dbd <dbd@ieee.org> 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.>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.Right. I agree with this. In the example I have given above this requires a very large DFT.>The trade-offs in bin width and sidelobe rejection available by >windowing are available in both approaches as well.I agree with this in principal (i.e. you can re-arrange the linear and non-linear parts of a system by making suitable changes); but I am wondering how exactly you would calculate typical windowed PSD values without doing the autocorrelation step. For example: autocorrelation -> Nuttall Window -> Cosine Transform What is the version of this that uses no autocorrelation? Steve
PSD vs. Fourier transform
Started by ●March 1, 2009
Reply by ●March 2, 20092009-03-02
Reply by ●March 3, 20092009-03-03
On 1 Mar, 20:50, dbd <d...@ieee.org> wrote:> On Mar 1, 11:08 am, Rune Allnor <all...@tele.ntnu.no> wrote: > > > > > > > On 1 Mar, 19:05, HardySpicer <gyansor...@gmail.com> wrote: > > > > On Mar 2, 5:10 am, Oli Charlesworth <ca...@olifilth.co.uk> wrote: > > > > > m26k9 wrote: > > > > > Hello, > > > > > > I have a confusion regarding Fourier transform of a time signal and the > > > > > PSD. > > > > > By definition, power spectral density (PSD) is the Fourier transform of > > > > > the autocorrelation function. > > > > > > Suppose my time-domain signal is a voltage signal. > > > > > I sample this at the correct rate, and taking the Fourier transform gives > > > > > me the maginitudes of of various frequency components present in the signal > > > > > (within the context of sampling rate). Taking the modulus squared of this > > > > > quantity will give me power/Hz. > > > > > > On the other hand, literature says, taking the Fourier transform of the > > > > > autocorrelation function (rather than the direct transform as discussed > > > > > earlier) gives the PSD. And confusingly gives the units as power/Hz. > > > > > > I am confused as what is the difference between these two? if both have > > > > > units of power/Hz, why/when should I do one over the other? > > > > > One difference is that the Fourier transform of a signal may not exist, > > > > for instance, a stochastic (i.e. random) process doesn't have a Fourier > > > > transform, but it will have an autocorrelation function. �Therefore, the > > > > PSD will exist even if you can't compute the Fourier transform of the > > > > signal itself. > > > > > -- > > > > Oli > > .> > .> > You can still compute the periodogram (PSD) of a random signal > (say > .> > speech) using the FFT directly of course > .> > and not via the autocorrelation. > .> > .> How would you do this if you want to use some window function? > .> > .> Rune > > How could you do this without using some window function? > > Dale B. DalrympleFunny. I'll get back to you on that in the very moment your Norwegian is at the level where it's worth splittng hairs on exact phrasings. Rune
Reply by ●March 3, 20092009-03-03
On 2 Mar, 03:18, HardySpicer <gyansor...@gmail.com> wrote:> On Mar 2, 8:08 am, Rune Allnor <all...@tele.ntnu.no> wrote: > > > > > > > On 1 Mar, 19:05, HardySpicer <gyansor...@gmail.com> wrote: > > > > On Mar 2, 5:10 am, Oli Charlesworth <ca...@olifilth.co.uk> wrote: > > > > > m26k9 wrote: > > > > > Hello, > > > > > > I have a confusion regarding Fourier transform of a time signal and the > > > > > PSD. > > > > > By definition, power spectral density (PSD) is the Fourier transform of > > > > > the autocorrelation function. > > > > > > Suppose my time-domain signal is a voltage signal. > > > > > I sample this at the correct rate, and taking the Fourier transform gives > > > > > me the maginitudes of of various frequency components present in the signal > > > > > (within the context of sampling rate). Taking the modulus squared of this > > > > > quantity will give me power/Hz. > > > > > > On the other hand, literature says, taking the Fourier transform of the > > > > > autocorrelation function (rather than the direct transform as discussed > > > > > earlier) gives the PSD. And confusingly gives the units as power/Hz. > > > > > > I am confused as what is the difference between these two? if both have > > > > > units of power/Hz, why/when should I do one over the other? > > > > > One difference is that the Fourier transform of a signal may not exist, > > > > for instance, a stochastic (i.e. random) process doesn't have a Fourier > > > > transform, but it will have an autocorrelation function. �Therefore, the > > > > PSD will exist even if you can't compute the Fourier transform of the > > > > signal itself. > > > > > -- > > > > Oli > > > > You can still compute the periodogram (PSD) of a random signal (say > > > speech) using the FFT directly of course > > > and not via the autocorrelation. > > > How would you do this if you want to use some window function? > > > Rune > > I normally just use magnitude squared/N for periodogram. I average of > course over successive periodograms if possible.But how would you modify your FFT algorithm to handle explicit window functions in lag domain? That is, if you use something like rxx[k] = w[k]*E[x[n]x[n+k]] where w[k] is an arbitrary window function, what will the equivalent expression be in terms of FFTs? The recipes I have seen for Welch's method [*] compute an average of window-weighted periodograms. Rune [*] I might have seen more, but I remember the recipe from Kay's 1988 book.
