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Normalized Butterworth and Chebyshev Functions
APPENDIX
NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
C
C.1 NORMALIZED BUTTERWORTH FUNCTION The normalized Butterworth squared magnitude function is given by jPn ðωÞj2 ¼
1 1 + ε2 ðωÞ2n
,
(C.1)
where n is the order and ε is the specified ripple on filter passband. The specified ripple in dB is expressed as εdB ¼ 10 log10(1 + ε2) dB. To develop the transfer function Pn(s), we first let s ¼ jω and then substitute ω2 ¼ s2 into Eq. (C.1) to obtain Pn ðsÞPn ðsÞ ¼
1 : 1 + ε2 ðs2 Þn
(C.2)
Eq. (C.2) has 2n poles, and Pn(s) has n poles on the left-hand half plane (LHHP) on the s-plane, while Pn( s) has n poles on the right-hand half plane (RHHP) on the s-plane. Solving for poles leads to ð1Þn s2n ¼ 1=ε2 :
(C.3)
If n is an odd number, Eq. (C.3) becomes s2n ¼ 1=ε2
and the corresponding poles are solved as 2πk
pk ¼ ε1=n ej 2n ¼ ε1=n ½ cos ð2πk=2nÞ + jsin ð2πk=2nÞ,
(C.4)
where k ¼ 0, 1, ⋯, 2n 1. Thus in the phasor form, we have r ¼ ε1=n , and θk ¼ 2πk=ð2nÞ for k ¼ 0,1, ⋯,2n 1:
(C.5)
When n is an even number, it follows that s2n ¼ 1=ε2 pk ¼ ε1=n ej
2πk + π 2n
¼ ε1=n ½ cos ðð2πk + π Þ=2nÞ + j sin ðð2πk + π Þ=2nÞ,
(C.6)
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APPENDIX C NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
where k ¼ 0, 1, ⋯, 2n 1. Similarly, the phasor form is given by r ¼ ε1=n , and θk ¼ ð2πk + π Þ=ð2nÞ for k ¼ 0, 1,⋯,2n 1:
(C.7)
When n is an odd number, we can identify the poles on the LHHP as pk ¼ r, k ¼ 0 and : pk ¼ r cos ðθk Þ + jr sin ðθk Þ,k ¼ 1,⋯, ðn 1Þ=2
(C.8)
Using complex conjugate pairs, we have p∗k ¼ r cos ðθk Þ jr sin ðθk Þ:
Note that ðs pk Þ s p∗k ¼ s2 + ð2r cos ðθk ÞÞs + r 2 ,
and a factor from the real pole (s + r), it follows that K
Pn ðsÞ ¼ ðs + r Þ
ðn1 QÞ=2
ðs2
(C.9)
+ ð2r cos ðθk
ÞÞs + r 2 Þ
k¼1
and θk ¼ 2πk/(2n) for k ¼ 1, ⋯, (n 1)/2. Setting Pn(0) ¼ 1 for the unit passband gain leads to K ¼ r n ¼ 1=ε:
When n is an even number, we can identify the poles on the LHHP as pk ¼ r cos ðθk Þ + jr sin ðθk Þ, k ¼ 0, 1,⋯, n=2 1:
(C.10)
Using complex conjugate pairs, we have p∗k ¼ r cos ðθk Þ jr sin ðθk Þ:
The transfer function is given by Pn ðsÞ ¼
K n=21 Y
s2 + ð2r cos ðθk ÞÞs + r2
k¼0
θk ¼ ð2πk + π Þ=ð2nÞ for k ¼ 0, 1,⋯,n=2 1
Setting Pn(0) ¼ 1 for the unit passband gain, we have K ¼ r n ¼ 1=ε:
:
(C.11)
C.1 NORMALIZED BUTTERWORTH FUNCTION
Let us examine the following examples. EXAMPLE C.1 Compute the normalized Butterworth transfer function for the following specifications: Ripple ¼ 3 dB n¼2
