
AR, Burg methods worse resolution than FFT
Hi, I've been told Autoregressive Process (AR) offers better frequency resolution than classic Periodogram methods.
So I used an example available in the MATLAB manual and compared with Periodogram. Results showed that Periodogram had better frequency resolution than AR via Burg method.
I also studied pyulear, pcov, pmcov, their spectral looks almost identical. I've two questions.
1. Why my Periodogram has better frequency resolution than ARbased methods?
2. Have you experienced very similar spectrum in your data analysis?
Thank you all,
Cheers
Code
randn('state',0)
fs = 1000; % Sampling frequency
t = (0:fs)/fs; % One second worth of samples
A = [1 2]; % Sinusoid amplitudes
f = [150;140]; % Sinusoid frequencies
xn = A*sin(2*pi*f*t) + 0.1*randn(size(t));
% [P1,f] = pwelch(xn,hamming(256),128,1024,fs);
[P1,f] = periodogram(xn,rectwin(length(xn)),1024,fs);
[P2,f] = pburg(xn,14,1024,fs);
plot(f,10*log10(P1),'r',f,10*log10(P2),'b'); grid
legend('Periodogram','ARBurg')
ylabel('PSD Estimates (dB/Hz)');
xlabel('Frequency (Hz)');


0




Reply

Dean

8/7/2010 10:56:03 PM 

See related articles to this posting
On Aug 7, 3:56=A0pm, "Dean " <jiangwei0...@gmail.com> wrote:
> Hi, I've been told Autoregressive Process (AR) offers better frequency re=
solution than classic Periodogram methods.
> ...
Perhaps you have rediscovered a quaint historical artifact that a
number of algorithms commonly labeled "high resolution" should have
been labeled "high SNR only".
Dale B. Dalrymple


0




Reply

dbd

8/8/2010 12:37:13 AM


dbd <dbd@ieee.org> wrote in message <c855b29ec1e0447da6d48a6ac14dbffe@m35g2000prn.googlegroups.com>...
> On Aug 7, 3:56 pm, "Dean " <jiangwei0...@gmail.com> wrote:
> > Hi, I've been told Autoregressive Process (AR) offers better frequency resolution than classic Periodogram methods.
> > ...
>
> Perhaps you have rediscovered a quaint historical artifact that a
> number of algorithms commonly labeled "high resolution" should have
> been labeled "high SNR only".
>
> Dale B. Dalrymple
Hi Dale,
I also found all these AR methods, such as Burg, Covariance, YW, Modified Covariance actually producing almost visually identical spectrum (even with noise added). Is that something textbook won't tell me?
Cheers,
Dean


