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ece538_e2_fa2016.pdf-NAME: 28 Oct. 2016 ECE 538
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ece538 e2 fa2016.pdf-NAME: 28 Oct. 2016 ECE 538
ece538_e2_fa2016.pdf-NAME: 28 Oct. 2016 ECE 538
ece538 e2 fa2016.pdf-NAME: 28 Oct. ...
ece538_e2_fa2016.pdf-NAME: 28 Oct. 2016 ECE 538
Page 11
NAME:
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Page 12
h
[n]
0
h
[n]
1
2
h
[n]
y[n]
3
h
[n]
interleaver
(counter-
clockwise)
y
[n]
0
y
[4n+3]
y
[4n]
y
[4n+2]
y
[4n+1]
y
[n]
1
y
[n]
2
y
[n]
3
+
+
+
+
Figure 3.
Problem 2.
This problem is about digital subbanding of the four DT signals
x
i
[
n
],
i
=0
,
1
,
2
,
3 from Problem 2. Digital subbanding of these four signals is effected in an
efficient way via filter bank in Figure 3. All of the quantities in Figure 1 are defined below:
the respective impulse responses of the polyphase component filters are defined in terms of
the ideal lowpass filter impulse response below.
h
LP
[
n
]=4
sin
π
4
n
πn
(1)
The polyphase component filters on the left side of Figure 3 are defined as
h
+
[
n
]=
h
LP
[4
n
+
]
,
=0
,
1
,
2
,
3
.
(2)
The respective signals at the inputs to these filters are formed from the input signals as
(from Problem 3) as described below.
There is only ONE part to this problem: plot
the magnitude of the DTFT
Y
(
ω
)
of the interleaved signal
y
[
n
]
.
y
0
[
n
]=
x
0
[
n
]+
x
1
[
n
] cos
π
2
0
-
ˆ
x
1
[
n
] sin
π
2
0
+
x
2
[
n
] cos
π
2
0
x
2
[
n
] sin
π
2
0
+
x
3
[
n
] cos(
π
0)
y
1
[
n
]=
x
0
[
n
]+
x
1
[
n
] cos
π
2
1
-
ˆ
x
1
[
n
] sin
π
2
1
+
x
2
[
n
] cos
π
2
1
x
2
[
n
] sin
π
2
1
+
x
3
[
n
] cos(
π
1)
y
2
[
n
]=
x
0
[
n
]+
x
1
[
n
] cos
π
2
2
-
ˆ
x
1
[
n
] sin
π
2
2
+
x
2
[
n
] cos
π
2
2
x
2
[
n
] sin
π
2
2
+
x
3
[
n
] cos(
π
2)
y
3
[
n
]=
x
0
[
n
]+
x
1
[
n
] cos
π
2
3
-
ˆ
x
1
[
n
] sin
π
2
3
+
x
2
[
n
] cos
π
2
3
x
2
[
n
] sin
π
2
3
+
x
3
[
n
] cos(
π
3)
(3)
12


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13


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NAME:
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14


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