The Fourier Transform and its Applications

31  Download (0)

Full text

(1)

The Fourier Transform and its Applications

Prof. Brad Osgood Stanford University

https://see.stanford.edu/materials/lsoftaee261/book-fall-07.pdf http://www.coursehero.org/lecture/fourier-series-0

https://www.youtube.com/view_play_list?p=B24BC7956EE040CD https://see.stanford.edu/Course/EE261

(2)

Chapter 2

Fourier Transform

2.1 A First Look at the Fourier Transform

We’re about to make the transition from Fourier series to the Fourier transform. “Transition” is the appropriate word, for in the approach we’ll take the Fourier transform emerges as we pass from periodic to nonperiodic functions. To make the trip we’ll view a nonperiodic function (which can be just about anything) as a limiting case of a periodic function as the period becomes longer and longer. Actually, this process doesn’t immediately produce the desired result. It takes a little extra tinkering to coax the Fourier transform out of the Fourier series, but it’s an interesting approach.1

Let’s take a specific, simple, and important example. Consider the “rect” function (“rect” for “rectangle”) defined by

Π(t) =

(1 |t| < 1/2 0 |t| ≥ 1/2 Here’s the graph, which is not very complicated.

0 1

1/2 1 3/2

−1 −1/2

−3/2

Π(t) is even — centered at the origin — and has width 1. Later we’ll consider shifted and scaled versions.

You can think of Π(t) as modeling a switch that is on for one second and off for the rest of the time. Π is also

1As an aside, I don’t know if this is the best way of motivating the definition of the Fourier transform, but I don’t know a better way and most sources you’re likely to check will just present the formula as a done deal. It’s true that, in the end, it’s the formula and what we can do with it that we want to get to, so if you don’t find the (brief) discussion to follow to your tastes, I am not offended.

(3)

called, variously, the top hat function (because of its graph), the indicator function, or the characteristic function for the interval (−1/2, 1/2).

While we have defined Π(±1/2) = 0, other common conventions are either to have Π(±1/2) = 1 or Π(±1/2) = 1/2. And some people don’t define Π at ±1/2 at all, leaving two holes in the domain. I don’t want to get dragged into this dispute. It almost never matters, though for some purposes the choice Π(±1/2) = 1/2 makes the most sense. We’ll deal with this on an exceptional basis if and when it comes up.

Π(t) is not periodic. It doesn’t have a Fourier series. In problems you experimented a little with periodiza- tions, and I want to do that with Π but for a specific purpose. As a periodic version of Π(t) we repeat the nonzero part of the function at regular intervals, separated by (long) intervals where the function is zero. We can think of such a function arising when we flip a switch on for a second at a time, and do so repeatedly, and we keep it off for a long time in between the times it’s on. (One often hears the term duty cycle associated with this sort of thing.) Here’s a plot of Π(t) periodized to have period 15.

0 1 1

5 10 15 20

−5 −1

−15 −10

−20

Here are some plots of the Fourier coefficients of periodized rectangle functions with periods 2, 4, and 16, respectively. Because the function is real and even, in each case the Fourier coefficients are real, so these are plots of the actual coefficients, not their square magnitudes.

−5 −4 −3 −2 −1 0 1 2 3 4 5

−0.2 0 0.2 0.4 0.6 0.8 1

n cn

(4)

2.1 A First Look at the Fourier Transform 67

−5 −4 −3 −2 −1 0 1 2 3 4 5

−0.2 0 0.2 0.4 0.6 0.8 1

n cn

−5 −4 −3 −2 −1 0 1 2 3 4 5

−0.2 0 0.2 0.4 0.6 0.8 1

n cn

We see that as the period increases the frequencies are getting closer and closer together and it looks as though the coefficients are tracking some definite curve. (But we’ll see that there’s an important issue here of vertical scaling.) We can analyze what’s going on in this particular example, and combine that with some general statements to lead us on.

Recall that for a general function f (t) of period T the Fourier series has the form f (t) =

X n=−∞

cne2πint/T

so that the frequencies are 0, ±1/T , ±2/T , . . . . Points in the spectrum are spaced 1/T apart and, indeed, in the pictures above the spectrum is getting more tightly packed as the period T increases. The n-th Fourier coefficient is given by

cn= 1 T

Z T 0

e−2πint/Tf (t) dt = 1 T

Z T /2

−T /2

e−2πint/Tf (t) dt .

(5)

We can calculate this Fourier coefficient for Π(t):

cn= 1 T

Z T /2

−T /2

e−2πint/TΠ(t) dt = 1 T

Z 1/2

−1/2

e−2πint/T · 1 dt

= 1 T

h 1

−2πin/Te−2πint/T it=1/2

t=−1/2= 1 2πin



eπin/T − e−πin/T



= 1 πnsin

πn T

 .

Now, although the spectrum is indexed by n (it’s a discrete set of points), the points in the spectrum are n/T (n = 0, ±1, ±2, . . .), and it’s more helpful to think of the “spectral information” (the value of cn) as a transform of Π evaluated at the points n/T . Write this, provisionally, as

(Transform of periodized Π)

n T



= 1 πnsin

πn T

 .

We’re almost there, but not quite. If you’re dying to just take a limit as T → ∞ consider that, for each n, if T is very large then n/T is very small and

1

πnsinπn T



is about size 1

T (remember sin θ ≈ θ if θ is small) . In other words, for each n this so-called transform,

1

πnsinπn T

 ,

tends to 0 like 1/T . To compensate for this we scale up by T , that is, we consider instead (Scaled transform of periodized Π)

n T



= T 1 πnsin

πn T



= sin(πn/T ) πn/T . In fact, the plots of the scaled transforms are what I showed you, above.

