The natural resonant frequency of a 2nd order low pass filter will have a pole zero diagram like this: -

If you are unfamiliar with pole-zero diagrams see if this helps: -

If you follow that, then go back to the first diagram and you should realize that the undamped natural resonant frequency (\$\omega_0\$) has a value anywhere on the semi circle and it is the Q of the filter how far round from the jw axis the two poles are. Given that any order of butterworth filter has all its poles on the same semi circle, the answer to your question is: -

Is the cut-off frequency for each high order butterworth filter the
same or different?

If you mean the natural resonant frequency then YES!

If you mean the 3dB point of each filter's response on the jw axis (the axis that pertains to "real-life" measurements on a spectrum analyser) then NO!

Butterworth pole-zero diagram: -

The above shows the poles of a 10th order butterworth filter - note that all the poles lie on the same circle and therefore all the individual filters (5 x 2nd order) have the same natural resonant frequency.

Taken from here.

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.

## Best Answer

At about 63% of the full peak value, the time = RC (if it were an RC filter): -

This is because: -

And when Tau = t the answer is 0.6321

Half-power frequency: -

That to me looks like it's happening at about 0.8 micro seconds and so,

F = 1000000/(2xPix0.8) = 199 kHz