Fast fourier transform on exponential decay function

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Hi all:
I have one confused question. one continuous exponential decay function f=exp(-lamda*t) start from t=0 to infinity. I sample 1024 data points from the decay function. time variable (t) ranges from 0 to 1 second. the tail data of this exponential function is zero. I apply discret FFT on this sampled data. Am I supposed to get frequency spectrum with ring side lobe? When we apply FFT on this sampled data, does FFT assume my sampled data repeat themselves from negative infinity to positive infinity? If so, in this repeated data, a discontinuity point in exponential function exists. We are supposed to see oscillating side lobe because abrupt amplitude change between the tail data and the first point of my sampled data. Am My understanding correct? thanks in advance.

xf
 

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boneh3ad
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An FFT will assume nothing. Your data is aperiodic and so taking the FFT is a fruitless endeavor.
 
xts
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FFT is applicable to periodic functions only. In case like yours the FFT result will be a transform of
[itex]e^{-\lambda\{t\}}[/itex] rather than of [itex]e^{-\lambda t}[/itex]
[itex]\{t\}[/itex] denotes a fractional part of [itex]t[/itex]
 
AlephZero
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When we apply FFT on this sampled data, does FFT assume my sampled data repeat themselves from negative infinity to positive infinity?
Yes, if you do an FFT numerically on a finite sample of data, then you are assuming the complete sample is one "cycle" of a repeating periodic waveform.

If so, in this repeated data, a discontinuity point in exponential function exists. We are supposed to see oscillating side lobe because abrupt amplitude change between the tail data and the first point of my sampled data. Am My understanding correct? thanks in advance.
Yes, that is correct.

If you wanted to do this in a practical situation, you would apply a windowing function to the data before you take the FFT, to miminise the effect of the discontinuity.

It is wrong to say this is "a fruitless endeavour". Engineers who make practical measurements the dynamic response of structures do ths sort of thing all the time. It is only "fruitless" if you don't understand how to interpret the results.
 
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Thank you all for the reply.

When we apply some window function on discrete sampled data, we can make the sampled tail data signal to be zero. Under the point of view of discrete FFT algorithm, This window-filtered sampled data will repeat themselves from minus infinity to positive infinity. Let's imagine we place a series of copy of window-filtered sampled data in the time axis. It doesn't form one "cycle" of a repeating periodic waveform. Because in minus infinity, signal magnitude is equal to f(0) (is equal to a constant) and f(t=+infinity)=f(t=1 second) =0 at plus infinity. At all connected points between two adjacent copy of window-filtered sampled, there is one dicontinuity point.there is huge jump on signal magnitude [from f(t)=0 to f(t)=f(t0)] at that connection point. Does this discontinuity make ring side lobe or not? If yes, however in the practice, I never saw this kind of ring side lobe of frequency spectrum on window-filted sampled data. I just don't understand why it doesn't make obvious thanks for your help.

XF


Yes, if you do an FFT numerically on a finite sample of data, then you are assuming the complete sample is one "cycle" of a repeating periodic waveform.



Yes, that is correct.

If you wanted to do this in a practical situation, you would apply a windowing function to the data before you take the FFT, to miminise the effect of the discontinuity.

It is wrong to say this is "a fruitless endeavour". Engineers who make practical measurements the dynamic response of structures do ths sort of thing all the time. It is only "fruitless" if you don't understand how to interpret the results.
 

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