| Line 15: | Line 15: | ||
<math>w(t)=\frac{1}{2}x(t)+\frac{1}{2}x(t)cos{2w_{c}t}</math> | <math>w(t)=\frac{1}{2}x(t)+\frac{1}{2}x(t)cos{2w_{c}t}</math> | ||
| + | |||
| + | Thus, w(t) consists of the sum of two terms, namely one-half the original signal and one-half the original signal modulated with a sinusoidal carrier at twice the original carrier frequency <math>w_c</math>. | ||
| + | |||
| + | Applying the lowpass filter to w(t) corresponds to retaining the first term, <math>\frac{1}{2}x(t)</math>. | ||
| + | Thus to recover x(t), a lowpass filter with gain of 2 is required. | ||
Latest revision as of 18:06, 17 November 2008
DEMODULATION FOR SINUSOIDAL AM
$ y(t) = x(t)cos{w_{c}t} $
the original signal can be recovered by modulating y(t) with the same sinusoidal carrier and applying a lowpass filter to the result.
consider,
$ w(t) = y(t)cos{w_{c}t} $
$ w(t) = x(t)cos^2{w_{c}t} $
using trigonometric identity,
$ cos^2{w_{c}t}=\frac{1}{2}+\frac{1}{2}cos{2w_{c}t} $
$ w(t)=\frac{1}{2}x(t)+\frac{1}{2}x(t)cos{2w_{c}t} $
Thus, w(t) consists of the sum of two terms, namely one-half the original signal and one-half the original signal modulated with a sinusoidal carrier at twice the original carrier frequency $ w_c $.
Applying the lowpass filter to w(t) corresponds to retaining the first term, $ \frac{1}{2}x(t) $. Thus to recover x(t), a lowpass filter with gain of 2 is required.
