# Analytical Selection of Leakage Inductance for Single-Phase Dual Active Bridge Converters
## 2 Steady State Analysis
It can be observed from Equations (1), (2) and (4) that the voltage on the primary and secondary sides of the transformer, as well as the inductor current, can be represented as infinite summations of odd sinusoids. The first, third, and fifth harmonics of the primary side voltage (\(v_{pr1}(t)\), \(v_{pr3}(t)\), and \(v_{pr5}(t)\)) alongside corresponding harmonics of the inductor current (\(i_{le1}(t)\), \(i_{le3}(t)\), and \(i_{le5}(t)\)) are graphically depicted in Figure 3. It is evident from Figure 3 that the first harmonics of both voltage and current contribute most significantly among all the frequency components. This indicates that the first harmonics carry the majority of the energy in these waveforms. Focusing primarily only on the first harmonics in a DAB DC/DC converter significantly simplifies the power flow analysis in the converter. In the modeling and analysis of the DAB converter, only the first harmonic components of both the voltages (\(v_{pr1}(t)\), \(v_{sr1}(t)\)), and current (\(i_{le1}(t)\)) are considered. This reduces the complexity of the mathematical model and enhances the understanding of the converter's fundamental operation without the complications introduced by higher-order harmonics, which typically have less impact on the overall system performance. To analyze the system in the \(dq\) rotating reference frame, the voltage and current phasors are projected on the \(d\) and \(q\) axes. Thus, the voltage vectors are expressed as  
\[\overrightarrow{v}_{pr1}= \begin{bmatrix}v_{pd}(t)\\ v_{pq}(t)\end{bmatrix} =\frac{4}{\pi}V_{in}(t)\cos(0.5\pi d_{1}) \begin{bmatrix} \sin(\pi d_{3})\\ \cos(\pi d_{3})\end{bmatrix}  \tag{5}\]  
\[\overrightarrow{v}_{sr1}= \begin{bmatrix}v_{sd}(t)\\ v_{sq}(t)\end{bmatrix} =\frac{4}{\pi}V_{o}(t)\cos(0.5\pi d_{2}) \begin{bmatrix} \sin\left(\frac{d_{2}-d_{1}}{2}\pi\right)\\ \cos\left(\frac{d_{2}-d_{1}}{2}\pi\right)\end{bmatrix}  \tag{6}\]  
Figure 3: Harmonic components of primary voltage and inductor current. The projection of the first harmonics of the inductor current vector \(i_{le1}(t)\) on the \(d\) and \(q\) axis of the \(dq\) rotating reference frame can be obtained via  
\[\overset{\rightarrow}{i}_{le1}= \begin{bmatrix}i_{led}(t)\\ i_{leq}(t)\end{bmatrix} =\frac{1}{\omega_{s}L_{e}} \begin{bmatrix}v_{pq}(t)-v_{ qq}(t)\\ v_{sd}(t)-v_{pd}(t)\end{bmatrix}  \tag{7}\]  
The active and reactive powers seen at the transformer's primary side can be expressed as  
\[ \begin{bmatrix}P_{1}(t)\\ Q_{1}(t)\end{bmatrix} =\frac{1}{2} \begin{bmatrix}v_{pd}(t)i_{led}(t)+&v_{pq}(t) i_{leq}(t)\\ v_{pq}(t)i_{led}(t)-&v_{pd}(t)i_{leq}(t)\end{bmatrix}  \tag{8}\]  
The aforementioned power equation can be represented in discrete differential form as  
\[ \begin{bmatrix}\dot{P}_{1}(k)\\ \dot{Q}_{1}(k)\end{bmatrix} =\frac{1}{2} \begin{bmatrix}v_{pd}(k)i_{led}(k)+v_{ pd}(k)i_{led}(k)+\dot{v_{pq}}(k)i_{leq}(k)+v_{pq}(k)i_{leq}(k)\\ v_{pq}(k)i_{led}(k)+v_{pq}(k)i_{led}(k)-\dot{v_{pd}}(k)i_{leq}(k)-v_{pd}(k)i_{ leq}(k)\end{bmatrix}  \tag{9}\]  
The rate of change of the \(dq\) voltages and current in discrete time can be expressed as  
\[ \begin{bmatrix}\Delta v_{pd}(k)\\ \Delta v_{pq}(k)\end{bmatrix} = \begin{bmatrix}\frac{v_{pd}^{*}(k)-v_{pd}(k)}{T_ {s}}\\ \frac{v_{pq}^{*}(k)-v_{pq}(k)}{T_{s}}\end{bmatrix}  \tag{10}\]  
\[ \begin{bmatrix}\Delta i_{led}(k)\\ \Delta i_{leq}(k)\end{bmatrix} = \begin{bmatrix}\frac{i_{led}^{*}(k)-i_{led}(k)}{ T_{s}}\\ \frac{i_{leq}^{*}(k)-i_{leq}(k)}{T_{s}}\end{bmatrix}  \tag{11}\]  
where the \(T_{s}\) denotes the sampling period and the superscript \"\({}^{*}\)\" indicates the corresponding command variables. Using Equations (9)-(11), the rate of change in discrete time of active and reactive power can be illustrated as  
\[\Delta P_{1}(k)=\frac{1}{2}\left( \begin{array}{c}\frac{v_{pd}^{*}(k)-v_{pd} (k)}{T_{s}}i_{led}(k)+\frac{i_{led}^{*}(k)-i_{led}(k)}{T_{s}}v_{pd}(k)\\ +\frac{v_{pq}^{*}(k)-v_{pq}(k)}{T_{s}}i_{leq}(k)+\frac{i_{leq}^{*}(k)-i_{leq}(k )}{T_{s}}v_{pq}(k)\end{array} \right) \tag{12}\]  
\[\Delta Q_{1}(k)=\frac{1}{2}\left(