Reply by ●March 3, 20092009-03-03
On Mar 2, 5:05 pm, spop...@speedymail.org (Steve Pope) wrote:> dbd <d...@ieee.org> 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..> >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. .> .> Right. I agree with this. In the example I have given above this .> requires a very large DFT. Yes, the same size for either approach under the conditions of a single data set transformed once. .> .> >The trade-offs in bin width and sidelobe rejection available by .> >windowing are available in both approaches as well. .> .> I agree with this in principal (i.e. you can re-arrange the .> linear and non-linear parts of a system by making suitable .> changes); but I am wondering how exactly you would calculate .> typical windowed PSD values without doing the autocorrelation .> step. For example: .> .> autocorrelation -> Nuttall Window -> Cosine Transform .> .> What is the version of this that uses no autocorrelation? .> .> Steve I said that windowing can be used for tradeoffs between mainlobe width and sidelobe rejection in both cases. The tradeoff need not be made by a window on the autocorrelation. My statement was that the input data was windowed. For long data sequences with multiple transforms performed on blocked subsets of data, Carter and Nuttall claim that the same estimate stability can be achieved by the two processes if they are constrained to the same frequency resolution and data size. On the weighted overlapped segment averaging method for power spectral estimation Carter, G.C. Nuttall, A.H. Yuen, C.K. Proceedings of the IEEE Oct. 1980, Vol: 68, Issue: 10 page(s): 1352- 1354 Dale B. Dalrymple
Reply by ●March 3, 20092009-03-03
dbd <dbd@ieee.org> wrote:>On Mar 2, 5:05 pm, spop...@speedymail.org (Steve Pope) wrote:>> I am wondering how exactly you would calculate >> typical windowed PSD values without doing the autocorrelation >> step. For example: >> >> autocorrelation -> Nuttall Window -> Cosine Transform >> >> What is the version of this that uses no autocorrelation?>I said that windowing can be used for tradeoffs between mainlobe width >and sidelobe rejection in both cases. The tradeoff need not be made by >a window on the autocorrelation. My statement was that the input data >was windowed.Okay, but windowing the data and windowing the ACF are not exactly mathematically interchageable.>For long data sequences with multiple transforms performed on blocked >subsets of data, Carter and Nuttall claim that the same estimate >stability can be achieved by the two processes if they are constrained >to the same frequency resolution and data size.>On the weighted overlapped segment averaging method for power spectral >estimation >Carter, G.C. Nuttall, A.H. Yuen, C.K. >Proceedings of the IEEE >Oct. 1980, Vol: 68, Issue: 10 >page(s): 1352- 1354Thanks. So, without looking up this article, I'm inferring it says that to answer my question above, you would block and window the data, then take DFT's, take the magnitudes, and combine. Per the Carter/Nuttall/Yuen you get the same basic accuracy. Seems reasonable. I haven't tried it, mostly because of my instinctual feel that this underweights data that happens to fall in the low-amplitude parts of an input window, whereas the ACF method does not. The counterargument being that you cannot resolve the corresponding low-frequency-resolution components in either case. Thanks for your replies, this has given me something to think about. Steve