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I have a vector of 736 temperature readings, one for every 3rd hour between jun-aug, therefore 8 per day for jun-aug.
I need to apply a Fourier analysis to this vector. I have normalized it. I have searched all over (because I really don't understand matlab) and all I see are these analysis with signs where you already have a sampling to compare it to (like 1000Hz keeps coming up). the sampling interval is 3hrs and the length of sampling is 90 days. I don't know how to do this. any help would REALLY help.
thank you, I am desperate.
The FFT operation is actually agnostic of the input sampling rate. The sampling rate is only used after the FFT is performed to interpret the frequencies of the output.
So, basically you can take the FFT of your vector and it will transform your 736 samples at discrete time instants into 736 samples at discrete frequencies. These frequencies are given by
F(k) = Fs * k / Nfft
Where Fs is the sampling rate and Nfft is the number of FFT points. The output vector from the FFT will be complex. In most cases, you will want to take the magnitude of the output to produce values proportional to the amount of energy at each frequency.
The MATLAB code may look like this:
x = %your input vector
Nfft = length(x);
Fs = 1 / (60 * 60 * 3); %1 / 3hrs
Xk = fft(x);
k = 0 : Nfft-1;
Fk = Fs * k / Nfft; %frequency vector
%plot frequency versus magnitude in dB scale
Plot(Fk, 20 * log10(abs(Xk)));
There are several other issues you need to be aware of:
Assuming your input is real-valued, the second half of the FFT output vector will be a mirror image of the first half. In this case, only the first half of the output is needed, giving you frequencies from 0 to Fs/2
Fs/2, the Nyquist rate, is the maximum frequency that can be represented by a signal sampled at Fs. So, for example, if there are temperature fluctuations within the 3 hours between samples, this information is not accurately captured by a 3hr sampling period.
I've left out a lot of details, but I tried to just give you enough info to get a frequency-magnitude plot without complicating things unnecessarily.
Related
As above lets Fs is sampling frequency, L is signal's length and t is time range.
As using mdwtdec in Matlab in order to decompose multi-raw signal into specific frequency band, I just notice that decomposed signal's length at 1st level is split into half, and keep slit into half of 1st level signal at 2nd level.
Raw signal's time range calculation: t = 0 --> (L/Fs)
My question is in every decomposition level the Sampling frequency Fs is still the same? and at every decomposition level how I can calculate the time range of each Detail and Approximation coefficient.
Also as verify the frequency band of Discrete Wavelet Transform I applied FFT at each level following this post: https://jp.mathworks.com/help/matlab/ref/fft.html?lang=en
According to this post my first question need to be answered.
Thank you very much.
I'm pretty positive that in discrete wavelet transform, the time series data or signals, if we wish, would be downsampled by a factor of 2, which means that if we would be having 2^10 or 1024 datapoints in our original time series data, in the first level, it would be divided into 2 and our level one sampling frequency would be 2^9 or 512, in second level would decrease to 256, and so on.
However, in continuous wavelet transform, it would most likely remain the same.
Based on the references, I copy some codes here that you might want to test and see, you might want to reduce the number of levels and Fs here and you can define your own x if you wish:
Fs = 1e6;
t = 0:1/Fs:1-1/Fs;
x = cos(2*pi*50*t);
[C,L] = wavedec(x,15,'db4');
details = detcoef(C,L,'cells');
d14recon = wrcoef('d',C,L,'db4',14);
plot(d14recon,'k');
d13recon = wrcoef('d',C,L,'db4',13);
hold on;
plot(d13recon,'r'); %look how small the amplitude is
a13recon = wrcoef('a',C,L,'db4',13);
plot(a13recon,'b');
Useful Links:
I'm not expert about it, you can likely read more about it and find out your desired information. There are also lots of YouTube videos about it.
Discrete wavelet transform relation to sampling frequency of the signal
Single-level discrete 2-D wavelet transform
This question already has answers here:
How do I obtain the frequencies of each value in an FFT?
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Closed 6 years ago.
I have a data set in a matrix in matlab. It contains 25,000 values taken every 0.5 ns; so the total time of the dataset is 1.25E-5 seconds.
The data set contains very high frequency noise that I am not interested in so I create another matrix is every 50th data point from the first matrix So the size of the matrix is 1000*.
I plot the absolute values from matlab's fft this matrix (I also normalise the amplitude and only plot the first half) and get the attached (two plots, second is a close up of the low frequencies I am interested in). How do I convert the x-axis to frequency?
