While determining Fourier coefficients we have this equation $$\int^{T}_{0} x(t) e^{-jn\omega_0t} dt = \sum^{+\infty}_{k\ =\ -\infty} a_k [\int^{T}_{0} e^{j(k-n)\omega_0t}dt]$$

I want to ask that how can we get final equation of $a_n$ as: $$a_n= 1/T \int^{T}_{0} x(t)e^{-jn\omega_0t}dt$$ from the above equation. please any one help to explain this

  • $\begingroup$ This answer to a related question also answers your question. $\endgroup$
    – Matt L.
    Oct 6, 2015 at 11:38
  • $\begingroup$ @MattL. while solving the integral part in equation 2 how $sin$ will be $\ 0 \ \forall k$ as we are solving integral for $T$, there is no $k$ in the integral $\endgroup$ Oct 7, 2015 at 4:44
  • $\begingroup$ @MattL. can you please answer this question $\endgroup$ Oct 19, 2015 at 7:37

1 Answer 1


For the periodic signal $x(t)$ with fundamental period $T=1/F_0$, proceeding from where you are:

$$ \Rightarrow \sum^{+\infty}_{k\ =\ -\infty} a_k \left[\int^{T}_{0} e^{j(k-n)\omega_0t}dt\right] = \sum^{+\infty}_{k\ =\ -\infty} a_k \left[\frac{e^{j(k-n)\omega_0t}}{j(k-n)\omega_0}\right]^{T}_{0} $$

  1. For $k \neq n$ you got $0$ (you can prove that)

  2. And for $k=n$ you see that $\int^{T}_{0}dt = T$

You therefore have:

$$\int^{T}_{0} x(t)e^{-jn\omega_0t} dt= a_kT = a_nT$$

You can proceed from there.


  • We do not assume the existence of the two possible solutions. The right-hand side of the equation evaluates to that. It can be written in compact form using the unit impulse which has two distinct values depending on the index. (see the Kronecker delta in digital signal processing). Think $\delta_{kn}$.
  • We still need the summation, but we can start with the integral. It's only that the integral is zero for all values of $k$ except at $k=n$ where it is $T$. So you remain with one term in your summation.

All in all, your initial equation equals: $\displaystyle \sum^{+\infty}_{k\ =\ -\infty} a_kT\delta_{kn}$

  • $\begingroup$ after the solving the integral, we assume two possible solution to solve the summation. 1) $k\neq n$ and 2) $k=n$... please tell me is it correct what I understand? Second thing I want to know is when we these two conditions then we don't need to summation? $\endgroup$ Oct 7, 2015 at 4:37
  • $\begingroup$ @AadnanFarooqA I have edited the answer. $\endgroup$
    – Gilles
    Oct 7, 2015 at 9:01
  • $\begingroup$ @AadnanFarooqA , any tick for the answer ? $\endgroup$
    – Gilles
    Oct 8, 2015 at 17:08

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