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partity-time_symmetric_wpt [2025/07/07 14:00] – [Output power and efficiency] klpartity-time_symmetric_wpt [2025/07/07 14:27] (current) – [Efficiency] kl
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 \begin{bmatrix} \begin{bmatrix}
 0 \\ 0 0 \\ 0
-\end{bmatrix}(1)+\end{bmatrix}~~~~~~\text{(eq. 1)} 
 $$  $$ 
  
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 $$R_1=R_2=0$$ $$R_1=R_2=0$$
 $$R_N=R_L$$ $$R_N=R_L$$
-Note in particulary that a PP topology is only exact PT-symmetric if no internal losses are present. In good WPT systems, the resistive losses should be small, resulting in a good approximate PT-symmetric PP system.+Note in particulary that a PP topology is only exact PT-symmetric if no internal losses are present. In well-designed WPT systems, the resistive losses should be small, resulting in a good approximate PT-symmetric PP system.
  
 Whereas it is possible to obtain an exact PP-symmetric circuit for SS and PP compenation topologies, this is not the case for SP and PS topologies. However, approximations are possible: if the coupling factor is low, and the coil quality factors are large (i.e. the resistances $R_1$ and $R_2$ are small), then SP and PS are in a good approximation also PT-symmetric, even though the circuits are obviously not parity symmetric (e.g., SP: serie in transmitter vs parallel compensation at receiver side == not symmetric). Whereas it is possible to obtain an exact PP-symmetric circuit for SS and PP compenation topologies, this is not the case for SP and PS topologies. However, approximations are possible: if the coupling factor is low, and the coil quality factors are large (i.e. the resistances $R_1$ and $R_2$ are small), then SP and PS are in a good approximation also PT-symmetric, even though the circuits are obviously not parity symmetric (e.g., SP: serie in transmitter vs parallel compensation at receiver side == not symmetric).
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 $$ $$
-[-R_N+R_1+j(\omega L_1-\frac{1}{\omega C_1})].[R_L+R_2+j(\omega L_2-\frac{1}{\omega C_2})]+\omega^2L_{12}^2=0       +[-R_N+R_1+j(\omega L_1-\frac{1}{\omega C_1})].[R_L+R_2+j(\omega L_2-\frac{1}{\omega C_2})]+\omega^2L_{12}^2=0   ~~~~~~\text{(eq. 2)}     
-$$ +$$ 
-(2)+
  
 ===Exact PT-symmetric state=== ===Exact PT-symmetric state===
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 $$L_1=L_2$$ $$L_1=L_2$$
 $$C_1=C_2$$ $$C_1=C_2$$
-$$-R_N+R_1=R_L+R_2$$ (3)+$$-R_N+R_1=R_L+R_2~~~~~~\text{(eq. 3)} $$
  
 Solving equation (2) with respect to $\omega$ results in two solutions: Solving equation (2) with respect to $\omega$ results in two solutions:
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 From the real part of equation (2), we find that the solution $\omega_{o3}=\omega_0$ only holds when the value of the negative resistor is given by: From the real part of equation (2), we find that the solution $\omega_{o3}=\omega_0$ only holds when the value of the negative resistor is given by:
  
-$$R_N=R_1+\frac{\omega_0 k^2L_1L_2}{R_2+R_L}$$ (4)+$$R_N=R_1+\frac{\omega_0 k^2L_1L_2}{R_2+R_L}~~~~~~\text{(eq. 4)} $$ 
  
-===Conclusion resonance frequency SS-circuit===+===Conclusion resonance frequency===
  
 When $k \geq k_c$, there are two resonance frequencies, $\omega_{o1}$ and $\omega_{o2}$. In this case, from equation (3), the negative resistor automatically compensates the other resistors in the network $R_1, R_2$ and $R_L$. When $k \geq k_c$, there are two resonance frequencies, $\omega_{o1}$ and $\omega_{o2}$. In this case, from equation (3), the negative resistor automatically compensates the other resistors in the network $R_1, R_2$ and $R_L$.
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 For $k<k_c$, the self resonant frequency, $\omega_{o3}=\omega_0$, remains the only resonance frequency in the system. From equation (4), it can be seen that the negative resistor is dependent on the coupling factor $k$ and does not automatically compensates the resistances in the network. For $k<k_c$, the self resonant frequency, $\omega_{o3}=\omega_0$, remains the only resonance frequency in the system. From equation (4), it can be seen that the negative resistor is dependent on the coupling factor $k$ and does not automatically compensates the resistances in the network.
  
