This problem illustrates one of the possible pitfalls of blindly applying numeri

ID: 3922690 • Letter: T

Question

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?

Explanation / Answer

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: ');
l2=input('Enter the lower limit: ');
u2=input('Enter the upper limit: ');
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

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

Comment only

28 Jan 2014 assad

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

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

23 Sep 2013 Aghil Vinayak

stem(y);

Comment only

30 Mar 2013 sonali 21 Feb 2013 Rama Krishna Tata 17 Feb 2013 Abdul-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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