I'm trying to filter a short audio file using a Butter worth filter and then a function to recursively solve the difference equation using the coefficients given by Butter.m.
My code is as follows.
[x,Fs]=wavread('bugsbunny1');
wn=2500/(Fs/2);
n=10;
[B,A]=butter(n,wn);
C=zeros(1,10);
for n=2:10
C(n-1)=-A(n); %rearrange A for the recur function
end
A=C;
x=x'; % make x a row vector
n=11:length(x);
y0=zeros(1,11);
x0=zeros(1,11);
y1=recur(A,B,n,x,x0,y0); %calculate approximation recursively
y1=[y0 y1]; %add initial conditions to vector
doing this results in y1 being a matrix of invalid data given by 'NaN', any help is appreciated thanks.
Code for the recur function:
function [ y ] = recur(a,b,n,x,x0,y0)
%Use Recursion to approximate solution to first order differential equation
% Detailed explanation goes here
N=length(a);
M=length(b)-1;
y=[y0 zeros(1,length(n))];
x=[x0 x];
a1=a(length(a):-1:1);
b1=b(length(b):-1:1);
for i=N+1:N+length(n),
y(i)=-a1*y(i-N:i-1)'+b1*x(i-N:i-N+M)';
end
y=y(N+1:N+length(n));
Related
I am trying to solve a system of differential equations by the Euler's method. First, I have encoded my system into a vector. Then I pass the initial conditions to the function ode_Euler.
However, there's somehting wrong about my attempt. I am getting this error:
>> nm06p03a
Unable to perform assignment because the size of the left side is 1-by-2 and the size of the right side is 2-by-2.
Error in ode_Euler (line 11)
y(k+1,:)= y(k,:) +h*feval(f,t(k),y(k,:));
Error in nm06p03a (line 12)
tic, [tE,xE]=ode_Euler(f,tspan,x0,N); t_Euler=toc;
This is my code so far:
clear, clf
f=#(t,x)[-x(2)+x(1)*x(2); x(1)-(0.5.*x(1).^2)+(0.5.*x(2).^2)]; %Encoding system of differential equations into a vector
t0=0; tf=10;
tspan=[t0 tf];
N=100;
x0s=[0.2 0]; % A matrix consisting of initial points
for iter=1:size(x0s,1)
x0=x0s(iter,:);
tic, [tE,xE]=ode_Euler(f,tspan,x0,N); t_Euler=toc;
subplot(220+iter),
plot(tE,xE,'r:')
legend('ode_ Euler')
end
Here is the Euler's method:
function [t,y]=ode_Euler(f,tspan,y0,N)
if nargin<4|N<=0, N=100; end
if nargin<3, y0=0; end
h=(tspan(2)-tspan(1))/N;
t=tspan(1)+[0:N]'*h;
y(1,:)=y0(:)'; %make it a row vector
for k=1:N
y(k+1,:)= y(k,:) +h*feval(f,t(k),y(k,:));
end
When I use this other method ode_Heun, I get the same error:
function [t,y]=ode_Heun(f,tspan,y0,N)
if nargin<4|N<=0, N=100; end
if nargin<3, y0=0; end
h=(tspan(2)-tspan(1))/N; % Step-size
t=tspan(1)+[0:N]'*h; % Time vector
y(1,:)=y0(:)'; % make the initial value a row vector
for k=1:N
fk= feval(f,t(k),y(k,:)); y(k+1,:)= y(k,:)+h*fk; % Eq.(6.2.3)
y(k+1,:)= y(k,:) +h/2*(fk +feval(f,t(k+1),y(k+1,:))); % Eq.(6.2.4)
end
Can I get some help to understand the problem with my code?
y(k,:) is a row vector, while the return value of f is a column vector. Per broadcasting rules the sum of a row and a column vector is a matrix as the sum of the matrices of repeated row and column vectors.
This is not very logical in the context of vector and matrix operations, but can make sense in the processing of (finite) sequences of vectors. Unfortunately that distinction is not realized and enforced in the type system.
