Suppose I have a stationary[1] signal sampled at some sample rate. As I
take the fourier transform of longer and longer intervals of this time
series, the resolution ("bandwidth"?) of the spectra increase. What I
want to know is, what is the relationship between (1) dividing a time
series into M equal-length subintervals, taking the DFT of each
subinterval individually, and then averaging these subintervals, and (2)
taking a DFT of the entire time series, and then "binning" it by averaging
groups of neighboring points?
[1] I don't actually know what this means, but my meaning is that if I
take a spectrum of one chunk of this time series, it will resemble the
spectrum of any other chunk (subinterval).
thanks,
Tobin
fft averaging and combining
Started by ●April 6, 2005
Reply by ●April 6, 20052005-04-06
"Tobin Fricke" <fricke@ocf.berkeley.edu> wrote in message news:Pine.SOL.4.62.0504061038430.28681@apocalypse.OCF.Berkeley.EDU...> Suppose I have a stationary[1] signal sampled at some sample rate. As I > take the fourier transform of longer and longer intervals of this time > series, the resolution ("bandwidth"?) of the spectra increase. What I > want to know is, what is the relationship between (1) dividing a time > series into M equal-length subintervals, taking the DFT of each > subinterval individually, and then averaging these subintervals, and (2) > taking a DFT of the entire time series, and then "binning" it by averaging > groups of neighboring points?When you divide the time series into M equal length sub-intervals, you've lost the original frequency resolution by a factor of M. By taking the DFT of each and averaging these subintervals, you still don't gain any more frequency resolution that you started with. When you take the DFT of the entire time series, you now have all of the frequency resolution. I assume that by 'binning' it, you are averaging groups of neighboring points to reduce your points to match your sub-interval of method 1. If you do this, then you could potentially lose amplitude accuracy of CW signals. I think both methods will result in similarly averaged noise floors. So if you your bin size was say 10 points and you had 2 tones separated by 4 points, your second method would've been able to distinguish the 2 tones, but when you average the 10 points to bin them, you get a result of just 1 tone whose ampltiude is the average of the 10 points (including the 2 tones). This amplitude would be less than the combined power of the 2 tones. Using method 1, you'd show the combined power of the tones and good averaging of your noise floor. I'm not sure if I have directly established a 'relationship' between method 1 and 2, but I hope this helps. Cheers Bhaskar> [1] I don't actually know what this means, but my meaning is that if I > take a spectrum of one chunk of this time series, it will resemble the > spectrum of any other chunk (subinterval). > > thanks, > Tobin
Reply by ●April 6, 20052005-04-06
Tobin Fricke wrote:> Suppose I have a stationary[1] signal sampled at some sample rate. As I > take the fourier transform of longer and longer intervals of this time > series, the resolution ("bandwidth"?) of the spectra increase. What I > want to know is, what is the relationship between (1) dividing a time > series into M equal-length subintervals, taking the DFT of each > subinterval individually, and then averaging these subintervals, and (2) > taking a DFT of the entire time series, and then "binning" it by > averaging groups of neighboring points? > > [1] I don't actually know what this means, but my meaning is that if I > take a spectrum of one chunk of this time series, it will resemble the > spectrum of any other chunk (subinterval). > > thanks, > TobinWell, there are a number of statistical properties, like the consistency of the estimate which come into play. A consistent estimator produces a lower error variance as the sample size increases. The long FFT approach is not a consistent estimator. Lets say you have 2 signal samples from the the same stationary stochastic process of equal length. If you use the shorter FFT method, the results will look more alike, while the long approach will be more dissimilar. When you do bin averaging (across adjacent bins), you're kind of implying that the spectrum is roughly equal in each bin. This might be an OK assumption if you have reason to believe it is true, but if there are abrupt transitions in the underlying spectrum, they would tend to be a bit more smeared. There are a number of issues related to window selection that influence this that I've just ignored. Another issue is that in practice there is an upper limit to the duration of the stationarity assumption. Most stuff tends to vary a bit. Fundamentally a long FFT is desirable, but not too long an FFT.