Next, if T is large then we can think of replacing the closely packed discrete points n/T by a continuous variable, say s, so that with s = n/T we would then write, approximately,

(Scaled transform of periodized Π)(s) = sin πs πs . What does this procedure look like in terms of the integral formula? Simply

(Scaled transform of periodized Π) n T



= T · cn

= T · 1 T

Z T /2

−T /2

e−2πint/Tf (t) dt = Z T /2

−T /2

e−2πint/Tf (t) dt .

If we now think of T → ∞ as having the effect of replacing the discrete variable n/T by the continuous variable s, as well as pushing the limits of integration to ±∞, then we may write for the (limiting) transform of Π the integral expression

b Π(s) =

Z

−∞

e−2πistΠ(t) dt . Behold, the Fourier transform is born!

Let’s calculate the integral. (We know what the answer is, because we saw the discrete form of it earlier.) Π(s) =b

Z

−∞

e−2πistΠ(t) dt = Z 1/2

−1/2

e−2πist· 1 dt = sin πs πs .

Here’s a graph. You can now certainly see the continuous curve that the plots of the discrete, scaled Fourier coefficients are shadowing.

(6)

2.1 A First Look at the Fourier Transform 69

−5 −4 −3 −2 −1 0 1 2 3 4 5

−0.2 0 0.2 0.4 0.6 0.8 1

s b Π(

s)

The function sin πx/πx (written now with a generic variable x) comes up so often in this subject that it’s given a name, sinc:

sinc x = sin πx πx pronounced “sink”. Note that

sinc 0 = 1 by virtue of the famous limit

lim

x→0

sin x x = 1 .

It’s fair to say that many EE’s see the sinc function in their dreams.

(7)

How general is this? We would be led to the same idea — scale the Fourier coefficients by T — if we had started off periodizing just about any function with the intention of letting T → ∞. Suppose f (t) is zero outside of |t| ≤ 1/2. (Any interval will do, we just want to suppose a function is zero outside some interval so we can periodize.) We periodize f (t) to have period T and compute the Fourier coefficients:

cn= 1 T

Z T /2

−T /2

e−2πint/Tf (t) dt = 1 T

Z 1/2

−1/2

e−2πint/Tf (t) dt .

(8)

2.1 A First Look at the Fourier Transform 71

How big is this? We can estimate

|cn| = 1 T

Z 1/2

−1/2

e−2πint/Tf (t) dt

≤ 1 T

Z 1/2

−1/2

|e−2πint/T| |f (t)| dt = 1 T

Z 1/2

−1/2

|f (t)| dt = A T ,

where

A = Z 1/2

−1/2

|f (t)| dt ,

which is some fixed number independent of n and T . Again we see that cntends to 0 like 1/T , and so again we scale back up by T and consider

(Scaled transform of periodized f ) n T



= T cn= Z T /2

−T /2

e−2πint/Tf (t) dt .

In the limit as T → ∞ we replace n/T by s and consider f (s) =ˆ

Z

−∞

e−2πistf (t) dt .

We’re back to the same integral formula.

Fourier transform defined There you have it. We now define the Fourier transform of a function f (t) to be

f (s) =ˆ Z

−∞

e−2πistf (t) dt .

For now, just take this as a formal definition; we’ll discuss later when such an integral exists. We assume that f (t) is defined for all real numbers t. For any s ∈ R, integrating f (t) against e−2πistwith respect to t produces a complex valued function of s, that is, the Fourier transform ˆf (s) is a complex-valued function of s ∈ R. If t has dimension time then to make st dimensionless in the exponential e−2πist s must have dimension 1/time.

While the Fourier transform takes flight from the desire to find spectral information on a nonperiodic function, the extra complications and extra richness of what results will soon make it seem like we’re in a much different world. The definition just given is a good one because of the richness and despite the complications. Periodic functions are great, but there’s more bang than buzz in the world to analyze.

The spectrum of a periodic function is a discrete set of frequencies, possibly an infinite set (when there’s a corner) but always a discrete set. By contrast, the Fourier transform of a nonperiodic signal produces a continuous spectrum, or a continuum of frequencies.

It may be that ˆf (s) is identically zero for |s| sufficiently large — an important class of signals called bandlimited — or it may be that the nonzero values of ˆf (s) extend to ±∞, or it may be that ˆf (s) is zero for just a few values of s.

The Fourier transform analyzes a signal into its frequency components. We haven’t yet considered how the corresponding synthesis goes. How can we recover f (t) in the time domain from ˆf (s) in the frequency domain?

(9)

Recovering f (t) from ˆf (s) We can push the ideas on nonperiodic functions as limits of periodic func- tions a little further and discover how we might obtain f (t) from its transform ˆf (s). Again suppose f (t) is zero outside some interval and periodize it to have (large) period T . We expand f (t) in a Fourier series,

f (t) = X n=−∞

cne2πint/T.

The Fourier coefficients can be written via the Fourier transform of f evaluated at the points sn= n/T . cn= 1

T Z T /2

−T /2

e−2πint/Tf (t) dt = 1 T

Z

−∞

e−2πint/Tf (t) dt

(we can extend the limits to ±∞ since f (t) is zero outside of [−T /2, T /2])

= 1 T

fˆ n T



= 1 T

f (sˆ n) . Plug this into the expression for f (t):

f (t) = X n=−∞

1 T

f (sˆ n)e2πisnt.

