5.2.1. Source Current Detection Control

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

(5.1)

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

(5.2)

$I_{\text{S}h}\hspace{0.17em}\approx \hspace{0.17em}0;\hspace{1em}k>>Z_{\text{S}h}$

(5.3)

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

(5.4)

Example 5.1

The strategy v_Ch = ki_Sh control will be applied to a HVS-type load. Figure 5.3 shows the circuit scheme to be simulated. The load consists of an uncontrolled three-phase rectifier with a capacitor of 2200 μF and a resistor of 50/3 Ω connected in parallel at the dc side, which is a typical HVS load. The source is sinusoidal, with an rms value of the phase voltage of 100 V and a frequency of 50 Hz. A resistance of 1.8 Ω and an inductance of 2.8 mH in series have been included to model the equivalent impedance from the PCC. The active filter is connected through three transformers of ratio 1:1 with an LC ripple filter of 13.5 mH inductance and 50 μF capacity. The inverter is a three-phase IGBT Bridge. At the dc side a constant source of 100 V has been connected. This example has been simulated in MATLAB–Simulink using device models of the SymPowerSystem library.

Figure 5.4 shows the waveforms of the voltage at the common connection point, v_PCC, and source current, i_S, when the active filter is not connected. The waveform THDs are: 20.95% for the current and 13.59% for the voltage. The most significant harmonics for the voltage and the current are the 5th, 7th, 9th, and 11th.

Figure 5.5 shows the waveforms when the SAF is connected and the control strategy is applied with a k constant value of 50. The current has a THD of 4.67% and the voltage at the PCC is 3.42%. The results corroborate the behavior expected by expression (5.2), where a high k value (k ≫Z_Sh) reduces the harmonic content of the source current.

On the other hand, the voltage at the PCC is practically sinusoidal. Expression (5.4) establishes a dependence on the source voltage and the voltage at the load side. If it is taken into account that the voltage at the source side is sinusoidal, the voltage at the common connection point will also be sinusoidal, if the k choice ensures that the APF eliminates voltage harmonics caused by the load. Therefore, this control strategy makes possible the elimination of the harmonic components of the voltage at the PCC produced by the load. Figure 5.6 shows the harmonic spectra of source current and voltage at the PCC, before and after the connection of the SAF.

In this compensation strategy it is possible to predict what k value will get the lowest THD. The Figure 5.7 shows a graphical representation of current THD for different values of the proportionality constant, k. It can be observed how from a k value of about 70, the THD deteriorates slightly, which is due to the appearance of a higher ripple on the current signal. This depends on the values of L and C of the ripple filter, which is connected to the output of the inverter. LC filter design becomes an important issue for the SAF. A detailed analysis will be left for a later section.

Another important issue is to analyze how the presence of voltage harmonics at the source side effects the compensation. To study this situation, a third harmonic of 8% with an initial phase of π rad and fifth harmonic of 5% with an initial phase of 0 rad have been added. So the applied voltage is

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

(5.5)

This waveform has been chosen as one of those included in the IEC 61000 standard, in its section on immunity to voltage harmonics.

Figure 5.8 shows the waveforms of the source current and voltage at the PCC for phase “a,” when the active filter is not connected. In this case, note the presence of a third harmonic in the source current before compensation, this is due to the presence of this harmonic in the source voltage.

When the active filter is connected with k = 50, the source current THD improves significantly; decreasing from 15.22% to 5.61%. As regards the voltage at the PCC, the THD decreases from 17.32% to 11.75%. This minor reduction compared to the case of sinusoidal supply voltage is because the active filter can only mitigate the harmonics generated by the load and not the harmonics present at the PCC due to the supply voltage. The waveforms of interest are shown in Figure 5.9.

Figure 5.10 shows the frequency spectra of the current and voltage before and after the filter connection, which allows comparison of the harmonic content in both situations. As was already indicated, SAF “isolates” the PCC of voltage harmonics produced by the load.

