I'm trying to obtain the transfer function which relates the input voltage with the inductor's voltage H(s) = Vl(S)/U(s)
so far I've tried to use Kirchoff's Laws in order to find it. Any suggestions? I'm not familiar with this circuit so much so I'm trying to get as much info as possible.
Thank you.
Edit: tried:
Edit: Vl/V(s):
I can't write this in a comment and it seems worthwhile to perform the algebra steps I took:
$$\begin{align*} \frac{V_{_\text{L}}-V\left(s\right)}{R_1}+\frac{V_{_\text{L}}}{R_2}+\frac{V_{_\text{L}}}{s\,L_1}+V_{_\text{L}} s C_1 &=0 \\\\ \frac{V_{_\text{L}}}{R_1}+\frac{V_{_\text{L}}}{R_2}+\frac{V_{_\text{L}}}{s\,L_1}+V_{_\text{L}} s C_1 &=\frac{V\left(s\right)}{R_1} \\\\ s R_1 R_2 L_1\left[\frac{V_{_\text{L}}}{R_1}+\frac{V_{_\text{L}}}{R_2}+\frac{V_{_\text{L}}}{s\,L_1}+V_{_\text{L}} s C_1\right] &=s R_1 R_2 L_1\left[\frac{V\left(s\right)}{R_1}\right] \\\\ s R_2 L_1V_{_\text{L}}+s R_1 L_1V_{_\text{L}}+R_1 R_2 V_{_\text{L}}+s^2 R_1 R_2 L_1 V_{_\text{L}} C_1 &=s R_2 L_1V\left(s\right) \\\\ \left(R_1 R_2 L_1 C_1 s^2 +s R_2 L_1+s R_1 L_1+R_1 R_2 \right)V_{_\text{L}}&=s R_2 L_1V\left(s\right) \\\\ \therefore \frac{V_{_\text{L}}}{V\left(s\right)}=\frac{R_2 L_1 s}{R_1 R_2 L_1 C_1 s^2 +\left(R_1+R_2 \right)L_1s+R_1 R_2} \end{align*}$$
You could just stop here. But there's a few standard forms for a bandpass like this that provide a little better information.
The first thing you'd want to do is to find that \$\omega_{_0}=\sqrt{\frac{b_0}{b_2}}=\sqrt{\frac{R_1 \,R_2}{R_1 \,R_2\,L_1\,C_1}}=\frac1{\sqrt{L_1\,C_1}}\$ (I use zero, not 'c' or 'p'.) You may want to highlight that fact in the resulting equation, somehow. (Yeah, \$b_2\$ is the coefficient for \$s^2\$, etc., in the denominator.)
The next thing you'd want to do is to find either the damping factor, \$\zeta\$, or else the \$Q\$. (Note: \$\zeta=\frac1{2 \,Q}\$.) It turns out that \$\zeta=\frac12\frac{b_1}{\sqrt{b_2\,b_0}}\$. (You can get \$Q=\frac{\sqrt{b_2\,b_0}}{b_1}\$, too.)
Then some standard forms are:
$$\begin{align*} \underbrace{\overbrace{K}^{\text{gain}}\cdot\overbrace{\left[\frac{2\zeta\,\omega_{_0}s}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:2}}=\frac{\frac1{Q}\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2+\frac1{Q}\left(\frac{s}{\omega_{_0}}\right)+1}\right]}^{\text{standard bandpass form}}}_{\text{standard bandpass form with gain}} \end{align*}$$
To get it into that form, you need to wrestle with the numerator in order to split it up into the \$K\$ part and the \$\frac1{Q}\$ part (or \$2\zeta\$ part.)
\$\zeta\$ might be preferred over \$Q\$ when \$Q\le 1\$ with \$Q\$ preferred, otherwise. It's a matter of what shapes are of more concern, which may be picked out.
But you may be able to get away with any correct form and don't need to find standard forms.
