(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...) |
(→Why does h_1(t) work when sampling with Zero-Order Hold) |
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<math> x_0(t) = \sum_{n = -\infty}^{\infty}x(nt) h_1(t-nT) </math> | <math> x_0(t) = \sum_{n = -\infty}^{\infty}x(nt) h_1(t-nT) </math> | ||
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| + | <math> h_1(t) </math> works because <math> p(t) </math> is a shifted delta function, and when you convolve a specific function with a shifted delta function, you get back the specific function but shifted in time. | ||
Latest revision as of 17:44, 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) $
$ h_1(t) $ works because $ p(t) $ is a shifted delta function, and when you convolve a specific function with a shifted delta function, you get back the specific function but shifted in time.