0




Reply

Dean

8/8/2010 3:14:06 PM


"Dean " <jiangwei0515@gmail.com> wrote in message <i3koa2$2nv$1@fred.mathworks.com>...
> Hi, I've been told Autoregressive Process (AR) offers better frequency resolution than classic Periodogram methods.
>
> So I used an example available in the MATLAB manual and compared with Periodogram. Results showed that Periodogram had better frequency resolution than AR via Burg method.
>
> I also studied pyulear, pcov, pmcov, their spectral looks almost identical. I've two questions.
> 1. Why my Periodogram has better frequency resolution than ARbased methods?
> 2. Have you experienced very similar spectrum in your data analysis?
>
> Thank you all,
> Cheers
> Code
> randn('state',0)
> fs = 1000; % Sampling frequency
> t = (0:fs)/fs; % One second worth of samples
> A = [1 2]; % Sinusoid amplitudes
> f = [150;140]; % Sinusoid frequencies
> xn = A*sin(2*pi*f*t) + 0.1*randn(size(t));
> % [P1,f] = pwelch(xn,hamming(256),128,1024,fs);
> [P1,f] = periodogram(xn,rectwin(length(xn)),1024,fs);
> [P2,f] = pburg(xn,14,1024,fs);
> plot(f,10*log10(P1),'r',f,10*log10(P2),'b'); grid
> legend('Periodogram','ARBurg')
> ylabel('PSD Estimates (dB/Hz)');
> xlabel('Frequency (Hz)');
Hi Dean, Typically resolution improvements with AR methods for sinusoids are seen with short data records. So for example:
reset(RandStream.getDefaultStream);
fs = 1000;
t = (0:fs/15)/fs;
A = [2 1];
f = [140;150];
xn = A*sin(2*pi*f*t) + 0.1*randn(size(t));
hburg = spectrum.burg(24);
psd(hburg,xn,'Fs',fs);
hper = spectrum.periodogram;
figure;
psd(hper,xn,'Fs',fs);
Note how the AR estimate resolves the two peaks, but the periodogram smears them together because they are closer than 14.92 hertz resolution of the periodogram (in this case).
Don't get me wrong, I'm a big fan of nonparametric methods and have used them much more frequently than parametric methods (which have their own issues), but the resolution superiority is usually brought up in the context of short data records. Of course, if you overmodel the data using parametric methods, you can get line splitting where one true sinusoid gets split into two separate ones. Kinda the opposite problem :)
Also, there are the socalled superresolution methods, which begin with a signal model that is a superposition of complex exponentials plus noise. These methods decompose the autocorrelation matrix into a signal subspace and a noise subspace. They don't yield a PSD estimate, so they're used when you want to estimate frequencies, see spectrum.music and spectrum.eigenvector for examples in MATLAB.
Wayne


0




Reply

Wayne

8/8/2010 3:32:07 PM


On 8 Aug, 00:56, "Dean " <jiangwei0...@gmail.com> wrote:
> Hi, I've been told Autoregressive Process (AR) offers better frequency resolution than classic Periodogram methods.
Don't believe what people tell you. There are plenty people around
who tell you all sort of stuff, that they  when challenged  turn
out to know nothing whatsoever about.
The AR methods don't provide frequency resolution. They characterize
the covariance structure of the signal in terms of an AR model.
The good part about this is that you can get good characteristics
with small ampunts of data (a novice or amateur might interpret
this as 'better frequency resolution'). The downside is that the
charactersitics might not be easily interpreted. The really bad side
is that if the the data do not comply to the assumed model (AR in
this case), the computed parameters might not make sense at all.
Rune


0




Reply

Rune

8/8/2010 5:29:37 PM


Hi All,
Thank you for all your comments, really useful and encouraging.
I feel I also like classic nonparametric ones better than AR methods.
To my understanding, the model order P should be selected equivalent to 2*number of sinusoidal assuming in an ideal case. However, I got an error message like shown.
The signal is synthesized without noise added. I don't understand what makes matlab think I should use smaller P? Of course, I will never see three peaks for P=4.
Cheers,
Dean
Error Message 
??? Error using ==> arspectra at 87
The variance estimate of the white noise input to the AR model is negative. Try to lower the order of the AR model.
END of Error Message 
Code
reset(RandStream.getDefaultStream);
fs = 1000;
t = (0:fs/15)/fs;
A = [2 1 1];
f = [140;150;155];
xn = A*sin(2*pi*f*t);
[Pxx_cov,f]=pcov(xn,6,2048,fs);


