Signals and Systems - Electrical Engineering

4.10 Other Properties of the Fourier Series 281

FIGURE 4.16
Nonsymmetric periodic signals.

2

− 2 − 1 0 1

− 2 − 1 01 2 3

2

(^23)
t
t
··· ···
··· ···
x(t)
y(t)
Solution
The given signalx(t)is neither even nor odd, but the advance signalx(t+0.5)is even with a period
ofT 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)e0.5s=

2

s

[

e0.5s−e−0.5s

]

which gives the Fourier coefficients

Xk=

1

2

2

jkπ

[

ejkπ/^2 −e−jkπ/^2

]

e−jkπ/^2

=

1

0.5πk

sin(0.5πk)e−jkπ/^2

after replacingsbyjko=jkπand dividing by the periodT 0 =2. These coefficients are complex

as corresponding to a signal that is neither even nor odd. The dc coefficient isX 0 =1.

The given signaly(t)is neither even nor odd, and cannot be made even or odd by shifting. The even

and odd components of a period ofy(t)are shown in Figure 4.17. The even and odd components

of a periody 1 (t)between−1 and 1 are

y 1 e(t)=[u(t+ 1 )−u(t− 1 )]

︸ ︷︷ ︸

rectangular pulse

+[r(t+ 1 )− 2 r(t)+r(t− 1 )]

︸ ︷︷ ︸

triangle

y 1 o(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 )