Signals and Systems - Electrical Engineering

456 C H A P T E R 8: Discrete-Time Signals and Systems

wish to have a sinusoid with periodN=10, thenTs=0.2kπforkchosen so the Nyquist sampling

rate condition is satisfied—that is,

0 <Ts=kπ/ 5 ≤π so that 0<k≤5.

From these possible values forkwe choosek=1 and 3 so thatNandkare not divisible by each

other and we get the desired periodN=10 (the valuesk=2 and 4 would give 5 as the period,

andk=5 would give a period of 2 instead of 10). Indeed, if we letk=1, thenTs=0.2πsatisfies

the Nyquist sampling rate condition, and we obtain the sampled signal

x[n]=cos(0.2nπ+π/ 4 )=cos

(

2 π

10

n+

π

4

)

which according to its frequency is periodic of period 10. This is the same fork=3. n

When sampling an analog sinusoid

x(t)=Acos( 0 t+θ) −∞<t<∞ (8.4)

of periodT 0 = 2 π/ 0 , 0 > 0 , we obtain aperiodic discrete sinusoid,

x[n]=Acos( 0 Tsn+θ)=Acos

(

2 πTs

T 0

n+θ

)

(8.5)

provided that

Ts

T 0

=

m

N

(8.6)

for positive integersNandm, which are not divisible by each other. To avoid frequency aliasing the sampling

period should also satisfy

Ts≤

π

 0

=

T 0

2

(8.7)

Indeed, sampling a continuous-time signalx(t)using as sampling periodTs, we obtain

x[n]=Acos( 0 Tsn+θ)

=Acos

(

2 πTs

T 0

n+θ

)

where the discrete frequency isω 0 = 2 πTs/T 0. For this signal to be periodic we should be able to

express this frequency as 2πm/Nfor nondivisible positive integersmandN. This requires that

T 0

Ts

=

N

m

be a rational number, or that

mT 0 =NTs (8.8)