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362 Chapter 7 FIR FILTER DESIGN
Then the weighted error E(ω) response is
Weighted Error Function
0.05
weight = 0.5 weight = 1.0
0.0
−0.05
0 0.3 0.5 1
frequency in π units
Thus the maximum error in both the passband and stopband is δ 2 . Therefore,
if we succeed in minimizing the maximum weighted error to δ 2 ,we
automatically also satisfy the specification in the passband to δ 1 . Substituting
H r (ω) from (7.40) into (7.42), we obtain
If we define
E (ω) =W (ω)[H dr (ω) − Q (ω) P (ω)]
[ ]
Hdr (ω)
= W (ω) Q (ω) − P (ω) , ω ∈S
Q (ω)
Ŵ (ω) △ = W (ω)Q(w) and Ĥ dr (ω) △ = H dr (ω)
Q (ω)
then we obtain
]
E(ω) =Ŵ [Ĥdr (ω) (ω) − P (ω) , ω ∈S (7.44)
Thus we have a common form of E(ω) for all four cases.
Problem statement
be defined as:
The Chebyshev approximation problem can now
Determine the set of coefficients a(n)or˜b(n)or˜c(n)or ˜d(n) [or equivalently
a(n) orb(n) orc(n) ord(n)] to minimize the maximum absolute
value of E (ω) over the passband and stopband, i.e.,
min
over coeff.
[ ]
max |E (ω)|
ω∈S
(7.45)
Now we have succeeded in specifying the exact ω p , ω s , δ 1 , and δ 2 .In
addition the error can now be distributed uniformly in both the passband
and stopband.
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