraic equation due to the time derivative property:

\begin{displaymath}{\cal L}[ \frac{d^n}{dt^n}x(t)]=s^n X(s) \end{displaymath}

We first consider how an LTI system can be represented in the Laplace domain.

Example 1: The transfer function of an LTI is

\begin{displaymath}H(s)=\frac{1}{a+s} \end{displaymath}

As shown before, without specifying the ROC, this $H(s)$ could be the Laplace transform of one of the two possible time signals $h(t)$.
  $a>0$ $a<0$
$Re[s]>-a$ $j\omega$-axis inside ROC $j\omega$-axis outside ROC,
$h(t)=e^{-at}u(t)$ causal, stable causal, unstable
$Re[s]<-a$ $j\omega$-axis outside ROC $j\omega$-axis inside ROC,
$h(t)=-e^{-at}u(-t)$ anti-causal, unstable anti-causal, stable

Example 2: The transfer function of an LTI is

\begin{displaymath}H(s)=\frac{e^{s\tau}}{s+1},\;\;\;\;\;\; Re[s]>-1 \end{displaymath}

This is a time-shifted version of ${\cal L}[e^{-t}u(t)]=1/(s+1)$, and the corresponding impulse response is:

\begin{displaymath}h(t)=e^{-(t+\tau)}u(t+\tau) \end{displaymath}

If $\tau>0$, then $h(t) \ne 0$ during the interval $-\tau < t <0$, the system is not causal, although its ROC is a right half plane. This example serves as a counter example showing it is not true that any right half plane ROC corresponds to a causal system, while all causal systems' ROCs are right half planes. However, if $X(s)$ is rational, then the system is causal if and only if its ROC is a right half plane.

Alternatively, as shown in Example 0, we have:

\begin{displaymath}e^{s\tau}=1+s\tau+\frac{1}{2}(s\tau)^2+\cdots \end{displaymath}

Now $H(s)$ can still be consider as a rational function of $s$ with a numerator polynomial of order $M=\infty$, which is greater than that of the denominator $N=1$, i.e., $H(s)$ has a pole at $s=\infty$, i.e., its ROC cannot be a right-sided half plane, therefore the system is not causal. On the other hand, if $\tau<0$, then this polynomial appears in denominator, there is no pole at $s=\infty$, the ROC is a right-sided half plane, the system is causal.


next up previous
Next: LTI Systems Characterized by Up: Laplace_Transform Previous: Laplace Transform of Typical
Ruye Wang 2012-01-28