Elliptic Filters
The magnitude squared frequency response of the normalized low-pass elliptic filter of
order n is defined by
2
Hn( jω ) =
1
1 + ε Rn2 (ω )
2
where Rn (ω ) is a Chebyshev rational function of ω determined from the specified ripple
characteristics
n odd
n even
Figure 1: Magnitude squared frequency response of elliptic LP filters of odd and
even orders.
Unlike the Butterworth and Chebyshev filters, ω =1 is calculated using
ω1ω2 = 1
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A parameter, ω r representing the sharpness of the transition region is defined as
ωr =
ω2
ω1
Thus a large value of ω r indicates a large transition band, while a small value of ω r
indicates a small transition band.
The general transfer function H n (s ) , for the normalized low-pass elliptic filter is given
from odd and even n by
n −1
2
H0
s 2 + A0i
H n (s) =
, odd n
∏
( s + s 0 ) i =1 s 2 + B1i s + B0i
n
2
s 2 + A0i
H n (s) = H 0 ∏ 2
, even n
i =1 s + B1i s + B0 i
To design a filter n, H 0 , S 0 , A0i , B1i , B0i have to be determined from the design
specifications
(1)
ε
(2)
A
(3)
ωr
or equivalently G1 ,G2 and ω r where


1
 = 20 log H n ( jω1 )
G1 = 20 log 
 (1 + ε 2 ) 
 1 
G2 = 20 log  2  = 20 log H n ( jω 2 )
A 
By finding the G1 and G2 in this fashion, the ω r requirement will not be satisfied
exactly; however, an ω r can be selected that exceeds the requirements.
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Determination of n for Normalized Elliptic Filters
The design of a low-pass normalized elliptic filter to satisfy the specifications G1 ,G2 and
ω r using Table 3.6 is straight forward.
(1)
(2)
(3)
(4)
Find G1 and G2 in dB.
Find the ω r portion of the table for which a value less than the determined ω r is
found.
Find the corresponding order n.
The values of ω1 and ω 2 are obtained by
ω1 =
1
ωr
and
ω2 = ωr
Thus it can be clearly verified that the requirements have been met.
The design of an un-normalized elliptic low-pass filter satisfying a G1 dB ripple, cutoff at
ω1' and a G2 dB gain at ω 2' can be obtained by LP ->LP transformation of a mutable
normalized elliptic filter
H ( s ) = H LP ( s ) |
s→
s
ω0
For the elliptic filter ω 0 is selected to be the geometric mean of ω1' and ω 2'
ω 0 = ω1'ω 2'
The corresponding LP requirements, ω1 , ω 2 and ω r (transition) are obtained as follows
ω1'
ω1 =
ω0
ω2 =
ω 2'
ω0
ωr =
ω 2 ω 2'
=
ω1 ω1'
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Example #1
Find the transfer function for an elliptic low-pass filter with –2 dB cutoff value at 10,000
rad/s and a stop band attenuation of 40 dB for all ω past 14,400 rad/s.
Solution
ω1' = 10,000rad / s
G1 = −2dB
ω 2' = 14,400rad / s
G2 = −40dB
Then
ω 0 = ω 2' ω1' = (1 *10 4 )(1.44 *10 4 ) = 12,000
ω1' 10,000 5
=
=
ω1 =
ω 0 12,000 6
ω2 =
ω 2' 14,400 6
=
=
ω 0 12,000 5
6
ω
ω r = 2 = 5 = 1.44
ω1 5
6
From the –2 dB and –40 dB part of the table it is seen that for n = 4 given ω r = 1.40542.
Thus
H LP ( s ) =
0.01( s 2 + 7.25202)( s 2 + 1.57676)
( s 2 + 0.467290s + 0.212344)( s 2 + 0.127954s + 0.677934)
The required LP is obtained by
H ( s ) = H LP ( s ) |
s→
s
12 , 000
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Example #2
Find the transfer function H(s) for a normalized elliptic filter that will satisfy the
following conditions:
G1 = −0.5dB
G2 = −30dB
ω r = 1.21
From table 3.6a with pass-band ripple of –0.5 dB and stop-band gain of –30 dB, the
smallest value of n that satisfies the ω r is n = 5, giving ω r = 1.12912.
H 5 ( s) =
0.118807( s 2 + 2.14490)( s 2 + 1.18122)
( s + 0.511701)( s 2 + 0.480774s + 0.648724)( s 2 + 0.088080s + 0.907216)
Where
ω1 =
1
ωr
= 0.941087
ω 2 = ω r = 1.12912 = 1.06260
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