Fourier Transformation

Introduction
Fourier’s theorem
Fourier series
Complex Fourier Series
Fourier Transform

Introduction

Joseph Fourier (3/21/1768 – 5/16/1830) realized that any function can be estimated as a sum sines and cosines, now known as Fourier’s theorem. This led to development of the mathematical functions known as the Fourier transformation and inverse Fourier transformation, functions that have widespread applications throughout mathematics and science. The goal of this article is to perform an as-simple-as-possible explanation of this theorem and these functions.

Fourier’s theorem

Suppose we have a function of time, $t$ , where $f(t) = 3\cos{t} + 2\sin{t}$ . A graph of this functions and the 2 individual functions that add together to create it are shown in the diagram below:

Figure 1
Blue – $3\sin{t}$
Green – $2\cos{2t}$
Red – *$3\sin{t}+2\cos{2t}$*

In figure 1, $t$ is represented on the axis (horizontal axis) and $f(t)$ is represented on the abscissa (vertical axis). Let’s consider the values of $f(t)$ for the 3 functions at 3 points: $t=-\frac{\pi}{2}$ , $t=0$ and $t=\frac{\pi}{2}$ :

For $t=-\frac{\pi}{2}$ ,

$f_{\text{blue}}(\frac{-\pi}{2})=3\sin{\frac{-\pi}{2}}=3(-1)=-3$
$f_{\text{green}}(\frac{-\pi}{2})=2\cos{2\frac{-\pi}{2}}=2\cos{-\pi}=2(-1)=-2$
$f_{\text{blue}}(\frac{-\pi}{2})+f_{\text{green}}(\frac{-\pi}{2})=-3+(-2)=-5=f_{\text{red}}(\frac{-\pi}{2})$

For $t=0$ ,

$f_{\text{blue}}(0)=3\sin{0}=3(0)=0$
$f_{\text{green}}(0)=2\cos{0}=2(1)=2$
$f_{\text{blue}}(0)+f_{\text{green}}(0)=0+2=2=f_{\text{red}}(0)$

For $t=\frac{\pi}{2}$ ,

$f_{\text{blue}}(\frac{\pi}{2})=3\sin{\frac{\pi}{2}}=3(1)=3$
$f_{\text{green}}(\frac{\pi}{2})=2\cos{2\frac{\pi}{2}}=2\cos{\pi}=2(-1)=-2$
$f_{\text{blue}}(\frac{-\pi}{2})+f_{\text{green}}(\frac{\pi}{2})=3+(-2)=1=f_{\text{red}}(\frac{\pi}{2})$

For the 3 points we considered, we can see that if we add the values of sine and cosine functions at those points, we get the value of a new periodic function (i.e., a function that repeats itself periodically, or more formally, one where $f(t + T)=f(t)$ , $T$ being the length of $t$ the function spans before it repeats itself). It turns out that, if we add the values of the functions depicted by the blue and green lines at every point, we get the value of the function represented by the red curve. The conclusion we can draw from this example is that, by adding simple sine and cosine functions with varying amplitudes and frequencies, we can create a more complex periodic function. It shouldn’t be too much of a stretch, then, to think that if we add up enough sines and cosines with the appropriate combination of amplitudes and frequencies, then we can approximate any periodic function we want to very closely. Furthermore, if we could add an infinite number of sines and cosines together with just the right amplitudes and frequencies, we could specify any periodic function exactly. And finally, if we consider the period of a complex function to be infinite (i.e., the function goes on forever without repeating itself) then any function can be represented by a weighted sum of cosines and sines. These are the ideas on which Fourier theory is based.

