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.
13
-
2$\begingroup$ riemann integration. then replace each skinny rectangle with an impulse having the same area. $\endgroup$– robert bristow-johnsonJan 5, 2020 at 15:37
-
1$\begingroup$ 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? $\endgroup$– Ashish BhigahJan 5, 2020 at 15:56
-
2$\begingroup$ @AshishBhigah: Well, the scaling in that answer is correct, just check the plot comparing the FT with its DFT approximation. $\endgroup$– Matt L.Jan 5, 2020 at 17:07
-
2$\begingroup$ @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 $1/N$). $\endgroup$– Matt L.Jan 5, 2020 at 17:32
-
2$\begingroup$ @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). $\endgroup$– Matt L.Jan 5, 2020 at 18:29
|
Show 8 more comments
1 Answer
$\begingroup$
$\endgroup$
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.
answered Jan 5, 2020 at 15:34