
#1
May3007, 01:44 AM

P: 1,772

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(t2k)[/tex] with period T = 2. It can be shown that [tex]\frac{dx(t)}{dt} = A_{1}g(tt_{1}) + A_{2}g(tt_{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(t2k)  u(t2k1)) + (2)(u(t2k1)  u(t2k2))[/tex] so [tex]x(t) = \sum_{k = \infty}^{\infty}u(t2k)  3\sum_{k = \infty}^{\infty}u(t2k1)) 2\sum_{k = \infty}^{\infty}u(t2k2) [/tex] so [tex]\frac{dx}{dt} = g(t)  3g(t1)  g(t2)[/tex] which is wrong... 



#2
May3007, 02:35 AM

P: 1,772

Okay I got it graphically, but what if I want to do it algebraically?



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