6.2.1. Control Strategy of Source Current Detection

$v_{\text{C}h}=\text{k}{i}_{\text{S}h}$

(6.3)

$I_{\text{S}h}=\frac{V_{\text{S}h}}{\left(Z_{\text{S}h}+k\right)}-\frac{V_{\text{L}h}}{\left(Z_{\text{S}h}+k\right)}$

(6.4)

$V_{\text{PCC}h}=\frac{k}{\left(Z_{\text{S}h}+k\right)}{V}_{\text{S}h}+\frac{Z_{\text{S}h}}{\left(Z_{\text{S}h}+k\right)}{V}_{\text{L}h}$

(6.5)

Example 6.2

Figure 6.10 shows a nonlinear load composed of a three-phase noncontrolled rectifier with a 2200 μF capacitor in the dc side and a 50/3 Ω resistor in parallel, which is a typical HVS load. It is powered by a three-phase sinusoidal source, 100 V rms, 50 Hz, with grounded neutral. The equivalent series impedance of the power supply network is modeled by a 1.8 Ω resistor and a 2.8 mH inductance. The active filter is connected through three 1:1 transformers, with a LC ripple filter of 13.5 mH inductance and 50 μF capacity. The inverter is a three-phase IGBT bridge with a 100 V constant source in the dc side. The passive LC filter is composed of three branches tuned to 250 Hz, with 13.5 mH inductance and 30 μF capacity, connected in star with the neutral of the network. The simulation has been done in MATLAB–Simulink using the device models of the SymPowerSystem library.

The active filter generates a voltage signal proportional to the source current harmonics. The proportionality constant used has a value of 50. As a simulation result the harmonic spectrum of the source current is obtained for three different situations: before compensation, only with the compensation of SAF, and with both active and passive filter connected.

With both kinds of filtering, the harmonic content of the source current is reduced. The current THD without compensation is 20.95%, and decreases to 4.24% when the active filter is connected. Compensating with the SAPPF the THD decreases to 2.2%. This is because the passive filter of the SAPPF eliminates the 5th harmonic, as it is shown in the harmonic spectrum of Figure 6.11. Regarding the fundamental harmonic, a reduction in the source current is observed when the active filter is connected. This is due to the voltage drop produced at the primary side of the coupling transformer, which reduces the voltage level at the load terminals and thus the load current. In the simulation, the transformer model includes the winding resistance and leakage reactance.

Regarding the voltage at the PCC, Figure 6.12 shows its harmonic spectrum. With the active filter the voltage THD is reduced, starting from a value of 13.59% before compensation it decreases to 3.00% with the SAF connection and improves slightly to 2.55% with the SAPPF compensation, in a similar way as the source current does.

The circuit model shown in Figure 6.9 has a voltage source V_Lh (that models a HVS load) in parallel with the LC branch. If the load generates a voltage harmonic at the resonance frequency of the passive filter, high values of current through the LC branch may be reached, theoretically an infinite value. In Example 6.2, the load has a voltage harmonic of 5th order of 9.5 V rms (Figure 6.12). However, the 5th harmonic of the current through the passive filter has a value of 2.98 A rms. Therefore, one would think that the load does not have the behavior of a HVS load, maybe because the capacity of the dc side of the rectifier is not high enough (2200 μF). It should be noted that the ripple factor¹ in the dc side is 0.39%. If the capacity is increased in a factor of 10, the ripple value reaches 0.039%; almost insignificant. Its behavior should be nearer to an ideal HVS load. However, the 5th order harmonic of the current in the LC branch is then 3.00 A. That is, it is almost unchanged although the capacity has increased ten times. A similar analysis can be done for other harmonics resulting in similar values for the current through the LC branch, with dc capacitors of 2200 μF and above. It can be concluded that the ideal model of HVS load is an approach that allows a comprehensive analysis of the system performance, but it is not possible to make a complete analysis of the system behavior. Later, an analysis will be made with state variables formulation, and using a more complete load model.

