(→Amplitude modulation with pulse-train carrier) |
(→Amplitude modulation with pulse-train carrier) |
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| − | === Amplitude modulation with pulse-train carrier === | + | |
| + | == === Amplitude modulation with pulse-train carrier === | ||
y(t)=x(t)c(t) with c(t) be a pulse train | y(t)=x(t)c(t) with c(t) be a pulse train | ||
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where <math>C(W)= \sum^{\infty}_{k = -\infty} a_k 2\pi\delta(\omega-\frac{2\pi}{T})</math> | where <math>C(W)= \sum^{\infty}_{k = -\infty} a_k 2\pi\delta(\omega-\frac{2\pi}{T})</math> | ||
| − | For k=0,<math>a_0 = \frac{\delta}{T} where \delta is the average of signal over 1 period. | + | For k=0,<math>a_0 = \frac{\delta}{T} </math> where <math>\delta</math> is the average of signal over 1 period. |
Else <math> a_k = \frac{1}{T} \int^{\frac{T}{2}}_{-\frac{T}{2}} c(t) e^{-jk\frac{T}{2}t} dt </math> | Else <math> a_k = \frac{1}{T} \int^{\frac{T}{2}}_{-\frac{T}{2}} c(t) e^{-jk\frac{T}{2}t} dt </math> | ||
| − | <math> | + | <math> = \frac{1}{T} \int^{\frac{\delta}{2}}_{-\frac{\delta}{2}} e^{-jk\frac{T}{2}t} dt </math> |
| − | <math> | + | <math> = \frac{2sin(k\frac{2\pi}{T}\frac{\delta}{2})}{TK\frac{T}{2}}</math> |
| − | <math> | + | <math> = \frac{sin(k\omega_0 \frac{\delta}{2})}{k\pi}</math> |
To recover x(t), use a low pass filter with gain <math> \frac {1}{a_0}=\frac {T}{\delta} </math>, and cut-off frequency <math> \omega_m < \omega_c < \omega_0 - \omega_m </math> | To recover x(t), use a low pass filter with gain <math> \frac {1}{a_0}=\frac {T}{\delta} </math>, and cut-off frequency <math> \omega_m < \omega_c < \omega_0 - \omega_m </math> | ||
Revision as of 16:10, 16 November 2008
=== Amplitude modulation with pulse-train carrier =
y(t)=x(t)c(t) with c(t) be a pulse train
$ C(t)= \sum^{\infty}_{k = -\infty} a_k e^{jk\frac{2\pi}{T}t} $
Thus Y(W)= $ \frac{1}{2\pi}X(W)*C(W) $
where $ C(W)= \sum^{\infty}_{k = -\infty} a_k 2\pi\delta(\omega-\frac{2\pi}{T}) $
For k=0,$ a_0 = \frac{\delta}{T} $ where $ \delta $ is the average of signal over 1 period.
Else $ a_k = \frac{1}{T} \int^{\frac{T}{2}}_{-\frac{T}{2}} c(t) e^{-jk\frac{T}{2}t} dt $
$ = \frac{1}{T} \int^{\frac{\delta}{2}}_{-\frac{\delta}{2}} e^{-jk\frac{T}{2}t} dt $
$ = \frac{2sin(k\frac{2\pi}{T}\frac{\delta}{2})}{TK\frac{T}{2}} $
$ = \frac{sin(k\omega_0 \frac{\delta}{2})}{k\pi} $
To recover x(t), use a low pass filter with gain $ \frac {1}{a_0}=\frac {T}{\delta} $, and cut-off frequency $ \omega_m < \omega_c < \omega_0 - \omega_m $
