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I always studied the DFT starting from his formula, but for some reasons I need to do comparison between the FT and the DFT. I found the pdf in this link very useful http://www.robots.ox.ac.uk/~sjrob/Teaching/SP/l7.pdf because it explains how the DFT can be obtained starting from the FT. After the first formula it says " We could regard each sample f[k] as an impulse having area f[k]". What I don't understand is how it possible that the sample, being an impulse, has an area f[k]. Shouldn't t have an area f[k] multiplied by a very low number? This passage for me is crucial because I cannot understand why, passing from the FT to the DFT, the time increment disappears.

Ashish Bhigah
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    riemann integration. then replace each skinny rectangle with an impulse having the same area. – robert bristow-johnson Jan 05 '20 at 15:37
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    I am not sure to have understood. So, the time increament disappears beacuse it is inside f[k]? If the area is known then how can I calculate the height of the impulse? – Ashish Bhigah Jan 05 '20 at 15:56
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    Does [this answer](https://dsp.stackexchange.com/a/62180/4298) help? – Matt L. Jan 05 '20 at 16:00
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    @MattL. this answer complicate my thoughts :) From this answer it seems that in order to obtain the FT you need to multiply the DFT by the time increment. But i thought it is not the same because the DFT needs also to be normalised in order to have the same DC of a FT or the same energy of a FT. – Ashish Bhigah Jan 05 '20 at 16:15
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    @AshishBhigah: Well, the scaling in that answer is correct, just check the plot comparing the FT with its DFT approximation. – Matt L. Jan 05 '20 at 17:07
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    @MattL. yesterday I have followed this discussion and it seemed to me reasonable to choose the normalization 1/N https://dsp.stackexchange.com/questions/63001/why-should-i-scale-the-fft-using-1-n . This is because I am confused – Ashish Bhigah Jan 05 '20 at 17:29
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    @AshishBhigah: You have to decide what you want. If you want to approximate the CT Fourier transform by the DFT you need to choose the scaling used in the answer I linked to. When you look at the plot of that answer, do the values at DC coincide or don't they? Did you follow the derivation? It is assumed that the DFT is defined without any scaling (and the corresponding inverse DFT would have a scaling of $1/N$). – Matt L. Jan 05 '20 at 17:32
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    @MattL. Yes, they coincide. but the DC shouldn't be the avarege of the time history? – Ashish Bhigah Jan 05 '20 at 17:59
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    @AshishBhigah: The DC value of the CT Fourier transform is just the integral $\int x(t)dt$ (which if the integration limits are finite is a *scaled* average). Now approximate that by a Riemann sum and you'll see that you get a $\Delta t$ (which just corresponds to the $dt$ in the integral). – Matt L. Jan 05 '20 at 18:29
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    So, if multiplicating by Ts is the scaling consistent with the FT why doing this with matlab, I obtain the right amplitudes of 0.7 and 1, as expected Fs = 1000; % Sampling frequency T = 1/Fs; % Sampling period L = 1500; % Length of signal t = (0:L-1)*T; % Time vector S = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t); f = Fs*(0:(L/2))/L; Y = fft(S); P2 = abs(Y/L); P1 = P2(1:L/2+1); P1(2:end-1) = 2*P1(2:end-1); plot(f,P1) title('Single-Sided Amplitude Spectrum of S(t)') xlabel('f (Hz)') ylabel('|P1(f)|') – Ashish Bhigah Jan 05 '20 at 20:27
  • Very hard to read the code. You may use $``$ for `code` – jomegaA Feb 06 '20 at 08:45
  • @MattL "It is assumed that the DFT is defined without any scaling (and the corresponding inverse DFT would have a scaling of 1/N)" Is this a general assumption or specific to this problem here? So the scaling is necessary when approximating CT Fourier transfrom by the DFT? In other post you have nice proof for it. – jomegaA Feb 06 '20 at 09:19
  • @jomegaA: There are different scaling conventions, and it doesn't matter which one you use as long as you know it and take it into account when approximating the CTFT. – Matt L. Feb 06 '20 at 10:08

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The area under an impulse function is one. So if you multiply a continuous time waveform with an impulse, the result will be a weighted impulse with the area equal to whatever the waveform was at that moment in time.

Dan Boschen
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