I am not doing signal processing. But in my area, I will use the spectral density of a matrix of data. I get quite confused at a very detailed level.
%matrix H is given.
corr=xcorr2(H); %get the correlation
spec=fft2(corr); % Wiener-Khinchin Theorem
In matlab, xcorr2 will calculate the correlation function of this matrix. The lag will range from -N+1 to N-1. So if size of matrix H is N by N, then size of corr will be 2N-1 by 2N-1. For discretized data, I should use corr or half of corr?
Another problem is I think Wiener-Khinchin Theorem is basically for continuous function. I have always thought that Discretized FT is an approximation to Continuous FT, or you can say it is a tool to calculate Continuous FT. If you use matlab build in function 'fft', you should divide the final result by \delta x.
Any kind soul who knows this area well there to share some matlab code with me?
Basically, approximating a continuous FT by a Discretized FT is the same as approximating an integral by a finite sum.
We will first discuss the 1D case, then we'll discuss the 2D case.
Let's look at the Wiener-Kinchin theorem (for example here).
It states that :
"For the discrete-time case, the power spectral density of the function with discrete values x[n], is :
where
Is the autocorrelation function of x[n]."
1) You can see already that the sum is taken from -infty to +infty in the calculation of S(f)
2) Now considering the Matlab fft - You can see (command 'edit fft' in Matlab), that it is defined as :
X(k) = sum_{n=1}^N x(n)*exp(-j*2*pi*(k-1)*(n-1)/N), 1 <= k <= N.
which is exactly what you want to be done in order to calculate the power spectral density for a frequency f.
Note that, for continuous functions, S(f) will be a continuous function. For Discretized function, S(f) will be discrete.
Now that we know all that, it can easily be extended to the 2D case. Indeed, the structure of fft2 matches the structure of the right hand side of the Wiener-Kinchin Theorem for the 2D case.
Though, it will be necessary to divide your result by NxM, where N is the number of sample points in x and M is the number of sample points in y.
Related
Suppose I have a continuous probability distribution, e.g., Normal, on a support A. Suppose that there is a Matlab code that allows me to draw random numbers from such a distribution, e.g., this.
I want to build a Matlab code to "approximate" this continuous probability distribution with a probability mass function spanning over r points.
This means that I want to write a Matlab code to:
(1) Select r points from A. Let us call these points a1,a2,...,ar. These points will constitute the new discretised support.
(2) Construct a probability mass function over a1,a2,...,ar. This probability mass function should "well" approximate the original continuous probability distribution.
Could you help by providing also an example? This is a similar question asked for Julia.
Here some of my thoughts. Suppose that the continuous probability distribution of interest is one-dimensional. One way to go could be:
(1) Draw 10^6 random numbers from the continuous probability distribution of interest and store them in a column vector D.
(2) Suppose that r=10. Compute the 10-th, 20-th,..., 90-th quantiles of D. Find the median point falling in each of the 10 bins obtained. Call these median points a1,...,ar.
How can I construct the probability mass function from here?
Also, how can I generalise this procedure to more than one dimension?
Update using histcounts: I thought about using histcounts. Do you think it is a valid option? For many dimensions I can use this.
clear
rng default
%(1) Draw P random numbers for standard normal distribution
P=10^6;
X = randn(P,1);
%(2) Apply histcounts
[N,edges] = histcounts(X);
%(3) Construct the new discrete random variable
%(3.1) The support of the discrete random variable is the collection of the mean values of each bin
supp=zeros(size(N,2),1);
for j=2:size(N,2)+1
supp(j-1)=(edges(j)-edges(j-1))/2+edges(j-1);
end
%(3.2) The probability mass function of the discrete random variable is the
%number of X within each bin divided by P
pmass=N/P;
%(4) Check if the approximation is OK
%(4.1) Find the CDF of the discrete random variable
CDF_discrete=zeros(size(N,2),1);
for h=2:size(N,2)+1
CDF_discrete(h-1)=sum(X<=edges(h))/P;
end
%(4.2) Plot empirical CDF of the original random variable and CDF_discrete
ecdf(X)
hold on
scatter(supp, CDF_discrete)
I don't know if this is what you're after but maybe it can help you. You know, P(X = x) = 0 for any point in a continuous probability distribution, that is the pointwise probability of X mapping to x is infinitesimal small, and thus regarded as 0.