Fourier series

Square function

To elucidate this theory, let’s start by deriving the so-called Fourier series for a simple periodic function – the square wave. This derivation is taken from Khan Academy. It begins at the following link:

https://www.khanacademy.org/science/electrical-engineering/ee-signals/ee-fourier-series/v/ee-fourier-series-intro

Fourier series derivation

The square function shown in figure 2 repeats itself over a period of $2\pi$ so we should be able to use a sum of cosines and sines to make an approximation. Such an approximation is expressed mathematically as

$\begin{array}{ccccc} f(t)&=&\displaystyle\sum_n&+ a_n\cos{nt}&+b_n\sin{nt}\\ &=&\,&+a_0\cos{0t}&+b_0\sin{0t}\\ \,&\,&\,&+a_1\cos{1t}&+b_1\sin{1t}\\ \,&\,&\,&+a_2\cos{2t}&+b_2\sin{2t}\\ \,&\,&\,&\vdots&\vdots\\ \,&\,&\,&+a_n\cos{nt}&+b_n\sin{nt} \end{array}$

Because

$a_0\cos{0t}=a_0\cos{0}=a_0\cdot 1= a_0$
$b_0\sin{0t}=b_0\sin{0}=b_0\cdot 0= 0$

many authors separate the $a_0$ and $b_0$ terms, replacing them with $a_0$ . If we do this, our equation becomes

$\begin{array}{ccccc} f(t)&=&\,&+a_0\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\,\\ \,&\,&\,&+a_1\cos{1t}&+b_1\sin{1t}\\ \,&\,&\,&+a_2\cos{2t}&+b_2\sin{2t}\\ \,&\,&\,&\vdots&\vdots\\ \,&\,&\,&+a_n\cos{nt}&+b_n\sin{nt} \end{array}$

Fourier coefficient derivation

What we’d like to do is find the coefficients $a_0,\, a_1,\, a_2\,\dots\,a_n$ and $b_1,\, b_2\,\dots\,b_n$ . Before we do, let’s take a moment and consider the meaning of these coefficients. The best way to visualize this is to consider $f(t)$ a signal in the time domain. That is, it’s a plot of the amplitude of a complex wave made up of a bunch of sine and cosine waves of differing frequencies and different amplitudes, versus time. The coefficients, $a$ and $b$ , are the amplitudes for each given frequency. They tell us, how much each frequency contributes to the time-domain signal.

Now on to the derivation of expressions for the coefficient values. For this derivation, we need the following information:

$\displaystyle\int_0^{2\pi}\sin{mt}\,dt = 0$ for all integers $\,m$
$\displaystyle\int_0^{2\pi}\cos mt \,dt = 0$ for all nonzero integers $\,m$
$\displaystyle\int_0^{2\pi}\sin{mt}\cos nt\,dt = 0$ for all integers $\,m$ and $\,n$
$\displaystyle\int_0^{2\pi}\sin{mt}\sin nt\,dt = 0$ for all integers $\,m\neq n$ or $\,m \neq -n$
$\displaystyle\int_0^{2\pi}\sin{mt}\sin nt\,dt = \pi$ for nonzero integers $\,m = n$
$\displaystyle\int_0^{2\pi}\cos{mt}\cos nt\,dt = 0$ for all integers $\,m\neq n$ or $\,m \neq -n$
$\displaystyle\int_0^{2\pi}\cos{mt}\cos nt\,dt = \pi$ for nonzero integers $\,m = n$

Proof of the above equations can be found here.

Let’s start with $a_0$ . Consider the equation:

Take the integral from $0$ to $2\pi$ of both sides of the equation:

$\begin{array}{ccccc} \displaystyle\int_0^{2\pi}f(t)\,dt&=&\,&\displaystyle\int_0^{2\pi}+a_0\,dt\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\,\\ \,&\,&\,&\displaystyle\int_0^{2\pi}+a_1\cos{1t}\,dt&+\displaystyle\int_0^{2\pi}b_1\sin{1t}\,dt\\ \,&\,&\,&+\displaystyle\int_0^{2\pi}a_2\cos{2t}\,dt&+\displaystyle\int_0^{2\pi}b_2\sin{2t}\,dt\\ \,&\,&\,&\vdots&\vdots\\ \,&\,&\,&+\displaystyle\int_0^{2\pi}a_n\cos{nt}\,dt&+\displaystyle\int_0^{2\pi}b_n\sin{nt}\,dt \end{array}$

We’ve already established that

$\displaystyle\int_0^{2\pi}\cos[(integer)]dt = 0$
$\displaystyle\int_0^{2\pi}\sin[(integer)]dt = 0$

Therefore, all of the terms on the righthand aspect of the equation above disappear except the $a_0$ term. We have:

$\begin{array}{rcl} \displaystyle\int_0^{2\pi}a_0\,dt &=& \eval{a_0\cdot t}_0^{2\pi}\\ &=&a_0(2\pi-0)\\ &=&a_0 2\pi \end{array}$

which means that

$\displaystyle\int_0^{2\pi}f(t)dt=a_0 2\pi \quad \Rightarrow \quad a_0=\frac{1}{2\pi}\displaystyle\int_0^{2\pi}f(t)dt$

But $\displaystyle\int_0^{2\pi}f(t)dt$ is the area under the curve $f(t)$ from $0$ to $2\pi$ and $2\pi$ is the value of the “base” of the complex shape we’re taking the area under. Thus, dividing $\displaystyle\int_0^{2\pi}f(t)dt$ by $2\pi$ is just the average value of the function $f(t)$ .

Next, lets calculate an expression for the coefficients $a_n$ . We begin again with our expression for $f(t)$ , multiply both sides by $\cos nt$ then take $\displaystyle\int_0^{2\pi}$ of both sides. We get

$\begin{array}{ccccc} \displaystyle\int_0^{2\pi}f(t)\cos nt\,dt&=&\,&\displaystyle\int_0^{2\pi}+a_0\cos nt\,dt\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\,\\ \,&\,&\,&\displaystyle\int_0^{2\pi}+a_1\cos{1t}\,\cos nt\,dt&+\displaystyle\int_0^{2\pi}b_1\sin{1t}\,\cos nt\,dt\\ \,&\,&\,&+\displaystyle\int_0^{2\pi}a_2\cos{2t}\,\cos nt\,dt&+\displaystyle\int_0^{2\pi}b_2\sin{2t}\,\cos nt\,dt\\ \,&\,&\,&\vdots&\vdots\\ \,&\,&\,&+\displaystyle\int_0^{2\pi}a_n\cos{nt}\,\cos nt\,dt&+\displaystyle\int_0^{2\pi}b_n\sin{nt}\,\cos nt\,dt \end{array}$

We know that

$\displaystyle\int_0^{2\pi}\sin{mt}\cos nt\,dt = 0$
$\displaystyle\int_0^{2\pi}\cos{mt}\cos nt\,dt = 0$ for all integers $\,m\neq n$ or $\,m \neq -n$
$\displaystyle\int_0^{2\pi}\cos{mt}\cos nt\,dt = \pi$ for nonzero integers $\,m = n$

Thus, all of the terms on the righthand side of the equation above evaluate to 0 except

$\displaystyle\int_0^{2\pi}a_n\cos{nt}\,\cos nt\,dt=\pi$

So,

$\begin{array}{rcl} \displaystyle\int_0^{2\pi}f(t)\cos nt\,dt&=&\displaystyle\int_0^{2\pi}a_n\cos{nt}\,\cos nt\,dt\\ &=&a_n\pi \end{array}$

Therefore,

$a_n = \frac{1}{\pi}\displaystyle\int_0^{2\pi}f(t)\cos nt\,dt$

You might be wondering why the integral $\displaystyle\int_0^{2\pi}f(t)\cos nt\,dt$ doesn’t evaluate to 0 since we’ve been saying that $\displaystyle\int_0^{2\pi}\cos[(integer)]dt = 0$ . If so, click .

