Transfer function of three cascaded RC filters?
From the following schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

I get the following set of expressions:

$$\begin{align*}
\frac{V_O}{R_3} + s\: C_3\: V_O &amp;= \frac{V_Y}{R_3}\tag{$V_O$}\\\\
\frac{V_Y}{R_2} + \frac{V_Y}{R_3} + s\: C_2\:V_Y &amp;= \frac{V_X}{R_2} + \frac{V_O}{R_3}\tag{$V_Y$}\\\\
\frac{V_X}{R_1} + \frac{V_X}{R_2} + s\: C_1\:V_X &amp;=\frac{V_I}{R_1} + \frac{V_Y}{R_2}\tag{$V_X$}
\end{align*}$$

Solving, I get this gnarly mess:

$$\begin{align*}
\tfrac{V_O}{V_I}=\tfrac{1}{\frac{R_1}{R_2 R_3}\left[R_2^2 R_3^2\left(C_1 s+\frac{1}{R_1}+\frac{1}{R_2}\right)\left(C_2 s+\frac{1}{R_2}+\frac{1}{R_3}\right)\left(C_3 s+\frac{1}{R_3}\right)-R_2^2\left(C_1 s+\frac{1}{R_1}+\frac{1}{R_2}\right)-R_3^2\left(C_3 s+\frac{1}{R_3}\right)\right]}
\end{align*}$$

Moving towards a characteristic form:

$$\begin{align*}
\tfrac{V_O}{V_I} &amp;=\frac{K}{s^3+A\cdot s^2+B\cdot s + C}, \quad where,\\\\
K&amp;=\frac{1}{C_3 C_2 C_1 R_3 R_2 R_1}\\\\
A &amp;= \frac{1}{C_1 R_1} + \frac{1}{C_1 R_2} + \frac{1}{C_2 R_2} + \frac{1}{C_2 R_3} + \frac{1}{C_3 R_3} \\\\
B &amp;= \frac{1}{C_2 C_1 R_2 R_1} + \frac{1}{C_2 C_1 R_3 R_1} + \frac{1}{C_2 C_1 R_3 R_2} \\&amp;\quad\quad + \frac{1}{C_3 C_1 R_3 R_1} + \frac{1}{C_3 C_1 R_3 R_2} + \frac{1}{C_3 C_2 R_3 R_2} \\\\
C&amp;=\frac{1}{C_3 C_2 C_1 R_3 R_2 R_1}
\end{align*}$$

At this point I think it moves on to a substitution to get the denominator into the form of \$x^3+p\cdot x + q\$ with \$x=s-\frac{A}{3}\$, \$p=B-\frac{A^2}{3}\$ and \$q=\frac{2 A^3}{27}-\frac{A B}{3}+C\$, then variable replacement using \$x=\sigma+j\omega\$, then sorting into real and imaginary parts. And so on. Enjoy.