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mathwurkz
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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.
f(t) = \frac{2t}{T} \ \ \ 0 < t < 2T\\
So if the period is 2l = 2T then l = T
\\ c_n = \frac{1}{2l} \int_{-l}^{l} f(x) e^{in\pi x / l} dx \\<br /> 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 \\
Then I used integration by parts.
\\ u = t \ \ \ du = dt\ \ \ dv = e^{-in\pit / T} \ \ \ v = -\frac{T}{in \pi }e^{-in\pit / T}\\
And here we go...
\\ = \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]\\
I skip a few steps too much nitty gritty and then I arrive at...
\\ \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]\\
and since I know:
\\ e^{+-iat} = \cos{at} +- i \sin{at} \\{
then it becomes
\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]\\
and the end result is...
c_n = -\frac{2(-1)^n}{in\pi }\\
and the Fourier series is...
f(t) = \sum_{-\infty}^{infty} -\frac{2(-1)^n}{in\pi } e^{in\pi t / T}
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 G(t) = ||G(t)||e^{i\phi t} where ||G(t)|| is the amplitude and \phi (t) 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: \omega _0 = \frac{\pi}{T} \ \ \ c_n = ||c_n||e^{i \phi}\\
So what I did was find ||c_n||...
||c_n|| = \frac{2}{\pi n} = \frac{2}{T \omega _0 n}}\\
Is this right though? And how do I go about finding the phase angle? and then plotting it vs omega. thanks for any help.
Q:Obtain the complex form of the Fourier series of the sawtooth function.
f(t) = \frac{2t}{T} \ \ \ 0 < t < 2T\\
So if the period is 2l = 2T then l = T
\\ c_n = \frac{1}{2l} \int_{-l}^{l} f(x) e^{in\pi x / l} dx \\<br /> 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 \\
Then I used integration by parts.
\\ u = t \ \ \ du = dt\ \ \ dv = e^{-in\pit / T} \ \ \ v = -\frac{T}{in \pi }e^{-in\pit / T}\\
And here we go...
\\ = \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]\\
I skip a few steps too much nitty gritty and then I arrive at...
\\ \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]\\
and since I know:
\\ e^{+-iat} = \cos{at} +- i \sin{at} \\{
then it becomes
\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]\\
and the end result is...
c_n = -\frac{2(-1)^n}{in\pi }\\
and the Fourier series is...
f(t) = \sum_{-\infty}^{infty} -\frac{2(-1)^n}{in\pi } e^{in\pi t / T}
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 G(t) = ||G(t)||e^{i\phi t} where ||G(t)|| is the amplitude and \phi (t) 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: \omega _0 = \frac{\pi}{T} \ \ \ c_n = ||c_n||e^{i \phi}\\
So what I did was find ||c_n||...
||c_n|| = \frac{2}{\pi n} = \frac{2}{T \omega _0 n}}\\
Is this right though? And how do I go about finding the phase angle? and then plotting it vs omega. thanks for any help.
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