Signals and Systems - Electrical Engineering

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

Fork6=0 we have|Yk|=|Xk|/(πk), so that askincreases the frequency components ofy(t)decrease

in magnitude faster than the corresponding ones ofx(t). Thus,y(t)is smoother thanx(t). The

magnitude line spectrum|Yk|, ignoring its average, goes faster to zero than the magnitude line

spectrum|Xk|, as seen in Figure 4.9.

Notice that in this casey(t)is even and its Fourier coefficientsYkare real, whilex(t)is odd and

its Fourier coefficientsXkare purely imaginary. If we subtract the average ofy(t), the signaly(t)

can be clearly approximated as a series of cosines, thus the need for real coefficients in its complex

exponential Fourier series. The signalx(t)is zero-average and as such it can be clearly approximated

by a series of sines requiring its Fourier coefficientsXkto be imaginary. n

nExample 4.8

Integration of a periodic signal, provided it has zero mean, gives a smoother signal. To see this,

find and compare the magnitude line spectra of a sawtooth signalx(t), of periodT 0 =2, and its

integral

y(t)=

∫

x(t)dt

shown Figure 4.10.

Solution

Before doing any calculations it is important to realize that the integral would not exist if the dc is

not zero. Using the following script we can compute the Fourier series coefficients ofx(t)andy(t).

A period ofx(t)is

x 1 (t)=tw(t)+(t− 2 )w(t− 1 ) 0 ≤t≤ 2

where w(t)=u(t)−u(t− 1 )is a rectangular window.

FIGURE 4.10

(a) Sawtooth signalx(t)

and (b) its integraly(t).

Notice thatx(t)is a

discontinuous function

whiley(t)is continuous. (a) (b)

x(t)

− 1 1

− 2

0

1

− 1

t

···

···

y(t) = ∫ x(t)dt

− 2 − 1

t

0.5

01 2

··· ···