86630-16-13P
AID: 1825 | 20/02/2020
(i)
The passband and stopband ripple values are,
1- δp = 0.8
δs = 0.2
Consider the formula to find the gain term (Gp ) .
Gp =
1
2
(1- δp )
- 1
Here, δp is the pass-band ripple.
Substitute 0.8 for 1- δp .
1
- 1
0.82
= 0.5625
Gp =
Consider the expression to find ripple control factor.
ε = Gp
Substitute 0.5625 for Gp .
ε=
0.5625
= 0.75
Consider the formula to find the gain term (Gs ) .
Gs =
1
2
(δs )
- 1
Here, δp is the pass-band ripple.
Substitute 0.2 for δs .
1
Gs =
- 1
0.22
= 25 - 1
= 24
Find order of the filter N of the Chebyshev polynomial.
1щ
й
2
жG ц
к
чъ
ъ
cosh - 1 кззз s ч
ч
кзиG ш
чъ
p
к
ъ
л
ы
N=
йω щ
cosh - 1 кк s ъ
ъ
кωp ы
ъ
л
Substitute 24 for Gs , 0.5625 for Gp , 0.25π for ωp , and 0.75π for ωs .
1щ
й
ж 24 ч
ц2 ъ
- 1к
cosh кзз
ъ
ч
з 0.5625 ч
шъ
ки
ъ
лк
ы
N=
й
щ
0.75
π
cosh - 1 к
ъ
кл0.25π ы
ъ
2.5639
1.76275
= 1.454
Round off the value to the nearest higher integer that is 2.
=
Consider the expression to find location of the 2N poles of H (S )H (- S ).
ж
ж
ж2n - 1 ц
цц
ж2n - 1 ц
цц
ч
ч
зз 1 sinh- 1 ж
ч
зз 1 sinh- 1 ж
зз 1 ч
зз
зз 1 ч
sn = sin зз
πч
sinh
+
jcos
π
cosh
ч
ч
ч
ч
ч
ч
ч
чш
ч
чш
зи 2 N ш
зεш
з 2N ш
зεш
зиN
зN
ч
ч
и
и
и
и
Here, N is order of the filter and ε is ripple control factor and the range of n is
0 Ј n Ј 2 N - 1.
Substitute 2 for N and 0.75 for ε .
ж2n - 1 ц
ж2n - 1 ц
ж1
ц
ж
ж1 ц
цц
ч
з
ч
ч
ч
зз 1 sinh - 1 ж
зз 1 ч
ч
ч
sn = sin ззз
πч
sinh зз sinh - 1 зз
+
jcos
з
π
cosh
ч
ч
ч
ч
ч
ч
зз 2 (2) ч
чш
чш
зи0.75 ш
з 0.75 ш
зи2
з2
ч
ч
ч
и
з 2 (2) ш
и
и
и
ш
ж
ц
ж
ц
ж2n - 1 ц
цч
ж2n - 1 ц
цч
зз 1 sinh - 1 ж
ч
зз 1 sinh - 1 ж
зз 1 ч
зз
зз 1 ч
= sin зз
πч
sinh
+
jcos
π
cosh
ч
ч
ч
ч
ч
ч
ч
ч
ч
чш
ч
чш
з 4
з 0.75 ш
з 4
з 0.75 ш
зи2
зи2
и
ш
и
и
ш
и
ж2n - 1 ц
ж2n - 1 ц
sn = 0.5773sin зз
πч
+ 1.1547 jcos зз
πч
…… (1)
ч
ч
ч
ч, 1 Ј n Ј 3
з 4
з 4
и
ш
и
ш
Substitute 0 for n in equation (1).
ж2 (0)- 1 ц
ж2 (0)- 1 ц
ч
ч
s0 = 0.5773sin ззз
πч
+ 1.1547 jcos ззз
πч
ч
ч
ч
ч
з 4
з 4
и
ш
и
ш
= - 0.4082 + 0.8165j
Substitute 1 for n in equation (1).
ж2 (1)- 1 ч
ц
ж2 (1)- 1 ч
ц
s1 = 0.5773sin ззз
πч
+ 1.1547 jcos ззз
πч
ч
ч
ч
ч
зи 4
з 4
ш
и
ш
= 0.4082 + 0.8165j
Substitute 2 for n in equation (1).
