Signals and Systems - Electrical Engineering

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

Assuming the maximum frequency ismax=N 0 forN≥2 (since the signal is not band limited

the maximum frequency is not known; to estimate it we could use Parseval’s result as indicated in

Chapter 7, instead we are assuming that it is a multiple of 0 ), if we letTs=π/N 0 after replacing

it in the above equations, we get

α=e−aπ/N^0

ω 0 = 0 π/N 0 =π/N

If we wantα=0.9 andω 0 =π/2, we have thatN=2 and

a=−

2  0

π

log 0.9

for any frequency 0 >0. For instance, if 0 = 2 π, thena=−4 log 0.9 andTs=0.25. Figure 8.4

displays the continuous- and the discrete-time signals generated using the above parameters. The

following script is used. The continuous-time and the discrete-time signals coincide at the sample

times.

%%%%%%%%%%%%%%%%%%%%%

% Example 8.12

%%%%%%%%%%%%%%%%%%%%%

a =− 4 ∗log(0.9);Ts = 0.25; % parameters

alpha = exp(−a∗Ts);

n = 0:30; y = alpha.ˆn.∗cos(pi∗n/2); % discrete-time signal

t = 0:0.001:max(n)∗Ts; x = exp(−a∗t).∗cos(2∗pi∗t); % analog signal

stem(n, y, ’r’); hold on

plot(t/Ts, x); grid; legend(’y[n]’, ’x(t)’); hold off

FIGURE 8.4

Determination of parameters for a

continuous-time signalx(t)that

when sampled gives a desired

discrete-time signaly[n].

0 5 10 15 20 25 30

− 1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t/Ts

x

(t

),

y[

n]

y[n]

x(t)

n