10.3 Fourier Series of Discrete-Time Periodic Signals 601
Solution
The period ofx[n] isN=4. Indeed,
x[n+4]= 1 +cos( 2 π(n+ 4 )/ 4 )+sin( 2 π(n+ 4 )/ 2 )
= 1 +cos( 2 πn/ 4 + 2 π)+sin( 2 πn/ 2 + 4 π)=x[n]
The frequencies inx[n] are: a DC frequency, corresponding to the constant, and frequencies
ω 0 = 2 π/ 4 =π/2 andω 1 = 2 π/ 2 = 2 ω 0 , corresponding to the cosine and the sine. No other
frequencies are present in the signal. The fundamental frequency isω 0 = 2 π/N=π/2, and the
complex exponential Fourier series can be obtained directly fromx[n] using Euler’s equation:
x[n]= 1 +0.5(ejπn/^2 +e−jπn/^2 )−0.5j(ejπn−e−jπn)
=X[0]+X[1]ejω^0 n+X[−1]e−jω^0 n+X[2]ej^2 ω^0 n+X[−2]e−j^2 ω^0 n ω 0 =
π
2
so that the Fourier series coefficients are X[0]=1, X[1]=X∗[−1]=0.5, and X[2]=
X∗[−2]=−0.5j. n
10.3.2 Connection with the Z-Transform
Recall that the Laplace transform was used to find the Fourier series coefficients. Likewise, for periodic
discrete-time signals we will show that the Z-transform of a period of the signal, which always exists,
can be connected with the Fourier series coefficients.
Ifx 1 [n]=x[n](u[n]−u[n−N])is a period of a periodic signalx[n]of periodN, its Z-transform
Z(x 1 [n])=
N∑− 1
n= 0
x[n]z−n
has the whole plane, except for the origin, as its region of convergence. The Fourier series coefficients ofx[n]
are thus determined as
X[k]=
1
N
N∑− 1
n= 0
x[n]e−j
2 π
Nkn
=
1
N
Z(x 1 [n])
∣
∣∣
∣z=ej^2 Nπk (10.29)
nExample 10.14
Consider a discrete pulsex[n] with a periodN=20, andx 1 [n]=u[n]−u[n−10] is the period
between 0 and 19. Find the Fourier series ofx[n].
