There are three things that should be understood before going into amplitude modulation:


1) A complex exponential in time domain corresponds to a shifted and scaled dirac delta in the frequency domain:


\begin{nicebox}
$$
e^{i\omega_0} \ftp 2\pi \delta (\omega -\omega_0) 
$$
\end{nicebox}

There exists a frequency $\omega_0$ in the signal such that all of the signals energy is at the one frequency:

% \img{images/dirac/singles/2pi.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-2,-1)(3,2)

\psline{->}(-2,0)(2,0)
\rput(2,-0.25){$\omega$}
\rput(2,1){$X(\omega)$}

\psline{->}(0,0)(0,1)
\rput(0.3,1){$(2\pi)$}
\rput(0,-0.25){$\omega_0$}

\end{pspicture}
\end{center}

2) Using the above formula we can determine the fourier transform pair for a cosine function in the time domain:

\begin{align*}
\cos(\omega_0 t) = \frac{1}{2}\lr{e^{i\omega_0} + e^{-i\omega_0}} &\ftp \frac{1}{2} \lr{ 2\pi\delta (\omega - \omega_0) + 2\pi \delta(\omega + \omega_0) } 
\end{align*}

\begin{nicebox}
$$
\cos(\omega_0 t) \ftp \pi\delta(\omega + \omega_0) + \pi\delta(\omega - \omega_0)
$$
\end{nicebox}

% \img{images/dirac/singles/doublefreq.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-2,-1)(3,2)

\psline{->}(-2,0)(2,0)
\rput(2,-0.25){$\omega$}
\rput(2,1){$X(\omega)$}

\psline{->}(-1,0)(-1,1)
\rput(-1.3,1){$(\pi)$}
\rput(-1,-0.25){$-\omega_0$}

\psline{->}(1,0)(1,1)
\rput(1.3,1){$(\pi)$}
\rput(1,-0.25){$\omega_0$}

\end{pspicture}
\end{center}

3) Multiplication in the time domain corresponds to convolution in the frequency domain.

We have shown that convolution in the time domain is multiplication in the frequency domain:

$$(f*g)(t) \ftp F(\omega)G(\omega)$$

Consider the Modulation Property of the Continuous-Time Fourier Transform:

\begin{nicebox}
\begin{align*}
f(t)g(t) \ftp \frac{1}{2\pi} (F*G)(\omega) 
\end{align*}
\end{nicebox}

\begin{proof}

\begin{align*}
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}(F*G)(\omega)e^{i\omega t}d\omega \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}F(W)G(\omega-W) dW e^{i\omega t}d\omega \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\int_{-\infty}^{\infty}G(\omega-W) e^{i\omega t} d\omega dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\int_{-\infty}^{\infty}G(\Omega) e^{i(\Omega+W) t} d\Omega dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)\lr{\int_{-\infty}^{\infty}G(\Omega) e^{i\Omega t} d\Omega }e^{iWt}dW \\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\int_{-\infty}^{\infty}F(W)e^{iWt}dW \lr{\int_{-\infty}^{\infty}G(\Omega) e^{i\Omega t} d\Omega}\\
\iftrans{(F*G)(\omega)} &= \frac{1}{2\pi}\lr{2\pi f(t)}\lr{2\pi g(t)} \\
\iftrans{(F*G)(\omega)} &= 2\pi f(t)g(t) \\
\iftrans{\frac{1}{2\pi}(F*G)(\omega)} &= f(t)g(t) \\
\end{align*}


Which shows that multiplication in the time domain is the product of a scalar and the convolution in the frequency domain.

\begin{nicebox}
$$
f(t)g(t) \ftp \frac{1}{2\pi} (F * G) (\omega)
$$
\end{nicebox}

\end{proof}


Useful Concepts for AM

Consider a voice signal that we need to transfer over a long distance, where its frequency spectrum lives within the frequency band $(-A,A)$.