$V_{\text{P}\text{C}\text{C}h}=V_{\text{S}h}-Z_{\text{S}h}{I}_{\text{L}h}$

(5.6)

5.2.2. Load Voltage Detection Control

$v_{\text{C}h}=-v_{\text{L}h}$

(5.7)

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

(5.8)

$V_{\text{P}\text{C}\text{C}h}=V_{\text{L}h}(1-k_{\text{v}})$

(5.9)

$I_{\text{S}h}=\frac{1}{Z_{\text{S}h}}{V}_{\text{S}h}-\frac{(1-k_{\text{v}})}{Z_{\text{S}h}}{V}_{\text{L}h}$

(5.10)

Example 5.2

Here, Example 5.1 will be analyzed according to the control strategy, v_Ch = −v_Lh.

Figure 5 shows the waveform before compensation. Figure 5.13 shows current and voltage waveforms, once the active filter is connected. For simulation it was considered that k_v = 0.95. The voltage THD and current THD are 2.79% and 3.47%, respectively. When the supply voltage is sinusoidal, the source current is practically sinusoidal also.

If it is considered that the voltage source is distorted and it is defined by the expression

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

The current source and the voltage at the PCC have the waveforms shown in Figure 5.14. The current THD is 5.61% and the voltage THD is 11.98%. Clearly there is significant improvement in current THD relative to the situation before the compensation, that was 15.22%, not so with the voltage THD where the reduction is less. This is because these harmonics are present in the supply voltage, thus cannot be eliminated by the active filter according to (5.10).

Figure 5.15 shows harmonic spectra before and after connecting the SAF.

5.2.3. Combined Control

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

(5.11)

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

(5.12)

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

(5.13)

Example 5.3

Same Example 5.1, where the hybrid strategy is applied to the filter, this is v_Ch = ki_Sh−k_v v_Lh.

First a sinusoidal supply voltage is considered. In the detection of load voltage harmonic, an error will be taken into account, for which is chosen a k_V = 0.95. Furthermore, the value of k has been set at 10. Waveforms when the filter is connected are shown in Figure 5.17. The current THD is reduced from 20.95% without filter to 0.95% with filter. Regarding the voltage at the PCC, the THD decreases from 13.59% to 0.90%. It is verified that a significant improvement is achieved in the THD of the current and voltage respect to the strategies by detection of the current source and by detection of the load voltage.

Current and voltage spectrum before and after connecting the active filter are shown in Figure 5.18.

Another test of interest is to consider that the supply voltage is distorted. A supply voltage defined by expression (5.5) is assumed. Figure 5.19 shows the waveforms after the SAF connection. The source current THD passes from 15.22% to 0.94%. This result verifies the behavior deduced from the expression (5.12). Therefore, it is possible to reduce the harmonic content of the source current with a k value smaller than with the strategy by detection of the source current. Furthermore, the voltage THD at the PCC decreases from 17.32% to 11.38%. This value is close to that of the source voltage defined by (5.5), which is 9.43%. The difference is due to the existence of high-frequency harmonics as a result of switching of the electronic devices, which makes the voltage waveform present a ripple, as shown in Figure 5.19b. The ripple reduction depends on the inductance and capacitance parameters, which constitute the inverter output filter.

Figure 5.20 shows harmonic spectrum before and after connecting the active filter.

Finally, in Table 5.1 the results of the voltage and current THD for the three compensation strategies and different source voltage conditions are summarized. This allows for easy comparison of the results obtained and consideration of the theoretical analysis.

Table 5.1

Voltage and Current THD for the Three Compensation Strategies and Different Source Voltage Conditions

	Without filter		VCh = 50 ISh		VCh = −0.95 VLh		VCh = 10 ISh−0.95 VLh
	THDv (%)	THDi (%)	THDv (%)	THDi (%)	THDv (%)	THDi (%)	THDv (%)	THDi (%)
Sinusoidal, VS	13.59	20.95	3.42	4.67	2.79	3.47	0.90	0.95
Nonsinusoidal, VS	17.32	15.22	11.75	5.61	11.98	5.61	11.38	0.94

5.3.1. State Space

$x=\left[x_1,x_2,\cdots ,x_n\right]$

(5.14)

$\begin{array}{l}{\dot{x}}_i=f_i(x_1,x_2,\dots ,x_n,u_1,\dots ,u_p),i=1,\dots ,n\\ y_j=g_j(x_1,x_2,\dots ,x_n,u_1,\dots ,u_p),j=1,\dots ,q\end{array}$

(5.15)

$\begin{array}{l}\dot{x}(t)=Ax(t)+Bu(t)\\ y(t)=Cx(t)+Du(t)\end{array}$

(5.16)

