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I have the Fourier transform $S(\nu)$. If I oversample by a factor $R$ in the frequency domain. Does the duration of the signal $s(t)=IFFT(S(\nu))$ change?

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    $\begingroup$ You seem to be mixing continuous-time and discrete-time concepts here. Could you try to clarify what you mean? At the same time, review the concept of "duration" of a discrete-time signal. $\endgroup$ – MBaz Feb 19 at 15:36
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If by "oversampling", you mean interpolating more frequency samples between your existing frequency samples, then in this case this would have the effect of zero padding your time domain waveform. Therefore the duration of your signal itself doesn't change, but the duration of time over which the DFT is performed would, with the addition of zero padding to fill out the added time duration (which could be before or after the waveform, changing the phase component of the resulting frequency spectrum accordingly).

To understand zero padding it is helpful to understand the difference between the Discrete Fourier Transform (DFT) and the Discrete Time Fourier Transform (DTFT). The Fast Fourier Transform (FFT) is an efficient algorithm for computing the DFT.

Zero padding a discrete waveform in time results in interpolation of DFT result in frequency. The DFT result is discrete, given by samples of the DTFT. The DTFT in contrast is a continuous function in frequency. By zero padding, we introduce more samples on this same DTFT.

Comparing the formulas for the DFT and DTFT will give you further insight into why this is so:

DTFT

DFT

Here the plots show how the DFT is samples of the DTFT, and by adding zeros to the time-domain waveform prior to taking the DFT (zero-padding), we are approximating the continuous DTFT by filling in more samples (interpolation)-- the more zeros we add the closer we can get to the continuous waveform DTFT. So zero padding and interpolating the frequency spectrum are Fourier Transform pairs.

Further notes: In the plots I say that the DFT is repeating in time, but in actuality it is only given over N samples. The repetition in time is a mathematical equivalence: similar to the Fourier Series Expansion which is defined over a finite time T, it's frequency components only exist at integer multiples of 1/T (discrete in frequency). If you allowed the time domain waveforms of those components to extend to $\pm \infty$, the base waveform that was defined over t=0 to T would repeat in time. The DFT as given above would be the same as the DFT of the same waveform repeating in time, and seeing that helps provide a lot of insight--- anything that repeats in time MUST be discrete in frequency, anything that is discrete in time MUST repeat in frequency (A/D sampling is an example). The DFT does both. Since the DTFT is performed on a single time domain waveform over all time, it never repeats in time and for this reason has a continuous frequency response.

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Oversampling a spectrum is a non-linear process. Specifically, since each bin is suppose to represent an integral number waves in the sampling window, oversampling must be done in accordance with that. But to answer your question, and assuming the spectrum is of some finite time domain window, the new time domain window duration is the wavelength of the lowest frequency bin in the FFT.

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