Reply by ●April 6, 20052005-04-06
On Wed, 6 Apr 2005 10:44:07 -0700, Tobin Fricke <fricke@ocf.berkeley.edu> wrote: Hi Tobin,>Suppose I have a stationary[1] signal sampled at some sample rate. As I >take the fourier transform of longer and longer intervals of this time >series, the resolution ("bandwidth"?) of the spectra increase.Well ..., no, the resolution decreases. The resolution, the spacing between FFT bins mausured in hertz, is the sample rate measured in hertz divided by the number of samples. If you increase the number of samples, the spacing between the FFT bins is *decreased*.>What I >want to know is, what is the relationship between (1) dividing a time >series into M equal-length subintervals, taking the DFT of each >subinterval individually, and then averaging these subintervals, and (2) >taking a DFT of the entire time series, and then "binning" it by averaging >groups of neighboring points? >[1] I don't actually know what this means, but my meaning is that if I >take a spectrum of one chunk of this time series, it will resemble the >spectrum of any other chunk (subinterval).Tobin, can you tell us what is the goal of your processing? What is it that you're trying to achieve? Are you trying to minimize the variance of of some spectral amplitude measurement? Are you trying to detect a very low-level spectral component? Regards, [-Rick-]
Reply by ●April 6, 20052005-04-06
"Rick Lyons" <R.Lyons@_BOGUS_ieee.org> wrote in message news:42548564.168632125@news.sf.sbcglobal.net...> On Wed, 6 Apr 2005 10:44:07 -0700, Tobin Fricke > <fricke@ocf.berkeley.edu> wrote: > > > Hi Tobin, > > >Suppose I have a stationary[1] signal sampled at some sample rate. As I > >take the fourier transform of longer and longer intervals of this time > >series, the resolution ("bandwidth"?) of the spectra increase. > > Well ..., no, the resolution decreases. > The resolution, the spacing between FFT bins > mausured in hertz, is > the sample rate measured in hertz divided by the number > of samples. If you increase the number of samples, the > spacing between the FFT bins is *decreased*.Hi Rick I don't like getting into debates on semantics...but in this case, yes - the bin width has *decreased* but the resolution has *increased*...because the bins are closer together, the resolving power has increased (not decreased). What you were probably thinking of is "the 'resolution bandwidth' - the spacing between FFT bins measured in Hz has decreased" ? A smaller resolution bandwidth provides improved resolution in finding signals on a spectrum. BTW - I'm ignoring any windowing since that complicates this discussion. Cheers Bhaskar> > >What I > >want to know is, what is the relationship between (1) dividing a time > >series into M equal-length subintervals, taking the DFT of each > >subinterval individually, and then averaging these subintervals, and (2) > >taking a DFT of the entire time series, and then "binning" it byaveraging> >groups of neighboring points? > >[1] I don't actually know what this means, but my meaning is that if I > >take a spectrum of one chunk of this time series, it will resemble the > >spectrum of any other chunk (subinterval). > > Tobin, can you tell us what is the goal of your > processing? What is it that you're trying to > achieve? Are you trying to minimize the variance of > of some spectral amplitude measurement? Are you > trying to detect a very low-level spectral component? > > Regards, > [-Rick-] >
Reply by ●April 6, 20052005-04-06
Rick Lyons wrote:> On Wed, 6 Apr 2005 10:44:07 -0700, Tobin Fricke > <fricke@ocf.berkeley.edu> wrote: > > > Hi Tobin, > > >>Suppose I have a stationary[1] signal sampled at some sample rate. As I >>take the fourier transform of longer and longer intervals of this time >>series, the resolution ("bandwidth"?) of the spectra increase. > > > Well ..., no, the resolution decreases. > The resolution, the spacing between FFT bins > mausured in hertz, is > the sample rate measured in hertz divided by the number > of samples. If you increase the number of samples, the > spacing between the FFT bins is *decreased*.Rick, There's a semantic problem with the interpretation of "resolution". High-resolution TV shows finer detail than NTSC. For graphics, a number proportional to the reciprocal of resolvable size -- one is line pairs per millimeter -- is a better measure of resolution. Without concerning ourselves with increase or decrease, I would say that a longer FFT /improves/ the resolution. Shortening an FFT /coarsens/ it. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●April 7, 20052005-04-07
On Thu, 7 Apr 2005, Rick Lyons wrote:> Well ..., no, the resolution decreases. The resolution, the spacing > between FFT bins mausured in hertz, is the sample rate measured in hertz > divided by the number of samples. If you increase the number of > samples, the spacing between the FFT bins is *decreased*.Right -- just depends on how we're interpreting the word 'resolution'. Your ability to resolve spectral lines certainly increases! But the hertz-per-point ("bandwidth"?) decreases. [Is that a common use of the word "bandwidth"?]> Tobin, can you tell us what is the goal of your processing? What is it > that you're trying to achieve? Are you trying to minimize the variance > of of some spectral amplitude measurement? Are you trying to detect a > very low-level spectral component?Yes, that's exactly it. We take time series, break it down into subintervals (of, say, 32 seconds), FFT each subinterval, and average these transforms. Right now I'm just averaging the magnitude of the Fourier Transform (the original time series is complex-valued). There is a tradeoff between longer intervals (greater spectral resolving ability) and more points in the averages. I'm just wondering if there is a more theoretical way to understand this situation. Another question is, how to average the complex spectra (not just their amplitudes). I imagine I should multiply the transform of an interval by exp(i t_0), where t_0 is the starting time of that interval... or something? thanks, Tobin Fricke http://web.pas.rochester.edu/~tobin/
