# Transfer function of high order filter

Hi I have tried to extract the transfer function of the filter above, as you see it a high order filter.

I have combined the $$\C_1\$$ in parallel with $$\R\$$ & $$\C_2\$$ then make a voltage divider in order to add the inductor. I always miss the capacitor or I don't get the same result. Where is my mistake?

• What do you mean by "always miss the capacitor"? Apr 8 '16 at 21:50
• The given transfer function is correct - you can derive it using the voltage divider rule. However, the form is a bit "uncommon" (factor D identical to a capacitor). Recommendet form: D=1 and factor C=time constant.
– LvW
Apr 9 '16 at 8:20
• above you can see the derivated transfer function, I have got completely different answers!!
Apr 9 '16 at 9:51
• Shouldn't the voltage gain actually be Z/sL+Z? Apr 9 '16 at 9:53
• it is a voltage divider, isnt it ?
Apr 9 '16 at 9:58

If you have access to MATLAB, I would recommend you try using a tool called scam. Using a netlist, MATLAB will spit out the parameter values for your transfer function.

That being said if you're still interested in doing it manually, I don't see any issue with combining the resistor and second capacitor in series, putting that in parallel with the first capacitor and doing a voltage division. Any way you could show us your calculations, I'm kind of flying blind without seeing what exactly you tried.

• please can you see above the calculation of transfer function
The transfer function should be $$T(s) = \frac{1 + C_2 R s}{1 + C_2 R s + (C_1 L + C_2 L) s^2 + C_1 C_2 L R s^3}$$ Your calculation of $$\Z\$$ is correct, however you made some mistake during the calculation of $$T(s) = \frac{Z}{sL + Z}$$