Signals and Systems - Electrical Engineering

8.2 Discrete-Time Signals 455

From their frequencies determine if these signals are periodic, and if so, determine their

corresponding periods.

Solution

The frequency ofx 1 [n] can be written as

ω 1 =π=

2 π

2

wherem=1 andN=2, so thatx 1 [n] is periodic of periodN 1 =2. Likewise, the frequency ofx 2 [n]

can be written as

ω 2 = 3 π=

2 π

2

3

wherem=3 andN=2, so thatx 2 [n] is also periodic of periodN 2 =2, which can be verified as

follows:

x 2 [n+2]=3 sin( 3 π(n+ 2 )+π/ 2 )=3 sin( 3 πn+ 6 π+π/ 2 )=x[n]

n

nExample 8.4

What is true for continuous-time sinusoids—that they are always periodic—is not true for discrete-

time sinusoids. These sinusoids can be nonperiodic even if they result from uniformly sampling a

continuous-time sinusoid. Consider the discrete signalx[n]=cos(n+π/ 4 ), which is obtained by

sampling the analog sinusoidx(t)=cos(t+π/ 4 )with a sampling periodTs=1 sec/sample. Isx[n]

periodic? If so, indicate its period. Otherwise, determine values of the sampling period, satisfying

the Nyquist sampling rate condition, that when used in samplingx(t)result in periodic signals.

Solution

The sampled signalx[n]=x(t)|t=nTs=cos(n+π/ 4 )has a discrete frequencyω=1 rad that cannot

be expressed as 2πm/Nfor any integersmandNbecauseπis an irrational number. Sox[n] is not

periodic.

Since the frequency of the continuous-time signalx(t)is=1 (rad/sec), then the sampling period,

according to the Nyquist sampling rate condition, should be

Ts≤

π



=π

and for the sampled signalx(t)|t=nTs=cos(nTs+π/ 4 )to be periodic of periodNor

cos((n+N)Ts+π/ 4 )=cos(nTs+π/ 4 ) is necessary that NTs= 2 kπ

for an integerk(i.e., a multiple of 2π). Thus,Ts= 2 kπ/N≤πsatisfies the Nyquist sampling con-

dition at the same time that it ensures the periodicity of the sampled signal. For instance, if we