(→Amplitude modulation with pulse-train carrier) |
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:<math>= \frac{1}{T}\int_{-\frac{d}{2}}^{\frac{d}{2}}1e^{-jk\frac{2\pi}{T}t}dt</math> | :<math>= \frac{1}{T}\int_{-\frac{d}{2}}^{\frac{d}{2}}1e^{-jk\frac{2\pi}{T}t}dt</math> | ||
| + | |||
| + | :<math>= \frac{1}{T}\left[\frac{e^{-jk\frac{2\pi}{T}t}}{-jk\frac{2\pi}{T}}\right]_{-\frac{d}{2}}^{\frac{d}{2}}</math> | ||
| + | |||
| + | :<math>= \frac{2sin(K\frac{2\pi}{T}\frac{d}{2})}{TK\frac{2\pi}{T}} | ||
| + | |||
| + | :<math>= \frac{2sin(Kw_o\frac{d}{2})}{K\pi}</math> | ||
Revision as of 12:13, 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}} :<math>= \frac{2sin(Kw_o\frac{d}{2})}{K\pi} $