Now, the points sn= n/T are spaced 1/T apart, so we can think of 1/T as, say ∆s, and the sum above as a Riemann sum approximating an integral

X n=−∞

1 T

f (sˆ n)e2πisnt= X n=−∞

f (sˆ n)e2πisnt∆s ≈ Z

−∞

f (s)eˆ 2πistds .

The limits on the integral go from −∞ to ∞ because the sum, and the points sn, go from −∞ to ∞. Thus as the period T → ∞ we would expect to have

f (t) = Z

−∞

f (s)eˆ 2πistds

and we have recovered f (t) from ˆf (s). We have found the inverse Fourier transform and Fourier inversion.

The inverse Fourier transform defined, and Fourier inversion, too The integral we’ve just come up with can stand on its own as a “transform”, and so we define the inverse Fourier transform of a function g(s) to be

ˇ g(t) =

Z

−∞

e2πistg(s) ds (upside down hat — cute) .

Again, we’re treating this formally for the moment, withholding a discussion of conditions under which the integral makes sense. In the same spirit, we’ve also produced the Fourier inversion theorem. That is

f (t) = Z

−∞

e2πistf (s) ds .ˆ Written very compactly,

( ˆf )ˇ= f .

The inverse Fourier transform looks just like the Fourier transform except for the minus sign. Later we’ll say more about the remarkable symmetry between the Fourier transform and its inverse.

By the way, we could have gone through the whole argument, above, starting with ˆf as the basic function instead of f . If we did that we’d be led to the complementary result on Fourier inversion,

g)ˆ= g .

(10)

2.1 A First Look at the Fourier Transform 73

A quick summary Let’s summarize what we’ve done here, partly as a guide to what we’d like to do next. There’s so much involved, all of importance, that it’s hard to avoid saying everything at once. Realize that it will take some time before everything is in place.

• The Fourier transform of the signal f (t) is f (s) =ˆ

Z

−∞

f (t)e−2πistdt .

This is a complex-valued function of s.

One value is easy to compute, and worth pointing out, namely for s = 0 we have f (0) =ˆ

Z

−∞

f (t) dt .

In calculus terms this is the area under the graph of f (t). If f (t) is real, as it most often is, then f (0) is real even though other values of the Fourier transform may be complex.ˆ

• The domain of the Fourier transform is the set of real numbers s. One says that ˆf is defined on the frequency domain, and that the original signal f (t) is defined on the time domain (or the spatial domain, depending on the context). For a (nonperiodic) signal defined on the whole real line we generally do not have a discrete set of frequencies, as in the periodic case, but rather a continuum of frequencies.2 (We still do call them “frequencies”, however.) The set of all frequencies is the spectrum of f (t).

◦ Not all frequencies need occur, i.e., ˆf (s) might be zero for some values of s. Furthermore, it might be that there aren’t any frequencies outside of a certain range, i.e.,

f (s) = 0ˆ for |s| large .

These are called bandlimited signals and they are an important special class of signals. They come up in sampling theory.

• The inverse Fourier transform is defined by ˇ g(t) =

Z

−∞

e2πistg(s) ds .

Taken together, the Fourier transform and its inverse provide a way of passing between two (equiva- lent) representations of a signal via the Fourier inversion theorem:

( ˆf )ˇ= f ,g)ˆ= g . We note one consequence of Fourier inversion, that

f (0) = Z

−∞

f (s) ds .ˆ

There is no quick calculus interpretation of this result. The right hand side is an integral of a complex-valued function (generally), and result is real (if f (0) is real).

2A periodic function does have a Fourier transform, but it’s a sum of δ functions. We’ll have to do that, too, and it will take some effort.

(11)

Now remember that ˆf (s) is a transformed, complex-valued function, and while it may be “equivalent”

to f (t) it has very different properties. Is it really true that when ˆf (s) exists we can just plug it into the formula for the inverse Fourier transform — which is also an improper integral that looks the same as the forward transform except for the minus sign — and really get back f (t)? Really? That’s worth wondering about.

• The square magnitude | ˆf (s)|2 is called the power spectrum (especially in connection with its use in communications) or the spectral power density (especially in connection with its use in optics) or the energy spectrum (especially in every other connection).

An important relation between the energy of the signal in the time domain and the energy spectrum in the frequency domain is given by Parseval’s identity for Fourier transforms:

Z

−∞

|f (t)|2dt = Z

−∞

| ˆf (s)|2ds . This is also a future attraction.

A warning on notations: None is perfect, all are in use Depending on the operation to be performed, or on the context, it’s often useful to have alternate notations for the Fourier transform. But here’s a warning, which is the start of a complaint, which is the prelude to a full blown rant. Diddling with notation seems to be an unavoidable hassle in this subject. Flipping back and forth between a transform and its inverse, naming the variables in the different domains (even writing or not writing the variables), changing plus signs to minus signs, taking complex conjugates, these are all routine day-to-day operations and they can cause endless muddles if you are not careful, and sometimes even if you are careful. You will believe me when we have some examples, and you will hear me complain about it frequently.

Here’s one example of a common convention:

If the function is called f then one often uses the corresponding capital letter, F , to denote the Fourier transform. So one sees a and A, z and Z, and everything in between. Note, however, that one typically uses different names for the variable for the two functions, as in f (x) (or f (t)) and F (s). This ‘capital letter notation’ is very common in engineering but often confuses people when ‘duality’ is invoked, to be explained below.

And then there’s this:

Since taking the Fourier transform is an operation that is applied to a function to produce a new function, it’s also sometimes convenient to indicate this by a kind of “operational” notation.

For example, it’s common to write F f (s) for ˆf (s), and so, to repeat the full definition F f (s) =

Z

−∞

e−2πistf (t) dt .

