Fourier Series
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Fourier Series When the French mathematician Joseph Fourier (1768–1830) was trying to solve a problem in heat heat conduction, he needed to express a function f as an infinite infinite series of sine and cosine functions:
f x
1
a0
a n cos nx
bn sin nx
n1
a0
a1 cos x
a2 cos 2 x
b1 sin x
b2 sin 2 x
a3 cos 3 x
b3 sin 3 x
Earlier, Daniel Bernoulli and Leonard Euler had used such series while investigating investigating problems concerning vibrating strings and astronomy. The series in Equation 1 is called a trigonometric series or Fourier series and it turns out that expressing a function as a Fourier series is sometimes more advantageous than expanding it as a power series. In particular, astronomical astronomical phenomena are usually periodic, as are heartbeats, tides, and vibrating strings, so it makes sense to express them in terms of periodic functions. We start by assuming that the trigonometric series converges and has a continuous function f x as its sum on the interval , , that is,
f x
2
a0
a n cos nx
bn sin nx
x
n1
Our aim is to find formulas for the coefficients a n and bn in terms of f . Recall that for a power series f x cn x a n we found a formula for the coefficients in terms of derivn atives: cn f an!. Here we use integrals. If we integrate both sides of Equation 2 and assume that it’s permissible permissible to integrate the series term-by-term, we get
y
f x dx
y
a 0 dx
y
2 a 0
an
n1
y
a n cos nx
bn sin nx dx
n1
cos nx dx
bn
n1
y
sin nx dx
But
y
cos nx dx
1 n
because n is an integer. Similarly,
sin nx
x
y
1
n
sin nx dx
f x dx
sin n sinn 0
0. So
2 a0
1
2
❙ ❙ ❙ ❙ FOURIER SERIES
and solving for a0 gives |||| Notice that a 0 is the average value of f over the interval , .
a0
3
1
2
y
f x dx
To determine an for n 1 we multiply both sides of Equation 2 by cos mx (where m is an integer and m 1) and integrate term-by-term from to :
y
4
y
f x cos mx dx
y
cos mx dx
a0
a0
an
n1
a n cos nx
bn sin nx
n1
y
bn
cos nx cos mx dx
n1
y
cos mx dx
sin nx cos mx dx
We’ve seen that the first integral is 0. With the help of Formulas 81, 80, and 64 in the Table Table of Integrals, it’ it ’s not hard to show that
y
y
sin nx cos mx dx
0
cos nx cos mx dx
for all n and m 0
for n
m
for n
m
So the only nonzero term in (4) is am and we get
y
f x cos mx dx
am
Solving for am , and then replacing m by n, we have
5
an
1
y
f x cos nx dx
n
1, 2, 3, . . .
Similarly, Similarly, if we multiply both sides of Equation 2 by sin mx and integrate from
to
,
we get
6
bn
1
y
f x sin nx dx
n
1, 2, 3, . . .
We have have derived derived Formulas Formulas 3, 5, and 6 assuming f is a continuous function such that Equation 2 holds and for which the term-by-term integration integration is legitimate. But we can still consider the Fourier series of a wider class of functions: A piecewise continuous function on a, b is continuous except perhaps for a finite number of removable or jump discontinuities. (In other words, the function has no infi in finite discontinuities. See Section 2.5 for a discussion of the different types of discontinuities.)
FOURIER SERIES
❙❙❙❙
3
7 Definition Let f be a piecewise continuous function on , . Then the Fourier series of f is the series
a0
a n cos nx
bn sin nx
n1
where the coef ficients an and bn in this series are defi de fined by a0
an
1
y
1
2
y
f x dx
f x cos nx dx
bn
1
y
f x sin nx dx
and are called the Fourier coefficients of f . Notice in Defi Definition 7 that we are not saying f x is equal to its Fourier series. Later we will discuss conditions under which that is actually true. For now we are just saying that associated with any piecewise continuous function f on , is a certain series called a Fourier series.
