Calculating the cut off frequency of a filter circuit if you know all the component values
This is an active Sallen-Key filter followed by a passive low-pass filter. The transfer function of the whole circuit is that of a 2nd-order polynomial form followed by a 1st-order \$RC\$ filter: this is a 3rd-order filter. If we consider the op amp a perfect element, its output impedance is 0 and thus eliminates the interaction with the \$RC\$ filter. We can thus first determine the transfer function of the S&amp;K filter, simply followed by the transfer function of the low-pass filter to obtain the whole expression linking \$V_{out}\$ to \$V_{in}\$. Let's start with the active filter.

There are plenty of options to determine the transfer function of this active filter. The fastest and most efficient to me are the fast analytical circuits techniques or FACTs described here. It consists of determining the time constants \$\tau\$ of the circuit when the excitation \$V_{in}\$ is reduced to 0 V. But first, we determine the dc gain of the circuit shown below with the adopted component labels:



To determine the dc gain for \$s=0\$, open the capacitors and determine the circuit gain \$H_0\$ in this configuration. You see an op amp wired in a non-inverting configuration whose gain is: \$H_0=\frac{R_f}{R_i}+1\$. The bias point simulation shows 2 V for a 1-V input confirming a gain of 2 with equal values for \$R_f\$ and \$R_i\$.

Now, reduce the excitation to 0 V (replace \$V_{in}\$ by a short circuit) and determine the resistance "seen" from the capacitor terminals when it is temporarily removed. The resistance multiplied by the capacitor forms the time constants we want. The below drawing will help you determine these:



To determine a resistance, inspection works in the upper left sketch. As \$C_2\$ is removed, you see the current generator biasing the two resistors \$R_1\$ and \$R_2\$ in series. So without writing a line of algebra, the first time constant is immediate: \$\tau_1=C_1(R_1+R_2)\$. For the second time constant, a few lines are necessary to determine the resistance. The second time constant is \$\tau_2=C_2R_1(1-A_v)\$ with \$A_v=\frac{R_f}{R_i}+1\$. The final time constant is obtained when \$C_2\$ is replaced by a short circuit and you determine the time constant involving \$C_1\$: \$\tau_{21}=C_1\frac{R_2}{1-A_v}\$. We can now assemble \$D(s)\$ by combining the time constants as follows: \$D(s)=1+s(\tau_1+\tau_2)+s^2\tau_2\tau_{21}\$. This can be advantageously rearranged under a canonical form \$D(s)=1+\frac{s}{Q\omega_0}+\left(\frac{s}{\omega_{0}}\right)^2\$ in which \$Q=\frac{\sqrt{b_2}}{b_1}\$ and \$\omega_0=\frac{1}{\sqrt{b_2}}\$. When you combine all these elements together, you have \$Q=\frac{\sqrt{R_1R_2C_1C_2}}{(R_1+R_2)C_1+R_1C_2(1-A_v)}\$ and the resonant frequency \$\omega_0=\frac{1}{\sqrt{R_1R_2C_1C_2}}\$. With the given component values, you have \$f_0=45.2\;Hz\$ and \$Q=1\$. All these expressions are computed in the below Mathcad sheet:



We have the denominator but do we know if there are zeros in this circuit? To check the presence of a zero, short the capacitor and check if a response still exists in this mode. If yes, there is a zero associated with the capacitor you have shorted, if not, there is no zero contributed by this element. Shorting any of the caps or both at the same time gives 0 V as confirmed by the below sketch: there is no zero and \$N(s)=1\$.



The final transfer function for the S&amp;K section is thus:

\$H(s)=H_0\frac{1}{1+\frac{s}{Q\omega_0}+\left(\frac{s}{\omega_{0}}\right)^2}\$. We can plot this transfer function as shown below:



We can check our calculation integrity with a SPICE simulation or by using the following site:



The resonant frequency and quality factor are identical to what I found and the frequency response is the same:



Ok, we now have the active filter part and the final result includes the passive low-pass section whose transfer function is simply:

\$H_{LP}(s)=\frac{1}{1+\frac{s}{\omega_p}}\$ with \$\omega_p=\frac{1}{R_LC_L}\$. The final transfer function of this 3rd-order filter is given by:

\$H_{3rd}(s)=H_0\frac{1}{1+\frac{s}{Q\omega_0}+\left(\frac{s}{\omega_{0}}\right)^2}\frac{1}{1+\frac{s}{\omega_p}}\$.

We can plot this expression and see the new downslope of 60 dB per decade confirming the 3rd-order behavior:



FACTs are well suited because you just calculate time constants via individual sketches you can later correct if necessary. You can also simulate these small schematics (as I did) in case you want to check you results. Proceeding step by step is the key to determining the transfer function in a swift and efficient manner. Highly recommended for designers dealing with transfer functions!