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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 c1 in parallel with R & C2 then make a voltage divider in order to add the inductor. I always miss the capacitor or i dont get the same result. please anyone can tell me where is my mistake.

enter image description here

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  • \$\begingroup\$ What do you mean by "always miss the capacitor"? \$\endgroup\$ – TisteAndii Apr 8 '16 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 '16 at 8:20
  • \$\begingroup\$ above you can see the derivated transfer function, I have got completely different answers!! \$\endgroup\$ – Fadi Apr 9 '16 at 9:51
  • \$\begingroup\$ Shouldn't the voltage gain actually be Z/sL+Z? \$\endgroup\$ – A. Darwin Apr 9 '16 at 9:53
  • \$\begingroup\$ it is a voltage divider, isnt it ? \$\endgroup\$ – Fadi Apr 9 '16 at 9:58
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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.

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  • \$\begingroup\$ please can you see above the calculation of transfer function \$\endgroup\$ – Fadi Apr 9 '16 at 9:52
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The transfer function should be $$ T(s) = \frac{1 + C2 R s}{1 + C2 R s + (C1 L + C2 L) s^2 + C1 C2 L R s^3} $$ Your calculation of Z is correct, however you made some mistake during the calculation of $$ T(s) = \frac{Z}{sL + Z} $$

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  • \$\begingroup\$ Thank you for your reply but it is different than the transfer function of the datasheet. \$\endgroup\$ – Fadi Apr 9 '16 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 '16 at 11:23

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