EXAMPLE 1 Find the Fourier coef ficients and Fourier series of the square-wave square-wave function f defi defined by f x
0
if x
1
if 0
x
0
f x
and
2
f x
So f is periodic with period 2 and its graph is shown in Figure 1. y
Engineers use the square-wave function in describing forces acting on a mechanical system and electromotive forces in an electric circuit (when a switch is turned on and off repeatedly). Strictly speaking, the graph of f is as shown in Figure 1(a), but it’s often represented as in Figure 1(b), where you can see why it’s called a square wave. ||||
1
0
_π
π
2π
x
π
2π
x
(a)
y 1
0
_π
FIGURE 1 Square-wave function
(b) SOLUTION
Using the formulas for the Fourier coef ficients in Defi Definition 7, we have a0
1 2
y
f x dx
1 2
y
0
0 dx
1 2
y
0
1 dx
0
1 2
1 2
4
❙ ❙ ❙ ❙ FOURIER SERIES
and, for n
1, an
1
y
f x cos nx dx
1
y
0
0 dx
1
y
0
cos nx dx
1 sin nx
0
n
bn
1
y
1
f x sin nx dx
sin n sin 0 0
n
0
1
y
0
0 dx
1
y
0
sin x dx
1 cos nx
n
Note that cos n equals 1 if n is even and 1 if n is odd. ||||
0
1 n
cos n cos 0
if n is even
0 2
if n is odd
n
The Fourier series of f is therefore a0
a1 cos x
1 2
0
a2 cos 2 x
b1 sin x
0 2
0
b2 sin 2 x
1 2
2
sin x
sin x
0 sin 2 x
2 3
sin 3 x
Since odd integers can be written as n Fourier series in sigma notation as 1 2
b3 sin 3 x
a3 cos 3 x
2 3 2 5
sin 3 x
sin 5 x
0 sin 4 x 2 7
sin 7 x
2 5
sin 5 x
2 k 1, where k is an integer, integer, we can write the
2
sin2 k 1 x k 1 2 k 1
In Example 1 we found the Fourier series of the square-wave function, but we don ’t know yet whether this function is equal to its Fourier series. Let ’s investigate this question graphically. Figure 2 shows the graphs of some of the partial sums Sn x
1 2
2
sin x
2 3
sin 3 x
2 n
sin nx
when n is odd, odd, together with the the graph of the square-wave function.
FOURIER SERIES
y
y
y
1
1
1 S£
S¡ _π
π
x
_π
_π
x
_π
y
y
y
1
1
1 S¡¡
π
x
_π
5
S∞
π
S¶
❙❙❙❙
π
x
π
x
S¡∞
x
π
_π
FIGURE 2 Partial sums of the Fourier series for the square-wave function We see that, as n increases, Sn x becomes a better approximation to the square-wave square-wave function. It appears that the graph of Sn x is approaching the graph of f x, except where x 0 or x is an integer multiple of . In other words, it looks as if f is equal to the sum of its Fourier series except at the points where f is discontinuous. discontinuous. The following theorem, which we state without proof, says that this is typical of the Fourier series of piecewise continuous functions. Recall that a piecewise continuous function has only a finite number of jump discontinuities on , . At a number a where f has a jump discontinuity, the one-sided limits exist and we use the notation f a
f a
lim f x
x l a
lim f x
x l a
8 Fourier Convergence Theorem If f is a periodic function with period 2 and f and f are piecewise continuous on , , then the Fourier series (7) is convergent. convergent. The sum of the Fourier series is equal to f x at all numbers x where f is continuous. At the numbers x where f is discontinuous, the sum of the Fourier series is the average of the right and left limits, that is 1 2
f x f x
If we apply the Fourier Convergence Convergence Theorem to the the square-wave square-wave function f in Example 1, we get what we guessed from the graphs. Observe that f 0
lim f x
x l 0
1
and
f 0
lim f x
x l 0
0
and similarly similarly for the other points at which f is discontinuous. discontinuous. The average average of these left and and 1 right limits is 2, so for any any integer integer n the Fourier Fourier Convergence Convergence Theorem says says that 1 2
2
x sin2k 1 x k 1 2k 1
(Of course, this equation is obvious for x
n .)
f x 1 2
if n
n
if x
n
6
❙ ❙ ❙ ❙ FOURIER SERIES
Functions with Period 2L
If a function f has period other than 2 , we can find its Fourier series by making a change of variable. Suppose f x has period 2 L, that is f x 2 L f x for all x. If we let t x L and Lt tt f x f Lt
then, as you can verify, verify, series of t is
t has period 2
L
and x
corresponds to t
.