Another point, if I take every data point (so I create an fft of the entire 25,000 points) then the x-axis is exactly the same; in other words, the size of my matrix seems to have no bearing on the x-axis returned by matlab. I've attached two links to the frequency spectrum, one of which is a close-up of the low frequencies I am interested in. It's axis goes from 0-50, so it is these values I need to convert to Hz.
Thankyou in advance!
Close up of frequency spectrum
frequency spectrum
From what I read on http://www.mathworks.com/help/matlab/math/fast-fourier-transform-fft.html#bresqop-1, it appears that the units on the x-axis of the plotted FFT are Hz if the first vector, f, you put into the plot(f,power), is defined as a sequence of n elements (n being the number of data points put into the FFT) increasing from zero to the sample frequency.
Thus, for the first plot, which used every 50th of points that were taken at a frequency of 2 GHz, the sample frequency would be 40 MHz. Thus, f = (0:n-1)*4*10^7/(25000/50)
It goes on to show how to use the fftshift function to put the center of the output of the fft function at 0, but it's clear you already did that and chopped off the negative part.
So, once you have the right separation of fs/n, sampling frequency divided by number of data points used, in the vector that supplies the x-axis to the plot function, then the units of the x-axis will be Hz.
(I hope you still have the numbers to graph again? If not, this question might help: Confusion in figuring out the relation between actual frequency values and FFT plot indexes in MATLAB)
I've a set of data from an EEG device from which I want to find the strength of different brain waves in Matlab. I tried to use EEGLAB but I wasn't really sure how, so at this point I'm simply using the dsp toolbox in Matlab.
For background: I've 15 epochs, 4 seconds in length. The device sampled at 256 Hz, and there are 264 sensors, so there are 1024 data points for each sensor for each epoch, i.e. my raw data is 264 x 1024 x 15. The baseline is removed. The data in each epoch is going to be used to train a classifier eventually, so I'm dealing with each epoch individually. I'll come up with more data samples later.
Anyways, what I've done so far is apply a Hann filter to the data and then run fft on the filtered data. So now I have the information in frequency domain. However, I'm not quite sure how to go from the power of the fft buckets to the power of certain frequency bands (e.g. alpha 8-13), to get the values I seek.
I know the answer should be straightforward but I can't seem to get find the answer I want online, and then there's further confusion by certain sources recommending using a wavelet transform? Here's the little bit of code I have so far, the input "data" is one epoch, i.e. 264 x 1024.
% apply a hann window
siz = size(data);
hann_window = hann(siz(2));
hann_window = repmat(hann_window.', siz(1), 1);
hann_data = data.' * hann_window;
% run fft
X = fft(hann_data, [], 2);
X_mag = abs(X);
X_mag = X_mag.';
Thanks for the assistance!
If I'm understanding your question correctly, you are wanting to scale the FFT output to get the correct power. To do this you need to divide by the number of samples used for the FFT.
X_mag = abs(X)/length(hann_data); % This gives the correct power.
See this question for more info.
Once the content is scaled correctly, you can find the power in a band (e.g. 8 - 13 Hz) by integrating the content from the start to the stop of the band. Since you are dealing with discrete values it is a discrete integration. For perspective, this is equivalent to changing the resolution bandwidth of a spectrum analyzer.
I am doing FFT with matlab.the time period i am doing the fft on it is 1 second and it consist of 50000 equlay spaces samples. I want to test the FFT results. so I have given an input as below (wave) which is a complex of sinosuidal waves (and I have samples it by sampling frequency of 50 KHZ)and I expect to have the frequency magnitude results as I have given in the input. the results are ok for low frequency ranges but for the higher frequency (5752 Hz and 7993 Hz) results are 5.87 and 6.7 respectively (instead of 6 and ). what Is the origin of this big mistake ? how can I improve my results ?!
Here is the code:
t = 0:1/50000:1;
wave = 100*sin(2*pi*50*t)+1*sin (2*pi*123*t)+2*sin (2*pi*203*t)+3*sin(2*pi*223*t)+4*sin(2*pi*331*t)+5*sin(2*pi*2812*t)+6*sin(2*pi*5752*t)+7*sin(2*pi*7993*t);
SPEC = fft(wave);
L = size(SPEC,2);
x= (0:L/2-1);
Half_SPEC = abs(SPEC(1:L/2))/(L/2); %% removing the mirror side and ranging the domain
plot(x,Half_SPEC);
As Oli Charlesworth has pointed out, you are taking the FFT of 50001 points, which means that the frequency spacing is 1/50001.
Typically the FFT will give you the exact magnitude of your sinusoid only if its frequency is an exact multiple of the frequency spacing. Otherwise, the energy will be spread over multiple FFT bins in a process called spectral leakage.
