This problem illustrates one of the possible pitfalls of blindly appl.pdf

This problem illustrates one of the possible pitfalls of blindly applying numerical methods
without paying attention to the theoretical aspects of the differential equation itself. Consider the
equation ty' + 3y - 9t^2 = 0. Use the MATLAB program myeuler.m from Chapter 8 to compute
the Euler Method approximation to the solution with initial condition y(-0.5) = 3.15, using step
size h = 0.2 and n = 10 steps The program will generate a lot of ordered pairs (x_i, .y_i). Use plot
to graph the piecewise linear function connecting the points (x_i, y_i). Now modify the program
to implement the Improved Euler Method. Can you make sense of your answers? Next, use
ode45 to find an approximate solution on the interval (-0.5, 0.5), and plot it with plot. Print out
the values of the solution at the points -0.06: 0.02: 0.06. What is the interval on which the
approximate solution is defined? Solve the equation explicitly and graph the solutions for the
initial conditions y(0) = 0, y(-0.5) = 3.15, y(0, 5) = 3.15, y(-0.5) = -3.45, and y(0.5) = -3.45. Now
explain your results in (a)-(c). Could we have known, without solving the equation, whether to
expect meaningful results in parts (a) and (b)? Why? Can you explain how ode45 avoids making
the same mistake?
Solution
clc
clear all
close all
x=input('Enter the first sequence: ');
l1=input('Enter the lower limit: ');
u1=input('Enter the upper limit: ');
x1=l1:1:u1;
h=input('Enter the second sequence: ');
h1=l2:1:u2;
l=l1+l2;
u=u1+u2;
n=l:1:u;
s=numel(n);
i=1;
for i=1:s
y(i)=0;
for k=1:numel(x)

if (i+1-k)<=0
y(i)=y(i)+(x(k)*0);
else if (i+1-k)>numel(h)
y(i)=y(i)+(x(k)*0);
else
y(i)=y(i)+(x(k)*h(i+1-k));
k=k+1;
end
end
end
i=i+1;
end
disp(y);
subplot(2,2,1);stem(x1,x);
title('First sequence');xlabel('n');ylabel('x(n)');
subplot(2,2,2);stem(h1,h);
title('Second Sequence');xlabel('n');ylabel('h(n)');
subplot(2,2,[3 4]);stem(n,y);
title('Convoluted sequence');xlabel('n');ylabel('y(n)');
Comment only
X = input('Enter x: '); %input vector X and H
H = input('Enter h: ') ;
LenX = length(X); %defining their lenghts
LenH = length(H);
y = zeros(1,LenX+LenH); %defing vector y of zeroes and of size
% lenth of X + length of H
t = zeros(1,LenH); % definign a vector t of same length as H
for i = 1:LenH+LenX-1 % Running a for loop from 1 to length of Y -1
if i<=LenX % till I IS Lesser then length of X i.e overlap about to begin
t(1)= X(i); % put x(i) on t(1) later it is shifted forwards in the vector t i.e. later t(2)=t(1)
for j = 1:LenH % in the if condition a for loop from 1 to length of H
y(i) = y(i) + H(j)*t(j); % summing for all H(j)*t(j) and putting it at y(1) or y(2) or Y(i) in first
iteration
% i.e. for i=1 only firt multiplication would
% be non zero rest all zeroes.
end

for k = LenH:-1:2 % shifting old value of t(i) to t(i+1) now there would me 1+ non zeroes values
in t
% this cycle would continue until i is lesser then
% length X i.e. overlap increasing every iteration less
% and less non zero vales in t every iteration
t(k) = t(k-1);
end
else % now when all of the T is non zero which means 100% overlap in else overlap would start
to decrease between T and H
% T is basically X
t(1)= 0;
for j = 1:LenH % Now we start filling up Zeroes in T i.e. overlap began to decrease now and
each iteration it would decrease
% i.e T moving to left until there is no more
% over lap
y(i) = y(i) + (H(j)*t(j)); % in this for loop we multiply all respective vales of h and t and add the
% putting it at y(1) or y(2) or Y(i) in first iteration
end
for k = LenH:-1:2 %% here just like similar loop above t where we were filling up t with vales of
x
%now we are filling up zeroos in t i.e. over lap decreasing
t(k) = t(k-1);
end
end
end
ly=length(y)
indices=[ly]
for i=1:ly
indices(i)=i;
end
disp (y); %displays vector y.
disp (indices); % displays vector indices.
stem(y);
ylabel('Y[n]');
xlabel('[n]');
title('Convolution without conv function');

