Difference between revisions of "Bavorndej Chanyasak: Amplitude Modulation ECE301Fall2008mboutin" - Rhea

Latest revision as of 12:14, 17 November 2008

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)e^{-jk\frac{2\pi}{T}t}dt $

$ = \frac{1}{T}\int_{-\frac{d}{2}}^{\frac{d}{2}}1e^{-jk\frac{2\pi}{T}t}dt $

$ = \frac{1}{T}\left[\frac{e^{-jk\frac{2\pi}{T}t}}{-jk\frac{2\pi}{T}}\right]_{-\frac{d}{2}}^{\frac{d}{2}} $

$ = \frac{2sin(K\frac{2\pi}{T}\frac{d}{2})}{TK\frac{2\pi}{T}} $

$ = \frac{sin(Kw_o\frac{d}{2})}{K\pi} $

Again, in the above expressions, $ d $ is Δ.