I am currently working on a project in MATLAB that compares two different numerical methods (direct and iterative), each for two different matrix sizes, to the actual analytic solution of the system. The first .m file included below provides an implementation of a tridiagonal matrix solver, where the main function plots a solution for two different sizes of tridiagonal matrices: a 2500x2500 matrix and 50x50 matrix.
function main
n=50;
for i=1:2
di=2*ones(n,1);
up=-ones(n,1);
lo=-ones(n,1);
b=ones(n,1)/(n*n);
[subl,du,supu]=tridiag_factor(lo,di,up);
usol = tridiag_solve(subl,du,supu,b);
%figure(i)
plot(usol);
hold on;
title(['tridiag ' num2str(n) ''])
n=2500;
end
function [subl,du,supu]=tridiag_factor(subd,d,supd)
n=length(d);
for i=1:n-1
subd(i)=subd(i)/d(i);
d(i+1)=d(i+1)-subd(i)*supd(i);
end
subl=subd; supu=supd; du=d;
function x = tridiag_solve(subl,du,supu,b)
% forward solve
n=length(b); z=b;
z(1)=b(1);
for i=2:n
z(i)=b(i)-subl(i)*z(i-1);
end
% back solve
x(n)=z(n);
for i=n-1:-1:1
x(i)=(z(i)-supu(i)*x(i+1))/du(i);
end
I have included a hold on statement after the first usol is plotted so I can capture the second usol on the same figure during the second iteration of the for loop. The next piece of code is a Jacobi method that iteratively solves a system for a tridiagonal matrix. In my case, I have constructed two tridiagonal matrices of different sizes which I will provide after this piece of code:
function [xf,r] = jacobi(A,b,x,tol,K)
%
% The function jacobi applies Jacobi's method to solve A*x = b.
% On entry:
% A coefficient matrix of the linear system;
% b right-hand side vector of the linear system;
% x initial approximation;
% eps accuracy requirement: sx=top when norm(dx) < eps;
% K maximal number of steps allowed.
% On return:
% x approximation for the solution of A*x = b;
% r residual vector last used to update x,
% if success, then norm(r) < eps.[
%
n = size(A,1);
fprintf('Running the method of Jacobi...\n');
for k = 1:K
% r = b - A*x;
for i=1:n
sum=0;
for j = 1:n
if j~=i
sum=sum+A(i,j)*x(j);
end
end
x(i)=(b(i)-sum)/(A(i,i));
end;
k=k+1;
r = b - A*x;
% fprintf(' norm(r) = %.4e\n', norm(r));
if (norm(r) < tol)
fprintf('Succeeded in %d steps\n', k);
return;
end;
end
fprintf('Failed to reached accuracy requirement in %d steps.\n', K);
xf=x;
%r(j) = r(j)/A(j,j);
%x(j) = x(j) + r(j);
Now, lastly, here is my code for the two tridiagonal matrices (and other related information for each system corresponding to the appropriate matrices) I wish to use in my two jacobi function calls:
% First tridiagonal matrix
n=50;
A_50=full(gallery('tridiag', n, -1, 2, -1));
% Second tridiagonal matrix
n=2500;
A_2500=full(gallery('tridiag', n, -1, 2, -1));
K=10000;
tol=1e-6;
b_A50=ones(length(A_50), 1);
for i=1:length(A_50)
b_A50(i,1)=0.0004;
end
x_A50=ones(length(A_50),1);
b_A2500=ones(length(A_2500), 1);
for i=1:length(A_2500)
b_A2500(i,1)= 1.6e-7;
end
x_A25000=ones(length(A_2500),1);
As stated in the question header, I need to plot the vectors produced from jacobi(A_50,b_A50,x_A50,tol,K), jacobi(A_2500,b_A2500,x_A25000,tol,K) and a mathematical function y=#(x) -0.5*(x-1)*x all on the same figure along with the two usol (for n=50 and n=2500) produced by the main function shown in the first piece of code, but I have had no luck due to the different lengths of vectors. I understand this is probably an easy fix, but of course my novice MATLAB skills are not sufficing.
