Signals and Systems - Electrical Engineering

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

Let us then show how the coefficientsckanddkcan be obtained directly from the signal. Using the

relationXk=X−∗kand the fact that for a complex numberz=a+jb, thenz+z∗=(a+jb)+(a−

jb)= 2 a= 2 Re(z), we have that

x(t)=X 0 +

∑∞

k= 1

[Xkejk^0 t+X−ke−jk^0 t]

=X 0 +

∑∞

k= 1

[Xkejk^0 t+X∗ke−jk^0 t]

=X 0 +

∑∞

k= 1

2 Re[Xkejk^0 t]

SinceXkis complex (verify this!),

2 Re[Xkejk^0 t]= 2 Re[Xk] cos(k 0 t)− 2 Im[Xk] sin(k 0 t)

Now, if we let

ck=Re[Xk]=

1

T 0

t (^0) ∫+T 0
t 0
x(t)cos(k 0 t)dt k=1, 2,...
dk=−Im[Xk]=

1

T 0

t (^0) ∫+T 0
t 0
x(t)sin(k 0 t)dt k=1, 2,...
we then have
x(t)=X 0 +

∑∞

k= 1

(

2 Re[Xk] cos(k 0 t)− 2 Im[Xk] sin(k 0 t)

)

=X 0 + 2

∑∞

k= 1

(ckcos(k 0 t)+dksin(k 0 t))

and since the averageX 0 =c 0 we obtain the second form of the trigonometric Fourier series shown

in Equation (4.19). Notice thatd 0 =0 and so it is not necessary to define it.

The coefficientsXk=|Xk|ejθkare connected with the coefficientsckanddkby

|Xk|=

√

c^2 k+d^2 k

θk=−tan−^1

[

dk

ck

]

This can be shown by adding the phasors corresponding tockcos(k 0 t)anddksin(k 0 t)and finding

the magnitude and phase of the resulting phasor.