Digital Signal Processing Reference
In-Depth Information
Example 5.6
Design a bandpass filter that approximates the ideal magnitude response given in
Figure 5.5(c), in which
ω
c
2
=
0
.
6
π
and
ω
c
1
=
0
.
2
π
. Let us select a Hamming
window of length
N
11 and plot the magnitude response of the filter.
The coefficients
c
BP
(n)
of the Fourier series for the magnitude response given
are computed from formula (5.35) given below:
=
⎨
⎩
(ω
c
2
−
ω
c
1
)
;
n
=
0
π
c
BP
(n)
=
sin
(ω
c
2
n)
πn
sin
(ω
c
1
n)
πn
|
|
≥
0
−
;
n
But since the Hamming window function has a length of 11, we need to compute
the coefficients
c
BP
(n)
also, from
n
5 only. So also we calculate
the 11 coefficients of the Hamming window, using the formula
=−
5to
n
=
=
0
.
54
+
0
.
46 cos
2
πn
N
w
H
(n)
;−
5
≤
n
≤
5
Their products
h
w
(n)
c
BP
(n)w
H
(n)
are computed next. The 11 coefficients
c
BP
(n)
,
w
H
(n)
and
h
w
(n)
for
−
5
≤
=
≤
5 are listed below. Next the coefficients
h
w
(n)
are delayed by five samples to get the coefficients of the FIR filter function
[i.e.,
h(n)
n
10 below. The
plot of the four sequences and the magnitude response of the FIR are shown in
Figures 5.10 and 5.11, respectively.
=
h
w
(n
−
5
)
], and these are also listed for 0
≤
n
≤
c
BP
(n)
=
0
.
00
0
.
0289
−
0
.
1633
−
0
.
2449
0
.
1156
0
.
400
0
.
1156
−
0
.
2449
−
0
.
1633
0
.
0289
0
.
000
w
H
(n)
=
0
.
08
0
.
1679
0
.
0379
0
.
6821
0
.
9121
1
.
00
.
9121
0
.
6821
0
.
0379
0
.
1679
0
.
0800
h
w
(n)
=
0
.
00
0
.
0049
−
0
.
0650
−
0
.
1671
0
.
1055
0
.
4000
0
.
1055
−
0
.
1671
−
0
.
0650
0
.
0049
0
.
000
h(n)
0.0000
0.0049
−
0.0650
−
0.1671
0.1055
0.4000
0.1055
−
0.1671
−
0.0650
0.0049
0.0000
Example 5.7
Design a lowpass FIR filter of length 11, with a cutoff frequency
ω
c
0
.
3
π
.
Using a Hamming window, find the value of the samples
h(
3
)
and
h(
9
)
of the
FIR filter given by
H(z
−
1
)
=
=
1
n
=
0
h(n)z
−
n
.
Since the length of the FIR filter is given as 11, its order is
N
=
10. The
coefficients
h
w
(n)
have to be known for
−
5
≤
n
≤
5 and delayed by five samples.
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