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Homework Help: Complex Fourier Series and Phase Spectra

  1. Dec 1, 2005 #1
    Please check my solution and I need help on understanding the second part of the question.
    Q:Obtain the complex form of the fourier series of the sawtooth function.
    [tex]f(t) = \frac{2t}{T} \ \ \ 0 < t < 2T\\[/tex]
    So if the period is 2l = 2T then l = T
    [tex]\\ c_n = \frac{1}{2l} \int_{-l}^{l} f(x) e^{in\pi x / l} dx \\
    c_n = \frac{1}{2T} \int_{-T}^{T} \frac{2t}{T} e^{in\pi t / T} dt = \frac{1}{T^2} \int_{-T}^{T} te^{-in\pi t / T} dt \\[/tex]
    Then I used integration by parts.
    [tex]\\ u = t \ \ \ du = dt\ \ \ dv = e^{-in\pit / T} \ \ \ v = -\frac{T}{in \pi }e^{-in\pit / T}\\[/tex]
    And here we go...
    [tex]\\ = \frac{1}{T^2}\left[-t\frac{T}{in\pi }e^{-in\pi t / T}|_{-T}^{T} + \frac{T}{in\pi } \int_{-T}^{T} e^{-in\pi t / T} dt\right]\\[/tex]
    I skip a few steps too much nitty gritty and then I arrive at...
    [tex]\\ \frac{1}{in\pi } \left[ \left(-e^{-in\pi } - e^{in\pi }\right) - \frac{1}{in \pi} \left( e^{-in\pi - e^{in\pi }\right)\right]\\[/tex]
    and since I know:
    [tex]\\ e^{+-iat} = \cos{at} +- i \sin{at} \\{[/tex]
    then it becomes
    [tex]\frac{1}{in\pi } \left[ \left(-\cos{n\pi } - \cos{n\pi }\right) - \frac{1}{in \pi} \left( \cos{n\pi } - \cos{n\pi }\right)\right]\\[/tex]
    and the end result is...
    [tex]c_n = -\frac{2(-1)^n}{in\pi }\\[/tex]
    and the fourier series is...
    [tex]f(t) = \sum_{-\infty}^{infty} -\frac{2(-1)^n}{in\pi } e^{in\pi t / T}[/tex]
    The second part the question, the one I do not understand is it asks to find and plot the discrete amplitude and phase spectra for f(t) above. In general, a complex quantity G(t), can be written [tex] G(t) = ||G(t)||e^{i\phi t}[/tex] where ||G(t)|| is the amplitude and [tex]\phi (t)[/tex] is the phase angle. When these quantities are function of angular frequency, w, the plots resulting from their graphs vs w are called spectra.
    hints: [tex]\omega _0 = \frac{\pi}{T} \ \ \ c_n = ||c_n||e^{i \phi}\\[/tex]
    So what I did was find ||c_n||...
    [tex]||c_n|| = \frac{2}{\pi n} = \frac{2}{T \omega _0 n}}\\[/tex]
    Is this right though? And how do I go about finding the phase angle? and then plotting it vs omega. thanks for any help.
    Last edited: Dec 1, 2005
  2. jcsd
  3. Dec 1, 2005 #2
    \frac{1}{in\pi } \left[ (-e^{-in\pi } &- e^{in\pi }) - \frac{1}{in \pi} ( e^{-in\pi} - e^{in\pi })\right]\notag\\
    &\neq \frac{1}{in\pi } \left[ \left(-\cos{n\pi } - \cos{n\pi }\right) - \frac{1}{in \pi}
    ( \cos{n\pi } - \cos{n\pi })\right]
    Last edited: Dec 1, 2005
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