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Fourier Transform and Parseval's Theorem

  1. Dec 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Using Parseval's theorem,

    $$\int^\infty_{-\infty} h(\tau) r(\tau) d\tau = \int^\infty_{-\infty} H(s)R(-s) ds$$

    and the properties of the Fourier transform, show that the Fourier transform of ##f(t)g(t)## is

    $$\int^\infty_{-\infty} F(s)G(\nu-s)ds$$

    2. Relevant equations
    Fourier transform for ##f(t)g(t)## is defined as:

    $$\int^\infty_{-\infty} f(t)g(t) e^{-2 \pi \nu t} dt$$

    3. The attempt at a solution
    So starting from the definition of Fourier transform:

    $$\int^\infty_{-\infty} f(t)g(t) e^{-2 \pi \nu t} dt$$

    So, do we need to ignore the exponential term here? If we ignore it, we can apply Parseval's theorem to get the frequency domain:

    $$\int^\infty_{-\infty} f(t)g(t) dt = \int^\infty_{-\infty} F(s) G(- s) d s$$

    Now, what property of the Fourier transform can I use to get ##G(-s) \implies G(\nu-s)##?

    I don't understand what the ##(\nu-s)## part means, does it indicates some sort of shifting or delay in the input? :confused:

    Any help is greatly appreciated.
     
  2. jcsd
  3. Dec 10, 2015 #2

    Samy_A

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    I think you forgot an ##i## in the exponent in the definition of Fourier transform.

    You cannot just ignore the exponential term in ##\int^\infty_{-\infty} f(t)g(t) e^{-2 \pi iv t} dt##

    Start with ##G(v-s)## for a fixed ##v##.
    Write out the formula for ##G(v-s)##, and you will note that ##G(v-s)=\mathcal F (g_v(-s))## for some function ##g_v## (##\mathcal F## denotes the Fourier transform).

    Using that knowledge about ##G(v-s)##, apply Parseval's theorem to ##\int^\infty_{-\infty} F(s)G(v-s)ds## and see what you get ...

    A little remark about terminology:
    It is clearer when you write something like:
    the Fourier transform of ##f(t)g(t)## in ##\nu## is $$\int^\infty_{-\infty} F(s)G(\nu-s)ds$$
     
    Last edited: Dec 10, 2015
  4. Dec 11, 2015 #3
    Thank you for your input. In ##g_v## what does the subscript ##v## mean? Should it not be ##g_t## since ##G(s)## is the Fourier transform of ##g(t)##? (In my course a function denoted by a capital letter is the Fourier transform of the function denoted by the corresponding small letter)

    So from the definition, the formula for the ##G(v-s)## part is:

    $$G(v-s) = \int^\infty_{-\infty} g(t) e^{-i 2 \pi st} dt$$

    Here I used the duality property of the Fourier transform i.e. ##F(t) \iff f(-s)##

    Is this the right formula?
     
  5. Dec 11, 2015 #4

    Samy_A

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    There is no ##v## in your right hand side expression for ##G(v-s)##, so that can't be correct.

    I really meant ##g_v##. Just pick one ##v## as a constant and let's work with that for a while.

    In general, the definition of the Fourier transform is:
    $$G(u) = \int^\infty_{-\infty} g(t) e^{-2 \pi iut} dt$$

    I deliberately used another name for the variable here.
    Now set ##u=v-s## in that definition. You get:
    $$G(v-s) = \int^\infty_{-\infty} g(t) e^{-2 \pi i(v-s)t} dt=\int^\infty_{-\infty} \Big(g(t)e^{- 2 \pi ivt}\Big)e^{-2 \pi i(-s)t} dt$$

    Compare that last expression with the definition of the Fourier transform. What we have there is the Fourier transform of the function between the big brackets at the point ##-s##. So now name the function between the big brackets ##g_v##, and what we have is that:
    $$g_v(t)=g(t)e^{-2 \pi ivt}$$
    $$G(v-s)=\mathcal F(g_v(-s))$$
    Or, if you prefer the convention of your course:
    $$G(v-s)=G_v(-s)$$
    Now apply this to evaluate ##\int^\infty_{-\infty} F(s)G(v-s)ds##, by using Parseval's theorem.
     
    Last edited: Dec 11, 2015
  6. Dec 12, 2015 #5

    Samy_A

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    Never mind. I thought something was wrong, but I was wrong about that.

    My sincere apologies for my confusion.
     
    Last edited: Dec 12, 2015
  7. Dec 12, 2015 #6
    I see. Thank you very much for the clear explanation.

    $$\int^\infty_{-\infty} F(s) G(\nu -s) ds = \int^\infty_{-\infty} F(s) G_\nu (-s) ds$$

    using Parseval's theorem this becomes:

    $$\int^\infty_{-\infty} f(t) g_\nu (t) dt = \int^\infty_{-\infty} f(t) g(t) e^{-2 \pi i \nu t} dt$$

    Is this then sufficient to 'show' that ##\int^\infty_{-\infty} F(s) G(\nu -s)ds## is the Fourier transform of ##f(t)g(t)##?
     
  8. Dec 12, 2015 #7

    Samy_A

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    Yes, because taking the two equations together, you now proved that for any ##\nu##: $$ \int^\infty_{-\infty} f(t) g(t) e^{-2 \pi i \nu t}=\int^\infty_{-\infty} F(s) G(\nu -s) ds$$
    The left hand side is by definition the Fourier transform of the function ##f.g## at point ##\nu##.

    As an aside, ##\int^\infty_{-\infty} F(s) G(\nu -s) ds## is called the convolution of F and G at ##\nu##. (More here and here)
     
    Last edited: Dec 12, 2015
  9. Dec 12, 2015 #8
    Thank you so much for your help. That was very helpful.
     
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