I have this circuit here which is a low pass chebyshev filter, I have been asked to find it's cutoff frequency. I have used matlab to plot an output response against angular frequency graph here. But I'm unsure of how to find the cut off frequency from this, is there an equation to do this?
Electrical – n equation for the cutoff frequency of a chebyshev filter
cutoff frequencyfilterlow pass
Related Solutions
What type of filter does this circuit represent?
It's a 2nd order RLC high pass filter
Also what would the cutoff frequency formula be in this case (of course that depends on what type of filter this is)?
Cutoff frequency = ~\$\dfrac{1}{2\pi\sqrt{LC}}\$
And no, if you have an L and a C forming a filter it's either high-pass, low-pass, band-pass or band-reject BUT in all cases the formula is the same i.e. a low pass 3 dB point is exactly same frequency as the high-pass 3 dB point. Notch and band-pass centre frequencies are identical too.
Low-pass example: -
High-pass example using the same values: -
For this circuit we have
$$H_{(s)} = - \frac{s R_2 C_1}{1 + s R_1 C_1} $$
So we have one Pole at
$$-\frac{1}{R_1 C_1}$$
And one Zero at the origin.
All this means that for low frequency the circuit behaves like an ordinary op-amp based differentiator.
With the gain $$A_V = \omega R_2C_1$$
And the gain reachs \$1 V/V\$ when sinal frequancy is equal to \$Fo=\frac{1}{2 \pi R_2 C_1}\$
As signal frequency increases \$Xc\$ drops and when \$Xc = R_1\$ we have a Pole:
$$\omega = \frac{1}{R_1 C_1}$$
$$Fp=\frac{1}{2 \pi R_1 C_1}$$.
And the the magnitude of a transfer function (voltage gain vs frequency) is equal $$A = \frac{\omega R_2 C_1}{\sqrt{1 + \left ( \omega R_1 C_1 \right )^2}}$$
Best Answer
The usual definition of the cut-off frequency of a (type I) Chebyshev filter is shown in the figure below:
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.