4.2.1. Control-To-Output Voltage Transfer Function

The output voltage and the duty cycle relationship are found by summing up the KVL equations in (46), considering the same assumptions as in Section 2.2.1, and substituting (42), (44), and (48).

$$\frac{V\_{in}}{K} \left( -\frac{\mathcal{K}R\_d}{V\_{in}} \left( \mathfrak{d}\_{out} \left( \mathbf{sC} + \frac{1}{R} \right) \right) + \sum\_{j=1}^{3} \frac{R\_d D\_{eff}}{3 \,\mathrm{kV}\_{in}} \mathfrak{d}\_{in} + \mathring{d}\_1 \right) = sL \Big( \mathfrak{d}\_{out} \left( \mathbf{sC} + \frac{1}{R} \right) \Big) + 3 \mathfrak{d}\_{out} \tag{49}$$

Simplifying (49) would result in (50):

$$G\_{vd} = \frac{\partial\_{out}}{\partial\_{j}} = \frac{\frac{V\_{\text{in}}}{K}}{s^{2}LC + s\left(\frac{l}{R} + R\_{d}C\right) + \frac{R\_{d}}{R} + 3} \tag{50}$$

### 4.2.2. Control-To-Filter Inductor Current Transfer Function

The filter inductor current and the duty cycle relationship is found by substituting *v*ˆ*out* in terms of ˆ*iLj* using (48) in (49), and considering the same assumptions as in Section 2.2.1.

$$\frac{V\_{\rm in}}{K}\hat{d}\_{1} - R\_{d}\sum\_{j=1}^{3}\hat{\mathfrak{i}}\_{Lj} = s\mathcal{L}\sum\_{j=1}^{3}\hat{\mathfrak{i}}\_{Lj} + \frac{3\mathcal{R}}{s\mathcal{R}\mathcal{C} + 1}\sum\_{j=1}^{3}\hat{\mathfrak{i}}\_{Lj} \tag{51}$$

Simplifying (51) would result in (52):

⎧ ⎪⎪⎪⎪⎪⎪⎨ ⎪⎪⎪⎪⎪⎪⎩

$$G\_{\rm id} = \frac{\hat{\mathbf{i}}\_{\rm L}}{\hat{d}\_{\rm j}} = \frac{\frac{V\_{\rm in}}{\mathcal{K}} (1 + s \text{RC})}{R \left(s^2 L \mathcal{C} + s \left(\frac{1}{\mathcal{R}} + R\_d \mathcal{C}\right) + \frac{R\_d}{\mathcal{R}} + 3\right)}\tag{52}$$

### 4.2.3. Control-To-Module Filter Inductor Current Transfer Function

The module filter inductor current and the duty cycle relationship is found by substituting (42), (44), and (48) in (46) assuming *v*ˆ*in* = 0.

$$\begin{aligned} \left(sL + R\_d + \frac{R}{sRC + 1}\right)\hat{\mathbf{i}}\_{L1} + \frac{R}{sRC + 1}\hat{\mathbf{i}}\_{L2} + \frac{R}{sRC + 1}\hat{\mathbf{i}}\_{L3} &= \frac{V\_{\text{in}}}{\mathcal{K}}\hat{\mathbf{d}}\_{1} \\ \frac{R}{s\mathcal{K}C + 1}\hat{\mathbf{i}}\_{L1} + \left(sL + R\_d + \frac{R}{s\mathcal{K}C + 1}\right)\hat{\mathbf{i}}\_{L2} + \frac{R}{s\mathcal{K}C + 1}\hat{\mathbf{i}}\_{L3} &= \frac{V\_{\text{in}}}{\mathcal{K}}\hat{\mathbf{d}}\_{2} \\ \frac{R}{s\mathcal{K}C + 1}\hat{\mathbf{i}}\_{L1} + \frac{R}{s\mathcal{K}C + 1}\hat{\mathbf{i}}\_{L2} + \left(sL + R\_d + \frac{R}{s\mathcal{K}C + 1}\right)\hat{\mathbf{i}}\_{L3} &= \frac{V\_{\text{in}}}{\mathcal{K}}\hat{\mathbf{d}}\_{3} \end{aligned} \tag{53}$$