ж2 (2)- 1 ц
ж2 (2)- 1 ц
зз
ч
ч
s2 = 0.5773sin ззз
πч
+
1.1547
jcos
πч
ч
ч
з
ч
ч
зи 4
з
4
ш
и
ш
= 0.4082 - 0.8165j
Substitute 3 for n in equation (1).
ж2 (3)- 1 ч
ц
ж2 (3)- 1 ц
ч
s3 = 0.5773sin ззз
πч
+ 1.1547 jcos ззз
πч
ч
ч
ч
ч
зи 4
з
ш
и 4
ш
= - 0.4082 - 0.8165j
The poles that are lying in the left-half s-plane are
S = [- 0.4082 + 0.8165j, - 0.4082 - 0.8165j]
The transfer function for the normalized Type I Chebyshev filter is,
K
H (S )=
(S + 0.4082 - 0.8165j)(S + 0.4082 + 0.8165j)
K
H (S )= 2
S + 0.8164S + 0.833
…… (2)
Find Gain K.
Find magnitude of the expression.
K
H (S ) = 2
S + 0.8164S + 0.833
Substitute jω for S.
H ( jω) =
K
2
(jω) + 0.8164 (jω)+ 0.833
Substitute 0 for ω .
K
H (0) =
0.833
Substitute 1 for H (0) .
K = 0.833
Substitute 0.833 for K in equation (2).
0.833
H (S )= 2
S + 0.8164S + 0.833
Consider the formula to find transfer function of the low pass filter.
H (s ) = H (S ) S = s
ωp
0.833
for H (S ) and 0.25π for ωp .
S + 0.8164S + 0.833
0.833
H (s ) = 2
S + 0.8164S + 0.833 S = s
Substitute
2
0.25 π
0.833
=
2
ж s ц
ж s ц
ч
ч
зз
зз
ч
ч
ч + 0.8164 и
ч+ 0.833
зи0.25π ш
з 0.25π ш
0.5138
H (s)= 2
s + 0.6412s + 0.5138
Consider the bilinear transformation.
2 z- 1
s=
T z+ 1
Here, T is the sampling time.
Consider T as 2.
Substitute 2 for T.
2 z- 1
s=
2 z+ 1
z- 1
=
z+ 1
Substitute
z- 1
for s in equation (3).
z+ 1
…… (3)
H (z ) =
0.5138
жz - 1ц
жz - 1ц
ч
ч
зз
зз
ч
ч
ч + 0.6412 и
ч+ 0.5138
зиz + 1ш
з z + 1ш
2
2
=
=
=
0.5138 (z + 1)
2
2
(z - 1) + 0.6412 (z - 1)(z + 1)+ 0.5138(z + 1)
0.5138(z 2 + 2 z + 1)
z 2 - 2 z + 1 + 0.6412 (z 2 - 1)+ 0.5138 (z 2 + 2 z + 1)
0.5138(z 2 + 2 z + 1)
z 2 - 2 z + 1 + 0.6412 (z 2 - 1)+ 0.5138 (z 2 + 2 z + 1)
Simplify further.
0.2384 z 2 + 0.4768 z + 0.2384
H (z ) =
z 2 - 0.4512 z + 0.4049
Consider the MATLAB code to draw the frequency response plot of the Chebyshev Type
I filter.
num=0.833;
den=[1 0.8164 0.833];
[numz,denz]=bilinear(num,den,2); % assume T=1
freqz(numz,denz)
title('DT filter');
Execute the MATLAB program and obtain the output.
Thus,
the
Chebyshev
Type
1
IIR
filter
using
bilinear
transformation
0.2384 z 2 + 0.4768z + 0.2384
and the frequency characteristics is shown in Figure 1.
z 2 - 0.4512 z + 0.4049
(ii)
Consider the expression to find ripple control factor.
1
ε=
Gs
Substitute 24 for Gs .
1
ε=
24
= 0.2041
Find order of the filter N of the Chebyshev polynomial.