$I_{\text{S}h}=\frac{Z_{\text{F}h}}{\left(Z_{\text{S}h}+Z_{\text{F}h}+k\right)}{I}_{\text{L}h}+\frac{1}{\left(Z_{\text{S}h}+Z_{\text{F}h}+k\right)}{V}_{\text{S}h}$

(6.6)

$k>>Z_{\text{F}h};Z_{\text{S}h}$

(6.7)

$V_{\text{PCC}h}=\frac{Z_{\text{F}h}+k}{\left(Z_{\text{S}h}+Z_{\text{F}h}+k\right)}{V}_{\text{S}h}-\frac{Z_{\text{S}h}{Z}_{\text{F}h}}{\left(Z_{\text{S}h}+Z_{\text{F}h}+k\right)}{I}_{\text{L}h}$

(6.8)

Example 6.3

The same circuit of Example 6.2 is considered, with a new load composed by an uncontrolled three-phase rectifier with a 55 mH inductance in series with a 50/3 Ω load resistor at the dc side, as shown in Figure 6.14.

In this case the proportionality constant k is set to 50 Ω. Figure 6.15 shows the harmonic spectrum of the source current, without compensation and with the SAPPF. The 5th order harmonic is eliminated with the passive filter. Being the most significant harmonic, this results in an improved THD of the source current, from 18.09% without filtering to 2.27% when the SAPPF is connected. The 7th harmonic of the source current is also reduced because the constant k is greater than the passive filter impedance at this frequency, so the first term of (6.6) is practically zero.

In respect to the voltage at the PCC, its THD is reduced from 12.60% to 2.15%, when the SAPPF is connected. Figure 6.16 shows the harmonic spectrum in both situations. In view of the expression (6.8), if the supply voltage has no harmonics, the first term is zero. Regarding the second term, when condition (6.7) is fulfilled, the load current is multiplied by a factor less than one, so that the harmonics values of the voltage at the PCC are attenuated.

A slight increase in the rms value of the voltage at the fundamental frequency is observed in Figure 6.16 because the current drawn by the load with SAPPF is also lower, and thus the voltage drop across the source impedance is reduced.

Next, a distorted supply voltage is assumed, with a voltage waveform defined by

$v(t)=V_p\left[\text{sin}\left({\omega }_{1}t\right)+0.08\text{sin}\left(3{\omega }_{1}t+\pi \right)+0.05\text{sin}\left(5{\omega }_{1}t\right)\right]$

(6.9)

In this case, the source current has a THD value of 3%, which is a slightly higher value than in the case of sinusoidal supply voltage. Since V_Sh is not zero, the second term on expression (6.6) is neither, although its value is reduced when condition (6.7) holds.

$I_{\text{S}h1}=\frac{Z_{\text{F}h}}{\left(Z_{\text{S}h}+Z_{\text{F}h}+k\right)}{I}_{\text{L}h}+\frac{Z_{\text{S}h}}{\left(Z_{\text{S}h}+Z_{\text{F}h}+k\right)}{I}_{\text{L}h2}+\frac{1}{\left(Z_{\text{S}h}+Z_{\text{F}h}+k\right)}{V}_{\text{S}h}$

(6.10)

6.2.2. Control Strategy of Load Voltage Detection

$V_{\text{C}h}=-k_v{V}_{\text{L}h}$

(6.11)

$I_{\text{S}h}=\frac{1-k_v}{{\text{Z}}_{\text{S}h}}{V}_{\text{L}h}+\frac{1}{{\text{Z}}_{\text{S}h}}{V}_{\text{S}h}$

(6.12)

$V_{\text{PCC}h}=\left(1-k_v\right)V_{\text{L}h}$

(6.13)

$I_{\text{S}h}=\frac{Z_{\text{F}h}\left(1-k_{\text{v}}\right)}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h}+\frac{1}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_v\right)\right)}{V}_{\text{S}h}$

(6.14)

$V_{\text{PCC}h}=\frac{Z_{\text{F}h}\left(1-k_v\right)}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{V}_{\text{S}h}-\frac{Z_{\text{F}h}{Z}_{\text{S}h}\left(1-k_{\text{v}}\right)}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h}$

(6.15)

Example 6.4

The control by load voltage detection to the circuit described in Example 6.2 is applied, and is reproduced in Figure 6.20.

