1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please explain what a Fourier transform is?

  1. Jun 14, 2010 #1
    I understand Fourier Series fairly well and how to use them to approximate functions (I even wrote a C program to do it) but Transforms are really confusing.

    If I was to take the value of each fourier coefficient and plot it on the y-axis against the angular frequency on x-axis (obv. there would be 2 graphs, one for the cos coefficients and one for sine unless you used the complex form which I guess is why the fourier transform uses the complex form and tends to plot the modulus of the real and imaginary parts as e^ix= cosx + isinx) would this essentially be the same as taking the fourier transform?

    It tells me how much of each frequency component I have right in the original function right? And isn't that what the fourier transform does?

    Any help would be greatly appreciated.

    Thank you for your time :)
  2. jcsd
  3. Jun 14, 2010 #2


    User Avatar
    Science Advisor

    "Fourier series" always produce functions that are periodic with some specific period. Even functions that are not periodic can be represented, over some bounded interval by a Fourier series by assuming periodicity off that interval.

    The "Fourier transform" is derived by taking the limit as the ends of the bounded interval go to infinity and negative infinity.

    It looks to me like you are talking about a specific application. In general, functions don't have "frequencies".
  4. Jun 14, 2010 #3
    Hmm, I noticed that the period goes to infinity and thus the discrete summation becomes a continuous integral, but what does that actually mean? I mean what is it that the Fourier transform actually tells us, what is it used for? I understand the usefulness of being able to approximate functions with trigonometric functions in Fourier series, but what is it that the Fourier transform does?
  5. Jun 14, 2010 #4
    What HallsofIvy told you is correct. But you seem to be looking for a deeper answer. So here goes. If this is too much, then don't get yourself too confused and just ask another question.

    Consider a periodic function with period [tex] \ 2pi [/tex]. You can think of it as function on the entire line that is periodic or you can think of it as function that is defined on [tex] [0, 2\pi] [/tex] and study it in that situation. Another way is to think of it as a function defined on a circle, and in particular on the unit circle in the complex plane via the correspondence [tex] x \rightarrow e^{ix} [/tex]. At that point you notice that the unit circle in the complex plane is a group under the usual operation of multiplication, which on the unit circle is just addiction of angles modulo [tex] \2 pi [/tex]. Fourier analysis on the circle produces the theory of the Fourier series, with which you seem to be familiar.

    Now consider a function defined on the real line. There is also an operation of translation on the line, just ordinary addition. And the real line is also a group under that operation. There is an interesting operation that comes along with that translation operation. It is called convolution and it is important for several things -- differential equations being one such thing.

    The convolution of two functions [tex]f,g[/tex], denoted [tex]f*g[/tex] is defined as

    [tex] f*g (x) = \int_{-\infty}^{\infty} f(x-t)g(t) dt[/tex]

    and is defined for functions that are absolutely integrable.

    Now that convolution multiplication, as you can see, is a complicated looking animal. But the Fourier transform help to understand it better. Convolution is complicated, but pointwise multiplication of functions is easy and

    [tex] F(f*g)(x) = F(f)(x)F(g)(x)[/tex] where [tex]F[/tex] denotes the Fourier transform.

    With some more work one can also show that the Fourier transform takes a differential equation and transforms it to an algebraic equation. So you can solve differential equations by just transforming, doing some elementary algebra and transforming back. That is the origin of what in electrical engineering are called "transfer functions" and those are extremely useful in signal analysis and in control theory. You will find several books dedicated to this subject.

    Now, if we go back to the Fourier series, there is also a notion there of translation -- multiplication of numbers on the complex unit circle or just addition of angles modulo [tex] 2 \pi [/tex]. You can also define a convolution on the circle in an analogous way to what is done on the real line. And the Fourier series takes convolution on the circle to multiplication of Fourier series. Again a complicated thing is related to a simple thing.

    These two examples illustrate a general notion due to Pontryagin. One starts with a group (technically a locally compact abelian group) and looks at what is called the "dual group". Then for a function on the original group the Fourier transform can be defined and is a function defined on the dual group. The dual group is the set of continuous homomorphisms of the original group onto the group that we discussed above -- the unit circle in the complex plane. It turns out that the dual of the circle is the integers -- hence Fourier series. The dual of the real line is the real line itself -- hence the Fourier integral transform. But in a general sense they are all part of one basic structure.

    So the idea is that Fourier analysis is the study of functions defined on something with a notion of translation which results in an operation called convolution and Fourier analysis is a way to study that operation. It is useful because convolution turns out to be extremely important to things that are described by differential equations.

    At the heart of all of this is solutions to differential equations. That was the original motivation for Fourier series. Differential equations involving sines and cosines are easier to solve than differential equations with more general forcing functions. So you solve the equation in the general case of sines and cosines and then use Fourier series to piece together a solution to your more general problem. The same basic thing can be done with Fourier transforms, and more since you are not limited to periodic functions.

    The main thing to recognize is that the Fourier transform and really also the Fourier series, are ways of looking at a function in another setting and not necessarily direct representations of the function itself -- the Fourier series is a bit misleading in that regard. One way of thinking about it is that the function is in the "time domain" while the transform is in the "frequency domain". There are some arbitrary constants also involved in "normalizing" things (factors of [tex] \pi [/tex] and you may see somewhat different conventions adopted by different authors in books on the subject.

    There is a huge area of mathematics dedicated to this study. It is called harmonic analysis, and involves some deep mathematics. You can study this stuff for a lifetime.
  6. Jun 15, 2010 #5
    The idea is to represent a function with the help of basis functions.
    Let's say you have a function g(x) and you have basis functions b1(x) and b2(x). The question is how do the coefficients const1 and const2 look like such that:

    g(x) = const1*b1(x) + const2*b2(x)

    The Fourier transform tells you how the coefficients look like. For more information have a look at this:
    http://learntofish.wordpress.com/2009/08/28/understanding-the-fourier-transform-intuitively/" [Broken]

    We also have a thread here on Fourier series tutorials:

    The Fourier transform has applications in communications, astronomy, Geology and Optics:
    http://grus.berkeley.edu/~jrg/ngst/fft/applicns.html [Broken]
    Last edited by a moderator: May 4, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook