Laplace Transform and Its Properties

The laplace transform is an extension to the fourier transform. Because for some common signals such as $u(t)$, the fourier transform does not converge.

Laplace Transform

Laplace transform use $s = \sigma + j\omega$ to replace $j\omega$. So the transformation can be extended from fourier transform,

$$X(\omega) = \int_{-\infty}^{+\infty} x(t) e^{-(j \omega + \sigma) t} \mathrm{d} \\ = \int_{-\infty}^{+\infty} x(t) e^{-st} \mathrm{d}t \\$$

And,

$$x(t) = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} X(s) e^{st} \mathrm{d}s$$

In signals and systems, we usually use single-side Laplace transform. This is because we don't usually care about the negative side.

$$L(s) = \int_{-0}^{+\infty} f(t) e^{-st} \mathrm{d}t$$

Please note that it is $-0$.

And inversely,

$$f(t) = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} L(s) e^{st} \mathrm{d}s$$

This is the laplace inverse transform.

It's obvious that laplace transform does not always converge. If we use single-side laplace transform, it is guaranteed to have the region of converge in form of,

$$Re(s) > \sigma_0$$

Laplace Transform Properties

In the following formula, unless the region of convergence is written, by default, it means that the region of convergence doesn't change.

Linear

Laplace transformation is linear. However, we must find the intersection of the two regions of converge.

$$af(t) + bg(t) \xrightarrow{\mathcal{L}} aL(s) + bL(s) \\ Re(s) > \max(\sigma_a, \sigma_b)$$

However, the region $Re(s) > \max(\sigma_a, \sigma_b)$ is the only part that guarantees convergence. In some case, the convergence may expand to the whole complex plane. For example, $f(t) - f(t)$.

Shifting Time is Shifting Phase

$$f(t)u(t) \xrightarrow{\mathcal{L}} L(s)$$

Then,

$$\int_{-0}^{+\infty} f(t - t_0) u(t-t_0) e^{-st} \mathrm{d} t \\ = e^{-st_0} \int_{-0}^{+\infty} f(t - t_0) e^{-s(t-t_0)} u(t-t_0) d(t - t_0) \\ = e^{-st_0} L(s)$$

Thus,

$$f(t - t_0) u(t-t_0) \xrightarrow{\mathcal{L}} e^{-st_0} L(s)$$

Shifting Phase is Shifting Frequency

$$f(t) \xrightarrow{\mathcal{L}} L(s)$$ $$\int_{-0}^{+\infty} f(t) e^{-\alpha t} e^{-st} \mathrm{d}t \\ = \int_{-0}^{+\infty} f(t) e^{-(\alpha + s)t} \mathrm{d}t \\ = L(\alpha + s)$$

Thus,

$$f(t) e^{-\alpha t} \xrightarrow{\mathcal{L}} L(\alpha + s)$$

Scale

$$f(t) \xrightarrow{\mathcal{L}} L(s) \\ Re(s) > \sigma_0$$ $$\int_{-0}^{+\infty} f(\alpha t) e^{-st} \mathrm{d}t \\ = \frac{1}{\alpha} \int_{-0}^{+\infty} f(\alpha t) e^{-\frac{s}{\alpha} \alpha t } \mathrm{d}\alpha t \\ = \frac{1}{\alpha} L(\frac{s}{\alpha})$$

$\alpha$ must be positive because we use single-side laplace transform.

