Revision as of 11:08, 14 February 2011

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) . \ $

Share your answers below

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 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 $

a₀ = 0,a₁ = 0

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

TA's comment: Yes. Your answer is correct. All coefficients are zero. You can check your answer by noticing that $ x[n]=cos(3\pi n+\pi/2)=-\sin(3\pi n)=0 \mbox{ for all } n $.

It doesn't look right intuitively.  From a_k, you are supposed to be able to get back your original signal. What you have is a_k = 0 for all values of k and therefore is a null signal.

How's this?

If $ x[n] = \cos \left(3\pi n + \frac{\pi}{2} \right) . \ $ then a_k must be something since you get x[n] by summing all the values of a_k multiplied by a factor of e. $ cos(3\pi*n+\pi/2) = \frac{e^(j(3\pi*n+\pi/2)+e^{-j*(3\pi*n+\pi/2)}}{2} =1/2e^{j3\pi*n}e^{j*\pi/2} + 1/2e^{-j3\pi*n}e^{-j\pi/2}= $

$ \frac{1}{2}*e^{j3\pi*n}*e^{j\pi/2}+\frac{1}{2}*e^{-j*3\pi*n}*e^{-j*\pi/2} $

$ a_3=1/2*e^{\pi/2} $

$ a_{-3}=1/2*e^{-j\pi/2} $ (Clarkjv 11:38, 11 February 2011 (UTC))

Back to ECE301 Spring 2011 Prof. Boutin