Hello, I have recorded a 1 minute signal from a sensor recorded at 20Hz. This is without an input to the sensor and hence the noise. http://tinypic.com/r/oje4c0/5 - (series 1 - axes on right) I wished to see the freq spectrum of the signal and this is what I get. http://tinypic.com/r/vy0d3l/5 Pardon my ignorance but this is not my area of expertise but what would the absence of a dominant peak mean? This looks like the lobes of a window function. so is it that there is no dominant noise source? I have tried the FFT on a 5Hz signal that I created in Excel and I can see the peak in FFT mag. Merry Christmas and thanks for your time! _____________________________ Posted through www.DSPRelated.com
What does this FFT mean
Started by ●December 26, 2013
Reply by ●December 26, 20132013-12-26
"coolarm" <99136@dsprelated> writes:> Hello, > > I have recorded a 1 minute signal from a sensor recorded at 20Hz. This is > without an input to the sensor and hence the noise. > http://tinypic.com/r/oje4c0/5 - (series 1 - axes on right) > I wished to see the freq spectrum of the signal and this is what I get. > http://tinypic.com/r/vy0d3l/5 > Pardon my ignorance but this is not my area of expertise but what would the > absence of a dominant peak mean? This looks like the lobes of a window > function. so is it that there is no dominant noise source?It means that your noise is pretty much uncorrelated sample-to-sample. It looks like you do have a bit of DC content, from the peak at 0 Hz. By the way, you usually compute 20*log10(mag(fft)) for a spectrum plot instead of just mag(fft). Note also that, for the FFT of real input data, the magnitude spectrum is symmetric and thus half is redundant and not usually plotted.> I have tried the FFT on a 5Hz signal that I created in Excel and I can see > the peak in FFT mag.Yes, then you are doing things right, it appears.> Merry Christmas and thanks for your time!Merry Christmas! -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●December 26, 20132013-12-26
On Thu, 26 Dec 2013 11:13:51 -0600, coolarm wrote:> Hello, > > I have recorded a 1 minute signal from a sensor recorded at 20Hz. This > is without an input to the sensor and hence the noise. > http://tinypic.com/r/oje4c0/5 - (series 1 - axes on right) > I wished to see the freq spectrum of the signal and this is what I get. > http://tinypic.com/r/vy0d3l/5 Pardon my ignorance but this is not my > area of expertise but what would the absence of a dominant peak mean? > This looks like the lobes of a window function. so is it that there is > no dominant noise source? > > I have tried the FFT on a 5Hz signal that I created in Excel and I can > see the peak in FFT mag. > > Merry Christmas and thanks for your time!Actually, I suspect that it's not the lobes of _a_ window function, but the lobes of _the_ window function -- the window function that you used, without removing the DC bias from your data. When you window sampled data before performing an FFT, you need to first remove the average value from the data, and if it shows any strong trends remove those, too. Otherwise, you end up with the FFT of bias * window, or trend * window, without much of the spectrum that you probably want. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
Reply by ●December 26, 20132013-12-26
On Thu, 26 Dec 2013 12:56:17 -0500, Randy Yates wrote:> "coolarm" <99136@dsprelated> writes: > >> Hello, >> >> I have recorded a 1 minute signal from a sensor recorded at 20Hz. This >> is without an input to the sensor and hence the noise. >> http://tinypic.com/r/oje4c0/5 - (series 1 - axes on right) >> I wished to see the freq spectrum of the signal and this is what I get. >> http://tinypic.com/r/vy0d3l/5 Pardon my ignorance but this is not my >> area of expertise but what would the absence of a dominant peak mean? >> This looks like the lobes of a window function. so is it that there is >> no dominant noise source? > > It means that your noise is pretty much uncorrelated sample-to-sample. > It looks like you do have a bit of DC content, from the peak at 0 Hz.Eh? Randy, take another look at that nice textbook abs(sinc) plot, and consider what you just said! -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●December 26, 20132013-12-26
Tim Wescott <tim@seemywebsite.really> writes:> On Thu, 26 Dec 2013 12:56:17 -0500, Randy Yates wrote: > >> "coolarm" <99136@dsprelated> writes: >> >>> Hello, >>> >>> I have recorded a 1 minute signal from a sensor recorded at 20Hz. This >>> is without an input to the sensor and hence the noise. >>> http://tinypic.com/r/oje4c0/5 - (series 1 - axes on right) >>> I wished to see the freq spectrum of the signal and this is what I get. >>> http://tinypic.com/r/vy0d3l/5 Pardon my ignorance but this is not my >>> area of expertise but what would the absence of a dominant peak mean? >>> This looks like the lobes of a window function. so is it that there is >>> no dominant noise source? >> >> It means that your noise is pretty much uncorrelated sample-to-sample. >> It looks like you do have a bit of DC content, from the peak at 0 Hz. > > Eh? Randy, take another look at that nice textbook abs(sinc) plot, and > consider what you just said!Yes, the sinc() with center at DC is clearly present, which implies there is a significant DC component. I suppose I meant that the lack of other peaks (which would still be obvious after convolving with the sinc()) means that the noise is fairly flat (uncorrelated) compared to DC. As per your other post where you suggest detrending the mean, what if that is part of what you wanted to estimate? If it is not, then obviously detrending before estimating the spectrum will give you better results. Or am I missing something? And what about averaging periodograms? .... -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●December 26, 20132013-12-26