$u(t)={\left[u_1,\dots ,u_p\right]}^T$

(5.17)

$y(t)={\left[y_1,\dots ,y_q\right]}^T$

(5.18)

$\begin{array}{l}sIX(s)=AX(s)+BU(s)\\ Y(s)=CX(s)+DU(s)\end{array}$

(5.19)

$X(s)={\left(s\text{I}-A\right)}^{-1}BU(s)$

(5.20)

$G(s)=Y(s)U^{-1}(s)=C{\left(sI-A\right)}^{-1}BU(s)+D$

(5.21)

5.3.2. SAF State Model

$\begin{array}{l}\dot{x}=Ax+B_{1}u+B_{2}v\\ y=Cx+D_{1}u+D_{2}v\end{array}$

(5.22)

$x={\left[i_{\text{S}} i_{\text{LL}}\right]}^T$

(5.23)

$v={\left[\begin{array}{cc}\hfill v_{\text{S}}\hfill & \hfill i_{\text{L}}\hfill \end{array}\right]}^T$

(5.24)

$A=\left[\begin{array}{cc}\hfill -\frac{\left(R_{\text{s}}+R_{\text{L}}\right)}{L_{\text{s}}}\hfill & \hfill \frac{R_{\text{L}}}{L_{\text{s}}}\hfill \\ \hfill \frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill \end{array}\right]$

(5.25)

$B_1={\left[\begin{array}{cc}\hfill -\frac{1}{L_{\text{S}}}\hfill & \hfill 0\hfill \end{array}\right]}^T$

(5.26)

$B_2=\left[\begin{array}{cc}\hfill \frac{1}{L_{\text{S}}}\hfill & \hfill \frac{R_{\text{L}}}{L_{\text{S}}}\hfill \\ \hfill 0\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill \end{array}\right]$

(5.27)

$C=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

(5.28)

$C=\left[\begin{array}{cc}\hfill R_{\text{L}}\hfill & \hfill -R_{\text{L}}\hfill \end{array}\right]$

(5.29)

$\begin{array}{l}{D}_1=\left[1\right]\\ D_2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill -R_{\text{L}}\hfill \end{array}\right]\end{array}$

(5.30)

5.3.3. Control Strategies

$u=k{i}_{\text{S}}$

(5.31)

$u=\left[\begin{array}{cc}\hfill k\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{c}\hfill i_{\text{S}}\hfill \\ \hfill i_{\text{LL}}\hfill \end{array}\right]$

(5.32)

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

(5.33)

$\dot{\text{x}}=(\text{A}+{\text{B}}_1\text{K})\text{x}+{\text{B}}_2\text{v}$

(5.34)

$A\text{'}=\left(A+B_{1}K\right)=\left[\begin{array}{cc}\hfill -\frac{\left(R_{\text{s}}+R_{\text{L}}+k\right)}{L_{\text{s}}}\hfill & \hfill \frac{R_{\text{L}}}{L_{\text{s}}}\hfill \\ \hfill \frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill \end{array}\right]$

(5.35)

$\varphi (s)=\left|sI-A\text{'}\right|$

(5.36)

$\varphi (s)=\frac{s^2{L}_{\text{S}}{L}_{\text{L}}+s\left(R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}+k{L}_{\text{L}}\right)+R_{\text{L}}\left(R_{\text{S}}+k\right)}{L_{\text{S}}{L}_{\text{L}}}$

(5.37)

$R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}+k{L}_{\text{L}}\ge 0$

(5.38)

$k\ge -\frac{\left(R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}\right)}{L_{\text{L}}}$

(5.39)

$\begin{array}{l}\dot{x}=\left(A+B_{1}K\right)x+B_{2}v=A\text{'}x+B_{2}v\\ y=Cx\end{array}$

(5.40)

$\begin{array}{l}\dot{x}=\left(A+B_{1}K\right)x+B_{2}v=A\text{'}x+B_{2}v\\ y=\left(C+D_{1}K\right)x+D_{2}v=C\text{'}c+D_{2}v\end{array}$

(5.41)

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

(5.42)

$u=-k_{\text{v}}{R}_{\text{L}}{i}_{\text{S}}+k_{\text{v}}{R}_{\text{L}}{i}_{\text{LL}}+k_{\text{v}}{R}_{\text{L}}{i}_{\text{L}}$

(5.43)

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

(5.44)

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

(5.45)

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

(5.46)

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

(5.47)

$A\text{'}=\left(A+B_1{K}_1\right)=\left[\begin{array}{cc}\hfill -\frac{\left(R_{\text{s}}+R_{\text{L}}\left(1-k_{\text{v}}\right)\right)}{L_{\text{s}}}\hfill & \hfill \frac{R_{\text{L}}\left(1-k_{\text{v}}\right)}{L_{\text{s}}}\hfill \\ \hfill \frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill \end{array}\right]$

(5.48)

$\varphi (s)=\left|sI-A\text{'}\right|=\frac{L_{\text{S}}{L}_{\text{L}}{s}^2+s\left[R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}\left(1-k_{\text{v}}\right)\right]+R_{\text{S}}{R}_{\text{L}}}{L_{\text{S}}{L}_{\text{L}}}$

(5.49)

$R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}\left(1-k_{\text{v}}\right)>0$

(5.50)

$k_{\text{v}}<\frac{R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}}{R_{\text{L}}{L}_{\text{L}}}$

(5.51)

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

(5.52)

$u=\left(k-k_{\text{v}}{R}_{\text{L}}\right)i_{\text{S}}+k_{\text{v}}{R}_{\text{L}}{i}_{\text{LL}}+k_{\text{v}}{R}_{\text{L}}{i}_{\text{L}}$

(5.53)

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

(5.54)

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

(5.55)

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

(5.56)

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

(5.57)

$A\text{'}=\left(A+B_1{K}_1\right)=\left[\begin{array}{cc}\hfill -\frac{\left(R_{\text{s}}+k+R_{\text{L}}\left(1-k_{\text{v}}\right)\right)}{L_{\text{s}}}\hfill & \hfill \frac{R_{\text{L}}\left(1-k_{\text{v}}\right)}{L_{\text{s}}}\hfill \\ \hfill \frac{R_{\text{L}}}{L_{\text{L}}}\hfill & \hfill -\frac{R_{\text{L}}}{L_{\text{L}}}\hfill \end{array}\right]$

(5.58)

$\varphi (s)=\left|sI-A\text{'}\right|=\frac{L_{\text{S}}{L}_{\text{L}}{s}^2+s\left[R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+k{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}\left(1-k_{\text{v}}\right)\right]+R_{\text{L}}\left[R_{\text{S}}+k\right]}{L_{\text{S}}{L}_{\text{L}}}$

(5.59)

$R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+k{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}\left(1-k_{\text{v}}\right)>0$

(5.60)

$k{L}_{\text{L}}-k_{\text{v}}{R}_{\text{L}}{L}_{\text{L}}>-R_{\text{L}}{L}_{\text{S}}-R_{\text{S}}{L}_{\text{L}}-R_{\text{L}}{L}_{\text{L}}$

(5.61)

$k{L}_{\text{L}}>-R_{\text{L}}{L}_{\text{S}}-R_{\text{S}}{L}_{\text{L}}$

(5.62)

$k_{\text{v}}<\frac{R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}+k{L}_{\text{L}}}{R_{\text{L}}{L}_{\text{L}}}$

(5.63)

$\underset{L_{\text{L}}\to 0}{\lim }\frac{R_{\text{L}}{L}_{\text{S}}+R_{\text{S}}{L}_{\text{L}}+R_{\text{L}}{L}_{\text{L}}+k{L}_{\text{L}}}{R_{\text{L}}{L}_{\text{L}}}=\frac{R_{\text{S}}+R_{\text{L}}+k}{R_{\text{L}}}$

(5.64)

Example 5.4

Figure 5.27 shows a SAF connected to an HVS-type load. The load consists of an uncontrolled three-phase rectifier with a capacitor of 2200 μF and a resistor of 50/3 Ω in parallel at the dc side. The source is sinusoidal, with an rms value of the phase voltage of 100 V and a frequency of 50 Hz. A resistance of 1.8 Ω and an inductance of 2.8 mH are included to model the equivalent impedance from connection point. The active filter is connected to the system by means of three-phase transformers with ratio 1:1, an LC ripple filter with inductance of 0.15 mH and a capacity and 50 μF in order to reduce the ripple of the output voltage. The inverter is a three-phase IGBT Bridge, the dc side has a connected constant source of 100 V. The simulation has been done in MATLAB–Simulink models with devices SymPowerSystem library.

To obtain the system model, the load without active filter is simulated and active and reactive power [21] are determined [22]. The values obtained by simulation are: 2125 W and 454 var, respectively. Considering that the rms value of the fundamental component of the load voltage across is 82.7 V, a value for the load resistor is obtained, R_L = 9.65 Ω and load inductance L_L = 144 mH. These values allow us to define the single-phase equivalent circuit model shown in Figure 5.22.

In the state model defined by the expression (5.22), matrix system without active filter is given by

$A=\left[\begin{array}{cc}\hfill -4089\hfill & \hfill 3446\hfill \\ \hfill 66.5\hfill & \hfill -66.5\hfill \end{array}\right]$

B₁ vector is given by

$B_1={\left[\begin{array}{cc}\hfill -357\hfill & \hfill 0\hfill \end{array}\right]}^T$

B₂ matrix is

$B_2=\left[\begin{array}{cc}\hfill 357\hfill & \hfill 3446\hfill \\ \hfill 0\hfill & \hfill -66.5\hfill \end{array}\right]$

When the current source is chosen as output signal, the matrix C is

$C=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \end{array}\right]$

With D₁ = [0] and D₂=[0 0]

Thereby state vector is formed

$\begin{array}{l}\dot{x}=Ax+B_{1}u+B_{2}v\\ y=Cx+D_{1}u+D_{2}v\end{array}$

Figure 5.28 shows the pole diagram of the uncompensated system, two real poles located at s = −10.3 and s = –4150 are observed.

On the other hand, Figure 5.29 shows the Bode magnitude when source current is considered as an output signal. When the input signal is the supply voltage harmonics, the system has a gain of −21.9 dB for a frequency of 250 Hz, corresponding to harmonic 5. Considering the distortion of the load current as input, the gain at that frequency is higher, −2.19 dB. This enables us to ensure that with this load type, the source current distortion is mainly due to the load rather than the supply voltage.

Subsequently, the active filter is connected to the system. First, the control strategy by detection of source current is applied. The new equation of state is defined by (5.34). The gain of the state vector is set to k = 20, which is obtained

$K=\left[\begin{array}{cc}\hfill 20\hfill & \hfill 0\hfill \end{array}\right]$

The new system matrix is defined in the form

$A+B_{1}K=\left[\begin{array}{cc}\hfill -11229\hfill & \hfill 3446\hfill \\ \hfill 66.5\hfill & \hfill -66.5\hfill \end{array}\right]$

Figure 5.30 shows the new pole diagram. The fed back system has two real poles located at s = −46 and s = −11200. As predicted, from the theoretical point of view, the system is stable for values of k > 0. Furthermore, the poles are moved away off the origin, allowing to provide greater system robustness.

Secondly, the control strategy by detection of the load voltage is applied. Matrices K₁ and K₂ as defined in (5.45) and (5.46), which applied to this example result in:

$\begin{array}{l}{K}_1=\left[\begin{array}{cc}\hfill -9.18\hfill & \hfill \text{9.18}\hfill \end{array}\right]\\ K_2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 9.18\hfill \end{array}\right]\end{array}$

Where k_v = 0,95. The new system matrix is

$A+B_1{K}_1=\left[\begin{array}{cc}\hfill -816.2\hfill & \hfill 173.2\hfill \\ \hfill 66.5\hfill & \hfill -66.5\hfill \end{array}\right]$

Furthermore, the matrix to be multiplied by the input vector is given by

$B_2+B_1{K}_2=\left[\begin{array}{cc}\hfill 357\hfill & \hfill 173.2\hfill \\ \hfill 0\hfill & \hfill -66.5\hfill \end{array}\right]$

In this case the system has the pole diagram shown in Figure 5.31. Two real poles located at s = –51.4 and s = –831 are observed. The two poles are located in the left half plane, which ensures system stability.