% \img{images/triangles/triangle.ps}

\begin{center}
\begin{pspicture}(-3.2,-3.2)(6,6)

  % axis
  \psline{->}(-3,0)(3,0)
  \psline(-2,0)(0,1)
  \psline(0,1)(2,0)


  \rput(-2,-0.25){$-A$}
  \rput(0,-0.25){0}
  \rput(2,-0.25){$A$}

  \rput(3.25,-0.25){$\omega$}
  \rput(-2,1){$X(\omega)$}

  \rput(1.12,1){$(1)$}


\end{pspicture}
\end{center}

This in reality is not possible to send. Generally an antenna must be approximately a fourth of the size of a cycle of the wave it transmits. ANy antenna shorter tha $\lambda/4$ is inefficient. The speed of light can be given in terms of a frequency $f$ in Hertz multiplied by the corresponding wavelength $\lambda$:

$$c=f\lambda = 2.9979 \times 10^8 m/s $$

Lets say the spectrum of ones voice fell along a range from $-3.3kHz$ to $3.3kHz$, then we can approximate the length of the antenna needed by first determining the wavelength:

\begin{align*}
\lambda &= c/f \\
\lambda &\approx 98 km\\
\end{align*}

This tells us we would need about $25km$ of antenna to transmit this persons voice. This is why we use amplitude modulation---to transmit a signal at higher frequencies.


Lets consider an amplitude modulator with a sinusoidal carrier given by 

%\img{images/modulators/amcos.ps}

The reason this is called amplitude modulation is because given an input signal $x(t)$, that signal multiplied with a $\cos(\omega_0 t)$ will give us a new signal $y(t)$, which is a cosine wave oscillating between $\pm x(t)$, where $x(t)$ serves as an envelope. \\ \\

Consider an arbitrary input signal, the info-bearing modulator signal:

% \img{images/modulators/ampmod/signal.ps}

Here we have an oscillator signal, the carrier signal whose amplitude will be modulated:

%\img{images/modulators/ampmod/cos.ps}

Here is the final result of the multiplication of the two above signals, where the amplitude of the carrier signal is modulated by the info-bearing signal:

% \img{images/modulators/ampmod/amp.ps}

The info-bearing input signal has a lower frequency and thus acts as an envelope for the carrier signal. \\ \\

$x(t)$ is the \emph{information-bearing} signal (a.k.a. the modulating signal), and $h(t) = \cos(\omega_0 t)$ is the \emph{carrier} signal (a.k.a. the modulated signal). The output signal $y(t)$ is the product of these two signals, $x(t)\cos(\omega_0 t)$. We can begin solving for the spectrum of $y$ by first finding the spectrum of $h$. 

\begin{align*}
h(t) &= \cos(\omega_0 t) \\
h(t) &= \frac{1}{2}e^{i\omega_0 t} + \frac{1}{2}e^{-i\omega_0 t} \\
\ftrans{h(t)} &= \frac{1}{2}\ftrans{e^{i\omega_0 t}} + \frac{1}{2}\ftrans{e^{-i\omega_0 t}} \\
H(\omega) &= \frac{1}{2}2\pi\delta(\omega-\omega_0) + \frac{1}{2}2\pi\delta(\omega+\omega_0) \\
H(\omega) &= \pi\delta(\omega-\omega_0) + \pi\delta(\omega+\omega_0) \\
\end{align*}

$H(\omega)$ is two dirac deltas symmetrically shifted each by $\omega_0$, 

% \img{images/dirac/doublefreq.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-1,-1)(5,5)

\psline{->}(-2,0)(2,0)
\rput(2,-0.25){$\omega$}
\rput(2,1){$H(\omega)$}

\psline{->}(-1,0)(-1,1)
\rput(-1.3,1){$(\pi)$}
\rput(-1,-0.25){$-\omega_0$}

\psline{->}(1,0)(1,1)
\rput(1.3,1){$(\pi)$}
\rput(1,-0.25){$\omega_0$}

\end{pspicture}
\end{center}

which gives us an output signal of shifted versions of the input signal with half of the original height:

% \img{images/triangles/double.ps}

\begin{center}
\begin{pspicture}(-3.2,-3.2)(6,6)

  % 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(-3,-0.25){$-\omega_0-A$}
  \rput(-2,-1.25){$-\omega_0$}
  \rput(-1,-0.25){$-\omega_0+A$}

  \rput(3,-0.25){$\omega_0+A$}
  \rput(2,-1.25){$\omega_0$}
  \rput(1,-0.25){$\omega_0-A$}

  \psline[linestyle=dashed](-2,-1)(-2,1)
  \psline[linestyle=dashed](2,-1)(2,1)

  \rput(5.25,-0.25){$\omega$}
  \rput(-3.5,1){$Y(\omega)$}

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


\end{pspicture}
\end{center}

where we must require that $-\omega_0+A \lt \omega_0-A$, so that $A\lt\omega_0$. This saves us a few complications when $\omega_0 \gt\gt A$.


Amplitude Modulation - Sending

What do we do at the receiver? How can we demodulate this signal? Once way would be to pass it through a amplitude modulator using a complex exponential $e^{-i\omega_0 t}$, and then we would get the signal shifted over to the left. We can then pass it through a low-pass filter with a pass band amplification of 2 to get the original height back. But, what if we had an oscillator using a cosine function at the receiving end instead of a complex exponential?

