Signals and Systems - Electrical Engineering

602 C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems

Solution

The Fourier series coefficients are found as (ω 0 = 2 π/20 rad)

X[k]=

1

20

Z(x 1 [n])

∣

∣

∣

z=ej

220 πk=

1

20

∑^9

n= 0

z−n

∣

∣

∣

z=ej

220 πk=

1

20

1 −z−^10

1 −z−^1

∣

∣

∣

z=ej

220 πk

A close expression forX[k] is obtained as follows:

X[k]=

z−^5 (z^5 −z−^5 )

20 z−0.5(z0.5−z−0.5)

∣

∣

∣z=ej^220 πk

=

e−jπk/^2 sin(πk/ 2 )

20 e−jπk/^20 sin(πk/ 20 )

=

e−j^9 πk/^20

20

sin(πk/ 2 )

sin(πk/ 20 ) n

10.3.3 DTFT of Periodic Signals

A discrete-time periodic signalx[n]of periodNwith a Fourier series representation of

x[n]=

∑

k

X[k]ej^2 πnk/N (10.30)

has a DTFT

X(ejω)=

∑

k

2 πX[k]δ(ω− 2 πk/N) −π≤ω < π (10.31)

If we letF(.)indicate the DTFT, the DTFT of a periodic signalx[n] is

X(ejω)=F

(

x[n]

)

=F

(

∑

k

X[k]ej^2 πnk/N

)

=

∑

k

F

(

X[k]ej^2 πnk/N

)

=

∑

k

2 πX[k]δ(ω− 2 πk/N) −π≤ω < π

whereδ(ω)is the analog delta function sinceωvaries continuously.

nExample 10.15

The periodic signal

δM[n]=

∑∞

m=−∞

δ[n−mM]

has a periodM. Find its DTFT.