I have the following passive circuit:

enter image description here

I want to calculate it's transfer function \$V_{out}(s)/V_{in}(s)\$.

My thoughts are: Part I: The voltage on the capacitor in the center (call it \$V_c(s)\$) is taken by the voltage divider created by the resistor and inductor in series to it's left and that capacitor i.e. \$V_c(s)=V_{in}(s)(1/sC)/((1/sC)(R//sL))\$ (these lines\$\\\$ mean that R and L are in parallel). Part II: Then since we have \$V_c(s)\$ which is of course a function of \$V_{in}(s)\$, we can calculate \$V_{out}(s)\$ from the voltage divider created by the rightmost resistor and the inductor on the left as\$V_{out}(s)= V_c(s)(R)/(R+sL)\$. Part III: Afterwards i substitude the expression of \$V_c(s)\$ that i calculated in part I and this way i find the expression for the transfer function \$V_{out}(s)/V_{in}(s)\$.

Are my ideas correct? If not what should i do? All ideas that can help me are welcome. Thanks in advance.


Sounds about right. However the voltage divider for Part one also needs to include the R and L of the right side of the cap.

Zc = (1/sC * (R + sL)) / (1/sC + R + sL)


Vc = Vin * Zc/(Zc + R + sL)

Unfortunately that's a bit of math to go through here. Since the Rs and Ls appear to the same, the whole thing will simplify quite a bit as you walk through the math.


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