*2.2. Injected Signal Cancellation*

By using the proposed IDM in [16] in the multi-PV system, cancellation can occur when the perturbation signals of PV1 and PV2 are in the opposite positions, as shown in Figure 3.


**Figure 3.** Perturbation signals in cancellation signal scenario.

Figure 4 describes in detail on how the injected signal cancellation can occur.

**Figure 4.** Controller perturbation signal circuit (*Vdc* is the measurement DC bus voltage; *Vpv*<sup>1</sup> and *Vpv*<sup>2</sup> are the PV1 and PV2 voltages, respectively; *Ipv*<sup>1</sup> and *Ipv*<sup>2</sup> are the PV1 and PV2 currents, respectively; *n*<sup>1</sup> and *n*<sup>2</sup> are the perturbation factors of IDM programs 1 and 2, respectively; *D*<sup>1</sup> and *D*<sup>2</sup> are the duty cycles of boost converter 1 and 2, respectively).

The cancellation problem is explained below: Based on the equation in [16], after islanding, the DC-link voltage is shown in Equation (1):

$$\upsilon\_{\rm dc}(t) = [\upsilon(0) - \mathcal{N}\_{\rm k} \mathcal{R}i\_{\rm pv}(t)]e^{-\frac{1}{\mathcal{R}\mathcal{C}}t} + \mathcal{N}\_{\rm k} \mathcal{R}i\_{\rm pv}(t) \tag{1}$$

and

$$N\_k = 1 - n\_k D \tag{2}$$

where:


The DC-link voltage at perturbation *N*<sup>1</sup> from PV1 is as follows:

$$v\_{\rm dc1}(t) = \mathcal{N}\_1 \mathcal{R} i\_{pv}(t) + [v(0) - \mathcal{N}\_1 \mathcal{R} i\_{pv}(t)]e^{-\frac{1}{\mathcal{R}\mathcal{C}}t} \tag{3}$$

where: *vdc*<sup>1</sup> (*t*) is the DC-link voltage at time *t* when perturbation signal 1 is injected and, *N*<sup>1</sup> is the perturbation signal of PV1.

The DC-link voltage at perturbation *N*<sup>2</sup> from PV2 is shown in Equation (4):

$$v\_{dc2}(t) = \mathcal{N}\_2 \mathcal{R}i\_{pv}(t) + [v(0) - \mathcal{N}\_2 \mathcal{R}i\_{pv}(t)]e^{-\frac{1}{\mathcal{R}C}t} \tag{4}$$

where: *vdc*<sup>2</sup> (*t*) is the DC-link voltage at time *t* when perturbation signal 2 is injected.

The DC-link voltage without injecting the perturbation signal is as follows:

$$v\_{\rm dc0}(t) = \mathcal{N}\_0 \mathcal{R}i\_{\rm pv}(t) + [v(0) - \mathcal{N}\_0 \mathcal{R}i\_{\rm pv}(t)]e^{-\frac{1}{\mathcal{R}\mathcal{C}}t} \tag{5}$$

where: *vdc*<sup>0</sup> (*t*) is the DC-link voltage without perturbation signal at time *t* after islanding occurs.

The fluctuation of DC-link voltage caused by the perturbation signal from PV1 is shown in Equation (6):

$$v\_{d\varepsilon1}(t) - v\_{d\varepsilon0}(t) = \left\{\mathcal{N}\_{\mathrm{I}}\mathrm{Ri}\_{p\upsilon}(t) + [v(0) - \mathcal{N}\_{\mathrm{I}}\mathrm{Ri}\_{p\upsilon}(t)]e^{-\frac{1}{\overline{\mathfrak{R}}^{\varepsilon}}t} \right\} - \left\{\mathcal{N}\_{\mathrm{I}}\mathrm{Ri}\_{p\upsilon}(t) + [v(0) - \mathcal{N}\_{\mathrm{I}}\mathrm{Ri}\_{p\upsilon}(t)]e^{-\frac{1}{\overline{\mathfrak{R}}^{\varepsilon}}t} \right\} \tag{6}$$

The Equation (6) is simplified

$$v\_{dc1}(t) - v\_{dc0}(t) = (N\_1 - N\_0)Ri\_{pv}(t)(1 - e^{-\frac{1}{RT}t})\tag{7}$$

Equation (7) becomes

$$v\_{dc1}(t) - v\_{dc0}(t) = (n\_0 - n\_1)DRi\_{pv}(t)(1 - e^{-\frac{1}{\hbar C}t})\tag{8}$$

or

$$
\Delta v\_{1dc}(t) = (n\_0 - n\_1)DRi\_{pv}(t)(1 - e^{-\frac{1}{RT}t})\tag{9}
$$

The fluctuation of DC-link voltage caused by the perturbation signal from PV2 is shown in Equation (10):

$$\sigma\_{d\mathcal{Q}}(t) - \sigma\_{d\mathcal{Q}}(t) = \left\{ \mathcal{N}\_2 \text{Ri}\_{\mathcal{P}^\psi}(t) + [\mathbf{v}(0) - \mathcal{N}\_2 \text{Ri}\_{\mathcal{P}^\psi}(t)]e^{-\frac{1}{\mathcal{R}\mathcal{C}}t} \right\} - \left\{ \mathcal{N}\_0 \text{Ri}\_{\mathcal{P}^\psi}(t) + [\mathbf{v}(0) - \mathcal{N}\_0 \text{Ri}\_{\mathcal{P}^\psi}(t)]e^{-\frac{1}{\mathcal{R}\mathcal{C}}t} \right\} \tag{10}$$

The Equation (10) is simplified

$$
\upsilon\_{dc2}(t) - \upsilon\_{dc0}(t) = (N\_2 - N\_0)Ri\_{pv}(t)(1 - e^{-\frac{1}{RT}t}) \tag{11}
$$

Equation (11) becomes

$$v\_{dc2}(t) - v\_{dc0}(t) = (n\_0 - n\_2) D R i\_{pv}(t) (1 - e^{-\frac{1}{RT}t}) \tag{12}$$

or

$$
\Delta v\_{2dc}(t) = (n\_0 - n\_2) D R i\_{p\upsilon}(t) (1 - e^{-\frac{1}{RT}t}) \tag{13}
$$

Based on Equations (9) and (13), the injected signal cancellation occurs when:

$$
\Delta v\_{1dc}(t) + \Delta v\_{2dc}(t) = 0\tag{14}
$$

$$2n\_0 = n\_1 + n\_2 \tag{15}$$

If *n*<sup>0</sup> = 1 (without perturbation signal), *n*<sup>1</sup> = 1.1, and *n*<sup>2</sup> = 0.9, then Equation (15) is satisfied. Consequently, the fluctuation of DC-link voltage is cancelled because the perturbation signals from PV1 and PV2 have the same value but opposite directions. Thus, the DC-link voltage fluctuations have been cancelled.

The simulation result verifies the problem.
