# Fourier Transform with Examples

## Definition of Fourier Transform

The Fourier transform decomposes a function of time (or a signal) into it frequency domain. Mathematically, it is defined as [1], [2], [3]: $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt$ Where $$F(\omega)$$ is the Fourier transform of the function $$f(t)$$ , $$\omega$$ is the angular frequency and $$j$$ is the imaginary unit defined by $$j = \sqrt {-1}$$ .
We also have the inverse Fourier transform that takes the Fourier $$F(\omega)$$ and transform back to the time domain: $f(t) = \dfrac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t} d\omega$

### Example 1: Fourier Transform of a Square Wave

Let us consider the square wave function $$f(t)$$ defined by: $f(t) = \begin{cases} 1, & \text{if } -\dfrac{T}{2} \leq t \leq \dfrac{T}{2} \\ 0, & \text{otherwise} \end{cases}$ The Fourier transform of $$f(t)$$ is defined by: $F(\omega) = \int_{-\frac{T}{2}}^{\frac{T}{2}} e^{-j\omega t} dt$ Evaluate this integral: $F(\omega) = \dfrac{1}{-j\omega} \left[ e^{-j\omega t}\right]^{T/2}_{-T/2}$ $= \dfrac{1}{-j\omega} \left( e^{-j\omega \dfrac{T}{2}} - e^{j\omega \dfrac{T}{2}} \right)$ Use Euler's Formula $$e^{j \; x} = \cos(x)+ j \; \sin(x)$$ to rewrite the above integral as $F(\omega) = \dfrac{1}{-j\omega} \left( \cos\left(\dfrac{\omega T}{2}\right) - j\sin\left(\dfrac{\omega T}{2}\right) - \cos\left(\dfrac{\omega T}{2}\right) - j\sin\left(\dfrac{\omega T}{2}\right) \right)$ $= \dfrac{2}{\omega} \sin\left(\dfrac{\omega T}{2}\right)$ So, the Fourier transform of the square wave function is: $F(\omega) = \dfrac{2}{\omega} \sin\left(\dfrac{\omega T}{2}\right)$ This is the amplitude spectrum of the square wave function. It shows how the amplitude of the different frequency components varies with frequency.

### Example 2: Fourier Transform of the Sine Function

Find the Fourier transform of a sine function defined by: $f(t) = A \sin( \omega_0 t)$ Where: - $$A$$ is the amplitude of the sine wave, - $$\omega_0$$ is the angular frequency of the sine wave, - $$t$$ is time. The Fourier transform of $$f(t)$$ is given by: $F(\omega) = \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt$ Substituting $$f(t) = A \sin(\omega_0 t)$$, we get: $F(\omega) = \int_{-\infty}^{\infty} A \sin( \omega_0 t) e^{-j\omega t} dt$ We can rewrite the sine function using Euler's formula: $\sin(x) = \dfrac{e^{jx} - e^{-jx}}{2\;j}$ So, we have: $F(\omega) = \dfrac{A}{2i} \int_{-\infty}^{\infty} (e^{j(\omega_0 - \omega t)} - e^{-j(\omega_0 t + \omega t)}) dt$ Evaluate these integrals separately: $F(\omega) = \dfrac{A}{2 \; j} \left( \int_{-\infty}^{\infty} e^{j(\omega_0 t - \omega t)} dt - \int_{-\infty}^{\infty} e^{-j(\omega_0 + \omega t)} dt \right)$ We can use the properties of the Dirac delta function to evaluate these integrals. When $$\omega = \omega_0$$, the first integral will yield $$2\pi \delta(\omega - \omega_0)$$, and the second integral will yield $$2\pi \delta(\omega + \omega_0)$$. Therefore, the Fourier transform of $$f(t) = A \sin(\omega_0 t)$$ is: $F(\omega) = -j \pi A (\delta(\omega - \omega_0) + j \pi A \delta(\omega + \omega_0))$ This result shows that the Fourier transform consists of two impulses located at frequencies $$\omega = \pm \omega_0$$, with amplitudes $$\pi A$$.