How to determine what kind of response this filter has

Question

I have the next Sallen Key type filter

I have determined the transfer function and the coefficients of \$a_{0}\$ and \$a_{1̣̣}\$, based on

\$N(s)=\frac{Ka_{0}}{s^{n}+a_{n-1}s^{n-1}+....a_{1}s+a_{0}}\$

But then how do I use this to determine (or approximate) analytically what kind of response the filter has? (Butterworth, Chebyshev etc.)

This is besides making a frequency sweep using a simulator.
UPDATE
Derivating the transfer function. I have named the node between the 47k resistances as \$V_{x}\$ and the node of the negative bias \$V_{y}\$
\$V_{out}=(1+\frac{R_{A}}{R_{B}})V_{y}\$

\$G=1+\frac{R_{A}}{R_{B}}=\frac{R_{A}+R_{B}}{R_{B}}\$
\$V_{y}=\frac{R_{B}}{R_{A}+R_{B}}V_{out}\$
Simplifying
\$V_{y}=\frac{V_{out}}{G}\$
using a voltage divider
\$V_{y}=\left(\frac{V_{x}}{R_{2}+\frac{1}{C_{2}s}}\right)\frac{1}{C_{2}s}=V_{x}\frac{1}{1+R_{2}C_{2}s}\$
\$V_{x}=(1+R_{2}C_{2}s)V_{y}\$
\$0=\frac{V_{x}-V_{in}}{R_{1}}+\frac{V_{x}-V_{y}}{R_{2}}+\frac{V_{x}-V_{out}}{\frac{1}{sC1}}\$
\$\frac{V_{x}-(1+R_{2}C_{2}s)V_{y}-V_{in}}{R_{1}}+\frac{(1+R_{2}C_{2}s)V_{y}-V_{y}}{R_{2}}+\frac{(1+R_{2}C_{2}s)V_{y}-V_{out}}{\frac{1}{sC1}}=0\$
\$\frac{(1+R_{2}C_{2}s)V_{y}}{R_{1}}-\frac{V_{in}}{R_{1}}+\frac{(1+R_{2}C_{2}s)V_{y}}{R_{2}}-\frac{V_{y}}{R_{2}}+\frac{(1+R_{2}C_{2}s)V_{y}}{\frac{1}{sC1}}-\frac{V_{o}}{\frac{1}{sC1}}=0\$
\$\frac{(1+R_{2}C_{2}s)V_{out}}{GR_{1}}-\frac{V_{in}}{R_{1}}+\frac{(1+R_{2}C_{2}s)V_{out}}{GR_{2}}-\frac{V_{out}}{GR_{2}}+\frac{(1+R_{2}C_{2}s)V_{out}}{\frac{G}{sC1}}-\frac{V_{o}}{\frac{1}{sC1}}=0\$
\$V_{out}\left[\frac{(1+R_{2}C_{2}s)}{GR_{1}}+\frac{(1+R_{2}C_{2}s)}{GR_{2}}-\frac{1}{GR_{2}}+\frac{(1+R_{2}C_{2}s)}{\frac{G}{sC1}}-\frac{1}{\frac{1}{sC1}}\right]=\frac{V_{in}}{R_{1}}\$
\$\frac{V_{out}}{G}\left[\frac{(1+R_{2}C_{2}s)}{R_{1}}+\frac{(1+R_{2}C_{2}s)}{R_{2}}-\frac{1}{R_{2}}+\frac{(1+R_{2}C_{2}s)}{\frac{1}{sC1}}-\frac{1}{\frac{1}{GsC1}}\right]=\frac{V_{in}}{R_{1}}\$
\$\frac{V_{out}}{G}\left[\frac{1}{R_{1}}+\frac{(R_{2}C_{2}s)}{R_{1}}+\frac{1}{R_{2}}+\frac{(R_{2}C_{2}s)}{R_{2}}-\frac{1}{R_{2}}+\frac{1}{\frac{1}{sC1}}+\frac{(R_{2}C_{2}s)}{\frac{1}{sC1}}-\frac{1}{\frac{1}{GsC1}}\right]=\frac{V_{in}}{R_{1}}\$
\$\frac{V_{out}}{G}\left[\frac{1}{R_{1}}+\frac{(R_{2}C_{2}s)}{R_{1}}+C_{2}s+sC1+sC1R_{2}C_{2}s-GsC_{1}\right]=\frac{V_{in}}{R_{1}}\$
\$\frac{V_{out}}{G}\left[\frac{1+R_{2}C_{2}s+R_{1}C_{2}s+R_{1}sC1+R_{1}sC1R_{2}C_{2}s-R_{1}GsC_{1}}{R_{1}}\right]=\frac{V_{in}}{R_{1}}\$
\$\frac{V_{out}}{G}\left[1+R_{2}C_{2}s+R_{1}C_{2}s+R_{1}sC1+R_{1}sC1R_{2}C_{2}s-R_{1}GsC_{1}\right]=V_{in}\$
\$\frac{V_{out}}{G}\left[s^{2}R_{1}R_{2}C1C_{2}+s(R_{1}C_{1}+R_{2}C_{2}+R_{1}C_{2}-GR_{1}C_{1})+1\right]=V_{in}\$
\$\frac{V_{out}}{G}\left[s^{2}R_{1}R_{2}C1C_{2}+s(R_{1}C_{1}-GR_{1}C_{1}+R_{2}C_{2}+R_{1}C_{2})+1\right]=V_{in}\$
\$\frac{V_{out}}{G}\left[s^{2}R_{1}R_{2}C1C_{2}+s(R_{1}C_{1}(1-G)+R_{2}C_{2}+R_{1}C_{2})+1\right]=V_{in}\$
\$\frac{V_{out}}{V_{i}}=\frac{G}{s^{2}R_{1}R_{2}C1C_{2}+s(R_{1}C_{1}(1-G)+R_{1}C_{2}+R_{2}C_{2})+1}\$
\$\frac{V_{out}}{V_{i}}=\frac{G}{s^{2}R_{1}R_{2}C1C_{2}+s(R_{1}C_{1}(1-G)+R_{1}C_{2}+R_{2}C_{2})+1}\$
\$\frac{V_{out}}{V_{i}}=\frac{\frac{G}{R_{1}R_{2}C1C_{2}}}{s^{2}+s(\frac{1}{R_{1}C_{1}}+\frac{1}{R_{1}C_{2}}+\frac{(1-G)}{R_{2}C_{2}})+\frac{1}{R_{1}R_{2}C1C_{2}}}\$
if \$R_{1}=R_{2}=R\$ and \$C_{1}=C_{2}=C\$
\$\frac{V_{out}}{V_{i}}=\frac{\frac{G}{R^{2}C^{2}}}{s^{2}+s(\frac{1}{RC}+\frac{1}{RC}+\frac{(1-G)}{RC})+\frac{1}{R^{2}C^{2}}}\$
\$\frac{V_{out}}{V_{i}}=\frac{\frac{G}{R^{2}C^{2}}}{s^{2}+s(\frac{(3-G)}{RC})+\frac{1}{R^{2}C^{2}}}\$
\$G=K\$

