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

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);

Answer 1

[Walter Roberson]

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);

[Walter Roberson]

By the way, the above can be rewritten without a for loop. R2015b or later, it would be

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; 
m = x(:) .* y;
plot(t, m)

[keerthana reddy]

Can you please share the graph you got after executing your code?

[Walter Roberson]

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; 
m = x(:) .* y;
plot(t, m)

So, 0's and 100's. Not very interesting.

You might be thinking the plot is wrong, but it is correct for what you asked for. You asked for a sampling frequency of 10*100 = 1000, so 1/1000 between entries, and you used cos(2*pi*1000*t) and 1000*1/1000 = 1, so you are doing cos(2*pi*1*integer) which is going to be a constant.

[keerthana reddy]

Yeah, but I am trying to get something like the sampled signal in below picture.

[Walter Roberson]

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; 
m = x(:) .* y;
stem(t, m)

This will be slow to draw, and will not look at all informative. You will need to zoom in a fair bit to see the signal.

Your code has little to do with your required output.

[Walter Roberson]

f = 10; %frequency of the impulse in Hz
fs = f * 7; % sample frequency is 7 times higher
t =-1/10:1/fs:1/10; % time vector
y = 100*cos(2*pi*1000*t);
stem(t, y)

Notice that we reduced the sampling frequency and the time interval.

Notice too that no loop was needed.

[keerthana reddy]

Wow!! This is great.Thank you for helping.

Answer 2

[Srivardhan Gadila]

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);

[keerthana reddy]

But,actually I want "x" values at different "t" values for "m". So, that is why I put for loop from "t=-1:1/fs:1". Whenever, "t" value changes "x" value also changes as there is "t" in function of "x". If for loop is not used then "x" value will be constant right. I hope I explained it clearly.

I am trying to write code for smapling theorem.

[Srivardhan Gadila]

You can Set Breakpoints at lines where the value of x and y are computed and you can see that they have been computed for all the values of t after executing that line itself. Also you can set the breakpoint within the for loop and see that the value of m will be same for every iteration.

[keerthana reddy]

Yeah, I just checked putting breakpoints. But, can you tell me how can I get different x at different t. But y value should be constant and when I execute m = x.*y I need to get discrete values.

[Srivardhan Gadila]

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.