(New page: ==Amplitude modulation with pulse-train carrier== First, we know that <math>y(t) = x(t)c(t)</math> with <math>c(t)</math> being a pulse-train. then: :<math>Y(w) = \frac{1}{2\pi}X(w)*C(w...) |
(→Amplitude modulation with pulse-train carrier) |
||
| Line 23: | Line 23: | ||
:<math>\sum_{K = -\infty}^\infty a_k\delta(w-\frac{K2\pi}{T})</math> | :<math>\sum_{K = -\infty}^\infty a_k\delta(w-\frac{K2\pi}{T})</math> | ||
| − | What we need to do now is to find <math>a_k's</math> knowing that <math>a_0</math> is the average of signal over one period, which is also equal to <math>\frac{ | + | What we need to do now is to find <math>a_k's</math> knowing that <math>a_0</math> is the average of signal over one period, which is also equal to <math>\frac{d}{T}</math>, where <math>d</math> is Δ. |
Revision as of 11:03, 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.
