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Fourier Series: Theory and Applications in Differential Equations, Slides of Fourier Transform and Series

The Open University (OU)Fourier Transform and Series

This simple procedure can be extended to any function that can be repre- sented as a sum of cosine (and sine) functions, even if that summation is not.

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Chapter 10

Fourier Series

10.1 Periodic Functions and Orthogonality Relations

The differential equation

𝑦′′ +𝛽2𝑦=𝐹cos 𝜔𝑡

models a mass-spring system with natural frequency 𝛽with a pure cosine

forcing function of frequency 𝜔. If 𝛽2∕=𝜔2a particular solution is easily

found by undetermined coefficients (or by using Laplace transforms) to be

𝑦𝑝=𝐹

𝛽2−𝜔2cos 𝜔𝑡.

If the forcing function is a linear combination of simple cosine functions, so

that the differential equation is

𝑦′′ +𝛽2𝑦=

𝑁

∑

𝑛=1

𝐹𝑛cos 𝜔𝑛𝑡

where 𝛽2∕=𝜔2

𝑛for any 𝑛, then, by linearity, a particular solution is obtained

as a sum

𝑦𝑝(𝑡) =

𝑁

∑

𝑛=1

𝐹𝑛

𝛽2−𝜔2

𝑛

cos 𝜔𝑛𝑡.

This simple procedure can be extended to any function that can be repre-

sented as a sum of cosine (and sine) functions, even if that summation is not

a finite sum. It turns out that the functions that can be represented as sums

in this form are very general, and include most of the periodic functions that

are usually encountered in applications.

723

Partial preview of the text

Download Fourier Series: Theory and Applications in Differential Equations and more Slides Fourier Transform and Series in PDF only on Docsity!

Chapter 10 Fourier Series

10.1 Periodic Functions and Orthogonality Relations

The differential equation

𝑦′′^ + 𝛽^2 𝑦 = 𝐹 cos 𝜔𝑡

models a mass-spring system with natural frequency 𝛽 with a pure cosine forcing function of frequency 𝜔. If 𝛽^2 ∕= 𝜔^2 a particular solution is easily found by undetermined coefficients (or by using Laplace transforms) to be

𝛽^2 − 𝜔^2

cos 𝜔𝑡.

If the forcing function is a linear combination of simple cosine functions, so that the differential equation is

𝑦′′^ + 𝛽^2 𝑦 =

X^ 𝑁

𝑛=

𝐹𝑛 cos 𝜔𝑛𝑡

where 𝛽^2 ∕= 𝜔^2 𝑛 for any 𝑛, then, by linearity, a particular solution is obtained as a sum

𝑦𝑝(𝑡) =

X^ 𝑁

𝑛=

𝛽^2 − 𝜔^2 𝑛

cos 𝜔𝑛𝑡.

This simple procedure can be extended to any function that can be repre- sented as a sum of cosine (and sine) functions, even if that summation is not a finite sum. It turns out that the functions that can be represented as sums in this form are very general, and include most of the periodic functions that are usually encountered in applications.

723

724 10 Fourier Series

Periodic Functions

A function 𝑓 is said to be periodic with period 𝑝 > 0 if

𝑓 (𝑡 + 𝑝) = 𝑓 (𝑡)

for all 𝑡 in the domain of 𝑓. This means that the graph of 𝑓 repeats in successive intervals of length 𝑝, as can be seen in the graph in Figure 10.1.

𝑦

𝑝 (^2) 𝑝 3 𝑝 4 𝑝 5 𝑝

Fig. 10.1 An example of a periodic function with period 𝑝. Notice how the graph repeats on each interval of length 𝑝.

The functions sin 𝑡 and cos 𝑡 are periodic with period 2 𝜋, while tan 𝑡 is periodic with period 𝜋 since

tan(𝑡 + 𝜋) = sin(𝑡 + 𝜋) cos(𝑡 + 𝜋)

− sin 𝑡 − cos 𝑡

= tan 𝑡.

The constant function 𝑓 (𝑡) = 𝑐 is periodic with period 𝑝 where 𝑝 is any positive number since 𝑓 (𝑡 + 𝑝) = 𝑐 = 𝑓 (𝑡).

Other examples of periodic functions are the square wave and triangular wave whose graphs are shown in Figure 10.2. Both are periodic with period 2.

