And here is how I found Fourier series for our voltage signal:
\(v_x(t)=C_0+\sum_{n=1}^{\infty}[A_n\cos{nw_0t}+B_n\sin{nw_0t}]\), where \(C_0=\frac{1}{T}\int_{\tau}^{\tau+T}f(t)dt=\frac{1}{T}V_p_k\left ( \int_{0}^{T/2}\sin{w_0t}dt+\int_{T/2}^{T} (-\sin{w_0t})dt\right )=...=\frac{2V_p_k}{\pi}\),\(A_n=\frac{2}{T}\int_{\tau}^{\tau+T}f(t)\cos{nw_0t}dt=\frac{2}{T}V_p_k\left ( \int_{0}^{T/2}\sin{w_0t}*\cos{nw_0t}dt+\int_{T/2}^{T} (-\sin{w_0t})*\cos{nw_0t}dt\right )=...=2V_p_k*\frac{(-1)^{n+1}-1}{\pi(n-1)(n+1)}\) and \(B_n=0\)
\(v_x(t)=C_0+\sum_{n=1}^{\infty}[A_n\cos{nw_0t}+B_n\sin{nw_0t}]\), where \(C_0=\frac{1}{T}\int_{\tau}^{\tau+T}f(t)dt=\frac{1}{T}V_p_k\left ( \int_{0}^{T/2}\sin{w_0t}dt+\int_{T/2}^{T} (-\sin{w_0t})dt\right )=...=\frac{2V_p_k}{\pi}\),\(A_n=\frac{2}{T}\int_{\tau}^{\tau+T}f(t)\cos{nw_0t}dt=\frac{2}{T}V_p_k\left ( \int_{0}^{T/2}\sin{w_0t}*\cos{nw_0t}dt+\int_{T/2}^{T} (-\sin{w_0t})*\cos{nw_0t}dt\right )=...=2V_p_k*\frac{(-1)^{n+1}-1}{\pi(n-1)(n+1)}\) and \(B_n=0\)
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