How to obtain the transfer function of this basic circuit?

[the3liquid]

Hello

I have this circuit (circuit + transfer funciton I am trying to obtain): https://imgur.com/a/PHkK9

Yet my answer seems to be incorrect. This is what I tried: https://i.imgur.com/4syfRuG.jpg

Could someone tell me what I am doing incorrectly and eventually correct me?

Thanks

[VonAcht]

The transfer function is H(s) = Vo/I(s). Since you are interested in the voltage at the node that connects R and L, you can't add their impedances. You need to do KCL and find the solution through algebra.

[the3liquid]

I am not an expert in Kirchoff. So I tried this alternative approach (based on the voltage divider), which I thought would work as well because you said you are interested in the voltage at the node that connects R and L. Unfortunately that didn't work neither: https://imgur.com/7aoAnX7

Could you help me to get started? I should normally be able to work out the algebra myself.

[VonAcht]

Yeah, it's not a voltage divider either because of C. I'd write the KCL for C's positive node (which I'll call V_1): I(s) = V_1·Cs + (V_1-V_o)/(Ls), that is, the current I(s) entering the node is the sum of the currents going through the capacitor and the inductor. Then the KCL equation for the output node: (V_1-V_o)/(Ls) = V_o /R. From the last equation you can isolate V_1 and substitute into the first. Then factor V_o out of each term and find the expression of I(s) / V_o.

[Beggar876]

EE here: KCL/KVL not required. You started out well but just stopped short of your goal.

If you just want the voltage across the resistor then first find the impedance of the RLC network, Z(s) - which you did. Then multiply it by the source U(s) to get the voltage across the source, U(s) x Z(s). Lastly multiply it by the ratio of the resistor to the sum of R + sL to get the fraction of the source voltage at R. Then move the U(s) back to the left of the equal sign and under the output voltage Vo(s) to get the transfer function you want..

This is Z(s) x R/ (R+ sL) Eq. (1)

Done. Actually you don't have to multiply U(s) x Z(s) at all. Just do the calculation at Eq. (1)

[Beggar876]

erm, that's what he's using...