\$a_{0}=\frac{1}{R^{2}C^{2}}\$

\$a_{1}=\frac{3-G}{RC}\$

Answer 1

Because you have the transfer function with all parts values you can compare this function with the general function - expressed NOT with coefficients (which have no direct relation to the kind of response) but with the pole data wp (pole frequency) and \$Q_p\$ (pole quality factor).

It is in particular the Qp value that gives you information on the kind of response.

Example: \$Q_p=0.7071\$ (Butterworth) and \$Q_p>0.7071\$ (Chebyshev).

Knowing the kind of filter response you can use filter tables to find the relation between wp and the corresponding cut-off frequency wc of the lowpass.

Transfer function (2nd order):

$$H(s)=\dfrac{A_o}{1+\dfrac{s}{\omega_pQ_p}+\dfrac{s^2}{\omega_p^2}}$$

Comment 1: In your circuit there are two equal \$R\$ and to equal \$C\$ values. In this special case ("equal component design") the pole \$Q\$ is \$Q_p=\dfrac{1}{3-v}\$ with \$v\$ = closed-loop opamp gain.

Comment 2: This "equal component design" has some disadvantages: It requires an "odd" gain value and the Qp value is very sensitive to gain variations. For this reason two other approaches are used very often (however, with "odd" and unequal values for R or C): Unity gain or gain of two (feedback with two equal resistors).