The Fourier
a0
a n cos nt bn sin nt
n 1
where a0
an
1
y
1
2
y
nt dt t t cos nt dt
t t dt bn
1
If we now use the Substitution Rule with x we have the following
Lt ,
y
nt dt t t sin nt dt
then t x L, dt L dx , and
9 If f is a piecewise continuous function on L, L , its Fourier series is
a0
n x
a n cos
n 1
L
bn sin
n x L
where Notice that when L the same as those in (7). ||||
these formulas are
a0
and, for n an
1 L
y
1
y 2 L
L
L
f x dx
1,
L L
f x cos
n x L
dx
bn
1 L
y
L
L
f x sin
n x L
dx
Of course, the Fourier Convergence Convergence Theorem (8) is also valid for functions with period 2 L.
EXAMPLE 2 Find the Fourier series of the triangular wave function de fined by f x x for 1 x 1 and f x 2 f x for all x. (The graph of f is shown in Figure 3.) For which values of x is f x equal to the sum of its Fourier series? y 1
FIGURE 3 The triangular wave function
_1
0
1
2
x
FOURIER SERIES
SOLUTION
We find the Fourier coef ficients by putting L 1
a0
2
Notice that a 0 is more easily calculated as an area. ||||
y
1
1
x dx
1
2 0
4 x
and for n
]
1
1
4 x
1 2
y
0
1
1 in (9): 1
x dx
2
y x dx 1
0
]
1 2 1 0 2
1, an
y
1
x cosn
x dx
1
y x cosn x dx 1
2
0
because y x cosn x is an even function. Here we integrate by parts with u and d v cosn x dx . Thus, an
Since y
❙❙❙❙
x sinn
2
0
1
x
sinn x
n
2
n
0
2 n
cosn x
n
y
1
0
sinn x dx
1
0
x
2 2
n
2
cos n 1
x is an odd function, we see that
bn
y
1
x sinn
x dx
1
0
We could therefore write the series as
1
2
2cos n 1
But cos n 1 if n is even and cos n
an
cosn x
n2 2
n 1
1
if n is odd, so
if n is even
0
2 n2 2
cos n 1
4 2
n
2
if n is odd
Therefore, the Fourier series is 1 2
4 2
1 2
cos x
4 2
9
cos3 x
4 25 2
cos5 x
4
cos2k 1 x 2 2 n 1 2k 1
The triangular wave function is continuous everywhere and so, according to the Fourier Convergence Convergence Theorem, we have f x
1 2
4
cos2k 1 x 2 2 n 1 2k 1
for all x
7
8
❙ ❙ ❙ ❙ FOURIER SERIES
In particular,
x
1 2
4
cos2k 1 x 2 2 k 1 2k 1
for 1
x
1
Fourier Series and Music
One of the main uses of Fourier series is in solving some of the differential equations that arise in mathematical physics, such as the wave equation and the heat equation. (This is covered in more advanced courses.) Here we explain briefl brie fly how Fourier series play a role in the analysis and synthesis of musical sounds. We hear a sound when our eardrums vibrate because of variations in air pressure. If a guitar string is plucked, or a bow is drawn across a violin string, or a piano string is struck, the string starts to vibrate. These vibrations are ampli fied and transmitted to the air. The resulting air pressure fluctuations arrive at our eardrums and are converted into electrical impulses that are processed by the brain. How is it, then, that we can distinguish between a note of a given pitch produced by two different musical instruments? The graphs in Figure 4 show these fluctuations (deviations from average air pressure) for a flute and a violin playing the same sustained note D (294 vibrations per second) as functions of time. Such graphs are called waveforms and we see that the variations in air pressure are quite different from each other. In particular, the violin waveform is more complex than that of the flute.
