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<math>w(t) = y(t)cos{w_{c}t}</math>
 
<math>w(t) = y(t)cos{w_{c}t}</math>
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<math>w(t) = x(t)cos^2{w_{c}t}</math>
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using trigonometric identity,
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<math>cos^2{w_{c}t}=\frac{1}{2}+\frac{1}{2}cos{2w_{c}t}</math>
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<math>w(t)=\frac{1}{2}x(t)+\frac{1}{2}x(t)cos{2w_{c}t}</math>
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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>.
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Applying the lowpass filter to w(t) corresponds to retaining the first term, <math>\frac{1}{2}x(t)</math>.
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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.

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009