− 1 0 1 2 3 4

𝑦

𝑡

(a) Square wave sw(𝑡)

− 2 − 1 0 1 2 3 4 5

𝑦

𝑡

(b) Triangular wave tw(𝑡)

Fig. 10.

726 10 Fourier Series

Applying these observations to the functions sin 𝑡 and cos 𝑡 with funda- mental period 2 𝜋 gives the following facts.

Theorem 1. For any 𝑎 > 0 the functions cos 𝑎𝑡 and sin 𝑎𝑡 are periodic with period 2 𝜋/𝑎. In particular, if 𝐿 > 0 then the functions

cos

𝑡 and sin

are periodic with fundamental period 𝑃 = 2𝐿/𝑛.

Note that since the fundamental period of the functions cos 𝑛𝜋 𝐿 𝑡 and sin 𝑛𝜋 𝐿 𝑡 is 𝑃 = 2𝐿/𝑛, it follows that 2 𝐿 = 𝑛𝑃 is also a period for each of these functions. Thus, a sum

X^ ∞

𝑛=

𝑎𝑛 cos

𝑡 + 𝑏𝑛 sin

will be periodic of period 2 𝐿. Notice that if 𝑛 is a positive integer, then cos 𝑛𝑡 and sin 𝑛𝑡 are periodic with period 2 𝜋/𝑛. Thus, each period of cos 𝑡 or sin 𝑡 contains 𝑛 periods of cos 𝑛𝑡 and sin 𝑛𝑡. This means that the functions cos 𝑛𝑡 and sin 𝑛𝑡 oscillate more rapidly as 𝑛 increases, as can be seen in Figure 10.3 for 𝑛 = 3.

− 1

−𝜋^ 𝜋

(a) The graphs of cos 𝑡 and cos 3𝑡.

− 1

−𝜋^ 𝜋

(b) The graphs of sin 𝑡 and sin 3𝑡.

Fig. 10.

Example 2. Find the fundamental period of each of the following periodic functions.

cos 2𝑡
sin 32 (𝑡 − 𝜋)
1 + cos 𝑡 + cos 2𝑡
sin 2𝜋𝑡 + sin 3𝜋𝑡

▶ Solution. 1. 𝑃 = 2𝜋/2 = 𝜋.

sin 32 (𝑡 − 𝜋) = sin( 32 𝑡 − 32 𝜋) = sin 32 𝑡 cos 32 𝜋 − cos 32 𝑡 sin 32 𝜋 = cos 32 𝑡. Thus, 𝑃 = 2𝜋/(3/2) = 4𝜋/ 3.

10.1 Periodic Functions and Orthogonality Relations 727

The constant function 1 is periodic with any period 𝑝, the fundamental period of cos 𝑡 is 2 𝜋 and all the periods are of the form 2 𝑛𝜋 for a positive integer 𝑛, and the fundamental period of cos 2𝑡 is 𝜋 with 𝑚𝜋 being all the possible periods. Thus, the smallest number that works as a period for all the functions is 2 𝜋 and this is also the smallest period for the sum. Hence 𝑃 = 2𝜋.
The fundamental period of sin 2𝜋𝑡 is 2 𝜋/ 2 𝜋 = 1 so that all of the periods have the form 𝑛 for 𝑛 a positive integer, and the fundamental period of sin 3𝜋𝑡 is 2 𝜋/ 3 𝜋 = 2/ 3 so that all of the periods have the form 2 𝑚/ 3 for 𝑚 a positive integer. Thus, the smallest number that works as a period for both functions is 2 and thus 𝑃 = 2. ◀

Orthogonality Relations for Sine and Cosine

The family of linearly independent functions 1 , cos

𝑡, cos

𝑡,... , sin

𝑡, sin

form what is called a mutually orthogonal set of functions on the interval [−𝐿, 𝐿], analogous to a mutually perpendicular set of vectors. Two functions 𝑓 and 𝑔 defined on an interval 𝑎 ≤ 𝑡 ≤ 𝑏 are said to be orthogonal on the interval [𝑎, 𝑏] if (^) Z 𝑏

𝑎

A family of functions is mutually orthogonal on the interval [𝑎, 𝑏] if any two distinct functions are orthogonal. The mutual orthogonality of the family of cosine and sine functions on the interval [−𝐿, 𝐿] is a consequence of the following identities.

