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.

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang