A Candle in the Dark

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Fourier transforms for the practical person

with 3 comments

In this series of posts, I plan to outline some basic ideas which I’ve learnt on the theory of Fourier transforms, and it’s practical applications in a non-rigorous manner. Once I’ve laid out the basics, I’ll then show you some interesting stuff from what I’m currently working on.

Let’s start with Fourier series. It’s actually a remarkable fact, that we can express any arbitrary periodic function, simply as the sum of the ordinary sine and cosine functions we’ve all studied at high school. If F(t) is a periodic function, then we have

F(t) = \sum_{n=-\infty}^{\infty} A_n \cos(2 \pi n \nu_0 t) + B_n \sin(2 \pi n \nu_0 t)

The Fourier transform is an extension of this, as the period of the function approaches infinity, and the gap between successive harmonics approaches 0. So, in some sense, the fourier transform decomposes a function into it’s frequency components.

For a non-periodic function F(t) which satisfies certain conditions, there are many conventions of describing the fourier transform. Following one such convention which is widely used, the forward fourier transform is

F(t)=\int\limits_{-\infty}^{\infty} \phi(\nu) e^{-2 \pi i \nu t} d\nu

While, the inverse fourier transform is

\phi(\nu)=\int\limits_{-\infty}^{\infty} F(t) e^{+2 \pi i \nu t} dt

Notice that, if F(t) is a continuous time signal, then it’s transformed into the frequency domain by the forward transform. One of the properties of the fourier transforms is that, \phi(\nu) and F(t) are transforms of each other, and form a fourier pair, and are represented by F(t) \rightleftharpoons \phi(\nu).

This means that, if f(x) \rightleftharpoons \phi(p) , then

f(x)=\int\limits_{-\infty}^{\infty} \left( \int\limits_{-\infty}^{\infty} f(x) e^{2 \pi i x p}dx \right) e^{-2 \pi i x p} dp

if f(x) isn’t discontinuous. If it is discontinuous, then the value at that point will be the average of the value around the discontinuity. So, we can simplify our terminology and say that the fourier transform of \phi(p) is f(x) and vice versa.

In the next post, I’ll look at the fourier transform of some useful functions, but before that, there’s one more nice result. For an Electromagnetic wave, or a signal in a wire, the fourier transform of the voltage can be complex. However the conjugate product \phi(\nu) \phi^{\ast}(\nu)=|\phi(\nu)|^2, is real and is proportional to the power density (or, power per unit frequency). This is know as the spectral power density.

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Written by parseval

June 16, 2007 at 8:35 am

Posted in mathematics

3 Responses

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  1. But what if your periodic function is highly discontinuous? Then the fourier-series might not even exist.

    Wilfred

    January 3, 2008 at 4:36 pm

  2. The answer to that is present in the theorem by Dirichlet which says that if f(x) is periodic in an interval, has a finite number of discontinuities, has a finite number of maxima and minima, then the Fourier series converges to the value of the function at the continuous points, and the midpoint of the “jump” at the points of discontinuity.

    This leads to interesting phenomena near the discontinuity known as the Gibbs Phenomenon.

    parseval

    January 3, 2008 at 6:46 pm

  3. First time I have seen someone from India tackling fourier transform on their blog. 🙂 Nice blog sir! I will be a frequent visitor!!!

    Hindu Atheist

    December 13, 2008 at 2:33 pm


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