1щ
й
2
жG ц
к
чъ
ъ
cosh - 1 кззз s ч
ч
кзиG ш
чъ
p
к
ъ
л
ы
N=
йω щ
cosh - 1 кк s ъ
ъ
клωp ы
ъ
Substitute 24 for Gs , 0.5625 for Gp , 0.25π for ωp , and 0.75π for ωs .
1щ
й
ж 24 ч
ц2 ъ
- 1к
cosh кзз
чъ
шъ
кзи0.5625 ч
ъ
лк
ы
N=
й
щ
0.75π
cosh - 1 к
ъ
ъ
лк0.25π ы
2.5639
1.76275
= 1.454
Round off the value to the nearest higher integer that is 2.
=
Consider the expression to find location of the 2N poles of H (S )H (- S ).
ж
ж
ж2n - 1 ц
цц
ж2n - 1 ц
цц
ч
ч
зз 1 sinh- 1 ж
ч
зз 1 sinh- 1 ж
зз 1 ч
зз
зз 1 ч
sn = sin зз
πч
sinh
+
jcos
π
cosh
ч
ч
ч
ч
ч
ч
ч
ч
ч
чш
ч
чш
зи 2 N ш
зεш
з 2N ш
зεш
зN
зN
и
и
и
и
и
is
Here, N is order of the filter and ε is ripple control factor and the range of n is
0 Ј n Ј 2 N - 1.
Substitute 2 for N and 0.2041 for ε .
ж2n - 1 ц
ж2n - 1 ц
ж1
ц
ж1
ж 1 ц
цц
ч
з
ч
ч
ч
зз sinh - 1 ж
зз 1 ч
ч
ч
sn = sin ззз
πч
sinh зз sinh- 1 зз
+
jcos
з
π
cosh
ч
ч
ч
ч
ч
ч
ч
зз 2 (2) ч
чш
ч
зи0.2041ш
з
зи2
з
ч
ч
и
ш
зи 2 (2) ш
2
0.2041
и
ш
и
ш
ж
ц
ж
ж2n - 1 ц
цч
ж2n - 1 ч
ц
цц
ч
зз 1 sinh- 1 ж
зз 1 sinh- 1 ззж 1 ч
зз 1 ч
зз
= sin зз
πч
sinh
+
jcos
π
cosh
ч
ч
ч
ч
ч
ч
ч
ч
ч
ч
зи 4
з
з
з
з
з
ч
ч
ш и2
и0.2041шш
и 4
ш
и0.2041шш
и2
ж2n - 1 ц
ж2n - 1 ц
зз
…… (4)
sn = 1.4143sin зз
πч
+
1.732
jcos
πч
0Ј nЈ 3
ч
ч
ч
ч,
з 4
з 4
и
ш
и
ш
Substitute 0 for n in equation (4).
ж2 (0)- 1 ц
ж2 (0)- 1 ц
ч
ч
s0 = 1.4143sin ззз
πч
+ 1.732jcos ззз
πч
ч
ч
ч
ч
зи 4
з
ш
и 4
ш
= - 1 + 1.2247 j
Substitute 1 for n in equation (4).
ж2 (1)- 1 ц
ж2 (1)- 1 ц
зз
ч
ч
s1 = 1.4143sin ззз
πч
+
1.732jcos
πч
ч
ч
з
ч
ч
зи 4
з
4
ш
и
ш
= 1 + 1.2247 j
Substitute 2 for n in equation (4).
ж2 (2)- 1 ц
ж2 (2)- 1 ц
зз
ч
ч
s2 = 1.4143sin ззз
πч
+
1.732jcos
πч
ч
ч
з
ч
ч
з 4
з 4
и
ш
и
ш
= 1- 1.2247 j
Substitute 3 for n in equation (4).
ж2 (3)- 1 ц
ж2 (3)- 1 ц
зз
ч
ч
s3 = 1.4143sin ззз
πч
+
1.732jcos
πч
ч
ч
з
ч
ч
зи 4
з
4
ш
и
ш
= - 1- 1.2247 j
The poles are,
S = [- 1+ 1.2247j,1+ 1.2247j,1- 1.2247j, - 1- 1.2247j]
The poles that are lying in the left-half s-plane are
S = [- 1+ 1.2247j, - 1- 1.2247j]
Take inverse on the poles to obtain poles of the normalized Type 2 filter.
