Here is an exercise from my textbook.
]1
At first I solve it by using Kirchhoff's laws directly and using complex impendance:
$$U_{in}=\frac{1}{Cs}(i_1-i_2)+i_1R_1$$
$$0=R_2i_2+\frac{1}{Cs}i_2+\frac{1}{Cs}(i_2-i_1)$$
By solving above equations, both $I_1$ and $I_2$ become function of $U_{in}$. Then I sub them into $U_o=i_1R_1+\frac{1}{Cs}i_2$ to get the relationship between $U_o$ and $U_{in}$. This method works and lead to the answer provided in the book.
However I found a solution looks quite different in solution manual with determinant:
Let,
$$\Delta=\begin{vmatrix}
G_2+Cs & -Cs & -G_2\\
-Cs & G_1 + 2Cs & -Cs \\
-G_2 & -Cs & Cs + G_2
\end{vmatrix}$$
then,
$$V_j=\dfrac {\Delta_{1j}} \Delta I_1~~~~~~or~~~~~~
\dfrac {V_3}{V_1}=\dfrac{\Delta_{13}I_1/\Delta}{\Delta_{11}I_1/\Delta}$$
Therefore the transfer function is
$$\dfrac {V_3}{V_1}=\dfrac{\Delta_{13}}{\Delta_{11}}=\dfrac{\begin{vmatrix}-Cs & 2Cs+G_1\\-G_2 & -Cs\end{vmatrix}}
{\begin{vmatrix}2Cs+G_1 & -Cs\\-Cs & Cs+G_2\end{vmatrix}}$$
$$=\dfrac{C^2 R_1 R_2 s^2+ 2CR_1 s+1}{C^2 R_1 R_2 s^2+(2R_1+R_2)Cs+1}$$
So far I am quite confused on what does the first determinant stands for and could this determinant be written down directly without any draft? (If so, I think it will be faster and easier to solve problems like this and I really want to learn about it.) Also, why do we need to use the Minor of that determinant?
Best Answer
I have added some theory before the recipe.
$s= \ j \omega$ where $\omega$ ids the angular frequency and $G$ is the conductance of a resistor of resistance $R$ with $G= \dfrac 1R$
There are three Kirchhoff's current law equations which can be set up for each of the nodes 1, 2 and 3.
For node 1 $(V_1-V_2)sC +(V_3-V_1) G_2 + I_1=0$ where $I_1$ is the current entering the network at node 1 from $+V_{\rm in}$.
Rearranging this equation gives $I_1=(G_2+sC) V_1-sCV_2-G_2V_3$
Repeating the process at nodes 2 and 3 gives $I_2 = -sCV_1+(G_1+2sC)V_2-sCV_3$ and $I_3= -G_2V_1 -sCV_2+(sC+G_2)V_3$
So the admittance matrix for this network is $\mathbf Y=\begin{bmatrix} G2+sC & -sC & -G_2 \\ -sC & G_1 + 2sC & -sC \\ -G_2 & -sC & sC+G_2\\ \end{bmatrix}$ and the matrix equation which links the currents $\mathbf I=\begin{bmatrix} I_1 \\ I_2 \\ I_3 \\ \end{bmatrix}$ and the voltages $\mathbf V=\begin{bmatrix} V_1 \\ V_2 \\ V_3 \\ \end{bmatrix}$ is $\mathbf I = \mathbf Y \,\mathbf V$
Here is the (cookery book) recipe for producing the admittance matrix.
You are dealing with admittances and four nodes which I have labelled 0, 1, 2 and 3 and three nodal voltages $V_1,\,V_2$ and $V_3$ which are relative to the reference node 0 which is at a potential of zero.
For element 11 you see that there are two circuit components, resistor $R_2$ and left-hand capacitor $C$, connected to node 1.
You add the two admittances to give $G_2 + sC$ and this is the value of element 11 of matrix $\Delta$.
For element 22 you see that there are three circuit components connected to node 2 and you add their admittances which produces $G_1+2sC$.
Finally element 33 have a value of $G_2+sC$.
Now you need to look at components which are common to two nodes to find elements $Y_{\rm ij}$ with $i\ne j$.
The left hand capacitor is common to node 1 and node 2 so the entry for element 12 and 21 is minus the admittance of the component which is $-sC$ in this case.
Elements 23 and 32 are also equal to $-sC$.
Elements 13 and 31 are equal to $-G_2$.
You have thus populated all the elements of the matrix $\mathbf Y$.
To finish off the problem Cramer's rule is used.
@PageDavid has provide a reference, Nodal Admittance Matrix.