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The sampling theorem gives us a sufficient but not a necessary condition

on the sampling frequency that we need to use with continuous time signal.

If we pick a sample in period so

that omega N is larger than the maximum positive frequency in the input signal,

we are guaranteed an alias free discrete time sequence out of the sampling process.

However, there are classes of signals, typically bandpass signals,

where the sampling frequency indicated by the sampling theorem is

way too large with respect to the amount of information that is actually contained

in the continuous-time signal.

A bandpass signal is typically the product of a modulation.

So you start it with a baseband signal, like so, whose frequency

content is say between minus omega 0 and omega 0 centered around 0.

And then via modulation, you bring this up to a center frequency omega C,

which is usually in the higher regions of the spectrum,

because you want to transmit this information over say, a wireless channel.

So the actual information content

of the signal has an effective bandwidth of 2 times omega 0.

And usually, this omega 0 is way, way smaller than omega c.

However, a literal interpretation of the sampling theorem tells you that you have

to sample at least at twice the maximum frequency which in this case is omega

c + omega 0, so a very large sampling frequency.

I could try a naive approach and say, well, since the effective bandwidth is 2

times omega 0, I'm going to just sample at 2 times omega 0 and see what happens.

And in most cases what will happen is that you will get a lot of copies because,

of course, now the periodization frequency is very small, and you will get overlap

which will destroy the information contained by the bandpass signal.

But if you play around with the sampling frequency while remaining in the low

frequency range so much smaller than 2 times omega C plus omega 0.

You can probably find for some lucky numbers,

you will get a situation in which the alias copies instead of overlapping,

actually interleave neatly without stepping onto each other.

In this case, the information contained in the bandpass signal is preserved and

the sampling operation goes under the name of bandpass sampling.

Let's now look at the conditions under which you can perform bandpass sampling,

and what is the result of the sampling operation.

So in order to use bandpass sampling,

first of all you have to have an input signal that is bandpass.

We indicate this by saying that the spectrum of the signal has to be 0 almost

everywhere except for a region centered around a modulation frequency omega C,

and whose width is 2 times omega 0.

So this is the support of the signal on the positive frequency axis.

And, of course,

this can be symmetric with a similar region centered around minus omega C.

Intuitively, in order to preserve the information contained in the bandpass

signal, your sampling frequency cannot be less than 2 times omega 0.

If your signal was baseband, so between minus omega 0 and

omega 0, then the sampling frequency would have to be twice omega 0, or

equivalently omega N should be equal to omega 0, there's no escaping that.

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Now, the second condition that allows you to perform bandpass sampling

is called the baseband condition.

This requires you to find a sampling frequency so that the modulation frequency

omega c can be expressed as an integer multiple of omega N.

If these two conditions are met, what you get is an alias-free,

baseband version of your bandpass signal.

So you're not really getting a sampled version of the bandpass signal.

You will now have the high-frequency localization of the spectrum, but

by sampling you will have preserved the information,

but automatically demodulated the signal back to baseband.

Graphically, it looks like so.

This is your bandpass signal.

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If you sample, respecting the two conditions, so omega N greater than

omega 0, and, of course, omega N = to omega c divided by some integer k,

in this case k is one, two, three, four, five, six, right?

Well, if you do that, the alias copies will not overlap, and

you will get one copy here centered around 0 which

contains exactly the same information as the passband spectrum, but

that has been automatically demodulated to baseband by the sampling operation.

So two birds with one stone.

So let's look at a practical example, and for a change, let's use hertz.

So let's forget about the 2 pi normalization factor in frequencies,

the AM radio band is a portion of the frequency spectrum that is located for

AM radio transmission.

And it usually goes, depends on the country, but

usually between 500kHz and 1.6MHz.

The portion of the spectrum is divided into independent channels that are then

allocated to different radio stations.

And the usual channel width is 9kHz.

So the F0, half the bandwidth,

half the effective bandwidth of each channel is 4.5KHz.

Let's pick a channel with a center frequency of 1.5MHz,

so pretty much at the end of the AM band.

In theory, if we wanted to sample the information contain into that

radio channel using the sampling theorem, would have to use a sampling frequency

which is two times the highest frequency contained in the channel.

So two times 1,504,500Hz, for

a sampling period of about one microsecond, which is pretty fast.

The antialias condition for

the single channel, however, imposes a much lower sampling frequency.

If the effective channel width is 9KHz, then

the sampling frequency is equal to the effective channel width, so we just need

to use a sampling frequency that is higher than 9KHz to preserve the information.

The baseband condition, on the other hand, requires us to pick a sampling frequency

so that we can express the modulation frequency, so

the centerband frequency, as a multiple of half the sampling frequency.

In other words, kFs/2 must be equal to 1.5MHz.