===== Resonance frequency of coupled coils =====

==== Circuit ====

We consider two coils with inductances $L_1$ and $L_2$, coupled by mutual inducance $M$. In series, capacitors $C_1$ and $C_2$ are added to the circuit. The resistive losses of the circuit are represented by the resistances $R_1$ and $R_2$.

{{:coupled-coils-with-c-with-input-and-output-voltage.png?500|}}

==== Impedance matix ====

The impedance matrix of the circuit is given by:

\begin{align}
	Z =
        \begin{bmatrix}
		z_{11}  & z_{12} \\
		z_{21}  & z_{22} \\
	\end{bmatrix}=
	\begin{bmatrix}
		R_1+j\omega L_1+\frac{1}{j \omega C_1} & j \omega M  \\
		j \omega M  & R_2+j\omega L_2 +\frac{1}{j \omega C_2}\\
	\end{bmatrix}
\end{align}

with $\omega$ the angular frequency.

==== Determination of the resonance frequency ====

We apply an input voltage $V_{in}$ at the left. The resulting voltage $V_{out}$ at the right is called the output voltage.

The relationships between voltage and current is given by:
\begin{align}
	Z =
        \begin{bmatrix}
		V_{in}  \\
		V_{out}  \\
	\end{bmatrix}=
        \begin{bmatrix}
		z_{11}  & z_{12} \\
		z_{21}  & z_{22} \\
	\end{bmatrix} .
        \begin{bmatrix}
		I_{1}  \\
		I_{2}  \\
	\end{bmatrix}
\end{align}

At the output port, no load is connected. As a result, $I_2=0$. The voltage-current relationships simplifies to

$$ \begin{cases}
V_{in}=z_{11}.I_1\\
V_{out}=z_{21}.I_1
\end{cases} $$ 

and thus:

$V_{out}=\frac{z_{21}}{z_{11}}V_{in}$

For our coupled coils circuit, we get:

$V_{out}=\frac{j\omega M}{R_1+j\omega L_1+\frac{1}{j \omega C_1}}V_{in}$

At very low frequency $(\omega \rightarrow 0)$, the magnitude of the output voltage is given by $\omega^2 MC_1 \rightarrow 0$. In other words, at low frequency, the output voltage is low.

At very high frequency $(\omega \rightarrow \infty)$, the magnitude of the output voltage is given by $\frac{M}{L_1}V_{in}$. In other words, at high frequency, the output voltage .......