(Amplitude Modulation)
(Amplitude Modulation)
 
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<math> Y(w) = F(e^{jw_{c}t}x(t)) = \frac{1}{2\pi}F(e^{jw_{c}t})X(w)</math>
 
<math> Y(w) = F(e^{jw_{c}t}x(t)) = \frac{1}{2\pi}F(e^{jw_{c}t})X(w)</math>
  
              \frac{1}{2\pi}2\pi\delta(w-w_{c}) * X(w) = X(w-w_{c})</math> , where * is convolution
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<math>      = \frac{1}{2\pi}2\pi\delta(w-w_{c}) * X(w) = X(w-w_{c})</math> , where * is convolution
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Thus, X(w) was delayed by <math> w_{c}</math>
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To demodulate, multiply <math> y(t) </math> by <math> e^{-jw_{c}t}</math>.
  
 
(2) Sinusidal type
 
(2) Sinusidal type
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The modulation method is same.
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To modulate the signal, multiply information bearing signal by <math> cosw_{c}t </math>
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<math> Y(w) = F(x(t)cosw_{c}t) = \frac{1}{2\pi}F(x(t))*F(cosw_{c}t) </math>
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<math> = \frac{1}{2\pi}X(w)*\pi(\delta(w+w_{c})+\delta(w-w_{c})) </math>
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<math> = \frac{1}{2}X(w+w_{c}) + \frac{1}{2}X(w-w_{c}) </math>
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Thus, X(w) was delayed by <math> w_{c}</math> with half of amplitude and X(w) was also advanced by  <math> w_{c}</math> with half of amplitude
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The important thing in this case, carrier frequency has to be greater than message frequency.
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If not, Y(w) will always overlap.
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To demodulate, multiply <math> cosw_{c}t </math> and use low pass filter with gain of 2.

Latest revision as of 11:35, 16 November 2008

Amplitude Modulation

The signal is transmitted to receiver by communication channel.

In this process, information bearing signal, x(t), is embeded by carrier signal, c(t) which has its amplitude.

So the modulated signal is the product of these two signals:

$ y(t) = x(t)c(t) $

Here are two types of carriers.

(1) complex exponential type

Suppoese $ c(t) $ is $ e^{jw_{c}t}. $

$ y(t) = e^{jw_{c}t} $

$ Y(w) = F(e^{jw_{c}t}x(t)) = \frac{1}{2\pi}F(e^{jw_{c}t})X(w) $

$ = \frac{1}{2\pi}2\pi\delta(w-w_{c}) * X(w) = X(w-w_{c}) $ , where * is convolution

Thus, X(w) was delayed by $ w_{c} $

To demodulate, multiply $ y(t) $ by $ e^{-jw_{c}t} $.

(2) Sinusidal type

The modulation method is same.

To modulate the signal, multiply information bearing signal by $ cosw_{c}t $

$ Y(w) = F(x(t)cosw_{c}t) = \frac{1}{2\pi}F(x(t))*F(cosw_{c}t) $

$ = \frac{1}{2\pi}X(w)*\pi(\delta(w+w_{c})+\delta(w-w_{c})) $

$ = \frac{1}{2}X(w+w_{c}) + \frac{1}{2}X(w-w_{c}) $

Thus, X(w) was delayed by $ w_{c} $ with half of amplitude and X(w) was also advanced by $ w_{c} $ with half of amplitude

The important thing in this case, carrier frequency has to be greater than message frequency.

If not, Y(w) will always overlap.

To demodulate, multiply $ cosw_{c}t $ and use low pass filter with gain of 2.

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