# Transfer Function RLC

I would calculate the transfer function of this circuit.

I have the following questions: I can do the series resistance (R + L) and then make the parallel with the capacitor. Finally make the voltage divider with the resistance R).

It would be changing the value of Vo(s)? In this case changing the value, someone could help me solve it?

I also attached a photo of Laplace transformed circuit.

Thank you.

• Your question is not clear. What do you think is changing the value of $V_o(s)$? What starting point do you think it is changing from? Commented Jan 18, 2016 at 21:33
• I am using zero initial conditions. The value vi(s) is the input value and the value vo (s) is the output. Commented Jan 18, 2016 at 21:48
• OK, then what did you change that you want to know whether it causes a change in $V_o(s)$? Commented Jan 18, 2016 at 21:58
• Determine the voltage across the series R/L (or across C), then use the voltage divider rule to find the voltage across L.
– Chu
Commented Jan 18, 2016 at 23:33

The series resistor and inductor $$Z_{RL} = R_1 + Ls$$ In parallel with the capacitor $$Z_{RLC} = (R_1 + Ls) || (\frac1 {Cs})$$ The voltage across the RL branch $$V_{RL} = (\frac{Z_{RLC}}{Z_{RLC} + R_2})V_i(s)$$

Taking $V_o(s)$ as the voltage across the inductor $$V_o(s) = (\frac{Ls}{Z_{RL}})V_{RL} = (\frac{Ls}{Z_{RL}})(\frac{Z_{RLC}}{Z_{RLC} + R_2})V_i(s)$$ Thus, the transfer function is: $$H(s) = \frac{V_o(s)}{V_i(s)} = (\frac{Ls}{Z_{RL}})(\frac{Z_{RLC}}{Z_{RLC} + R_2})$$

Simplify away :)

The easiest way is to convert each component into a Z component, and then treat them like resistors and break it down to simple algebra before re-introducing the components...

Lets call the first resistor R1 and the second R2, than call all of them Z underscore component name... The || means use the resistors in parallel equation or 'X*Y/(X+Y)'...

Z_final = Z_R1 + [(Z_R2+Z_L)||Z_C]

Reintroducing the values, you would have

Z_final = R + [(R2+Ls)||(1/Cs)]

Further algebra would give you the simplest answer.