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Change of Sampling Rate in the DFT

Consider a signal x(k) with sampling time Tc =1, then its Discrete Fourier Transorm is X(f).
If now I have to consider y(k) that is the upsapled version of the signal x(k),  with sampling time Tc = 1/4,  (meaning y(k) = x(k) for T=4*k, y(k)=0 otherwise).
What does it change in the DFT of y(k) comparing it with X(f)? 
POSTED BY: Francesca Rossi
3 years ago
If this happens to be the case, you will introduce some distorsion of your spectrum. 
data = Table[Sin[5*x] + RandomReal[]*0.5, {x, 0, 10, 0.02}];
ListPlot[data, Filling -> Axis]
ListPlot[Abs[Fourier[data]], Filling -> Axis, PlotRange -> All]

To include the gap, we can use the following code to get the new spectrum: 
datay = Flatten@Map[{#, 0., 0., 0.} &, data];
ListPlot[(Norm /@ Fourier[datay]), Filling -> Axis, PlotRange -> All]

These are two useful links: 
POSTED BY: Shenghui Yang
3 years ago

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