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Practice Question on Computing the Fourier Series coefficients of a sine wave
Obtain the Fourier series coefficients of the CT signal
$ x(t) = \sin \left(3\pi t + \frac{\pi}{2} \right) . \ $
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Answer 1
for $ sin(t) $, the coefficients are $ a_1=\frac{1}{2j},a_{-1}=\frac{-1}{2j}, a_k=0 \mbox{ for }k\ne 1,-1 $
Time shift property: $ x(t-t_0) \to e^{-jkw_0t_0}a_k $
Thus with $ w_0=3\pi\, $ and $ t_0=\frac{-\pi}{2} $,
$ a_1=\frac{e^{j 3 \pi \frac{\pi}{2}}}{2j},a_{-1}=\frac{-e^{-j 3 \pi \frac{\pi}{2}}}{2j}, a_k=0 \mbox{ for }k\ne 1,-1 $
Is that right? I'm not sure about the time shift property.
--Cmcmican 21:09, 7 February 2011 (UTC)
- Instructor's comment: we will see the time shifting property later. Can you solve the problem without it? Perhaps you could write sin(u) as a sum of two exponentials, and then replace u by what is inside the sine. You should be able to factor out the phase as a separate exponential (a constant) in front of a complex exponential function. -pm
So like this?
$ sin(t)=\frac{1}{2j}e^{jkw_0t}-\frac{1}{2j}e^{-jkw_0t} $
$ x(t)=\frac{1}{2j}e^{jk3\pi(t+\frac{\pi}{2})}-\frac{1}{2j}e^{-jk3\pi(t+\frac{\pi}{2})} $
therefore,
$ a_1=\frac{e^{j 3 \pi \frac{\pi}{2}}}{2j},a_{-1}=\frac{-e^{-j 3 \pi \frac{\pi}{2}}}{2j}, a_k=0 \mbox{ for }k\ne 1,-1 $
--Cmcmican 08:23, 8 February 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.