0




Reply

Dean

8/8/2010 5:50:08 PM


"Dean " <jiangwei0515@gmail.com> wrote in message <i3mqog$j8q$1@fred.mathworks.com>...
> Hi All,
>
> Thank you for all your comments, really useful and encouraging.
>
> I feel I also like classic nonparametric ones better than AR methods.
>
> To my understanding, the model order P should be selected equivalent to 2*number of sinusoidal assuming in an ideal case. However, I got an error message like shown.
>
> The signal is synthesized without noise added. I don't understand what makes matlab think I should use smaller P? Of course, I will never see three peaks for P=4.
>
> Cheers,
>
> Dean
>
> Error Message 
> ??? Error using ==> arspectra at 87
> The variance estimate of the white noise input to the AR model is negative. Try to lower the order of the AR model.
> END of Error Message 
>
>
>
> Code
> reset(RandStream.getDefaultStream);
> fs = 1000;
> t = (0:fs/15)/fs;
> A = [2 1 1];
> f = [140;150;155];
> xn = A*sin(2*pi*f*t);
> [Pxx_cov,f]=pcov(xn,6,2048,fs);
Hi Dean, Fitting an AR(p) models explicitly models the output, y(n), as
y(n)a(1)y(n1)a(2)y(n2) a(p)y(np)=e(n)
where e(n) is white noise. There should be a white noise input.
To be fair to AR spectral methods, they do have a lot of important uses. There are applications where the physical nature of the signal generation mechanism makes an AR model a good a priori choice. If you look at human speech analysis, you find a number of applications where AR models are quite useful. It just so happens that the things I've used spectral analysis for, nonparametric methods have been more appropriate.
Another thing you might find interesting about AR processes given your other posts, AR processes have nice closedform expressions for their PSDs. Therefore, you can generate a realization of an AR process, determine its true PSD, and then test spectral estimators against it. Note for the example the use of an AR(4) process to illustrate the problem of bias in the periodogram:
reset(RandStream.getDefaultStream);
x = randn(1024,1);
%AR(4) Process
A = [1 2.7670 3.8106 2.65335 0.9238];
y = filter(1,A,x);
sigma = 1;
f =0:(1/length(x)):0.5;
tmp = zeros(length(2:length(A)),length(f));
for k = 2:length(A)
tmp(k1,:) = A(k)*exp(1j*(k1)*2*pi*f);
end
tmp = sum(tmp);
denom = abs(1+tmp).^2;
psdtrue = sigma^2./denom;
h = spectrum.periodogram;
psdest = psd(h,y,'Fs',1,'NFFT',length(x));
figure;
plot(f,10*log10(psdest.Data)); hold on;
plot(f,10*log10(psdtrue),'r','linewidth',2);
See how the periodogram blows the estimate above about 0.2 Hz? That is a demonstration of the known bias effect in the periodogram. Note how the use of a Hamming window alleviates that problem.
hwin = spectrum.periodogram('Hamming');
psdwin = psd(hwin,y,'Fs',1,'NFFT',length(x));
figure;
plot(f,10*log10(psdtrue),'r','linewidth',2);
hold on;
plot(f,10*log10(psdwin.Data));
Wayne


0




Reply

Wayne

8/8/2010 8:07:07 PM


On 8 Aug, 19:50, "Dean " <jiangwei0...@gmail.com> wrote:
> Hi All,
>
> Thank you for all your comments, really useful and encouraging.
>
> I feel I also like classic nonparametric ones better than AR methods.
Agreed. Better than that: There exists a scientific argument to
support that subjective feeling. See below.
> To my understanding, the model order P should be selected equivalent to 2*number of sinusoidal assuming in an ideal case.
Yep indeed!
Congratulations! You are one of the extremely few who are able
to figure that one out on your own! Seriously. For once none of
my irony or sarcasm; merely sheer joy and happiness that somebody
are able to actually figure this one out.
Yes, the model order is the CATCH 22 of DSP. The feasible range of
model orders depends on the number data points available (the
exact details depend on the method in use). For simplicity, say
P = N/2 where N is the number of data points and P is the maximum
allowable order. Denote the *true* number of sinusoidals as D.
For the ARtype predictors to work, D < P. If one uses the Levinson
recursion one can estimate D on the fly, using some sort of
order estimator.
If one contemplates this, one finds that one necessary criterion
for ARtype methods to work, is that the relation
D < P = N/2
holds.
In other words, the designer / analyst who configures the ARtype
methods can not choose N arbitrarily, but must know in advance what
scenario he is working in. Feed the method too few data points and
one ends up with
D >= N/2
and the method unconditionally breaks down.
Just try it. Use N = 12 data points, no noise, and try and feed
the AR method first data from an AR(1) process, then and AR(2),
then AR(3), and so on. With the 'usual' model, you will be able
to recover the AR parameters (possibly with errors due to estimator
bias) for models up to and including AR(5).From AR(6) and beyond,
the computations need not make any sense at all.
All this forms the argument in support of nonparametric methods:
They might not produce the best results, but they have no such
Achilles heels either. The nonparametrics work where the
parametrics break down.
Rune


