Signals and Systems - Electrical Engineering

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

TheXkare purely imaginary. Indeed, for an oddx(t),

Xk=

1

T 0

∫

T 0

x(t)e−jk^0 tdt=

1

T 0

∫

T 0

x(t)[cos(k 0 t)−jsin(k 0 )]dt

=

−j

T 0

∫

T 0

x(t)sin(k 0 t)dt

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

x(t)= 2

∑∞

k= 1

(jXk)sin(k 0 t) (4.39)

According to the even and odd decomposition, any periodic signalx(t)can be expressed as

x(t)=xe(t)+xo(t)

wherexe(t)is the even andxo(t)is the odd component ofx(t). Finding the Fourier coefficients of

xe(t), which will be real, and those ofxo(t), which will be purely imaginary, we would then have

Xk=Xek+Xoksince

xe(t)=0.5[x(t)+x(−t)] ⇒ Xek=0.5[Xk+X−k]

xo(t)=0.5[x(t)−x(−t)] ⇒ Xok=0.5[Xk−X−k] (4.40)

n Reflection: If the Fourier coefficients of a periodic signalx(t)are{Xk}then those ofx(−t), the time-reversed

signal with the same period asx(t), are{X−k}.

n Even periodic signalx(t): Its Fourier coefficientsXkare real, and its trigonometric Fourier series is

x(t)=X 0 + 2

∑∞

k= 1

Xkcos(k 0 t) (4.41)

n Odd periodic signalx(t): Its Fourier coefficientsXkare imaginary, and its trigonometric Fourier series is

x(t)= 2

∑∞

k= 1

jXksin(k 0 t) (4.42)

For any periodic signalx(t)=xe(t)+xo(t)wherexe(t)andxo(t)are the even and odd component ofx(t), then

Xk=Xek+Xok (4.43)

where{Xek}are the Fourier coefficients ofxe(t)and{Xok}are the Fourier coefficients ofxo(t).

nExample 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.