## 2 Harmonic Analysis of DC-Link Capacitor Current for One VSI
\begin{split} S_{B}(t)=\frac{A_{00}}{2}+A_{01}\text{cos}\left(w_{ 0}t-\frac{2\pi}{3}\right)\\ \qquad\qquad\qquad+\sum\limits_{n=2}^{n=\infty}A_{0n}\text{cos} \left(n\left(w_{01}-\frac{2\pi}{3}\right)\right)+\sum\limits_{m=1}^{m=\infty} \sum\limits_{n=-\infty}^{n=\infty}A_{mn}\text{cos}\left(m(w_{c}t+\theta_{c})+n \left(w_{0}t-\frac{2\pi}{3}\right)\right)\end{split}  \tag{13}\]  
\[ \begin{split} S_{C}(t)=\frac{A_{00}}{2}+A_{01}\text{cos}\left(w_{ 0}t+\frac{2\pi}{3}\right)\\ \qquad\qquad+\sum\limits_{n=2}^{n=\infty}A_{0n}\text{cos}\left(n \left(w_{01}+\frac{2\pi}{3}\right)\right)+\sum\limits_{m=1}^{m=\infty}\sum \limits_{n=-\infty}^{n=\infty}A_{mn}\text{cos}\left(m(w_{c}t+\theta_{c})+n \left(w_{0}t+\frac{2\pi}{3}\right)\right)\end{split}  \tag{14}\]  
For a clearer analysis, the input current of the inverter in Equation (1) is divided into four parts, which are represented as follows:  
\[i_{inv}=i_{inv\_1 st}+i_{inv\_2 nd}+i_{inv\_3 rd}+i_{inv\_4 th} \tag{15}\]  
Substituting the first terms of Equations (12)-(14) together with (3)-(5) into Equation (1) gives the first term of the inverter input current, which is represented by  
\[i_{inv\_1 st}=\frac{1}{2}\times I_{o}\text{cos}(w_{0}t-\mathcal{O})+\frac{1}{2 }\times I_{o}\text{cos}\left(w_{0}t-\frac{2\pi}{3}-\mathcal{O}\right)+\frac{1 }{2}\times I_{o}\text{cos}\left(w_{0}t-\frac{4\pi}{3}-\mathcal{O}\right)=0 \tag{16}\]  
where the result is zero because the sum of the output phase currents is zero in a three-phase balanced system. Substituting the second terms of Equations (12)-(14) along with (3)-(5) into Equation (1) gives the second term of the inverter input current, which is represented as  
\[ \begin{split} i_{inv\_2 nd}=\frac{M}{2}\text{cos}w_{0}t\times I_{0 }\text{cos}(w_{0}t-\mathcal{O})+\frac{M}{2}\text{cos}\left(w_{0}t-\frac{2\pi} {3}\right)\times I_{0}\text{cos}\left(w_{0}t-\frac{2\pi}{3}-\mathcal{O}\right) \\ \qquad\qquad+\frac{M}{2}\text{cos}\left(w_{0}t+\frac{2\pi}{3} \right)\times I_{0}\text{cos}\left(w_{0}t-\frac{4\pi}{3}-\mathcal{O}\right)= \frac{3}{4}MI_{0}\text{cos}\mathcal{O}\end{split}  \tag{17}\]  
where the result is the average input current of the three-phase inverter which is directly related to the modulation index and power factor angle. Substituting the first summation terms of Equations (12)-(14) along with (3)-(5) into Equation (1) gives the third term of the inverter input current, which is given by  
\[ \begin{split} i_{inv\_3 rd}=\frac{3\sqrt{3}M}{2\pi(\mu^{2}-1)} \text{sin}\left(\frac{n\pi}{6}\right)\text{sin}\left(\frac{n\pi}{2}\right) \text{cos}(nw_{0}t)\times I_{0}\text{cos}(w_{0}t-\mathcal{O})\\ \qquad\qquad+\frac{3\sqrt{3}M}{2\pi(\mu^{2}-1)}\text{sin}\left( \frac{n\pi}{6}\right)\text{sin}\left(\frac{n\pi}{2}\right)\text{cos}\left(n(w_ {0}t-\frac{2\pi}{3}\right)\right)\times I_{0}\text{cos}\left(w_{0}t-\frac{2 \pi}{3}-\mathcal{O}\right)\\ \qquad\qquad+\frac{3\sqrt{3}M}{2\pi(\mu^{2}-1)}\text{sin}\left( \frac{n\pi}{6}\right)\text{sin}\left(\frac{n\pi}{2}\right)\text{cos}\left(n(w_ {0}t+\frac{2\pi}{3}\right)\right)\times I_{0}\text{cos}\left(w_{0}t-\frac{4 \pi}{3}-\mathcal{O}\right)=0\ for\ n=3,\ 9,\ 15\ldots\end{split}  \tag{18}\]  
where the result is equal to zero. Substituting the second summation terms of Equations (12)-(14) along with (3)-(5) into Equation (1) gives the fourth term of the inverter input current represented by the following equation:  
\[ \begin{split} i_{inv\_4 th}=A_{mn}\text{cos}(m(w_{c}t+\theta_{c})+n (w_{0}t))*I_{0}\text{cos}(w_{0}t-\mathcal{O})+A_{mn}\text{cos}\left(m(w_{c}t+ \theta_{c})+n\left(w_{0}t-\frac{2\pi}{3}\right)\right)\\ *I_{0}\text{cos}(w_{0}t-\frac{2\pi}{3}-\mathcal{O})+A_{mn}\text{ cos}\left(m(w_{c}t+\theta_{c})+n\left(w_{0}t-\frac{4\pi}{3}\right) \right)*I_{0}\text{cos}\left(w_{0}t-\frac{4\pi}{3}-\mathcal{O}\right)\\ =\left\{\begin{array}{cc}\frac{3}{2}A_{mn}I_{0}\text{cos}(m(w_{c}t+ \theta_{c})+(n+1)w_{0}t-\mathcal{O}),&n=3h-1\\ \frac{3}{2}A_{mn}I_{0}\text{cos}(m(w_{c}t+\theta_{c})+(n-1)w_{0}t+\mathcal{O}),& n=3h+1\\ 0\,&n=3h\end{array}\right.\end{split}