The Fourier series (5.2) then reduces to a sine series :

∞

bn sin nx, (5.22) n=1 X

with π 2 bn = π f(x) sin nx dx. 0 Z Thus any integrable function f on 0 < x < π has a sine series (5.22). This sine series can be thought of as the full Fourier series for an odd function fodd on −π < x < π that coincides with f on 0 < x < π .

π Symmetry about the line x = 2 : When working with cosine and sine series it is helpful to think of symmetry about the π line x = 2 .

π i) f is even with respect to the line x = 2 means

f(π − x) = f(x) for all x.

π ii) f is odd with respect to the line x = 2 means

f(π − x) = −f(x) for all x.

a) y b) y π− x π 0 x π − x x 0 x π x π x = 2 π x = 2

π π Figure 5.9: a) f is even with respect to x = 2 . b) f is odd with respect to x = 2 .

Exercise 5.5: Show that

π i) cos 2 n x and sin(2 n − 1) x are even with respect to x = 2 .

π ii) cos(2 n − 1) x and sin 2 nx are odd with respect to x = 2 .

130 π Symmetry with respect to x = 2 and the cosine series:

i) If f(π − x) = f(x) on 0 < x < π , then by (5.21),

π 2 2 1 a n− = π f(x) cos (2 n − 1) x dx = 0 , 0 Z ↑ ↑ even odd and the cosine series is ∞ 1 2 a0 + a2n cos 2 nx. (5.23) n=1 X ii) If f(π − x) = −f(x) on 0 < x < π , then

π 2 2 a n = π f(x) cos 2 nx dx = 0 , 0 Z ↑ ↑ odd even

and the cosine series is ∞

a2n−1 cos(2 n − 1) x. (5.24) n=1 X

π Symmetry with respect to x = 2 and the sine series:

i) If f(π − x) = f(x) on 0 < x < π , then by (5.22),

π 2 2 b n = π f(x) sin 2 nx dx = 0 , 0 Z ↑ ↑ even odd and the sine series is ∞

b2n−1 sin(2 n − 1) x.  (5.25) n=1 X ii) If f(π − x) = −f(x) on 0 < x < π , then

π 2 2 1 b n− = π f(x) sin (2 n − 1) x dx = 0 , 0 Z

↑ ↑ odd even and the sine series is ∞

b2n sin 2 nx. (5.26) n=1 X 131 Example 5.5: Find the Fourier sine series of

π f(x) = 4 , 0 < x < π.

π Use the fact that f is even with respect to x = 2 .

(This series is # iii) in Table 5.1.)

Solution: Since we want the Fourier sine series, we first extend f to be an odd function fodd on −π < x < π :

π 4 , 0 < x < π fodd (x) =    π − 4 , −π < x < 0.  The Fourier sine coefficients are then giving by (5.11): 1 π b = fodd (x) sin( nx )dx. n π Z−π Since fodd (x) sin( nx ) is even (the product of two odd functions), we can write

π 2 odd bn = π f (x) sin( nx )dx 0 Z π 2 π = π 4 sin( nx )dx. 0 Z
π Next, since the given function is symmetric about x = 2 , we have

b2n = 0 , (since sin(2 nx ) is antisymmetric)

and π 2 π 2 1 b n− = π 4 sin(2 n − 1) x dx 0 Z π 2 4 π = π 4 sin(2 n − 1) x dx 0 Z π cos(2 n − 1) x 2 = − (2 n − 1) 0 1 = . 2n − 1 The Fourier sine series is thus ∞ sin(2 n − 1) x . 2n − 1 n=1 X

132 = y f p ( x) y = f (x) y = f_(x)

x x x π − π π − π π

1 Figure 5.10: The odd, period 2 π extension of f(x) = 2 (π − x), 0 < x < π .

5.1.4 Complex form of the Fourier series The general Fourier series (5.2) can be written more concisely by using complex numbers, i.e. by writing the nth term in terms of einx . This complex representation leads naturally to the Fourier integral and Fourier transform, and is the most convenient representation for applications to signal processing. For a τ-periodic function f, the complex Fourier series is

∞ inω 0t f(t) = cne , (5.27) n=−∞ X where 2π ω0 = (5.28) τ is the fundamental frequency. The complex Fourier coefficients cn, n = 0 , ±1, ±2,..., are

given by the formula τ 2 1 −inω 0t cn = f(t)e dt, (5.29) τ τ Z− 2 and satisfy c−n =c ¯ n, (5.30) where the overbar denotes complex conjugation. This condition ensures that the function f(t) in (5.27) is real-valued. The complex basis functions

inω 0t en(t) = e (5.31) satisfy the orthogonality condition

τ/ 2 0, if m =6 n en(t)em(t)dt =  (5.32) 2  Z−τ/  τ if m = n  (verify as an exercise). 

133 Derivation of equation (5.29) :

As in section 5.1.1, we use the fact that Fourier series can be integrated term-by-term (to be justified in Section 5.3.2). Using (5.31), write (5.27) as

∞

f(t) = cnen(t). n=−∞ X τ τ Multiply by em(t) and integrate from − 2 to 2 :

τ τ ∞ 2 2 f(t)em(t)dt = cnen(t)em(t) dt τ τ Z− 2 Z− 2 !n=−∞ " ∞ X τ 2 = cn en(t)em(t)dt (integrate term-by-term) τ n=−∞ − 2 X Z = τc m (by (5.32)) .

−imω 0t Since em(t) = e (see (5.31)), we obtain (5.29), with n replaced by m. Relation between the real and complex forms

Write cn, n = 0 , 1, 2,... in terms of real and imaginary parts: 1 cn = 2 (an − ib n). (5.33) Then by (5.30), 1 c−n = 2 (an + ib n). (5.34) Setting n = 0 in these equations shows that

1 b0 = 0 , c 0 = 2 a0. (5.35) By Euler’s formula ( eiθ = cos θ + i sin θ), the basis vectors (5.31) have the form

en(t) = cos( nω 0t) + i sin( nω 0t). (5.36) We can write (5.27) as a series summed from 1 to ∞:

∞ 1 inω 0t −inω 0t f(t) = 2 a0 + cne + c−ne , (5.37) n=1 X
where we have made use of (5.35). It now follows, using (5.33), (5.34), (5.31) and (5.36) that inω 0t −inω 0t cne + c−ne = an cos( nω 0t) + bn sin( nω 0t) (verify as an exercise). Thus (5.37) becomes the real form of the Fourier series of a τ-periodic function f(t): ∞ 1 f(t) = 2 a0 + [an cos( nω 0t) + bn sin( nω 0t)] , (5.38) n=1 X 134