Solution: n=2 ¼ 1 θk ¼ ð2π 0 + π Þ=ð2 2Þ ¼ 0:25π ε2 ¼ 100:13 1, r ¼ 1, and K ¼ 1:
Applying Eq. (C.11) leads to P2 ðsÞ ¼
1 1 : ¼ s2 + 2 1 cos ð0:25π Þs + 12 s2 + 1:4141s + 1
EXAMPLE C.2 Compute the normalized Butterworth transfer function for the following specifications: Ripple ¼ 3 dB n¼3
Solution: ðn 1Þ=2 ¼ 1 ε ¼ 100:13 1, 2
r ¼ 1, and K ¼ 1 θk ¼ ð2π 1Þ=ð2 3Þ ¼ π=3:
From Eq. (C.9), we have 1 ðs + 1Þ s2 + 2 1 cos ðπ=3Þs + 12 1 : ¼ ðs + 1Þðs2 + s + 1Þ
P3 ðsÞ ¼
For the unfactored form, we can carry out P3 ðsÞ ¼
s3
1 : + 2s + 1
+ 2s2
827
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APPENDIX C NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
EXAMPLE C.3 Compute the normalized Butterworth transfer function for the following specifications: Ripple ¼ 1.5 dB n¼3
Solution: ðn 1Þ=2 ¼ 1 ε2 ¼ 100:11:5 1, r ¼ 1:1590, and K ¼ 1:5569 θk ¼ ð2π 1Þ=ð2 3Þ ¼ π=3:
Applying Eq. (C.9), we achieve the normalized Butterworth transfer function as 1 ðs + 1:1590Þ s2 + 2 1:1590 cos ðπ=3Þs + 1:15902 1 ¼ : ðs + 1Þðs2 + 1:1590s + 1:3433Þ
P3 ðsÞ ¼
For the unfactored form, we can carry out P3 ðsÞ ¼
1:5569 : s3 + 2:3180s2 + 2:6866s + 1:5569
C.2 NORMALIZED CHEBYSHEV FUNCTION Similar to analog Butterworth filter design, the transfer function is derived from the normalized Chebyshev function, and the result is usually listed in a table for design reference. The Chebyshev magnitude response function with an order of n and the normalized cutoff frequency ω ¼ 1 radian per second is given by 1 jBn ðωÞj ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ,n 1, 1 + ε2 C2n ðωÞ
where the function Cn(ω) is defined as Cn ðωÞ ¼
cos ðncos 1 ðωÞÞ ω 1 , cosh ðncosh 1 ðωÞÞ ω > 1
(C.12)
(C.13)
where ε is the ripple specification on the filter passband. Note that pffiffiffiffiffiffiffiffiffiffiffiffi cosh 1 ðxÞ ¼ ln x + x2 1 :
(C.14)
To develop the transfer function Bn(s), we let s ¼ jω and substitute ω2 ¼ s2 into Eq. (C.12) to obtain Bn ðsÞBn ðsÞ ¼
1 : 1 + ε2 C2n ðs=jÞ
(C.15)
C.2 NORMALIZED CHEBYSHEV FUNCTION
829
The poles can be found from 1 + ε2 C2n ðs=jÞ ¼ 0
or Cn ðs=jÞ ¼ cos n cos 1 ðs=jÞ ¼ j1=ε:
(C.16)
Introduce a complex variable v ¼ α + jβ such that v ¼ α + jβ ¼ cos 1 ðs=jÞ,
(C.17)
s ¼ jcos ðvÞ:
(C.18)
we can then write
Substituting Eq. (C.17) into Eq, (C.16) and using trigonometric identities, it follows that Cn ðs=jÞ ¼ cos ðn cos 1 ðs=jÞÞ ¼ cos ðnvÞ ¼ cos ðnα + jnβÞ ¼ cos ðnαÞ cosh ðnβÞ jsin ðnαÞ sinh ðnβÞ ¼ j1=ε:
(C.19)
To solve Eq. (C.19), the following conditions must be satisfied: cos ðnαÞcosh ðnβÞ ¼ 0
(C.20)
sin ðnαÞ sinh ðnβÞ ¼ 1=ε:
(C.21)
Since cosh(nβ) 1 in Eq. (C.20), we must let cos ðnαÞ ¼ 0,
(C.22)
αk ¼ ð2k + 1Þπ=ð2nÞ,k ¼ 0,1, 2,⋯,2n 1:
(C.23)
which therefore leads to With Eq. (C.23), we have sin(nαk) ¼ 1. Then Eq. (C.21) becomes sinh ðnβÞ ¼ 1=ε:
(C.24)
β ¼ sinh 1 ð1=εÞ=n:
(C.25)
s ¼ j cos ðvÞ ¼ j½ cos ðαk Þcosh ðβÞ j sin ðαk Þ sinh ðβÞ : for k ¼ 0, 1,⋯, 2n 1
(C.26)
Solving Eq. (C.24) gives
Again from Eq. (C.18),
The poles can be found from Eq. (C.26): pk ¼ sin ðαk Þ sinh ðβÞ + j cos ðαk Þcosh ðβÞ : for k ¼ 0, 1,⋯, 2n 1
(C.27)
Using Eq. (C.27), if n is an odd number, the poles on the left-hand side are solved to be pk ¼ sin ðαk Þ sinh ðβÞ + jcos ðαk Þ cosh ðβÞ for k ¼ 0,1,⋯, 2n 1:
(C.28)
830
APPENDIX C NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
Using complex conjugate pairs, we have p∗k ¼ sin ðαk Þ sinh ðβÞ jcos ðαk Þ cosh ðβÞ
(C.29)
pk ¼ sinh ðβÞ,k ¼ ðn 1Þ=2:
(C.30)
ðs pk Þ s p∗k ¼ s2 + bk s + ck
(C.31)
and a real pole
Note that
and a factor from the real pole [s + sinh(β)], it follows that K
Bn ðsÞ ¼ ½s + sinh ðβÞ
ðn1Q Þ=21
, ðs2
(C.32)
+ bk s + ck Þ
k¼0
where bk ¼ 2 sin ðαk Þ sinh ðβÞ
(C.33)
ck ¼ ½ sin ðαk Þsinh ðβÞ2 + ½ cos ðαk Þ cosh ðβÞ2
(C.34)
αk ¼ ð2k + 1Þπ=ð2nÞ for k ¼ 0,1, ⋯, ðn 1Þ=2 1:
(C.35)
For the unit passband gain and the filter order as an odd number, we set Bn(0) ¼ 1. Then K ¼ sinh ðβÞ
ðn1 Þ=21 Y
ck
(C.36)
k¼0
β ¼ sinh 1 ð1=εÞ=n pffiffiffiffiffiffiffiffiffiffiffi sinh 1 ðxÞ ¼ ln x + x2 + 1 :
(C.37) (C.38)
Following the similar procedure for the even number of n, we have Bn ðsÞ ¼
K n=21 Q
(C.39)
ðs2 + bk s + ck Þ
k¼0
bk ¼ 2 sin ðαk Þ sinh ðβÞ
(C.40)
ck ¼ ½ sin ðαk Þsinh ðβÞ2 + ½ cos ðαk Þ cosh ðβÞ2
(C.41)
where αk ¼ ð2k + 1Þπ=ð2nÞ for k ¼ 0,1, ⋯,n=2 1:
(C.42)
pffiffiffiffiffiffiffiffiffiffiffi For the unit passband gain and the filter order as an even number, we require that Bn ð0Þ ¼ 1= 1 + ε2 , so that the maximum magnitude of the ripple on passband equals 1. Then we have K¼
n=21 Y
pffiffiffiffiffiffiffiffiffiffiffi ck = 1 + ε 2
(C.43)
k¼0
β ¼ sinh 1 ð1=εÞ=n
(C.44)
C.2 NORMALIZED CHEBYSHEV FUNCTION pffiffiffiffiffiffiffiffiffiffiffi sinh 1 ðxÞ ¼ ln x + x2 + 1 :
831
(C.45)
Eqs. (C.32) to (C.45) are applied to compute the normalized Chebyshev transfer function. Now let us look at the following illustrative examples. EXAMPLE C.4 Compute the normalized Chebyshev transfer function for the following specifications: Ripple ¼ 0.5 dB n¼2
Solution: n=2 ¼ 1:
