Electrical – Butterworth Filter RLC LPF

$$ V_o = V_s * \frac{Z}{Z+R1} $$ where $$ Z = \frac{R_2}{j\omega R_2C +1} $$ Thus

$$ \frac{V_o}{V_s} = \frac{Z}{Z+R_1} $$

$$ \frac{V_o}{V_s} = \frac{R_2}{j\omega R_1 R_2C+R_1+R_2} $$

Perhaps the following information can help to clarify the problem:

1) The expression SQRT[1/(1+x^4) is a second-order function we WANT to realize because this is a function which has a "maximally flat" magnitude response. That means: If compared with all other possible 2nd-order functions the magnitude deviates as less a s possible from its value at x=0 (which in this case is "1"). Note that for a frequency-dependent filter the variable x is identical to a frequency variable w=2*Pi*f which is normalized to the 3dB-cutoff frequency wc. Therefore we have x=w/wc.

2.) We can REALIZE only filter functions with lumped elements which can comply with the general second-order lowpass function H(s)=Ao/(1+as+bs²). If we calculate the zeros of the denominator (poles of the whole function H) we find that a=1/(wp*Qp) and b=1/(wp²). Both pole parameters wp and Qp are definded using the pole location of the pole pair in the complex s-domain.

3.) Now, the results from 1.) and 2) are equalized (what we want = what we can realize). For this purpose, we use the squares of both functions. After some mathematical manipulations we arrive at the expression:

H(s)=Ao/(1+0.7071x+x²) with x=w/wc

4.) This is the general transfer function for a 2nd-order Butterworth lowpass with DC gain Ao. There are different methods (topologies) for realizing this function as an electronic circuit. One simple solution (passive) is to use a suitable RLC combination (as in your example with Ao=1). To find the corresponding expressions (parts values) you have nothing to do than to compare the corresponding coefficients of both functions (from the circuit and the given H(s) for the desired 3dB-frequency wc).