1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [tex]
    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=
    ...=0
    [/tex]

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

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

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

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member


    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fourier coefficients of cos(4πt)
  1. Fourier Coefficient (Replies: 0)

  2. Fourier coefficients (Replies: 1)

  3. Fourier coefficients (Replies: 8)

  4. Fourier coefficients (Replies: 1)

  5. Fourier coefficients (Replies: 2)

Loading...