(→Amplitude modulation with pulse-train carrier) |
(→Amplitude modulation with pulse-train carrier) |
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if <math>K</math> is not equal to 0, then: | if <math>K</math> is not equal to 0, then: | ||
| − | :<math>a_k = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}c(t)</math> | + | :<math>a_k = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}c(t)e^{-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> | ||
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 $
