As I see in this slides, Fourier series representation for discrete-time signal $s[n]$ with period $N$ is $\sum_{k = 0}^{N-1} c_k e^{j2\pi k n / N}$

According to Wiki, Fourier series representation for continuous-time signal $s(x)$ with period $P$ is $s_{\infty}(x) = \sum_{n = -\infty}^\infty c_n e^{j 2\pi n x / P}$

My question is that why in discrete-time Fourier series representation, they only sum up from $0$ to $N - 1$ instead of from $-\infty$ to $\infty$ ? Is the slides above gives a wrong formula ?

P.s: in my opinion, the more terms we have, the more accurate our Fourier series is

  • $\begingroup$ Hi! not a duplicate but the answer is highly related with this one... $\endgroup$ – Fat32 Dec 18 '18 at 10:50

Because the discrete-time basis functions $e^{j2\pi kn/N}$ are not only $N$-periodic in $n$ (as is the signal $s[n]$), but also in $k$. So adding terms for $k\ge N$ does not add anything new, simply because

$$e^{j2\pi (k+N)n/N}=e^{j2\pi kn/N}e^{j2\pi n}=e^{j2\pi kn/N}$$

  • $\begingroup$ According to your answer, in case of continuous-time signals, if the period $P$ is an integer, we only have to sum up from $0$ to $P$, is it right ? Because $e^{j 2\pi n (x + P) / P} = e^{j 2\pi n x / P} e^{j 2\pi n}$ $\endgroup$ – HOANG GIANG Dec 18 '18 at 10:55
  • $\begingroup$ @HOANGGIANG: My answer is about the discrete-time basis functions, so I'm not sure I understand your question. Integrate what? $\endgroup$ – Matt L. Dec 18 '18 at 10:56
  • $\begingroup$ I've edited my comment $\endgroup$ – HOANG GIANG Dec 18 '18 at 10:57
  • $\begingroup$ @HOANGGIANG: In order to compute the Fourier coefficients of a continuous-time $P$-periodic function, we have to integrate the function (multiplied by the basis functions) over an interval of length $P$, no matter whether $P$ is an integer or not. $\endgroup$ – Matt L. Dec 18 '18 at 10:57
  • $\begingroup$ @HOANGGIANG: No, because $e^{j2\pi (n+P)x/P}\neq e^{j2\pi nx/P}$ since $x$ is real-valued. $\endgroup$ – Matt L. Dec 18 '18 at 11:01

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