Figure 6.21 shows the harmonic spectrum of the source current. A value of k_v = 0.95 for the simulations is considered. With the hybrid filter can improve the source current waveform because the THD decreases from 18.09% to 2.03%. This is justified as the first term of the expression (6.14) is almost zero, and on the other hand, since the supply voltage is sinusoidal, the second term of (6.14) is zero.

Relative to the voltage at the PCC, Figure 6.22 shows the frequency spectrum. The hybrid filter THD is improved, passing from 12.60% to 1.89%, which justifies their behavior according to (6.15).

Then it is considered that the supply voltage is distorted. Thus, the waveform of the voltage source is defined by the expression (6.9). This includes two voltage harmonics: one of order 3 of 8% and other of order 5 of 5%. Figure 6.23 shows the harmonic spectrum of source current before and after connecting the hybrid filter. The presence of a harmonic of order 5 is observed to be nonzero. This is because the second term of (6.14) is nonzero V_Sh also. A harmonic that was not present in the previous case also appears, this is the harmonic of order 3. Due to this harmonic component, THD increases going from 16.14% to 27.84%.

With respect to the voltage at the PCC, the most significant order harmonics are the 3rd and 5th order (Figure 6.24). When the SAPPF is connected, voltage THD decreases from 19.78% to 3.19%. From expression (6.15), the voltage harmonic content at the PCC depends on the k_v value and source impedance Z_Sh. As k_v≠1, the two terms tend to decrease depending on the value of the source impedance, which justifies the reduction of THD.

$I_{\text{S}h1}=\frac{Z_{\text{F}h}\left(1-k_v\right)}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h}+\frac{1}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{V}_{\text{S}h}+\frac{Z_{\text{S}h}}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h2}$

(6.16)

$V_{\text{PCC}h}=\frac{Z_{\text{F}h}\left(1-k_{\text{v}}\right)}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{V}_{\text{S}h}-\frac{Z_{\text{F}h}{Z}_{\text{S}h}\left(1-k_{\text{v}}\right)}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h}+\frac{Z_{\text{F}h}\left(1-k_{\text{v}}\right)}{\left(Z_{\text{S}h}+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h2}$

(6.17)

6.2.3. Combined Control

$v_{\text{C}h}=k{i}_{\text{S}h}-k_{\text{v}}{v}_{\text{L}h}$

(6.18)

$I_{\text{S}h}=\frac{Z_{\text{F}h}\left(1-k_{\text{v}}\right)}{\left(Z_{\text{S}h}+k+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h}+\frac{1}{\left(Z_{\text{S}h}+k+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{V}_{\text{S}h}$

(6.19)

$V_{\text{P}\text{C}\text{C}h}=\frac{{\text{Z}}_{\text{F}h}\left(1-k_v\right)+\text{k}}{\left({\text{Z}}_{\text{S}h}+{\text{Z}}_{\text{F}h}\left(1-k_{\text{v}}\right)+\text{k}\right)}{V}_{\text{S}h}-\frac{{\text{Z}}_{\text{S}h}{\text{Z}}_{\text{F}h}\left(1-k_{\text{v}}\right)}{\left({\text{Z}}_{\text{S}h}+{\text{Z}}_{\text{F}h}\left(1-k_v\right)+\text{k}\right)}{I}_{\text{L}h}$

(6.20)

Example 6.5

A hybrid control shall be applied to the circuit presented in Example 6.4.

In this case it has been considered a proportionality constant of 50 (k = 50) for the current source term and k_v = 0.95 to the corresponding term of the load voltage. Figure 6.28 shows the frequency spectrum of the source current before and after connecting the compensation system. A significant improvement of the source current waveform is observed, as its THD is reduced from 18.19% to 1.79%. This confirms the behavior according to expression (6.19).

The voltage at the PCC before and after connecting the SAPPF also shows an improvement in its THD, as it is seen in the harmonic spectrum that is shown in Figure 6.29. Its value is reduced from 12.60% to 2.30%.

Then in order to verify the functionality of SAPPF in presence of harmonics in the supply voltage, a source voltage waveform defined by expression (6.9) is considered. The distortion of the source current depends on the value of k selected (6.19), the higher value does lower value of the harmonics due to the supply voltage; however, as a result, the rated power of the filter increases. Relative to the voltage at the PCC, according to expression (6.20), the harmonic content will be virtually the same as the supply voltage.

$I_{\text{S}h1}=\frac{Z_{\text{F}h}\left(1-k_{\text{v}}\right)}{\left(Z_{\text{S}h}+k+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h}+\frac{1}{\left(Z_{\text{S}h}+k+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{V}_{\text{S}h}+\frac{Z_{\text{S}h}}{\left(Z_{\text{S}h}+k+Z_{\text{F}h}\left(1-k_{\text{v}}\right)\right)}{I}_{\text{L}h2}$

(6.21)

6.3.1. Source Current Detection

$u=\left[\begin{array}{cccc}\hfill k\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \begin{array}{cc}\hfill 0\hfill & \hfill \begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \end{array}\hfill \end{array}\hfill \end{array}\right]{\left[\begin{array}{cccccc}\hfill i_{\text{S}}\hfill & \hfill i_5\hfill & \hfill i_7\hfill & \hfill i_{\text{LL}}\hfill & \hfill v_5\hfill & \hfill v_7\hfill \end{array}\right]}^T$

(6.30)

$K=\left[\begin{array}{cc}\hfill k\hfill & \hfill \begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill \begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\hfill \end{array}\hfill \end{array}\right]$

(6.31)

$\dot{x}=\left(A+B_{1}K\right)x+B_{2}v$

(6.32)

$A\text{'}=\left(A+B_{1}K\right)=\left[\begin{array}{cccccc}\hfill -\frac{\left(R_{\text{S}}+R_L+k\right)}{L_{\text{S}}}\hfill & \hfill \frac{R_L}{L_{\text{S}}}\hfill & \hfill \frac{R_{\text{L}}}{L_{\text{S}}}\hfill & \hfill \frac{R_L}{L_{\text{S}}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{\left(R_{\text{L}}+R_5\right)}{L_5}\hfill & \hfill -\frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{1}{L_5}\hfill & \hfill 0\hfill \\ \hfill \frac{R_{\text{L}}}{L_7}\hfill & \hfill -\frac{R_{\text{L}}}{L_7}\hfill & \hfill -\frac{\left(R_{\text{L}}+R_7\right)}{L_7}\hfill & \hfill -\frac{R_{\text{L}}}{L_7}\hfill & \hfill 0\hfill & \hfill -\frac{1}{L_7}\hfill \\ \hfill \frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_L}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{C_5}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{C_7}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

(6.33)

6.3.2. Load Voltage Detection

$u=-k_v{v}_L$

(6.34)

$v_{\text{L}}=R_{\text{L}}{i}_S-R_{\text{L}}{i}_5-R_{\text{L}}{i}_7-R_{\text{L}}{i}_{\text{LL}}-R_{\text{L}}{i}_{\text{L}}$

(6.35)

$u=-k_{\text{v}}{R}_{\text{L}}{i}_S+k_{\text{v}}{R}_{\text{L}}{i}_5+k_{\text{v}}{R}_{\text{L}}{i}_7+k_v{R}_{\text{L}}{i}_{\text{LL}}+k_v{R}_{\text{L}}{i}_{\text{L}}$

(6.36)

$u=K_{1}x+K_{2}v$

(6.37)

$K_1=\left[\begin{array}{cc}\hfill -k_{\text{v}}{R}_L\hfill & \hfill \begin{array}{cccc}\hfill k_v{R}_{\text{L}}\hfill & \hfill k_{\text{v}}{R}_L\hfill & \hfill k_{\text{v}}{R}_{\text{L}}\hfill & \hfill \begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \end{array}\hfill \end{array}\hfill \end{array}\right]$

(6.38)

$K_2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill k_{\text{v}}{R}_{\text{L}}\hfill \end{array}\right]$

(6.39)

$\dot{x}=\left(A+B_1{K}_1\right)x+\left(B_2+B_1{K}_2\right)v$

(6.40)