This is often the most unambiguous notation. Similarly, the operation of taking the inverse Fourier transform is then denoted by F−1, and so

F−1g(t) = Z

−∞

e2πistg(s) ds .

We will use the notation F f more often than not. It, too, is far from ideal, the problem being with keeping variables straight — you’ll see.

(12)

2.2 Getting to Know Your Fourier Transform 75

Finally, a function and its Fourier transform are said to constitute a “Fourier pair”, ; this is concept of

‘duality’ to be explained more precisely later. There have been various notations devised to indicate this sibling relationship. One is

f (t) F (s) Bracewell advocated the use of

F (s) ⊃ f (t) and Gray and Goodman also use it. I hate it, personally.

A warning on definitions Our definition of the Fourier transform is a standard one, but it’s not the only one. The question is where to put the 2π: in the exponential, as we have done; or perhaps as a factor out front; or perhaps left out completely. There’s also a question of which is the Fourier transform and which is the inverse, i.e., which gets the minus sign in the exponential. All of the various conventions are in day-to-day use in the professions, and I only mention this now because when you’re talking with a friend over drinks about the Fourier transform, be sure you both know which conventions are being followed. I’d hate to see that kind of misunderstanding get in the way of a beautiful friendship.

Following the helpful summary provided by T. W. K¨orner in his book Fourier Analysis, I will summarize the many irritating variations. To be general, let’s write

F f (s) = 1 A

Z

−∞

eiBstf (t) dt .

The choices that are found in practice are A =

B = ±1

A = 1 B = ±2π

A = 1 B = ±1

The definition we’ve chosen has A = 1 and B = −2π.

Happy hunting and good luck.

2.2 Getting to Know Your Fourier Transform

In one way, at least, our study of the Fourier transform will run the same course as your study of calculus.

When you learned calculus it was necessary to learn the derivative and integral formulas for specific functions and types of functions (powers, exponentials, trig functions), and also to learn the general principles and rules of differentiation and integration that allow you to work with combinations of functions (product rule, chain rule, inverse functions). It will be the same thing for us now. We’ll need to have a storehouse of specific functions and their transforms that we can call on, and we’ll need to develop general principles and results on how the Fourier transform operates.

2.2.1 Examples

We’ve already seen the example

Π = sincb orF Π(s) = sinc s using the F notation. Let’s do a few more examples.

(13)

The triangle function Consider next the “triangle function”, defined by Λ(x) =

(1 − |x| |x| ≤ 1

0 otherwise

0 1

1

−1 −1/2 1/2

For the Fourier transform we compute (using integration by parts, and the factoring trick for the sine function):

F Λ(s) = Z

−∞

Λ(x)e−2πisxdx = Z 0

−1

(1 + x)e−2πisxdx + Z 1

0

(1 − x)e−2πisxdx

=

1 + 2iπs

2s2e2πis 2s2



2iπs − 1

2s2 +e−2πis 2s2



= −e−2πis(e2πis− 1)2

2s2 = −e−2πis(eπis(eπis− e−πis))2 2s2

= −e−2πise2πis(2i)2sin2πs

2s2 =

sin πs πs

2

= sinc2s.

It’s no accident that the Fourier transform of the triangle function turns out to be the square of the Fourier transform of the rect function. It has to do with convolution, an operation we have seen for Fourier series and will see anew for Fourier transforms in the next chapter.

The graph of sinc2s looks like:

(14)

2.2 Getting to Know Your Fourier Transform 77

−3 −2 −1 0 1 2 3

0 0.2 0.4 0.6 0.8 1

s b Λ(s)

The exponential decay Another commonly occurring function is the (one-sided) exponential decay, defined by

f (t) =

(0 t ≤ 0 e−at t > 0

where a is a positive constant. This function models a signal that is zero, switched on, and then decays exponentially. Here are graphs for a = 2, 1.5, 1.0, 0.5, 0.25.

−2 −1 0 1 2 3 4 5 6

0 0.2 0.4 0.6 0.8 1

t

f(t)

(15)

Which is which? If you can’t say, see the discussion on scaling the independent variable at the end of this section.

Back to the exponential decay, we can calculate its Fourier transform directly.

F f (s) = Z

0

e−2πiste−atdt = Z

0

e−2πist−atdt

= Z

0

e(−2πis−a)tdt =

e(−2πis−a)t

−2πis − a

t=∞

t=0

= e(−2πis)t

−2πis − ae−at

t=∞e(−2πis−a)t

−2πis − a

t=0= 1 2πis + a

In this case, unlike the results for the rect function and the triangle function, the Fourier transform is complex. The fact that F Π(s) and F Λ(s) are real is because Π(x) and Λ(x) are even functions; we’ll go over this shortly. There is no such symmetry for the exponential decay.

The power spectrum of the exponential decay is

|F f (s)|2= 1

|2πis + a|2 = 1 a2+ 4π2s2 .

Here are graphs of this function for the same values of a as in the graphs of the exponential decay function.

−0.6 −0.4 −0.2 0 0.2 0.4 0.6

0 2 4 6 8 10 12 14 16

s

|ˆ f(s)|2

Which is which? You’ll soon learn to spot that immediately, relative to the pictures in the time domain, and it’s an important issue. Also note that |F f (s)|2 is an even function of s even though F f (s) is not.

We’ll see why later. The shape of |F f (s)|2is that of a “bell curve”, though this is not Gaussian, a function we’ll discuss just below. The curve is known as a Lorenz profile and comes up in analyzing the transition probabilities and lifetime of the excited state in atoms.