Note: I understand x_A50 and x_A2500 are trivial choices for x, but I figured while I ask for help I better keep it simple for now and not create any more issues in my code.
In MATLAB traces sent to same plot must have same length.
I have allocated just one variable containing all traces, for the respective tridiagonals and resulting traces out of your jacobi function.
I have shortened from 2500 to 250, and the reason is that with 2500, compared to 50, the tridiag traces, and the short jacobi are so flat compared to the last trace than one has to recur to dB scale to find them on same graph window as you asked.
1st generate all data, and then plot all data.
So here you have the script plotting all traces in same graph :
clear all;close all;clc
n=[50 250];
us_ol=zeros(numel(n),max(n));
% generate data
for k=[1:1:numel(n)]
di=2*ones(n(k),1);
up=-ones(n(k),1);
lo=-ones(n(k),1);
b=ones(n(k),1)/(n(k)*n(k));
[subl,du,supu]=tridiag_factor(lo,di,up);
us_ol0 = tridiag_solve(subl,du,supu,b);
us_ol(k,[1:numel(us_ol0)])=us_ol0;
end
n_us_ol=[1:1:size(us_ol,2)];
str1=['tridiag ' num2str(n(1))];
str2=['tridiag ' num2str(n(2))];
legend(str1,str2);
grid on
% the jacobi traces
nstp=1e3;
tol=1e-6;
A1=zeros(max(n),max(n),numel(n));
for k=1:1:numel(n)
A0=full(gallery('tridiag', n(k), -1, 2, -1));
A1([1:1:size(A0,1)],[1:1:size(A0,2)],k)=A0;
end
b_A1=ones(max(n),max(n),numel(n));
for k=1:1:numel(n)
for i=1:n(k)
b_A1(i,1,k)=0.0004;
end
end
n_A1=[1:1:size(A1,1)];
jkb=zeros(numel(n),max(n));
for k=1:1:numel(n)
A0=A1([1:n(k)],[1:n(k)],k);
b_A0=b_A1([1:n(k)],[1:n(k)],k);
n0=[1:1:n(k)];
jkb0=jacobi(A0,b_A0,n0',tol,nstp)
jkb(k,[1:numel(jkb0)])=jkb0';
end
% plot data
figure(1)
ax1=gca
plot(ax1,n_us_ol,us_ol(1,:),n_us_ol,us_ol(2,:));
hold(ax1,'on')
plot(ax1,n_A1,jkb(1,:),n_A1,jkb(2,:))
grid on
legend('3/1','3/2','jkb1','jkb2')
title('3diags and jakobians graph')
As mentioned above, one has to zoom to find some of the traces
One way to combine really small traces with large traces is to use Y log scale
figure(2)
ax2=gca
plot(ax2,n_us_ol,10*log10(us_ol(1,:)),n_us_ol,10*log10(us_ol(2,:)));
hold(ax2,'on')
plot(ax2,n_A1,10*log10(jkb(1,:)),n_A1,10*log10(jkb(2,:)))
grid on
legend('3/1[dB]','3/2[dB]','jkb1[dB]','jkb2[dB]')
title('3diags and jakobians graph in dB')
let us consider following Page :
http://djj.ee.ntu.edu.tw/S_Transform.pdf
paragraph 2.3 The Discrete S Transform
let say that we have sampled version of signal x, and given sampling frequency fs,i have calculated discrete Fourier transform using following code
function y=DFT(x);
N=length(x);
D=zeros(N,N);
for k=1:N
for n=1: N
D(k,n)=exp((-j*(k-1)*2*pi*(n-1))/N);
end
end
y=D*x'/N;
end
and started to estimate discrete S transform
function [S]=discrete_s_transform(x,fs);
%compute discrete s transform
%fs-sampling frequency
N=length(x); % length of signal
T=1/fs; % sampling period
Y=DFT(X);
how can i continue related to this part ?
clearly loops are not problem to implement,just they go from 1 to N instead of 0 to N-1 because of matlab vectors are 1 based,but what about main code?multiplication to exponential? coudl you please help me to finish S transform ?
