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Ben
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1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

3 - Notice the error equation of your PLL is not linear

$error = sin(\theta_{grid} - \theta_{PLL})$

The equation is approximately linear if $\theta_{grid} \approx \theta_{PLL}$ then $error \approx \theta_{grid} - \theta_{PLL}$

If the initial phase of your PLL is 90 degrees out of phase compared to your input, your theoretical settling time will not be accurate as you are in a non-linear zone of operation. Try your PLL with a 10-degree difference, your settling time should be closer to the theoretical performance.

I noticed that in your first schematic, you use the sine of the phase. You should use the cosine instead. Remember, when the phase of your PLL is 0, the amplitude of $V_a$ should be reach the positive maximum.

4 - I worked with a designer of the 3-phase PLL in the Matlab toolbox, he told me that this DQ-PLL has a metastable state with the PLL being 180 degrees out of phase. The $U_q$ component will be 0 while the $U_d$ component will be -1, instead of +1.

1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

3 - Notice the error equation of your PLL is not linear

$error = sin(\theta_{grid} - \theta_{PLL})$

The equation is approximately linear if $\theta_{grid} \approx \theta_{PLL}$ then $error \approx \theta_{grid} - \theta_{PLL}$

If the initial phase of your PLL is 90 degrees out of phase compared to your input, your theoretical settling time will not be accurate as you are in a non-linear zone of operation. Try your PLL with a 10-degree difference, your settling time should be closer to the theoretical performance.

I noticed that in your first schematic, you use the sine of the phase. You should use the cosine instead. Remember, when the phase of your PLL is 0, the amplitude of $V_a$ should be reach the positive maximum.

1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

3 - Notice the error equation of your PLL is not linear

$error = sin(\theta_{grid} - \theta_{PLL})$

The equation is approximately linear if $\theta_{grid} \approx \theta_{PLL}$ then $error \approx \theta_{grid} - \theta_{PLL}$

If the initial phase of your PLL is 90 degrees out of phase compared to your input, your theoretical settling time will not be accurate as you are in a non-linear zone of operation. Try your PLL with a 10-degree difference, your settling time should be closer to the theoretical performance.

I noticed that in your first schematic, you use the sine of the phase. You should use the cosine instead. Remember, when the phase of your PLL is 0, the amplitude of $V_a$ should be reach the positive maximum.

4 - I worked with a designer of the 3-phase PLL in the Matlab toolbox, he told me that this DQ-PLL has a metastable state with the PLL being 180 degrees out of phase. The $U_q$ component will be 0 while the $U_d$ component will be -1, instead of +1.

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Ben
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1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

3 - Notice the error equation of your PLL is not linear

$error = sin(\theta_{grid} - \theta_{PLL})$

The equation is approximately linear if $\theta_{grid} \approx \theta_{PLL}$ then $error \approx \theta_{grid} - \theta_{PLL}$

If the initial phase of your PLL is 90 degrees out of phase compared to your input, your theoretical settling time will not be accurate as you are in a non-linear zone of operation. Try your PLL with a 10-degree difference, your settling time should be closer to the theoretical performance.

I noticed that in your first schematic, you use the sine of the phase. You should use the cosine instead. Remember, when the phase of your PLL is 0, the amplitude of $V_a$ should be reach the positive maximum.

1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

I noticed that in your first schematic, you use the sine of the phase. You should use the cosine instead. Remember, when the phase of your PLL is 0, the amplitude of $V_a$ should be reach the positive maximum.

1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

3 - Notice the error equation of your PLL is not linear

$error = sin(\theta_{grid} - \theta_{PLL})$

The equation is approximately linear if $\theta_{grid} \approx \theta_{PLL}$ then $error \approx \theta_{grid} - \theta_{PLL}$

If the initial phase of your PLL is 90 degrees out of phase compared to your input, your theoretical settling time will not be accurate as you are in a non-linear zone of operation. Try your PLL with a 10-degree difference, your settling time should be closer to the theoretical performance.

I noticed that in your first schematic, you use the sine of the phase. You should use the cosine instead. Remember, when the phase of your PLL is 0, the amplitude of $V_a$ should be reach the positive maximum.

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Ben
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1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

I noticed that in your first schematic, you use the sine of the phase. You should use the cosine instead. Remember, when the phase of your PLL is 0, the amplitude of $V_a$ should be reach the positive maximum.

1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

1 There's a mistake in the PID connection. You must feed the quadrature component, i.e $U_q$ to the PID, not $U_d$. The setpoint of your PLL is $U_q = 0$ because you want your PLL to be in phase with your 3-phase input i.e. $U_d = 1, U_q = 0$.

2 - Perhaps there are hidden delays in the block you instantiated ?

I noticed that in your first schematic, you use the sine of the phase. You should use the cosine instead. Remember, when the phase of your PLL is 0, the amplitude of $V_a$ should be reach the positive maximum.

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Ben
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Ben
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