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