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How do we like our Fourier

Transform so far, class?

Student:Whoo.

Instructor (Brad Osgood):Yeah, all right. That’s the spirit. All right. So let me remind

you – as a matter of fact, let me reintroduce the star of the show. So let’s recall the

Fourier Transform and its inverse, and I want to make a couple of general remarks before

plunging back into specific properties, specific transforms and some properties, and its

inverse.

So, again, f of t is a signal and the Fourier Transform or function, same thing, the Fourier

Transform, I use this notation. I want to comment about that, again, in just a second.

Integral from my infinity – infinity of either the -2p I ST, F of T, VT and the inverse

Fourier Transform looks very similar except for a change in sign in the exponential. So

the inverse Fourier Transform of – I use a different function, although it doesn’t matter.

We’re gonna go from -8 to 8 of either the +2p I ST, G of S, DS, okay?

All right. Now, I want to make a few general comments here. This isn’t one I have to

write down, but it’s actually quite a complicated operation, all right? Integration is not

such a simple operation. Integrating a function against the complex exponential is gonna

make the thing oscillate, and computing an infinite integral from -8 to 8 brings with it all

sorts of peril, all right?

Now, the rigor police are off duty, for the most part, all right? So we are not gonna be so

concerned about the existence of those integrals. That is an issue, all right, and it is

something we’re gonna have to deal with but not right away. Right now, I just want to get

a little practical, hands-on know how with using the formulas and being comfortable with

the formulas.

So I’m not gonna worry about the convergence right now. We will talk a little bit more

about later, although, it’s never gonna be a big issue for us, or rather, we’re gonna – well,

we’re gonna talk about it in a number of different ways, all right, but it is something to

worry about.

So when I write down the definition of the Fourier Transform, what I really should have

said was the Fourier Transform is defined by that integral whenever the integral exists, all

right? So the question is the existence of the integral on the existence of the integral, that

is to say convergence of the integral – of the integral. All right, so more on this, to some

extent, later.

The set of points but just to introduce a bit of terminology here, something that I’m sure

you’ve said, and I actually have used, I think, somewhat informally. The set of points for

which the integral does exist, that is to say, for which the Fourier Transform exists, is

called the spectrum, all right? So the set of S of S in R for which the Fourier Transform is

defined, that is to say for which the integral exists, is called the spectrum.

So when I made the bold statement last time that every signal has a spectrum, and the

signal is determined by a spectrum, I was thinking of exactly this – that is that, in many

cases, the integral is not an issue; that is it will exist, all right? And the cases when it

doesn’t exist, of course, that poses an additional problem; you have to do some further

analysis.

All right. Now, that’s one comment that I wanted to make. As I said, the rigor police are

off duty. I’ll tell you when we have to worry about those kinds of questions. Now, that

Transform so far, class?

Student:Whoo.

Instructor (Brad Osgood):Yeah, all right. That’s the spirit. All right. So let me remind

you – as a matter of fact, let me reintroduce the star of the show. So let’s recall the

Fourier Transform and its inverse, and I want to make a couple of general remarks before

plunging back into specific properties, specific transforms and some properties, and its

inverse.

So, again, f of t is a signal and the Fourier Transform or function, same thing, the Fourier

Transform, I use this notation. I want to comment about that, again, in just a second.

Integral from my infinity – infinity of either the -2p I ST, F of T, VT and the inverse

Fourier Transform looks very similar except for a change in sign in the exponential. So

the inverse Fourier Transform of – I use a different function, although it doesn’t matter.

We’re gonna go from -8 to 8 of either the +2p I ST, G of S, DS, okay?

All right. Now, I want to make a few general comments here. This isn’t one I have to

write down, but it’s actually quite a complicated operation, all right? Integration is not

such a simple operation. Integrating a function against the complex exponential is gonna

make the thing oscillate, and computing an infinite integral from -8 to 8 brings with it all

sorts of peril, all right?

Now, the rigor police are off duty, for the most part, all right? So we are not gonna be so

concerned about the existence of those integrals. That is an issue, all right, and it is

something we’re gonna have to deal with but not right away. Right now, I just want to get

a little practical, hands-on know how with using the formulas and being comfortable with

the formulas.

So I’m not gonna worry about the convergence right now. We will talk a little bit more

about later, although, it’s never gonna be a big issue for us, or rather, we’re gonna – well,

we’re gonna talk about it in a number of different ways, all right, but it is something to

worry about.

So when I write down the definition of the Fourier Transform, what I really should have

said was the Fourier Transform is defined by that integral whenever the integral exists, all

right? So the question is the existence of the integral on the existence of the integral, that

is to say convergence of the integral – of the integral. All right, so more on this, to some

extent, later.

The set of points but just to introduce a bit of terminology here, something that I’m sure

you’ve said, and I actually have used, I think, somewhat informally. The set of points for

which the integral does exist, that is to say, for which the Fourier Transform exists, is

called the spectrum, all right? So the set of S of S in R for which the Fourier Transform is

defined, that is to say for which the integral exists, is called the spectrum.

So when I made the bold statement last time that every signal has a spectrum, and the

signal is determined by a spectrum, I was thinking of exactly this – that is that, in many

cases, the integral is not an issue; that is it will exist, all right? And the cases when it

doesn’t exist, of course, that poses an additional problem; you have to do some further

analysis.

All right. Now, that’s one comment that I wanted to make. As I said, the rigor police are

off duty. I’ll tell you when we have to worry about those kinds of questions. Now, that

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