0




Reply

Rune

8/9/2010 8:07:01 AM


On Aug 8, 10:50=A0am, "Dean " <jiangwei0...@gmail.com> wrote:
> Hi All,
>
> Thank you for all your comments, really useful and encouraging.
>
> I feel I also like classic nonparametric ones better than AR methods.
>
> ...
> Dean
The classical nonparametric method for tones is really a two part
process. First the peaks of the periodogram and then an interpolation
method applied to the dft coefficients about the peaks. See:
D. C. Rife, and R. R. Boorstyn, =93Single tone parameter estimation
from discretetime observations,=94 IEEE Trans. Inform. Theory, vol.
IT20, pp591598, 1974
D. C. Rife, and R. R. Boorstyn, =93Multiple tone parameter estimation
from discretetime observations,=94 Bell Sysr. Tech. J., vol. 55, pp.
13891410, 1976.
The stationary tone case has been frequently presented in the IEEE
Trans on Instrumentation and Measurement:
http://wwwtw.vub.ac.be/elec/Papers%20on%20web/Papers/JohanSchoukens/IM92Sch=
oukensThe%20Interpolated.pdf
For the nonstationary case, one of the approaches can be found under
the label "frequency reasignment", for example:
http://perso.enslyon.fr/patrick.flandrin/0065Ch05.pdf
Anyone who limits their comparisons to nonparametric techniques to
peakpicking the periodogram is either uninformed or intentionally
misleading. You can find both in the literature.
Dale B. Dalrymple


0




Reply

dbd

8/9/2010 6:00:17 PM



8 Replies
711 Views
Similar Articles
[PageSpeed]
42

Similar Artilces:
Simplex resolution of the method tab by the Method of M.Hi!!! I am looking for a software for Hp48 that solve by Simplex
method tab by the Method of M. I have downloaded mm.lib of hpcalc.org
but in some iteration do a error. Could somebody help me? I can send
my dump rom, exercise to solve, and mm.lib.
fran.roy.morales@gmail.com
... FFT ResolutionI am trying to increase the resolution on an FFT plot.
Currently I am using the following:
y=fft(Vout);
x=y.*conj(y);
Vout is a large 1x4001 double array which contains a module of a AM
modulated signal.
The resulting Spectrum plot shows the spikes in the correct location,
but they look more like somewhat skinny triangles. I am looking to
make these mole like spikes like I would see on a spectrum analyzer
display. Does anyone know how to do this??
L. Ciotti wrote:
>
>
> I am trying to increase the resolution on an FFT plot.
>
> Currently I am using the following:
>
> ... FFT Spectrogram ResolutionHello all,
My name is Nick and I'm new to this forum.
Currently, I am learning DSP and, specifically, trying to figure out Wav
theory and Fourier's calculus. This is all necessary since I'd like t
write a program that can _accurately_ convert a sound into a bitma
spectrogram, and then back again, without any loss in quality (if that'
even physically possible!).
As many of the professionals here probably know, the digital spectrogra
feature which is a component to many modern audio software is driven b
the fast fourier transform. My question is: How is it that one ca
increas... Accuracy and Resolution of an FFTIs there a difference? ie suppose I sample at 10kHz with 1024 points  the
resolution is then 10000/1024Hz.
Is this the same as the accuracy ie isuppose the resolution is 1Hz can we
say the accuracy is + or  0.5Hz?Also, when we average FFTs I assume the
resoltion stays teh same but does teh accuracy increase and if so by how
much depending on the SNR and the number of averages,type of average.
M.

