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The Real Purpose of Laplace/Fourier Transform?

  1. May 11, 2010 #1
    Well, I do under stand *what* they do but I have trouble understanding the purpose of these transforms in simple words. Laplace transform converts an arbitrary single into exponentials while Fourier transform converts a given signal into sinusoids.

    As for the purpose, is the following right?:

    We use these techniques, basically, to convert a signal with *no form* into signals with can be specified with certain equations and into specific portions with form. Is that right?

    Thanks.
     
  2. jcsd
  3. May 11, 2010 #2
    From a purely selfish point of view, taking the transform of a differential equation changes it into an algebraic one.

    Which would you prefer to solve?
     
  4. May 11, 2010 #3

    sophiecentaur

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    The question doesn't have a proper answer. Maths is a set of symbolic operations and it stands on its own. Sometimes the operations have a correspondence to some physical reality and sometimes not. The two transforms you quote happen to be of use in solving problems in the 'real world' so I would say they have 'uses' rather than "purposes".
    Btw, the Fourier transform is not only used to transform between frequency and time domains - it can be used in spacial / angular domains too (Antenna design, optics and X ray diffraction). Same sums - different applications.
     
  5. May 17, 2010 #4
    I can only tell the uses of Laplace/Fourier transform which I know. There may be plenty too which I don't know. But there can't be an answer if you ask "purpose" of those transforms. It depends on people who use it for the given application.
    In general any transform if used will represent the given signal in to an equivalent form which is convenient for the user to manipulate.
    Well, Laplace transform can be used solve differential equations in a simple manner. The way it is done is once you take the Laplace transform of given differential equation, it will be converted to algebraic equation which can be easily manipulated and taking inverse Laplace transform will give solution to the equation.
    Fourier transform represents the signal by its frequency content. If I want to know the frequency spectrum of my signal in hand, FT is the best choice.
    Note: Any good transform wont alter the signal. It just converts the signal in to its equivalent form in another domain. Nothing lost or nothing gained.
     
  6. May 17, 2010 #5
    Have you ever used phasors before? Take the fourier transform of [tex]i=C \frac{d v}{dt}[/tex], and rewrite this in the form of [tex]Z(\omega)=\frac{V(\omega)}{I(\omega)}[/tex]. What do you get?
     
  7. May 17, 2010 #6
    Sorry I pressed submit too soon...

    Note: There are many more uses of the fourier/laplace transform and mine was just a very very small example.
     
  8. May 18, 2010 #7
    Maybe a real world example would help make it clear. Think of the way that you hear sound. Let's say that someone is pressing two piano keys at the same time. If you have some musical training then you should be able to figure out which keys were pressed but anyone will know that they're hearing one high-pitched key and one low-pitched key. How can you know that? You're ear is only getting one signal, the mixed sound of two keys. Somehow, you're brain is able to separate the notes.

    Intuitively, we know that we should be able to take a signal and separate out it's frequency components just like our brain does. The Fourier transform gives us a mathematical way to do it. It takes a time signal and transforms it into a spectrum of frequency components. It allows us to work in the frequency domain to understand how systems will behave.
     
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