й
щ
1
1
ъ
S= к
,
кл- 1 + 1.2247 j - 1- 1.2247 j ы
ъ
= [- 0.4 - 0.4899 j, - 0.4 + 0.4899 j ]
Find zeros of the normalized Chebyshev filter 2.
j
s=
,
0Ј nЈ N - 1
ж2n + 1 ч
ц
cos зз
πч
ч
зи 2 N ш
Substitute 2 for N.
j
s=
,
0Ј nЈ 1
ж2n + 1 ч
ц
cos зз
πч
ч
зи 4
ш
Substitute 0 for n in equation (5).
j
s=
ж2 (0)+ 1 ч
ц
cos ззз
πч
ч
ч
и 4
ш
…… (5)
= 1.414 j
Substitute 1 for n in equation (5).
j
s=
ж2 (1)+ 1 ч
ц
cos ззз
πч
ч
ч
и 4
ш
= - 1.414 j
The transfer function for the normalized Type 2 Chebyshev filter is,
K (S + 1.414 j)(S - 1.414 j)
H (S ) =
(S + 0.4 + 0.4899 j)(S + 0.4 - 0.4899 j)
H (S ) =
K (S 2 + 2)
S 2 + 0.8S + 0.4
Find Gain K.
Find magnitude of the expression.
K
H (S ) = 2
S + 0.8164S + 0.833
Substitute jω for S.
H ( jω) =
K
2
(jω) + 0.8164 (jω)+ 0.833
Substitute 0 for ω .
K
H (0) =
0.833
Substitute 1 for H (0) .
K = 0.833
Substitute 0.833 for K in equation (2).
…… (6)
H (S )=
0.833
S + 0.8164S + 0.833
2
Consider the formula to find transfer function of the low pass filter.
H (s ) = H (S ) S = s
ωp
0.833
for H (S ) and 0.25π for ωp .
S + 0.8164S + 0.833
0.833
H (s ) = 2
S + 0.8164S + 0.833 S = s
Substitute
2
0.25 π
0.833
ж s ц
ж s ц
ч
ч
зз
зз
+
0.8164
ч
ч
ч
ч+ 0.833
з 0.25π ш
з 0.25π ш
и
и
0.5138
H (s)= 2
s + 0.6412s + 0.5138
=
2
…… (3)
Consider the bilinear transformation.
2 z- 1
s=
T z+ 1
Here, T is the sampling time.
Consider T as 2.
Substitute 2 for T.
2 z- 1
s=
2 z+ 1
z- 1
=
z+ 1
z- 1
for s in equation (3).
z+ 1
0.5138
H (z ) =
2
жz - 1ц
жz - 1ц
ч
ч
зз
зз
+
0.6412
ч
ч
ч
ч+ 0.5138
з z + 1ш
з z + 1ш
и
и
Substitute
2
=
=
=
0.5138 (z + 1)
2
2
(z - 1) + 0.6412 (z - 1)(z + 1)+ 0.5138(z + 1)
0.5138(z 2 + 2 z + 1)
z 2 - 2 z + 1 + 0.6412 (z 2 - 1)+ 0.5138 (z 2 + 2 z + 1)
0.5138(z 2 + 2 z + 1)
z 2 - 2 z + 1 + 0.6412 (z 2 - 1)+ 0.5138 (z 2 + 2 z + 1)
Simplify further.
0.2384 z 2 + 0.4768 z + 0.2384
H (z ) =
z 2 - 0.4512 z + 0.4049
Consider the MATLAB code to draw the frequency response plot of the Chebyshev Type
I filter.
num=0.833;
den=[1 0.8164 0.833];
[numz,denz]=bilinear(num,den,2); % assume T=1
freqz(numz,denz)
title('DT filter');
Execute the MATLAB program and obtain the output.
Thus,
the
2
Chebyshev
Type
1
IIR
filter
using
bilinear
transformation
0.2384 z + 0.4768z + 0.2384
and the frequency characteristics is shown in Figure 1.
z 2 - 0.4512 z + 0.4049
is