in MFCCs i have specified f_low and f_high which are my frequency min and max bands, and i am about to compute N equally distanced mel values between these two frequency values. So i wrote
f_low=1000;
f_high=fs/2;
filt_num=26; % number of filters
stp=round(f_high/filt_num); % step
f=f_low:stp:f_high; % my frequency vector
but i can't divide equally my f vector, maybe there is a function in matlab that does it , or am i missing something? Please help and thanks in advance.
A bit of digging around leads me to believe you want a linearly spaced vector with filt_num entries, starting at f_low and ending at f_high. You should use linspace for that as follows:
f = linspace(f_low,f_high,filt_num);
This is essentially the same as your last two lines of code. Keep in mind your code only works when f_high is larger than f_low. linspace does not have this issue, as it also supports descending vectors.
Related
I am using the function findchangepts and use 'linear' which detects changes in mean and slope. How does it note a change? Is it by consecutive points until the next point has a different mean and slope?
Mathworks has the following explanation:
If x is a vector with N elements, then findchangepts partitions x into two regions, x(1:ipt-1) and x(ipt:N), that minimize the sum of the residual (squared) error of each region from its local mean.
How does the function get ipt?
Thanks in advance!
I am working with a single vector with N elements.
I've got an arbitrary probability density function discretized as a matrix in Matlab, that means that for every pair x,y the probability is stored in the matrix:
A(x,y) = probability
This is a 100x100 matrix, and I would like to be able to generate random samples of two dimensions (x,y) out of this matrix and also, if possible, to be able to calculate the mean and other moments of the PDF. I want to do this because after resampling, I want to fit the samples to an approximated Gaussian Mixture Model.
I've been looking everywhere but I haven't found anything as specific as this. I hope you may be able to help me.
Thank you.
If you really have a discrete probably density function defined by A (as opposed to a continuous probability density function that is merely described by A), you can "cheat" by turning your 2D problem into a 1D problem.
%define the possible values for the (x,y) pair
row_vals = [1:size(A,1)]'*ones(1,size(A,2)); %all x values
col_vals = ones(size(A,1),1)*[1:size(A,2)]; %all y values
%convert your 2D problem into a 1D problem
A = A(:);
row_vals = row_vals(:);
col_vals = col_vals(:);
%calculate your fake 1D CDF, assumes sum(A(:))==1
CDF = cumsum(A); %remember, first term out of of cumsum is not zero
%because of the operation we're doing below (interp1 followed by ceil)
%we need the CDF to start at zero
CDF = [0; CDF(:)];
%generate random values
N_vals = 1000; %give me 1000 values
rand_vals = rand(N_vals,1); %spans zero to one
%look into CDF to see which index the rand val corresponds to
out_val = interp1(CDF,[0:1/(length(CDF)-1):1],rand_vals); %spans zero to one
ind = ceil(out_val*length(A));
%using the inds, you can lookup each pair of values
xy_values = [row_vals(ind) col_vals(ind)];
I hope that this helps!
Chip
I don't believe matlab has built-in functionality for generating multivariate random variables with arbitrary distribution. As a matter of fact, the same is true for univariate random numbers. But while the latter can be easily generated based on the cumulative distribution function, the CDF does not exist for multivariate distributions, so generating such numbers is much more messy (the main problem is the fact that 2 or more variables have correlation). So this part of your question is far beyond the scope of this site.
Since half an answer is better than no answer, here's how you can compute the mean and higher moments numerically using matlab:
%generate some dummy input
xv=linspace(-50,50,101);
yv=linspace(-30,30,100);
[x y]=meshgrid(xv,yv);
%define a discretized two-hump Gaussian distribution
A=floor(15*exp(-((x-10).^2+y.^2)/100)+15*exp(-((x+25).^2+y.^2)/100));
A=A/sum(A(:)); %normalized to sum to 1
%plot it if you like
%figure;
%surf(x,y,A)
%actual half-answer starts here
%get normalized pdf
weight=trapz(xv,trapz(yv,A));
A=A/weight; %A normalized to 1 according to trapz^2
%mean
mean_x=trapz(xv,trapz(yv,A.*x));
mean_y=trapz(xv,trapz(yv,A.*y));
So, the point is that you can perform a double integral on a rectangular mesh using two consecutive calls to trapz. This allows you to compute the integral of any quantity that has the same shape as your mesh, but a drawback is that vector components have to be computed independently. If you only wish to compute things which can be parametrized with x and y (which are naturally the same size as you mesh), then you can get along without having to do any additional thinking.
