## 3 Proposed Power Losses Model
### Proposed Conduction Losses Model
#### 3.1.2 The Solution of the Duty Cycle Based on SVPWM  
Two different methods are adopted to calculate the A-phase voltage, and through the comparison of these two equations, the solution of duty cycle can be obtained. It should be pointed out that the calculation of the phase voltages is based on the middle point M of the dc-link voltage source. As shown in Figure 1, supposing that the duty cycle of T1 is \(D(t)\), the average voltage of \(V_{AM}\) can be calculated in a switching period. \[V_{AM}=D(t)\times\frac{V_{L}}{2}-(1-D(t))\times\frac{V_{L}}{2} \tag{9}\]  
Furthermore, \(V_{AM}\) can be obtained according to the SVPWM modulation. The duty cycle of switching tubes based on SVPWM is not obtained by comparing the modulated sinusoidal wave with the triangular wave, but by the FOC algorithm. The modulation principle diagram is shown in Figure 2. Taking the first section as an example, it is assumed that the operating time of two basic vectors \(U_{110}\) and \(U_{100}\) is \(T_{\texttt{b6}}\) and \(T_{\texttt{b4}}\), respectively. The voltage in the first section can be expressed:  
\[V_{AM}=\frac{T_{V4}}{T_{sw}}\times\frac{V_{L}}{2}+\frac{T_{V6}}{T_{sw}}\times \frac{V_{L}}{2} \tag{10}\]Furthermore, assuming that the reference voltage vector \(V_{r}\) is synthesized from the two basic vectors, the operating time of two basic vectors can be calculated according to the volt-second equilibrium principle. \[\frac{\left(\frac{\pi}{3}V_{L}\right)T_{V\alpha}/T_{\infty}}{\sin(\frac{\pi}{3}- \alpha)}=\frac{V_{r}}{\sin\frac{2\pi}{3}}=\frac{\left(\frac{2}{3}V_{L}\right)T _{V6}/T_{\infty}}{\sin\alpha} \tag{11}\]  
where \(\alpha\) represents the phase angle between the reference voltage vector and X-axis. Equation (11) can be solved as:  
\[ \begin{array}{c}\frac{T_{V\alpha}}{T}=\frac{V_{r}\sin(\frac{\pi}{3}-\alpha)} {\frac{2}{3}V_{L}\sin\frac{2\pi}{3}}=\frac{\sqrt{3}}{2}\frac{V_{r}}{\frac{1}{2 }V_{L}}\sin(\frac{\pi}{3}-\alpha)\\ \frac{T_{V6}}{T}=\frac{\sqrt{3}}{2}\frac{V_{r}}{\frac{1}{2}V_{L}}\sin\alpha \end{array}  \tag{12}\]  
Substitute Equation (12) into Equation (10):  
\[V_{AM}=\frac{1}{2}V_{L}\frac{\sqrt{3}}{2}m\sin\alpha+\frac{1}{2}V_{L}\frac{ \sqrt{3}}{2}m\sin(\frac{\pi}{3}-\alpha) \tag{13}\]  
where \(m\) represents the voltage modulation ratio and it can be expressed as \(m=\frac{2V_{r}}{V_{L}}\). Similarly, the A-phase voltage in other sectors can be calculated as:  
\[V_{AM}=\left\{ \begin{array}{ll}\frac{1}{2}V_{L}\frac{\sqrt{3}}{2}m\cos( \alpha-\frac{\pi}{2})+\frac{1}{2}V_{L}\frac{\sqrt{3}}{2}m\cos(\alpha+\frac{ \pi}{6})&0\leq\alpha\leq\frac{\pi}{3}\\ -\frac{1}{2}V_{L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{5\pi}{6})+\frac{1}{2}V_{ L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{\pi}{6})&\frac{\pi}{3}\leq\alpha\leq \frac{2\pi}{3}\\ -\frac{1}{2}V_{L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{\pi}{2})-\frac{1}{2}V_{ L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{7\pi}{6})&\frac{2\pi}{3}\leq\alpha\leq \pi\\ -\frac{1}{2}V_{L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{3\pi}{2})-\frac{1}{2}V_{ L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{5\pi}{6})&-\pi\leq\alpha\leq-\frac{2\pi}{3}\\ -\frac{1}{2}V_{L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{7\pi}{6})+\frac{1}{2}V_{ L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{11\pi}{6})&-\frac{2\pi}{3}\leq\alpha\leq-\frac{ \pi}{3}\\ \frac{1}{2}V_{L}\frac{\sqrt{3}}{2}m\cos(\alpha+\frac{\pi}{6})+\frac{1}{2}V_{ L}\frac{\sqrt{3}}{2}m\cos(\alpha-\frac{\pi}{6})&-\frac{\pi}{3}\leq\alpha\leq 0 \end{array} \right. \tag{14}\]  
Figure 2: The modulation principle of space vector pulse width modulation (SVPWM). By comparing Equations (9) and (14), the duty cycle can be solved as:  
\[D(t)=\left\{