Question 5.2: Find the Fourier expansion of the function F(t) shown in Fig...

The function in the figure is defined as follows
F(t) = \begin{cases} 0 & −T_{f}/2 < t < −T_{f}/4 \\F_{0} & −T_{f}/4 ≤ t ≤ T_{f}/4 \\ 0 & T_{f}/4 < t ≤ T_{f}/2 \end{cases}

This function is periodic since F(t) = F(t +T_f). Furthermore, the function is an even function since F(t) = F(−t). Therefore, b_m = 0 for m = 1, 2, . . ..
The coefficient a₀ can be obtained using Eq. 5.9 as follows

a_0 =\frac{2}{T_f} \int_{-T_f/2}^{T_f/2} F(t)   dt = F_{0}
The coefficient a_m, m = 1, 2, . . ., can be obtained by using Eq. 5.14 as follows

that is,
a_m = \begin{cases} 0 & \text{if m is even} \\ (−1)^{(m−1)/2}(\frac{2F_0}{mπ}) & \text{if m is odd} \end{cases}