The following scheme is called \emph{synchronous demodulation}, and is when we have an identical cosine at the receiving end. Looking at what happened upon transmission, the signal was shifted $\pm \omega_0$ and the amplitude was cut in half. Applying this again, we would get a signal $q(t) = y(t)\cos(\omega_0 t)$, where we shift the shifted signal $\pm \omega_0$. This gives us three disjoint triangular components in the frequency spectrum. The first shift to the left $\omega_0$ gives us two triangles with a height of $1/4$ centered at $-2\omega_0$ and $0$. The shift to the right gives us two triangles with a height of $1/4$ centered at $2\omega_0$ and $0$. The two triangles centered at 0 add to form the original triangle halfed. So we can now put this signal $q(t)$ through a low-pass filter and amplify the signal by 2 to retrieve the original signal.

% \img{images/triangles/three.ps}

\begin{center}
\begin{pspicture}(-5,-3.2)(6,6)

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


  \psline(-4,0)(-3,0.5)
  \psline(-3,0.5)(-2,0)

  \psline(2,0)(3,0.5)
  \psline(3,0.5)(4,0)

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


  \rput(-3,-0.25){$-2\omega_0$}
  \rput(0,-0.25){$0$}
  \rput(3,-0.25){$2\omega_0$}


  \rput(5.25,-0.25){$\omega$}
  \rput(-3.5,1){$Q(\omega)$}

  \rput(0.5,1){$(1/2)$}
  \rput(3.7,0.5){$(1/4)$}
  \rput(-2.3,0.5){$(1/4)$}


\end{pspicture}
\end{center}


At the receiving end the local oscillator has potentially a phase difference and frequency difference relative to the transmitter. This demodulation scheme assumes the same frequency and phase at the sender and receiver. What happens if at the receiver we have $q(t) = y(t)\cos((\omega_0+\Delta\omega)t + \theta)$?

% \img{images/modulators/receiver/amcos.ps}

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

% in from x
\rput(-3.1,0.3){$x(t)\cos(\omega_0t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(-2.7,0)(-0.25,0)
% out to y
\rput(4.2,0.3){$x(t)\cos(\omega_0t)\cos((\omega_0+\Delta\omega) t+\theta)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0.25,0)(2.7,0)
% up arrow
\rput(0,-2){$\cos((\omega_0+\Delta\omega) t+\theta)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0,-1.65)(0,-0.25)

% multiplier
\pscircle(0,0){0.25}
\psline(-0.175,0.175)(0.175,-0.175)
\psline(0.175,0.175)(-0.175,-0.175)

% box
\pspolygon(-1.6,-2.65)(-1.6,1)(1.6,1)(1.6,-2.65)

\end{pspicture}
\end{center}

\begin{align*}
y(t) &= x(t)\cos(\omega_0 t) \\
r(t) &= y(t)\cos((\omega_0+\Delta\omega) t + \theta) \\
r(t) &= x(t)\cos(\omega_0 t)\cos((\omega_0+\Delta\omega) t + \theta) \\
\end{align*}

We can use the trigonometric identity:

$$ \cos\alpha \cos\beta = \frac{1}{2}\lr{\cos\lr{\alpha+\beta}+\cos\lr{\alpha-\beta}} $$

\begin{align*}
r(t) &= x(t)\cos(\omega_0 t)\cos((\omega_0+\Delta\omega) t + \theta) \\
r(t) &= \frac{x(t)}{2} \lr{\cos((2\omega_0 + \Delta\omega)t + \theta) + \cos(\Delta \omega t + \theta)} \\
\end{align*}

After we pass this signal through a low pass filter, we get rid of the first term containing the frequency $\omega_0+\Delta\omega)$. So if $g(t)$ denotes our signal after the low pass filter, then $g(t) = x(t)\cos(\Delta\omega t + \theta)$. We hope this is equal to $x(t)$.

Lets consider the different cases:

1) no frequency jitter ($\Delta\omega = 0$), such that $g(t) = x(t)\cos(\theta)$


If $\theta \approx \frac{\pi}{2}$, you get an extremely weak baseband. When $\theta = 0$, you have a synchronous demodulation scheme.


2) the phase is zero, but there is frequency jitter, then $g(t) = x(t)\cos(\Delta\omega t)$.


This is amplitude modulation with a very slow carrier signal, where we will have overlap in the frequency domain, causing echos and other problems.