On Thu, 26 Dec 2013 11:13:51 -0600, "coolarm" <99136@dsprelated> wrote:>Hello, > >I have recorded a 1 minute signal from a sensor recorded at 20Hz. This is >without an input to the sensor and hence the noise. >http://tinypic.com/r/oje4c0/5 - (series 1 - axes on right) >I wished to see the freq spectrum of the signal and this is what I get. >http://tinypic.com/r/vy0d3l/5 >Pardon my ignorance but this is not my area of expertise but what would the >absence of a dominant peak mean? This looks like the lobes of a window >function. so is it that there is no dominant noise source? > >I have tried the FFT on a 5Hz signal that I created in Excel and I can see >the peak in FFT mag. > >Merry Christmas and thanks for your time!There's a notable DC value around 1.5+, so the big spike at DC is understandable. The sinx/x response is probably because the data was zero-padded quite a bit (probably 8x or so), i.e., the FFT size was bigger than the data size. You said 20Hz for about a minute, so that'd yield about 1200 samples. The nulls are at the sample rate, so it looks like (if I'm seeing correctly), the FFT is supporting about 8x past that, so it was probably an 8k FFT or so. The scale suggests maybe a 10k DFT, but either way, it appears to me that what you're seeing is the sinx/x from the rectangular window that shows up due to a lot of zero-padding. The FFT output is also not "folded", so DC is at the left edge and the "negative" frequencies start at 10 and run up to 20. If you cut the 10-20 region and paste it to the left of the DC spike, you'll see the symmetric spectrum from -f to +f. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●December 26, 20132013-12-26
On Thu, 26 Dec 2013 22:55:10 GMT, eric.jacobsen@ieee.org (Eric Jacobsen) wrote:>On Thu, 26 Dec 2013 11:13:51 -0600, "coolarm" <99136@dsprelated> >wrote: > >>Hello, >> >>I have recorded a 1 minute signal from a sensor recorded at 20Hz. This is >>without an input to the sensor and hence the noise. >>http://tinypic.com/r/oje4c0/5 - (series 1 - axes on right) >>I wished to see the freq spectrum of the signal and this is what I get. >>http://tinypic.com/r/vy0d3l/5 >>Pardon my ignorance but this is not my area of expertise but what would the >>absence of a dominant peak mean? This looks like the lobes of a window >>function. so is it that there is no dominant noise source? >> >>I have tried the FFT on a 5Hz signal that I created in Excel and I can see >>the peak in FFT mag. >> >>Merry Christmas and thanks for your time! > >There's a notable DC value around 1.5+, so the big spike at DC is >understandable. > >The sinx/x response is probably because the data was zero-padded quite >a bit (probably 8x or so), i.e., the FFT size was bigger than the data >size. You said 20Hz for about a minute, so that'd yield about 1200 >samples. The nulls are at the sample rate, so it looks like (if I'm >seeing correctly), the FFT is supporting about 8x past that, so it was >probably an 8k FFT or so. The scale suggests maybe a 10k DFT, but >either way, it appears to me that what you're seeing is the sinx/x >from the rectangular window that shows up due to a lot of >zero-padding. > >The FFT output is also not "folded", so DC is at the left edge and the >"negative" frequencies start at 10 and run up to 20. If you cut the >10-20 region and paste it to the left of the DC spike, you'll see the >symmetric spectrum from -f to +f. > >Eric Jacobsen >Anchor Hill Communications >http://www.anchorhill.comBleah...I think I had an inversion in my brain while doing this. With 1013 samples (from your time-domain plot), and only three sidelobes in your frequency plot, that would suggest oversampling in the FFT by about 500/4 or 125. I doubt you zero-padded that much, but the sinx/x you're seeing is probably an artifact of how the data was processed rather than something in the signal. As has been suggested, removing the DC and zooming on the unambiguous portion of the support spectrum (i.e., between DC and fs/2), should look a lot more random than what you're seeing now. Eric Jacobsen Anchor Hill Communications http://www.anchorhill.com
Reply by ●December 27, 20132013-12-27
