 Consider an AM modulator where $y(t) = x(t)g(t)$, where $g(t) = e^{i\omega_0 t}$, where $\omega_0$ is large and positive:

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

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

% in from x
\rput(-3.2,0){ $x(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(-2.7,0)(-0.25,0)
% out to y
\rput(3.2,0){$y(t)$}
\psline[linewidth=1.25 pt, arrowscale=1.1]{->}(0.25,0)(2.7,0)
% up arrow
\rput(0,-2){ $g(t)$}
\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(-1.5,-2.65)(-1.5,1)(1.5,1)(1.5,-2.65)

\end{pspicture}
\end{center}

Lets say the spectrum of the input signal is given by the following figure:


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

\begin{center}
\begin{pspicture}(-3.2,-3.2)(3,3)

% 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}


We can determine that $Y(\omega) = \frac{1}{2\pi}(X*G)(\omega)$. Since $g(t) = e^{i\omega_0 t}$, it represents a single frequency in the frequency domain. In other words, the spectrum of $g(t)$, namely, $G(\omega)$, is a single dirac delta shifted by $\omega_0$ with an amplitude of $2\pi$ (see previous sections for derivation).

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


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

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

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


\end{pspicture}
\end{center} 

In other words, 

$$ g(t) = e^{i\omega_0 t} \ftp 2\pi \delta(\omega-\omega_0) $$

When doing amplitude modulation with a single complex exponential of frequency $\omega_0$, we find that the frequency response $Y(\omega) = X(\omega-\omega_0)$. This is because $(X*G)(\omega)=2\pi X(\omega - \omega_0)$, and the $2\pi$ cancels out. Here is how it works out:

\begin{align*}
Y(\omega) &= \frac{1}{2\pi}(X*G)(\omega) \\
Y(\omega) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\Omega)G(\omega-\Omega)d\Omega \\
Y(\omega) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X(\Omega)2\pi\delta(\omega-\Omega-\omega_0)d\Omega \\
Y(\omega) &= \int_{-\infty}^{\infty}X(\Omega) \delta(\omega-\omega_0-\Omega)d\Omega \\
Y(\omega) &= X(\omega-\omega_0) 
\end{align*}

Thus we have a spectrum for $Y(\omega)$ that looks like

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

\begin{center}
\begin{pspicture}(-3.2,-3.2)(3,3)

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


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

\rput(3.25,-0.25){$\omega$}
\rput(-2,1){$Y(\omega)=X(\omega-\omega_0)$}

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


\end{pspicture}
\end{center}

Complex Exponential Carrier