## 2 Small-Signal Analysis
Figure 1:2  
Equations (1), (3), and (5) hold good for inductor currents \(i_{L1},\ i_{L2}\), and output capacitor \(c_{o}\). Identically, state equations for the other half-cycle can also be derived. State equations are averaged over an HF cycle. The average value for the rate of change of \(i_{Ls}\) over one complete HF cycle is zero, and the averaged state equation is:  
Figure 1:3  
So, the state variable \(i_{Ls}\) is omitted for the following analysis. Define: \(d_{1}T_{s}=t_{2}-t_{1}\), \(d_{2}T_{s}=t_{3}-t_{2}\), \(d_{3}T_{s}=t_{4}-t_{3}\), \(d_{4}T_{s}=t_{5}-t_{4}\), \(d_{5}T_{s}=t_{6}-t_{5}\), \(d_{6}T_{s}=t_{7}-t_{6}\), \(d_{7}T_{s}=t_{8}-t_{7}\), \(d_{8}T_{s}=t_{9}-t_{8}\), \(d_{9}T_{s}=t_{10}-t_{9}\), \(d_{10}T_{s}=t_{8}-t_{10}\). The averaged state equations of defined state variables over an HF cycle are given:  
Figure 1:4  
Figure 1:5  
Figure 1:6  
where \(i_{average}\) is the average current feeding the output capacitor and load from secondary side H-bridge switches and is given by:  
Figure 1:7  
Substituting Equation (12) into Equation (11) gives:  
Figure 1:8  
The duty ratio of the main switches, including conduction of the anti-parallel diodes, is defined as:  
Figure 1:9  
Figure 3:0  
Perturbation is introduced around the steady-state values of the state variables and input voltage such that \(i_{L1}=I_{L}+i_{L1}\), \(i_{L2}=I_{L}+i_{L2}\), \(v_{in}=V_{in}+\phi_{in}\), \(v_{o}=V_{o}+\phi_{o}\), \(d_{s1}=D+\hat{d}_{s1}\), and \(d_{s2}=D+\hat{d}_{s2}\). The state equations are modified to the following:  
Figure 3:1Figure 3:2  
Figure 3:3  
Neglecting the second order terms and steady-state or DC terms results in the following equations:  
Figure 3:4  
Figure 3:5  
Figure 3:6  
Taking Laplace transform, and then solving results in:  
Figure 3:7  
Figure 3:8  
Figure 3:9  
Writing in matrix form:  
\[\left[ \begin{array}{c}i_{L1}(s)\\ i_{L2}(s)\\ \vartheta_{o}(s)\end{array} \right]=\ [A(s)]\cdot\left[ \begin{array}{c}\frac{V_{o}}{n} \\ 0\\ \frac{l_{L1}}{n}\end{array} \right]\hat{d}_{s1}(s)+[A(s)]\cdot\left[ \begin{array} []{c}0\\ \frac{V_{o}}{n}\\ \frac{l_{L2}}{n}\end{array} \right]\hat{d}_{s2}(s)+[A(s)]\cdot\left[ \begin{array} []{c}1\\ 1\\ 0\end{array} \right]\hat{o}_{in}(s) \tag{25}\]  
where  
\[A(s)=\left[ \begin{array}{ccc}sL_{1}&0&\frac{(1-D)}{n}\\ 0&sL_{2}&\frac{(1-D)}{n}\\ \frac{(1-D)}{n}&\frac{(1-D)}{n}&-\Big{(}sC_{o}+\frac{1}{R_{L}}\Big{)}\end{array}  \right]^{-1} \tag{26}\]  
\[sL(i_{L1}(s)+i_{L2}(s))+\frac{2(1-D)\vartheta_{o}(s)}{n}=\frac{V_{o}}{n} \Big{(}\hat{d}_{s1}(s)+\ \hat{d}_{s2}(s)\Big{)}+2\ \hat{v}_{in}(s) \tag{27}\]  
Writing in matrix form:  
\[\left[ \begin{array}{c}i_{L1}(s)+i_{L2}(s)\\ \vartheta_{o}(s)\end{array} \right]=\ [A(s)]\cdot\left[ \begin{array}{c}\frac{v_{o}}{n} \\ \frac{l_{L}}{n}\end{array} \right]\left[\begin{array}{c}\hat{d}_{s1}(s)+\hat{d} _{s2}(s)\Big{)}+[A(s)]\cdot\left[ \begin{array}{c}2\\ 0\end{array} \right]\hat{o}_{in}(s) \tag{28}\]  
where  
\[A(s)=\left[ \begin{array}{ccc}sL&\frac{2(1-D)}{n}\\ \frac{(1-D)}{n}&-\Big{(}sC_{o}+\frac{1}{R_{L}}\Big{)}\end{array} \right]^{-1} \tag{29}\]  
The control-to-output transfer function is obtained from Equation (28) by keeping \(\vartheta_{in}=0\), resulting in the following equation:  
\[\frac{\vartheta_{o}(s)}{\hat{d}_{s1}(s)+\hat{d}_{s2}(s)}=\frac{\frac{(1-D)V_{o} }{n^{2}}-s\frac{L_{L1}}{n}}{(LC_{o})s^{2}+\frac{L}{R_{L}}s+\frac{2(1-D)^{2}}{n^ {2}}} \tag{30}\]