Revision as of 11:06, 17 November 2008 by Bchanyas (Talk)

Amplitude modulation with pulse-train carrier

First, we know that $ y(t) = x(t)c(t) $ with $ c(t) $ being a pulse-train.

then:

$ Y(w) = \frac{1}{2\pi}X(w)*C(w) $

with $ C(w) $ being:

$ \sum_{K = -\infty}^\infty a_k2\pi\delta(w-\frac{2\pi}{T}K) $

From the expression above, we know that:

$ c(t) = \sum_{K = -\infty}^\infty a_ke^{jk\frac{2\pi}{T}t} $

which also means:

$ C(w) = \sum_{K = -\infty}^\infty a_kF(e^{jk\frac{2\pi}{T}t}) $

which is equal to:

$ \sum_{K = -\infty}^\infty a_k\delta(w-\frac{K2\pi}{T}) $

What we need to do now is to find $ a_k's $ knowing that $ a_0 $ is the average of signal over one period, which is also equal to $ \frac{d}{T} $, where $ d $ is Δ.

if $ K $ is not equal to 0, then:

$ a_k = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}c(t) $

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

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