Proposition 3 (Orthogonality Relations). Let 𝑚 and 𝑛 be positive in- tegers, and let 𝐿 > 0. Then

Z (^) 𝐿

−𝐿

cos

Z 𝐿

−𝐿

sin

Z 𝐿

−𝐿

cos

𝑡 sin

Z 𝐿

−𝐿

cos

𝑡 cos

𝐿, if 𝑛 = 𝑚, 0 if 𝑛 ∕= 𝑚.

Z 𝐿

−𝐿

sin

𝑡 sin

𝐿, if 𝑛 = 𝑚, 0 if 𝑛 ∕= 𝑚.

10.1 Periodic Functions and Orthogonality Relations 729

Since 𝑔(−𝑡) = 3(−𝑡) − (−𝑡)^2 sin 3(−𝑡) = − 3 𝑡 + 𝑡^2 sin 3𝑡 = −𝑔(𝑡) for all 𝑡, it follows that 𝑔 is an odd function.
Since ℎ(−𝑡) = (−𝑡)^2 + (−𝑡) + 1 = 𝑡^2 − 𝑡 + 1 = 𝑡^2 + 𝑡 + 1 = ℎ(𝑡) ⇐⇒ −𝑡 = 𝑡 ⇐⇒ 𝑡 = 0, we conclude that ℎ is not even. Similarly, ℎ(−𝑡) = −ℎ(𝑡) ⇐⇒ 𝑡^2 −𝑡+1 = −𝑡^2 −𝑡− 1 ⇐⇒ 𝑡^2 +1 = (−𝑡^2 +1) ⇐⇒ 𝑡^2 +1 = 0, which is not true for any 𝑡. Thus ℎ is not odd, and hence it is neither even or odd. ◀

The graph of an even function is symmetric with respect to the 𝑦-axis, while the graph of an odd function is symmetric with respect to the origin, as illustrated in Figure 10.4.

−𝑡 𝑡

𝑓 (−𝑡) 𝑓 (𝑡)

(a) The graph of an even function.

𝑡

−𝑡

𝑓 (𝑡)

𝑓 (−𝑡) = −𝑓 (𝑡)

(b) The graph of an odd function.

Fig. 10.

Here is a list of basic properties of even and odd functions that are useful in applications to Fourier series. All of them follow easily from the definitions, and the verifications will be left to the exercises.

Proposition 5. Suppose that 𝑓 and 𝑔 are functions defined on the interval −𝐿 ≤ 𝑡 ≤ 𝐿.

If both 𝑓 and 𝑔 are even then 𝑓 + 𝑔 and 𝑓 𝑔 are even.
If both 𝑓 and 𝑔 are odd, then 𝑓 + 𝑔 is odd and 𝑓 𝑔 is even.
If 𝑓 is even and 𝑔 is odd, then 𝑓 𝑔 is odd.
If 𝑓 is even, then (^) Z 𝐿

−𝐿

Z 𝐿

If 𝑓 is odd, then (^) Z 𝐿

−𝐿

Since the integral of 𝑓 computes the signed area under the graph of 𝑓 , the integral equations can be seen from the graphs of even and odd functions in Figure 10.4.

730 10 Fourier Series

Exercises

1–9. Graph each of the following periodic functions. Graph at least 3 periods.

3 if 0 < 𝑡 < 3 − 3 if − 3 < 𝑡 < 0

− 3 if − 2 ≤ 𝑡 < − 1 0 if − 1 ≤ 𝑡 ≤ 1 3 if 1 < 𝑡 < 2

𝑓 (𝑡) = sin 𝑡, 0 < 𝑡 ≤ 𝜋; 𝑓 (𝑡 + 𝜋) = 𝑓 (𝑡).

0 if −𝜋 ≤ 𝑡 < 0 sin 𝑡 if 0 < 𝑡 ≤ 𝜋

−𝑡 if − 1 ≤ 𝑡 < 0 1 if 0 ≤ 𝑡 < 1

8. 𝑓 (𝑡) = 𝑡^2 , − 1 < 𝑡 ≤ 1 ; 𝑓 (𝑡 + 2) = 𝑓 (𝑡).

9. 𝑓 (𝑡) = 𝑡^2 , 0 < 𝑡 ≤ 2 ; 𝑓 (𝑡 + 2) = 𝑓 (𝑡).

