Thank you Steve. Yes there will be leakage but I was wondering if there will be 'better' information obtainable from it over using DFT. I think I get this phenomena, although not quite sure how FT/DFT is less superior than using autocorrelation. After all its the same set of samples that we correlate. Thank you for all the replies.
PSD vs. Fourier transform
Started by ●March 1, 2009
Reply by ●March 2, 20092009-03-02
Reply by ●March 2, 20092009-03-02
m26k9 <maduranga.liyanage@gmail.com> wrote:> I think I get this phenomena, although not quite sure how > FT/DFT is less superior than using autocorrelation. After all > its the same set of samples that we correlate.In my experience, the autocorrelation method is superior if it is advantageous to apply a window between the autocorrelation and the transform. A common situation would be trying to measure the amount of energy in the stopband of a signal. You want a window with more stopband rejection than you are trying to observe. Otherwise, you would need a much longer dataset (corresponding to a rectangular window) to get the same accuracy in the stopband. If you're only looking at the spectrum in the strong parts of the signal, it may not matter very much. The above statements are not intended to be rigorous, but they hold in my experience. Steve
Reply by ●March 2, 20092009-03-02
>If you're only looking at the spectrum in the strong parts >of the signal, it may not matter very much. >Thank you Steve. That is what I was looking for.
Reply by ●March 2, 20092009-03-02
m26k9 wrote:> Thank you. > > >> Which goes back to my earlier comment -- you cannot compute >> the PSD with a discrete fourier transform. The PSD is defined >> at any frequency, whereas the DFT only gives you bins. >> >> You might get something that is PSD-like, it may serve your >> purposes, but it is not the real deal. >> >> Steve > > If PSD is defined at any frequency, what is the meaning of getting a value > with units power/Hz?Consider a piece of wood. Its density varies from place to place. (The darker parts are usually denser than the light. Knots, especially with resin are even denser.) Yet we measure the density -- even at a point -- as mass per unit volume. Energy density is energy per unit frequency.> And how to calculate PSD of discrete signals (when working in digital > domain) if DFT cannot be used?It can be used, but it provides only a (probably) useful approximation. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 2, 20092009-03-02
On Mar 2, 5:57�pm, spop...@speedymail.org (Steve Pope) wrote:> Jerry Avins �<j...@ieee.org> wrote: > > > > >m26k9 wrote: > >> What I gather from the discussion is that: > > >> 1) Both methods give essentially the same amount of information. > > >> 2) Direct FFT method is easier, but the FFT{autocorrelation} gives a > >> averaged value, which can better characterize a stochastic process. > >> What I am confused/interested in now is; > >> 1) How to prove that the FFT{autcorrelation} method gives better results > >> over the FFT{time-signal} ? > >> 2) What is the advantage of doing FFT{autcorrelation} over averaging > >> several FFT{time-signal} calculations ? > >> 3) How is it possible for me to evaluate the PSD at f= F Hz, and get the > >> value with units of power/Hz? this is like a rate/slope? �If I am actually > >> measuring the energy content at F Hz, I should get a value with units > >> power(watts), not power/Hz. What is this dilemma? > >You are measuring the total power in a band ("bin"), not at a point. > > Which goes back to my earlier comment -- you cannot compute > the PSD with a discrete fourier transform. �The PSD is defined > at any frequency, whereas the DFT only gives you bins. > > You might get something that is PSD-like, it may serve your > purposes, but it is not the real deal. > > SteveIt's the Periodogram but considered by most to be good enough. I see what you are saying though.
