cutoff frequency
I want to calculate the cutoff frequency for a specific filter, but I can't find any formula for that.
I know the formula for the cutoff frequency of a low pass filter:
$$f_c=\frac{1}{2\pi RC}$$
But how is that derived in the first place? I don't have a regular low pass filter, but something similar that I want to calculate the cutoff frequency of.
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: -
simulate this circuit – Schematic created using CircuitLab
Applying KVL,
$$ V_{in} = V_{R} + V_{C} = I\cdot Z_C + I\cdot Z_R \\ \implies I = \frac{V_{in}}{Z_R+Z_C} $$
Setting \$f=\frac{1}{2\pi RC}\$, we have
$$ Z_C=\frac{1}{j2\pi fC}=\frac{1}{j2\pi \frac{1}{2\pi RC} C} = \frac{R}{j} = -jR $$
In the low-pass filter, the output voltage is taken across the capacitor:
$$ V_{out,lpf}=I\cdot Z_C = V_{in} \cdot \frac{Z_C}{Z_R+Z_C} = V_{in} \cdot \frac{-jR}{R-jR} = V_{in} \cdot \frac{-j}{1-j} \\ \implies \left|\frac{V_{out,lpf}}{V_{in}}\right| = \left|\frac{-j}{1-j}\right| = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} $$
In the high-pass filter, the output voltage is taken across the resistor:
$$ V_{out,hpf}=I\cdot Z_R = V_{in} \cdot \frac{Z_R}{Z_R+Z_C} = V_{in} \cdot \frac{R}{R-jR} = V_{in} \cdot \frac{1}{1-j} \\ \implies \left|\frac{V_{out,hpf}}{V_{in}}\right| = \left|\frac{1}{1-j}\right| = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}} $$
Best Answer
The specific formula applies only for a first order RC low pass filter. This is derived from its frequency response:
$$H(j\omega)=\frac{1}{1+j\omega RC}$$
The cutoff frequency is defined as the frequency where the amplitude of \$H(j\omega)\$ is \$1\over\sqrt2\$ times the DC amplitude (approximately -3dB, half power point).
$$|H(j\omega_c)|=\frac{1}{\sqrt{1^2+\omega_c^2R^2C^2}}=\frac{1}{\sqrt{2}}\cdot|H(j0)|=\frac{1}{\sqrt{2}}$$
Solve it for \$\omega_c\$ (cutoff angular frequency), you'll get \$1\over RC\$. Divide that by \$2\pi\$ and you get the cutoff frequency \$f_c\$.
If you know the frequency response of your filter, you can apply this method (given that the cutoff frequency is defined as above). Obviously, for high-pass filters for example, you calculate with the value for \$\omega\to \infty\$ as opposed to the DC value (always the maximum of the amplitude response, relative to which there is a 3dB decrease in amplitude at the cutoff frequency.)