Amplitude Modulation - Receiving

The previous method of transmission was known as \emph{Coherent Detection}, or \emph{Coherent Demodulation}, or \emph{Synchronous Detection/Demodulation}. The transmitter and receiver may be even hundreds of miles and in the case of short wave radio, thousands of miles away. What you transmit is not what you receive. The atmosphere degrades the signal in various ways---all kinds of interference, climate issues can affect the integrity of the signal. This major assumption that we made in the last discussion in regards to transmitting signals is that we send $y(t)$ and receive $y(t)$. In particular, in the city, the signal you transmit gets reflected off buildings. At the receiver you not only get the original signal but echoes of that signal with attenuated amplitudes. This level of generalization by allowing frequency jitter and the phase difference is just a small level of generalization. There are all kinds of things that can go wrong. Can we use a better method? Lets look at \emph{Quadrature Multiplexing}, otherwise known as \emph{Quadrature Amplitude Modulation}:

%\img{images/modulators/quadmult/quad.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-1,-1)(6,4)
    
%text
\rput(0.8,3){$x_1(t)$}

%text
\rput(0.8,1.5){$x_2(t)$}

%text
\rput(5.5,2.25){$y(t)$}

%text
\rput(2.75,0.5){$sin(\omega_0t)$}

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


% wire with arrow 
\psline{->}(1.2,3)(2.5,3)

% wire with arrow 
\psline{->}(1.2,1.5)(2.5,1.5)

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

% wire with arrow 
\psline{}(3,1.5)(4.5,1.5)

% wire with arrow 
\psline{->}(4.5,3)(4.5,2.5)

% wire with arrow 
\psline{->}(4.5,1.5)(4.5,2)

% wire with arrow 
\psline{->}(4.75,2.25)(5.25,2.25)

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

% wire with arrow 
\psline{->}(2.75,0.75)(2.75,1.25)

% 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)


% multiplier
\pscircle(2.75,1.5){0.25}
\psline(2.575,1.675)(2.925,1.325)
\psline(2.925,1.675)(2.575,1.325)


% adder
\pscircle(4.5,2.25){0.25}
\psline(4.25,2.25)(4.75,2.25)
\psline(4.5,2.5)(4.5,2)

\end{pspicture}
\end{center}

This is used to transmit color signals in the standard television system before HDTV.  It works in the following way: The system transmits two signals, carrying them on to the same frequency and at the receiver separates the signals:

% \img{images/modulators/quadmult/rec.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-1,-1)(6,4)
    

%frame
\rput(4.5,3){$\mbox{LPF}$}
\psframe(5,2.75)(4,3.25)


%frame
\rput(4.5,1.5){$\mbox{LPF}$}
\psframe(5,1.25)(4,1.75)

%text
\rput(6,3){$g_1(t)$}

%text
\rput(6,1.5){$g_2(t)$}

%text
\rput(0.8,2.25){$y(t)$}

%text
\rput(2.75,0.5){$sin(\omega_0t)$}

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


% wire with arrow 
\psline{->}(1.2,2.25)(1.5,2.25)

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

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

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

% wire with arrow 
\psline{->}(5,1.5)(5.5,1.5)

% wire with arrow 
\psline{}(3,3)(4,3)

% wire with arrow 
\psline{}(3,1.5)(4,1.5)

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

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

% wire with arrow 
\psline{->}(2.75,0.75)(2.75,1.25)

% 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)


% multiplier
\pscircle(2.75,1.5){0.25}
\psline(2.575,1.675)(2.925,1.325)
\psline(2.925,1.675)(2.575,1.325)

\end{pspicture}
\end{center}

Hopefully $g_1(t),g_2(t)$ are equal to $x_1(t),x_2(t)$ respectively. Lets do the math for a received signal $y(t) = x_1(t)\cos(\omega_0t) + x_2(t)\sin(\omega_0t)$, by passing it through the cosine portion of the receiving end system:

\begin{align*}
y(t) &= x_1(t)\cos(\omega_0t) + x_2(t)\sin(\omega_0t) \\
y(t)\cos(\omega_0t) &= x_1(t)\cos^2(\omega_0t) + x_2(t)\sin(\omega_0t)\cos(\omega_0t) \\
r_1(t) &= x_1(t)\lr{\frac{1+\cos(2\omega_0t)}{2}} + x_2(t)\frac{1}{2}\sin(2\omega_0t) \\
r_1(t) &= \frac{x_1(t)}{2} + \frac{x_1(t)}{2}\cos(2\omega_0t) + \frac{x_2(t)}{2}\sin(2\omega_0t) \\
\end{align*}

Once we pass $r_1(t)$ through a low-pass filter with a gain of 2 to get $g_1(t)$, we can recover the signal $x_1(t)$ because all terms with frequency $2\omega_0$ will be cut out.


