Differences

This shows you the differences between two versions of the page.

--- ipt_resonance_frequency [2024/04/05 07:18] – admin
+++ ipt_resonance_frequency [2024/04/05 21:09] (current) – [Determination of the resonance frequency] 2a02:1812:40f:9800:b9c3:9c84:7c6b:4f8e
@@ Line 5: / Line 5: @@
 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?400|}}
+{{:coupled-coils-with-c-with-input-and-output-voltage.png?500|}}
 ==== Impedance matix ====
@@ Line 27: / Line 28: @@
 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 .......