Reply by ●March 2, 20092009-03-02
On Mar 2, 6:05�pm, "m26k9" <maduranga.liyan...@gmail.com> wrote:> Thank you. > > >Which goes back to my earlier comment -- you cannot compute > >the PSD with a discrete fourier transform. �The PSD is defined > >at any frequency, whereas the DFT only gives you bins. > > >You might get something that is PSD-like, it may serve your > >purposes, but it is not the real deal. > > >Steve > > If PSD is defined at any frequency, what is the meaning of getting a value > with units power/Hz? > > And how to calculate PSD of discrete signals (when working in digital > domain) if DFT cannot be used? > > Thank you.You can say the same about any discrete process. Everything sampled is an approximation to analogue. Hardy
Reply by ●March 2, 20092009-03-02
On Mar 2, 6:49�pm, spop...@speedymail.org (Steve Pope) wrote:> m26k9 <maduranga.liyan...@gmail.com> wrote: > > But what did you mean exactly when you said DFT cannot be used to > > calculate the PSD? Suppose if we only want to find the frequency > > content of discrete bins, will CT-cosine transform give anything > > better than the plain DFT? > > I think it might. �Let's say you have a billion samples at 1 MHz, > and there is a frequency component at 400.13 Hz. �The PSD can > be evaluated at exactly this frequency, whereas if you choose > a reasonable-sized DFT to use instead of the continuous-time > transform, it might not have a bin at this frequency, so the energy > associated with this component bleeds into adjacent bins. �You do > not get the same information, at least not very directly. � > > SteveSurely the cc is also an approximation? It is found from samples and must be?
Reply by ●March 2, 20092009-03-02
HardySpicer <gyansorova@gmail.com> wrote:>On Mar 2, 6:49�pm, spop...@speedymail.org (Steve Pope) wrote:>> Let's say you have a billion samples at 1 MHz, >> and there is a frequency component at 400.13 Hz. �The PSD can >> be evaluated at exactly this frequency, whereas if you choose >> a reasonable-sized DFT to use instead of the continuous-time >> transform, it might not have a bin at this frequency, so the energy >> associated with this component bleeds into adjacent bins. �You do >> not get the same information, at least not very directly. �>Surely the cc is also an approximation? It is found from samples and >must be?I think for the PSD calculation, your frequency resolution/accuracy is on the order of the inverse of the original sample length (in this case, 0.001 Hz) whereas for the case using the DFT it's only on the order of the inverse of the DFT size. Whether this means you've lost information with the DFT I'm not entirely sure. At least superficially you don't have the same information. Steve
Reply by ●March 2, 20092009-03-02
HardySpicer wrote:> On Mar 2, 6:05 pm, "m26k9" <maduranga.liyan...@gmail.com> wrote: >> Thank you. >> >>> Which goes back to my earlier comment -- you cannot compute >>> the PSD with a discrete fourier transform. The PSD is defined >>> at any frequency, whereas the DFT only gives you bins. >>> You might get something that is PSD-like, it may serve your >>> purposes, but it is not the real deal. >>> Steve >> If PSD is defined at any frequency, what is the meaning of getting a value >> with units power/Hz? >> >> And how to calculate PSD of discrete signals (when working in digital >> domain) if DFT cannot be used? >> >> Thank you. > > You can say the same about any discrete process. Everything sampled is > an approximation to analogue.So you reject the sampling theorem? :-) Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●March 2, 20092009-03-02
On Mar 2, 10:22 am, spop...@speedymail.org (Steve Pope) wrote:> HardySpicer <gyansor...@gmail.com> wrote: > >On Mar 2, 6:49 pm, spop...@speedymail.org (Steve Pope) wrote: > >> Let's say you have a billion samples at 1 MHz, > >> and there is a frequency component at 400.13 Hz. The PSD can > >> be evaluated at exactly this frequency, whereas if you choose > >> a reasonable-sized DFT to use instead of the continuous-time > >> transform, it might not have a bin at this frequency, so the energy > >> associated with this component bleeds into adjacent bins. You do > >> not get the same information, at least not very directly. > >Surely the cc is also an approximation? It is found from samples and > >must be? >. I think for the PSD calculation, your frequency resolution/accuracy . is on the order of the inverse of the original sample length (in this . case, 0.001 Hz) whereas for the case using the DFT it's only . on the order of the inverse of the DFT size. In the DFT case its proportional to the inverse of the window (and thus data) size. . . Whether this means you've lost information with the DFT I'm . not entirely sure. At least superficially you don't have . the same information. . . Steve As has already been pointed out by others in this thread, for the discrete case, by zero extending the data to twice the (windowed) data size and performing this larger DFT, the result is exactly the same as the discrete version of the PSD-via-autocorrelation. You have exactly the same information. There's nothing superficial about it. The trade-offs in bin width and sidelobe rejection available by windowing are available in both approaches as well. Dale B. Dalrymple