Applying Eqs. (C.39) to (C.45), we obtain α0 ¼ ð2 0 + 1Þπ=ð2 2Þ ¼ 0:25π ε ¼ 100:10:5 1 ¼ 0:1220,1=ε ¼ 2:8630 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi β ¼ sinh 1 ð2:8630Þ=n ¼ ln 2:8630 + 2:86302 + 1 =2 ¼ 0:8871 2
b0 ¼ 2 sin ð0:25π Þ sinh ð0:8871Þ ¼ 1:4256 c0 ¼ ½ sin ð0:25π Þ sinh ð0:8871Þ2 + ½ cos ð0:25π Þ cosh ð0:8871Þ2 ¼ 1:5162 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K ¼ 1:5162= 1 + 0:1220 ¼ 1:4314:
Finally, the transfer function is derived as B2 ðsÞ ¼
1::4314 : s2 + 1:4256s + 1:5162
EXAMPLE C.5 Compute the normalized Chebyshev transfer function for the following specifications: Ripple ¼ 1 dB n¼3
Solution: ðn 1Þ=2 ¼ 1:
Applying Eqs. (C.32) to (C.38) leads to α0 ¼ ð2 0 + 1Þπ=ð2 3Þ ¼ π=6 ε2 ¼ 100:11 1 ¼ 0:2589, 1=ε ¼ 1:9653 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi β ¼ sinh 1 ð1:9653Þ=n ¼ ln 1:9653 + 1:96532 + 1 =3 ¼ 0:4760 b0 ¼ 2 sin ðπ=6Þ sinh ð0:4760Þ ¼ 0:4942
832
APPENDIX C NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
c0 ¼ ½ sin ðπ=6Þ sinh ð0:4760Þ2 + ½ cos ðπ=6Þcosh ð0:4760Þ2 ¼ 0:9942 sinh ðβÞ ¼ sinh ð0:4760Þ ¼ 0:4942 K ¼ 0:4942 0:9942 ¼ 0:4913:
We can conclude the transfer function as B3 ðsÞ ¼
0:4913 : ðs + 0:4942Þðs2 + 0:4942s + 0:9942Þ
Finally, the unfactored form is found to be B3 ðsÞ ¼
0:4913 : s3 + 0:9883s2 + 1:2384s + 0:4913
NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
C
C.1 NORMALIZED BUTTERWORTH FUNCTION The normalized Butterworth squared magnitude function is given by jPn ðωÞj2 ¼
1 1 + ε2 ðωÞ2n
,
(C.1)
where n is the order and ε is the specified ripple on filter passband. The specified ripple in dB is expressed as εdB ¼ 10 log10(1 + ε2) dB. To develop the transfer function Pn(s), we first let s ¼ jω and then substitute ω2 ¼ s2 into Eq. (C.1) to obtain Pn ðsÞPn ðsÞ ¼
1 : 1 + ε2 ðs2 Þn
(C.2)
Eq. (C.2) has 2n poles, and Pn(s) has n poles on the left-hand half plane (LHHP) on the s-plane, while Pn( s) has n poles on the right-hand half plane (RHHP) on the s-plane. Solving for poles leads to ð1Þn s2n ¼ 1=ε2 :
(C.3)
If n is an odd number, Eq. (C.3) becomes s2n ¼ 1=ε2
and the corresponding poles are solved as 2πk
pk ¼ ε1=n ej 2n ¼ ε1=n ½ cos ð2πk=2nÞ + jsin ð2πk=2nÞ,
(C.4)
where k ¼ 0, 1, ⋯, 2n 1. Thus in the phasor form, we have r ¼ ε1=n , and θk ¼ 2πk=ð2nÞ for k ¼ 0,1, ⋯,2n 1:
(C.5)
When n is an even number, it follows that s2n ¼ 1=ε2 pk ¼ ε1=n ej
2πk + π 2n
¼ ε1=n ½ cos ðð2πk + π Þ=2nÞ + j sin ðð2πk + π Þ=2nÞ,
(C.6)
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APPENDIX C NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
where k ¼ 0, 1, ⋯, 2n 1. Similarly, the phasor form is given by r ¼ ε1=n , and θk ¼ ð2πk + π Þ=ð2nÞ for k ¼ 0, 1,⋯,2n 1:
(C.7)
When n is an odd number, we can identify the poles on the LHHP as pk ¼ r, k ¼ 0 and : pk ¼ r cos ðθk Þ + jr sin ðθk Þ,k ¼ 1,⋯, ðn 1Þ=2
(C.8)
Using complex conjugate pairs, we have p∗k ¼ r cos ðθk Þ jr sin ðθk Þ:
Note that ðs pk Þ s p∗k ¼ s2 + ð2r cos ðθk ÞÞs + r 2 ,
and a factor from the real pole (s + r), it follows that K
Pn ðsÞ ¼ ðs + r Þ
ðn1 QÞ=2
ðs2
(C.9)
+ ð2r cos ðθk
ÞÞs + r 2 Þ
k¼1
and θk ¼ 2πk/(2n) for k ¼ 1, ⋯, (n 1)/2. Setting Pn(0) ¼ 1 for the unit passband gain leads to K ¼ r n ¼ 1=ε:
When n is an even number, we can identify the poles on the LHHP as pk ¼ r cos ðθk Þ + jr sin ðθk Þ, k ¼ 0, 1,⋯, n=2 1:
(C.10)
Using complex conjugate pairs, we have p∗k ¼ r cos ðθk Þ jr sin ðθk Þ:
The transfer function is given by Pn ðsÞ ¼
K n=21 Y
s2 + ð2r cos ðθk ÞÞs + r2
k¼0
θk ¼ ð2πk + π Þ=ð2nÞ for k ¼ 0, 1,⋯,n=2 1
Setting Pn(0) ¼ 1 for the unit passband gain, we have K ¼ r n ¼ 1=ε:
:
(C.11)
C.1 NORMALIZED BUTTERWORTH FUNCTION
Let us examine the following examples. EXAMPLE C.1 Compute the normalized Butterworth transfer function for the following specifications: Ripple ¼ 3 dB n¼2
Solution: n=2 ¼ 1 θk ¼ ð2π 0 + π Þ=ð2 2Þ ¼ 0:25π ε2 ¼ 100:13 1, r ¼ 1, and K ¼ 1:
Applying Eq. (C.11) leads to P2 ðsÞ ¼
1 1 : ¼ s2 + 2 1 cos ð0:25π Þs + 12 s2 + 1:4141s + 1
EXAMPLE C.2 Compute the normalized Butterworth transfer function for the following specifications: Ripple ¼ 3 dB n¼3
Solution: ðn 1Þ=2 ¼ 1 ε ¼ 100:13 1, 2
r ¼ 1, and K ¼ 1 θk ¼ ð2π 1Þ=ð2 3Þ ¼ π=3:
From Eq. (C.9), we have 1 ðs + 1Þ s2 + 2 1 cos ðπ=3Þs + 12 1 : ¼ ðs + 1Þðs2 + s + 1Þ
P3 ðsÞ ¼
For the unfactored form, we can carry out P3 ðsÞ ¼
s3
1 : + 2s + 1
+ 2s2
827
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APPENDIX C NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
EXAMPLE C.3 Compute the normalized Butterworth transfer function for the following specifications: Ripple ¼ 1.5 dB n¼3
Solution: ðn 1Þ=2 ¼ 1 ε2 ¼ 100:11:5 1, r ¼ 1:1590, and K ¼ 1:5569 θk ¼ ð2π 1Þ=ð2 3Þ ¼ π=3:
Applying Eq. (C.9), we achieve the normalized Butterworth transfer function as 1 ðs + 1:1590Þ s2 + 2 1:1590 cos ðπ=3Þs + 1:15902 1 ¼ : ðs + 1Þðs2 + 1:1590s + 1:3433Þ
P3 ðsÞ ¼
For the unfactored form, we can carry out P3 ðsÞ ¼
1:5569 : s3 + 2:3180s2 + 2:6866s + 1:5569
C.2 NORMALIZED CHEBYSHEV FUNCTION Similar to analog Butterworth filter design, the transfer function is derived from the normalized Chebyshev function, and the result is usually listed in a table for design reference. The Chebyshev magnitude response function with an order of n and the normalized cutoff frequency ω ¼ 1 radian per second is given by 1 jBn ðωÞj ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ,n 1, 1 + ε2 C2n ðωÞ
where the function Cn(ω) is defined as Cn ðωÞ ¼
cos ðncos 1 ðωÞÞ ω 1 , cosh ðncosh 1 ðωÞÞ ω > 1
(C.12)
(C.13)
where ε is the ripple specification on the filter passband. Note that pffiffiffiffiffiffiffiffiffiffiffiffi cosh 1 ðxÞ ¼ ln x + x2 1 :
(C.14)
To develop the transfer function Bn(s), we let s ¼ jω and substitute ω2 ¼ s2 into Eq. (C.12) to obtain Bn ðsÞBn ðsÞ ¼
1 : 1 + ε2 C2n ðs=jÞ
(C.15)
C.2 NORMALIZED CHEBYSHEV FUNCTION
829
The poles can be found from 1 + ε2 C2n ðs=jÞ ¼ 0
or Cn ðs=jÞ ¼ cos n cos 1 ðs=jÞ ¼ j1=ε:
(C.16)
Introduce a complex variable v ¼ α + jβ such that v ¼ α + jβ ¼ cos 1 ðs=jÞ,
(C.17)
s ¼ jcos ðvÞ:
(C.18)
we can then write
Substituting Eq. (C.17) into Eq, (C.16) and using trigonometric identities, it follows that Cn ðs=jÞ ¼ cos ðn cos 1 ðs=jÞÞ ¼ cos ðnvÞ ¼ cos ðnα + jnβÞ ¼ cos ðnαÞ cosh ðnβÞ jsin ðnαÞ sinh ðnβÞ ¼ j1=ε:
(C.19)
To solve Eq. (C.19), the following conditions must be satisfied: cos ðnαÞcosh ðnβÞ ¼ 0
(C.20)
sin ðnαÞ sinh ðnβÞ ¼ 1=ε:
(C.21)
Since cosh(nβ) 1 in Eq. (C.20), we must let cos ðnαÞ ¼ 0,
(C.22)
αk ¼ ð2k + 1Þπ=ð2nÞ,k ¼ 0,1, 2,⋯,2n 1:
(C.23)
which therefore leads to With Eq. (C.23), we have sin(nαk) ¼ 1. Then Eq. (C.21) becomes sinh ðnβÞ ¼ 1=ε:
(C.24)
β ¼ sinh 1 ð1=εÞ=n:
(C.25)
s ¼ j cos ðvÞ ¼ j½ cos ðαk Þcosh ðβÞ j sin ðαk Þ sinh ðβÞ : for k ¼ 0, 1,⋯, 2n 1
(C.26)
Solving Eq. (C.24) gives
Again from Eq. (C.18),
The poles can be found from Eq. (C.26): pk ¼ sin ðαk Þ sinh ðβÞ + j cos ðαk Þcosh ðβÞ : for k ¼ 0, 1,⋯, 2n 1
(C.27)
Using Eq. (C.27), if n is an odd number, the poles on the left-hand side are solved to be pk ¼ sin ðαk Þ sinh ðβÞ + jcos ðαk Þ cosh ðβÞ for k ¼ 0,1,⋯, 2n 1:
(C.28)
830
APPENDIX C NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
Using complex conjugate pairs, we have p∗k ¼ sin ðαk Þ sinh ðβÞ jcos ðαk Þ cosh ðβÞ
(C.29)
pk ¼ sinh ðβÞ,k ¼ ðn 1Þ=2:
(C.30)
ðs pk Þ s p∗k ¼ s2 + bk s + ck
(C.31)
and a real pole
Note that
and a factor from the real pole [s + sinh(β)], it follows that K
Bn ðsÞ ¼ ½s + sinh ðβÞ
ðn1Q Þ=21
, ðs2
(C.32)
+ bk s + ck Þ
k¼0
where bk ¼ 2 sin ðαk Þ sinh ðβÞ
(C.33)
ck ¼ ½ sin ðαk Þsinh ðβÞ2 + ½ cos ðαk Þ cosh ðβÞ2
(C.34)
αk ¼ ð2k + 1Þπ=ð2nÞ for k ¼ 0,1, ⋯, ðn 1Þ=2 1:
(C.35)
For the unit passband gain and the filter order as an odd number, we set Bn(0) ¼ 1. Then K ¼ sinh ðβÞ
ðn1 Þ=21 Y
ck
(C.36)
k¼0
β ¼ sinh 1 ð1=εÞ=n pffiffiffiffiffiffiffiffiffiffiffi sinh 1 ðxÞ ¼ ln x + x2 + 1 :
(C.37) (C.38)
Following the similar procedure for the even number of n, we have Bn ðsÞ ¼
K n=21 Q
(C.39)
ðs2 + bk s + ck Þ
k¼0
bk ¼ 2 sin ðαk Þ sinh ðβÞ
(C.40)
ck ¼ ½ sin ðαk Þsinh ðβÞ2 + ½ cos ðαk Þ cosh ðβÞ2
(C.41)
where αk ¼ ð2k + 1Þπ=ð2nÞ for k ¼ 0,1, ⋯,n=2 1:
(C.42)
pffiffiffiffiffiffiffiffiffiffiffi For the unit passband gain and the filter order as an even number, we require that Bn ð0Þ ¼ 1= 1 + ε2 , so that the maximum magnitude of the ripple on passband equals 1. Then we have K¼
n=21 Y
pffiffiffiffiffiffiffiffiffiffiffi ck = 1 + ε 2
(C.43)
k¼0
β ¼ sinh 1 ð1=εÞ=n
(C.44)
C.2 NORMALIZED CHEBYSHEV FUNCTION pffiffiffiffiffiffiffiffiffiffiffi sinh 1 ðxÞ ¼ ln x + x2 + 1 :
831
(C.45)
Eqs. (C.32) to (C.45) are applied to compute the normalized Chebyshev transfer function. Now let us look at the following illustrative examples. EXAMPLE C.4 Compute the normalized Chebyshev transfer function for the following specifications: Ripple ¼ 0.5 dB n¼2
Solution: n=2 ¼ 1:
Applying Eqs. (C.39) to (C.45), we obtain α0 ¼ ð2 0 + 1Þπ=ð2 2Þ ¼ 0:25π ε ¼ 100:10:5 1 ¼ 0:1220,1=ε ¼ 2:8630 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi β ¼ sinh 1 ð2:8630Þ=n ¼ ln 2:8630 + 2:86302 + 1 =2 ¼ 0:8871 2
b0 ¼ 2 sin ð0:25π Þ sinh ð0:8871Þ ¼ 1:4256 c0 ¼ ½ sin ð0:25π Þ sinh ð0:8871Þ2 + ½ cos ð0:25π Þ cosh ð0:8871Þ2 ¼ 1:5162 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi K ¼ 1:5162= 1 + 0:1220 ¼ 1:4314:
Finally, the transfer function is derived as B2 ðsÞ ¼
1::4314 : s2 + 1:4256s + 1:5162
EXAMPLE C.5 Compute the normalized Chebyshev transfer function for the following specifications: Ripple ¼ 1 dB n¼3
Solution: ðn 1Þ=2 ¼ 1:
Applying Eqs. (C.32) to (C.38) leads to α0 ¼ ð2 0 + 1Þπ=ð2 3Þ ¼ π=6 ε2 ¼ 100:11 1 ¼ 0:2589, 1=ε ¼ 1:9653 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi β ¼ sinh 1 ð1:9653Þ=n ¼ ln 1:9653 + 1:96532 + 1 =3 ¼ 0:4760 b0 ¼ 2 sin ðπ=6Þ sinh ð0:4760Þ ¼ 0:4942
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APPENDIX C NORMALIZED BUTTERWORTH AND CHEBYSHEV FUNCTIONS
c0 ¼ ½ sin ðπ=6Þ sinh ð0:4760Þ2 + ½ cos ðπ=6Þcosh ð0:4760Þ2 ¼ 0:9942 sinh ðβÞ ¼ sinh ð0:4760Þ ¼ 0:4942 K ¼ 0:4942 0:9942 ¼ 0:4913:
We can conclude the transfer function as B3 ðsÞ ¼
0:4913 : ðs + 0:4942Þðs2 + 0:4942s + 0:9942Þ
Finally, the unfactored form is found to be B3 ðsÞ ¼
0:4913 : s3 + 0:9883s2 + 1:2384s + 0:4913