# Cut off frequency of a passive second order low pass filter

I was trying to do a simulation for second order low pass passive filter, and the equation of the cut off frequency is $$f_c=\frac{1}{2\pi\sqrt{R_1C_1R_2C_2}}$$ when I calculate it on the values of this circuit I get $$f_c=1591Hz$$ but when I use OrCad to approximate the value of it, I find that it's around $$f_c=608Hz$$ can any one explain how ?

• That's not a great 2nd order configuration because the 1st RC is significantly loaded by the 2nd RC. You should get better (but still imperfect) results by increasing the 2nd resistor (and decreasing the 2nd capacitor) by a factor of 10. That way the 2nd stage doesn't significantly load the 1st and the two sections can both implement single order filters at the same frequency. Commented Apr 26, 2023 at 5:01
• You are simply stating: "...the equation of the cut-off frequency is...". How can you state this? How can you be sure? Did you read this formula somewhere?
– LvW
Commented Apr 26, 2023 at 9:39
• @LvW Yes I did, I found it here : electronics-tutorials.ws/filter/filter_2.html Commented Apr 26, 2023 at 20:21
• @td127 Okay, I'll try it Commented Apr 26, 2023 at 20:22
• Ahmed - this is a good example for a wrong internet contribution. One never should blindly trust anonyme internet articles without any justification/derivation.
– LvW
Commented Apr 26, 2023 at 21:01

What the OP refers to as the corner/cutoff frequency is actually the natural frequrncy $$\f_N=\frac{\omega_N}{2\pi}\$$ of the system. The corner frequency $$\f_c=\frac{\omega_c}{2\pi}\$$ takes more work;

Using nodal analysis the transfer function can be derived: $$H(s)=\frac{V_{C_{2}}(s)}{V_{1}(s)}=\frac{\frac{1}{R_{1}R_{2}C_{1}C_{2}}}{s^{2}+\left(\frac{1}{R_{1}C_{1}}+\frac{1}{R_{2}C_{1}}+\frac{1}{R_{2}C_{2}}\right)s+\frac{1}{R_{1}R_{2}C_{1}C_{2}}}$$

Comparing this to the standard 2nd order form: $$H(s)=\frac{\omega_{N}^{2}}{s^{2}+2\zeta\omega_{N}s+\omega_{N}^{2}}\tag{equ 1}$$ reveals that $$2\zeta\omega_{N}=\left(\frac{1}{R_{1}C_{1}}+\frac{1}{R_{2}C_{1}}+\frac{1}{R_{2}C_{2}}\right)\text{, and that }\omega_{N}=\frac{1}{\sqrt{R_{1}R_{2}C_{1}C_{2}}}$$

These two equations are used to solve for $$\\zeta\$$ and $$\\omega_N\$$

Then solve (equ 1) $$\H(j\omega)=\frac{1}{\sqrt{2}}\$$ for $$\\omega=\omega_c\$$

This then results in the -3dB corner frequency $$f_c=f_N \sqrt{\left(1-2\zeta^{2}\right)+\sqrt{4\zeta^{4}-4\zeta^{2}+2}}$$

The confusion of the corner frequency with the natural frequency is the basic issue. The natural frequency is not a vibration or oscillation. It is a system parameter only.

You should work through the derivation steps that I have skipped.