First - I need to plot this radially averaged spectrum 1-D, against the wavenumber |k|. But early I calculate the |k| using two matrix, so I have a matrix that represent |k|. So I have to radial average this matrix too?
Second - If I remove a mean value from my data before using the FFT2 function, I expected to see only a difference in the amplitude of the values, but actually I am noting a difference in the shape of the radially averaged spectrum.
I am trying to use this to analyse magnetic field data.
Thanks for any help!!
The explanation for your second question follows from the additive property of the FFT:
FFT(c) = FFT(a+b) = FFT(a) + FFT(b)
If b is an offset - namely the mean value - which you remove from signal c, then the spectrum of c-b will equal your original spectrum minus the spectrum of a constant, b, but the FFT of a constant results in sinc wiggles! So by removing the mean value of the signal you are removing a sinc glitch at zero frequency.
Related
Good evening guys,
I wanna ask you a question regarding the analysis of a function in the domain of frequencies (Fourier). I have two vectors: one containing 7700 values for pressure, and the other one containing 7700 values (same number) for time.
For example, I call the firt vector "a" and the second one "b". With the command "figure(1),plot(a,b)" I obtain the curve in the domain of time.
How can I do to plot this curve in the domain of frequency, to make Fourier transform?
I've read about the function "fft", but I've not understood very well how it can be used...can anyone help me?
Thanks in advance for your attention!
fft returns spectrum as complex numbers. In order to analyze it you have to use its absolute value or phase. In general, it should look like this (let's assume that t is vector containing time and y is the one with actual signal, N is the number of samples):
fY = fft(y) / (N/2) % scale it to amplitude, typically by N/2
amp_fY = abs(fY)
phs_fY = angle(fY)
Additionally, it would be nice to have FFT with known frequency resolution. For that, you need sampling period/frequency. Let's call that frequency fs:
fs = 1/(t(1) - t(0))
and the vector of frequencies for FFT (F)
should be:
F = (0:fs/N:(N-1)*fs/N)
and finally plots:
plot(F, amp_fY)
% or plot(F, phs_fy) according to what you need
I you can use stem instead of plot to get some other type of chart.
Note that the DC component (the average value) will be doubled on the plot.
Hope it helps
Today I have stumbled upon a strange outcome in matlab. Lets say I have a sine wave such that
f = 1;
Fs = 2*f;
t = linspace(0,1,Fs);
x = sin(2*pi*f*t);
plot(x)
and the outcome is in the figure.
when I set,
f = 100
outcome is in the figure below,
What is the exact reason of this? It is the Nyquist sampling theorem, thus it should have generated the sine properly. Of course when I take Fs >> f I get better results and a very good sine shape. My explenation to myself is that Matlab was having hardtime with floating numbers but I am not so sure if this is true at all. Anyone have any suggestions?
In the first case you only generate 2 samples (the third input of linspace is number of samples), so it's hard to see anything.
In the second case you generate 200 samples from time 0 to 1 (including those two values). So the sampling period is 1/199, and the sampling frequency is 199, which is slightly below the Nyquist rate. So there is aliasing: you see the original signal of frequency 100 plus its alias at frequency 99.
In other words: the following code reproduces your second figure:
t = linspace(0,1,200);
x = .5*sin(2*pi*99*t) -.5*sin(2*pi*100*t);
plot(x)
The .5 and -.5 above stem from the fact that a sine wave can be decomposed as the sum of two spectral deltas at positive and negative frequencies, and the coefficients of those deltas have opposite signs.
The sum of those two sinusoids is equivalent to amplitude modulation, namely a sine of frequency 99.5 modulated by a sine of frequency 1/2. Since time spans from 0 to 1, the modulator signal (whose frequency is 1/2) only completes half a period. That's what you see in your second figure.
To avoid aliasing you need to increase sample rate above the Nyquist rate. Then, to recover the original signal from its samples you can use an ideal low pass filter with cutoff frequency Fs/2. In your case, however, since you are sampling below the Nyquist rate, you would not recover the signal at frequency 100, but rather its alias at frequency 99.
Had you sampled above the Nyquist rate, for example Fs = 201, the orignal signal could ideally be recovered from the samples.† But that would require an almost ideal low pass filter, with a very sharp transition between passband and stopband. Namely, the alias would now be at frequency 101 and should be rejected, whereas the desired signal would be at frequency 100 and should be passed.