Posted via a free Usenet account from http://www.teranews.com
>Is there a difference? ie suppose I sample at 10kHz with 1024 points
the
>resolution is then 10000/1024Hz.
>Is this ... FFT Resolution bandwidthIf I have sampled a signal at say fs=1MHz sampling rate and I capture N=10
samples of that signal, my resolution bandwidth in frequency domain will b
fs/N = 10kHz.
100 samples correspond to 100us of capture.
If sampling frequency is doubled and if the capture duration is als
doubled, I would expect the resolution bandwidth also to increase by
factor of x.
But it is not the case. Resolution bandwidth remains the same. Why i
that?
What is the intuition behind this?
Sumeer,
The resolution should change. If it is not happening
then we have to see the reason why?
case 1 : Fs... Super Resolution FFTHi there,
I am working on a method of peak detection, and after I run my fft
process with a suitable N length I am out by about 15Hz on the target
freq (I know what it is because I embedded it). (Fs = 16000Hz,
Ftarget = 5000Hz, N = 1024. Ftarget is embedded in to an audio wav
file with freqs upto 8000Hz, most nonembedded power is around 2kHz
at about 70dB the power of Ftarget)
Zeropadding isn't going to help in this case as far as I can tell
(not that I can figure out how to actually do it in MATLAB)
A quick search on here show this thread:
skt, "Getting the true frequencies&quo... FFT resolution linesThe FFT function returns Nyquist (half) of input points. How do I get
400 lines of resolution out of 1024 points? Do I just disregard other
112?
Thanks
In an ideal world, a 1024point FFT results in 512 valid values, at
frequencies from 0/N * Fsamp up to (N/21) / N * Fsamp.
Of course, an ideal world would also have a filter with 100% pass at
the nyquist frequency, and 0% at any frequency above it.
T'ain't no such beast.
The usual reason for using 400 lines instead of 512 (or 800 instead of
1024, etc.) is the imprecision of the input filters  you don't really
know if their cuto... PSD with FFT methodDear Reader,
I am posting my question again with a more clear problem.
I was calculating PSD with fft method with the help of Tech note 1702
and with periodogram method.
<http://www.mathworks.com/support/technotes/1700/1702.html>
Please see the following code:
fu=10.6e9; N=64;
Tm=1e9;
%Tm=1;
hold all
t=linspace(Tm/2,Tm/2,N+1);
% x=2*fu*sinc(2*fu*t)2*fl*sinc(2*fl*t);
x=sin(2*fu*t);
Fs=(N+1)/Tm;
% Use next highest power of 2 greater than or equal to
% length(x) to calculate FFT.
NFFT= 2^(nextpow2(length(x)));
% Take fft, padding with zeros so that length(FFTX) is equal to
% NFFT
F... dynamic method resolutionHello,
What does dynamic method resolution mean? Thanks.
Frank
Frank wrote:
> Hello,
>
> What does dynamic method resolution mean? Thanks.
Try asking on a Smalltalk or ObjectiveC newsgroup.
Chris

