I'm designing a two-stage RC LPF (Low-Pass Filter) and I want a cutoff frequency \$f_c=1000\mathrm{Hz}\$.
So, I choose \$R_1=3\mathrm{k\Omega}, R_2=33\mathrm{k\Omega}, C_1=4.7\mathrm{nF}, C_2=47\mathrm{nF}\$, and when I calculate \$f_c\$ with these values, it results \$1000\mathrm{Hz}\$.
But when I feed the input with a white noise signal and measure the output, after plotting the bode plot, I see that \$f_c\$ is actually \$650\mathrm{Hz}\$.
I know that the calculated value \$f_c=1000\mathrm{Hz}\$ is exact only in an ideal situation, and I think the tolerance of circuit elements cause \$f_c\$ fail to reach the sought for value.
But how can I choose the RC element values in order to get \$f_c = 1000\mathrm{Hz}\$ in a real situation???
Best Answer
If you consider 2 cascaded \$RC\$ filters, the second network loads the first one and the cutoff frequency of the resulting filter needs to be derived. It is not that complicated though using the fast analytical techniques or FACTs that I described in my last book. The principle is simple: determine the time constants of the filter in two different conditions: when the excitation is turned off and when the response is nulled. The first approach deals with the poles determination while the second one deals with the determination of zeroes.
Let's have a look with the simple cascaded \$RC\$ filter shown below:
First, you look at the circuit for \$s=0\$ in which you open the capacitors. As you can see, the gain is 1 in this configuration (we considered an unloaded filter). Then, turn the excitation off and replace the source by a short circuit: a 0-V source is a short circuit while a 0-A current source is an open circuit. Now, "look" through the terminals of the capacitors to determine the resistance in this mode while the caps are temporarily removed. You do it by inspecting the circuit, no equation. When you have the resistance, multiply by the capacitor value to form the time constant. For instance, you should find \$\tau_1=R_1C_1\$ and \$\tau_2=(R_1+R_2)C_2\$. As this is a second-order network, you can now write \$b_1=\tau_1+\tau_2=C_1R_1+(R_1+R_2)C_2=3RC\$ if we consider \$R_1=R_2=R\$ and \$C_1=C_2=C\$.
For the second-order term, place capacitor \$C_1\$ in its high-frequency state (a short circuit) and "look" through \$C_2\$'s connecting terminals to determine the resistance: \$\tau_{12}=C_2R_2\$. This is it, you can write \$b_2=\tau_1\tau_{12}=R_1C_1R_2C_2=(RC)^2\$ if we consider equal resistances and capacitors.
The transfer function is assembled as follows: \$H(s)=H_0\frac{1}{1+s3RC+s^2R^2C^2}\$. It could be reorganized under a low-entropy form with a quality factor \$Q\$ and a resonant frequency \$\omega_0\$ but no need here. What we want is to determine the cutoff frequency at which the magnitude is reduced by 3 dB or equal \$\frac{1}{\sqrt{2}}\$. By replacing \$s\$ by \$j\omega\$ in the transfer function, we find a magnitude expression defined as: \$|H(\omega)|=\frac{1}{\sqrt{(1-(RC\omega)^2)^2+(3RC\omega)^2}}=\frac{1}{\sqrt{2}}\$. If we determine the value of \$\omega\$ which satisfies this equation, after rearranging, we have: \$f_0\approx \frac{0.05956}{RC}\$.
Now assume you want a cutoff frequency of 1 kHz. Arbitrarily select \$R\$ (or \$C\$) and set it to 1 k\$\Omega\$ then applying the formula, \$C\$=59.6 nF. The below Mathcad sheet shows the steps and the resulting magnitude and phase curves: