How do I arrive from the Fourier series to discrete Fourier transform?

Alright. So let's lay some ground rules first ! Fourier Series - The continuous periodic function s(x)s(x) s(x) is being re-imagined as series of infinite/finite sinusoids, with a set of coefficients an,bnan,bna_n, b_n guarding sine & cosine wave respectively. This can be mathematically seen in the following equation - If you need to visualise this for a square wave or a saw-tooth wave, see the following page - http://en.wikipedia.org/wiki/Fourier_series  Fourier Transform - The continuous aperiodic function f(t)f(t) f(t) is transformed into its frequency domain equivalent F(ω)F(ω) F(\omega) using the following mathematical relation - F(ω)=∫∞−∞f(t)e−iωtdtF(ω)=∫−∞∞f(t)e−iωtdt F(\omega) = \int_{-\infty}^{\infty} f(t) \mathrm{e}^{-i \omega t} \mathrm{d}t This is also called CTFT sometimes. If you need an intuitive explanation of how FT works I would recommend reading the following answer - Now, lets continue to expand into the discrete version of the transforms ! Sampling f(t)f(t)f(t) at a rate of Ts=1FsTs=1FsT_s = \frac{1}{F_s} to yield fnfnf_n sequence, which may/maynot be finite in nature. This sequence is then transformed into the frequency domain equivalent by the following relation - X(jω)=∑∞n=−∞fne−iωnX(jω)=∑n=−∞∞fne−iωn X(j\omega) = \sum_{n=-\infty}^{\infty} f_n \mathrm{e}^{-i \omega n} The frequency spectrum yielded X(jω)X(jω)X(j\omega) is continuous in nature, but the input signal fnfnf_n is discrete in kind. Hence, this is called DTFT. If you visualise spectrum, it is F(ω)F(ω)F(\omega) repeated periodically across the frequency axis. In the above picture, S(f)≡F(ω)S(f)≡F(ω)S(f) \equiv F(\omega). The BOTTOM LEFT figure is the DTFT spectrum i.e X(jω)X(jω)X(j\omega). Discretising X(jω)X(jω)X(j\omega) would give us BOTTOM RIGHT figure Sn(k)≡X(k)Sn(k)≡X(k)S_n(k) \equiv X(k) as per our definition. This discretised fourier spectrum X[k]X[k]X[k] is called DFT of the sequence x[n]orfnx[n]orfn x[n] or f_n as per our definition. The first repeated portion (centered about ω=0ω=0\omega = 0) of the infinitely repeating spectrum is considered. Imagine if you will a band-pass filter with a response that passes only frequencies between −FstoFs−FstoFs - F_s   to   F_s Hz and hence, X[k]≡S[k]X[k]≡S[k]X[k] \equiv S[k] in the TOP RIGHT figure is a finite sequence. The http://en.wikipedia.org/wiki/Sequence of N http://en.wikipedia.org/wiki/Complex_number is transformed into an N-periodic sequence of complex numbers - Let me define ωl=2Ï€Nωl=2Ï€N\omega_l = \frac{2\pi}{N}. To compute the original sequence, you use the IDFT relation - You asked about physical significance - let me explain it like this - A DFT converts a finite list of equally spaced http://en.wikipedia.org/wiki/Sampling_(signal_processing) of a http://en.wikipedia.org/wiki/Function_(mathematics) into the list of http://en.wikipedia.org/wiki/Coefficient of a finite combination of http://en.wikipedia.org/wiki/Complex_number http://en.wikipedia.org/wiki/Sine_wave, ordered by their http://en.wikipedia.org/wiki/Frequency, that has those same sample values. What this means is that, these coefficients measure the magnitude of the existence of a particular frequency in the input signal. For eg - X[0]X[0]X[0], also called the DC component, captures how dominant is the 0th0th0^{th} frequency component in the input signal. X[1]X[1]X[1], also called the 1st1st1^{st} AC component, captures how dominant is the 1st1st1^{st} frequency component in the input signal, and so on. You would now ask, what are these frequency components and what are they measured against - The frequencies of the output sinusoids are integer multiples of a fundamental frequency, whose corresponding period is the length of the sampling interval (ωlωl\omega_l). Hope this clears it up !

You can arrive at the DFT from the Fourier series in three steps:1. Take the Fourier series equation and make the time period infinity - you get the Fourier  Transform equation.2. Take the Fourier transform equation and sample it the time domain - you get the DTFT equation.3. Take the DTFT equation and sample it in the frequency domain and you get the DFT equation.