Question

How did they find the derivative for the triangular function x(t)

Can you show me please

I’m trying to understand

5.10. Consider the triangular wave x(t) shown in Fig. 5-14(a). Using the differentiation technique, find the triangular Fourier series of x(t) From Fig. 5-14(a) the derivative x'() of the triangular wave x(t) is, as shown in Fig. 5-14(b), (5.125) Using Eq. (5.117), Eq. (5.125) becomes COS cos k ?0t (5.126) To Ti Equating Eqs. (5.126 and (5.122), we have 2 A To a- 0, k 0 kwobkor From Fig. 5-14(a) and Eq. (5.9a) we have Thus, substituting these values into Eq. (5.8), we get x( , ) _ 3 + ? ?' ksin k@gt (S.127) oh- To

Answer 1

Consider triangular pulse, -A/T_0 (t) between (-T_0 to 0) implies x_1 (t) = -A/T_0 t(u(t + 10^bar) - u(t)) between (0 to T_0) implies x_2 (t) = [u(t) - u(t - T_0)] (-A/T_0 (t - T_0)] similarly for remaining (Because d/dx (uv) = u dv + v du) consider x_1 (t) implies d/dt x_1 (t) = [u(t) - u(t - T_0)] d/dt (-A/T_0 t) - A/T_0 t d/dt [(u(t) - u(t - T_0)] d/dt x_1 (t) = [u(t) - u(t - T_0)] (-A/T_0) + (-A/T_0) + (-A/T_0 t)[delta (t) - delta(t - T_0)] = -A/T_0 [u(t) - u(t - T_0)] - 0 + A/T_0 t delta(t - T_0) d/dt x_1 (t) = -A/T_0 [u(t) - u(t - T_0)] + A/T_0 delta(t - T_0) [Because p(t) delta(t - t_0) = p(t_0) delta(t - t_0)] similarly consider x_2 (t) = [u(t) - u(t - T_0)] [-AT_0 (t - T_0)] \r\n d/dt x_2 (t) = [u(t) - u(t - T_0)] d/dt [-A/T_0 (t - T_0)] + -A/T_0 (t - T_0) d/dt [u(t) - u(t - T_0)] = [u(t) - u(t - T_0)] [-A/T_0] - A/T_0 (t - T_0) [delta(t) - delta(t - T_0)]