Design of Chebyshev Low-Pass Filter

DEEPAK KOTA

800544828

Given,

Max Attenuation, A1 = (log(b+10)) dB

A1 = log( 10* 4 + 8 + 10) = log(58) = 1.7634 dB

Given, The order of the filter , n = 3

The ɛ is given by, ɛ = √ 10𝐴

10 − 1

ɛ = √ 101.7634

10 − 1

ɛ = 0.7077

Given, The filter has to have an equal-ripple pass-band in 0 < f < (c) Hz

The C is given by, C = 2*1000 + 800

C = 2800 Hz

The Cn+1(Ѡ) for the Chebyshev filter is given by,

Cn+1(Ѡ) = 2ѠCn(Ѡ) – Cn-1(Ѡ)

C0(Ѡ) = 1

C1(Ѡ) = Ѡ

As the filter is of order 3,

C2(Ѡ) = 2 ѠC1(Ѡ) – C0(Ѡ) = 2 Ѡ2 - 1

C3(Ѡ) = 2 Ѡ2(Ѡ) – C1(Ѡ)

C3(Ѡ) = 4Ѡ3 - 3Ѡ

The transfer function is given by,

|𝐺(𝑗Ѡ)|2 = 1

1+ ɛ2∗𝐶𝑛( Ѡ)

|𝐺(𝑗Ѡ)|2 = 1

1 + ɛ2 ∗ 𝐶3( Ѡ)

|𝑮(𝒋Ѡ)|𝟐 = 𝟏

𝟏 + (𝟎. 𝟕𝟎𝟕𝟕)𝟐 ∗ ( 𝟒Ѡ𝟑 – 𝟑Ѡ)𝟐

The magnitude characteristic should have an equi-ripple between the frequency 0 < f < 2.8 KHz.

The ripple should not decrease over 1

√ ( 1+ɛ2 ) = 0.8162

The value of the poles of the above transfer function is given by,

Sk = -Sin [(2K+1)𝜋

2𝑛] Sinh[

1

𝑛 Sinh-11

∈ ] + j Cos [ (2K+1)

𝜋

2𝑛] Cosh[

1

𝑛 Sinh-11

∈ ]

The values of K are 0,1 and 2.

When K = 0,

S0 = 0.1956 + j 0.93

When K = 1,

S1 = 0.3911 + j 0

When K = 2,

S3 = S1*

The transfer function is given by

G(s) = |𝑆1|

𝑆− 𝑆1

|𝑆0|2

𝑆2− 2𝑅𝑒[𝑆0]𝑆+|𝑆0|2

G(s) = 𝟎.𝟑𝟗𝟏𝟏

𝑺+ 𝟎.𝟑𝟗𝟏𝟏

𝟎.𝟗𝟎𝟑

𝑺𝟐− 𝟎.𝟑𝟗𝟏𝟐𝑺+𝟎.𝟗𝟎𝟑

The third order chebyshev circuit is obtained by cascading an first order and second order circuits.

The Design of the first order equation is given by comparing it with the basic first order

transfer function,

G1(s) = 1

𝑅𝐶𝑆+1

RC = 1

0.3911

RC = 2.5568

Let us choose, C = 1F

R = 2.5568 Ω

Using the frequency scaling with 2.8KHz, when C = 10nF

R = 2.5568

2𝜋∗2800∗10𝑛

R = 14.5331 KΩ

The below is the second order of Sallen-Key Circuit :

And the transfer function for the second order is

G2(s) = 0.903

𝑆2+ 0.3912𝑆+0.903

G2(s) = 0.95022

𝑆2+ 0.9502

2.4289𝑆+0.95022

Ѡ0 = 0.9502

Q = 2.4289

K = 3 - 1

𝑄

K = 3 - 1

2.4289 = 2.5882

And,

a = 1/K

a = 1/ 2.5882

a = 0.3863

The values of the components are given by,

R2 = 1

Ѡ0𝐶 =

1

2𝜋∗2800∗10𝑛∗0.9502 = 5.982K

R1 = R2 / a

And R4 = R2 / (1-a)

R1 = 5.982/0.3863 = 15.485KΩ

R4 = 5.982/ (1 – 0.3863) = 9.7474KΩ

The value of a results in the dc gain being equal to 0 dB. R5 is a arbitrary parameter and can be

chosen of choice.

R5 = 5KΩ

And R3 = 5(K-1) KΩ

R3 = 5(2.5882 – 1) = 7.941KΩ

These two stages must not be cascaded directly because the first stage cannot be loaded without

affecting its performance.

Circuit Diagram:

1. Magnitude characteristic Vs frequency on lin-lin Scale.

2. Magnitude characteristic Vs frequency f on log scale:

3. Phase Characteristics Vs Frequency on linear scale:

4. Group Delay: