7th Order Chebyshev Low Pass Filter Design and Simulation

Background on Chebyshev Filters and Polynomials

    The design of filters with op-amps can be done using Chebyshev polynomials. Such a filter designed using these polynomials is called a Chebyshev filter. Filters of nth order can be designed by using their corresponding nth order polynomials and using an amplifier's transfer functions to determine each component's value. These type of filters have a steep roll-off from the passband to the stopband compared to other filters. They have one main drawback which is that they have some ripple along the bandpass. Depending on the application, this might not be ideal.

Specifications 

In this post, I will be designing and simulating a 7th order low-pass filter with the following specifications:

-1dB ripple

-7th Order

-280Hz cutoff

-passband gain of 30

Amplifier Building Blocks

For filters with order higher than 4, the design process doesn't change much. The nth order filter can be constructed using a combination of 1st order and 2nd order LPFs.  Below are the two configurations for a 1st order and 2nd order LPF that are building blocks to design an nth order Chebyshev filter. 

Figure 1: First order configuration

Figure 2: Second order configuration

Calculations

Transfer functions and passband gain

The generalized transfer function of a Chebyshev LPF with a cutoff ω0 in rad/s is given by EQ. 1:

 EQ. 1

Where Qn is the Chebyshev polynomial of the nth degree and K is related to the passband gain Ap as shown in EQ. 2:

EQ. 2

To meet the specifications, a 7th order polynomial must be used. Using a table of Chebyshev polynomials, Q7 is as shown in EQ. 3:

 EQ. 3

After selecting the appropriate polynomial, K can be found by computing the value of Q7(0) since the passband gain Ap = 30.

Since Q7(0) is now known, its value can be substituted into EQ. 2 to solve for K

Next, we re-write equation Q7(s) as:

Before proceeding, since I don't like dealing with large decimal values when manipulating equations algebraically by hand, I will shorten the notation by making the following substitutions to the previous decimal values:

Now the equation becomes EQ. 4:

EQ. 4

Then, we put the equation in standard form by multiplying by the highest order common denominator. In this case by:

and we get our standard form in EQ. 5:

EQ. 5

Using EQ. 5, we can substitute it into EQ. 1 to get our expanded transfer function below in EQ. 6:

EQ. 6

To further simplify the notation, I will assign each product term in the denominator to a letter as follows:

Then the form of the transfer function in EQ. 6 can be generalized as EQ. 7 below: 

EQ. 7

    In other words, it's just the product of the transfer functions with their respective denominators. Additionally, from here we see that the design will have a total of 4 stages which consist of 1 first order LPF as denoted by A, and 3 second order LPFs denoted by B-D. 

The transfer function of a first-order configuration as shown in Figure 1 is shown below in EQ. 8

 EQ. 8

The transfer function of a second-order configuration shown in Figure 2 is shown below in EQ. 9

EQ. 9

Substituting EQ. 8 for the first order term corresponding to A, and EQ. 9 for the remaining second order terms in EQ. 7 makes our transfer function in terms of components the one in EQ. 10 below:

EQ. 10

To design the circuit, all the values of the unknowns must be found. Since capacitors are harder to finely tune or get in exact values I decided to choose a standard value for them. To further simplify the design process I also chose all capacitor values to be equal. 

Which means that now the only values that need to be solved for are :

Solving for RA:

To solve for Ra we compare both version of our transfer functions term by term. Comparing EQ. 6 to EQ. 10 we can see by inspection the following relation:

This leaves the numerator term 

 

which we will leave until the end to solve for. 

Solving for RB and AOB
To solve for RB and AOB we compare the transfer functions again and look at the terms corresponding to B to see the relation:

Thus, we solve for the values

Solving for AOC and RC

To solve for AOC and RC we follow the same process of comparing terms and see the following relation:

Then

Solving for AOD and RD

Again, we repeat the same process and see:

Which results in 

Solving for AOA

Finally, we can solve for AOA by recognizing that the product of the numerators of each transfer function stage has the following relation:

Schematic 

We can now construct the schematic in Tina-TI with the calculated values

Using the following MATLAB we can simulate get the Bode Magnitude plot of the system and get the expected cutoff frequency at 280Hz.

We can confirm by running a frequency response simulation on Tina-Ti:

Conclusion

In conclusion, the process for higher order nth degree polynomials is tedious to do by hand but it still proves to be accurate. Automation of calculations can be done purely through symbolic mathematics, but I refrained from using it in this application as I enjoy long derivations. The frequency response of the circuit was simulated using MATLAB and Tina-TI which showed very similar results.