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Introduction

Why do we want to find the fourier transform function? This question should be answered first so that we can have motivation for learning how to perform this transformation. Once we see why we want the learn about the fourier transform it will be useful to work through the derivation of the transform. This will give us a greater understanding of the transform and greatly increase the chances that we will remember how to use it in the future.

Why

The reasoning behind performing the fourier transform is fairly simple compared to actually performing the transform. We want this transform so that for any given real signal x(t) we can easily see the frequency and amplititude of each sine and cosine component. These components when added together using the principle of superposition sum to exactly x(t). This transform is very useful in many applications from radio signal processing to image processing, and this is why we want to learn about it.

How

Fourier Series

The fourier transform is a generalization of the fourier series. The fourier series is given by $\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}$, where $\alpha _{n}=1/T\int _{-T/2}^{T/2}x(t)e^{-j2\pi nt/T}dt$, and can be used to represent periodic functions.

Fourier Transform Derivation

Now from the Fourier Series we want to derive the Fourier Transform, which is very similar except that it can be used for nonperiodic funtions. This can be done by simply letting the period $T$ go to infinity.

So for $x(t)=\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}$

We'll let:

$T$ -> $\infty$

$n/T$ -> $f$

$1/T$ -> $df$

$\sum _{n=-\infty }^{\infty }\alpha _{n}e^{j2\pi nt/T}$ -> $\int _{-\infty }^{\infty }\alpha _{n}e^{j2\pi ft}df$

$\alpha _{n}$ -> $X(f)=\int _{-\infty }^{\infty }x(u)e^{-j2\pi ft}du$

Now we have:

$x(t)=\int _{-\infty }^{\infty }(\int _{-\infty }^{\infty }x(u)e^{-j2\pi ft}du)e^{j2\pi ft}df$ -> $x(t)={\mathcal {F}}^{-1}[X(f)]$

Where $\int _{-\infty }^{\infty }x(u)e^{-j2\pi ft}du=X(f)={\mathcal {F}}[x(t)]$

and

$\int _{-\infty }^{\infty }X(f)e^{j2\pi ft}df=x(t)={\mathcal {F}}^{-1}[X(f)]$

Some Fourier Transform Identities

Differentiation

${\frac {dx}{dt}}=\int _{-\infty }^{\infty }j2\pi X(f)e^{j2\pi ft}df={\mathcal {F}}^{-1}[j2\pi fX(f)]$

Time Shift

$x(t-t_{o})=\int _{-\infty }^{\infty }j2\pi X(f)e^{j2\pi f(t-t_{o})}df$

$=\int _{-\infty }^{\infty }e^{-j2\pi ft_{o}}X(f)e^{j2\pi ft}df$

$={\mathcal {F}}^{-1}[e^{-j2\pi ft_{o}}X(f)]$