$1$ All $s$ $u(t)$ (Unit Step) $\frac{1}{s}$ $\text{Re}(s) > 0$ $t^n u(t)$ $\frac{n!}{s^{n+1}}$ $\text{Re}(s) > 0$ $e^{at}u(t)$ $\frac{1}{s-a}$ $\text{Re}(s) > \text{Re}(a)$ $\sin(\omega t)u(t)$ $\frac{\omega}{s^2+\omega^2}$ $\text{Re}(s) > 0$ $\cos(\omega t)u(t)$ $\frac{s}{s^2+\omega^2}$ $\text{Re}(s) > 0$ $e^{at}\sin(\omega t)u(t)$ $\frac{\omega}{(s-a)^2+\omega^2}$ $\text{Re}(s) > \text{Re}(a)$ $e^{at}\cos(\omega t)u(t)$ $\frac{s-a}{(s-a)^2+\omega^2}$ $\text{Re}(s) > \text{Re}(a)$ $t e^{at}u(t)$ $\frac{1}{(s-a)^2}$ $\text{Re}(s) > \text{Re}(a)$

Fourier Transform

The Fourier Transform decomposes a function (signal) into its constituent frequencies. It is particularly useful for analyzing signals in the frequency domain and solving partial differential equations.

Continuous-Time Fourier Transform (CTFT)

For a function $f(t)$ integrable over $(-\infty, \infty)$, its CTFT $\mathcal{F}\{f(t)\}$ is given by:

$$F(\omega) = \mathcal{F}\{f(t)\} = \int_{-\infty}^\infty f(t) e^{-j\omega t} dt$$

where $\omega$ is the angular frequency (in rad/s).

Inverse Continuous-Time Fourier Transform (ICTFT)

The inverse CTFT $\mathcal{F}^{-1}\{F(\omega)\}$ recovers the original time-domain function:

$$f(t) = \mathcal{F}^{-1}\{F(\omega)\} = \frac{1}{2\pi} \int_{-\infty}^\infty F(\omega) e^{j\omega t} d\omega$$

Properties of Fourier Transform

Linearity: $\mathcal{F}\{af(t) + bg(t)\} = aF(\omega) + bG(\omega)$
Time Shifting: $\mathcal{F}\{f(t-t_0)\} = e^{-j\omega t_0}F(\omega)$
Example: If $\mathcal{F}\{\text{rect}(t)\} = \text{sinc}(\omega/2)$, then $\mathcal{F}\{\text{rect}(t-1)\} = e^{-j\omega} \text{sinc}(\omega/2)$.
Frequency Shifting: $\mathcal{F}\{e^{j\omega_0 t}f(t)\} = F(\omega-\omega_0)$
Example: $\mathcal{F}\{\cos(\omega_0 t)\} = \mathcal{F}\left\{\frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2}\right\} = \frac{1}{2}\mathcal{F}\{e^{j\omega_0 t}\} + \frac{1}{2}\mathcal{F}\{e^{-j\omega_0 t}\}$.
Using frequency shifting with $f(t)=1$ (whose FT is $2\pi\delta(\omega)$):

$\mathcal{F}\{e^{j\omega_0 t}\} = 2\pi\delta(\omega-\omega_0)$

$\mathcal{F}\{e^{-j\omega_0 t}\} = 2\pi\delta(\omega+\omega_0)$

So, $\mathcal{F}\{\cos(\omega_0 t)\} = \pi[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)]$.
Time Scaling: $\mathcal{F}\{f(at)\} = \frac{1}{|a|}F\left(\frac{\omega}{a}\right)$
Example: If $\mathcal{F}\{e^{-|t|}\} = \frac{2}{1+\omega^2}$, then $\mathcal{F}\{e^{-2|t|}\} = \frac{1}{2}\frac{2}{1+(\omega/2)^2} = \frac{1}{1+\omega^2/4} = \frac{4}{4+\omega^2}$.
Time Differentiation: $\mathcal{F}\left\{\frac{d^n f(t)}{dt^n}\right\} = (j\omega)^n F(\omega)$
Example: Let $f(t) = e^{-at}u(t)$, then $F(\omega) = \frac{1}{a+j\omega}$.

$\mathcal{F}\{f'(t)\} = \mathcal{F}\{-ae^{-at}u(t) + \delta(t)\} = -a\frac{1}{a+j\omega} + 1 = \frac{-a + a+j\omega}{a+j\omega} = \frac{j\omega}{a+j\omega}$.

From the property: $\mathcal{F}\{f'(t)\} = j\omega F(\omega) = j\omega \frac{1}{a+j\omega}$. This matches.
Convolu