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364 Chapter 7 FIR FILTER DESIGN
1.07
1.0
0.93
Amplitude Response
Error Function
0.07
0
−0.07
0.1
L + 3 = 6
extrema
0.4 1
ω/π
0.04
0.0
−0.04
0.1
L − 1 = 2
extrema
0.4 1
FIGURE 7.34 Amplitude response and the error function in Example 7.22
ω/π
a maximum at ω = π). Now if we include the end points ω =0and ω = π,
then P (ω) has at most (L +1)local extrema in the closed interval 0 ≤ ω ≤ π.
Finally, we would like the filter specifications to be met exactly at band edges
ω p and ω s. Then the specifications can be met at no more than (L +3)extremal
frequencies in the 0 ≤ ω ≤ π interval.
Conclusion
The error function E(ω) has at most (L +3)extrema in S. □
□ EXAMPLE 7.22 Let us plot the amplitude response of the filter given in Example 7.21 and count
the total number of extrema in the corresponding error function.
Solution
The impulse response is
h(n) = 1 [1, 2, 3, 4, 3, 2, 1], M =7 or L =3
15
and α(n) = 1 [4, 6, 4, 2] and β(n) =[ 0, 0, 8 , 8 15 15 15]
from Example 7.21. Hence
P (ω) = 8
15 cos2 ω + 8
15 cos3 ω
which is shown in Figure 7.34. Clearly, P (ω) has (L − 1) = 2 extrema in the
open interval 0