Signals and Systems - Electrical Engineering

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

In a dual way, when we multiply a signal by a complex exponentialejω^0 nfor some frequencyω 0 ,

the spectrum of the signal is shifted in frequency. So ifx[n] has a DTFTX(ejω), the modulated signal

x[n]ejω^0 nhas as DTFTX(ej(ω−ω^0 )). Indeed, the DTFT ofx 1 [n]=x[n]ejω^0 nis

X 1 (ejω)=

∑

n

x 1 [n]e−jωn=

∑

n

x[n]e−j(ω−ω^0 )n=X(ej(ω−ω^0 ))

The following pairs illustrate the duality in time and frequency shifts: if the DTFT ofx[n]isX(ejω)then

x[n−N]⇔X(ejω)e−jωN

x[n]ejω^0 n⇔X(ej(ω−ω^0 )) (10.19)

RemarkThe signal x[n]ejω^0 nwas calledmodulatedbecause x[n]modulates the complex exponential or

discrete-time sinusoids. It can be written as

x[n] cos(ω 0 n)+jx[n] sin(ω 0 n)

nExample 10.7

The DTFT ofx[n]=cos(ω 0 n),−∞<n<∞, cannot be found from the Z-transform or from the

sum defining the DTFT asx[n] is not a finite-energy signal. Use the frequency-shift and the time-

shift properties to find the DTFTs ofx[n]=cos(ω 0 n)andy[n]=sin(ω 0 n).

Solution

Using Euler’s identity we have that

x[n]=cos(ω 0 n)=

ejω^0 n+e−jω^0 n

2

and so the DTFT ofx[n] is given by

X(ejω)=F[0.5ejω^0 n]+F[0.5e−jω^0 n]

=F[0.5]ω−ω 0 +F[0.5]ω+ω 0

=π[δ(ω−ω 0 )+δ(ω+ω 0 )]

Since

y[n]=sin(ω 0 n)=cos(ω 0 n−π/ 2 )=cos(ω 0 (n−π/( 2 ω 0 ))=x[n−π/( 2 ω 0 )]

we have that according to the time-shift property its DTFT is given by

Y(ejω)=X(ejω)e−jωπ/(^2 ω^0 )

=π

[

δ(ω−ω 0 )e−jωπ/(^2 ω^0 )+δ(ω+ω 0 )e−jωπ/(^2 ω^0 )

]