Fourier Transform with Examples
Table of Contents
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 \( \pi A \).
More References and links
[1] - University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
[2] - Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
[3] - Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8
Formulas For Fourier Series and Transform