(New page: == Why does <math> h_1(t) </math> work when sampling with Zero-Order Hold == <math> x_0(t) = h_1(t) * (x(t)p(t)) </math> <math> x_0(t) = h_1(t) * \sum_{n = -\infty}^{\infty}x(t) \delta(t...) |
(No difference)
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Revision as of 17:40, 10 November 2008
Why does $ h_1(t) $ work when sampling with Zero-Order Hold
$ x_0(t) = h_1(t) * (x(t)p(t)) $
$ x_0(t) = h_1(t) * \sum_{n = -\infty}^{\infty}x(t) \delta(t-nT) $
$ x_0(t) = h_1(t) * \sum_{n = -\infty}^{\infty}x(nt) \delta(t-nT) $
$ x_0(t) = \sum_{n = -\infty}^{\infty}x(nt) h_1(t-nT) $