$A\text{'}=\left(A+B_1{K}_1\right)=\left[\begin{array}{cccccc}\hfill -\frac{\left(R_{\text{s}}+R_{\text{L}}\left(1-k_{\text{v}}\right)\right)}{L_{\text{S}}}\hfill & \hfill \frac{R_L\left(1-k_{\text{v}}\right)}{L_{\text{S}}}\hfill & \hfill \frac{R_{\text{L}}\left(1-k_{\text{v}}\right)}{L_{\text{S}}}\hfill & \hfill \frac{R_L\left(1-k_{\text{v}}\right)}{L_S}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{(R_{\text{L}}+R_5)}{L_5}\hfill & \hfill -\frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{1}{L_5}\hfill & \hfill 0\hfill \\ \hfill \frac{R_{\text{L}}}{L_7}\hfill & \hfill -\frac{R_{\text{L}}}{L_7}\hfill & \hfill -\frac{(R_{\text{L}}+R_7)}{L_7}\hfill & \hfill -\frac{R_{\text{L}}}{L_7}\hfill & \hfill 0\hfill & \hfill -\frac{1}{L_7}\hfill \\ \hfill \frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_L}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_L}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{C_5}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{C_7}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

(6.41)

6.3.3. Hybrid Control

$u=k{i}_{\text{S}}-k_{\text{v}}{v}_{\text{L}}$

(6.42)

$u=\left(k-k_{\text{v}}{R}_{\text{L}}\right)i_S+k_{\text{v}}{R}_{\text{L}}{i}_5+k_{\text{v}}{R}_{\text{L}}{i}_7+k_{\text{v}}{R}_{\text{L}}{i}_{\text{LL}}+k_v{R}_L{i}_{\text{L}}$

(6.43)

$u=K_{1}x+K_{2}v$

(6.44)

$K_1=\left[\begin{array}{cc}\hfill \left(k-k_v{R}_{\text{L}}\right)\hfill & \hfill \begin{array}{cccc}\hfill k_{\text{v}}{R}_{\text{L}}\hfill & \hfill k_{\text{v}}{R}_{\text{L}}\hfill & \hfill k_v{R}_{\text{L}}\hfill & \hfill \begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \end{array}\hfill \end{array}\hfill \end{array}\right]$

(6.45)

$K_2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill k_{\text{v}}{R}_{\text{L}}\hfill \end{array}\right]$

(6.46)

$\dot{x}=\left(A+B_1{K}_1\right)x+\left(B_2+B_1{K}_2\right)v$

(6.47)

$A\text{'}=\left(A+B_1{K}_1\right)=\left[\begin{array}{cccccc}\hfill -\frac{\left(R_{\text{s}}+R_{\text{L}}\left(1-k_{\text{v}}\right)+k\right)}{L_{\text{S}}}\hfill & \hfill \frac{R_{\text{L}}\left(1-k_{\text{v}}\right)}{L_S}\hfill & \hfill \frac{R_{\text{L}}\left(1-k_{\text{v}}\right)}{L_{\text{S}}}\hfill & \hfill \frac{R_{\text{L}}\left(1-k_{\text{v}}\right)}{L_{\text{S}}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{(R_{\text{L}}+R_5)}{L_5}\hfill & \hfill -\frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{R_{\text{L}}}{L_5}\hfill & \hfill -\frac{1}{L_5}\hfill & \hfill 0\hfill \\ \hfill \frac{R_{\text{L}}}{L_7}\hfill & \hfill -\frac{R_{\text{L}}}{L_7}\hfill & \hfill -\frac{(R_{\text{L}}+R_7)}{L_7}\hfill & \hfill -\frac{R_{\text{L}}}{L_7}\hfill & \hfill 0\hfill & \hfill -\frac{1}{L_7}\hfill \\ \hfill \frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_L}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{C_5}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{C_7}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$

(6.48)

Example 6.6

Figure 6.32 shows a sinusoidal voltage source with 1.8 Ω resistance and 2.8 mH inductance. The load is formed by an uncontrolled three-phase rectifier with a resistance of 50/3 Ω connected in series with an inductance of 55 mH. This load is connected in parallel with a passive filter comprised of two LC branches tuned to the 5th and 7th harmonic. Resistors of 2.1 Ω and 1.1 Ω model the resistive effect of the two passive filter coils. The active filter is connected in series with the voltage source through three coupling transformers of ratio 1:1. A dc voltage source of 100 V is connected at the dc side of the inverter.