Let’s move on now to calculation of the coefficients $b_n$ . The derivation is similar to that for $a_n$ . Again, we start with the expression for $f(t)$ . But his time we multiply all terms by $\sin nt$ and take the integral $\displaystyle\int_0^{2\pi}$ of each term. This gives us

$\begin{array}{ccccc} \displaystyle\int_0^{2\pi}f(t)\sin nt\,dt&=&\,&\displaystyle\int_0^{2\pi}+a_0\sin nt\,dt\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\,\\ \,&\,&\,&\displaystyle\int_0^{2\pi}+a_1\cos{1t}\,\sin nt\,dt&+\displaystyle\int_0^{2\pi}b_1\sin{1t}\,\sin nt\,dt\\ \,&\,&\,&+\displaystyle\int_0^{2\pi}a_2\cos{2t}\,\sin nt\,dt&+\displaystyle\int_0^{2\pi}b_2\sin{2t}\,\sin nt\,dt\\ \,&\,&\,&\vdots&\vdots\\ \,&\,&\,&+\displaystyle\int_0^{2\pi}a_n\cos{nt}\,\sin nt\,dt&+\displaystyle\int_0^{2\pi}b_n\sin{nt}\,\sin nt\,dt \end{array}$

We know that

$\displaystyle\int_0^{2\pi}\sin{mt}\cos nt\,dt = 0$
$\displaystyle\int_0^{2\pi}\sin{mt}\sin nt\,dt = 0$ for all integers $\,m\neq n$ or $\,m \neq -n$
$\displaystyle\int_0^{2\pi}\sin{mt}\sin nt\,dt = \pi$ for nonzero integers $\,m = n$

Thus, all of the terms on the righthand side of the equation above evaluate to 0 except

$\displaystyle\int_0^{2\pi}b_n\sin{nt}\,\sin nt\,dt$

So,

$\begin{array}{rcl} \displaystyle\int_0^{2\pi}f(t)\sin nt\,dt&=&\displaystyle\int_0^{2\pi}b_n\sin{nt}\,\sin nt\,dt\\ &=&b_n\pi \end{array}$

Therefore,

$b_n = \frac{1}{\pi}\displaystyle\int_0^{2\pi}f(t)\sin nt\,dt$

In summary,

$a_0=\frac{1}{2\pi}\displaystyle\int_0^{2\pi}f(t)dt$
$a_n = \frac{1}{\pi}\displaystyle\int_0^{2\pi}f(t)\cos nt\,dt$
$b_n = \frac{1}{\pi}\displaystyle\int_0^{2\pi}f(t)\sin nt\,dt$

Fourier series of square function

We’ll now use the formulas for the Fourier coefficients we just derived to calculate the Fourier series of the square function pictured in figure 2.

We’ll start by calculating $a_0$ . From the diagram, note that the value of $f(t)$ from $0$ to $\pi$ is 3 and from $\pi$ to $2\pi$ , it’s 0. Therefore,

$\begin{array}{rcl} a_0&=&\frac{1}{2\pi}\displaystyle\int_0^{2\pi}f(t)dt\\ &=&\frac{1}{2\pi}\displaystyle\int_0^{\pi}3\,dt + \displaystyle\int_\pi^{2\pi}0\,dt\\ &=&\eval{\frac{1}{2\pi}(3t)}_0^\pi\\ &=&\frac{3\cancel{\pi}}{2\cancel{\pi}}\\ &=&\frac32 \end{array}$

$\begin{array}{rcl} a_n&=&\frac{1}{\pi}\displaystyle\int_0^{2\pi}f(t)\cos nt\,dt\\ &=&\frac{1}{1\pi}\displaystyle\int_0^{\pi}3\cos nt\,dt + \displaystyle\int_\pi^{2\pi}0\cos nt\,dt\text{;}\quad \text{multiply right side by }1= \frac{n}{n}\\ &=&\frac{3}{n\pi}\displaystyle\int_0^{\pi}n\cos nt\,dt + 0\\ &=&\eval{\frac{3}{n\pi}(\sin nt)}_0^\pi\\ &=&\frac{3}{n\pi}(0-0)\\ &=&0 \end{array}$