t
t
FIGURE 4 Waveforms
(a) Flute
(b) Violin
We gain insight into the differences between waveforms if we express them as sums of Fourier series: Pt
a0
a1 cos
t
L
t
b1 sin
L
a2 cos
2 t L
b2 sin
2 t L
In doing so, we are expressing the sound as a sum of simple pure sounds. The difference in sounds between two instruments can be attributed to the relative sizes of the Fourier coef ficients of the respective waveforms. The n th term of the Fourier series, that is,
a n cos
n t L
bn
n t L
is called the nth harmonic of P. The amplitude of the n th harmonic is A n
and its square, A2n
a 2n
s a 2n b2n
b2n , is sometimes called energy of the the n th harm harmoni onic. c. (Noti (Notice ce that that
FOURIER SERIES
❙❙❙❙
9
for a Fourier series with only sine terms, as in Example 1, the amplitude is A n bn and the energy is A2n b 2n.) The graph of the sequence A2n is called the energy spectrum of P and shows at a glance the relative sizes of the harmonics. Figure 5 shows the energy spectra for the flute and violin waveforms in Figure 4. Notice that, for the flute, A2n tends to diminish rapidly as n increases whereas, for the violin, the higher harmonics are fairly strong. This accounts for the relative simplicity of the flute waveform in Figure 4 and the fact that the flute produces relatively pure sounds when compared with the more complex violin tones. A@ n
A@ n
0
FIGURE 5
2
4
Energy spectra
6
8
0
n
10 10
2
4
(a) Flute
6
8
n
10 10
(b) Violin
In addition to analyzing the sounds of conventional conventional musical instruments, Fourier series enable us to synthesize sounds. The idea behind music synthesizers is that we can combine various pure tones (harmonics) to create a richer sound through emphasizing certain harmonics by assigning larger Fourier coef fi coef ficients (and therefore higher corresponding energies).
||||
A function f is given on the interval , and f is periodic with period 2 . (a) Find the Fourier Fourier coef ficients of f . (b) Find the Fourier series of f . For what values values of x is f x equal to its Fourier series? (c) Graph f and the partial sums S2, S4, and S6 of the Fourier series.
1–6
;
Exercises ||||
1. f x 2. f x
1
if x
1
if 0
x
0
if x
x
if 0
x
5. f x
if x
0
if 1
0
if 2
0
if x
cos x
if 0
1
x
if x if 2
0
if 0 s
s
x 2
x
1
2
0
x
if 4
0
if 0
s
s
s
4
f x
f x
4
f x
f x f x
1
8
2
f x
f x
t 1
s
s
s
s
s
s
s
so-called half-wave recti fier that clips the negative part of the wave. Find the Fourier series of the resulting periodic function
2
0
f t
0
if
E sin t t if 0
s
0
4
1 x 1
0
x
x
f x
s
12. A voltage E sin t t, where t represents time, is passed through a
0
x
x s
1
10. f x 1 x,
if 1 x
0
s
s
1
11. f t sin3 t t ,
s
8. f x 1 if 0 x
9. f x
2
6. f x 1 s
Find the Fourier series of the function.
7. f x
||||
0
3. f x x 4. f x x
7–11
s
s
s
s
s
t 0
t
f t 2
f t
10
❙ ❙ ❙ ❙ FOURIER SERIES
13–16
18. Use the result of Example 2 to show that
Sketch the graph of the sum of the Fourier series of f without actually calculating the Fourier series. ||||
13. f x
14. f x
1
if 4
3
if 0
x
if 1
x
1
x
x
if 0
1
16. f x e x,
2 x
2
s
s
1 3
1
1
1
3
5
7
2
2
2
2
8
19. Use the result of Example 1 to show that 0 1
1
s
s
s
s
1, then
1 3
1 5
1 7
4
20. Use the given graph of f and Simpson ’s Rule with n 8 to
s
17. (a) Show Show that, that, if 1 x x 2
4
x
1 x
s
0
x
15. f x x 3,
s
1
s
s
s
estimate the Fourier coef ficients a 0, a1, a 2, b1, and b2. Then use them to graph the second partial sum of the Fourier series and compare with the graph of f . y
4
1 n 2 2 cosn x n 1 n
1 (b) By substituting substituting a speci speci fic value of x, show that
1
n
n 1
2
2
6
0.25
x
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