Repeated Pole RC Filters
This is a follow-up to last week's post on Passive RC Filters.
I wanted to investigate whether or not you could design a cascaded RC low-pass filter - such as the one discussed - that has the same response as a single-stage design repeated twice with a buffer in between.
This is a repeated-pole filter with a transfer-function in the form:
$$ T(s) = \frac{1}{(As + B)^2} = \frac{1}{A^2 s^2 + 2ABs + B^2} $$
Which has two repeated poles at $s= -\frac{B}{A}$.
To design our filter, we must first find the transfer function of a generalised cascaded RC low-pass filter (with different values of R and C).
The analysis is almost identical to before, and again I've hidden it with this cut
We start with basic potential-divider equations: $$ V_2 = V_1 \frac{X_{C2}}{R_2 + X_{C2}} $$ and $$ V_1 = V_0 \frac{X_{C1} // (X_{C2} + R_2)}{R_1 + X_{C1} // (X_{C2} + R_2)} $$ Where $R_1$, $R_2$, $C_1$, and $C_2$ denote our first and second stage's resistors and capacitors. Again, we can split our analysis into three parts: $$ \begin{align*} \frac{X_{C2}}{R_2 + X_{C2}} &= \frac{\frac{1}{sC_2}}{\frac{1}{sC_2} + R_2} \newline &= \frac{1}{sR_2 C_2 + 1} \end{align*} $$ Next: $$ \begin{align*} X_{C1} // (X_{C2} + R_2) &= \Bigg(sC_1 + \frac{1}{\frac{1}{sC_2} + R_2}\Bigg)^{-1} \newline &= \Bigg( sC_1 + \frac{sC_2}{s R_2 C_2 + 1} \Bigg)^{-1} \newline &= \Bigg( \frac{sC_1(sR_2 C_2 + 1) + sC_2}{sR_2 C_2 + 1} \Bigg)^{-1} \newline &= \frac{sR_2 C_2 + 1}{sC_1(sR_2 C_2 + 1) + sC_2} \end{align*} $$ And finally: $$ \begin{align*} R_1 + X_{C1} // (X_{C2} + R_2) &= R_1 + \frac{s R_2 C_2 + 1}{s C_2 (s R_2 C_2 + 1) + sC_2} \newline &= \frac{sR_1 C_1 (sR_2 C_2 + 1) + sR_1 C_2 + s R_2 C_2 + 1}{sC_1(sR_2 C_2 + 1) + sC_2} \end{align*} $$ Now we can see that: $$ \frac{X_{C1} // (X_{C2} + R_2)}{R_1 + X_{C1} // (X_{C2} + R_2)} = \frac{sR_2 C_2 + 1}{s R_1 C_1 (s R_2 C_2 + 1) + s R_1 C_2 + sR_2 C_2 + 1} $$ And finally, we can write that: $$ \begin{align*} T(s) = \frac{V_2}{V_0} &= \frac{1}{s R_1 C_1 (s R_2 C_2 + 1) + s R_1 C_2 + sR_2 C_2 + 1} \newline &= \frac{1}{s^2(R_1 R_2 C_1 C_2) + s(R_1 C_1 + R_1 C_2 + R_2 C_2) + 1} \end{align*} $$ To find the solution for this system that corresponds to our ideal repeated-pole transfer-function, we can simply equate coefficients in the denominators of both equations: $$ A^2 s^2 + 2ABs + B^2 = s^2(R_1 R_2 C_1 C_2) + s(R_1 C_1 + R_1 C_2 + R_2 C_2) + 1 $$ By inspection, $B=1$, and: $$ A^2 = R_1 R_2 C_1 C_2 $$ $$ 2A = R_1 C_1 + R_1 C_2 + R_2 C_2 $$ Squaring the second equation gives: $$ 4A^2 = R_1^2 C_1^2 + R_1^2 C_2^2 + R_2^2 C_2^2 + 2R_1^2 C_1 C_2 + 2 R_1 R_2 C_2^2 + 2 R_1 R_2 C_1 C_2 $$ We can now combine our two equations for $A^2$ to give: $$ \begin{align*} R_1^2 C_1^2 + R_1^2 C_2^2 + R_2^2 C_2^2 + 2R_1^2 C_1 C_2 + 2 R_1 R_2 C_2^2 - 2 R_1 R_2 C_1 C_2 &= 0 \newline R_1^2 \big(C_1^2 + C_2^2 + \frac{R_2^2}{R_1^2} C_2^2 + 2 C_1 C_2 + 2 \frac{R_2}{R_1} C_2^2 - 2 \frac{R_2}{R_1}C_1 C_2\big) &= 0 \newline R_1^2 \Bigg( \Big(\frac{R_2}{R_1}\Big)^2 C_2^2 + \frac{R_2}{R_1} \Big( 2C_2^2 - 2 C_1 C_2 \Big) + \Big( C_1^2 + C_2^2 + 2C_1 C_2 \Big) \Bigg) &= 0 \newline R_1^2 \Bigg( \Big(\frac{R_2}{R_1}\Big)^2 C_2^2 + 2 \frac{R_2}{R_1} C_2 \Big( C_2 - C_1 \Big) + \Big( C_1 + C_2 \Big)^2 \Bigg) &= 0 \end{align*} $$ The null result is $R_1 = 0$ - we don't care about this as we are only interested in positive, real values for our components. This leaves us with a quadratic in $\frac{R_2}{R_1}$ with the solution $$ \frac{R_2}{R_1} = \frac{(\sqrt{C_1} \pm j \sqrt{C_2})^2}{C_2} $$The result of the analysis is that our transfer function is:
$$ T(s) = \frac{1}{s^2(R_1 R_2 C_1 C_2) + s(R_1 C_1 + R_1 C_2 + R_2 C_2) + 1} $$
And in order to make an cascaded RC filter with the desired transfer function, we must satisfy the equation:
$$ R_2 = \frac{R_1 (\sqrt{C_1} \pm j \sqrt{C_2})^2}{C_2} $$
We can see that there are no solutions to this equation if $\{R_1, R_2, C_1, C_2\} \in \mathbb{R}_{>0}$. No solutions to this equation means that there is no solution for $A \in \mathbb{R}$. Which means that you cannot find a combination of $R$s and $C$s that produces a repeated-pole filter.
So there you have it: if you want to simply create a filter with a response twice as steep as a single RC stage, you will need to look into a more complex topology, or accept the effects that our equal-value filter from before has on the pass-band.
As a final note, it is possible to get from the transfer function here to the $-3dB$ point, but the algebra is tedious and does not arrive at anything approaching an elegant solution.
The $-3dB$ point can be found at the roots of the equation:
$$ (1-\omega^2 R_1 R_2 C_1 C_2)^2 + \omega^2 (R_1 C_1 + R_1 C_2 + R_2 C_2)^2 = 2 $$
Which is a quadratic in $\omega^2$:
$$ a (\omega^2)^2 + b (\omega^2) + c = 0 $$
where
$$ \begin{align*} a &= R_1^2 R_2^2 C_1^2 C_2^2 \newline b &= R_1^2 C_1^2 + R_1^2 C_2^2 + R_2^2 C_2^2 + 2R_1^2 C_1 C_2 + 2 R_1 R_2 C_2^2 \newline c &= -1 \end{align*} $$