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}

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  \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){$(1)$}


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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}

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\rput(-1,-0.25){$-\omega_0$}

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

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

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

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  \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){$\omega_0-A$}

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  \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