Your sum depends only on m so I assume that other parameters as well as function H((m+n)/(NT)) are defined. Via for-loops the simplest one is:
function [S]=discrete_s_transform(x);
N=length(x); % length of signal
S=0;
m=0;
for i=1:N
S=S+H((m+n)/(NT))*exp(a)*exp(b*m/N);
m=m+1;
end
Hope that helps!
My intention is to plot the original mathematical function values from the differential equation of the second order below:
I(thetadbldot)+md(g-o^2asin(ot))sin(theta)=0
where thetadbldot is the second derivative of theta with respect to t and m,d,I,g,a,o are given constants. Initial conditions are theta(0)=pi/2 and thetadot(0)=0.
My issue is that my knowledge and tutoring is limited to storing the values of the derivatives and returning them, not values from the original mathematical function in the equation. Below you can see a code that calculates the differential in Cauchy-form and gives me the derivatives. Does anyone have suggestions what to do? Thanks!
function xdot = penduluma(t,x)
% The function penduluma(t,x) calculates the differential
% I(thetadbldot)+md(g-o^2asin(ot))sin(theta)=0 where thetadbldot is the second
% derivative of theta with respect to t and m,d,I,g,a,o are given constants.
% For the state-variable form, x1=theta and x2=thetadot. x is a 2x1 vector on the form
% [theta,thetadot].
m=1;d=0.2;I=0.1;g=9.81;a=0.1;o=4;
xdot = [x(2);m*d*(o^2*a*sin(o*t)-g)*sin(x(1))/I];
end
options=odeset('RelTol', 1e-6);
[t,xa]=ode45(#penduluma,[0,20],[pi/2,0],options);
% Then the desired vector from xa is plotted to t. As it looks now the desired
% values are not found in xa however.
Once you have the angle, you can calculate the angular velocity and acceleration using diff:
options=odeset('RelTol', 1e-6);
[t,xa]=ode45(#penduluma,[0,20],[pi/2,0],options);
x_ddot = zeros(size(t));
x_ddot(2:end) = diff(xa(:,2))./diff(t);
plot(t,xa,t,x_ddot)
legend('angle','angular velocity','angular acceleration')
which gives the following plot in Octave (should be the same in MATLAB):
Alternatively, you can work it out using your original differential equation:
x_ddot = -m*d*(o^2*a*sin(o*t)-g).*sin(xa(:,1))/I;
which gives a similar result:
I'm totally new to MATLAB and I know only few basic commands. My task is to plot a function of this kind:
I(T) = ((2*J(k*r*sin(T))/(k*r*sin(T))))^2
where
T = angle
k = (2*pi*f)/c (f= frequency in Hz and c is speed of sound)
r = radius
J = bessel function first kind
I explain a bit: the function represent the power of a soundwave in the space. I've tried many times to plot this but i get always a single point in the plot.
I assume you've defined your Bessel function in J. If not, the MATLAB command for a Bessel function of the first kind is besselj. You'll also have to specify the order of the Bessel function.
You can define your anonymous function as
f=#(t,k,r)(2*besselj(0,k*r*sin(t))./(k*r*sin(t))).^2
and plot it as
T=linspace(0,pi,100);%# a sample angle vector
plot(T,f(T,k,r)) %# where k and r are values you'll have to provide
what about
I = ((2*J(k*r*sin(T))./(k*r*sin(T)))).^2
I finally found how to manage the problem described above, here's the solution that i found, in case that other people may need it.
% MATLAB Istruction for graph generation of bessel like function %
% Variables (fixed values)%
k = 912.91
r = 0.0215
% Set the range for the angle theta %
theta = (-(2/3)*pi:pi/180:(2/3)*pi)
% Calculation of bessel function of first kind %
J = besselj(1,k*r*sin(theta))
% Calculation I function %
% notice the ./ and .^ operators
I = ((2*J)./(k*r*sin(theta))).^2
% now plot the results with the plot command
plot(theta,I)