$\begin{array}{rcl} a_n&=&\frac{1}{\pi}\displaystyle\int_0^{2\pi}f(t)\sin nt\,dt\\ &=&\frac{1}{1\pi}\displaystyle\int_0^{\pi}3\sin nt\,dt + \displaystyle\int_\pi^{2\pi}0\cos nt\,dt\text{;}\quad \text{multiply right side by }1= \frac{-n}{-n}\\ &=&\frac{3}{-n\pi}\displaystyle\int_0^{\pi}-n\sin nt\,dt + 0\\ &=&\eval{\frac{-3}{n\pi}(\cos nt)}_0^\pi\\ &=&\frac{-3}{n\pi}[(\cos n\pi)-(\cos n\cdot 0)] \end{array}$

If $n$ is even, then

$\begin{array}{rcl} b_n &=&\frac{-3}{n\pi}(1-1)\\ &=&\frac{-3}{n\pi}(0)\\ &=&0 \end{array}$

If $n$ is odd, then

$\begin{array}{rcl} b_n &=&\frac{-3}{n\pi}(-1-1)\\ &=&\frac{-3}{n\pi}(-2)\\ &=&\frac{6}{n\pi} \end{array}$

Putting everything together, the Fourier series for a square wave is:

$f(t) = \frac32 + \frac{6}{\pi}\sin t + \frac{6}{3\pi}\sin 3t + \frac{6}{5\pi}\sin 5t + \dots$

Notice that the Fourier series above contains only sine terms. This is because it is an odd function. Let me explain.

Functions can be classified as even, odd or neither even nor odd. Even functions are functions that are symmetric around the y-axis (in Cartesian coordinates). That is, $f(t)=f(-t)$ . $f(t)=cos{t}$ and $f(t)=t^2$ are examples of even functions. On the other hand, odd functions are functions that are symmetric about the origin (in Cartesian coordinates). That is, $-f(t)=f(-t)$ . $f(t)=sin{t}$ and $f(t)=t^3$ are examples of odd functions. Even functions can be approximated by a sum of cosines. Odd functions can be approximated by as sum of sines. A proof of this can be found at the following link (although information presented later in this article is needed to understand this proof):

http://mathonline.wikidot.com/trigonometric-fourier-series-of-even-and-odd-functions

Most functions, however, fall into the “neither” category. To represent these, a sum of both cosines and sines are needed. This is why the expression for the Fourier series we started with contains both cosine and sine terms.

Complex Fourier series

Series derivation

We can express the Fourier series in a more complex form using exponential functions. Specifically, from Euler’s formula, we have

$\cos x = \frac{e^{ix}+e^{-ix}}{2}\quad \sin x = \frac{e^{ix}-e^{-ix}}{2i}$

It will be most useful to us if we multiply the righthand part of the righthand equation by $1=\frac{i}{i}$ :

$\sin x = \frac{e^{ix}-e^{-ix}}{2i}\frac{i}{i} = i\frac{e^{ix}-e^{-ix}}{2(-1)} = -i\frac{e^{ix}-e^{-ix}}{2}$

We plug these into our expression for the Fourier series:

$\begin{array}{rcl} f(t) &=& a_0 + \displaystyle\sum_{n=1}^\infty a \cos nt + b \sin nt \\ \, &\,& \, \\ &=& a\left( \frac{{e^{int}}+e^{-int}}{2} \right) + b\left( -i\frac{{e^{int}}-e^{-int}}{2} \right)\\ \, &\,& \, \\ &=& a_0 + \displaystyle\sum_{n=1}^\infty \frac12\left( ae^{int} + ae^{-int} - ibe^{int} + ibe^{-int} \right)\\ \, &\,& \, \\ &=& a_0 + \displaystyle\sum_{n=1}^\infty \frac12\left( a - ib \right)e^{int} + \frac12\left( a + ib \right)e^{-it} \\ \, &\,& \, \\ &=& a_0 + \displaystyle\sum_{n=1}^\infty \frac{a-ib}{2}e^{int} + \frac{a+ib}{2}e^{-int} \end{array}$

Therefore,

$f(t) = a_0 + \displaystyle\sum_{n=1}^\infty a \cos nt + b \sin nt = \displaystyle\sum_{n=-\infty}^{+\infty} c_n e^{int}$ where