Figure 6.33 shows the system pole diagram without an active filter. The system has two real poles and four complex poles. All poles are located on the left semi plane.

On the other hand, Figure 6.34 shows the Bode gain when the current source is considered as an output signal. In the left diagram, where the input signal is the supply voltage, two peaks appear with gains of: −12.1 dB at 225 Hz and −14.6 dB at 323 Hz. These peaks represent two series resonances between the network and the passive filter. They are close to the resonance frequency of the LC branches, and depend on the impedance of the network from the PCC. In the ideal case of zero impedance, network will correspond to the tuning frequency of the LC branches.

In the second diagram of Figure 6.34, the input signal is the current source, which represents the load Norton equivalent. In this situation, the gain decreases to −10.1 dB at 249 Hz and −17.3 dB at 352 Hz, therefore, the system has two resonances at these frequencies. This does not occur exactly at 250 Hz and 350 Hz, due to tolerance of passive elements in the harmonic filter. Moreover, it shows that the gain value is not too low for a situation of resonance, however, must take into account that the coil resistive effect is modeled by a resistance of 2.1 Ω for tuned branch to the 5th harmonic, and 1.1 Ω for tuned branch to the 7th harmonic.

When the control strategy of source current detection is applied, with k = 40, the system pole position is modified. Figure 6.35 shows the new location of the poles. When the results shown in Figure 6.33 and Figure 6.35 are compared, a shift of the system poles on the left semiplane is observed. The implementation of this strategy causes the system poles to move away from the imaginary axis so that system stability is ensured.

To check what k values could destabilize the system, the Routh–Hurwitz criterion is used. The system characteristic polynomial with the connected active filter is analyzed. When the control strategy of source current detection is applied the obtained polynomial is

$\begin{array}{l}\varphi (s)=s^6+(6314.10+357.10k)s^5+(1.03\times {10}^7+8.50\times {10}^{5}k)s^4+(3.75\times {10}^{10}+2.77\times {10}^{9}k)s^3+\\ +(2.11\times {10}^{13}+2.86\times {10}^{12}k)\hspace{0.17em}{s}^2+(4.82\times {10}^{16}+4.35\times {10}^{15}k)s+(9.87\times {10}^{16}+5.48\times {10}^{16}k)\end{array}$

For k =−1.8, the resulting table is

$\left.\begin{array}{c}\hfill s^6\hfill \\ \hfill s^5\hfill \\ \hfill s^4\hfill \\ \hfill s^3\hfill \\ \hfill s^2\hfill \\ \hfill s^1\hfill \\ \hfill s^0\hfill \end{array}\right|\begin{array}{cccc}\hfill 1\hfill & \hfill 8.78\times {10}^6\hfill & \hfill 1.6\times {10}^{13}\hfill & \hfill 1.04\times {10}^3\hfill \\ \hfill 5.67\times {10}^3\hfill & \hfill 3.25\times {10}^{10}\hfill & \hfill 4.03\times {10}^{16}\hfill & \hfill 0\hfill \\ \hfill 3.05\times {10}^6\hfill & \hfill 8.89\times {10}^{12}\hfill & \hfill 1.04\times {10}^3\hfill & \hfill \hfill \\ \hfill 1.59\times {10}^{10}\hfill & \hfill 4.03\times {10}^{16}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill 1.18\times {10}^{12}\hfill & \hfill 1.04\times {10}^3\hfill & \hfill \hfill & \hfill \hfill \\ \hfill 4.03\times {10}^{16}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill 1.04\times {10}^3\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \end{array}$

The absence of any change of sign in the first column indicates that there are poles with negative real part, therefore, all are located in the left semiplane.