On Thu, 26 Dec 2013 15:19:23 -0500, Randy Yates wrote:> Tim Wescott <tim@seemywebsite.really> writes: > >> On Thu, 26 Dec 2013 12:56:17 -0500, Randy Yates wrote: >> >>> "coolarm" <99136@dsprelated> writes: >>> >>>> Hello, >>>> >>>> I have recorded a 1 minute signal from a sensor recorded at 20Hz. >>>> This is without an input to the sensor and hence the noise. >>>> http://tinypic.com/r/oje4c0/5 - (series 1 - axes on right) >>>> I wished to see the freq spectrum of the signal and this is what I >>>> get. >>>> http://tinypic.com/r/vy0d3l/5 Pardon my ignorance but this is not my >>>> area of expertise but what would the absence of a dominant peak mean? >>>> This looks like the lobes of a window function. so is it that there >>>> is no dominant noise source? >>> >>> It means that your noise is pretty much uncorrelated sample-to-sample. >>> It looks like you do have a bit of DC content, from the peak at 0 Hz. >> >> Eh? Randy, take another look at that nice textbook abs(sinc) plot, and >> consider what you just said! > > Yes, the sinc() with center at DC is clearly present, which implies > there is a significant DC component. I suppose I meant that the lack of > other peaks (which would still be obvious after convolving with the > sinc()) means that the noise is fairly flat (uncorrelated) compared to > DC. > > As per your other post where you suggest detrending the mean, what if > that is part of what you wanted to estimate? > > If it is not, then obviously detrending before estimating the spectrum > will give you better results. > > Or am I missing something? > > And what about averaging periodograms? ....I considered going into a whole riff about how you have to tailor what you do with your data to what you're trying to measure, but instead I made the -- unfounded -- assumption that the OP wanted to measure high frequency noise. I see three things in the time-domain plot that one may be interested in. I think I mentioned these, but didn't explicitly ask the OP which he wanted to measure. They are the DC bias, the low-frequency noise (I call it noise; I assume it is something random and not a desired measurement), and some high-frequency noise. I suspect that if you could plot the spectrum of the output of that sensor, you'd see flat, high frequency noise, possibly with some small- amplitude tones, then as you went down in frequency you'd see a 1/f or 1/ f^2 component (or both), leveling off at the low frequencies, then finally you'd see the DC bias. Getting that much detail in a reliable way would require a much longer sample, and if I'm wrong about the 1/f^2 noise leveling off then you'd never get one long enough. With the FFT, if you want to window and not end up with the FFT of the window function, you need to remove bias. If you want to know what the bias is -- well, it has to be estimated to remove it, so you can examine the number when you do bias removal. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●December 27, 20132013-12-27
Tim Wescott <tim@seemywebsite.really> writes:> [...] > With the FFT, if you want to window and not end up with the FFT of the > window function, you need to remove bias.Tim, I'm confused along a couple of points here. Correct me if I'm wrong or looking at things catawampous: 1. You HAVE to window - i.e., there isn't a choice about it. The choice you have is the type of window; not doing anything but grabbing N from infinite samples of a stream of data is choosing a rectangular window. 2. Any spectral energy is smeared by the window function. That holds for energy at DC or anywhere else. So tell me again why we need to remove DC? I can see that if any one spectral component (whether at DC or otherwise) is large relative to the other components, removing it will give us a better view of the other energy. Is that all this is about? -- Randy Yates Digital Signal Labs http://www.digitalsignallabs.com
Reply by ●December 27, 20132013-12-27
On Fri, 27 Dec 2013 13:38:40 -0500, Randy Yates wrote:> Tim Wescott <tim@seemywebsite.really> writes: >> [...] >> With the FFT, if you want to window and not end up with the FFT of the >> window function, you need to remove bias. > > Tim, > > I'm confused along a couple of points here. Correct me if I'm wrong or > looking at things catawampous: > > 1. You HAVE to window - i.e., there isn't a choice about it. The > choice you have is the type of window; not doing anything but grabbing > N from infinite samples of a stream of data is choosing a rectangular > window.If your FFT is exactly as long as the data that you have collected then you really aren't windowing. Or, at least, that's what I meant by "if you window".> 2. Any spectral energy is smeared by the window function. That holds > for energy at DC or anywhere else. > > So tell me again why we need to remove DC? I can see that if any one > spectral component (whether at DC or otherwise) is large relative to the > other components, removing it will give us a better view of the other > energy. Is that all this is about?If you have a lot of DC bias, then after windowing (even if it's just an unintended rectangular window from padding), then your signal ends up being signal = (dc bias) * (window) + (desired signal). If the DC bias is significantly bigger than the signal strength, then the 'signal' from the window swamps the desired signal. So you have to think about what you want out of your analysis. Usually you find you need to correct the measured signal before doing the FFT (almost always by detrending and windowing with some non-rectangular window). -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
