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Homework Help: Why is Fourier Transform of a Real Function Complex?

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the Fourier transform F(w) of the function f(x) = [e-2x (x>0), 0 (x ≤ 0)]. Plot approximate curves using CAS by replacing infinite limit with finite limit.

    2. Relevant equations
    F(w) = 1/√(2π)*∫ f(x)*e-iwxdx, with limits of integration (-∞,∞).

    3. The attempt at a solution
    I solved the integral and found the Fourier transform of f(x) to be:

    F(w) = 1/√(2π) * (2+iw)/(4+w2).

    I am pretty confident in my solution (but feel free to correct me if I'm wrong!). Where I have an issue is this.. F(w) is complex, so how do I plot it? Do I only plot the real part? How do the real and complex parts of F(w) relate to f(x)?

    I may have this question because I am still having a hard time fundamentally understanding the relationship between F(w) and f(x), so any information on this topic would be welcome. Describing F(w) as a function that determines the coefficient (contribution) of eiwx in f(x) makes some sense, (as explained in the below link):

    http://math.stackexchange.com/questions/1002/fourier-transform-for-dummies (answer #2, with plot of sines).

    However I still am confused how a complex-valued F(w) is tied in.

  2. jcsd
  3. Oct 13, 2015 #2


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    It would probably be more useful to plot the magnitude and phase of ##F(\omega)##.

    The expression for a sine wave of frequency ##\omega## is ##A \sin(\omega x + \phi)##. You need to specify two numbers, ##A## and ##\phi##, to describe the wave. ##|F(\omega)|## corresponds to ##A##, and the complex phase of ##F(\omega)## basically corresponds to ##\phi##.
  4. Oct 13, 2015 #3

    Buzz Bloom

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    Hi LunaFly:

    Here is a hint.

    e-iwx = cos wx + i sin wx

  5. Oct 13, 2015 #4
    Thank you Vela for the insight. So the Fourier transform F(w) is more of a phasor relating the "amount" of oscillatory function e-iwt of frequency w present in the function f(x) than a constant coefficient.. Interesting!
  6. Oct 13, 2015 #5
    Thanks for the hint Buzz. I am guessing you are pointing to the fact that when applying a Fourier transform to a function f(x), we are introducing a function with non-real terms, namely e-iwx. Or you may be hinting at the fact that the inverse Fourier transform is a continuous linear combination of complex terms weighted by F(w) that is equal to f(x), meaning the weight function F(w) must have complex terms in order to equal a real function f(x). Maybe? Thanks regardless.
  7. Oct 13, 2015 #6

    Buzz Bloom

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    Hi LunaFly:

    Actually I had a different thought than either of the two ideas in your post. Here is a second hint.

    Any real function f(x) over [-∞,+∞] can be separated into two parts: f(x) = s(x) + a(x)
    s(x) is symmetric, i.e., s(-x) = s(x)
    a(x) is anti-symmetric, i.e., a(-x) = -a(x).
    Now consider the transform: F(w) = S(w) + A(w).

    Now combine that idea with my first hint.
    Good luck.

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