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partity-time_symmetric_wpt [2025/07/07 13:58] – [Circuit analysis] klpartity-time_symmetric_wpt [2025/07/07 14:27] (current) – [Efficiency] kl
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 The name "parity-symmetric" refers to the fact that the circuit does not change when //both// parity and time are changed:  The name "parity-symmetric" refers to the fact that the circuit does not change when //both// parity and time are changed: 
   * **Changing parity**: the position is changed, i.e., the spatial coordinate is flipped, e.g., $x$ becomes $-x$. In the context of WPT (due to the fact that the wavelengths are long compared to the circuit), it corresponds to switching the labels of transmitter (1) and receiver (2), e.g., a coil $L_1$ at the transmitter side becomes $L_2$, the coil at receiver side, and vice versa. In other words, there has to be symmetry present in Kirchoff's laws.   * **Changing parity**: the position is changed, i.e., the spatial coordinate is flipped, e.g., $x$ becomes $-x$. In the context of WPT (due to the fact that the wavelengths are long compared to the circuit), it corresponds to switching the labels of transmitter (1) and receiver (2), e.g., a coil $L_1$ at the transmitter side becomes $L_2$, the coil at receiver side, and vice versa. In other words, there has to be symmetry present in Kirchoff's laws.
-  * **Changing time**: the direction of time is flipped from $t$ to $-t$. Since current is the time derivative of charge, it results in flipping the current in the context of WPT. If we apply this on Ohm's law, this results in changing the sign of a resistor: we get negative resistors! Instead of a linear current decrease when the voltages lowers, we now have a lineary increasing current for a voltage reduction. Energy is now not dissipated any more in the resistor, but the negative resistor generates energy. The negative resistance is not any more a passive component, but an active component, supplying energy to the system. For inductors and capacitors, nothing changes when we apply time reversal. Applying Ohm's law also inverts the current, but the time derivative in the voltage-current relationship flips the sign again, resulting in a cancellation of the minus sign.+  * **Changing time**: the direction of time is flipped from $t$ to $-t$. Since current is the time derivative of charge, it results in flipping the current in the context of WPT. If we apply this on Ohm's law, this results in changing the sign of a resistor: we get negative resistors! Instead of a linear current decrease when the voltages lowers, we now have a lineary increasing current for a voltage reduction. Energy is now not dissipated any more in the resistor, but the negative resistor generates energy. The negative resistance is not any more a passive component, but an active component, supplying energy to the system. In other words, the supply of the system is now represented by a negative resistance. For inductors and capacitors, nothing changes when we apply time reversal. Applying Ohm's law also inverts the current, but the time derivative in the voltage-current relationship flips the sign again, resulting in a cancellation of the minus sign.
  
  
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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).
  
  
-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 
  
-The supply is now represented by a negative resistance. In essence, this is an active source whose output voltage and source are in phase.+
  
 ====Resonance frequencies==== ====Resonance frequencies====
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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.
  
 +{{ :0:pt-symmetric_power_and_efficiency.png?600 |}}
  
-====Efficiency==== 
  
  
partity-time_symmetric_wpt.1751896739.txt.gz · Last modified: by kl