-FIGURE k vs frequency (e..g, FIG 3.5) HERE?+{{ :0:pt-symmetric_frequency.png?600 |}} 
 + 
  
 The same conclusions can be drawn for the PP compensated circuit. The same conclusions can be drawn for the PP compensated circuit.
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 ====Current and voltage ratio==== ====Current and voltage ratio====
  
-From equation (1), both the current ratio $I_1/I_2$ and the voltage ratio $U_{in}/U_L$ can be determined, with $U_{in}$ the voltage over the negative resistor, and $U_L$ the voltage over the load. +From equation (1), both the current ratio $I_1/I_2$ and the voltage ratio $V_1/V_2$ can be determined, with $V_{1}$ the voltage over the negative resistor, and $V_2$ the voltage over the load.
- +
-MOETEN DIE WAARDEN IN DE FIGUUR KOMEN?+
  
 ===Exact PT-symmetric state=== ===Exact PT-symmetric state===
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 $$ \frac{I_1}{I_2}=1$$ $$ \frac{I_1}{I_2}=1$$
-$$ \frac{U_{in}}{U_{L}}=\frac{R_1+R_2+R_L}{R_L}$$+$$ \frac{V_{1}}{V_{2}}=\frac{R_1+R_2+R_L}{R_L}$$
  
 We notice that, in the exact PT-symmetric region, the current and voltage ratios are constant: they do NOT depend on the value of the coupling factor $k$. We notice that, in the exact PT-symmetric region, the current and voltage ratios are constant: they do NOT depend on the value of the coupling factor $k$.
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 Analogously, we get for the power transfer efficiency $\eta$: Analogously, we get for the power transfer efficiency $\eta$:
  
-$$ \eta=\frac{L_1R_L}{L_2R_1+L_1(R_2+R_L} $$+$$ \eta=\frac{L_1R_L}{L_2R_1+L_1(R_2+R_L)} $$
  
 These two equations indicitate the important aspect of parity-time symmetric WPT: both the output power and efficiency are constant, i.e., independent on the coupling factor $k$, in the exact symmetric PT-region ($k \geq k_c$). These two equations indicitate the important aspect of parity-time symmetric WPT: both the output power and efficiency are constant, i.e., independent on the coupling factor $k$, in the exact symmetric PT-region ($k \geq k_c$).
 +
 +The underlying explanation is that in a PT-symmetric region, the eigenvalues of the system are pure imaginary. As a result, even when losses are present, there is no change in the eigenstates of the system. 
  
 ===Broken PT-symmetric state=== ===Broken PT-symmetric state===
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 This is not the case in the broken PT-symmetric region, when $k<k_c$. Taken into account the value for the negative resistor in the broken PT-symmetric region, equation (4), and the self-resonance frequency $\omega_{o3}=\omega_0$, it can be calculated that This is not the case in the broken PT-symmetric region, when $k<k_c$. Taken into account the value for the negative resistor in the broken PT-symmetric region, equation (4), and the self-resonance frequency $\omega_{o3}=\omega_0$, it can be calculated that
  
-$$ P_L=\frac{\omega_0^2k^2L_1L_2R_LU_{in}^2}{[\omega_0^2k^2L_1L_2+R_1(R_2+R_L)]^2}$$+$$ P_L=\frac{\omega_0^2k^2L_1L_2R_LV_{1}^2}{[\omega_0^2k^2L_1L_2+R_1(R_2+R_L)]^2}$$
  
 $$ \eta=\frac{\omega_0^2k^2L_1L_2R_L}{R_1(R_2+R_L)^2+\omega_0^2k^2L_1L_2(R_2+R_L)}$$ $$ \eta=\frac{\omega_0^2k^2L_1L_2R_L}{R_1(R_2+R_L)^2+\omega_0^2k^2L_1L_2(R_2+R_L)}$$
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 We can see that now both the output power and efficiency are NOT constant; they change with varying coupling factor $k$. We can see that now both the output power and efficiency are NOT constant; they change with varying coupling factor $k$.
  
-FIGUUR 3.6 HIER? +In the broken PT-symmetric region, the eigenstate amplitude increases or decays exponentially, i.e., no stable region is obtained.
- +
-PT-symmetric: eigenvalues are pure imaginary => even when there are losses present, no change in the eigenstates in time + energy distribution of the mode on both transmitting and receiving sides is mirror-symmetrical. +
- +
-Broken PT-symmetric: eigenstate amplitude increases or decays exponentially+
  
 +{{ :0:pt-symmetric_power_and_efficiency.png?600 |}}
  
-====Efficiency==== 
  
  
partity-time_symmetric_wpt.1751896839.txt.gz · Last modified: by kl · Currently locked by: 216.73.216.38