enter image description hereHi 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?

enter image description here

  • \$\begingroup\$ What do you mean by "always miss the capacitor"? \$\endgroup\$ Apr 8, 2016 at 21:50
  • \$\begingroup\$ 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. \$\endgroup\$
    – LvW
    Apr 9, 2016 at 8:20
  • \$\begingroup\$ above you can see the derivated transfer function, I have got completely different answers!! \$\endgroup\$
    – Fadi
    Apr 9, 2016 at 9:51
  • \$\begingroup\$ Shouldn't the voltage gain actually be Z/sL+Z? \$\endgroup\$
    – A. Darwin
    Apr 9, 2016 at 9:53
  • \$\begingroup\$ it is a voltage divider, isnt it ? \$\endgroup\$
    – Fadi
    Apr 9, 2016 at 9:58

2 Answers 2


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.

  • \$\begingroup\$ please can you see above the calculation of transfer function \$\endgroup\$
    – Fadi
    Apr 9, 2016 at 9:52

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 $$ T(s) = \frac{Z}{sL + Z} $$

  • \$\begingroup\$ Thank you for your reply but it is different than the transfer function of the datasheet. \$\endgroup\$
    – Fadi
    Apr 9, 2016 at 11:14
  • \$\begingroup\$ I'm confident that the derivation is correct. Maybe there is some typo in the datasheet. You can check by simulation. Just use some arbitrary values. \$\endgroup\$
    – Mario
    Apr 9, 2016 at 11:23

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