Comment only
x = input('Enter x: ');
h = input('Enter h: ') ;
Ni = length(x);
Nh = length(h);
y = zeros(1,Ni+Nh);
t = zeros(1,Nh);
for i = 1:Ni+Nh-1
if i<=Ni
t(1)= x(i);
for j = 1:Nh
y(i) = y(i) + h(j)*t(j);
end
for k = Nh:-1:2
t(k) = t(k-1);
end
else
t(1)= 0;
for j = 1:Nh
y(i) = y(i) + (h(j)*t(j));
end
for k = Nh:-1:2
t(k) = t(k-1);
end
end
end
stem(y);
Comment only
function [y] = myconv( x,h )
m=length(x);
n=length(h);

x=[x,zeros(1,n)];
h=[h,zeros(1,m)];
for i=1:n+m-1
y(i)=0;
for j=1:m
if(i-j+1>0)
y(i)=y(i)+x(j)*h(i-j+1);
end
end
end
clc
clear all
close all
x=input('Enter the first sequence: ');
x1=l1:1:u1;
h=input('Enter the second sequence: ');
h1=l2:1:u2;
l=l1+l2;
u=u1+u2;
n=l:1:u;
s=numel(n);
i=1;
for i=1:s
y(i)=0;
for k=1:numel(x)
if (i+1-k)<=0
y(i)=y(i)+(x(k)*0);
else if (i+1-k)>numel(h)
y(i)=y(i)+(x(k)*0);
else
y(i)=y(i)+(x(k)*h(i+1-k));
k=k+1;

end
end
end
i=i+1;
end
disp(y);
subplot(2,2,1);stem(x1,x);
title('First sequence');xlabel('n');ylabel('x(n)');
subplot(2,2,2);stem(h1,h);
title('Second Sequence');xlabel('n');ylabel('h(n)');
subplot(2,2,[3 4]);stem(n,y);
title('Convoluted sequence');xlabel('n');ylabel('y(n)');
Comment only28 Jan 2014assad
X = input('Enter x: '); %input vector X and H
H = input('Enter h: ') ;
LenX = length(X); %defining their lenghts
LenH = length(H);
y = zeros(1,LenX+LenH); %defing vector y of zeroes and of size
% lenth of X + length of H
t = zeros(1,LenH); % definign a vector t of same length as H
for i = 1:LenH+LenX-1 % Running a for loop from 1 to length of Y -1
if i<=LenX % till I IS Lesser then length of X i.e overlap about to begin
t(1)= X(i); % put x(i) on t(1) later it is shifted forwards in the vector t i.e. later t(2)=t(1)
for j = 1:LenH % in the if condition a for loop from 1 to length of H
y(i) = y(i) + H(j)*t(j); % summing for all H(j)*t(j) and putting it at y(1) or y(2) or Y(i) in first
iteration
% i.e. for i=1 only firt multiplication would
% be non zero rest all zeroes.
end
for k = LenH:-1:2 % shifting old value of t(i) to t(i+1) now there would me 1+ non zeroes values
in t
% this cycle would continue until i is lesser then
% length X i.e. overlap increasing every iteration less
% and less non zero vales in t every iteration
t(k) = t(k-1);
end

else % now when all of the T is non zero which means 100% overlap in else overlap would start
to decrease between T and H
% T is basically X
t(1)= 0;
for j = 1:LenH % Now we start filling up Zeroes in T i.e. overlap began to decrease now and
each iteration it would decrease
% i.e T moving to left until there is no more
% over lap
y(i) = y(i) + (H(j)*t(j)); % in this for loop we multiply all respective vales of h and t and add
the
% putting it at y(1) or y(2) or Y(i) in first iteration
end
for k = LenH:-1:2 %% here just like similar loop above t where we were filling up t with vales
of x
%now we are filling up zeroos in t i.e. over lap decreasing
t(k) = t(k-1);
end
end
end
ly=length(y)
indices=[ly]
for i=1:ly
indices(i)=i;
end
disp (y); %displays vector y.
disp (indices); % displays vector indices.
stem(y);
ylabel('Y[n]');
xlabel('[n]');
title('Convolution without conv function');
Comment only23 Sep 2013Aghil Vinayak
x = input('Enter x: ');
h = input('Enter h: ') ;
Ni = length(x);
Nh = length(h);
y = zeros(1,Ni+Nh);

t = zeros(1,Nh);
for i = 1:Ni+Nh-1
if i<=Ni
t(1)= x(i);
for j = 1:Nh
y(i) = y(i) + h(j)*t(j);
end
for k = Nh:-1:2
t(k) = t(k-1);
end
else
t(1)= 0;
for j = 1:Nh
y(i) = y(i) + (h(j)*t(j));
end
for k = Nh:-1:2
t(k) = t(k-1);
end
end
end
stem(y);
Comment only30 Mar 2013sonali21 Feb 2013Rama Krishna Tata17 Feb 2013Abdul-Rauf
Mujahid
function [y] = myconv( x,h )
m=length(x);
n=length(h);
x=[x,zeros(1,n)];
h=[h,zeros(1,m)];
for i=1:n+m-1
y(i)=0;
for j=1:m

if(i-j+1>0)
y(i)=y(i)+x(j)*h(i-j+1);
end
end
end

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