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From: David Painter on 4 Aug 2010 07:32
Hi all,

I've been using the short-time Fourier transform (STFT, spectrogram function) to examine the time course of specific frequency amplitudes over time.

How confident can I be that the power values reflect the signal in the intended time bin and not adjacent bins (i.e., can I work out if there is smearing of frequency components across time bins)? For a given frequency, to what extent are the amplitude estimates for a time bin dependent on the strength of that frequency in adjacent bins? If this is a problem, does the Hamming window reduce the influence of the frequency signal from adjacent bins?

Finally, I've read quite a few forum entries written by people struggling with STFT. I found the following link particularly helpful:

http://note.sonots.com/SciSoftware/STFT.html

Thanks, :-)

David

....

STFT settings:

data = x;
wind = 1575;
noverlap = 875;
nfft = wind;
fs.new = 1000;

[S, F, T, P] = spectrogram(x, wind, noverlap, nfft, fs.new, 'yaxis');
From: Wayne King on 4 Aug 2010 11:04
"David Painter" <david.ross.painter(a)gmail.com> wrote in message <i3bj3k$ivc$1(a)fred.mathworks.com>...
> Hi all,
>
> I've been using the short-time Fourier transform (STFT, spectrogram function) to examine the time course of specific frequency amplitudes over time.
>
> How confident can I be that the power values reflect the signal in the intended time bin and not adjacent bins (i.e., can I work out if there is smearing of frequency components across time bins)? For a given frequency, to what extent are the amplitude estimates for a time bin dependent on the strength of that frequency in adjacent bins? If this is a problem, does the Hamming window reduce the influence of the frequency signal from adjacent bins?
>
> Finally, I've read quite a few forum entries written by people struggling with STFT. I found the following link particularly helpful:
>
> http://note.sonots.com/SciSoftware/STFT.html
>
> Thanks, :-)
>
> David
>
> ...
>
> STFT settings:
>
> data = x;
> wind = 1575;
> noverlap = 875;
> nfft = wind;
> fs.new = 1000;
>
> [S, F, T, P] = spectrogram(x, wind, noverlap, nfft, fs.new, 'yaxis');

Hi David, Your time resolution is the difference between the window length and the number of samples that you overlap your windows, (wind-noverlap)*1/fs.new in your case. You can see this by:

plot(diff(T)); % equal to 0.700 (wind-noverlap)*1/fs.new. Your frequency resolution is fs.new/wind in each of your windows. Again, you can see this with

plot(diff(F));

In the example below I create two signals one with a periodic component (200 Hz) in the first time window [0,1.57] seconds (although it only lasts for the first .700 seconds) and one that has a periodic component in part of both the first and second windows. You see within these windows, you have no time resolution.

t = linspace(0,10,10000);
t = t(:);
x = zeros(1e4,1);
y = x;
x(1:700) = cos(2*pi*200*t(1:700));
y(100:1200) = cos(2*pi*200*t(1:1101));
x = x +0.2*randn(size(t));
y = y +0.2*randn(size(t));

[Sx, F, T, Px] = spectrogram(x, wind, noverlap, nfft, fs.new);
[Sy, F, T, Py] = spectrogram(y, wind, noverlap, nfft, fs.new);
% Note
subplot(211)
plot(F,Px(:,1));
title('P_x');
subplot(212);
plot(F,Px(:,2));

% versus
figure;
subplot(211);
plot(F,Py(:,1));
title('P_y');
subplot(212);
plot(F,Py(:,2));

