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<math>Y(\omega) = \frac{1}{2\pi}(\pi\delta(\omega + \omega_ct)+\pi\delta(\omega - \omega_ct))X(\omega)</math> | <math>Y(\omega) = \frac{1}{2\pi}(\pi\delta(\omega + \omega_ct)+\pi\delta(\omega - \omega_ct))X(\omega)</math> | ||
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+ | <math>Y(\omega) = \frac{1}{2}(X(\omega + \omega_ct)+X(\omega + \omega_ct))</math> | ||
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+ | With this modulation <math>x(t)</math> is being copied and halved, one copy is shifted <math>\omega</math> while the other one is shifted <math>-\omega</math> | ||
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+ | ===Demodulation=== | ||
+ | |||
+ | <math>y(t)\cos(\omega_ct) = x(t)\cos^2(\omega_ct)</math> | ||
+ | |||
+ | <math>x(t)\cos^2(\omega_ct)</math> --> a lowpass filter with a height of 2 and <math>\omega_m<\omega_c<2\omega_c-\omega_m</math> --> <math>x(t)</math> | ||
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+ | ==Honorable Mention: Amplitude Modulation with pulse-train== | ||
+ | |||
+ | <math>c(t) = \sum_{k = -\infty}^\infty a_k e^{jk\frac{2\pi}{T}t}</math> |
Latest revision as of 17:26, 17 November 2008
Contents
How it works
$ x(t)c(t)=y(t) $
Where $ x(t) $ is the "information signal" and $ c(t) $ is the "carrier"
Two Major Carriers
Complex Exponential
$ c(t) = e^{j(\omega_ct+\theta_c)} $
Sinusoidal
$ c(t) = cos(\omega_ct+\theta_c) $
Where $ \omega_c $ is the frequency and $ \theta_c $ is the phase
Complex Exponential Modulation
$ y(t) = e^{j\omega_ct}x(t) $
$ Y(\omega)=F(e^{j\omega_ct}x(t)) $
$ Y(\omega)=\frac{1}{2\pi}F(e^{j\omega_ct})X(\omega) $
$ Y(\omega)=\frac{1}{2\pi}(2\pi \delta(\omega-\omega_c)X(\omega) $
$ Y(\omega)=X(\omega-\omega_c) $
What happens with this modulation is that the original signal $ x(t) $ and shifted in the frequency domain by $ \omega_c $
Demodulation ie. How the Heck do I get back my original signal
$ y(t)e^{-j\omega_ct}=x(t) $
Sinusodial Modulation
$ y(t) = \cos(\omega_ct)x(t) $
$ Y(\omega) = F(\cos(\omega_ct)x(t)) $
$ Y(\omega) = \frac{1}{2\pi}F(\cos(\omega_ct))F(x(t)) $
$ Y(\omega) = \frac{1}{2\pi}(\pi\delta(\omega + \omega_ct)+\pi\delta(\omega - \omega_ct))X(\omega) $
$ Y(\omega) = \frac{1}{2}(X(\omega + \omega_ct)+X(\omega + \omega_ct)) $
With this modulation $ x(t) $ is being copied and halved, one copy is shifted $ \omega $ while the other one is shifted $ -\omega $
Demodulation
$ y(t)\cos(\omega_ct) = x(t)\cos^2(\omega_ct) $
$ x(t)\cos^2(\omega_ct) $ --> a lowpass filter with a height of 2 and $ \omega_m<\omega_c<2\omega_c-\omega_m $ --> $ x(t) $
Honorable Mention: Amplitude Modulation with pulse-train
$ c(t) = \sum_{k = -\infty}^\infty a_k e^{jk\frac{2\pi}{T}t} $