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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 $R_2$.

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.

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 …….