Finally, the hybrid control strategy is applied, which is a combination of the two previous approaches. It is considered k_v = 0.95 and k = 20, so that according to (5.55) and (5.56), the following vectors are obtained

$\begin{array}{l}K{\text{'}}_1=\left[\begin{array}{cc}\hfill 10.83\hfill & \hfill 9.16\hfill \end{array}\right]\\ K{\text{'}}_2=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 9.16\hfill \end{array}\right]\end{array}$

Therefore, the system matrix is

$A+B_{1}K{\text{'}}_1=\left[\begin{array}{cc}\hfill -7956\hfill & \hfill 173.2\hfill \\ \hfill 66.5\hfill & \hfill -66.5\hfill \end{array}\right]$

The matrix that multiplies the input vector is

$B_2+B_{1}K{\text{'}}_2=\left[\begin{array}{cc}\hfill 357\hfill & \hfill 173.2\hfill \\ \hfill 0\hfill & \hfill -66.5\hfill \end{array}\right]$

This result shows how this matrix is the same as the previously obtained, with strategy by detection of the load voltage.

Figure 5.32 presents the system poles diagram when this control strategy is applied. Two real poles appear in the left half plane located at s = −65 and s = −7960.

From the point of view of the system gain is of interest for analyzing the behavior of the three strategies for the two harmonic input sources: the load distortion and source voltage. When the supply voltage is distorted, the strategy that presents a lower gain is that based on the detection of the current source, Figure 5.33. This is almost constant with a value of about −30 dB in a frequency range between 50 Hz and 1 kHz (this involves considering the first 20 harmonics). By contrast, the strategy for detection of the load voltage has highest gain. For the same range of frequencies, the gain varies from −8 dB to −30 dB. These values are higher than those obtained when the system has no active filter. Therefore, when the supply voltage is distorted, strategy by detection of the load voltage amplifies source current harmonics.

Control hybrid strategy presents a practically constant gain, though slightly higher than that achieved with the strategy by detection of the source current.

When harmonics are generated by the load (Figure 5.33), the lowest gain is achieved by the hybrid strategy. In the frequency range of the most significant harmonics of the system (100–1000 Hz) the gain is −33.4 dB. On the opposite side is the strategy by detection of the source current, which has a gain of −10 dB.

In the strategy by detection of the current source, the system gain value depends on the k value. Figure 5.34 shows how the gain varies with the k constant. When k increases the gain decreases. However, this variation is not linear because the gain reduction is not proportional to the k constant. This occurs regardless of the input source that is considered. As design criterion, a k value, which allows obtaining a desired gain value can be adopted. The Figure 5.35 shows the absolute gain versus k when the load is considered as distortion source. Thus, when the goal is to have a gain of 0.1, the value of k must be 90. However, this is a cost of the control signal, which is in many cases unnecessary, since we will always have an error about the tension filter output due to the switching of the power devices and the measurement system itself. Thereby, it can produce acceptable results with lower gain values.

The hybrid strategy requires less gain from the point of view of the harmonics elimination of source current. In regard to the most appropriate k value, with smaller k values in strategy by detection of the source current, it is possible to obtain smaller gain. Figure 5.36 shows the gain for different k values when the hybrid strategy is applied. In the case that the supply voltage is distorted, k must be high, since this strategy has a gain slightly greater than the strategy by detection of the source current. If the supply voltage is sinusoidal with a value of k = 10 it reaches −28.2 dB at a frequency of 250 Hz, which means reducing this harmonic at 95.3%.

In what follows, the system when the output is the voltage at the PCC to the network is analyzed. In this situation, the matrix C is

$\text{C}=\left[\begin{array}{cc}\hfill 9.65\hfill & \hfill -9.65\hfill \end{array}\right]$

In addition, D₁ = [1] and D₂=[0 −9.65]

Figure 5.37 shows the gains for different strategies when the output is the voltage at the PCC. When the input are the supply voltage harmonics, the strategy by detection of the current source and the hybrid strategy have a gain of the same order as that of the system without the active filter. Therefore, these control strategies do not improve the THD of this voltage. Moreover, the strategy of detecting the load voltage shows a similar gain to the system without compensating, so that the voltage at the PCC remains the same total harmonic distortion.

Finally, when the distortion of the load source current is taken into account, the control strategy with lower gain is the hybrid strategy. At the same time, the voltage at the PCC is less influenced by the harmonic distortion produced by the load.

5.4.1. Results of Practical Cases