Question 5.1: Find the Fourier series of the periodic function F(t) shown ...
Find the Fourier series of the periodic function F(t) shown in Fig. 5.2.
The function F(t) shown in the figure is defined over the interval (−T_{f}/2, T_{f}/2) by
F(t) = \begin{cases} 0 & − T_{f}/2 < t < 0\\ F_{0} & 0 ≤ t ≤ T_{f}/2 \end{cases}
Therefore, the coefficients a_{0}, a_{m}, and b_{m} in the Fourier series of Eq. 5.3 are obtained as follows
F(t)=\frac{a_0}{2}+\sum_{n=1}^{\infty} a_n \cos n \omega_{\mathrm{f}} t+\sum_{n=1}^{\infty} b_n \sin n \omega_{\mathrm{f}} t (5.3)
\begin{aligned} a_0 &=\frac{2}{T_{\mathrm{f}}} \int_{-T_{\mathrm{f}} / 2}^{T_{\mathrm{f}} / 2} F(t) d t=\frac{2}{T_{\mathrm{f}}}\left[\int_{-T_{\mathrm{f}} / 2}^0(0) d t+\int_0^{T_{\mathrm{f}} / 2} F_0 d t\right] \\ &=\frac{2 F_0}{T_{\mathrm{f}}} \frac{T_{\mathrm{f}}}{2}=F_0 \\ a_m &=\frac{2}{T_{\mathrm{f}}} \int_{-T_{\mathrm{f}} / 2}^{T_{\mathrm{f}} / 2} F(t) \cos m \omega_{\mathrm{f}} t d t \\ &=\frac{2}{T_{\mathrm{f}}}\left[\int_{-T_{\mathrm{f}} / 2}^0(0) \cos m \omega_{\mathrm{f}} t d t+\int_0^{T_{\mathrm{f}} / 2} F_0 \cos m \omega_{\mathrm{f}} t d t\right] \\ &=\left.\frac{2 F_0}{m \omega_{\mathrm{f}} T_{\mathrm{f}}} \sin m \omega_{\mathrm{f}} t\right|_0 ^{T_{\mathrm{f}} / 2}=\frac{2 F_0}{m \omega_{\mathrm{f}} T_{\mathrm{f}}} \sin \left(m \omega_{\mathrm{f}} T_{\mathrm{f}} / 2\right) \\ &=\frac{F_0}{\pi m} \sin m \pi=0 \\ b_m &=\frac{2}{T_{\mathrm{f}}} \int_{-T_{\mathrm{f}} / 2}^{T_{\mathrm{f}} / 2} F(t) \sin m \omega_{\mathrm{f}} t d t \\ &=\frac{2}{T_{\mathrm{f}}}\left[\int_{-T_{\mathrm{f}} / 2}^0(0) \sin m \omega_{\mathrm{f}} t d t+\int_0^{T_{\mathrm{f}} / 2} F_0 \sin m \omega_{\mathrm{f}} t d t\right] \end{aligned}\begin{aligned} &=-\left.\frac{2 F_0}{m \omega_{\mathrm{f}} T_{\mathrm{f}}} \cos m \omega_{\mathrm{f}} t\right|_0 ^{T_{\mathrm{f}} / 2}=-\frac{F_0}{\pi m}[\cos m \pi-1] \\ &=\left\{\begin{array}{ll} \frac{2 F_0}{\pi m} & \text { if } m \text { is odd } \\ 0 & \text { if } \quad m \text { is even } \end{array}=\frac{2 F_0}{m \pi}, \quad m=1,3,5, \ldots\right.\end{aligned}
Therefore, the Fourier series of the periodic forcing function shown in Fig. 5.2 is given by
\begin{aligned} F(t) &=\frac{F_0}{2}+\frac{2 F_0}{\pi} \sin \omega_{\mathrm{f}} t+\frac{2 F_0}{3 \pi} \sin 3 \omega_{\mathrm{f}} t+\frac{2 F_0}{5 \pi} \sin 5 \omega_{\mathrm{f}} t+\ldots \\ &=\frac{F_0}{2}+\sum_{n=1,3,5}^{\infty} \frac{2 F_0}{n \pi} \sin n \omega_{\mathrm{f}} t=F_0\left[\frac{1}{2}+\sum_{n=1,3,5}^{\infty} \frac{2}{n \pi} \sin n \omega_{\mathrm{f}} t\right] \end{aligned}
where \omega_{\mathrm{f}} t=2 \pi / T_{\mathrm{f}}