Here is the code:
fs = 22050;
x = rand(fs,1);
x = x - mean(x); % get rid of DC offeset
% set comb-filter coefficients
f = 220; % fundamental
L = round(fs/f); % delay length
coef = 0.99; % IIR coefficient
% build delay vector and filter
b = [1 zeros(1,L-1) coef];
y = filter(1, b, x);
% create amplitude envelope for output
decay = 8;
expEnv = exp ((0:(length(y)-1))/length(y));
expEnv = (1./expEnv).^ decay;
fprintf('%d\n',length(expEnv));
fprintf('%d\n', length(y));
% envelope output signal
z = y .* expEnv;
sound(z , fs); % play sound
It refuses to execute the z = y.*expEnv line. The printf lines shows that both y and expEnv have the same length (22050)
Yes...the dangers of the length command.
although they have the same number of elements, they do not have the same shape:
>> size(y)
ans =
22050 1
>> size(expEnv)
ans =
1 22050
This will fix that:
z = y .* expEnv.';
Related
function bvp()
solinit = bvpinit(linspace(0,1,10),[0, 0.3, 0, 0.5]);
sol = bvp4c(#myprojec,#mybounds, solinit);
plot(sol.x,sol.y(1,:),'.');
%Hold on can be used to suspend the graph so as to plot another graph on
%the same axis with different parameters
xlabel('\c');
ylabel('\theta');
function dydx = myprojec(x,y)
% enter the parameter
% x represents c
% y(i) represents f,theta
% f = y(1), f' = y(2), theta = y(4), theta' = y(5)
% star (*) replaced by 0
% infinity =5
% Max infinity = 10
% B = input('input the value of Beta:');
B = 5;
% M = input('input the value of M:');
M = 1;
% Km = input('input the value of Km:');
Km = 1.5;
% Br = input ('input the value of Br:');
Br = 12;
dydx=[ y(2);(B^2-M)*y(1)-Km;y(4);Br*((M- B^2)*y(1)-y(2)^2)];
% Now d bcs
function res = mybounds(ya,yb)
% where to denote initial values of y at zero(0)
% finf denotes initial vlue of y at infinity (5 or 10)
% n = input('input the value of n:');
res = [ya(1)
yb(1)-1
ya(4)
yb(4)-(1)];
The error I got is that Matlab cannot solve the collocation equations. A singular Jacobian encountered. I would appreciate your suggestions.
I am trying to use Matlab's ifftn in 3-dimensions to get solution in physical space. In particular I am trying to use ifftn on 1/k^2. The analytical solution to that in physical space is 1/(4*pi*r). However I am not recovering that. Please note: $r = sqrt(x^2 + y^2 + z^2)$ and $k = sqrt(kx^2 + ky^2 + kz^2)$.
clc; clear;
n = 128; % no. of points for ifft
L = 2*pi; % size of the periodic domain
x = linspace(-L/2,L/2,n); y = x; z = x; % creating vectors for x-y-z direction
[X,Y,Z] = meshgrid(x,y,z); % creating meshgrid for physical space
R = sqrt(X.^2 + Y.^2 + Z.^2); % use for 1/(4*pi*r)
k = (2*pi/L)*[0:n/2 -n/2+1:-1]; % wave vector;
[k1,k2,k3] = meshgrid(k,k,k);
denom = (k1.^2 + k2.^2 + k3.^2); % This is k^2
F = 1./denom; F(1,1,1) = 0; % The first value is set to zero as it is infinite
phi = 1./(4*pi*R); % physical domain solution
phys = fftshift(ifftn(F)); % Using ifftn
ph_abs = abs(phys);
mid = ph_abs(n/2,:,:); % looking at the midplane of the output
mid = permute(mid,[3 2 1]); % permuting for contourplot.
PHI = phi(n/2,:,:); %looking at the midplane of the physical space.
PHI = permute(PHI,[3 2 1]);
figure(1)
surf(x,z,log(mid))
shading flat
colorbar();
figure(2)
surf(x,z,log10(abs(PHI)))
shading flat
colorbar();
I am trying to implement the finite difference method in matlab. I did some calculations and I got that y(i) is a function of y(i-1) and y(i+1), when I know y(1) and y(n+1). However, I don't know how I can implement this so the values of y are updated the right way. I tried using 2 fors, but it's not going to work that way.
EDIT
This is the script and the result isn't right
n = 10;
m = n+1;
h = 1/m;
x = 0:h:1;
y = zeros(m+1,1);
y(1) = 4;
y(m+1) = 6;
s = y;
for i=2:m
y(i) = y(i-1)*(-1+(-2)*h)+h*h*x(i)*exp(2*x(i));
end
for i=m:-1:2
y(i) = (y(i) + (y(i+1)*(2*h-1)))/(3*h*h-2);
end
The equation is:
y''(x) - 4y'(x) + 3y(x) = x * e ^ (2x),
y(0) = 4,
y(1) = 6
Thanks.
Consider the following code. The central differential quotient is discretized.
% Second order diff. equ.
