# Oversampling/undersampling of a continuous time signal and preventing aliasing

Hi everyone, I want to understand how these concepts work.

Suppose that we have a signal x(t) which has a maximum frequency component of 3 Hz. So let the DTFT of this signal be like that:

[PLAIN]http://img341.imageshack.us/img341/1134/31096081.png [Broken]

Also let y[n] be the digital signal that we get from sampling on x(t).

If we want to be able to get the analog signal back, after the sampling, we need to sample with a rate of 6 Hz. I think until here we're correct.

Let's suppose that we sample with a rate of 8 Hz. The new DTFT will be like that:

[PLAIN]http://img401.imageshack.us/img401/1138/60749513.png [Broken]

Now, oversampling typically gives us more information about the continuous signal which is pretty much useless, because we're basically using more power in order to get a result that would be possible to get with less resources(less sampling frequency).

Why is it that if I sample with EXACTLY 2FMAX frequency, in this case 6 Hz the space between the two periodic components is becoming less and less? Why is that changing? See this DTFT for 6 Hz:

[PLAIN]http://img809.imageshack.us/img809/2876/53245096.png [Broken]

Let's go on under sampling. If I sample with a frequency of 4 Hz I will get a DTFT like that:

[PLAIN]http://img843.imageshack.us/img843/3771/36854781.png [Broken]

We can see that the frequencies kind of collude each other, so for example for the frequency of 3 we have an amplitude of zero or something else, so basically this is what's called aliasing right?

3. Question about anti-aliasing with a low pass filter.

If I use a low pass filter for a continuous time signal before applying the sampling method, how would that prevent me from getting an aliased result? A low pass filter let's some frequency components to pass, and others just disappear, so how would I use that in order to prevent aliasing?

Sorry for my english, thanks a lot :)

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rude man
Homework Helper
Gold Member
Sampling is essentially multiplying your input (sine) signal of frequency f by a time sequence of delta functions. It's very straightforward to determine the Fourier series of such a pulse train. It contains a dc term and integer multiples of the sampling frequency F, all of equal amplitude.

Thus, when you sample an input signal f with a sampling signal F you get sum & difference frequency outputs. So, taking each sampling frequency component one at a time we get:
dc term --> (f+0) and (f-0) = f --> wanted at output
F --> (F-f) and (F+f) --> neither one wanted
2F --> (2F-f) and (2F+f) --> neither one wanted
etc.

So you see that the lowest output component is (F-f). If F is exactly 2f the lowest aliased component is at f. If we make F < 2f we get band overlaps since then a signal at f shows up at < f. If F > 2f the distance between the bands increases, with the distance between bands now > 0.

If we sample at F = 2f and post-filter at a cutoff frequency of F = 2f we get only dc to f at the output, thus reconstructing the input signal perfectly.

Since there is no such thing as a perfect cutoff filter, we need F > 2f to allow for the fact that the filter rolls off its gain gradually. So it's not only not a waste to sample at > 2f but an absolute necessity.

(There is such a thing as deliberate undersampling to reduce a high signal frequency to a more manageable level but let's not go into that now).