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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)?

EDIT:

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.

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A continuous-time signal $x(t)$ is sampled to create an ideal discrete time signal $x[n]$ by just taking the instantaneous value of the continuous-time signal at the sampling instants. If you sample the signal at an interval $T_s$, then:

$$ x[n] = x(nT_s) = [0, x(T_s), x(2T_s),\ \ldots \ ] $$

In your case, $T_s = \frac{1}{8000}\text{sec}$. In practice, there are no ideal samplers, so this model doesn't strictly hold, but it is close enough for most applications and is the model that you would typically see used in introductory signal processing textbooks and courses.

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  • $\begingroup$ Thanks for the answer. This sounds like the idea I had in my head. I may be screwing up the Matlab calculation by just dividing by t. I'm going to accept your answer and try to figure that part out. $\endgroup$ – nico_c May 7 '13 at 22:02
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    $\begingroup$ @nickecarlo: just remember that this only works if your signal doesn't contain any frequencies above half the sampling rate. you can't use it on square waves, for instance. $\endgroup$ – endolith May 9 '13 at 16:04

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