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:
=
c 2
ω c 1 )
;
n
= 0
π
c BP (n)
=
sin c 2 n)
πn
sin 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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