# Correct form of discrete-time Fourier series representation

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

• Hi! not a duplicate but the answer is highly related with this one... – 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
$$e^{j2\pi (k+N)n/N}=e^{j2\pi kn/N}e^{j2\pi n}=e^{j2\pi kn/N}$$
• 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 $e^{j 2\pi n (x + P) / P} = e^{j 2\pi n x / P} e^{j 2\pi n}$ – HOANG GIANG Dec 18 '18 at 10:55
• @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. – Matt L. Dec 18 '18 at 10:57
• @HOANGGIANG: No, because $e^{j2\pi (n+P)x/P}\neq e^{j2\pi nx/P}$ since $x$ is real-valued. – Matt L. Dec 18 '18 at 11:01