10–17. Determine if the given function is periodic. If it is periodic find the fundamental period.

1
sin 2𝑡
1 + cos 3𝜋𝑡
cos 2𝑡 + sin 3𝑡
𝑡 + sin 2𝑡
sin^2 𝑡
cos 𝑡 + cos 𝜋𝑡
sin 𝑡 + sin 2𝑡 + sin 3𝑡

18–26. Determine if the given function is even, odd, or neither.

732 10 Fourier Series

where the coefficients 𝑎 0 , 𝑎 1 ,.. ., 𝑏 1 , 𝑏 2 ,.. ., are to be determined. Since the individual terms in the series (1) are periodic with periods 2 𝐿, 2 𝐿/ 2 , 2 𝐿/ 3 ,

.. ., the function 𝑓 (𝑡) determined by the sum of the series, where it converges, must be periodic with period 2 𝐿. This means that only periodic functions of period 2 𝐿 can be represented by a series of the form (1). Our first problem is to find the coefficients 𝑎𝑛 and 𝑏𝑛 in the series (1). The first term of the series is written 𝑎 0 / 2 , rather than simply as 𝑎 0 , to make the formula to be derived below the same for all 𝑎𝑛, rather than a special case for 𝑎 0. The coefficients 𝑎𝑛 and 𝑏𝑛 can be found from the orthogonality relations of the family of functions cos(𝑛𝜋𝑡/𝐿) and sin(𝑛𝜋𝑡/𝐿) on the interval [−𝐿, 𝐿] given in Proposition 3 of Sect. 10.1. To compute the coefficient 𝑎𝑛 for 𝑛 = 1, 2 , 3 ,.. ., multiply both sides of the series (1) by cos(𝑚𝜋𝑡/𝐿), with 𝑚 a positive integer and then integrate from −𝐿 to 𝐿. For the moment we will assume that the integrals exist and that it is justified to integrate term by term. Then using (1), (2), and (3) from Sect. 10.1, we get

Z (^) 𝐿

−𝐿

𝑓 (𝑡) cos

Z 𝐿

−𝐿

cos

= 0

X^ ∞

𝑛=

h 𝑎𝑛

Z 𝐿

−𝐿

cos

𝑡 cos

⎧ ⎨ ⎩

0 if 𝑛 ∕= 𝑚 𝐿 if 𝑛 = 𝑚

Z 𝐿

−𝐿

sin

𝑡 cos

= 0

i = 𝑎𝑚𝐿.

Thus,

𝑎𝑚 =

Z 𝐿

−𝐿

𝑓 (𝑡) cos

or, replacing the index 𝑚 by 𝑛,

Z 𝐿

−𝐿

𝑓 (𝑡) cos

To compute 𝑎 0 , integrate both sides of (1) from −𝐿 to 𝐿 to get Z (^) 𝐿

−𝐿

Z 𝐿

−𝐿

=2𝐿

X^ ∞

𝑛=

h 𝑎𝑛

Z 𝐿

−𝐿

cos

Z 𝐿

−𝐿

sin

= 0

Thus,

𝑎 0 =

Z 𝐿

−𝐿

10.2 Fourier Series 733

Hence, 𝑎 0 is two times the average value of the function 𝑓 (𝑡) over the interval −𝐿 ≤ 𝑡 ≤ 𝐿. Observe that the value of 𝑎 0 is obtained from (2) by setting 𝑛 = 0. Of course, if the constant 𝑎 0 in (1) were not divided by 2, we would need a separate formula for 𝑎 0. It is for this reason that the constant term in (1) is labeled 𝑎 0 / 2. Thus, for all 𝑛 ≥ 0 , the coefficients 𝑎𝑛 are given by a single formula

Z 𝐿

−𝐿

𝑓 (𝑡) cos

To compute 𝑏𝑛 for 𝑛 = 1, 2 , 3 ,.. ., multiply both sides of the series (1) by sin(𝑚𝜋𝑡/𝐿), with 𝑚 a positive integer and then integrate from −𝐿 to 𝐿. Then using (1), (2), and (4) from Sect. 10.1, we get

Z (^) 𝐿

−𝐿

𝑓 (𝑡) sin

Z 𝐿

−𝐿

sin

= 0

X^ ∞

𝑛=

h 𝑎𝑛

Z 𝐿

−𝐿

cos

𝑡 sin

= 0

Z 𝐿

−𝐿

sin

𝑡 sin

⎧ ⎨ ⎩

0 if 𝑛 ∕= 𝑚 𝐿 if 𝑛 = 𝑚

i = 𝑏𝑚𝐿.