Chris Gray chris@kiffer.eunet.be
/k/ Embedded Java Solutions
... how do I increase fft resolution?I'm currently running an fft calc on a real time accelerometer DAQ signal. By tweaking the DAQ settings or setting my own time average using shift registers I can reduce the dt which does increase the resolution but the number of data points seems to be limited to 50 regardless of what I do, i.e. reducing the dt decreases range of frequencies visible in the freqspectrum.
Thanks,
Rob
Here are simple guidelines
1) the number of lines (answers in the FFT) are equal to number of dat samples passed to the FFT divided by two. Most vibration experts use 2.56 as t... Virtual method resolution
How does Java resolve a virtual method at runtime? If a particular
virtual call site can have many potential targets because the object
could possibly be one of many types, what exactly goes on in the
machine to resolve this call? Is there a lot of overhead associated
with this resolving process? I have heard of devirtualization as a
technique to overcome these overheads..what exactly is that?
Thanks!
KL
One object can be an instance of only one class. The class may have a
superclass, the superclass another superclass and so on, until we reach
java.lang.Object.
During the loa... inheritanceWhere is the rule that explains why this will not compile? I've always
expected this to work, but it would appear that I haven't run into
this problem yet.
To resolve the problem, do I really need to override every single
method from the Base with the same name as the specific method I am
interested in overriding? I have a good 20 of them in production code.
A simple test case to reproduce what I am experiencing in more
complicated code:
class Base
{
public:
virtual void Foo()
{
}
void Foo(int x)
{
}
};
class Derived : public Base
{
public:
void Foo()
{
... Spectrogram/FFT Frequency ResolutionWhen computing the Power spectrum from the spectrogram or fft functions, the frequency range is in integers. Our situation is that we are running [S,F,T,P] = spectrogram(...) with a sample rate of 1kHz. By hand we computed the most prominant oscillation in the signal to be 8.3Hz. However, the power spectral density output (P) is in integers and rounds to 8Hz. Is there a way to get the prerounded amount?
Thanks in advance
Travis
"Travis " <meyert11@gmail.com> wrote in message <hv304b$c56$1@fred.mathworks.com>...
> When computing the Power spectrum from the s... any smart methodsHi,
I am trying to optimize a piece of code where am stopped at the
following point:
I have
Y(F) = FFT[ x(n) * w(n)] where * is multiplication. 0<= n <=N
With Y(F) in hand (already computed) I would like to know if anyone
could think of a smart way of finding out Z(F) without much complexity
where,
Z(F) = FFT[ x(n) * w(Nn)]
On Wed, 20 May 2009 05:13:59 0700, hurry wrote:
> Hi,
>
> I am trying to optimize a piece of code where am stopped at the
> following point:
>
> I have
> Y(F) = FFT[ x(n) * w(n)] where * is multiplication. 0<= n <=N
>
> Wi... frequency resolution in FFT function?Hello Guys
This is not a quite Matlab question, but hopfully someone can give me
any hint.
I have 1000 piont signal sampled at 10kHz. The fundamental frequency
is 50Hz. It also has harmonics ranges from 55~75Hz which are
important to me to measure them. When I used FFT in Matlab, the
frequency resolution was 10Hz. Is there any way to imporve the
resolution without changing the sampling frequency or number of
points?
Thank you very much
Mike
Mike skrev:
> Hello Guys
>
> This is not a quite Matlab question, but hopfully someone can give me
> any hint.
>
> I have 1000 pion... FFT frequency resolution questionHi all,
i've got a signal which is I know is periodic. However the frequency
resolution is of my FFT is too low.
T=0.1s
f_sample=10000 Hz
N=500
df=20
Now I thought it should be possible ( as the signal is periodic) to
put this signal into an array, say, 5 times to increase N to 2500
(decreasing df to 4). However when I do this and perform an FFT I get
the same results as in the first case only with 4 zeros in between
each result. Padding with zeros is also not possible because my
signal clearly does not decay.
Does anybody have an idea, or is the only solution to increase N by
taking a... stft vs fft resolutionDoes a shorttime fourier transform have greater frequency resolution
that a normal fft?
the code at
http://ccrma.stanford.edu/~jos/sasp/STFT_Matlab.html
would suggest not since it's only using a normal fft on a frame of
