Laplace Transform for C.T. System Analysis

Because of the linearity property of linear time-invariant (LTI) systems, we can find the response of these systems by breaking the input $x(t)$ into several components and then summing the system response to all the components of $x(t)$. We have already used this procedure in time-domain analysis, in which the input $x(t)$ is broken into impulsive components.

In the frequency domain, we break up the input $x(t)$ into exponentials of the form $e^{st}$, where the parameter $s$ is the complex frequency of the signal $e^{st}$. This method offers an insight into the system behavior complementary to that seen in the time-domain analysis. The tool to represent an arbitrary input $x(t)$ in terms of exponential components is the Laplace transform.

Definition

For a signal $x(t)$, its Laplace transform $X(s)$ is given by:

$$X(s)={\int }_{-\infty }^{\infty }x(t)e^{-st}dt$$

where $s$ is a complex number such that $s=\sigma +j\omega$.

The signal $x(t)$ is said to be the inverse Laplace transform of $X(s)$. It can be shown that:

$$x(t)=\frac{1}{2\pi j}{\int }_{c-j\infty }^{c+j\infty }X(s)e^{st}ds$$

where $c$ is a constant chosen to ensure the convergence of the integral.

In terms of Laplace transforms as a function, these are written as:

$$X(s)=\mathcal{L}[x(t)]\text{and}x(t)={\mathcal{L}}^{-1}[X(s)]$$

Eigenfunction and Transfer Function

Let’s say we want to find the response of an LTI system with impulse response $h(t)$ to the input signal $x(t)=e^{st}$. Then, we have:

$$\begin{array}{rl}y(t)& =x(t)\ast h(t)={\int }_{-\infty }^{\infty }h(\tau )x(t-\tau )d\tau \\ & ={\int }_{-\infty }^{\infty }h(t)e^{s(t-\tau )}d\tau \\ & =e^{st}{\int }_{-\infty }^{\infty }h(\tau )e^{-s\tau }d\tau \\ y(t)& =e^{st}H(s)\end{array}$$

where $H(s)={\int }_{-\infty }^{\infty }h(\tau )e^{-s\tau }d\tau$. This is just the Laplace transform of $h(t)$!

Since we defined $x(t)=e^{st}$, we just have:

$$y(t)=e^{st}H(s)=x(t)H(s)$$

For a given $s$, $H(s)$ is constant; output is the input multiplied by a constant.

Since the input $x(t)$ appears at the output (passes through the system), we say it is an eigenfunction of the system.

We also have:

$$H(s)=\frac{y(t)}{x(t)}{|}_{x(t)=e^{st}}$$

so $H(s)$ is called the transfer function of the system. (Seen this in MTE 220 Sensor Transfer Function).

Like we said before, $H(s)$ is just the Laplace transform of $h(t)$, so we have:

$$H(s)=\mathcal{L}[h(t)]={\int }_{-\infty }^{\infty }h(t)e^{-st}dt$$