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Homework Help: Fourier coefficients of cos(4πt)

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    suppose that we have the continuous time signal

    x(t) = cos(4πt) with fundamental period of T=1/2

    2. Relevant equations

    [tex]a_{k} = \frac{1}{T} \int_{T}{}x(t)e^{-j\omega_{0}kt}dt[/tex]

    where [tex]\omega_{0}[/tex] is obviously [tex]\frac{2\pi}{1/2} = 4\pi[/tex]

    well the problem is that this integration becomes

    a_{k} = \frac{1}{T} \int_{T}{}x(t)e^{-j\omega_{0}kt}dt =
    2\int_{0}^{\frac{1}{2}}cos(4 \pi t)e^{-j4 \pi kt}dt=

    I find as a result 0, the calculation is complicated but using wolfram alpha's calculator I have the same result

    here is the picture

    [PLAIN]http://img52.imageshack.us/img52/6729/36554157.gif [Broken]

    as you can see

    we have

    [tex] 1 - e^{-2\pi j k} = 1 - ( cos(2\pi k) - jsin(2\pi k)) = 1 - 1 = 0[/tex]

    hence it must be 0
    now my book says the following

    The nonzero FS coefficients of x(t) are [tex]a_{1} = a_{-1} = 1/2 [/tex]

    but I have 0 as a result, and if I plug 1 or -1 I will get 0 again, what's wrong here?

    thanks in advance
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 17, 2011 #2


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    Hi Jncik! :smile:

    πj212 + π = πj2(-1)2 + π = 0 :wink:
  4. Jul 17, 2011 #3
    thanks tiny-tim, I didn't notice that :)
  5. Jul 17, 2011 #4


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    And the numerator is 0 also, giving an indeterminate form. You can best see the problem if you calculate the sine/cosine coefficients. You will see that the integrals for an and bn have a different form when n = 1, requiring a separate and easier integration. If you work out the complicated formulas for other n and plug n = 1 in those answers, you will get the same of indeterminate answer.
  6. Jul 18, 2011 #5
    thanks what I did for n = 1 first I tried to find out the result using the integration but I used these two formulas

    [tex]2\int_{0}^{\frac{1}{2}}cos(4 \pi t)e^{-4 \pi t} dt = 2\int_{0}^{\frac{1}{2}}cos(4 \pi t)cos(4 \pi t) dt - 2j\int_{0}^{\frac{1}{2}}cos(4 \pi t)sin(4 \pi t) dt = ... [/tex]

    because as you said it wouldnt work using just the integration without expanding the form of [tex]e^{-j 4 \pi t} [/tex]

    sadly our professor doesnt want the trigonometric expansion but prefers the series with the e^...

    after solving this I thought it would be better if I just used Euler's formula about the cos(4πt)

    it is known that

    [tex] cos(4\pi t) = \frac{e^{j 4 \pi t} + e^{-j 4 \pi t}}{2} [/tex]

    hence the coefficients are [tex]a_{1} = a_{-1} = \frac{1}{2} [/tex]

    but im not sure if this was what the exercise wanted cause this solution takes less than 5 seconds to finish
  7. Jul 18, 2011 #6


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    :rofl: :rofl: :rofl: :rofl:

    ruined your day, did it? :rolleyes:
  8. Jul 18, 2011 #7
    haha yes :p :rofl:
  9. Jul 18, 2011 #8


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    On a quiz I would expect that is exactly what the professor wanted. The point being for the student to recognize that trig functions like that are their own (finite) Fourier Series.
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