Practice Question on Computing the Fourier Series coefficients of a discrete-time (sampled) cosine wave

Obtain the Fourier series coefficients of the DT signal

$ x[n] = \cos \left(3\pi n + \frac{\pi}{2} \right) . \ $

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Answer 1

for c'o's(n), the coefficients are $ a_1=\frac{1}{2},a_{-1}=\frac{1}{2}, a_k=0 \mbox{ for }k\ne 1,-1 $

Time shift property: $ x(n-n_0) \to e^{-jkw_0n_0}a_k $

Thus with $ w_0=3\pi\, $ and $ n_0=\frac{-\pi}{2} $,

$ a_1=\frac{e^{j 3 \pi \frac{\pi}{2}}}{2},a_{-1}=\frac{e^{-j 3 \pi \frac{\pi}{2}}}{2}, a_k=0 \mbox{ for }k\ne 1,-1 $

Is that right? I'm not sure about the time shift property.

--Cmcmican 21:53, 7 February 2011 (UTC)

Student Question: Since this is DT and not CT, shouldn't the focus be on N=2 and not w_o? (Clarkjv 20:36, 8 February 2011 (UTC))

Student Response: Yeah, it should be. I did this before today's lecture, and made some mistakes. I'm posting a new answer in answer 2. --Cmcmican 19:46, 9 February 2011 (UTC)

Answer 2

Based on lecture today, I am changing my answer to the following:

$ N=\frac{2\pi}{3\pi}k=2 $ so there will be two coefficients.

$ x[n]=\frac{1}{2}e^{j(3\pi n + \frac{\pi}{2})}+\frac{1}{2}e^{-j(3\pi n + \frac{\pi}{2})}=\frac{1}{2}e^{j\frac{\pi}{2}}e^{j3\pi n}+\frac{1}{2}e^{-j\frac{\pi}{2}}e^{-j3\pi n}=\frac{j}{2}e^{j3\pi n}-\frac{j}{2}e^{-j3\pi n} $

and $ e^{j3\pi n}=e^{-j3\pi n}=e^{j\pi n}\, $

so $ x[n]=\frac{j}{2}e^{j\pi n}-\frac{j}{2}e^{j\pi n}=0 $

$ a_0=0, a_{1}=0 $

is that right? --Cmcmican 20:01, 9 February 2011 (UTC)

Answer 3

Write it here.

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