Electrical – How to calculate cut-off frequency for T section high pass filter

The -3dB point is your cutoff frequency. It's just standard practice to define it that way. In order to find what your values should be, I'd go with equal element implementation (it's simpler, and you can correct for gain with a simple gain stage later if you need to). Choose R1=R2, C1=C2, and pick a value for either R or C. This yields the following formula for the cutoff frequency: $$f_0=\frac{1}{2\pi RC}$$ I generally choose a value of C initially, as it's easier to find or make a resistor with a strange value, whereas it's more difficult with capacitors. So, set your cutoff frequency equal to f0, and solve for R.

Here's an example: let's say I want a LPF with f0=250 Hz. I'll choose C to be 0.1 micro and solve for R. $$250=\frac{1}{2\pi RC} \rightarrow 250=\frac{1}{2\pi R(0.1x10^{-6})}\rightarrow R\approx6400\Omega.$$

From there, all you need to do is implement your circuit. Once you know what your value for R is supposed to be, you can use a dual-channel potentiometer that has the correct resistance within it's range in place of the two resistors (for the above example, something like a 10k ohm potentiometer would do the trick). This will allow you to change your cutoff frequency, since it's based upon both R and C.

Edit: As Matt Young suggested in the comments, adding a resistor in series with the potentiometer will set the maximum cutoff, and prevent shorts. It's an excellent addition to the circuit, and will keep some sanity when adding the potentiometers.

The usual definition of the cut-off frequency of a (type I) Chebyshev filter is shown in the figure below:

The common practice of defining the cutoff frequency at −3 dB is usually not applied to Chebyshev filters; instead the cutoff is taken as the point at which the gain falls to the value of the ripple for the final time.

Knowing the characteristics of a Chebyshev filter helps in computing the cut-off frequency (as defined above) without explicitly solving the equation \$|H(j\omega_0)|=c\$, where the constant \$c\$ is chosen according to the definition of the cut-off frequency.

The squared magnitude of the frequency response of an \$n^{th}\$ order type I Chebyshev filter is given by

$$|H(j\omega)|^2=\frac{1}{1+\epsilon^2T^2_n(\frac{\omega}{\omega_0})}\tag{1}$$

where \$T_n(\omega)\$ is the \$n^{th}\$-order Chebyshev polynomial of the first kind, \$\omega_0\$ is the cut-off frequency as defined above, and the constant \$\epsilon\$ specifies the pass band ripple, as shown in above figure. You should know the exact value of \$\epsilon\$ from your design specifications, but I can estimate it from your figure: \$\epsilon\approx 0.23403\$ (note that you need to take into account that the maximum of your transfer function is \$\frac12\$ instead of \$1\$, so the smallest (linear) pass band value is given by \$0.5/\sqrt{1+\epsilon^2}\$).

In order to find \$\omega_0\$ we need to compare the expression for the actual transfer function to the one given by (1). It's a basic exercise to show that the transfer function of your filter is

$$H(s)=\frac{\frac12}{\frac{L^2C}{2R}s^3+LCs^2+(\frac{L}{R}+\frac{RC}{2})s+1}\tag{2}$$

Knowing that \$T_3(x)\$ is given by

$$T_3(x)=4x^3-3x\tag{3}$$

we can compare the factors of the highest power of \$\omega\$ (which is \$\omega^6\$) of the denominators of (1) and of the squared magnitude of (2) for \$s=j\omega\$:

$$\frac{16\epsilon^2}{\omega_0^6}=\left(\frac{L^2C}{2R}\right)^2\tag{4}$$

From (4) \$\omega_0\$ can be expressed as

$$\omega_0=\sqrt[3]{\frac{8R\epsilon}{L^2C}}\approx 3829.7\text{ rad/s}\tag{5}$$

where I've used the approximate value of \$\epsilon\$ given above.