Signals and Systems - Electrical Engineering

8.2 Discrete-Time Signals 469

nExample 8.13

Show how to obtain the discrete-time exponentialx[n]=(− 1 )nforn≥0 and zero otherwise, by

sampling a continuous-time signalx(t).

Solution

Because the values ofx[n] are 1 and−1,x[n] cannot be generated by sampling a real exponential

signale−atu(t); indeed,e−at>0 for any values ofaandt. The discrete signal can be written as

x[n]=(− 1 )n=cos(πn)

forn≥0. If we sample an analog signalx(t)=cos( 0 t)u(t)with a sampling periodTs, we get

x[n]=x(nTs)=cos( 0 nTs)=cos(πn) n≥ 0

and zero otherwise. Thus, 0 Ts=π, givingTs=π/ 0. For instance, for 0 = 2 π, thenTs=0.5. n

Discrete-Time Sinusoids
Discrete-time sinusoids are a special case of the complex exponential. Lettingα=ejω^0 andA=|A|ejθ,
we have according to Equation (8.16),

x[n]=Aαn=|A|ej(ω^0 n+θ)=|A|cos(ω 0 n+θ)+j|A|sin(ω 0 n+θ) (8.17)

so the real part ofx[n] is a cosine, while the imaginary part is a sine. As indicated before, discrete
sinusoids of amplitudeAand phase shiftθare periodic if they can be expressed as

Acos(ω 0 n+θ)=Asin(ω 0 n+θ+π/ 2 ) −∞<n<∞ (8.18)

wherew 0 = 2 πm/Nrad is the discrete frequency for integersmandN>0, which are not divisible
by each other. Otherwise, discrete-time sinusoids are not periodic.

Becauseωis given in radians, it repeats periodically with 2πas the period—that is,

ω=ω+ 2 πk kinteger (8.19)

To avoid this ambiguity, we will let−π < ω≤πas the possible range of discrete frequencies. This is
possible since

ω=

{

ω− 2 πk whenω > 2 π, for somek>0 integer

ω− 2 π 0 ≤ω≤ 2 π

(8.20)

See Figure 8.5. Thus, sin( 3 πn)equals sin(πn), and sin(1.5πn)equals sin(−0.5πn)=−sin(0.5πn).