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All right. I want to spend a little more time today [Audio breaks up] theorem – sampling,

and interpolation, and some of the phenomena that was associated with it. I'll remind you

what the setup was from last time. And I'll almost carry out the proof again – or at least

I'll give you the setup for the proof again because as I said many times in this – last time,

and as I will say again today, for me the sampling [Audio breaks up] the derivation of the

– the sampling theorem, the formula is identical with a proof of the sampling formula. I

think the two are so closely related that to understand one you really have to think in

terms of the other. All right.

So here's the setup. We look at a signal which is bandlimited. That is to say, its Fourier

transform is identically zero from a certain point on. All right. So suppose – say F of T is

a signal. F of T is a bandlimited signal. So I'll remind you again what that means. This

was the key definition really that got the whole thing rolling. So it means that the Fourier

transform is a condition on the [Audio breaks up] signal. The Fourier transform is

identically zero say for S bigger than equal to sum P over two [Audio breaks up] – I'm

writing it. All right.

I'm writing as [Audio breaks up] with respect to the or [Audio breaks up] and there's

[Audio breaks up] such P is called the bandwidth. There may be many P's for which that

is true, but the smallest one is called the band [Audio breaks up] the small such P is

called the bandwidth. Bandwidth is sort of a ubiquitous [Audio breaks up] in signal

processing, but I think this is – I don't know if this is the first definition of it, but I think

this is the definition that's most commonly accepted. It's the smallest – you're assuming

the Fourier transform is zero – identically zero from some point on, and the smallest

number for which that's true is called the bandwidth.

So the picture is this. Here's zero. Here's minus P over two. Here's plus P over two. And

from there on out going up to plus infinity/minus infinity the Fourier transform is

identically zero. Now, remember you can't really plug a Fourier transform because the

Fourier transform is complex, so – and the Fourier transform, if it's a real signal, is

symmetric, and all the rest of that jazz. So this is not to be taken literally as a picture, but

it's something [Audio breaks up] mind. That's the general picture you should keep in

[Audio breaks up] next step [Audio breaks up] what is [Audio breaks up] the clever key

idea.

You periodize the Fourier transform – FFS by comb [Audio breaks up] by [Audio breaks

up] that spacing. All right. That is you look at this. The Fourier transform of F convolved

with the Shah function. And what that does is because it's identically zero outside going

from P over two to infinity minus P over two minus infinity, this just shifts it, and adds it

up, and you get a bunch of copies of the original signal – non-overlapping copies of the

original signal. So here's zero. Here is P. Here is minus P. Here is, again, P over two – I'll

get out of the way in just a second. Here is minus P over two, and so on, and so on. All

right. So this is a picture of the periodized version of – hello? Is something wrong?

Student:This one's cutting out.

Instructor (Brad Osgood):Oh, it's cutting out? I wonder if it's because I dropped it.

Okay. Back to live action.

And then the main trick is to cut off.

and interpolation, and some of the phenomena that was associated with it. I'll remind you

what the setup was from last time. And I'll almost carry out the proof again – or at least

I'll give you the setup for the proof again because as I said many times in this – last time,

and as I will say again today, for me the sampling [Audio breaks up] the derivation of the

– the sampling theorem, the formula is identical with a proof of the sampling formula. I

think the two are so closely related that to understand one you really have to think in

terms of the other. All right.

So here's the setup. We look at a signal which is bandlimited. That is to say, its Fourier

transform is identically zero from a certain point on. All right. So suppose – say F of T is

a signal. F of T is a bandlimited signal. So I'll remind you again what that means. This

was the key definition really that got the whole thing rolling. So it means that the Fourier

transform is a condition on the [Audio breaks up] signal. The Fourier transform is

identically zero say for S bigger than equal to sum P over two [Audio breaks up] – I'm

writing it. All right.

I'm writing as [Audio breaks up] with respect to the or [Audio breaks up] and there's

[Audio breaks up] such P is called the bandwidth. There may be many P's for which that

is true, but the smallest one is called the band [Audio breaks up] the small such P is

called the bandwidth. Bandwidth is sort of a ubiquitous [Audio breaks up] in signal

processing, but I think this is – I don't know if this is the first definition of it, but I think

this is the definition that's most commonly accepted. It's the smallest – you're assuming

the Fourier transform is zero – identically zero from some point on, and the smallest

number for which that's true is called the bandwidth.

So the picture is this. Here's zero. Here's minus P over two. Here's plus P over two. And

from there on out going up to plus infinity/minus infinity the Fourier transform is

identically zero. Now, remember you can't really plug a Fourier transform because the

Fourier transform is complex, so – and the Fourier transform, if it's a real signal, is

symmetric, and all the rest of that jazz. So this is not to be taken literally as a picture, but

it's something [Audio breaks up] mind. That's the general picture you should keep in

[Audio breaks up] next step [Audio breaks up] what is [Audio breaks up] the clever key

idea.

You periodize the Fourier transform – FFS by comb [Audio breaks up] by [Audio breaks

up] that spacing. All right. That is you look at this. The Fourier transform of F convolved

with the Shah function. And what that does is because it's identically zero outside going

from P over two to infinity minus P over two minus infinity, this just shifts it, and adds it

up, and you get a bunch of copies of the original signal – non-overlapping copies of the

original signal. So here's zero. Here is P. Here is minus P. Here is, again, P over two – I'll

get out of the way in just a second. Here is minus P over two, and so on, and so on. All

right. So this is a picture of the periodized version of – hello? Is something wrong?

Student:This one's cutting out.

Instructor (Brad Osgood):Oh, it's cutting out? I wonder if it's because I dropped it.

Okay. Back to live action.

And then the main trick is to cut off.

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