% y'' - 4*y' + 3*y = x*exp(2*x)
% (y(i+1)-2*y(i)+y(i-1))/h^2-4*(y(i+1)-y(i-1))/(2*h) + 3*y(i) = x(i)*exp(2*x(i));
The solution region is specified.
x = (0:0.01:1)'; % Solution region
h = min(diff(x)); % distance
As said in my comment, using this method, all points have to be solved simultaneously. Therefore, above numerical approximation of the equation is transformed in a linear system of euqations.
% System of equations
% Matrix of coefficients
A = zeros(length(x));
A(1,1) = 1; % known solu for first point
A(end,end) = 1; % known solu for last point
% y(i) y'' y
A(2:end-1,2:end-1) = A(2:end-1,2:end-1)+diag(repmat(-2/h^2+3,[length(x)-2 1]));
% y(i-1) y'' -4*y'
A(1:end-1,1:end-1) = A(1:end-1,1:end-1)+diag(repmat(1/h^2+4/(2*h),[length(x)-2 1]),-1);
% y(i+1) y'' -4*y'
A(2:end,2:end) = A(2:end,2:end)+diag(repmat(1/h^2-4/(2*h),[length(x)-2 1]),+1);
With the rhs of the differential equation. Note that the known values are calculated by 1 in the matrix and the actual value in the solution vector.
Y = x.*exp(2*x);
Y(1) = 4; % known solu for first point
Y(end) = 6; % known solu for last point
y = A\Y;
Having an equation to approximate the first order derivative (see above) you can verify the solution. (note, ddx2 is an own function)
f1 = ddx2(x,y); % first derivative (own function)
f2 = ddx2(x,f1); % second derivative (own function)
figure;
plot(x,y);
saveas(gcf,'solu1','png');
figure;
plot(x,f2-4*f1+3*y,x,x.*exp(2*x),'ko');
ylim([0 10]);
legend('lhs','rhs','Location','nw');
saveas(gcf,'solu2','png');
I hope the solution shown below is correct.
I used Harmonic Product Spectrum to find the fundamental note present when a number of harmonics are present. This is the code I've implemented;
[song,FS] = wavread('C major.wav');
%sound(song,FS);
P = 20000;
N=length(song); % length of song
t=0:1/FS:(N-1)/FS; % define time period
song = sum(song,2);
song=abs(song);
%----------------------Finding the envelope of the signal-----------------%
% Gaussian Filter
w = linspace( -1, 1, P); % create a vector of P values between -1 and 1 inclusive
sigma = 0.335; % standard deviation used in Gaussian formula
myFilter = -w .* exp( -(w.^2)/(2*sigma.^2)); % compute first derivative, but leave constants out
myFilter = myFilter / sum( abs( myFilter ) ); % normalize
% fft convolution
myFilter = myFilter(:); % create a column vector
song(length(song)+length(myFilter)-1) = 0; %zero pad song
myFilter(length(song)) = 0; %zero pad myFilter
edges =ifft(fft(song).*fft(myFilter));
tedges=edges(P:N+P-1); % shift by P/2 so peaks line up w/ edges
tedges=tedges/max(abs(tedges)); % normalize
%---------------------------Onset Detection-------------------------------%
% Finding peaks
maxtab = [];
mintab = [];
x = (1:length(tedges));
min1 = Inf;
max1 = -Inf;
min_pos = NaN;
max_pos = NaN;
lookformax = 1;
for i=1:length(tedges)
peak = tedges(i:i);
if peak > max1,
max1 = peak;
max_pos = x(i);
end
if peak < min1,
min1 = peak;
min_pos = x(i);
end
if lookformax
if peak < max1-0.07
maxtab = [maxtab ; max_pos max1];
min1 = peak;
min_pos = x(i);
lookformax = 0;
end
else
if peak > min1+0.08
mintab = [mintab ; min_pos min1];
max1 = peak;
max_pos = x(i);
lookformax = 1;
end
end
end
max_col = maxtab(:,1);
peaks_det = max_col/FS;
No_of_peaks = length(peaks_det);
[song,FS] = wavread('C major.wav');
song = sum(song,2);
%---------------------------Performing STFT--------------------------------%
h = 1;
%for i = 2:No_of_peaks
song_seg = song(max_col(7-1):max_col(7)-1);
L = length(song_seg);
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
seg_fft = fft(song_seg,NFFT);%/L;
f = FS/2*linspace(0,1,NFFT/2+1);
seg_fft_2 = 2*abs(seg_fft(1:NFFT/2+1));
L5 = length(song_seg);
figure(6)
plot(f,seg_fft_2)
%plot(1:L/2,seg_fft(1:L/2))
title('Frequency spectrum of signal (seg_fft)')
xlabel('Frequency (Hz)')
xlim([0 2500])
ylabel('|Y(f)|')
ylim([0 500])
%----------------Performing Harmonic Product Spectrum---------------------%
% In harmonic prodcut spectrum, you downsample the fft data several times and multiply all those with the original fft data to get the maximum peak.