data rather than the whole dataset.
On Jan 4, 12:30=A0pm, Adam Chapman
<adam.chap...@student.manchester.ac.uk> wrote:
> Does a shorttime fourier transform have greater frequency resolution
> that a normal fft?
>
> the code athttp://ccrma.stanford.edu/~jos/sasp/STFT_Matlab.html
> would suggest not since it's only using a normal fft on a frame of
... FFT of audio signal resolution??Hi,
I have a question which may be very basic. I am receiving an audio input of 1.5 second at fs=44100 from a mic. Now i want to perform fft of the sample and analyse it at 1hz or lesser resolution. My interest is only to find the frequencies at which the peaks occur the actual value of magnitude is irrelevant. Currently I am using a hanning window of size 128 and pwelch function.
I thought i could increase the size of the hanning window but i was not sure if that would increase the resolution or distort the given data.
Any help along these lines would be very useful. I tried reading... How to debug method resolution problem?Good morning,
I'm faced with a tricky problem and the C++ gurus I work with couldn't
help. I apologize for not posting actual code, I simply can't for IP
reasons. I have the following inheritance diagram:
class A
{
virtual A *foo( ... ) const;
};
template< class T >
class B : public A
{
virtual A *foo( ... ) const;
};
template< class T >
inline A *B::foo( ... ) const
{
throw( "B's template specializations must override foo!" );
return NULL;
}
I have specializations of B for 8 distinct classes T. I have
explicitly overridden foo for ... How do I change my fft display resolutionI'm sampling at 6000 hz but when I display my fft it only has a
resolution of 1hz. How can I increase this resolution?
FFT resolution depends on the number of samples as well as the
sampling rate. More samples => better resolution. Check the help
files. I think they have some discussion of the resolution. Also any
textbook on Fourier analysis will give the theory (not for the
mathematically fainthearted).
Lynn
Try using the FFT Zoom. I also came across this problem recently. I
increased the resolution by using the fft zoom to meet my needs.
The following links should help in th... Frequency resolution of FFT after windowingWhen I normally do a FFT, the frequency resolution = samplin
frequency/number of samples.
When I use windowing (like BH4) to prevent smearing, this should increas
the frequency resolution of the FFT, because of the convolution in th
frequency domain. I have seen tables which show the frequency resolutio
increases by a factor of 1.9 for a Blackman Harris window, so if th
frequency resolution was 100 Hz without windowing, it should now be ~19
Hz.
1) How can I calculate what this factor is, or what the new frequenc
resolution is?
2) I can only do FFTs when the number of samples is a power... FFT with subHz resolution?Newbie question:
I am learning FFT and wandering how to achieve subHz resolution. If
the sampling rate is Fs, and you take N points out of the samples to
do FFT, then the maximum resolution is 1Hz. Or can you take more
points than sampled in a second to achieve subHz resolution?
Maybe the zoom FFT can achieve subHz resolution?
Can somebody enlighten me on this? Thanks!
me4dtrade asked:
> can you take more
> points than sampled in a second
Yes
jim
On Jan 7, 11:56=A0am, me4dtrade <me4dtr...@gmail.com> wrote:
> Newbie question:
>
> I am learning FFT and wanderi... Looking for an FFT with degraded resolutionPoorly worded title, but it comes close.
My goal is to *SKETCH* the frequency content of a 10 second signal
sampled at 44 kHz out to ~20 kHz with a resolution of roughly 100 Hz.
For that resolution a .1 second long window would work.
HOWEVER, that would ignore 99% of what I wish to represent.
A 10 second window would get all the data, with orders of magnitude
excessive computation.
Are my desires as incompatible as they appear. Or might there be a
reasonable means of estimation? Whatever approach taken, the output plot
will have ~1000 points.
On Dec 8, 10:59 am, Richard Owlett <rowl.... resolution of eeg signal FFti have taken fft of eeg signal, but i want to good resolution of that fft. for that parpose what i can do..please tell me.
On Monday, April 7, 2014 7:12:06 PM UTC+12, Ganesh Vasekar wrote:
> i have taken fft of eeg signal, but i want to good resolution of that fft. for that parpose what i can do..please tell me.
Get a longer signal.
The longer the signal in the time domain, the better the resolution in the frequency domain.
...



