How do I find the equivalent circuit of a compensated attenuator?
A direct way to analyze this is to define \$Z_1\$ as the equivalent impedance of \$R_1\$ // \$C_1\$; and \$Z_2\$ as the equivalent impedance of \$R_2\$ // \$C_2\$. This leaves us with a simple impedance divider, where:
$$
\frac{V_o(s)}{V_i(s)}=\frac{Z_2(s)}{Z_1(s)+Z_2(s)}
$$
Converting to admittance to simplify the calculation, we get the equivalent formula:
$$
\frac{V_o(s)}{V_i(s)}=\frac{1}{\frac{Y_2(s)}{Y_1(s)}+1}
$$
where:
$$
Y_x=sC_x+\frac{1}{R_x}
$$
So, we get:
$$
\frac{V_o(s)}{V_i(s)}=\frac{1}{\frac{sC_2+\frac{1}{R_2}}{sC_1+\frac{1}{R_1}}+1}
$$
What can be rewritten as:
$$
\frac{V_o(s)}{V_i(s)}=\frac{R_2}{R_1+R_2}\frac{sR_1C_1+1}{sR_PC_P+1}
$$
or
$$
\frac{V_o(s)}{V_i(s)}=\frac{C_1}{C_1+C_2}\frac{s+\frac{1}{R_1C_1}}{s+\frac{1}{R_PC_P}}
$$
where \$R_P\$ and \$C_P\$ are the equivalent parallel of \$R_1\$//\$R_2\$ and \$C_1\$//\$C_2\$ respectively:
$$
\frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}
$$
$$
C_P=C_1+C_2
$$
Notice that this transfer function has one pole and one zero.
At DC (\$s=0\$), the gain is determined by the resistor divider defined by \$R_2\$ and \$R_1\$:
$$
\frac{V_o}{V_i}=\frac{R_2}{R_1+R_2}
$$
At higher frequencies (above the greater between the zero and the pole frequencies), the gain is defined by the capacitor divider:
$$
\frac{V_o}{V_i}=\frac{C_1}{C_1+C_2}
$$
This makes the Bode plot look something like this:

Or like this, depending on the component values:

This circuit is commonly used for compensation in oscilloscope probes. The idea is to have one variable component (typically \$C_2\$) properly tuned so that the frequency response becomes flat (\$f_{zero}=f_{pole}\$). In this case, \$R_2\$ and \$C_2\$ are the equivalent parallel of the external components combined with the input impedance of the scope.