Subtracting the 2nd Equation in (53) from the 1st Equation in (53):

$$
\hat{i}\_{L2} = \hat{i}\_{L1} - \frac{V\_{in}}{K(sL + R\_d)}(\hat{d}\_1 - \hat{d}\_2) \tag{54}
$$

Similarly, subtracting the 3rd Equation in (53) from the 1st Equation in (53):

$$
\hat{\mathbf{i}}\_{L3} = \hat{\mathbf{i}}\_{L1} - \frac{V\_{\text{in}}}{K(sL + R\_d)}(\hat{d}\_1 - \hat{d}\_3) \tag{55}
$$

Substituting (54) and (55), in the first Equation in (53):

$$\begin{split} \frac{\left(sL + R\_d + \frac{3R}{sRC + 1}\right)\hat{t}\_{L1}}{\frac{V\_{in}}{K}\left[\left(\frac{2R}{(sL + R\_d)(sRC + 1)} + 1\right)\hat{d}\_1 - \frac{R}{(sL + R\_d)(sRC + 1)}\hat{d}\_2 - \frac{R}{(sL + R\_d)(sRC + 1)}\hat{d}\_3\right]} \tag{56} \end{split} \tag{57}$$

Rearranging (56), the following equation is obtained:

$$
\hat{d}\_{L1} = A(s)\hat{d}\_1 + B(s)\hat{d}\_2 + B(s)\hat{d}\_3 \tag{57}
$$

where

$$A(s) = \frac{V\_{in}(s^2LCR + s(L + CR\_dR) + R\_d + 2R)}{K(sL + R\_d)(s^2LCR + s(L + CR\_dR) + 3R)}\tag{58}$$

$$B(\mathbf{s}) = \frac{-V\_{in}R}{K(\mathbf{s}L + R\_d)(\mathbf{s}^2 L \mathbf{C}R + \mathbf{s}(L + \mathbf{C}R\_d R) + 3\mathbf{R})} \tag{59}$$

Similar steps can be done for the filter inductor current for module 2 and module 3 such that:

$$\begin{cases} \dot{l}\_{L1} = A(s)\hat{d}\_1 + B(s)\hat{d}\_2 + B(s)\hat{d}\_3\\ \dot{l}\_{L2} = B(s)\hat{d}\_1 + A(s)\hat{d}\_2 + B(s)\hat{d}\_3\\ \dot{l}\_{L3} = B(s)\hat{d}\_1 + B(s)\hat{d}\_2 + A(s)\hat{d}\_3 \end{cases} \tag{60}$$

### 4.2.4. Converter Output Impedance

Similarly, the IPOP converter output impedance can be found by modifying (48) such that:

$$\sum\_{j=1}^{3} \hat{\imath}\_{Lj} + \hat{\imath}\_{out} = \vartheta\_{out} \left( \frac{sRC + 1}{R} \right) \tag{61}$$

The output voltage and the output current relationship is found by considering the same assumptions as in Section 2.2.4, summing Equations in (46), and substituting (42), (44), and (61).

$$-R\_d \left( \mathfrak{d}\_{out} \left( \frac{sRC + 1}{R} \right) - \mathfrak{f}\_{out} \right) = sL \left( \mathfrak{d}\_{out} \left( \frac{sRC + 1}{R} \right) - \mathfrak{f}\_{out} \right) + 3\mathfrak{d}\_{out} \tag{62}$$

Simplifying (62) would result in (63):

$$Z\_{\rm out} = \frac{\mathfrak{d}\_{\rm out}}{\mathfrak{d}\_{\rm out}} = \frac{(R\_d + sL)}{s^2LC + s\left(\frac{L}{R} + R\_dC\right) + \frac{R\_d}{R} + 3} \tag{63}$$