What you could do instead, in order to approximate it to a discrete probability space, is to define some points (x_1, x_2, ..., x_n), and let their discrete probabilities be the integral of some range of the PDF (from your continuous probability distribution), that is
P(x_1) = P(X \in (-infty, x_1_end)), P(x_2) = P(X \in (x_1_end, x_2_end)), ..., P(x_n) = P(X \in (x_(n-1)_end, +infty))
:-)
As part of a course in signal processing at university, we have been asked to write an algorithm in Matlab to calculate the single sided spectrum of our signal using DFT, without using the fft() function built in to matlab. this isn't an assessed part of the course, I'm just interested in getting this "right" for myself. I am currently using the 2018b version of Matlab, should anyone find this useful.
I have built a signal of a 1 KHz and 2KHz sinusoid, phase shifted by 135 degrees (2*pi/3 rad).
then using the equations in 9.1 of Discrete time signal processing (Allan V. Oppenheim) and Euler's formula to simplify the exponent, I produce this code:
%%DFT(currently buggy)
n=0;m=0;
for m=1:DFT_N-1 %DFT_Fmin;DFT_Fmax; %scrolls through DFT m values (K in text.)
for n=1:DFT_N-1;%;(DFT_N-1);%<<redundant code? from Oppenheim eqn. 9.1 % eulers identity, K=m and n=n
X(m)=x(n)*(cos((2*pi*n*m)/DFT_N)-j*sin((2*pi*n*m)/DFT_N));
n=n+1;
end
%m=m+1; %redundant code?
end
This takes x as the input, in this case the signal mentioned earlier, as well as the resolution of the transform, as represented by the DFT_N, which has been initialized to 100. The output of this function, X, should be something in the frequency domain, but plotting X yields a circular plot slightly larger than the unit circle, and with a gap on the left hand edge.
I am struggling to see how I am supposed to convert this to the stem() plots as given by the in-built DFT algorithm.
Many thanks, J.
This is your bug:
replace X(m)=x(n)*(cos.. with X(m)=X(m)+x(n)*(cos..
For a given m, it does not integrate over the variable n, but overwrites X(m) only the last calculation for n = DFT_N-1.
Notice that integrating over n=1:DFT_N-1 omits one harmonic, i.e., the first one, exp(-j*2*pi). Replace
n=1:DFT_N-1 with n=1:DFT_N to include that. I would also replace m=1:DFT_N-1 with m=1:DFT_N for plotting concerns.
Also replace any 2*pi*n*m with 2*pi*(n-1)*(m-1) to get the phase right, since the first entry of X should correspond to zero-frequency, yielding sum_n x(n) * (cos(0) + j sin(0)) = sum_n x(n). If your signal x is real-valued then the zero-frequency component X(1) should be real-valued, angle(X(1))=0.
Last remark, don't forget to shift zero-frequency component to the center of the spectrum for better visibility, X = circshift(X,floor(size(X)/2));
If you are interested in the single-sided spectrum only, than you can just calculate X(m) for m=1:DFT_N/2 since X it is conjugate symmetric around m=DFT_N/2, i.e., X(DFT_N/2+m) = X(DFT_N/2-m)', due to exp(-j*(pi*n+2*pi/DFT_N*m)) = exp(-j*(pi*n-2*pi/DFT_N*m))'.
