This may be a very straightforward question but I've been trying to understand this for a long time and can't wrap my head around it.
How would one go about discretizing the following continuous time signal:
signal_continuous = 1/6 * sin(880*pi*t) + 1/15*sin(1760*pi*t);
As far as I understand, if I wanted to sample the above signal with a sampling rate of 8Khz over two seconds. I would do something like this (in Matlab speak):
t = [0:1/8000:2-1/8000]; signal_discrete = signal_continuous./t;
Is this the whole picture or have I completely missed the point here?
My rationale behind doing the above is that since the continuous time signal is, well, continuous and if I want to sample it at 8000 cycles per second over a two second period. I would just divide the signal by that (t in this case). However, what happens to the rest of the signal? Or rather, does this not distort the whole signal (if the original was longer than 2 seconds, for example)?
I was complicating things for no reason. In Matlab I would just need to do the following:
t = [0:1/8000:2-1/8000]; discrete_signal = 1/6 * sin(880*pi*t) + 1/15*sin(1760*pi*t);
This would create a two second discrete-time signal out of the original continuous signal.
The theory provided in the answer, I think, is beneficial so I'm leaving the question up.