You could also define a function for the integration:
function res=trapz2(xv,yv,A,arg)
if ~isscalar(arg) && any(size(arg)~=size(A))
error('Size of A and var must be the same!')
end
res=trapz(xv,trapz(yv,A.*arg));
end
This way you can compute stuff like
weight=trapz2(xv,yv,A,1);
mean_x=trapz2(xv,yv,A,x);
NOTE: the reason I used a 101x100 mesh in the example is that the double call to trapz should be performed in the proper order. If you interchange xv and yv in the calls, you get the wrong answer due to inconsistency with the definition of A, but this will not be evident if A is square. I suggest avoiding symmetric quantities during the development stage.
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
When I use [f,xi] = ksdensity(x) in Matlab, I get the probability density estimate, f, and xi evaluation points at which ksdensity calculates f.
My question is: How is each xi point calculated/determined? Is there a formula?
The documentation center says: Default is 100 equally spaced points that cover the range of data in x. So, they cover the range, but this does not explain how are calculated.
Thank you very much!
Juan
The standard method of achieving equally spaced points in MATLAB is using the linspace command. linspace(a,b,n) generates n linearly spaced points between and including a and b.
So it most probably is equivalent to xi = linspace(min(x),max(x)) (default number of points is 100 inlinspace).
I have two variables in a .mat file here:
https://www.yousendit.com/download/UW13UGhVQXA4NVVQWWNUQw
testz is a vector of cumulative distance (in meters, monotonically and regularly increasing)
testSDT is a vector of integrated (cumulative) sound wave travel time (in milliseconds) generated using the distance vector and a vector of velocities
(there is an intermediate step of creating interval travel times)
Since velocity is a continuously variable function the resulting interval travelt times and also the integrated travel times are non integers and variable in magnitude
What I want is to resample the distance vector at regular time intervals (e.g. 1 ms, 2 ms, ..., n ms)
What makes it difficult is that the maximum travel time, 994.6659, is less than the number of samples in the 2 vectors, therefore it is not straightforward to use interp1.
i.e.:
X=testSDT -> 1680 samples
Y=testz -> 1680 samples
XI=[1:1:994] -> 994 samples
This is the code I've come up with. It is a working code and it is not too bad I think.
%% Initial chores
M=fix(max(testSDT));
L=(1:1:M);
%% Create indices
% this loops finds the samples in the integrated travel time vector
% that are closest to integer milliseconds and their sample number
for i=1:M
[cl(i) ind(i)] = min(abs(testSDT-L(i)));
nearest(i) = testSDT(ind(i));
end
%% Remove duplicates
% this is necessary to remove duplicates in the index vector (happens in this test).
% For example: 2.5 ms would be the closest to both 2 ms and 2 ms
[clsst,ia,ic] = unique(nearest);
idx=(ind(ia));
%% Interpolation
% this uses the index vectors to resample the depth vectors at
% integer times
newz=interp1(clsst,testz(idx),[1:1:length(idx)],'cubic')';
As far as I can see there is one issue with this code:
I rely on the vector idx as my XI for interpolation. Vector idx is 1 sample shorter than vector ind (one duplicate was removed).
Therefore my new times will stop one millisecond short. This is a very small issue, and duplicate are unlikely but I am wondering if anybody can think of a workaround, or of a different way to approach the problem altogether.
Thank you
If I understand you correctly, you want to extrapolate to that extra point.
you can do this is many ways, one is to add that extra point to the interp1 line.
If you have some function you expect to follow your data you can use it by fitting it to the data and then obtaining that extra point or with a tool like fnxtr.
But I have a problem understanding what you want because of the way you used the line. The third argument you use, [1:1:length(idx)], is just the series [1 2 3 ...], usually when interpolating, one uses some vector x_i of points of interest, though I doubt your points of interest happen to be the series of integers 1:length(idx), what you want is just [1:length(idx) xi], where xi is that extra point x-axis value.
EDIT:
Instead of the loop just produce matrix forms out of L and testSDT, then matrix operation is somewhat faster in doing the min(abs(...:
MM=ones(numel(testSDT),1)*L;
TT=testSDT*ones(1,numel(L));
[cl ind]=(min(abs(TT-MM)));
nearest=testSDT(ind);