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 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\'); 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=l.

1. 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)');

2. 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

3. 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');
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)

4. 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

5. % 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');
Solution

6. 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)');

7. 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

8. 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');
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)

9. 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

10. % 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');