As a side note, for a given m this program calculates an inner product between the array x and another array of complex exponentials, i.e., exp(-j*2*pi/DFT_N*m*n), for n = 0,1,...,N-1. MATLAB syntax is very convenient for such calculations, and you can avoid this inner loop by the following command
exp(-j*2*pi/DFT_N*m*(0:DFT_N-1)) * x
where x is a column vector. Similarly, you can avoid the first loop too by expanding your complex exponential vector row-wise for every m, i.e., build the matrix exp(-j*2*pi/DFT_N*(0:DFT_N-1)'*(0:DFT_N-1)). Then your DFT is simply
X = exp(-j*2*pi/DFT_N*(0:DFT_N-1)'*(0:DFT_N-1)) * x
For single-sided spectrum, instead use
X = exp(-j*2*pi/DFT_N*(0:floor((DFT_N-1)/2))'*(0:DFT_N-1)) * x
In my homework, I am required to depict that a method can generate an Gaussian Distribution. the matlab program is shown below:
n=100;
b=25;
len=200000;
X=rand(n,len);
x=sum(X-0.5)*b/n;
[ps2,t2]=hist(x,50);
ps2=ps2/len;
bar(t2,ps2,'y');
hold on;
sigma_2=b^2/(12*n);
R=normrnd(0,sqrt(sigma_2),1,len);
[ps2,t2]=hist(R,50);
ps2=ps2/len;
plot(t2,ps2,'bo-','linewidth',1.5);
x is the sum of n uniformly distributed variables multiplying by b/n. And x is gaussian distributed with zero-mean and sigma^2=b^2/12n.
Then I got the image where the two distribution matched.
However, when I substituted the t2 inside the normal distibution density function f(x)=exp(-x.^2/(2*sigma_2))/sqrt(2*pi*sigma_2), the output is quite larger than the first one, although the shape is similar.
I wander why this occurs?
Its because you did not normalize discrete histograms. We know that in a continuous distributions the integral of probability functions are one. For solving this issue you should divide histogram to its integral. An approximate integral of a discrete function is rectangular integral:
integral (f) = sum(f)* LengthStep
so you should change your code this way :
n=100;
b=25;
len=200000;
X=rand(n,len);
x=sum(X-0.5)*b/n;
[ps2,t2]=hist(x,50);
ps2=ps2/(sum(ps2)*(t2(2)-t2(1))); % normalize discrete distribution
bar(t2,ps2,'y');
hold on;
sigma_2=b^2/(12*n);
R=normrnd(0,sqrt(sigma_2),1,len);
[ps2,t2]=hist(R,50);
ps2=ps2/(sum(ps2)*(t2(2)-t2(1))); % normalize discrete distribution
plot(t2,ps2,'bo-','linewidth',1.5);
hold on
plot(t2,exp(-t2.^2/(2*sigma_2))/sqrt(2*pi*sigma_2),'r'); %plot continuous distribution
and this is the result :
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 need to create a diagonal matrix containing the Fourier coefficients of the Gaussian wavelet function, but I'm unsure of what to do.
Currently I'm using this function to generate the Haar Wavelet matrix
http://www.mathworks.co.uk/matlabcentral/fileexchange/33625-haar-wavelet-transformation-matrix-implementation/content/ConstructHaarWaveletTransformationMatrix.m
and taking the rows at dyadic scales (2,4,8,16) as the transform:
M= 256
H = ConstructHaarWaveletTransformationMatrix(M);
fi = conj(dftmtx(M))/M;
H = fi*H;
H = H(4,:);
H = diag(H);
etc
How do I repeat this for Gaussian wavelets? Is there a built in Matlab function which will do this for me?
For reference I'm implementing the algorithm in section 4 of this paper:
http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=04218361
I maybe would not being answering the question, but i will try to help you advance.
As far as i know, the Matlab Wavelet Toolbox only deal with wavelet operations and coefficients, increase or decrease resolution levels, and similar operations, but do not exposes the internal matrices serving to doing the transformations from signals and coefficients.
Hence i fear the answer to this question is no. Some time ago, i did this for some of the Hart Class wavelet, and i actually build the matrix from the scratch, and then i compared the coefficients obtained with the Built-in Matlab Wavelet Toolbox, hence ensuring your matrices are good enough for your algorithm. In my case, recursive parameter estimation for time varying models.
