Signals and Systems - Electrical Engineering

4.10 Other Properties of the Fourier Series 287

nExample 4.17

Letg(t)be the derivative of a triangular train of pulsesf(t), of periodT 0 =1. The period off(t),

0 ≤t≤1, is

f 1 (t)= 2 r(t)− 4 r(t−0.5)+ 2 r(t− 1 )

Use the Fourier series ofg(t)to find the Fourier series off(t).

Solution

According to the derivative property we have that

Fk=

Gk

jk 0

k6= 0

are the Fourier coefficients off(t). The signalg(t)=df(t)/dthas a corresponding periodg 1 (t)=

df 1 (t)/dt= 2 u(t)− 4 u(t−0.5)+ 2 u(t− 1 ). The Fourier series coefficients ofg(t)are

Gk=

2 e−0.5s

s

(

e0.5s− 2 +e−0.5s

)

|s=j 2 πk= 2 (− 1 )k

cos(πk)− 1

jπk

k6= 0

which are used to obtain the coefficientsFkfork6=0. The dc component off(t)is found to be 0.5

from its plot asg(t)does not provide it. n

nExample 4.18

Consider the reverse of Example 4.17. That is, given the periodic signalg(t)of periodT 0 =1 and

Fourier coefficients

Gk= 2 (− 1 )k

cos(πk)− 1

jπk

k6= 0

andG 0 =0. Find the integral

z(t)=

∫t

−∞

g(τ)dτ

Solution

As shown above,z(t)is also periodic of the same period asg(t)(i.e.,T 0 =1). The Fourier

coefficients ofz(t)are

Zk=

Gk

j 0 k

=(− 1 )k

4 (cos(πk)− 1 )

(j 2 πk)^2

=(− 1 )(k+^1 )

cos(πk)− 1

π^2 k^2

k6= 0