To relax the filter requirements you need can sample well above the Nyquist rate. That way the aliases are further appart from the signal and the filter has an easier job separating signal from aliases.
† That doesn't mean the graph looks like your original signal (see SergV's answer); it only means that after ideal lowpass filtering it will.
Your problem is not related to the Nyquist theorem and aliasing. It is simple problem of graphic representation. You can change your code that frequency of sine will be lower Nyquist limit, but graph will be as strange as before:
t = linspace(0,1,Fs+2);
plot(sin(2*pi*f*t));
Result:
To explain problem I modify your code:
Fs=100;
f=12; %f << Fs
t=0:1/Fs:0.5; % step =1/Fs
t1=0:1/(10*Fs):0.5; % step=1/(10*Fs) for precise graphic representation
subplot (2, 1, 1);
plot(t,sin(2*pi*f*t),"-b",t,sin(2*pi*f*t),"*r");
subplot (2, 1, 2);
plot(t1,sin(2*pi*f*t1),"g",t,sin(2*pi*f*t),"r*");
See result:
Red star - values of sin(2*pi*f) with sampling rate of Fs.
Blue line - lines which connect red stars. It is usual data representation of function plot() - line interpolation between data points
Green curve - sin(2*pi*f)
Your eyes and brain can easily understand that these graphs represent the sine
Change frequency to more high:
f=48; % 2*f < Fs !!!
See on blue lines and red stars. Your eyes and brain do not understand now that these graphs represent the same sine. But your "red stars" are actually valid value of sine. See on bottom graph.
Finally, there is the same graphics for sine with frequency f=50 (2*f = Fs):
P.S.
Nyquist-Shannon sampling theorem states for your case that if:
f < 2*Fs
You have infinite number of samples (red stars on our plots)
then you can reproduce values of function in any time (green curve on our plots). You must use sinc interpolation to do it.
copied from Matlab Help:
linspace
Generate linearly spaced vectors
Syntax
y = linspace(a,b)
y = linspace(a,b,n)
Description
The linspace function generates linearly spaced vectors. It is similar to the colon operator ":", but gives direct control over the number of points.
y = linspace(a,b) generates a row vector y of 100 points linearly spaced between and including a and b.
y = linspace(a,b,n) generates a row vector y of n points linearly spaced between and including a and b. For n < 2, linspace returns b.
Examples
Create a vector of 100 linearly spaced numbers from 1 to 500:
A = linspace(1,500);
Create a vector of 12 linearly spaced numbers from 1 to 36:
A = linspace(1,36,12);
linspace is not apparent for Nyquist interval, so you can use the common form:
t = 0:Ts:1;
or
t = 0:1/Fs:1;
and change the Fs values.
The first Figure is due to the approximation of '0': sin(0) and sin(2*pi). We can notice the range is in 10^(-16) level.
I wrote the function reconstruct_FFT that can recover critically sampled data even for short observation intervals if the input sequence of samples is periodic. It performs lowpass filtering in the frequency domain.
I create a matlab gui and have some element in it with some axes. I plot one of my desire plot in ploter1 ( first axes ) using
plot(handles.ploter1,xx); title(handles.ploter1,'Waveform');
and it is ok,but I want use specgram and plot specgram result in another axes by I dont know how can do it :(
I test
specgram(wav,N,fs,hamming(N/4),round(0.9*N/4));xlabel('time, s');
or
specgram(handles.ploter2,wav,N,fs,hamming(N/4),round(0.9*N/4));xlabel('time, s');
but return me error or nothing !!!
please help me. thank you very much
EDIT
as mentioned in the comments by bdecaf, what should work, is to set the current axes:
axes(handles.ploter2);
now, when using just
spectrogram(x,window,noverlap,F)]
the plot should be on the specified axes. If not, try:
hold on
before!
OLD
specgram or spectogram does not have a parameter for the plot. You have to define it later on.
I suggest to get the result first by:
[S,F,T]=spectrogram(x,window,noverlap,F)]
and then plot it on a specific axes:
plot(handles.ploter2, S,F)
But I am not sure about which parameter you want to plot. Please take a look at the docs.
From the docs:
[S,F,T] = spectrogram(...) returns a vector of frequencies, F, and a vector of times, T, at which the spectrogram is computed. F has length equal to the number of rows of S. T has length k (defined above) and the values in T correspond to the center of each segment.
