Signals and Systems - Electrical Engineering

8.2 Discrete-Time Signals 467

− 6 − 4 − 2 0 2 4 6

0

1

2

3

4

n

x^1

[n

]=

(0.8)

n

x^2

[n

]=

1.25

n

− 6 − 4 − 2 0246

n

0

1

2

3

4

− 5 0 5

− 4

− 2

0

2

4

n

y^1

[n

]

− 5 05

− 4

− 2

0

2

4

n

y^2

[n

]

(a)

(b)

FIGURE 8.3
(a) Real exponentialx 1 [n]=0.8n,x 2 [n]=1.25n, and (b) modulated exponentialy 1 [n]=x 1 [n] cos(πn)and
y 2 [n]=x 2 [n] cos(πn).

and zero otherwise. Consider the case whenα=0.9 andω 0 =π/2. Finda, 0 , andTsthat will

permit us to obtainy[n] fromx(t)by sampling. Plotx[n] andy[n] using MATLAB.

Solution

Comparing the sampled continuous-time signalx(nTs)=(e−aTs)ncos(( 0 Ts)n)u[n] withy[n] we

obtain the following two equations:

α=e−aTs

ω 0 = 0 Ts

with three unknowns (a, 0 , andTs), so there is no unique solution. According to the Nyquist

sampling rate condition,

Ts≤

π

max