6.2.2. Control-To-Filter Inductor Current Transfer Function

The filter inductor current and the duty cycle relationship is found by using (102) to find the following equation:

$$\psi\_{\rm out} = \frac{R(1 + sR\_c\mathcal{C})}{s\mathcal{R}\mathcal{C} + sbR\_c\mathcal{C} + b} \sum\_{i=1}^{a} \sum\_{j=1}^{b} \hat{\imath}\_{Lij} \tag{109}$$

Substituting (109) in (107) and considering the same assumptions as in Section 7.2.

$$\frac{V\_{in}}{\rho K}d\_1 - R\_d \sum\_{j=1}^n \mathbf{\hat{i}}\_{Lj} = sL \sum\_{j=1}^n \mathbf{\hat{i}}\_{Lj} + \frac{cR(1+sR\_cC)}{sRC + sbR\_cC + b} \sum\_{j=1}^n \mathbf{\hat{i}}\_{Lj} \tag{110}$$

Simplifying (110) would result in (111).

$$G\_{\rm id} = \frac{\hat{\mathbf{i}}\_L}{\hat{d}} = \frac{\frac{V\_{\rm in}}{\rho \mathcal{K}} \left(b + s \mathcal{R} \mathcal{C} + sb \mathcal{R}\_c \mathcal{C}\right)}{R \left(s^2 L \mathcal{C} \left(1 + \frac{b \mathcal{R}\_c}{\mathcal{R}}\right) + s \left(\frac{b \mathcal{L}}{\mathcal{R}} + \mathcal{R}\_d \mathcal{C} \left(1 + \frac{b \mathcal{R}\_c}{\mathcal{R}}\right) + c \mathcal{R}\_c \mathcal{C}\right) + \frac{b \mathcal{R}\_d}{\mathcal{R}} + c}\right)} \tag{111}$$
