Recall we can take a signal $x(t)$:

% \img{images/modulators/envelope/justx.ps}

and use amplitude modulation to send the signal across a network:

% \img{images/modulators/envelope/combined.ps}

If you recall, we made assumptions that the transmitter and receiver were in sync. If we want to bypass all of that oscillator synchronization issue at the receiver, we can hookup a receiver that somehow tracks this envelope at the receiver. Consider the following circuit:
  
% \img{images/modulators/envelope/circuit/combo.ps}

What happens is $y(t)$ comes in, and we have capacitor and a resister and something proportional to $x(t)$ shows up at the other end, what does this do? A diode is device where if the voltage after the diode is greater than the voltage before it, the diode acts like a short circuit, cutting off all voltage. The moment the voltage preceeding the diode is greater, this diode opens up, thus charging the capacitor. This allows us to track the envelope of the incoming signal.

% \img{images/modulators/envelope/track/combined.ps}

If the incoming signal $y(t)$ is near one of those peaks, it is positive. This is in the positive part of the signals cycles, and the diode shorts. When it shorts, the capacitor starts charging up until the capacitor voltage exceeds the modulated signal. So as the signal is coming on its way down, the capacitor has been charging up. At some point they become equal and the diode opens. So initially, the capacitor starts to track the peak and as the voltage drops its voltage drops. But at some point, the diode opens and nothing gets transmitted until the next cycle.



Note that this assumes that $x(t) \gt 0$ $\forall t$, so if the signal is ever negative, this method breaks down. We need to shift the signal up. This brings us to \emph{large carrier amplitude modulation}. Basically you also send a copy of the carrier signal with the modulated signal. If $\abs{x(t)} \leq k \in \R_{\oplus}$, then modulate the carrier with $A + x(t)$, where $A \gt 0$. Then we can utilize the envelope tracking and the receiver must shift the signal down after the tracking. This system works as follows:

% \img{images/modulators/largecarrier/largec.ps} 



\begin{center}
\begin{pspicture}(1,1)(6.5,4.5)
    
%text
\rput(1.3,3){$x(t)$}

%text
\rput(5.75,3){$y(t)$}
\rput(3.75,4){$y(t) = \left( A + x(t) \right) \cos(\omega_0t)$}

%text
\rput(2.75,2){$cos(\omega_0t)$}


\pspolygon(4.5,1.8)(5,1.8)(4.75,2.25)
\rput(4.75,2){$A$}

% wire with arrow 
\psline{->}(1.5,3)(2.5,3)

% wire with arrow 
\psline{->}(3,3)(4.5,3)

% wire with arrow 
\psline{->}(5,3)(5.5,3)

% wire with arrow 
\psline{->}(2.75,2.25)(2.75,2.75)

% wire with arrow 
\psline{->}(3.25,2)(4.5,2)

% wire with arrow 
\psline{->}(4.75,2.25)(4.75,2.75)

% multiplier
\pscircle(2.75,3){0.25}
\psline(2.575,3.175)(2.925,2.825)
\psline(2.925,3.175)(2.575,2.825)


% adder
\pscircle(4.75,3){0.25}
\psline(4.5,3)(5,3)
\psline(4.75,3.25)(4.75,2.75)

\end{pspicture}
\end{center}

Given a generic signal with bandwidth of $2A$, lets determine what $Y(\omega)$ looks like.

% \img{images/modulators/largecarrier/freqdomain/double.ps}

\begin{center}
\begin{pspicture}(-4,-1)(4,3)

  % axis
  \psline{->}(-5,0)(5,0)


  \psline(-3,0)(-2,1)
  \psline(-2,1)(-1,0)

  \psline(1,0)(2,1)
  \psline(2,1)(3,0)


  \rput(-2, 2.3){$(A\pi)$}
  \rput(2, 2.3){$(A\pi)$}
  \psline{->}(-2,0)(-2,2)
  \psline{->}(2,0)(2,2)


  \rput(-3,-0.35){$-\omega_0-A$}
  \rput(-2,-0.25){$-\omega_0$}
  \rput(0,-0.35){$0$}
  \rput(-1,-0.35){$-\omega_0+A$}

  \rput(3,-0.35){$\omega_0+A$}
  \rput(2,-0.25){$\omega_0$}
  \rput(1,-0.35){$\omega_0-A$}

  \psline(-3,-0.1)(-3,0.1)
  \psline(-2,-0.1)(-2,0.1)
  \psline(-1,-0.1)(-1,0.1)
  \psline(0,-0.1)(0,0.1)
  \psline(1,-0.1)(1,0.1)
  \psline(2,-0.1)(2,0.1)
  \psline(3,-0.1)(3,0.1)

  \rput(5.25,-0.25){$\omega$}
  \rput(-3.5,2){$Y(\omega)$}

  \rput(2.6,1){$(1/2)$}
  \rput(-1.4,1){$(1/2)$}


\end{pspicture}
\end{center}


Large Carrier AM