Thus, replacing the index 𝑚 by 𝑛, we find that

Z 𝐿

−𝐿

𝑓 (𝑡) sin

We have arrived at what are known as the Euler Formulas for a function 𝑓 (𝑡) that is the sum of a trigonometric series as in (1):

Z 𝐿

−𝐿

Z 𝐿

−𝐿

𝑓 (𝑡) cos

Z 𝐿

−𝐿

𝑓 (𝑡) sin

The numbers 𝑎𝑛 and 𝑏𝑛 are known as the Fourier coefficients of the func- tion 𝑓. Note that while we started with a periodic function of period 2 𝐿, the formulas for 𝑎𝑛 and 𝑏𝑛 only use the values of 𝑓 (𝑡) on the interval [−𝐿, 𝐿].

10.2 Fourier Series 735

Z 𝐿

−𝐿

𝑓 (𝑡) cos

Z 2 𝐿

𝑓 (𝑡) cos

We now consider some examples of the calculation of Fourier series.

Example 2. Compute the Fourier series of the odd square wave function of period 2 𝐿 and amplitude 1 given by

See Figure 10.5 for the graph of 𝑓 (𝑡).

𝑦

𝑡

− 1

− 3 𝐿 − 2 𝐿 −𝐿^ 𝐿^2 𝐿^3 𝐿

Fig. 10.5 The odd square wave of period 2 𝐿

▶ Solution. Use the Euler formulas for 𝑎𝑛 (Equations (6) and (7)) to con- clude

𝑎𝑛 =

Z 𝐿

−𝐿

𝑓 (𝑡) cos

for all 𝑛 ≥ 0. This is because the function 𝑓 (𝑡) cos 𝑛𝜋 𝐿 𝑡 is the product of an odd and even function, and hence is odd, which implies by Proposition 5 (Part 5) of Section 10.1 that the integral is 0. It remains to compute the coefficients 𝑏𝑛 from (8).

Z 𝐿

−𝐿

𝑓 (𝑡) sin

Z 0

−𝐿

𝑓 (𝑡) sin

Z 𝐿

𝑓 (𝑡) sin

Z 0

−𝐿

(−1) sin

Z 𝐿

(+1) sin

cos

−𝐿

cos

[(1 − cos(−𝑛𝜋)) + (1 − cos(𝑛𝜋))]

(1 − cos 𝑛𝜋) =

Therefore,

736 10 Fourier Series

0 if 𝑛 is even, 4 𝑛𝜋 if^ 𝑛^ is odd,

and the Fourier series is

sin

Example 3. Compute the Fourier series of the even square wave function of period 2 𝐿 and amplitude 1 given by

See Figure 10.6 for the graph of 𝑓 (𝑡).

𝑦

𝑡

− 1

𝐿 2 3 𝐿 2 5 𝐿 − (^2) 𝐿 − (^2) 3 𝐿 − (^2) 5 𝐿 2

Fig. 10.6 The even square wave of period 2 𝐿

▶ Solution. Use the Euler formulas for 𝑏𝑛 (Equation (8)) to conclude

Z 𝐿

−𝐿

𝑓 (𝑡) sin

for all 𝑛 ≥ 1. As in the previous example, this is because the function 𝑓 (𝑡) sin 𝑛𝜋 𝐿 𝑡 is the product of an even and odd function, and hence is odd, which implies by Proposition 5 (Part 5) of Section 10.1 that the integral is

It remains to compute the coefficients 𝑎𝑛 from (6) and (7). For 𝑛 = 0, 𝑎 0 is twice the average of 𝑓 (𝑡) over the period [−𝐿, 𝐿], which is easily seen to be 0 from the graph of 𝑓 (𝑡). For 𝑛 ≥ 1 ,

Z 𝐿

−𝐿

𝑓 (𝑡) cos

Z −𝐿/ 2

−𝐿

𝑓 (𝑡) cos

Z 𝐿/ 2

−𝐿/ 2

𝑓 (𝑡) cos

Z 𝐿

𝐿/ 2

𝑓 (𝑡) cos

738 10 Fourier Series

Z 𝜋

−𝜋

Z 𝜋

𝑡^2

𝜋

For 𝑛 ≥ 1 , using the fact that 𝑓 (𝑡) is even, and taking advantage of the integration by parts formula Z 𝑥 cos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 + cos 𝑥 + 𝐶,

