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===Answer 1=== | ===Answer 1=== | ||
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| + | for <math>cos(n)</math>, the coefficients are <math>a_1=\frac{1}{2},a_{-1}=\frac{1}{2}, a_k=0 \mbox{ for }k\ne 1,-1</math> | ||
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| + | Time shift property: <math>x(n-n_0) \to e^{-jkw_0n_0}a_k</math> | ||
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| + | Thus with <math>w_0=3\pi\,</math> and <math>n_0=\frac{-\pi}{2}</math>, | ||
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| + | <math>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</math> | ||
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| + | Is that right? I'm not sure about the time shift property. | ||
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| + | --[[User:Cmcmican|Cmcmican]] 21:53, 7 February 2011 (UTC) | ||
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===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 17:53, 7 February 2011
Contents
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) . \ $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
for $ cos(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)
Answer 2
Write it here.
Answer 3
Write it here.