$c_n$ is a coefficient (a complex number)
$c_n=\frac{a-ib}{2}$ when $n$ is positive
$c_n=\frac{a+ib}{2}$ when $n$ is negative
$c_n=a_0$ when $n=0$

Coefficient derivation

Now we need to find the complex Fourier coefficients in terms of $f(t)$ . The procedure is similar to that used to find the coefficients for the trigonometric Fourier series. We multiply both sides of the complex Fourier series equation by $e^{imt}$ . We get:

$f(t)e^{imt} = \displaystyle\sum_{n=-\infty}^{+\infty} c_n e^{int}e^{-imt}$

Next, take the definite integral of both sides from $0$ to $2\pi$ :

$\displaystyle\int_0^{2\pi}f(t)e^{imt}dt = \displaystyle\int_0^{2\pi}\displaystyle\sum_{n=-\infty}^{+\infty} c_n e^{int}e^{-imt}dt$

We’ve shown (or at least accepted) previously that $\int\sum= \sum\int$ . Therefore, make that change and bring the constant, $c_n$ outside of the integral:

$\displaystyle\int_0^{2\pi}f(t)e^{imt}dt = \displaystyle\sum_{n=-\infty}^{+\infty} c_n \displaystyle\int_0^{2\pi} e^{int}e^{-imt}dt=\displaystyle\sum_{n=-\infty}^{+\infty} c_n \displaystyle\int_0^{2\pi} e^{i(n-m)t}dt$

Using Euler’s formula, we know that

$e^{i(n-m)t} = \cos (n-m)t + i\sin (n-m)t$

Thus,

$\displaystyle\int_0^{2\pi}e^{i(n-m)t}\,dt = \displaystyle\int_0^{2\pi} \cos (n-m)t \,dt+ \displaystyle\int_0^{2\pi}i\sin (n-m)t\,dt$

We’ve previously written down, in this article (and proven elsewhere on this site) that

if $n \neq m$ then

$\cos (n-m)t$ and $\sin (n-m)t$ will both be nonzero

$\displaystyle\int_0^{2\pi} \cos (n-m)t\,dt = 0$ and $\displaystyle\int_0^{2\pi} \sin (n-m)t\,dt = 0$

$\displaystyle\int_0^{2\pi}e^{i(n-m)t}\,dt = \displaystyle\int_0^{2\pi} \cos (n-m)t }\,dt+ \displaystyle\int_0^{2\pi}i\sin (n-m)t}\,dt = 0+0=0$

if $n = m$ then

$\cos (n-m)t = 1$ and $\sin (n-m) = 0$

$\displaystyle\int_0^{2\pi} \cos (n-m)t\,dt = \displaystyle\int_0^{2\pi} \cos 0\,dt = \displaystyle\int_0^{2\pi} 1\,dt=\eval{ t}_0^{2\pi}=2\pi-0=2\pi$

$\begin{array}{ccccc} \displaystyle\int_0^{2\pi}e^{i(n-m)t}\,dt &=& \displaystyle\int_0^{2\pi} \cos (n-m)t }\,dt &+& \displaystyle\int_0^{2\pi}i\sin (n-m)t}\,dt\\ \,&=& 2\pi&+&0\\ \,&=&2\pi&\,&\, \end{array}$

We said that

$\displaystyle\int_0^{2\pi}f(t)e^{imt}dt = \displaystyle\sum_{n=-\infty}^{+\infty} c_n \displaystyle\int_0^{2\pi} e^{i(n-m)t}dt$

The only term we get from the sum on the righthand side of the equation occurs when $m=n$ . When this occurs, $\displaystyle\int_0^{2\pi} e^{i(n-m)t}dt = 2\pi$ . All of the other terms in the sum equal zero. Also, when $m=n$ , $\displaystyle\int_0^{2\pi}f(t)e^{imt}dt$ becomes $\displaystyle\int_0^{2\pi}f(t)e^{int}dt$ . So we wind up with

$\displaystyle\int_0^{2\pi}f(t)e^{int}dt = c_n 2\pi$

and

$c_n=\frac{1}{2\pi}\displaystyle\int_0^{2\pi}f(t)e^{int}dt$

Derivation of the Fourier transformation

This derivation is taken largely from

Video: Derivation of Fourier Transformation

The trigonometric and complex Fourier series we have been working with are useful in representing periodic functions, i.e., functions that repeat themselves. But how do we represent aperiodic functions, i.e., functions with a period of infinity? We use the Fourier Transformation.

