# Finding corner frequency of RC bandpass filter

I'm working with a RC bandpass filter of the following topology: I am trying to find the corner frequencies. It has been explained to me that this is just a low pass RC filter and high pass RC filter cascaded together. For instance this site suggests that to find the corner frequencies you just need to use $1/(2\pi RC)$ to find the upper and lower corners. But this doesn't seem right. To find the corners like that I would expect a circuit like this: simulate this circuit – Schematic created using CircuitLab

Is there some simplifying trick to get to this from the previous circuit? Right now my solution is to find the transfer function and set it equal to $1/\sqrt{2}$ and solve for $\omega$. This is a pain to do without MATLAB and I am wondering if there is another, simpler method.

This type of passive circuit can be easily solved and expressed in a so-called low-entropy format using the fast analytical circuits techniques or FACTs. The principle is to apply the generalized transfer function formula for a second-order system. It is defined as:

$H(s)=\frac{H_0+s(H_1\tau_1+H_2\tau_2)+s^2H_{12}\tau_1\tau_{12}}{1+s(\tau_1+\tau_2)+s^2\tau_1\tau_{12}}$

The $\tau$ are the natural time constants of the circuits determined when the excitation (the stimulus, $V_{in}$, is reduced to $0\;V$). Here, short the input source, implying that $R_1$ left terminal is grounded. Now, "look" at the resistance offered by the terminals of $C_1$ and $C_2$ in this condition: $\tau_1=C_1(R_1+R_2)$ and $\tau_2=C_2R_2$. Then, do the same but shorting $C_2$ and "looking" at the resistance offered by $C_1$. You should find $\tau_{21}=C_1R_1$. We have $D(s)$ now:

$D(s)=1+s(C_1(R_1+R_2)+C_2R_2)+s^2C_1C_2R_1R_2$

The high-frequency gains $H$ are found by setting the corresponding energy-storing elements in their high-frequency states. For $H_1$ and $H_2$, respectively replace $C_1$ and $C_2$ by short circuits and find: $H_1=\frac{R_2}{R_1+R_2}$ while $H_2=0$. As $H_{12}$ implies that both caps are shorted, $H_{12}=0$. We have:

$N(s)=sH_1\tau_1=s\frac{R_2}{R_2+R_1}C_1(R_1+R_2)=sR_2C_1$

The complete transfer function involving the zero at the origin is then:

$H(s)=\frac{sR_2C_1}{1+s(C_1(R_1+R_2)+C_2R_2)+s^2C_1C_2R_1R_2}=\frac{\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}$

If I now factor the term $\frac{s}{\omega_z}$ in the numerator and $\frac{s}{\omega_0Q}$ in the denominator then rearrange, you obtain a true low-entropy transfer function defined as:

$H(s)=H_{00}\frac{1}{1+Q(\frac{s}{\omega_0}+\frac{\omega_0}{s})}$

in which:

$Q=\frac{\sqrt{C_1R_1C_2R_2}}{C_2R_2(\frac{C_1(R_1+R_2)}{C_2R_2}+1)}$

$\omega_0=\frac{1}{\sqrt{C_1C_2R_1R_2}}$

$H_{00}=\frac{1}{\frac{R_1}{R_2}+\frac{C_2}{C_1}+1}$

I have captured these equations in a Mathcad sheet to show how the reference equations kindly suggested by Marcus compares with the low-entropy format. They perfectly match. The difference is that you now have a transfer function letting you calculate the values for all components depending on how you want to tune this filter and what attenuation you want at the peak. What truly matters is the low-entropy well-ordered form which tells you what terms contribute gains (attenuation), poles and zeros. Without this arrangement, there is no way you can design your circuit to meet a certain goal. To my opinion, the FACTs are unbeatable to obtain these results in one clean shot (you would need to rework the raw reference function to obtain the form I gave). If you are designing circuits (passive or active) and need to determine transfer functions, I encourage you to acquire that skill because once you have it, you won't go back to the classical approach. If you start slowly step by step, it is quite simple actually. Complicate expressions when you master 1st-order circuits.

You can discover FACTs further here

and also through examples published in the introductory book

The second circuit you have proposed as a replacement differs from the original one though it also performs a high-pass function. If you follow the flow I described earlier, you should find the following expression:

$H(s)=\frac{sR_2C_2}{1+s(C_2(R_1+R_2)+C_1R_1)+s^2C_1C_2R_1R_2}=\frac{\frac{s}{\omega_z}}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}$

in which

$Q=\frac{\sqrt{C_1R_1C_2R_2}}{C_2(R_1+R_2)+C_1R_1}$

$\omega_0=\frac{1}{\sqrt{C_1C_2R_1R_2}}$

$H_{00}=\frac{1}{\frac{C_1R_1}{C_2R_2}+\frac{R_1}{R_2}+1}$

The reference transfer function requires to include the Thévenin output resistance made of $C_1$ in parallel with $R_1$. The curves are given below: These are the joys of FACTs!

The "trick" usually is doing the full math. Once you've done that, things do get easier, I promise :)

So, you'd go and fuse R1 and C1 into one complex component Z1 ($Z1=R1+\frac1{j\omega C1}$), and do the same for R2, C2 forming Z2 ($Z2 = R2 || \frac1{j\omega C2}$).

Then you'd consider the whole thing as bog-normal voltage divider, set Vout = 0.5 Vin, and solve for $\omega$ which will give you the cutoff-$\omega$ values.

Circuit analysis works for any topology!

• this is kind of what i figured, I'll wait and see if anyone else has a trick otherwise I'll make sure to come back and pick your answer
– user146139
Apr 12 '17 at 22:08