# Confused with network analysis

Can someone please explain me their workflow in solving the following transferfunction. I have to find the output-voltage Vo over RL, for an input-voltage of Vi. I'm not used to network analysis, and it's pretty confusing.

As far is got, this is how I tried to solve it: - change every component by it's impedance - calculate the parallel equivalence impedance Z, of L1 & C1 (Z=L1//C1) - serial sommation of Z & L2 - voltage divider, for the voltage Vo over RL

However, i don't seem to get the correct result

• Show us you intermediate and final results: What Z do you get for L1 || C, what Z do you get after adding L2, what do you get for vo? – Curd May 6 '15 at 14:44

It's not clear to me the procedure you followed, but it seems to me you got it somewhat reversed.

Using obvious notations and calling $V_{C1}$ the voltage across C1:

$V_o = V_{C1} \; \dfrac{R_L}{R_L+Z_{L2}}$

where:

$V_{C1} = V_i \; \dfrac{Z_{eq}}{Z_{L1}+Z_{eq}} \qquad$ and $\qquad Z_{eq} = Z_{C1} \parallel (R_L+Z_{L2})$

put these equations together and you should have the solution.

• I think the OP was trying to find the Thevenin equvialent for the node between L1 and C and then go on to find vo as the result of a voltage devider whose lower resistor is RL. – Curd May 6 '15 at 15:09
• @Curd Yep! Now that you point it out, it makes sense. I didn't think of Thevenin. – Lorenzo Donati May 6 '15 at 15:27
• I'm sorry for the late response, but thanks for your answer! So it seems you've used voltage dividers. I was confusing myself by trying to use several network theorema's at the same time. Thanks!And also @Curd thanks for your help! – c0s1n3 Jun 4 '15 at 14:13

Write A as the L1/C/L2 node voltage, and replace the components by their Laplace equivalent impedances: sL1, 1/sC, sL2, R (=RL)

Node A: (A-V0)/sL2 + AsC +(A-Vi)/sL1 =0 ...(1)

Node Vo: Vo/R + (Vo-A)/sL2 = 0 ...(2)

Solve for A from (2); then substitute into (1) and solve the resultant equation for Vo(s)/Vi(s)

Answer is a standard 2nd order transfer function