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Math Help - Exponential Fourier coefficients problem

  1. #1
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    Question Exponential Fourier coefficients problem

    Hi everyone,
    My problem is to do with calculating the fourier coefficients C_k of x(t)=cos(2t),
    where C_k = {1 \over T} \int_0^Tx(t)e^{-j\omega_0kt} dt where k \in \bold Z and \omega_0 = 2 \to T = \pi
    Everytime I try to evaluate the integral by letting cos(2t) = {1 \over 2}(e^{j2t} + e^{-j2t}) I get C_k = 0 which I am sure is incorrect as the fourier series of cos(2t) \not= 0. If some one could show me the working to this problem so I can compare it with mine to see where I am going wrong it would be greatly appreciated.
    Thank you,
    Mat
    Last edited by matyoun89; October 11th 2010 at 03:51 AM.
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  2. #2
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    Quote Originally Posted by matyoun89 View Post
    Hi everyone,
    My problem is to do with calculating the fourier coefficients C_k of x(t)=cos(2t),
    where C_k = {1 \over T} \int_0^Tx(t)e^{-j\omega_0kt} dt where k \in \bold Z and \omega_0 = 2 \to T = \pi
    Everytime I try to evaluate the integral by letting cos(2t) = {1 \over 2}(e^{j2t} + e^{-j2t}) I get C_k = 0 which I am sure is incorrect as the fourier series of cos(2t) \not= 0. If some one could show me the working to this problem so I can compare it with mine to see where I am going wrong it would be greatly appreciated.
    You have already found this Fourier series without realising it!

    In fact, it is correct to say that C_k = 0 for almost every value of k. But you should look more carefully to see what happens if k=2 or k=2. In those cases, the calculation of the integral is a bit different, and you should find that C_2 = C_{-2} = \frac12. Those are the only nonzero Fourier coefficients, so the Fourier series of \cos(2t) is \cos(2t) = \frac12(e^{j2t} + e^{-j2t}) (an equation that you had already written down without noticing that it gives the answer to the problem).
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  3. #3
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    Thanks for the Response however I'm still getting zero. I'll post my working for the integration when k = 2 below.

    X_2 = {1 \over \pi} \int_0^\pi {1 \over 2}(e^{2tj} + e^{-2tj})e^{-4tj} dt

    X_2 = {1 \over 2\pi} \int_0^\pi e^{2tj}e^{-4tj} + e^{-2tj}e^{-4tj} dt

    X_2 = {1 \over 2\pi} \int_0^\pi e^{-2tj} + e^{-6tj} dt

    X_k = {1 \over 2\pi}([{1 \over -2j}e^{-2tj}]_0^\pi + [{1 \over -6j} e^{-6tj}]_0^\pi)

    X_2 = {1 \over 2\pi}(-{1 \over 2j}e^{-2\pi j} + {1 \over 2j} - {1 \over 6j} e^{-6\pi j} + {1 \over 6j})

     e^{-2\pi j} = 1, e^{-6\pi j} = 1 \to X_2 = {1 \over 2\pi}(0) = 0

    any help spotting my problem?
    Thanks, Mat
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  4. #4
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    Quote Originally Posted by matyoun89 View Post
    Thanks for the Response however I'm still getting zero.

    [snip]

    any help spotting my problem?
    My fault, I was using the wrong scaling (I forgot that \omega_0 = 2). In fact, the critical values of k are 1 and 1, not 2 and 2. When you integrate \int_0^\pi\frac12\bigl(e^{2tj} + e^{-2tj}\bigr)e^{2tj}dt or \int_0^\pi\frac12\bigl(e^{2tj} + e^{-2tj}\bigr)e^{-2tj}dt, the answer comes out differently from other values of k.
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  5. #5
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    Thanks I worked it out,
    Turns out I had to take the limit of the integral when k=-1 or 1 in order to get the answer due to the division by (1-k) for one term and (1+k) for another.
    Mat.
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