Hi The following question is from Oppenheim/Wilsky/Nawab chapter 1. Consider a periodic signal [tex]x(t) = 1[/tex] for [tex]0 \leq t \leq 1[/tex] [tex]x(t) = -2 [/tex] for [tex]1 < t <2[/tex] with period [itex]T = 2[/itex]. The derivative of this signal is related to the impulse train [tex]g(t) = \sum_{k = -\infty}^{\infty}\delte(t-2k)[/tex] with period T = 2. It can be shown that [tex]\frac{dx(t)}{dt} = A_{1}g(t-t_{1}) + A_{2}g(t-t_{2})[/tex] Determine the values of [itex]A_{1}[/itex], [itex]t_{1}[/itex], [itex]A_{2}[/itex], and [itex]t_{2}[/itex]. I got stuck with this one. Anyway here's my solution. Would appreciate any help in solving the problem. [tex]x(t) = \sum_{k = -\infty}^{\infty}(u(t-2k) - u(t-2k-1)) + (-2)(u(t-2k-1) - u(t-2k-2))[/tex] so [tex]x(t) = \sum_{k = -\infty}^{\infty}u(t-2k) - 3\sum_{k = -\infty}^{\infty}u(t-2k-1)) -2\sum_{k = -\infty}^{\infty}u(t-2k-2) [/tex] so [tex]\frac{dx}{dt} = g(t) - 3g(t-1) - g(t-2)[/tex] which is wrong...