For the function ConstructHaarWaveletTransformationMatrix it is really simple to create the matrix, because the Hart Class could be really simple expressed as Kronecker products.
The Gaussian Wavelet case as i fear should be done from the scratch too...
THe steps i suggest would be;
Although MATLAB dont include explicitely the matrices, you can use the Matlab built-in functions to recover the Gaussian Wavelets, and thus compose the matrix for your algorithm.
Build every column of the matrix with every Gaussian Wavelet, for every resolution levels you are requiring (the dyadic scales). Use the Matlab Wavelets toolbox for recover the shapes.
After this, compare the coefficients obtained by you, with the coefficients of the toolbox. This way you will correct the order of the Matrix row.
Numerically, being fj the signal projection over Vj (the PHI signals space, scaling functions) at resolution level j, and gj the signal projection over Wj (the PSI signals space, mother functions) at resolution level j, we can write:
f=fj0+sum_{j0}^{j1-1}{gj}
Hence, both fj0 and gj will induce two matrices, lets call them PHIj and PSIj matrices:
f=PHIj0*cj0+sum_{j0}^{j1-1}{PSIj*dj}
The PHIj columns contain the scaled and shifted scaling wavelet signal (one, for j0 only) for the approximation projection (the Vj0 space), and the PSIj columns contain the scaled and shifted mother wavelet signals (several, from j0 to j1-1) for the detail projection (onto the Wj0 to Wj1-1 spaces).
Hence, the Matrix you need is:
PHI=[PHIj0 PSIj0... PSIj1]
Thus you can express you original signal as:
f=PHI*C
where C is a vector of approximation and detail coefficients, for the levels:
C=[cj0' dj0'...dj1']'
The first part, for addressing the PHI build can be achieved by writing:
function PHI=MakePhi(l,str,Jmin,Jmax)
% [PHI]=MakePhi(l,str,Jmin,Jmax)
%
% Build full PHI Wavelet Matrix for obtaining wavelet coefficients
% (extract)
%FILTER
[LO_R,HI_R] = wfilters(str,'r');
lf=length(LO_R);
%PHI BUILD
PHI=[];
laux=l([end-Jmax end-Jmax:end]);
PHI=[PHI MakeWMatrix('a',str,laux)];
for j=Jmax:-1:Jmin
laux=l([end-j end-j:end]);
PHI=[PHI MakeWMatrix('d',str,laux)];
end
the wfilters is a MATLAB built in function, giving the required signal for the approximation and or detail wavelet signals.
The MakeWMatrix function is:
function M=MakeWMatrix(typestr,str,laux)
% M=MakeWMatrix(typestr,str,laux)
%
% Build Wavelet Matrix for obtaining wavelet coefficients
% for a single level vector.
% (extract)
[LO_R,HI_R] = wfilters(str,'r');
if typestr=='a'
F_R=LO_R';
else
F_R=HI_R';
end
la=length(laux);
lin=laux(2); lout=laux(3);
M=MakeCMatrix(F_R,lin,lout);
for i=3:la-1
lin=laux(i); lout=laux(i+1);
Mi=MakeCMatrix(LO_R',lin,lout);
M=Mi*M;
end
and finally the MakeCMatrix is:
function [M]=MakeCMatrix(F_R,lin,lout)
% Convolucion Matrix
% (extract)
lf=length(F_R);
M=[];
for i=1:lin
M(:,i)=[zeros(2*(i-1),1) ;F_R ;zeros(2*(lin-i),1)];
end
M=[zeros(1,lin); M ;zeros(1,lin)];
[ltot,lin]=size(M);
lmin=floor((ltot-lout)/2)+1;
lmax=floor((ltot-lout)/2)+lout;
M=M(lmin:lmax,:);
This last matrix should include some interpolation routine for having better general results in each case.
I expect this solve part of your problem.....
Hyp