Digital Signal Processing Reference
In-Depth Information
0.8333
0.3
≈
0.2
G
(
W
)
0.1
1
0
−0.1
k
−0.2
W
−25
−20
−15
−10
−5
0
5 0 5 0 5
−
p
−
p
/2
0
p
/2
p
(a)
(b)
stop band
pass band
LTID systems with the following transfer functions:
Fig. 11.20. (a) Impulse response
h
[
k
] and (b) magnitude
spectrum
H
(
Ω
) of an ideal
highpass filter specified in
Example 11.20(ii). The phase
response is zero for all
frequencies.
2
(i)
H
1
(
Ω
)
=
;
(11.65)
1
−
3
−
j
Ω
+
1
−
j2
Ω
4
e
8
e
≤
π
6
1
Ω
(ii)
H
2
(
Ω
)
=
(11.66)
π
6
0
<
≤π,
Ω
≤
π
6
0
Ω
(iii)
H
3
(
Ω
)
=
(11.67)
π
6
1
<
≤π.
Ω
Solution
The DT sequence
x
[
k
]isgivenby
x
[
k
]
=
x
(
kT
s
)
=
3 cos(1000
π
kT
s
)
+
5 cos(2000
π
kT
s
)
.
Substituting
T
s
=
1
/
8000, we obtain
π
k
8
π
k
4
x
[
k
]
=
3 cos
+
5 cos
,
which implies that
x
[
k
] consist of two frequency components,
Ω
1
= π/
8 and
= π/
4
.
This is also apparent from the DTFT of
x
[
k
]
,
given by
Ω
2
−
π
8
+
π
8
−
π
4
+
π
4
X
(
Ω
)
=
3
π
δ
+ δ
+
5
π
δ
+ δ
,
Ω
Ω
Ω
Ω
which consists of impulses at frequencies
Ω
1
=π/
8 and
Ω
2
=π/
4
.
As the DTFT is 2
π
-periodic, in the above equation we showed
X
(
Ω
) only
in the frequency range
−π
≤ π
. This simplifies the analysis, and hence
we will use the same approach to express the DTFTs in the following.
If the transfer function of an LTID system is
H
(
Ω
), the DTFT
Y
(
Ω
) of the
output sequence is given by
Y
(
Ω
)
=
H
(
Ω
)
X
(
Ω
)
=
H
(
Ω
)3
π
≤
Ω
−
π
8
+
π
8
−
π
4
+
π
4
δ
+ δ
+
5
π
δ
+ δ
Ω
Ω
Ω
Ω
Ω
−
π
8
π
8
Ω
+
π
8
−
π
8
=
3
π
δ
+ δ
H
H
−
π
4
π
4
+
π
4
−
π
4
+
5
π
δ
H
+ δ
H
.
Ω
Ω
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