Quadrature Multiplexing

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

The FCC controls something on the order of 10 $kHz$ and 2.75 $GHz$. The frequency band is in the control of the FCC like public land. The FCC sells frequency ranges. Radio stations and cable companies pay for that, and in the end, that frequency spectrum belongs to you and me.


\emph{Frequency Division Multiplexing} enables simultaneous transmission of multiple signals. 


% \img{images/modulators/freqdiv/freq.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(0,-3)(5.5,4)
    
%text
\rput(0.8,3){$x_1(t)$}

%text
\rput(0.8,1.5){$x_2(t)$}

%text
\rput(0.8,-1){$x_n(t)$}

%text
\rput(0.8,0){$\vdots$}

%text
\rput(2.75,0){$\vdots$}

%text
\rput(5.5,2.25){$y(t)$}

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

%text
\rput(2.75,0.5){$cos(\omega_2t)$}

%text
\rput(2.75,2){$cos(\omega_1t)$}


% wire with arrow 
\psline{->}(1.2,3)(2.5,3)

% wire with arrow 
\psline{->}(1.2,1.5)(2.5,1.5)

% wire with arrow 
\psline{->}(1.2,-1)(2.5,-1)

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

% wire with arrow 
\psline{}(3,1.5)(4.5,1.5)

% wire with arrow 
\psline{}(3,-1)(4.5,-1)

% wire with arrow 
\psline{->}(4.5,3)(4.5,2.5)

% wire with arrow 
\psline{->}(4.5,-1)(4.5,2)

% wire with arrow 
\psline{->}(4.75,2.25)(5.25,2.25)

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

% wire with arrow 
\psline{->}(2.75,0.75)(2.75,1.25)

% wire with arrow 
\psline{->}(2.75,-1.75)(2.75,-1.25)

% 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)


% multiplier
\pscircle(2.75,1.5){0.25}
\psline(2.575,1.675)(2.925,1.325)
\psline(2.925,1.675)(2.575,1.325)


% multiplier
\pscircle(2.75,-1){0.25}
\psline(2.575,-0.825)(2.925,-1.175)
\psline(2.925,-0.825)(2.575,-1.175)


% adder
\pscircle(4.5,2.25){0.25}
\psline(4.25,2.25)(4.75,2.25)
\psline(4.5,2.5)(4.5,2)

\end{pspicture}
\end{center}


We can multiply each incoming signal by a mutually orthogonal functions, in this case we use the cosine function with frequencies $\omega_1, \omega_2, \dots, \omega_n$. So, you have a frequency spectrum, you split it into different ranges and distribute different frequency bands to different stations. Let's say we are talking about three radio stations. So, you have multiple signals awaiting transmission:

% \img{images/modulators/freqdiv/stations/justd.ps}
% \img{images/modulators/freqdiv/stations/justamadou.ps}
% \img{images/modulators/freqdiv/stations/justbabak.ps}

What happens when we use frequency division multiplexing? We get the following spectra for $Y(\omega)$:

% \img{images/modulators/freqdiv/stations/all.ps}

How would we get a particular station? Lets suppose we want to get the station at $\omega_2$. A crude solution would be to use a band pass filter. At the receiver, lets say you have a knob, and by turning that knob you determine which frequency you are interested in, which station. So, with that knob you would have to control, not only this frequency, but also this band-pass filter. For practical reasons building a band pass filter with that high frequency cutoff when these frequency bands are so close together, it's like city of Berkeley, there is no empty land. Lots of homes are small, and they are all tightly connected to each other---there is not a whole lot of wiggle room. This is an ideal band pass filter. Things have to roll out. The higher you go on frequency band the more difficult it is for circuit designers to build band pass fillers.


There has got to be another solution. Your knob will control the local oscillator, and maps whatever part of the spectrum you are interested in to one intermediate frequency. Then you have only one band pass filter for which you can design a filter and pass that through. So, regardless of where you are in the spectrum, by tuning the knob you mapped and grab this part and map it to that intermediate frequency. Grab the region around that intermediate frequency, and you have got a built-in band pass filter already. Then you demodulate. When the knob is tuned to a particular frequency, the circuit shifts it to a fixed intermediate frequency that is well within the range of our filter---not necessarily baseband but maps it to something close to zero.  


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