# Simple question [Signals and Systems]

by maverick280857
Tags: signals, simple, systems
 P: 1,770 Hi The following question is from Oppenheim/Wilsky/Nawab chapter 1. Consider a periodic signal $$x(t) = 1$$ for $$0 \leq t \leq 1$$ $$x(t) = -2$$ for $$1 < t <2$$ with period $T = 2$. The derivative of this signal is related to the impulse train $$g(t) = \sum_{k = -\infty}^{\infty}\delte(t-2k)$$ with period T = 2. It can be shown that $$\frac{dx(t)}{dt} = A_{1}g(t-t_{1}) + A_{2}g(t-t_{2})$$ Determine the values of $A_{1}$, $t_{1}$, $A_{2}$, and $t_{2}$. I got stuck with this one. Anyway here's my solution. Would appreciate any help in solving the problem. $$x(t) = \sum_{k = -\infty}^{\infty}(u(t-2k) - u(t-2k-1)) + (-2)(u(t-2k-1) - u(t-2k-2))$$ so $$x(t) = \sum_{k = -\infty}^{\infty}u(t-2k) - 3\sum_{k = -\infty}^{\infty}u(t-2k-1)) -2\sum_{k = -\infty}^{\infty}u(t-2k-2)$$ so $$\frac{dx}{dt} = g(t) - 3g(t-1) - g(t-2)$$ which is wrong...
 P: 1,770 Okay I got it graphically, but what if I want to do it algebraically?

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