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Phase Spectra from Fourier Transform

  1. Oct 13, 2011 #1
    How can you read the phase spectra from a Fourier Transform?

    if [tex]g(t) = Sin(2\pi f_{c}t)[/tex]

    then for the single sided spectrum, you have one frequency component at [tex]f=f_{c}[/tex] with a height of [tex]\frac{1}{j}[/tex] which from looking at the complex plane, corresponds to a phase of [tex]\frac{\pi }{2}[/tex] (ie. [tex]g(t) = Sin(2\pi f_{c}t)[/tex] is made up of a cosine component with [tex]f=f_{c}[/tex] and phase = [tex]\frac{\pi }{2}[/tex].

    But, if you consider [tex]\frac{1}{j} = -j[/tex], then the phase would correspond to [tex]\frac{3\pi }{2}[/tex] which would in effect negate the amplitude ([tex]Cos(x - \frac{3\pi }{2}) = -Cos(x - \frac{\pi }{2})[/tex].

    So which complex amplitude should be considered correct?[tex]\frac{1}{j} or -j[/tex] ?
  2. jcsd
  3. Oct 13, 2011 #2
    I've worked it out (for anyone whose found this in a google search or something).

    To find the phase spectra you need to take the arctangent of the ratio of imaginary to real components, and NOT merely look at the position of the vector on the complex plane.

    So first you need to get those imaginary components to have j on the numerator instead of on the denominator, that way you know the coefficient of it.

    So for Sin(sPIfct) the impulse at f=fc has an imaginary coefficient of -1/2, so the ratio of imaginary to real is -infinity (since there is no real part, real=0), arctan of -infinity = -PI/2 radians.

    For the impulse at f=-fc, the coefficient of the imaginary component is 1/2, the ratio is +infinity, arctan of +infinity = +pi/2 :)
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