%HPS
seg_fft = seg_fft(1 : size(seg_fft,1)/2 );
seg_fft = abs(seg_fft);
a = length(seg_fft);
seg_fft2 = ones(size(seg_fft));
seg_fft3 = ones(size(seg_fft));
seg_fft4 = ones(size(seg_fft));
seg_fft5 = ones(size(seg_fft));
for i = 1:((length(seg_fft)-1)/2)
seg_fft2(i,1) = seg_fft(2*i,1);%(seg_fft(2*i,1) + seg_fft((2*i)+1,1))/2;
end
%b= size(seg_fft2)
L1 = length(seg_fft2);
NFFT1 = 2^nextpow2(L1); % Next power of 2 from length of y
f1 = FS/2*linspace(0,1,NFFT1/2+1);
seg_fft12 = 2*abs(seg_fft2(1:NFFT1/2+1));
figure(7);
plot(f1,seg_fft12)
title('Frequency spectrum of signal (seg_fft2)')
xlabel('Frequency (Hz)')
xlim([0 2500])
ylabel('|Y(f)|')
ylim([0 500])
This is the plot for Figure 6
So in the real scenario once I perform HPS (downsample by 2) the peak at 440.1 should shift down to 220 while the peak at 881 should shift down to around 440. But when I plot the graph that is not what I get. Insetad this is the graph I get,
Why don't I get the correct graph???? I don't seem to understand what Im doing wrong here... Can someone please have a look and let me know.. Thank you.....
The problem with your downsampling is that you trim the vector by 2x before doing the downsampling, instead of after. You do
seg_fft = seg_fft(1 : size(seg_fft,1)/2 );
% [... other stuff ...]
for i = 1:((length(seg_fft)-1)/2)
seg_fft2(i,1) = seg_fft(2*i,1);%(seg_fft(2*i,1) + seg_fft((2*i)+1,1))/2;
end
Instead you need to downsample first, then trim:
for i = 1:((length(seg_fft)-1)/2)
seg_fft2(i,1) = seg_fft(2*i,1);%(seg_fft(2*i,1) + seg_fft((2*i)+1,1))/2;
end
seg_fft = seg_fft(1 : size(seg_fft,1)/2 );
UPDATE you asked why this does not preserve the peaks. The short answer is that you may not be "looking at" the peaks. If you want to keep (nearest) peaks during downsampling by n, you can do the following:
n = 3; % degree of decimation or downsampling we want to do
N = size(seg_fft, 1); % number of samples in original FFT
Nn = n * floor(N/n); % number of samples that can be divided by n
fftBlock = reshape(seg_fft(1:Nn, 1), n, N);
fftResampled = max(fftBlock);
How does this work? Let's use a simple example of 10 x 1 points:
seg_fft = [0 1 10 5 4 3 6 12 4 3];
We want "every 3rd". Naive algorithm would give
fftResampled = [2 3 7];
But we would have liked the "peaks" [10 3 12] - unfortunately they are not in the right location.
After reshaping the array (and losing the last element; if it might be an interesting value we could append and pad with zeros) we get:
fftBlock = [0 5 6;
1 4 3;
10 3 4];
(Remember that Matlab matrices are rows-first)
Now taking the max (the function will operate along the first dimension unless we tell it otherwise) you get
fftResampled = [10 5 6];
i.e. always the peak. While this preserves the peaks, it does mean that your "valleys" are getting filled in a bit.
Bottom line: there is no way not to destroy "some" information in the process of downsampling - you are, after all, throwing half the samples away. What you keep, and how you account for the information content in the data you discard, is something only you can decide, as it will depend on your application, and what is important for you.
I would like to fit a curve on the form y=a+b*sin(2*pi*x)+c*cos(2*pi*x) to some data points in Matlab. I've been trying to use 'fit' but then I only get this message 'if isa( fittypeobj, 'fittype' )'
This is my code:
L = load('file.mat');
x = filedata(:,1);
ft = fittype('a+b*sin(2*pi*x)+c*cos(2*pi*x)');
fit(x, filedata(:,3), ft)
Can somebody please tell me what I'm doing wrong?
Here is how to do the fit 'by hand' in a least-squares way:
x = x(:); %make sure we have column vectors
y = y(:);
f0 = 1;
M = [ones(size(x)), sin(2*pi*f0*x), cos(2*pi*f0*x)];
%design matrix, columns are base vectors
% least square approximation of x = k(1)*M(:,1) + k(2)*M(:,2) + k(3)*M(:,3);
% see help mldivide
k = M \ y;
a = k(1);
b = k(2);
c = k(3);
Quick test to see if it works:
>> x = linspace(0,10,1000)'; % note transpose to make column
>> y = 3 + 1.5 * sin(2*pi*x) + 8 * cos(2*pi*x) + randn(size(x)); % add some noise
>> f0 = 1;
>> M = [ones(size(x)), sin(2*pi*f0*x), cos(2*pi*f0*x)];
>> k = M \ y
k =
3.0383
1.5264
7.9385
>> plot(x, y, x, M*k, 'r'); legend('measurement', 'fit')