What are the final steps to arrive at the transfer function for this RCL circuit?
I am attempting to arrive at this transfer function $$\dfrac{V_o(s)}{V_i(s)}=\dfrac{sL}{CL(R_1+R_2)s^2+(R_1R_2C+L)s+R_1}$$ for the following RCL circuit.
My initial problem was not knowing the correct steps to take to get from the circuit to the final TF, using the complex impedance method.
So, I have now used the KVL method, which has got me to a final transfer function. But, not the one that I desire.
$$\dfrac{I_2(s)}{V_1(s)}=\dfrac{s^2CL}{CL(R_1+R_2)s^2+(R_1R_2C+L)s+R_1}$$
Therefore, I would just like to know whether or not the last remaining step is to take \$V_o = I_2(s)Z_c\$, where \$Z_c = \frac{1}{sC}\$ and therefore multiply \$I_2(s)\$ by \$\frac{1}{sC}\$ and thus reduce the numerator from \$s^2CL\$ to \$sL\$
So, these are the steps taken using KVL:
Loop 1:
$$V_1(s) - I_1(s)R_1 - L(sI_1(s)-sI_2(s)) = 0$$
$$I_1(s)(R_1 + sL)- sLI_2(s) = V_1(s)\tag{1}$$
Loop 2:
$$I_2(s)R_2 + \frac{I_2(s)}{sC} + L(sI_2(s)-sI_1(s)) = 0$$
$$-sLI_1(s)+I_2(s)(sL+R_2+\frac{1}{sC}) = 0\tag{2}$$
From (2) I get that \$I_1\$
$$I_1=I_2(s)(sL+R_2+\frac{1}{sC})\frac{1}{sL}$$
$$I_1=I_2(s)(1+\frac{R_2}{sL}+\frac{1}{s^2CL})\tag{3}$$
which, when substituted into (1), gives
$$I_2(s)(1+\frac{R_2}{sL}+\frac{1}{s^2CL})(sL+R_1) - sLI_2(s) = V_1(s)\tag{4}$$
Multiplying through by \$(sL+R_1)\$
$$I_2(s)(sL+\frac{sLR_2}{sL}+\frac{sL}{s^2CL}+R_1+\frac{R_1R_2}{sL}+\frac{R_1}{s^2CL})-sLI_2(s)= V_1(s)\tag{5}$$
Multiplying through by \$I_2(s)\$
$$sLI_2(s)+\frac{sLR_2}{sL}I_2(s)+\frac{sL}{s^2CL}I_2(s)+R_1I_2(s)+\frac{R_1R_2}{sL}I_2(s)+\frac{R_1}{s^2CL}I_2(s)-sLI_2(s)= V_1(s)\tag{6}$$
cancelling out like terms, I get
$$\frac{sLR_2}{sL}I_2(s)+\frac{sL}{s^2CL}I_2(s)+R_1I_2(s)+\frac{R_1R_2}{sL}I_2(s)+\frac{R_1}{s^2CL}I_2(s)= V_1(s)$$
factoring out \$I_2(s)\$, I get
$$I_2(s)(\frac{R_1}{s^2CL}+\frac{R_1R_2}{sL}+\frac{1}{sC}+R_1+R_2)= V_1(s)\tag{7}$$
obtaining a common denominator, I get the following
$$I_2(s)(\frac{R_1}{s^2CL}+\frac{sR_1R_2C+sL}{s^2CL}+R_1+R_2)= V_1(s)\tag{8}$$
$$I_2(s)(\frac{R_1}{s^2CL}+\frac{sR_1R_2C+sL+s^2CL(R_1+R_2)}{s^2CL})= V_1(s)\tag{9}$$
factoring out the \$1/s^2CL\$, I get
$$\frac{I_2(s)}{s^2CL}(R_1+sR_1R_2C+sL+s^2CL(R_1+R_2)= V_1(s)\tag{10}$$
Rearranging in terms of \$I_2(s)\$ and \$V_1(s)\$, I get
$$\dfrac{I_2(s)}{V_1(s)}=\dfrac{s^2CL}{CL(R_1+R_2)s^2+(R_1R_2C+L)s+R_1}\tag{11}$$
So, is the last step to take \$V_o = I_2(s)Z_c\$, where \$Z_c = \frac{1}{sC}\$ and therefore multiply \$I_2(s)\$ by \$\frac{1}{sC}\$ whereby reducing the numerator from \$s^2CL\$ to \$sL\$?
UPDATE:
Following on from the answer provided (see below), I am trying to follow the steps as outlined and have hit a stumbling block.
$$\therefore \frac{V_\text{OUT}}{V_\text{IN}}=\frac{1}{1+\frac{R_1}{Z_1}}\cdot\frac{1}{1+\frac{R_2}{Z_2}}$$
From that, I get:
$$H\left(s\right)=\frac{L\,s}{L\,C\left(R_1+R_2\right)s^2+\left(R_1\,R_2\,C+L\right)s+R_1}$$
If I substitute in the expressions for \$Z_1\$ and \$Z_2\$, I arrive at the following (which is where I get stuck):
$$\frac{V_\text{OUT}}{V_\text{IN}}=\frac{1}{1+\frac{R_1}{\frac{s^2R_2CL+sL}{s^2CL+sR_2C+1}}}\cdot\frac{1}{1+\frac{R_2}{\frac{1}{sC}}}$$
$$\frac{V_\text{OUT}}{V_\text{IN}}=\frac{1}{1+\frac{s^2R_1CL+sR_1R_2C+R_1}{s^2R_2CL+sL}}\cdot\frac{1}{1+sR_2C}$$
$$\frac{V_\text{OUT}}{V_\text{IN}}=\frac{s^2R_2CL+sL}{L\,C\left(R_1+R_2\right)s^2+\left(R_1\,R_2\,C+L\right)s+R_1}\cdot\frac{1}{1+sR_2C}$$