We’ll stick with the complex Fourier representations because they tend to be easier to work with. The complex Fourier series we derived is

$\begin{array}{rcl} f(t)&=&\displaystyle\sum_{n=0}^{\infty}(a_n cos nt + b_n \sin nt)dt\\ &=&\displaystyle\sum_{n=-\infty}^{+\infty}}c_n e^{int}dt \end{array}$

Recall that the argument of sine and cosine functions is an angle. So $nt$ represents frequency x time, $n$ being the frequency, either cycles per second or radians per second. We didn’t talk about this earlier, but for the Fourier series to work, the frequencies used to represent a function need to be integer multiples of some base frequency, and that base frequency has to make one cycle during the period of the periodic function we’re trying to estimate. A good explanation of why this can be found in this video:

Justification for Fourier series

So the $n$ in the equation for the Fourier series actually means $n\cdot\frac{2\pi}{T}$ where $T$ is the period of the periodic function we’re trying to represent with our series.

Let

$\frac{2\pi}{T}=\omega_0$ called the primary frequency
$\frac{1}{T}=\frac{\Delta \omega}{2\pi}$ , the difference between primary frequencies that $n$ multiplies
$k\frac{2\pi}{T}$ is the overall frequency of the wave where $k$ , like $n$ , is an integer

Now if a function is aperiodic, that means the the period of the function is infinity which means that $T\to\infty$ . As $T$ gets larger and larger:

$\frac{\Delta\omega}{2\pi} \to \frac{d\omega}{2\pi}$
Sums become integrals

We start off with our Fourier series for $f(t)$

$f(t)=\displaystyle\sum_{n=-\infty}^{+\infty}c_ne^{int}$

We proved previously that

$c_n = \frac{1}{2\pi}\displaystyle\int_0^{2\pi} f(t)e^{-int}\,dt$

We change the limits of integration to the more generic limits $T_0$ to $T_0+T$ which allows us to evaluate over a period of any length starting at any point:

$c_n = \frac{1}{T}\displaystyle\int_{T_0}^{T_0+T} f(t)e^{-int}\,dt$

We combine these 2 equations to get

$f(t)=\displaystyle\sum_{n=-\infty}^{+\infty}\frac{1}{T}\displaystyle\int_{T_0}^{T_0+T} f(t)e^{-int}\,dt \cdot e^{int}$

Into this equation, we substitute the values

$\frac{1}{T}=\frac{\Delta\omega}{2\pi}$
$n=\frac{k2\pi}{T}=\frac{k\cancel{2\pi} \Delta \omega t}{\cancel{2\pi}}=k\Delta\omega$

$\begin{array}{rcl} f(t)&=&\displaystyle\sum_{k=-\infty}^{+\infty}\frac{\Delta\omega}{2\pi}\displaystyle\int_{T_0}^{T_0+T} f(t)e^{-i\frac{k\cancel{2\pi} \Delta \omega t}{\cancel{2\pi}}}\,dt \cdot e^{i\frac{k\cancel{2\pi} \Delta \omega t}{\cancel{2\pi}}}\\ &=&\displaystyle\sum_{k=-\infty}^{+\infty}\frac{\Delta\omega}{2\pi}\displaystyle\int_{T_0}^{T_0+T} f(t)e^{-ik\Delta\omega t}\,dt \cdot e^{ik\Delta\omega t}\\ &=&\displaystyle\sum_{k=-\infty}^{+\infty}\frac{1}{2\pi}\displaystyle\int_{T_0}^{T_0+T} f(t)e^{-ik\Delta\omega t}\,dt \cdot e^{ik\Delta\omega t}\,\Delta\omega \end{array}$

