# When I executed below code it got executed perfectly but in the graph I am getting just the coordinates but not the plot.

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keerthana reddy on 4 Jun 2021
Commented: keerthana reddy on 9 Jun 2021
f=10; %frequency of the impulse in Hz
fs=f*100; % sample frequency is 10 times higher
t=-1:1/fs:1; % time vector
x= 100*cos(2*pi*1000*t);
y=zeros(size(t));
y(1:fs/f:end)=1;
for t= -1:1/fs:1
m = x.*y;
end
plot(t,m);

Walter Roberson on 7 Jun 2021
f=10; %frequency of the impulse in Hz
fs=f*100; % sample frequency is 10 times higher
t=-1:1/fs:1; % time vector
numt = length(t);
y=zeros(size(t));
y(1:fs/f:end)=1;
for tidx = 1 : numt
x = 100*cos(2*pi*1000*t(tidx));
m(tidx, :) = x.*y;
end
plot(t,m);
keerthana reddy on 9 Jun 2021
Wow!! This is great.Thank you for helping.

Srivardhan Gadila on 7 Jun 2021
Based on the above code it is clear that there is no use of the for loop by redefining the variable t to iterate over the loop and the value of m = x.*y; will be same for all the iterations. I think what you are looking for is may be the following:
f=10; %frequency of the impulse in Hz
fs=f*100; % sample frequency is 10 times higher
t=-1:1/fs:1; % time vector
x= 100*cos(2*pi*1000*t);
y=zeros(size(t));
y(1:fs/f:end)=1;
% Removing the for loop
m = x.*y;
plot(t,m);
Srivardhan Gadila on 7 Jun 2021
If your equation is m(tidx) = x(tidx) * y(tidx) then the above code in my answer should work fine because with the operator times, .* element wise multiplication is performed so you don't have to iterate over individual elements of the vectors x and y, refer to the documentation of times, .* for more information. If your equation is m(tidx) = x(tidx) * y then you can refer to the below code posted by Walter.