We can obtain the transfer function much easier by applying the voltage divider formula:
\begin{equation} \begin{split} \mathscr{H}\left(\text{s}\right)&=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}\\ \\ &=\frac{\displaystyle\text{R}_2\space\text{||}\space\text{sL}\space\text{||}\space\frac{1}{\text{sC}}}{\displaystyle\text{R}_1+\left(\text{R}_2\space\text{||}\space\text{sL}\space\text{||}\space\frac{1}{\text{sC}}\right)}\\ \\ &=\frac{\displaystyle\displaystyle\frac{\displaystyle1}{\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\frac{1}{\frac{1}{\text{sC}}}}}{\displaystyle\text{R}_1+\displaystyle\frac{\displaystyle1}{\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\frac{1}{\frac{1}{\text{sC}}}}}\\ \\ &=\frac{\displaystyle\displaystyle\frac{\displaystyle1}{\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\text{sC}}}{\displaystyle\text{R}_1+\displaystyle\frac{\displaystyle1}{\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\text{sC}}}\\ \\ &=\frac{\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\text{sC}}{\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\text{sC}}\cdot\frac{\displaystyle\displaystyle\frac{\displaystyle1}{\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\text{sC}}}{\displaystyle\text{R}_1+\displaystyle\frac{\displaystyle1}{\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\text{sC}}}\\ \\ &=\frac{1}{\displaystyle\text{R}_1\left(\displaystyle\frac{1}{\text{R}_2}+\frac{1}{\text{sL}}+\text{sC}\right)+1}\\ \\ &=\frac{\text{R}_2}{\displaystyle\text{R}_1\left(\displaystyle\frac{\text{R}_2}{\text{R}_2}+\frac{\text{R}_2}{\text{sL}}+\text{sCR}_2\right)+1}\\ \\ &=\frac{\text{R}_2}{\displaystyle\text{R}_1\left(\displaystyle1+\frac{\text{R}_2}{\text{sL}}+\text{sCR}_2\right)+1}\\ \\ &=\frac{\text{sLR}_2}{\displaystyle\text{R}_1\left(\displaystyle\text{sL}+\frac{\text{sLR}_2}{\text{sL}}+\text{s}^2\text{CLR}_2\right)+1}\\ \\ &=\frac{\text{sLR}_2}{\displaystyle\text{R}_1\left(\displaystyle\text{sL}+\text{R}_2+\text{s}^2\text{CLR}_2\right)+1}\\ \\ &=\frac{\text{sLR}_2}{\displaystyle\displaystyle\text{sLR}_1+\text{R}_1\text{R}_2+\text{s}^2\text{CL}\text{R}_1\text{R}_2+1}\\ \\ &=\frac{\text{sLR}_2}{\displaystyle\displaystyle\text{s}^2\text{CL}\text{R}_1\text{R}_2+\text{sLR}_1+\text{R}_1\text{R}_2+1} \end{split}\tag1 \end{equation}
Where \$\displaystyle\alpha\space\text{||}\space\beta\space\text{||}\space\gamma:=\displaystyle\frac{\displaystyle1}{\displaystyle\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}}\$.
When using sinusoidal functions we can use \$\text{s}:=\text{j}\omega\$. Now, we can obtain the absolute value:
\begin{equation} \begin{split} \left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|&=\left|\frac{\text{j}\omega\text{LR}_2}{\displaystyle\displaystyle\left(\text{j}\omega\right)^2\text{CL}\text{R}_1\text{R}_2+\text{j}\omega\text{LR}_1+\text{R}_1\text{R}_2+1}\right|\\ \\ &=\frac{\left|\text{j}\omega\text{LR}_2\right|}{\left|\displaystyle\text{j}\omega\text{LR}_1+\text{R}_1\text{R}_2-\omega^2\text{CL}\text{R}_1\text{R}_2+1\right|}\\ \\ &=\frac{\omega\text{LR}_2}{\displaystyle\sqrt{\left(\text{R}_1\text{R}_2-\omega^2\text{CL}\text{R}_1\text{R}_2+1\right)^2+\left(\omega\text{LR}_1\right)^2}}\\ \\ &=\frac{\omega\text{LR}_2}{\displaystyle\sqrt{\left(\text{R}_1\text{R}_2\left(1-\omega^2\text{CL}\right)+1\right)^2+\left(\omega\text{LR}_1\right)^2}} \end{split}\tag2 \end{equation}
In order to find the argument, lets define \$\alpha:=\text{R}_1\text{R}_2\left(1-\omega^2\text{CL}\right)+1\$ and \$\beta:=\omega\text{LR}_1\$. So, we get:
