Ok, sorry if this may sound a bit strange, but I\'m unsure as to how the DFT app

Question

Ok, sorry if this may sound a bit strange, but I'm unsure as to how the DFT approximates the CFT. So basically the fourier series components of a periodic function will be components of the CFT. ck = F(jkw0). So if we had say a periodic function with T = 1000, we would have a step size of k*2pi/1000.

Now, I'm curious exactly for the case of the DFT. Why exactly will we end up with more components in the fourier series coefficients vector (fft(x)) than just computing the fourier series coefficients directly from the CFT as stated above?

Any help greatly appreciated. I'm really thinking of this in the following order for the DFT.

one period of a periodic signal
sample that signal at rate nT
leads to periodic F(jw) in the frequency domain
convolve x(t) at rate nT to sample the frequency domain
arrive at the fourier series components.

Not sure if this makes much sense.

Explanation / Answer

think about it in terms of the DTFT (which is just a DFT with an infinite block length). Because with an infinite block length there is no lower bound to the minimum frequency that can be represented, the frequency resolution of the DTFT is effectively infinite. So provided the original signal is bandlimited to Nyquist etc, the output of the DTFT is exactly the same as the output of the CFT on a discrete signal. RATE ME LIFE SAVER!!!

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