Hope that helps,
Wayne
From: David Painter on 5 Aug 2010 22:29
"Wayne King" <wmkingty(a)gmail.com> wrote in message <i3bvh4$4nn$1(a)fred.mathworks.com>...
> "David Painter" <david.ross.painter(a)gmail.com> wrote in message <i3bj3k$ivc$1(a)fred.mathworks.com>...
> > Hi all,
> >
> > I've been using the short-time Fourier transform (STFT, spectrogram function) to examine the time course of specific frequency amplitudes over time.
> >
> > How confident can I be that the power values reflect the signal in the intended time bin and not adjacent bins (i.e., can I work out if there is smearing of frequency components across time bins)? For a given frequency, to what extent are the amplitude estimates for a time bin dependent on the strength of that frequency in adjacent bins? If this is a problem, does the Hamming window reduce the influence of the frequency signal from adjacent bins?
> >
> > Finally, I've read quite a few forum entries written by people struggling with STFT. I found the following link particularly helpful:
> >
> > http://note.sonots.com/SciSoftware/STFT.html
> >
> > Thanks, :-)
> >
> > David
> >
> > ...
> >
> > STFT settings:
> >
> > data = x;
> > wind = 1575;
> > noverlap = 875;
> > nfft = wind;
> > fs.new = 1000;
> >
> > [S, F, T, P] = spectrogram(x, wind, noverlap, nfft, fs.new, 'yaxis');
>
> Hi David, Your time resolution is the difference between the window length and the number of samples that you overlap your windows, (wind-noverlap)*1/fs.new in your case. You can see this by:
>
> plot(diff(T)); % equal to 0.700 (wind-noverlap)*1/fs.new. Your frequency resolution is fs.new/wind in each of your windows. Again, you can see this with
>
> plot(diff(F));
>
> In the example below I create two signals one with a periodic component (200 Hz) in the first time window [0,1.57] seconds (although it only lasts for the first .700 seconds) and one that has a periodic component in part of both the first and second windows. You see within these windows, you have no time resolution.
>
> t = linspace(0,10,10000);
> t = t(:);
> x = zeros(1e4,1);
> y = x;
> x(1:700) = cos(2*pi*200*t(1:700));
> y(100:1200) = cos(2*pi*200*t(1:1101));
> x = x +0.2*randn(size(t));
> y = y +0.2*randn(size(t));
>
> [Sx, F, T, Px] = spectrogram(x, wind, noverlap, nfft, fs.new);
> [Sy, F, T, Py] = spectrogram(y, wind, noverlap, nfft, fs.new);
> % Note
> subplot(211)
> plot(F,Px(:,1));
> title('P_x');
> subplot(212);
> plot(F,Px(:,2));
>
> % versus
> figure;
> subplot(211);
> plot(F,Py(:,1));
> title('P_y');
> subplot(212);
> plot(F,Py(:,2));
>
> Hope that helps,
> Wayne

Hi Wayne,

That was really helpful! So the short-time Fourier transform will pick of frequencies within the fft analysis window, though the results returned create a misleading impression that the fft has (in effect) only been applied to the returned time bins.

In the following example there is a 100 Hz signal during the first bin (0-700 ms) and a 200 Hz signal during the second bin (701-1400 ms). The first fft is run on data from 1-1575 ms, and the second is run on data from 701-2276 ms. Consequently, the 100 Hz signal is present only during the first time segment, and the 200 Hz signal is present in both time segments. The answer to my original question is then that the frequency spectrum in any one bin includes information from adjacent times.

Seems like a series of ffts (with zero-padding for Hz resolution and a Hamming window for the padding) on binned data will be a better solution for me since these will not be affected by frequency signals at adjacent times.

Thanks,

David

win = 1575;
noverlap = 875;
nfft = 1575;
fs = 1000;

t = linspace(0,10,10000);
t = t(:);

x = zeros(1e4,1);

x(1:700) = cos( 2*pi*100*t(1:700) );
x(701:1400) = cos( 2*pi*200*t(701:1400) );

