Reflection and Even and Odd Periodic Signals

4.10.1 Reflection and Even and Odd Periodic Signals

If the Fourier series of x(t), periodic with fundamental frequency  0 , is

x(t) X

= jk X

then the one for its reflected version x( −t) is

X jk x( t −t) = X

m e −jm 0 =

X −k e 0 (4.37)

so that the Fourier coefficients of x( −t) are X −k (remember that m and k are just dummy variables). This can be used to simplify the computation of Fourier series of even and odd signals.

For an even signal x(t), we have that x(t) = x(−t), and as such X k =X −k and therefore x(t) is naturally represented in terms of cosines and a dc term. Indeed, its Fourier series is

X −1

x(t) =X 0 +

k [e o +e −jk o ]

X k cos(k 0 t)

k =1

indicating that X k are real-valued. This is also seen from

x(t)e −jk 0 t dt =

x(t)[cos(k 0 t) − j sin(k 0 ) ]dt

x(t) cos(k 0 t)dt

because x(t) sin(k 0 t) is odd and their integral is zero. It will be similar for an odd function for which x(t) = −x(−t), or X k = −X −k , in which case the Fourier series has a zero dc value and sine harmonics.

C H A P T E R 4: Frequency Analysis: The Fourier Series

The X k are purely imaginary. Indeed, for an odd x(t),

x(t)e −jk 0 dt =

x(t)[cos(k 0 t) − j sin(k 0 ) ]dt

−j Z

x(t) sin(k 0 t)dt

since x(t) cos(k 0 t) is odd. The Fourier series of an odd function can thus be written as

x(t) =2

( jX k ) sin(k 0 t)

k =1

According to the even and odd decomposition, any periodic signal x(t) can be expressed as x(t) =x e ( t) +x o ( t)

where x e ( t) is the even and x o ( t) is the odd component of x(t). Finding the Fourier coefficients of x e ( t), which will be real, and those of x o ( t), which will be purely imaginary, we would then have

X k =X ek +X ok since

x e ( t) = 0.5[x(t) + x(−t)] ⇒

X ek = 0.5[X k +X −k ]

x o ( t) = 0.5[x(t) − x(−t)]

⇒ X ok = 0.5[X k −X −k ]

Reflection: If the Fourier coefficients of a periodic signal x(t) are {X k } then those of x(−t), the time-reversed signal with the same period as x(t), are {X −k }.

Even periodic signal x(t): Its Fourier coefficients X k are real, and its trigonometric Fourier series is

x(t) =X 0 +2

X k cos(k 0 t)

k =1

Odd periodic signal x(t): Its Fourier coefficients X k are imaginary, and its trigonometric Fourier series is

x(t) =2

jX k sin(k 0 t)

k =1

For any periodic signal x(t) = x e ( t) +x o ( t) where x e ( t) and x o ( t) are the even and odd component of x(t), then X k =X ek +X ok

(4.43) where {X ek } are the Fourier coefficients of x e ( t) and {X ok } are the Fourier coefficients of x o ( t).

■ Example 4.14

Consider the periodic signals x(t) and y(t) shown in Figure 4.16. Determine their Fourier coefficients by using the symmetry conditions and the even–odd decomposition.

4.10 Other Properties of the Fourier Series 281

t Nonsymmetric periodic signals.

The given signal x(t) is neither even nor odd, but the advance signal x(t + 0.5) is even with a period of T 0 = 2,  0 = π. Then between −1 and 1 the shifted period is

x 1 ( t + 0.5) = 2[u(t + 0.5) − u(t − 0.5)]

so that its Laplace transform is

X 1 ( s)e

0.5s

2 h 0.5s

e −e −0.5s

which gives the Fourier coefficients

= jkπ/2 e −e −jkπ/2 e −jkπ/2

2 jkπ

sin(0.5πk)e −jkπ/2

0.5πk

after replacing s by jk o = jkπ and dividing by the period T 0 = 2. These coefficients are complex as corresponding to a signal that is neither even nor odd. The dc coefficient is X 0 = 1.

The given signal y(t) is neither even nor odd, and cannot be made even or odd by shifting. The even and odd components of a period of y(t) are shown in Figure 4.17. The even and odd components

of a period y 1 ( t) between −1 and 1 are y 1e ( t) = [u(t + 1) − u(t − 1)] + [r(t + 1) − 2r(t) + r(t − 1)]

{z

{z

rectangular pulse

triangle

y 1o ( t) = t[u(t + 1) − u(t − 1)] = [(t + 1)u(t + 1) − u(t + 1)] − [(t − 1)u(t − 1) + u(t − 1)] = r(t + 1) − r(t − 1) − u(t + 1) − u(t − 1)

C H A P T E R 4: Frequency Analysis: The Fourier Series

y 1e (t )

y 1o (t)

−1 t FIGURE 4.17

0 1 1 Even and odd components of the period of y(t),

−1 ≤ t ≤ 1. −1

Thus, the mean value of y e ( t) is the area under y 1e ( t) divided by 2 or 1.5, and for k 6= 0,

ek =

Y 1e ( s) s =jk 0 =

( e −e −s ) + ( e −2+e −s )

s 2 s =jkπ

sin(kπ )

1 − cos(kπ)

1 1 − cos(kπ) k − (−1)

The mean value of y o ( t) is zero, and for k 6= 0,

T 0 =jk 0 2 s 2 s

s =jkπ

sin(kπ )

cos(kπ )

cos(kπ )

kπ Finally, the Fourier series coefficients of y(t) are

ek +Y ok = (1 − (−1) )/( kπ ) 2 + j(−1) k /( kπ ) k 6= 0 ■