Z 𝜋

−𝜋

𝑓 (𝑡) cos 𝑛𝑡 𝑑𝑡 =

Z 𝜋

𝑓 (𝑡) cos 𝑛𝑡 𝑑𝑡

Z 𝜋

𝑡 cos 𝑛𝑡 𝑑𝑡

let 𝑥 = 𝑛𝑡 so 𝑡 =

and 𝑑𝑡 =

Z 𝑛𝜋

cos 𝑥

𝑛^2 𝜋

[𝑥 sin 𝑥 + cos 𝑥]𝑥𝑥==0𝑛𝜋

𝑛^2 𝜋

[cos 𝑛𝜋 − 1] =

𝑛^2 𝜋

[(−1)𝑛^ − 1]

Therefore,

𝑎𝑛 =

0 if 𝑛 is even, − (^) 𝑛^42 𝜋 if 𝑛 is odd

and the Fourier series is

cos 𝑡 12

cos 3𝑡 32

cos 5𝑡 52

cos 7𝑡 72

X^ ∞

𝑘=

cos(2𝑘 + 1)𝑡 (2𝑘 + 1)^2

Example 5. Compute the Fourier series of the sawtooth wave function of period 2 𝐿 given by

𝑓 (𝑡) = 𝑡 for −𝐿 ≤ 𝑡 < 𝐿; 𝑓 (𝑡 + 2𝐿) = 𝑓 (𝑡).

See Figure 10.8 for the graph of 𝑓 (𝑡).

▶ Solution. As in Example 2, the function 𝑓 (𝑡) is odd, so the cosine terms 𝑎𝑛 are all 0. Now compute the coefficients 𝑏𝑛 from (8). Using the integration by parts formula

10.2 Fourier Series 739

𝑦

𝑡

𝐿

−𝐿

− 3 𝐿 − 2 𝐿 −𝐿 𝐿 2 𝐿 3 𝐿

Fig. 10.8 The sawtooth wave of period 2 𝐿

Z

𝑥 sin 𝑥 𝑑𝑥 = sin 𝑥 − 𝑥 cos 𝑥 + 𝐶,

Z 𝐿

−𝐿

𝑓 (𝑡) sin

Z 𝐿

𝑡 sin

let 𝑥 =

𝑡 so 𝑡 =

𝑥 and 𝑑𝑡 =

Z 𝑛𝜋

𝑥 sin 𝑥

𝑛^2 𝜋^2

Z 𝑛𝜋

𝑥 sin 𝑥 𝑑𝑥

𝑛^2 𝜋^2

[sin 𝑥 − 𝑥 cos 𝑥]𝑥𝑥==0𝑛𝜋

𝑛^2 𝜋^2

(𝑛𝜋 cos 𝑛𝜋) = −

Therefore, the Fourier series is

sin

X^ ∞

𝑛=

sin

All of the examples so far have been of functions that are either even or odd. If a function 𝑓 (𝑡) is even, the resulting Fourier series will only have cosine terms, as in the case of Examples 3 and 4, while if 𝑓 (𝑡) is odd, the resulting Fourier series will only have sine terms, as in Examples 2 and 5. Here are some examples where both sine an cosine terms appear.

Example 6. Compute the Fourier series of the function of period 4 given by

10.2 Fourier Series 741

Z 𝑛𝜋

sin 𝑥

𝑛^2 𝜋^2

Z 𝑛𝜋

𝑥 sin 𝑥 𝑑𝑥

𝑛^2 𝜋^2

[sin 𝑥 − 𝑥 cos 𝑥]𝑥𝑥==0𝑛𝜋

𝑛^2 𝜋^2

(−𝑛𝜋 cos 𝑛𝜋) = −

Thus,

𝑏𝑛 =

for all 𝑛 ≥ 1.

Therefore, the Fourier series is

𝜋^2

X^ ∞

𝑘=

(2𝑘 + 1)^2

cos

X^ ∞

𝑛=

sin

Example 7. Compute the Fourier series of the square pulse wave function of period 2 𝜋 given by

See Figure 10.10 for the graph of 𝑓 (𝑡).