[S,F,T] = spectrogram(x,window,noverlap,F) uses a vector F of frequencies in Hz. F must be a vector with at least two elements. This case computes the spectrogram at the frequencies in F using the Goertzel algorithm. The specified frequencies are rounded to the nearest DFT bin commensurate with the signal's resolution. In all other syntax cases where nfft or a default for nfft is used, the short-time Fourier transform is used. The F vector returned is a vector of the rounded frequencies. T is a vector of times at which the spectrogram is computed. The length of F is equal to the number of rows of S. The length of T is equal to k, as defined above and each value corresponds to the center of each segment.
[S,F,T] = spectrogram(x,window,noverlap,F,fs) uses a vector F of frequencies in Hz as above and uses the fs sampling frequency in Hz. If fs is specified as empty [], it defaults to 1 Hz.
In MATLAB I need to generate a second derivative of a gaussian window to apply to a vector representing the height of a curve. I need the second derivative in order to determine the locations of the inflection points and maxima along the curve. The vector representing the curve may be quite noise hence the use of the gaussian window.
What is the best way to generate this window?
Is it best to use the gausswin function to generate the gaussian window then take the second derivative of that?
Or to generate the window manually using the equation for the second derivative of the gaussian?
Or even is it best to apply the gaussian window to the data, then take the second derivative of it all? (I know these last two are mathematically the same, however with the discrete data points I do not know which will be more accurate)
The maximum length of the height vector is going to be around 100-200 elements.
Thanks
Chris
I would create a linear filter composed of the weights generated by the second derivative of a Gaussian function and convolve this with your vector.
The weights of a second derivative of a Gaussian are given by:
Where:
Tau is the time shift for the filter. If you are generating weights for a discrete filter of length T with an odd number of samples, set tau to zero and allow t to vary from [-T/2,T/2]
sigma - varies the scale of your operator. Set sigma to a value somewhere between T/6. If you are concerned about long filter length then this can be reduced to T/4
C is the normalising factor. This can be derived algebraically but in practice I always do this numerically after calculating the filter weights. For unity gain when smoothing periodic signals, I will set C = 1 / sum(G'').
In terms of your comment on the equivalence of smoothing first and taking a derivative later, I would say it is more involved than that. As which derivative operator would you use in the second step? A simple central difference would not yield the same results.
You can get an equivalent (but approximate) response to a second derivative of a Gaussian by filtering the data with two Gaussians of different scales and then taking the point-wise differences between the two resulting vectors. See Difference of Gaussians for that approach.
I am new to Matlab and FFT.
I need to extract the dominant frequency from a signal which is varying in magnitude and frequency. I tried to perform a detrend and then an FFT to obtain the frequency but couldn't get rid of the large peak at 0Hz (DC component?). I used diff function on the signal and the resulting signal was processed through FFT. In this case, the FFT output didn't have the peak at zero. I compared the two FFT curves and it seems that except the peak at zero, the two show similar (not same) spectrum. I am wondering if diff function is a valid (and very effective) detrending method or am I losing some information here? In other words, does differentiating a signal have any effect on its frequency: [diff(sin(omega.t))= cos(omega.t) - no change in frequency]?
Thanks a lot!
By differentiating the signal you actually apply a a type of a highpass filter, a not so smart highpass which corrupts your signal. Instead you could try some other filter such as the following:
b=fir1(32,2*0.01/fs,'high');
a=1;
FilteredX=filtfilt(x,a,b)
where:
x is your original signal,
FilteredX is the filtered signal.
fs - your sample frequency.
This way you will filter out any frequencies lower then 0.01 and will almost not hurt the rest of the spectrum allowing you to detect peaks as you wished.
The discrete fourier transform X of a signal x is defined (up to scaling) by
X(k) = Sum[ exp(2*pi*i*k*n/N) * x(n) ]
If we take first differences of the signal you get
Sum[ exp(2*pi*i*k*n/N) * (x(n) - x(n-1)) ]
which you can rearrange to give
(1 - exp(2*pi*i*k/N)) * Sum[ exp(2*pi*i*k*n/N) * x(n) ]
i.e. it is a multiple of the original fourier transform. In the k = 0 case (i.e. the zero frequency component) the multiplier is zero, which explains why this removes the spike at k = 0.
Note however that the multiplier depends on the value of k, so you won't get the same signal out - you can't just take first differences to remove a spike. There is a comparison to be made with the continuous fourier transform, where if g = F(f) and h = F(df/dx) then
h(k) = i * k * g(k)
i.e. the fourier transform of the derivative is the fourier transform of the original function, multiplied by ik.