periodogram function
in [Matlab]
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From: EUIX Wang on 12 Mar 2010 01:52 "Wayne King" <wmkingty(a)gmail.com> wrote in message <hnbp8s$23t$1(a)fred.mathworks.com>... > "EUIX Wang" <kuhanw(a)gmail.com> wrote in message <hnbksd$jcv$1(a)fred.mathworks.com>... > > When using the periodogram function matlab states in its help file that > > > > "periodogram uses an nfft-point FFT of the windowed data (x.*window) to compute the periodogram. If the value you specify for nfft is less than the length of x, then x.*window is wrapped modulo nfft. If the value you specify for nfft is greater than the length of x, then x.*window is zero-padded to compute the FFT." > > > > Now what exactly does the first part mean? I understand how zero padding works what is wrapped modulo? > > > > Also does this work if your value of nfft is not a integer multiple of your number of data points? For example if I have 1000 data points and I pick nfft=210. Then does matlab zero pad the data with 50 extra zeros when executing periodogram? > > Hi, what it means is that it divides the data into nonoverlapping segments of length NFFT, possibly padding the last one with zeros if a case like you describe holds. Then it stacks those segments and adds them together to form a length NFFT segment. So for example: > > X=1:20; > NFFT=10; > Y = buffer(X,NFFT); > Y=sum(Y,2); > > Hope that helps, > Wayne Hey thanks for the reply but I think I am still missing something. Tell me if I am understanding this correctly. If for example I have a dataset of 1000 points and using nfft=210. Is matlab using datawrap on the 1000 points, thus creating 210 points and then applying the periodogram function to that 210 point dataset (that is periodogram(data210,[],210,Fs))? Because I tried that and compared it to taking the periodogram directly (periodogram(data1000,[],210,Fs)) and while the shape seems to be the same the two data sets seem to differ by some factor. I am posting this pretty late but perhaps it has to do some with the sampling frequency Fs?
From: EUIX Wang on 12 Mar 2010 02:04 I did a little bit of tinker and I think the factor is N/nfft, where N is the total number of points. I say I think because the exact decimal places don't always match up. Although I am not sure if that is because of precision issues in matlab. However I am not where to come up with this extra factor just yet. It's getting late so I will think about it tomorrow.
From: Wayne King on 12 Mar 2010 07:40
"EUIX Wang" <kuhanw(a)gmail.com> wrote in message <hncp15$k6m$1(a)fred.mathworks.com>... > I did a little bit of tinker and I think the factor is N/nfft, where N is the total number of points. I say I think because the exact decimal places don't always match up. Although I am not sure if that is because of precision issues in matlab. However I am not where to come up with this extra factor just yet. It's getting late so I will think about it tomorrow. Hi, It sounds like it's definitely just the scaling factor from your description. In your code are you using 1/Fs as part of the scaling factor? I think you'll find that the scaling is 1/(Fs*N) where N is the length of the input series (before wrapping). Then, for the one-sided PSD estimate, you scale everything except DC by 2. For example: reset(RandStream.getDefaultStream); x=randn(10,1); N = length(x); nfft=5; Fs=1000; %Periodogram with nfft less than segment length [Pxx,Fxx]=periodogram(x,[],nfft,Fs); % Pxx is 0.0039 0.0061 0.0077 % now the brute force way % wrapping the windowed time data with datawrap() % same as sum(buffer(x,5),2) X=datawrap(x,nfft); xdft=fft(X,5); Pwr = xdft.*conj(xdft); % Scaling Pwr = (1/(Fs*N))*Pwr; % select the bins for one-sided estimate Pwr=Pwr(1:(nfft+1)/2); % scale for one-sided estimate 2*everything but DC Pwr(2:end)=2*Pwr(2:end); % compare Pwr and Pxx Hope that helps, Wayne |