Chapter 2
Discrete-Time Signals and Systems
2.1 Sampling frequency F
s
= 100 sam/sec
(a) Continuous-time signal x
c
(t) = 2cos(40πt + π/3) has frequency of 20 Hz. Hence
x(n) = x
c
(t)
|
t=n/F
s
= 2cos
40π n
100
+ π/3
which implies that ω
0
=
40π
100
=
2π
5
.
(b) Steady-state response y
c,ss
(t): Given that h(n) = 0.8
n
u(n), the frequency response function is
H(e
jω
) =
1
1 −0.8e
−jω
Since ω
0
= 2π/5, the system response at ω
0
is
H(e
jω
) =
1
1 − 0.8e
−j2π/5
= 0.9343 e
−j0.2517π
Hence y
ss
(n) = 2(0.9343) cos(2πn/5 +π/3 − 0.2517π),or
y
c,ss
(t) = 1. 8686 cos(40πt + 0.585π)
(c) Any x
c
(t) that has the same digital frequency ω
0
after sampling and the same phase shift as above will
have the same steady state response. Since F
s
= 100 sam/sec, the two other frequencies are 120 and 220
Hz.
2.2 The discrete-time signal is
x(n) = A cos(ω
0
n)w
R
(n)
where w
R
(n) is an N-point rectangular window.
(a) The DTFT of x(n) is determined as
X(e
jω
) = F [A cos(ω
0
n)w
R
(n)]
=
(
A/2
)
F
e
jω
0
w
R
(n)
+
(
A/2
)
F
e
−jω
0
w
R
(n)
(1)
Using the DTFT of w
R
(n) as
F [w
R
(n)] =
N−1
n=0
e
−jωn
= e
−jω(N−1)/2
sin(ω N/2)
sin(ω/2)
(2)
and the fact that complex exponential causes a translation in the frequency domain (1) can be written
after a fair amount of algebra and trigonometry as
X(e
jω
) = X
R
(e
jω
) + jX
I
(e
jω
)
1
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