You may confirm this by changing the number of samples such that the frequency of your sinusoids are a multiple of the frequency spacing:
t = 0:1/50000:1-1/50000;
Windowing the input signal can also help control the amount of leakage.
I am confused about the terminology used in scipy.signal.periodogram, namely:
scaling : { 'density', 'spectrum' }, optional
Selects between computing the power spectral density ('density')
where Pxx has units of V*2/Hz if x is measured in V and computing
the power spectrum ('spectrum') where Pxx has units of V*2 if x is
measured in V. Defaults to 'density'
(see: http://docs.scipy.org/doc/scipy/reference/generated/scipy.signal.periodogram.html)
1) a few tests show that result for option 'density' is dependent on signal and window length and sampling frequency (grows when signal length increases). How come? I would say that it is exactly density that should be not dependent on these things. If I take a longer signal I should just get more accurate estimation, not different result. Not to mention that dependence on window length is also very surprising.
Result diverges in the limit of infinite signal, which could be a feature of energy, but not power. Shouldn't the periodogram converge to real theoretical PSD when length increases? If, so, am I supposed to perform another normalisation outside of the signal.periodogram method?
2) to the contrary I see that alternative option 'spectrum' gives what I would previously call Power Spectrum Density, that is, it gives a resuls independent on window segment and window length and consistent with theoretical calculation. For instance for Asin(2PIft) a two sided solution yields two peaks at -f and f, each of height 0.25*A^2.
There is a lot of literature on this subject, but I get an impression that also there is a lot of incompatibile terminology, so I will be thankful for any clarification. The straightforward question is how to interpret these options and their units. (I am used to seeing V^2/Hz which are labeled "Power Spectrum Density").
Let's take a real array called data, of length N, and with sampling frequency fs. Let's call the time bin dt=1/fs, and T = N * dt. In frequency domain, the frequency bin df = 1/T = fs/N.
The power spectrum PS (scaling='spectrum' in scipy.periodogram) is calculated as follow:
import numpy as np
import scipy.fft as fft
dft = fft.fft(data)
PS = np.abs(dft)**2 / N ** 2
It has the units of V^2. It can be understood as follow. By analogy to the continuous Fourier transform, the energy E of the signal is:
E := np.sum(data**2) * dt = 1/N * np.sum(np.abs(dft)**2) * dt
(by Parseval's theorem). The power P of the signal is the total energy E divided by the duration of the signal T:
P := E/T = 1/N**2 * np.sum(np.abs(dft)**2)
The power P only depends on the Discrete Fourier Transform (DFT) and the number of samples N. Not directly on the sampling frequency fs or signal duration T. And the power per frequency channel, i.e., power spectrum SP, is thus given by the formula above:
PS = np.abs(dft)**2 / N ** 2
For the power spectrum density PSD (scaling='density' in scipy.periodogram), one needs to divide the PS by the frequency bin of the DFT, df:
PSD := PS/df = PS * N * dt = PS * N / fs
and thus:
PSD = np.abs(dft)**2 / N * dt
This has the units of V^2/Hz = V^2 * s, and now depends on the sampling frequency. That way, integrating the PSD over the frequency range gives the same result as summing the individual values of the PS.
This should explain the relations that you see when changing the window, sampling frequency, duration.
scipy.signal.peridogram uses the scipy.signal.welch function with 0 overlap. Therefore, the scaling is similar to the one provided by the welch function, density or spectrum.
In case of the density scaling, the amplitude will vary with window length, as the longer the window the higher the frequency resolution e.g. the \Delta_f is smaller. Since the estimated density is the average one, the smaller the \Delta_f the less zero energy is considered in the averaging.
As you have mentioned spectrum scaling is an integration of the energy density over the spectrum to produce the energy. Therefore, the integration over zero values does not affect the final value.
Fourier transform actually requires finite energy in an infinite duration of time series (like a decay). So, If you just make your time series sample longer by "duplicating", the energy will be infinite with an infinite duration.
My main confusion was on the "spectrum" option for scipy.signal.periodogram, which seems to create a constant energy spectrum even when the time series become longer.
Normally, 0.5*A^2=S(f)*delta_f, where S(f) is the power density spectrum. S(f)*delta_f, representing energy is constant if A is constant. But when using a longer duration of time series, delta_f (i.e. incremental frequency) is reduced accordingly, based on FFT procedure. For example, 100s time series will lead to a delta_f=0.01Hz, while 1000s time series will have a delta_f=0.001Hz. S(f) representing density will accordingly change.