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| − | = Practice Question on Computing the Fourier Series coefficients of a discrete-time (sampled) cosine wave = | + | = [[:Category:Problem_solving|Practice Question]] on Computing the Fourier Series coefficients of a discrete-time (sampled) cosine wave = |
Obtain the Fourier series coefficients of the DT signal | Obtain the Fourier series coefficients of the DT signal | ||
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| − | == Share your answers below == | + | == 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! | 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! | ||
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| − | === Answer 1 === | + | === Answer 1 === |
for <span class="texhtml">''c''''o''''s''(''n'')</span>, the coefficients are <math>a_1=\frac{1}{2},a_{-1}=\frac{1}{2}, a_k=0 \mbox{ for }k\ne 1,-1</math> | for <span class="texhtml">''c''''o''''s''(''n'')</span>, 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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Student Question: Since this is DT and not CT, shouldn't the focus be on N=2 and not <span class="texhtml">''w''<sub>''o''</sub></span>? ([[User:Clarkjv|Clarkjv]] 20:36, 8 February 2011 (UTC)) | Student Question: Since this is DT and not CT, shouldn't the focus be on N=2 and not <span class="texhtml">''w''<sub>''o''</sub></span>? ([[User:Clarkjv|Clarkjv]] 20:36, 8 February 2011 (UTC)) | ||
| − | Student Response: Yeah, it should be. | + | 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. --[[User:Cmcmican|Cmcmican]] 19:46, 9 February 2011 (UTC) |
| − | === Answer 2 === | + | === Answer 2 === |
| − | Based on lecture today, I am changing my answer to the following: | + | Based on lecture today, I am changing my answer to the following: |
| − | <math>N=\frac{2\pi}{3\pi}k=2</math> so there will be two coefficients. | + | <math>N=\frac{2\pi}{3\pi}k=2</math> so there will be two coefficients. |
| − | <math>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}</math> | + | <math>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}</math> |
| − | and <math>e^{j3\pi n}=e^{-j3\pi n}=e^{j\pi n}\,</math> | + | and <math>e^{j3\pi n}=e^{-j3\pi n}=e^{j\pi n}\,</math> |
| − | so <math>x[n]=\frac{j}{2}e^{j\pi n}-\frac{j}{2}e^{j\pi n}=0</math> | + | so <math>x[n]=\frac{j}{2}e^{j\pi n}-\frac{j}{2}e^{j\pi n}=0</math> |
| − | < | + | <span class="texhtml">''a''<sub>0</sub> = 0,''a''<sub>1</sub> = 0</span> |
| − | is that right? | + | is that right? --[[User:Cmcmican|Cmcmican]] 20:01, 9 February 2011 (UTC) |
| − | --[[User:Cmcmican|Cmcmican]] 20:01, 9 February 2011 (UTC) | + | |
| + | <span style="color:green"> | ||
| + | TA's comment: Yes. Your answer is correct. All coefficients are zero. You can check your answer by noticing that <math>x[n]=cos(3\pi n+\pi/2)=-\sin(3\pi n)=0 \mbox{ for all } n</math>. | ||
| + | </span> | ||
| − | |||
| − | + | It doesn't look right intuitively. From a<sub>k</sub>, you are supposed to be able to get back your original signal. What you have is a<sub>k</sub> = 0 for all values of k and therefore is a null signal. | |
| − | + | How's this? | |
| − | [ | + | If <math>x[n] = \cos \left(3\pi n + \frac{\pi}{2} \right) . \ </math> then a<sub>k</sub> must be something since you get x[n] by summing all the values of a<sub>k</sub> multiplied by a factor of e. |
| + | <math>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}=</math> | ||
| − | [[ | + | <math>\frac{1}{2}*e^{j3\pi*n}*e^{j\pi/2}+\frac{1}{2}*e^{-j*3\pi*n}*e^{-j*\pi/2}</math> |
| + | |||
| + | <math> | ||
| + | a_3=1/2*e^{\pi/2} | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | a_{-3}=1/2*e^{-j\pi/2} | ||
| + | </math> ([[User:Clarkjv|Clarkjv]] 11:38, 11 February 2011 (UTC)) | ||
| + | |||
| + | ---- | ||
| + | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
Latest revision as of 10:24, 11 November 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 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 wo? (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 ak, you are supposed to be able to get back your original signal. What you have is ak = 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 ak must be something since you get x[n] by summing all the values of ak 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))
