Been trying this for quite a few hours and no success anyone understand how this is meant to be solved?
Given the electric circuit below, find it's transfer function considering v(t) as an input and i₂ as the output. Initial conditions are null. Data: R1 = 0.5 ohms. R2 = 1 ohm, C = 1 FARAD and L = 1 Henry. Transfer function should be output divided by input.
Electric circuit for the problem at hand.
Answer should be something like this: G(s) = (s²)/(1.5s²+1.5s+0.5)
Let us list all the possible loops and physical laws. Then, we analyze the circuit. For this case, I prefer to use $p$ as my Laplace transformation, for $p=j\omega$.
Physical laws in time and $p$-domain: \begin{align} v_L(t) & = L(\dot{i_1} - \dot{i_2})(t) \space \leftrightarrow \space V_L(p) = pL(I_1 - I_2)(p) & \text{(P1)} \\ i_2(t) & = C\dot{v_C}(t) \space \leftrightarrow \space I_2(p) = pCV_C(p) & \text{(P2)} \\ \end{align}
KVL loops in $p$-domain: \begin{align} V(p) & = R_1I_1(p)+V_L(p) & \text{(KVL1)} \\ V_C(p) & = -R_2I_2(p)+V_L(p) & \text{(KVL2)} \\ \end{align}
VDR in $p$-domain: \begin{align} V_L(p) & = \frac{pL}{R_1+pL}V(p) = \frac{pL}{R_2+pL}V_C(p) & \text{(VDR1)} \\ \end{align}
Let us consider VDR1. We express $V_C(p)$ as a function of $V(p)$. \begin{align} V_C(p) & = \frac{R_2 + pL}{pL} V_L(p) \\ & = \frac{R_2 + pL}{pL} \frac{pL}{R_1 + pL}V(p) \\ & = \frac{R_2 + pL}{R_1 + pL} V(p) \\ \end{align}
Next, let us consider law P2 and KVL2 in $p$-domain. \begin{align} I_2(p) & = pCV_C(p) \\ & = pC \left( -R_2I_2(p) + \frac{pL}{R_1+pL}V(p) \right) \\ & = -pR_2CI_2(p) + \frac{p^2LC}{R_1+pL}V(p) \\ <=> (1+pR_2C)I_2(p) & = \frac{p^2LC}{R_1+pL}V(p) \\ <=> I_2(p) & = \frac{p^2LC}{(1+pR_2C)(R_1+pL)}V(p) \\ <=> \frac{I_2(p)}{V(p)} & = \frac{p^2LC}{R_1+p(R_1R_2C+L) + p^2(R_2LC)} \\ \end{align}
If you plug in the values, the transfer function is: $$ \frac{I_2(p)}{V(p)} = \frac{p^2}{0.5 + 1.5p + p^2} $$
Please verify it on your own to ensure it is correct. For circuit questions, it is more appropriate to place in the Electrical Engineering stackexchange as this section is mainly for math problems. Hope it helps.