Results 1 to 6 of 6

Math Help - Findind the complex Fourier Series and deducing the real one

  1. #1
    Junior Member
    Joined
    Feb 2011
    Posts
    25

    Findind the complex Fourier Series and deducing the real one

    I'm trying to find the complex form for this fourier series and then transform it into the real one.I don't know exactly how this transformation is suppose to happen..I'm thinking it must have something to do with e^{-ikt}=\cos (kt)+i\sin (kt).Also when I try to calculate the complex form I get zero which means I"m doing something wrong.
    F(t) is defined as:
    f(t)=-1  \ldots  -(\pi)=<t<0  and f(t)=-1 \ldots   0<t<\pi
    Now c_{k}=\frac{1}{2/pi}*\int {f(x) e^{-ikt}}(don't know how to define a definite integral on latex but this one is from -pi to pi)
    => c_{k}=\frac{1}{2/pi}*(\int { -e^{-ikt}}+\int{e^{-ikt}})
    (the first integral is from -pi->0 and the second from 0->pi)
    => \frac{1}{2(/pi)ik}[1-e^{-ikt}+e^{ikt}-1]=0.
    This is not the correct answer.Can someone please help me?
    Last edited by StefanM; April 30th 2011 at 11:58 AM. Reason: Fixed the LaTeX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by StefanM View Post
    I'm trying to find the complex form for this fourier series and then transform it into the real one.I don't know exactly how this transformation is suppose to happen..I'm thinking it must have something to do with e^{-ikt}=\cos (kt)+i\sin (kt).Also when I try to calculate the complex form I get zero which means I"m doing something wrong.
    F(t) is defined as:
    f(t)=\begin{cases}-1  & ( -\pi\leqslant t<0), \\ 1&(0<t<\pi).\end{cases}
    Now c_{k}=\int_{-\pi}^\pi f(t) e^{-ikt}dt

    \Rightarrow\;c_{k}=\frac{1}{2\pi}\Bigl(\int_{-\pi}^0  -e^{-ikt}dt+\int_0^\pi e^{-ikt}dt\Bigr)
    . . . .. . . = \frac{1}{2\pi ik}\Bigl[1-e^{-ikt}+e^{ikt}-1\Bigr]=0.
    This is not the correct answer.Can someone please help me?
    (I have tweaked the TeX a bit there and corrected minor errors.)

    The last line is wrong. you should have \int_{-\pi}^0  -e^{-ikt}dt = -\frac{1-e^{ik\pi}}{-ik} = \frac{1-(-1)^k}{ik} (remembering that e^{ik\pi} = (-1)^k). Similarly, \int_0^\pi e^{-ikt}dt = \frac{e^{-ik\pi}-1}{-ik}, which is also equal to \frac{1-(-1)^k}{ik}. To complete the calculation, you probably need to notice that 1-(-1)^k is 0 if k is even, and 2 if k is odd.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2011
    Posts
    25
    I'm such an idiot...it was supposed to be c_{k}=\frac{1}{2/pi}*\int {f(x) e^{-ikt}},sorry about that...thanks for the reply.
    How can I transforme the series from complex to real form?According to the answer book the result should be...(4/pi)*SUM[(2m+1)^(−1) *sin((2m+1)t)]
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by StefanM View Post
    How can I transform the series from complex to real form? According to the answer book the result should be...(4/pi)*SUM[(2m+1)^(−1) *sin((2m+1)t)]
    Use the facts that \cos kt = \tfrac12(e^{ikt}+e^{-ikt}) and \sin kt = \tfrac1{2i}(e^{ikt}-e^{-ikt}). That tells you that the cosine coefficients are given by a_k = c_k+c_{-k}, and the sine coefficients are given by b_k = i(c_k-c_{-k}).

    In this example, you should find that the complex Fourier coefficients c_k have the property that c_{-k} = -c_k, so all the cosine terms in the real Fourier series vanish. Also, c_k=0 when k is odd, so the only nonzero sine terms occur for odd values of k. Write k = 2m+1 for those terms, and you should get the given solution.
    Last edited by Opalg; May 2nd 2011 at 03:40 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2011
    Posts
    25
    I found the right answer....but the problem is that I was under the impression that I did that by accident...so I tried another ex. and unfortunately this was the case.....can you please give me a full example of a fourier transformation from complex to real?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by StefanM View Post
    I found the right answer....but the problem is that I was under the impression that I did that by accident...so I tried another ex. and unfortunately this was the case.....can you please give me a full example of a fourier transformation from complex to real?
    The formulas I gave in my previous comment were wrong (I have now edited it to correct them). They should be a_k = c_k+c_{-k} (for the cosine coefficients) and b_k = i(c_k-c_{-k}) (for the sine coefficients).

    For the square wave function in this example, you should have found that the complex Fourier coefficients are given by

    c_k = \begin{cases}0& (k \text{ even,}) \\ \tfrac2{k\pi i}& (k \text{ odd.}) \end{cases}

    Here, you can see that c_{-k} = c_k = 0 if k is even, and c_{-k} = -c_k if k is odd. So the cosine coefficients are all zero. For the sine coefficients, b_k = 0 if k is even, and if k is odd then b_k = i(c_k-c_{-k}) = 2ic_k = \tfrac4{k\pi}. Since k is odd, we can put k=2m+1. Then you see that the coefficient of \sin((2m+1)t) is \tfrac4{(2m+1)\pi}, for m=1,2,3,..., and all the other terms in the Fourier series are 0.

    That gives you the formula f(t) = \frac4\pi\sum_{m=1}^\infty (2m+1)^{-1}\sin((2m+1)t), just as the book says.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex Fourier Series
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 19th 2011, 11:42 PM
  2. Complex Fourier Series
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 10th 2011, 08:45 PM
  3. Complex fourier series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 6th 2010, 03:02 PM
  4. Complex Fourier Series & Full Fourier Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 9th 2009, 05:39 AM
  5. Complex Fourier series
    Posted in the Calculus Forum
    Replies: 0
    Last Post: April 4th 2009, 04:02 PM

Search Tags


/mathhelpforum @mathhelpforum