− 4 𝜋 − 2 𝜋 ℎ 2 𝜋 4 𝜋

Fig. 10.10 A square pulse wave of period 2 𝜋.

▶ Solution. For this function, it is more convenient to compute the 𝑎𝑛 and 𝑏𝑛 using integration over the interval [0, 2 𝜋] rather than the interval [−𝜋, 𝜋]. Thus,

Z 2 𝜋

Z ℎ

cos 𝑛𝑡 𝑑𝑡 = sin 𝑛ℎ 𝜋𝑛

, and

Z ℎ

sin 𝑛𝑡 𝑑𝑡 =

1 − cos 𝑛ℎ 𝜋𝑛

742 10 Fourier Series

and the Fourier series is

X^ ∞

𝑛=

sin 𝑛ℎ 𝑛

cos 𝑛𝑡 + 1 − cos 𝑛ℎ 𝑛

sin 𝑛𝑡

If the square pulse wave 𝑓 (𝑡) is divided by ℎ, then one obtains a function whose graph is a series of tall thin rectangles of height 1 /ℎ and base ℎ, so that each of the rectangles with the bases starting at 2 𝑛𝜋 has area 1, as in Figure 10.11. Now consider the limiting case where ℎ approaches 0. The

1 ℎ

− 4 𝜋 − 2 𝜋 ℎ 2 𝜋 4 𝜋

Fig. 10.11 A square unit pulse wave of period 2 𝜋.

graph becomes a series of infinite height spikes of width 0 as in Figure 10.12. This looks like an infinite sum of Dirac delta functions, which is the regular delta function extended to be periodic of period 2 𝜋. That is,

lim ℎ→ 0

X^ ∞

𝑛=−∞

Now compute the Fourier coefficients 𝑎𝑛/ℎ and 𝑏𝑛/ℎ as ℎ approaches 0.

sin 𝑛ℎ 𝜋𝑛ℎ

and

𝑏𝑛 =

1 − cos 𝑛ℎ 𝜋𝑛ℎ

→ 0 as ℎ → 0

Also, 𝑎 0 /ℎ = 1/𝜋. Thus, the 2 𝜋-periodic delta function has a Fourier series

X^ ∞

𝑛=−∞

X^ ∞

𝑛=

cos 𝑛𝑡. (12)

It is worth pointing out that this is one series that definitely does not converge for all 𝑡. In fact it does not converge for 𝑡 = 0. However, there is a sense of convergence in which convergence makes sense. We will discuss this in the next section.

Fourier Series: Theory and Applications in Differential Equations, Slides of Fourier Transform and Series

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Chapter 10

Fourier Series

10.1 Periodic Functions and Orthogonality Relations

𝛽^2 − 𝜔^2

𝑦′′^ + 𝛽^2 𝑦 =

X^ 𝑁

X^ 𝑁

𝛽^2 − 𝜔^2 𝑛

Periodic Functions

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

Exercises

8. 𝑓 (𝑡) = 𝑡^2 , − 1 < 𝑡 ≤ 1 ; 𝑓 (𝑡 + 2) = 𝑓 (𝑡).

9. 𝑓 (𝑡) = 𝑡^2 , 0 < 𝑡 ≤ 2 ; 𝑓 (𝑡 + 2) = 𝑓 (𝑡).

Z 𝐿

X^ ∞

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

X^ ∞

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

X^ ∞

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

Z 𝐿

Z 2 𝐿

Z 𝐿

Z 𝐿

Z 0

Z 𝐿

Z 0

Z 𝐿

Z 𝐿

Z 𝐿

Z −𝐿/ 2

Z 𝐿/ 2

Z 𝐿

Z 𝜋

Z 𝜋

Z 𝜋

𝑡^2

Z 𝜋

Z 𝜋

Z 𝜋

Z 𝑛𝜋

𝑛^2 𝜋

𝑛^2 𝜋

𝑛^2 𝜋

[(−1)𝑛^ − 1]

X^ ∞

Z

Z 𝐿

Z 𝐿

Z 𝑛𝜋

𝑛^2 𝜋^2

Z 𝑛𝜋

𝑛^2 𝜋^2

𝑛^2 𝜋^2

X^ ∞

Z 𝑛𝜋

𝑛^2 𝜋^2

Z 𝑛𝜋

𝑛^2 𝜋^2