spectrogram(x, win, noverlap, nfft, fs)
From: Wayne King on 6 Aug 2010 01:25
"David Painter" <david.ross.painter(a)gmail.com> wrote in message <i3fs1j$jpe$1(a)fred.mathworks.com>...
> "Wayne King" <wmkingty(a)gmail.com> wrote in message <i3bvh4$4nn$1(a)fred.mathworks.com>...
> > "David Painter" <david.ross.painter(a)gmail.com> wrote in message <i3bj3k$ivc$1(a)fred.mathworks.com>...
> > > Hi all,
> > >
> > > I've been using the short-time Fourier transform (STFT, spectrogram function) to examine the time course of specific frequency amplitudes over time.
> > >
> > > How confident can I be that the power values reflect the signal in the intended time bin and not adjacent bins (i.e., can I work out if there is smearing of frequency components across time bins)? For a given frequency, to what extent are the amplitude estimates for a time bin dependent on the strength of that frequency in adjacent bins? If this is a problem, does the Hamming window reduce the influence of the frequency signal from adjacent bins?
> > >
> > > Finally, I've read quite a few forum entries written by people struggling with STFT. I found the following link particularly helpful:
> > >
> > > http://note.sonots.com/SciSoftware/STFT.html
> > >
> > > Thanks, :-)
> > >
> > > David
> > >
> > > ...
> > >
> > > STFT settings:
> > >
> > > data = x;
> > > wind = 1575;
> > > noverlap = 875;
> > > nfft = wind;
> > > fs.new = 1000;
> > >
> > > [S, F, T, P] = spectrogram(x, wind, noverlap, nfft, fs.new, 'yaxis');
> >
> > Hi David, Your time resolution is the difference between the window length and the number of samples that you overlap your windows, (wind-noverlap)*1/fs.new in your case. You can see this by:
> >
> > plot(diff(T)); % equal to 0.700 (wind-noverlap)*1/fs.new. Your frequency resolution is fs.new/wind in each of your windows. Again, you can see this with
> >
> > plot(diff(F));
> >
> > In the example below I create two signals one with a periodic component (200 Hz) in the first time window [0,1.57] seconds (although it only lasts for the first .700 seconds) and one that has a periodic component in part of both the first and second windows. You see within these windows, you have no time resolution.
> >
> > t = linspace(0,10,10000);
> > t = t(:);
> > x = zeros(1e4,1);
> > y = x;
> > x(1:700) = cos(2*pi*200*t(1:700));
> > y(100:1200) = cos(2*pi*200*t(1:1101));
> > x = x +0.2*randn(size(t));
> > y = y +0.2*randn(size(t));
> >
> > [Sx, F, T, Px] = spectrogram(x, wind, noverlap, nfft, fs.new);
> > [Sy, F, T, Py] = spectrogram(y, wind, noverlap, nfft, fs.new);
> > % Note
> > subplot(211)
> > plot(F,Px(:,1));
> > title('P_x');
> > subplot(212);
> > plot(F,Px(:,2));
> >
> > % versus
> > figure;
> > subplot(211);
> > plot(F,Py(:,1));
> > title('P_y');
> > subplot(212);
> > plot(F,Py(:,2));
> >
> > Hope that helps,
> > Wayne
>
> Hi Wayne,
>
> That was really helpful! So the short-time Fourier transform will pick of frequencies within the fft analysis window, though the results returned create a misleading impression that the fft has (in effect) only been applied to the returned time bins.
>
> In the following example there is a 100 Hz signal during the first bin (0-700 ms) and a 200 Hz signal during the second bin (701-1400 ms). The first fft is run on data from 1-1575 ms, and the second is run on data from 701-2276 ms. Consequently, the 100 Hz signal is present only during the first time segment, and the 200 Hz signal is present in both time segments. The answer to my original question is then that the frequency spectrum in any one bin includes information from adjacent times.
>
> Seems like a series of ffts (with zero-padding for Hz resolution and a Hamming window for the padding) on binned data will be a better solution for me since these will not be affected by frequency signals at adjacent times.
>
> Thanks,
>
> David
>
> win = 1575;
> noverlap = 875;
> nfft = 1575;
> fs = 1000;
>
> t = linspace(0,10,10000);
> t = t(:);
>
> x = zeros(1e4,1);
>
> x(1:700) = cos( 2*pi*100*t(1:700) );
> x(701:1400) = cos( 2*pi*200*t(701:1400) );
>
> spectrogram(x, win, noverlap, nfft, fs)

Hi David, Glad it's making more sense. I think the easiest way to think about it is whatever your window length, the Fourier transform sums (integrates) over all time within that window. Therefore you lose all time information within that window. You can see from your example:

[S, F, T, P] = spectrogram(x, win, noverlap, nfft, fs);
plot(F,P(:,1));

that both 100 and 200 Hz are evident in your first window of 1.57 seconds, but we don't get the sense that 100 Hz occurred in the first 1/2 (roughly) of that window and 200 Hz occurred in the 2nd half. That information is lost to us because of the window length. Of course, you can reduce your window length to improve your time resolution, but then you lose frequency resolution, because the spacing between your Fourier transform bins is inversely proportional to the window length. The danger here is that two "closely" spaced periodic components get resolved as only one.

And so it goes... the dilemma that is time-frequency analysis :)

Wayne
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