Now we take the limit $T\to\infty$ . As we noted above, when we do

$\Delta\omega \to d\omega$
$k\Delta\omega$ becomes the continuous variable $\omega$
The summation becomes an integral

In addition, we are saying that our wave never repeats itself. That means that our period extends from $-\infty$ to $+\infty$ . Therefore, our limits on the integral change, as follows:

$T_0 \to -\infty$ and $T_0+T \to +\infty$

The results are:

$\begin{array}{rcl} f(t) &=& \displaystyle\lim_{T\to\infty} \displaystyle\sum_{k=-\infty}^{+\infty} \frac{1}{2\pi} \displaystyle\int_{T_0}^{T_0 + T} f(t)e^{-i k2\pi\Delta\omega t}\,dt \cdot e^{ik2\pi\Delta\omega t}\Delta\omega \\ &=& \frac{1}{2\pi}\displaystyle\int_{-\infty}^{+\infty} \underbrace{\displaystyle\int_{-\infty}^{\infty} f(t)e^{-i \omega t}\,dt}_{X(\omega)} \cdot e^{i \omega t}\,d\omega \end{array}$

As suggested in the last equation, as $T\to\infty$ , the expression $\displaystyle\int_{-\infty}^{\infty} f(t)e^{-i \omega t}\,dt$ becomes a continuous function which we’ll call $X(\omega)$ .

We end up with 2 important equations:

Inverse Fourier transformation (also termed the Synthesis equation)

$f(t) = \frac{1}{2\pi}\displaystyle\int_{-\infty}^{+\infty} X(\omega) \cdot e^{i \omega t}\,d\omega$

Forward Fourier transformation (also called the Analysis equation)

$X(\omega) = \displaystyle\int_{-\infty}^{\infty} f(t)e^{-i \omega t}\,dt$

Other forms

There are two other common ways that the Fourier transformation is formulated:

Hertz Frequency

If we work with units of cycles/sec (instead of radians/sec) in our exponentials (i.e., $\Delta f = \frac{2\pi}{T}$ ), then

the exponents of our exponential functions would be $\pm i 2\pi f$ and
$f$ would go to $df$ without division by $2\pi$ .

Therefore, there would be no $\frac{1}{2\pi}$ outside the integral in the equation for the inverse Fourier transformation

Our equations would be

$f(t) = \displaystyle\int_{-\infty}^{+\infty} X(f) \cdot e^{i 2\pi f t}\,df$

and

$X(f) = \displaystyle\int_{-\infty}^{\infty} f(t)e^{-i 2\pi f t}\,dt$

Scaled (Balanced) Version

We can scale the equations we originally derived by multiplying by $X(\omega)$ by $\sqrt{2\pi}$ . It would look like this:

$\begin{array}{rcl} f(t) &=& \frac{1}{2\pi}\displaystyle\int_{-\infty}^{+\infty} \sqrt{2\pi}X(\omega) \cdot e^{i \omega t}\,d\omega\\ &=& \frac{\sqrt{2\pi}}{2\pi}\displaystyle\int_{-\infty}^{+\infty} X(\omega) \cdot e^{i \omega t}\,d\omega\\ &=& \frac{1}{\sqrt{2\pi}}\displaystyle\int_{-\infty}^{+\infty} X(\omega) \cdot e^{i \omega t}\,d\omega\\ \end{array}$

$\begin{array}{rcl} \sqrt{2\pi}X(\omega) &=& \displaystyle\int_{-\infty}^{\infty} f(t)e^{-i \omega t}\,dt \\ X(\omega) &=& \frac{1}{\sqrt{2\pi}} \displaystyle\int_{-\infty}^{\infty} f(t)e^{-i \omega t}\,dt \end{array}$

We can apply this scaling factor because, although it effects the absolute amplitudes of $f(t)$ and $X(\omega)$ , the shapes of these functions are unchanged.