How does the graph of f (ax) compare with the graph of f (x)? Let me remind you of some elementary lore on scaling the independent variable in a function and how scaling affects its graph. The

(16)

2.2 Getting to Know Your Fourier Transform 79

question is how the graph of f (ax) compares with the graph of f (x) when 0 < a < 1 and when a > 1; I’m talking about any generic function f (x) here. This is very simple, especially compared to what we’ve done and what we’re going to do, but you’ll want it at your fingertips and everyone has to think about it for a few seconds. Here’s how to spend those few seconds.

Consider, for example, the graph of f (2x). The graph of f (2x), compared with the graph of f (x), is squeezed. Why? Think about what happens when you plot the graph of f (2x) over, say, −1 ≤ x ≤ 1.

When x goes from −1 to 1, 2x goes from −2 to 2, so while you’re plotting f (2x) over the interval from −1 to 1 you have to compute the values of f (x) from −2 to 2. That’s more of the function in less space, as it were, so the graph of f (2x) is a squeezed version of the graph of f (x). Clear?

Similar reasoning shows that the graph of f (x/2) is stretched. If x goes from −1 to 1 then x/2 goes from

−1/2 to 1/2, so while you’re plotting f (x/2) over the interval −1 to 1 you have to compute the values of f (x) from −1/2 to 1/2. That’s less of the function in more space, so the graph of f (x/2) is a stretched version of the graph of f (x).

2.2.2 For Whom the Bell Curve Tolls

Let’s next consider the Gaussian function and its Fourier transform. We’ll need this for many examples and problems. This function, the famous “bell shaped curve”, was used by Gauss for various statistical problems. It has some striking properties with respect to the Fourier transform which, on the one hand, give it a special role within Fourier analysis, and on the other hand allow Fourier methods to be applied to other areas where the function comes up. We’ll see an application to probability and statistics in Chapter 3.

The “basic Gaussian” is f (x) = e−x2. The shape of the graph is familiar to you.

−3 −2 −1 0 1 2 3

0 0.2 0.4 0.6 0.8 1

x

f(x)

For various applications one throws in extra factors to modify particular properties of the function. We’ll

(17)

do this too, and there’s not a complete agreement on what’s best. There is an agreement that before anything else happens, one has to know the amazing equation3

Z

−∞

e−x2dx =π.

Now, the function f (x) = e−x2 does not have an elementary antiderivative, so this integral cannot be found directly by an appeal to the Fundamental Theorem of Calculus. The fact that it can be evaluated exactly is one of the most famous tricks in mathematics. It’s due to Euler, and you shouldn’t go through life not having seen it. And even if you have seen it, it’s worth seeing again; see the discussion following this section.

The Fourier transform of a Gaussian In whatever subject it’s applied, it seems always to be useful to normalize the Gaussian so that the total area is 1. This can be done in several ways, but for Fourier analysis the best choice, as we shall see, is

f (x) = e−πx2.

You can check using the result for the integral of e−x2 that Z

−∞

e−πx2dx = 1 .

Let’s compute the Fourier transform

F f (s) = Z

−∞

e−πx2e−2πisxdx .

Differentiate with respect to s:

d

dsF f (s) = Z

−∞

e−πx2(−2πix)e−2πisxdx .

This is set up perfectly for an integration by parts, where dv = −2πixe−πx2dx and u = e−2πisx. Then v = ie−πx2, and evaluating the product uv at the limits ±∞ gives 0. Thus

d

dsF f (s) = − Z

−∞

ie−πx2(−2πis)e−2πisxdx

= −2πs Z

−∞

e−πx2e−2πisxdx

= −2πsF f (s) So F f (s) satisfies the simple differential equation

d

dsF f (s) = −2πsF f (s) whose unique solution, incorporating the initial condition, is

F f (s) = F f (0)e−πs2.

3Speaking of this equation, William Thomson, after he became Lord Kelvin, said: “A mathematician is one to whom that is as obvious as that twice two makes four is to you.” What a ridiculous statement.

(18)

2.2 Getting to Know Your Fourier Transform 81

But

F f (0) = Z

−∞

e−πx2dx = 1 .

Hence

F f (s) = e−πs2.

We have found the remarkable fact that the Gaussian f (x) = e−πx2 is its own Fourier transform!

Evaluation of the Gaussian Integral We want to evaluate I =

Z

−∞

e−x2dx .

It doesn’t matter what we call the variable of integration, so we can also write the integral as I =

Z

−∞

e−y2dy .

Therefore

I2 =

Z

−∞

e−x2dx

 Z

−∞

e−y2 dy

 .

Because the variables aren’t “coupled” here we can combine this into a double integral4 Z

−∞

Z

−∞

e−x2dx



e−y2dy = Z

−∞

Z

−∞

e−(x2+y2)dx dy .

Now we make a change of variables, introducing polar coordinates, (r, θ). First, what about the limits of integration? To let both x and y range from −∞ to ∞ is to describe the entire plane, and to describe the entire plane in polar coordinates is to let r go from 0 to ∞ and θ go from 0 to 2π. Next, e−(x2+y2) becomes e−r2 and the area element dx dy becomes r dr dθ. It’s the extra factor of r in the area element that makes all the difference. With the change to polar coordinates we have

I2= Z

−∞

Z

−∞

e−(x2+y2)dx dy = Z

0

Z 0

e−r2r dr dθ

Because of the factor r, the inner integral can be done directly:

Z 0

e−r2r dr = −12e−r2i 0

= 12. The double integral then reduces to

I2 = Z

0 1

2dθ = π ,

whence Z

−∞

e−x2dx = I =π .

Wonderful.

4We will see the same sort of thing when we work with the product of two Fourier transforms on our way to defining convolution in the next chapter.

(19)

2.2.3 General Properties and Formulas

We’ve started to build a storehouse of specific transforms. Let’s now proceed along the other path awhile and develop some general properties. For this discussion — and indeed for much of our work over the next few lectures — we are going to abandon all worries about transforms existing, integrals converging, and whatever other worries you might be carrying. Relax and enjoy the ride.

2.2.4 Fourier transform pairs and duality

One striking feature of the Fourier transform and the inverse Fourier transform is the symmetry between the two formulas, something you don’t see for Fourier series. For Fourier series the coefficients are given by an integral (a transform of f (t) into ˆf (n)), but the “inverse transform” is the series itself. The Fourier transforms F and F−1 are the same except for the minus sign in the exponential.5 In words, we can say that if you replace s by −s in the formula for the Fourier transform then you’re taking the inverse Fourier transform. Likewise, if you replace t by −t in the formula for the inverse Fourier transform then you’re taking the Fourier transform. That is

F f (−s) = Z

−∞

e−2πi(−s)tf (t) dt = Z

−∞

e2πistf (t) dt = F−1f (s)

F−1f (−t) = Z

−∞

e2πis(−t)f (s) ds = Z

−∞

e−2πistf (s) ds = F f (t)

This might be a little confusing because you generally want to think of the two variables, s and t, as somehow associated with separate and different domains, one domain for the forward transform and one for the inverse transform, one for time and one for frequency, while in each of these formulas one variable is used in both domains. You have to get over this kind of confusion, because it’s going to come up again.

Think purely in terms of the math: The transform is an operation on a function that produces a new function. To write down the formula I have to evaluate the transform at a variable, but it’s only a variable and it doesn’t matter what I call it as long as I keep its role in the formula straight.

Also be observant what the notation in the formula says and, just as important, what it doesn’t say. The first formula, for example, says what happens when you first take the Fourier transform of f and then evaluate it at −s, it’s not a formula for F (f (−s)) as in “first change s to −s in the formula for f and then take the transform”. I could have written the first displayed equation as (F f )(−s) = F−1f (s), with an extra parentheses around the F f to emphasize this, but I thought that looked too clumsy. Just be careful, please.

The equations

F f (−s) = F−1f (s) F−1f (−t) = F f (t)

5Here’s the reason that the formulas for the Fourier transform and its inverse appear so symmetric; it’s quite a deep mathematical fact. As the general theory goes, if the original function is defined on a group then the transform (also defined in generality) is defined on the “dual group”, which I won’t define for you here. In the case of Fourier series the function is periodic, and so its natural domain is the circle (think of the circle as [0, 1] with the endpoints identified). It turns out that the dual of the circle group is the integers, and that’s why ˆf is evaluated at integers n. It also turns out that when the group is R the dual group is again R. Thus the Fourier transform of a function defined on R is itself defined on R. Working through the general definitions of the Fourier transform and its inverse in this case produces the symmetric result that we have before us. Kick that one around over dinner some night.

(20)

2.2 Getting to Know Your Fourier Transform 83

are sometimes referred to as the “duality” property of the transforms. One also says that “the Fourier transform pair f and F f are related by duality”, meaning exactly these relations. They look like different statements but you can get from one to the other. We’ll set this up a little differently in the next section.

Here’s an example of how duality is used. We know that F Π = sinc and hence that

F−1sinc = Π . By “duality” we can find F sinc:

F sinc(t) = F−1sinc(−t) = Π(−t) .

(Troubled by the variables? Remember, the left hand side is (F sinc)(t).) Now with the additional knowl- edge that Π is an even function — Π(−t) = Π(t) — we can conclude that

F sinc = Π .

Let’s apply the same argument to find F sinc2. Recall that Λ is the triangle function. We know that F Λ = sinc2

and so

F−1sinc2 = Λ . But then

F sinc2(t) = (F−1sinc2)(−t) = Λ(−t) and since Λ is even,

F sinc2 = Λ .

Duality and reversed signals There’s a slightly different take on duality that I prefer because it suppresses the variables and so I find it easier to remember. Starting with a signal f (t) define the reversed signal f by

f(t) = f (−t) . Note that a double reversal gives back the original signal,

(f)= f .

Note also that the conditions defining when a function is even or odd are easy to write in terms of the reversed signals:

f is even if f= f f is odd if f= −f

In words, a signal is even if reversing the signal doesn’t change it, and a signal is odd if reversing the signal changes the sign. We’ll pick up on this in the next section.

Simple enough — to reverse the signal is just to reverse the time. This is a general operation, of course, whatever the nature of the signal and whether or not the variable is time. Using this notation we can rewrite the first duality equation, F f (−s) = F−1f (s), as

(F f )= F−1f

(21)

and we can rewrite the second duality equation, F−1f (−t) = F f (t), as (F−1f )= F f .

This makes it very clear that the two equations are saying the same thing. One is just the “reverse” of the other.

Furthermore, using this notation the result F sinc = Π, for example, goes a little more quickly:

F sinc = (F−1sinc) = Π= Π . Likewise

F sinc2 = (F−1sinc2)= Λ= Λ .

A natural variation on the preceding duality results is to ask what happens with F f, the Fourier transform of the reversed signal. Let’s work this out. By definition,

F f(s) = Z

−∞

e−2πistf(t) dt = Z

−∞

e−2πistf (−t) dt .

There’s only one thing to do at this point, and we’ll be doing it a lot: make a change of variable in the integral. Let u = −t so that du = −dt , or dt = −du. Then as t goes from −∞ to ∞ the variable u = −t goes from ∞ to −∞ and we have

Z

−∞

e−2πistf (−t) dt = Z −∞

e−2πis(−u)f (u) (−du)

= Z

−∞

e2πisuf (u) du (the minus sign on the du flips the limits back)

= F−1f (s) Thus, quite neatly,

F f= F−1f

Even more neatly, if we now substitute F−1f = (F f )from earlier we have F f= (F f ).

Note carefully where the parentheses are here. In words, the Fourier transform of the reversed signal is the reversed Fourier transform of the signal. That one I can remember.

To finish off these questions, we have to know what happens to F−1f. But we don’t have to do a separate calculation here. Using our earlier duality result,

F−1f= (F f)= (F−1f ).

In words, the inverse Fourier transform of the reversed signal is the reversed inverse Fourier transform of the signal. We can also take this one step farther and get back to F−1f= F f

And so, the whole list of duality relations really boils down to just two:

F f = (F−1f ) F f= F−1f

(22)

2.2 Getting to Know Your Fourier Transform 85

Learn these. Derive all others.

Here’s one more:

F (F f )(s) = f (−s) or F (F f ) = f without the variable.

This identity is somewhat interesting in itself, as a variant of Fourier inversion. You can check it directly from the integral definitions, or from our earlier duality results.6 Of course then also

F (F f) = f .

2.2.5 Even and odd symmetries and the Fourier transform

We’ve already had a number of occasions to use even and odd symmetries of functions. In the case of real-valued functions the conditions have obvious interpretations in terms of the symmetries of the graphs;

the graph of an even function is symmetric about the y-axis and the graph of an odd function is symmetric through the origin. The (algebraic) definitions of even and odd apply to complex-valued as well as to real- valued functions, however, though the geometric picture is lacking when the function is complex-valued because we can’t draw the graph. A function can be even, odd, or neither, but it can’t be both unless it’s identically zero.

How are symmetries of a function reflected in properties of its Fourier transform? I won’t give a complete accounting, but here are a few important cases.

• If f (x) is even or odd, respectively, then so is its Fourier transform.

Working with reversed signals, we have to show that (F f )= F f if f is even and (F f ) = −F f if f is odd. It’s lighting fast using the equations that we derived, above:

(F f )= F f=

(F f, if f is even F (−f ) = −F f if f is odd

Because the Fourier transform of a function is complex valued there are other symmetries we can consider for F f (s), namely what happens under complex conjugation.

• If f (t) is real-valued then (F f )= F f and F (f) = F f .

This is analogous to the conjugate symmetry property possessed by the Fourier coefficients for a real-valued periodic function. The derivation is essentially the same as it was for Fourier coefficients, but it may be helpful to repeat it for practice and to see the similarities.

(F f )(s) = F−1f (s) (by duality)

= Z

−∞

e2πistf (t) dt

=

Z

−∞

e−2πistf (t) dt



(f (t) = f (t) since f (t) is real)

= F f (s)

6And you can then also then check that F (F (F (F f )))(s) = f (s), i.e., F4is the identity transformation. Some people attach mystical significance to this fact.

(23)

We can refine this if the function f (t) itself has symmetry. For example, combining the last two results and remembering that a complex number is real if it’s equal to its conjugate and is purely imaginary if it’s equal to minus its conjugate, we have:

• If f is real valued and even then its Fourier transform is even and real valued.

• If f is real valued and odd function then its Fourier transform is odd and purely imaginary.

We saw this first point in action for Fourier transform of the rect function Π(t) and for the triangle function Λ(t). Both functions are even and their Fourier transforms, sinc and sinc2, respectively, are even and real. Good thing it worked out that way.

2.2.6 Linearity

One of the simplest and most frequently invoked properties of the Fourier transform is that it is linear (operating on functions). This means:

F (f + g)(s) = F f (s) + F g(s)

F (αf )(s) = αF f (s) for any number α (real or complex).

The linearity properties are easy to check from the corresponding properties for integrals, for example:

F (f + g)(s) = Z

−∞

(f (x) + g(x))e−2πisxdx

= Z

−∞

f (x)e−2πisxdx + Z

−∞

g(x)e−2πisxdx = F f (s) + F g(s) .

We used (without comment) the property on multiples when we wrote F (−f ) = −F f in talking about odd functions and their transforms. I bet it didn’t bother you that we hadn’t yet stated the property formally.

2.2.7 The shift theorem

A shift of the variable t (a delay in time) has a simple effect on the Fourier transform. We would expect the magnitude of the Fourier transform |F f (s)| to stay the same, since shifting the original signal in time should not change the energy at any point in the spectrum. Hence the only change should be a phase shift in F f (s), and that’s exactly what happens.

To compute the Fourier transform of f (t + b) for any constant b, we have Z

−∞

f (t + b)e−2πistdt = Z

−∞

f (u)e−2πis(u−b)du

(substituting u = t + b; the limits still go from −∞ to ∞)

= Z

−∞

f (u)e−2πisue2πisbdu

= e2πisb Z

−∞

f (u)e−2πisudu = e2πisbf (s).ˆ

(24)

2.2 Getting to Know Your Fourier Transform 87

The best notation to capture this property is probably the pair notation, f F .7 Thus:

2πisbF (s).

◦ A little more generally, f (t ± b) e±2πisbF (s).

Notice that, as promised, the magnitude of the Fourier transform has not changed under a time shift because the factor out front has magnitude 1:

e±2πisbF (s) =

e±2πisb

|F (s)| = |F (s)| .

2.2.8 The stretch (similarity) theorem

How does the Fourier transform change if we stretch or shrink the variable in the time domain? More precisely, we want to know if we scale t to at what happens to the Fourier transform of f (at). First suppose a > 0. Then

Z

−∞

f (at)e−2πistdt = Z

−∞

f (u)e−2πis(u/a)1 adu

(substituting u = at; the limits go the same way because a > 0)

= 1 a

Z

−∞

f (u)e−2πi(s/a)udu = 1 aF f s

a



If a < 0 the limits of integration are reversed when we make the substitution u = ax, and so the resulting transform is (−1/a)F f (s/a). Since −a is positive when a is negative, we can combine the two cases and present the Stretch Theorem in its full glory:

1

|a|F s a

 .

This is also sometimes called the Similarity Theorem because changing the variable from x to ax is a change of scale, also known as a similarity.

There’s an important observation that goes with the stretch theorem. Let’s take a to be positive, just to be definite. If a is large (bigger than 1, at least) then the graph of f (at) is squeezed horizontally compared to f (t). Something different is happening in the frequency domain, in fact in two ways. The Fourier transform is (1/a)F (s/a). If a is large then F (s/a) is stretched out compared to F (s), rather than squeezed in. Furthermore, multiplying by 1/a, since the transform is (1/a)F (a/s), also squashes down the values of the transform.

The opposite happens if a is small (less than 1). In that case the graph of f (at) is stretched out horizon- tally compared to f (t), while the Fourier transform is compressed horizontally and stretched vertically.

The phrase that’s often used to describe this phenomenon is that a signal cannot be localized (meaning

7This is, however, an excellent opportunity to complain about notational matters. Writing F f (t+b) invites the same anxieties that some of us had when changing signs. What’s being transformed? What’s being plugged in? There’s no room to write an s. The hat notation is even worse — there’s no place for the s, again, and do you really want to writef (t + b) with such[ a wide hat?

(25)

concentrated at a point) in both the time domain and the frequency domain. We will see more precise formulations of this principle.8

To sum up, a function stretched out in the time domain is squeezed in the frequency domain, and vice versa. This is somewhat analogous to what happens to the spectrum of a periodic function for long or short periods. Say the period is T , and recall that the points in the spectrum are spaced 1/T apart, a fact we’ve used several times. If T is large then it’s fair to think of the function as spread out in the time domain — it goes a long time before repeating. But then since 1/T is small, the spectrum is squeezed. On the other hand, if T is small then the function is squeezed in the time domain — it goes only a short time before repeating — while the spectrum is spread out, since 1/T is large.

Careful here In the discussion just above I tried not to talk in terms of properties of the graph of the transform — though you may have reflexively thought in those terms and I slipped into it a little — because the transform is generally complex valued. You do see this squeezing and spreading phenomenon geometrically by looking at the graphs of f (t) in the time domain and the magnitude of the Fourier transform in the frequency domain.9

Example: The stretched rect Hardly a felicitous phrase, “stretched rect”, but the function comes up often in applications. Let p > 0 and define

Πp(t) =

(1 |t| < p/2 0 |t| ≥ p/2

Thus Πp is a rect function of width p. We can find its Fourier transform by direct integration, but we can also find it by means of the stretch theorem if we observe that

Πp(t) = Π(t/p) . To see this, write down the definition of Π and follow through:

Π(t/p) =

(1 |t/p| < 1/2 0 |t/p| ≥ 1/2 =

(1 |t| < p/2

0 |t| ≥ p/2 = Πp(t) . Now since Π(t) sinc s, by the stretch theorem

Π(t/p) p sinc ps , and so

F Πp(s) = p sinc ps . This is useful to know.

Here are plots of the Fourier transform pairs for p = 1/5 and p = 5, respectively. Note the scales on the axes.

8In fact, the famous Heisenberg Uncertainty Principle in quantum mechanics is an example.

9We observed this for the one-sided exponential decay and its Fourier transform, and you should now go back to that example and match up the graphs of |F f | with the various values of the parameter.

(26)

2.2 Getting to Know Your Fourier Transform 89

0 0.2 0.4 0.6 0.8 1

1

−0.2

−0.4

−0.6

−0.8

−1

Π1/5(t)

−20 −15 −10 −5 0 5 10 15 20

−0.1

−0.05 0 0.05 0.1 0.15 0.2 0.25

s

b Π1

/5(s)

0 1

1

2 3

−1

−2

−3

Π5(t)

(27)

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2

−1 0 1 2 3 4 5 6

s

b Π5

(s)

2.2.9 Combining shifts and stretches

We can combine the shift theorem and the stretch theorem to find the Fourier transform of f (ax + b), but it’s a little involved.

Let’s do an example first. It’s easy to find the Fourier transform of f (x) = Π((x−3)/2) by direct integration.

F (s) = Z 4

2

e−2πisxdx

= − 1

2πise−2πisx ix=4

x=2 = − 1

2πis(e−8πis− e−4πis) . We can still bring the sinc function into this, but the factoring is a little trickier.

e−8πis− e−4πis= e−6πis(e−2πis− e2πis) = e−6πis(−2i) sin 2πs . Plugging this into the above gives

F (s) = e−6πissin 2πs

πs = 2e−6πissinc 2s .

The Fourier transform has become complex — shifting the rect function has destroyed its symmetry.

Here’s a plot of Π((x − 3)/2) and of 4sinc22s, the square of the magnitude of its Fourier transform. Once again, looking at the latter gives you no information about the phases in the spectrum, only on the energies.

0 1

1

3 5

−3 −1

−5

Figure

Updating...

References

Related subjects :