For a value of k = −1.85, the application of the stability criterion of Routh–Hurwitz allows obtaining the next table

$\left.\begin{array}{c}\hfill s^6\hfill \\ \hfill s^5\hfill \\ \hfill s^4\hfill \\ \hfill s^3\hfill \\ \hfill s^2\hfill \\ \hfill s^1\hfill \\ \hfill s^0\hfill \end{array}\right|\begin{array}{cccc}\hfill 1\hfill & \hfill 8.74\times {10}^6\hfill & \hfill 1.59\times {10}^{13}\hfill & \hfill -2.74\times {10}^{15}\hfill \\ \hfill 5.65\times {10}^3\hfill & \hfill 3.24\times {10}^{10}\hfill & \hfill 4.01\times {10}^{16}\hfill & \hfill 0\hfill \\ \hfill 3.01\times {10}^6\hfill & \hfill 8.76\times {10}^{12}\hfill & \hfill -2.74\times {10}^{15}\hfill & \hfill \hfill \\ \hfill 1.59\times {10}^{10}\hfill & \hfill 4.01\times {10}^{16}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill 1.17\times {10}^{12}\hfill & \hfill -2.74\times {10}^{15}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill 4.02\times {10}^{16}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill -2.74\times {10}^{15}\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \end{array}$

As can be seen, there is a sign change in the first column, which indicates the existence of a positive real pole, therefore, the system is unstable.

Figure 6.36 shows the graphical representation of the real parts of the system poles for k values between −100 and 100. The system will be stable for values of k > −1.8.

On the other hand, it is appropriate to establish the most suitable k value from the point of view of harmonic mitigation. The goal is that the system presents a low-input gain for a given harmonic content. Figure 6.37 represents the Bode diagram magnitude for k = 0, 10, 20, 40, 80, and 120. Regardless of the input or output variable chosen, a larger value of k means that the gain is less. This fact is evident when the current source is taken as output signal and the gain is plotted for k value different. Highlights the fact that the increase in k is not proportional to the decrease in the gain.

Figure 6.38 reflects more clearly this nonproportionality between gain and k constant. This shows the gain at the frequency of 250 Hz for different k values. A gain of −30 dB (about 0.030 in absolute value) implies that the magnitude of 5 order harmonic drops to 3%. So, when the distortion source is the supply voltage, a value of k = 25 will be sufficient, whereas if it is considered that the distortion source is the load, it is necessary that k = 50.

Subsequently, the strategy control of load voltage detection is applied. The system matrix changes and therefore the system poles. The stability criterion application of Routh–Hurwitz for k_v values between −2 and 2 allows to represent the graph shown in Figure 6.39. This shows the actual values of the system poles in terms of k_v. It is noted that when the condition k_v ≥ 1.198 is accomplished the system has a pole in the right semiplane.

Figure 6.40 shows the system gain for different values of the k_v parameter. When the load harmonic source is considered as input signal (Figure 6.40 right) the minimum value of the gain is obtained for k_v = 1. This does not happen when the input signal is the supply voltage, as it is shown in the Bode diagram in the Figure 6.40 left. In this situation it is clear that the implementation of this strategy can even increase the harmonic distortion of the current source.

Finally, results are analyzed when the control strategy that combines the previous two are applied. A stability analysis must take into account two parameters: k_v and k. In a first analysis, it is considered that k_v = 1. The stability criterion application of Routh–Hurwitz allows us to determine the minimum value of k for which the system is stable. This occurs for k > −1.8. This value is the same as obtained with the control strategy of source current detection. If the value of k_v is set at 1.2, the system is stable whenever k > 0. It must be remembered that the strategy of load voltage detection, with k_v = 1.2 the system was unstable. Therefore, the hybrid strategy provides a more robust system. For k_v < 1, the stability limit is located at values of k < −1.8. Therefore, it can be concluded that if k_v has a value close to the unit, the system will be stable as long as k is greater than zero. Thus the fulfillment of these two premises will be the design goal.

With known minimum values of k, it remains to establish what would be its appropriate value from the point of view of system gain. When set to target the low-frequency gain is −30 dB, with a k_v value of 0.95, the proportionality constant must be k = 29. As shown in Figure 6.41, with this k value the maximum gain will be 30 dB, regardless of the input signal. In case of a system where the supply voltage is slightly distorted, the k value can be reduced to 12, having a maximum gain of −30dB at low frequencies, as shown in the Bode plot on the Figure 6.41 right.