Thus,

$$f(\alpha t) \xrightarrow{\mathcal{L}} \frac{1}{\alpha} L(\frac{s}{\alpha})$$

Derivative

$$f(t) \xrightarrow{\mathcal{L}} L(s)$$ $$\int_{-0}^{+\infty} f'(t) e^{-st} \mathrm{d}t \\ = -f(-0) + s \int_{-0}^{+\infty} f(t) e^{-st} \mathrm{d}t \\ = s L(s) - f(-0)$$

Thus,

$$f'(t) \xrightarrow{\mathcal{L}} s L(s) - f(-0)$$

Integration

$$f(t) \xrightarrow{\mathcal{L}} L(s)$$

We note $F(t)$ as,

$$F(t) = \int_{-\infty}^{t} f(\tau) \mathrm{d}\tau$$ $$\int_{-0}^{+\infty} F(t) e^{-st} \mathrm{d}t \\ = F(-0) + \frac{1}{s} \int_{-0}^{+\infty} f(t) e^{-st} \mathrm{d}t \\ = \int_{-\infty}^{-0} f(t) \mathrm{d}t + \frac{L(s)}{s}$$

Or,

$$\int_{-\infty}^{t} f(\tau) \mathrm{d}\tau \xrightarrow{\mathcal{L}} \int_{-\infty}^{t} f(\tau) \mathrm{d}\tau + \frac{L(s)}{s}$$

Convolution

The proof here is exactly the same, done by splitting $e^{-st}$ into the product of two parts. So we omit it here. We only need to know,

$$f(t) \ast g(t) \xrightarrow{\mathcal{L}} L_f(s) L_g(s)$$

Initial and Final Value Theorems

Consider,

$$f(t) \xrightarrow{\mathcal{L}} L(s)$$ $$\lim_{s \to +\infty} sL(s) \\ = \lim_{s \to +\infty} \int_{-0}^{+\infty} f(t) se^{-st} \mathrm{d}t \\ = \int_{-0}^{+\infty} f(t) \delta(t-0) \mathrm{d}t \\ = f(+0)$$

tip

$se^{-st}$ for $s > 0$ is a valid PDF. It is obviously positive and,

$$\int_{+0}^{+\infty} se^{-st} \mathrm{d}t = 1$$

And the mean is,

$$\int_{+0}^{+\infty} s t e^{-st} \mathrm{d}t = \frac{1}{s^2}$$

Please note that this mean converge to positive zero. This distinguish is important.

In addition that,

$$\lim_{s \to +\infty} \frac{s}{e^{st}} = \begin{cases} 0 & \text{if} \quad t > 0 \\ +\infty & \text{if} \quad t=0 \end{cases}$$

So,

$$\lim_{s \to +\infty}se^{-st} = \delta(t-0)$$

And, similarly

$$\lim_{s \to +0} sL(s) \\ = \lim_{s \to +0} \int_{-0}^{+\infty} f(t) se^{-st} \mathrm{d}t \\ = \int_{+0}^{+\infty} f(t) \delta(t-\infty) \mathrm{d}t \\ = f(+\infty)$$

In all,

$$\lim_{s \to +\infty} sL(s) = f(+0) \\ \lim_{s \to +0} sL(s) = f(+\infty)$$

info

Another way to get the final result is,

Consider the derivative property,

$$f'(t) \xrightarrow{\mathcal{L}} sL(s) - f(-0)$$

That is to say,

$$sL(s) - f(-0) = \int_{-0}^{+\infty}f'(t) e^{-st} \mathrm{d}t \\ = \int_{-0}^{+0} f'(t) e^{-st} + \int_{+0}^{+\infty}f'(t) e^{-st} \mathrm{d}t \\ = f(+0) - f(-0) + \int_{+0}^{+\infty}f'(t) e^{-st} \mathrm{d}t$$

That is to say,

$$sL(s) = f(+0) + \int_{+0}^{+\infty}f'(t) e^{-st} \mathrm{d}t$$

So,

$$\lim_{s \to +\infty} sL(s) = f(+0)$$

And,

$$\lim_{s \to +0} sL(s) = f(+\infty)$$

Laplace Transformation Table

We will not calculate them here, but just list them. They are mostly the same as the fourier transform, except $j\omega$ is replaced by $s$. You also need to remember the region of convergence.

Original Function	Transformed Function	Region of Convergence
$c$	$-\frac{c}{s}$	$Re(s)>0$
$e^{-at}$	$\frac{1}{s+a}$	$Re(s) > -a$
$\delta(t)$	$1$	Anywhere