\begin{equation} \begin{split} \arg\left(\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right)&=\arg\left(\frac{\text{j}\omega\text{LR}_2}{\displaystyle\text{j}\omega\text{LR}_1+\text{R}_1\text{R}_2\left(1-\omega^2\text{CL}\right)+1}\right)\\ \\ &=\arg\left(\text{j}\omega\text{LR}_2\right)-\arg\left(\text{j}\omega\text{LR}_1+\text{R}_1\text{R}_2\left(1-\omega^2\text{CL}\right)+1\right)\\ \\ &=\frac{\pi}{2}-\displaystyle\begin{cases} \displaystyle0&\space\text{if}\space\displaystyle\alpha=0\space\wedge\space\beta=0\\ \\ \displaystyle\frac{\pi}{2}&\space\text{if}\space\displaystyle\alpha=0\space\wedge\space\beta>0\\ \\ \displaystyle\pi&\space\text{if}\space\displaystyle\alpha<0\space\wedge\space\beta=0\\ \\ \displaystyle\frac{3\pi}{2}&\space\text{if}\space\displaystyle\alpha=0\space\wedge\space\beta<0\\ \\ \displaystyle\arctan\left(\frac{\beta}{\alpha}\right)&\space\text{if}\space\displaystyle\alpha>0\space\wedge\space\beta>0\\ \\ \displaystyle\frac{\pi}{2}+\arctan\left(\frac{\left|\alpha\right|}{\beta}\right)&\space\text{if}\space\displaystyle\alpha<0\space\wedge\space\beta>0\\ \\ \displaystyle\pi+\arctan\left(\frac{\left|\beta\right|}{\left|\alpha\right|}\right)&\space\text{if}\space\displaystyle\alpha<0\space\wedge\space\beta<0\\ \\ \displaystyle\frac{3\pi}{2}+\arctan\left(\frac{\alpha}{\left|\beta\right|}\right)&\space\text{if}\space\displaystyle\alpha>0\space\wedge\space\beta<0 \end{cases}\\ \\ &=\displaystyle\begin{cases} \displaystyle\frac{\pi}{2}&\space\text{if}\space\displaystyle\alpha=0\space\wedge\space\beta=0\\ \\ \displaystyle0&\space\text{if}\space\displaystyle\alpha=0\space\wedge\space\beta>0\\ \\ \displaystyle-\frac{\pi}{2}&\space\text{if}\space\displaystyle\alpha<0\space\wedge\space\beta=0\\ \\ \displaystyle\frac{\pi}{2}-\arctan\left(\frac{\beta}{\alpha}\right)&\space\text{if}\space\displaystyle\alpha>0\space\wedge\space\beta>0\\ \\ \displaystyle-\arctan\left(\frac{\left|\alpha\right|}{\beta}\right)&\space\text{if}\space\displaystyle\alpha<0\space\wedge\space\beta>0 \end{cases} \end{split}\tag3 \end{equation}
Without getting into graph theory you can solve this problem by either node voltages method or loop currents. There are 2 nodes and 3 loops and the 1'st node voltage is readily defined by the source. So it is advantageous to use node voltages. Using this method you write down the current balance for each node wihout forgetting to include the current through the voltage source current. You then replace currents with their equivalents interms of node voltages using ohms law (using Ls and 1/Cs for the inductor and the capacitor). Now you should end up with 2 unknowns 1- The current through the source at node 1 and the voltage at node 2 (Where the inductor and the capacitor are connected). You also have 2 independent sets of equations which can be solved. İf you prefer using mesh currents you must take the principal current that defines the mesh. There should be no more mesh currents then the number of meshes. Then you use the voltage equations round 3 meshes (taking care of currents being common to more than one meshes) You end up with 3 euations in 3 unknowns (3 principal mesh currents). Please note that this method is more advantageous if current sources are used. Also please take a note that picking up mesh currents and node voltages this way arent the only way. In general you use fundamental cut set equations (current balance) for solving in terms of node voltages and Use fundamental loop equations for solving in terms of loop currents. For that you will have to establish a tree of the garph (a set of branches that goes to every node of the circuit without makin a loop). But this is a more theoretical approach and involve notions of graph theory. My suggestion is the use of node voltage equations. İf you short out (combine the ends of R2, L and C) I3 in your equations you get I1 = I2 + I4 + I5 (there will be no I3). If this node is assigned as 2 then you have : (V1-V2)/R1 = V2(1/R2 + 1/Ls + Cs) and V1 = Vs(s) (given in time domain as V(t) in your diagram). It should be fairly easy to solve for the final equation V2(1/R1 + 1/R2 + 1/Ls + Cs) = V1/R1 for V2 in terms of component values and source voltage V1 then I1, I2,I4,I5 may easily be calculated. In fact the equation may automatically written by summing up the conductances emanating from a node with a (+) sign and those shared with another node with (-) sign and equate to zero. (Please note that R1 is shared here but we assighned I1 in opoposite direction to bring current into node 2 whereas other current directions draw current away).