Reply by ●April 7, 20052005-04-07
in article Pine.SOL.4.62.0504062013140.28681@apocalypse.OCF.Berkeley.EDU, Tobin Fricke at fricke@ocf.berkeley.edu wrote on 04/06/2005 23:18:> Another question is, how to average the complex spectra (not just their > amplitudes). I imagine I should multiply the transform of an interval by > exp(i t_0), where t_0 is the starting time of that interval... or > something?the something should be exp(i 2 pi f t_0). f*t_0 is dimensionless and i think should be the same as n_0*k/N that you would get in the FFT (where t_0 = n_0 / Fs). -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by ●April 13, 20052005-04-13
On Wed, 6 Apr 2005 18:26:47 -0700, "Bhaskar Thiagarajan" <bhaskart@deja.com> wrote:>"Rick Lyons" <R.Lyons@_BOGUS_ieee.org> wrote in message >news:42548564.168632125@news.sf.sbcglobal.net... >> On Wed, 6 Apr 2005 10:44:07 -0700, Tobin Fricke >> <fricke@ocf.berkeley.edu> wrote: >> >> >> Hi Tobin, >> >> >Suppose I have a stationary[1] signal sampled at some sample rate. As I >> >take the fourier transform of longer and longer intervals of this time >> >series, the resolution ("bandwidth"?) of the spectra increase. >> >> Well ..., no, the resolution decreases. >> The resolution, the spacing between FFT bins >> mausured in hertz, is >> the sample rate measured in hertz divided by the number >> of samples. If you increase the number of samples, the >> spacing between the FFT bins is *decreased*. > >Hi Rick > >I don't like getting into debates on semantics...but in this case, yes - the >bin width has *decreased* but the resolution has *increased*...because the >bins are closer together, the resolving power has increased (not decreased). >What you were probably thinking of is "the 'resolution bandwidth' - the >spacing between FFT bins measured in Hz has decreased" ? >A smaller resolution bandwidth provides improved resolution in finding >signals on a spectrum. >BTW - I'm ignoring any windowing since that complicates this discussion. > >Cheers >BhaskarHi Bhaskar, I just saw your and Jerry's posts. Of course you are both correct. This is definitely a "semantic" issue. That word "resolution" surely does require careful definition before it's used in any meaningful discussion. [Ha ha. How many hot debates have occurred here because two guys didn't carefully define their terminology?] I've always thought of DFT resolution as being directly related to DFT bin spacing (in hertz). As the DFT size increases, the resolution decreases (by my definition). Kinda like the "resolution" of a microscope or a telescope. I think it's also reasonable to say, "as the DFT size increases our ability to "resolve" two signals improves." Anyway, I clearly understand you and Jerry. I sure hope my post didn't confuse Tobin. Thanks, [-Rick-]
Reply by ●April 13, 20052005-04-13
On Wed, 6 Apr 2005 20:18:41 -0700, Tobin Fricke <fricke@ocf.berkeley.edu> wrote: (snippped)> >> Tobin, can you tell us what is the goal of your processing? What is it >> that you're trying to achieve? Are you trying to minimize the variance >> of of some spectral amplitude measurement? Are you trying to detect a >> very low-level spectral component?Hi Tobin,>Yes, that's exactly it. We take time series, break it down into >subintervals (of, say, 32 seconds), FFT each subinterval, and average >these transforms. Right now I'm just averaging the magnitude of the >Fourier Transform (the original time series is complex-valued). There is >a tradeoff between longer intervals (greater spectral resolving ability) >and more points in the averages. I'm just wondering if there is a more >theoretical way to understand this situation.Ah. I'm no expert in this area but there's a couple of schemes for reducing the variance of your spectral magnitude measurements when you have a fixed number of time samples (and you are unable to collect more time samples). Those techniques are called "Welch's Method" and "Bartlett's Method". Searching the Internet will yield all sorts of info on these. Oppenheim & Schafer discuss the Welch technique in their Chapter 10. Proakis & Manolakis discuss the both techniques in their Chapter 12.>Another question is, how to average the complex spectra (not just their >amplitudes).Oops. You're averaging spectral magnitudes, and not the real & imag parts separately, right? Averaging complex amplitudes (real & imag parts separately) is only sensible if the signal is periodic in time and you've carefully time- synchronized your data collection process.>I imagine I should multiply the transform of an interval by >exp(i t_0), where t_0 is the starting time of that interval... or >something?Humm, it sounds like you're thinking about changing the phase of some DFT results in order to implement an equivalent time-shift of your signal samples in the time domain. That's possible to do, but it's only sensible, as far as I know, if your signal is periodic in time. Is that the case with your signal? (Are you testing A/D converters, or maybe processing radar signals?) To implement an equivalent time shift of K samples, you can multiply your N-point DFT's X(m) complex spectral samples by e^(j*2*pi*m*K/N) Tobin, I liked your website. (Micah's photo is funny.) Here's a quote for Ryah: Major 'Dutch' Schaeffer, Predator "If it bleeds, we can kill it." Good Luck, [-Rick-]>thanks, >Tobin Fricke > >http://web.pas.rochester.edu/~tobin/
