In old movies with carriage wheels, sometimes it appears that the wheel is moving backwards at a slower rate, or maybe slower in the correct direction. Suppose you flash a strobe at a particular frequency on a carriage wheel. If the strobe light and the carriage wheel both share the frequency $\omega_0$, then carriage wheel would seem to stand still, you would see a ``frozen'' image. Lets see what happens when we sample at other rates. We can model the carriage wheel as $e^{i\omega_o t}$ or $\sin(\omega_0 t)$. 

%\img{images/sampling/aliasing/carriagewheel/wheel.ps}

We can sample $x(t)$ by multiplying it with $g(t) = \sum \limits_{k \in \Z} \delta(t - kT_s)$, where $\omega_s = \frac{2\pi}{T_s}$:

%\img{images/sampling/system/sys.ps}

The spectra of our signal $x$ is two complex valued diracs:

\begin{align*}
x(t) &= \sin(\omega_0 t) \\
x(t) &= \frac{e^{i\omega_0 t} - e^{-i\omega_0 t}}{2i} \\
X(\omega) &= \ftrans{\frac{e^{i\omega_0 t} - e^{-i\omega_0 t}}{2i}} \\
X(\omega) &= \frac{\pi}{i}\delta(\omega-\omega_0) - \frac{\pi}{i}\delta(\omega+\omega_0) \\
\end{align*}

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

Note that the Nyquist rate tells us we must sample at $\omega_s \geq 2\omega_0$ to recover the signal, since $2\omega_0$ is the bandwidth of the signal. Lets use $\omega_s = \frac{3}{2}\omega_0$. The frequency spectra of the signal $g$ is an impulse train where each impulse has a strength of $\omega_s$ and is shifted by $\omega_s$, given by: $G(\omega) = \omega_s \sum \limits_{k\in\Z} \delta(\omega - k\omega_s)$.

%\img{images/sampling/aliasing/sin/gw.ps}

Multiplication in the time domain corresponds to convolution in the frequency domain, hence we have $\ftrans{x(t)g(t)} = \frac{1}{2\pi} \lr{ X * G } (\omega)$. This gives us shifted versions of the frequency spectra of $x$ at every dirac in the frequency spectra of $g$:

\begin{align*}
\ftrans{x(t)g(t)} &= \frac{1}{2\pi} \lr{ X * G } (\omega) = X_g(\omega) \\
x_g(\omega) &= \frac{1}{2\pi} \infint X(\Omega) G(\omega - \Omega) d\Omega \\
x_g(\omega) &= \frac{1}{2\pi} \infint X(\Omega) \omega_s \sum \limits_{k \in Z} \delta(\omega - \Omega - k\omega_s) d\Omega \\
x_g(\omega) &= \frac{\omega_s}{2\pi} \sum \limits_{k \in Z} \infint X(\Omega) \delta(\omega - \Omega - k\omega_s) d\Omega \\
x_g(\omega) &= \frac{1}{T_s} \sum \limits_{k \in Z} X(\omega - k\omega_s) \\
x_g(\omega) &= \frac{1}{T_s} \sum \limits_{k \in Z} \lr{\frac{\pi}{i}\delta(\omega - k\omega_s -\omega_0) - \frac{\pi}{i}\delta(\omega - k\omega_s+\omega_0)} \\
x_g(\omega) &= \frac{1}{T_s} \sum \limits_{k \in Z} \lr{\frac{\pi}{i}\delta(\omega - \omega_0(3k/2+1)) - \frac{\pi}{i}\delta(\omega - \omega_0(3k/2-1))} \\
\end{align*}

%\img{images/sampling/aliasing/sin/xg.ps}

So what is it that we see? The human visual system behaves like a low pass filter. We have spatial and temportal frequency responses that look something like a family of curves:

%\img{images/hvs/hvs.ps}

This captures the lower frequencies of $X_g(\omega)$, where $\omega \in \left( -\omega_0 , \omega_0 \right)$.


%\img{images/sampling/aliasing/sin/xgframed.ps}

What does the final signal look like? Let $y(t)$ be what you see after the the signal is passed into the human visual system:

%\img{images/hvs/sys/sys.ps} 

\begin{align*}
x_g(\omega) &= \frac{1}{T_s} \sum \limits_{k \in Z} \lr{\frac{\pi}{i}\delta(\omega - \omega_0(3k/2+1)) - \frac{\pi}{i}\delta(\omega - \omega_0(3k/2-1))} \\
Y(\omega) &= \frac{\pi}{i}\delta(\omega + \omega_0/2) - \frac{\pi}{i}\delta(\omega - \omega_0/2) \\
\iftrans{Y(\omega)} &= \frac{1}{2i} e^{-i\omega_0/2} - \frac{1}{2i} e^{i\omega_0/2 t} \\
y(t) &= -\sin(\frac{\omega_0}{2} t) = \sin(-\frac{\omega_0}{2}t) \\
\end{align*}

The human visual system filters the higher frequencies out, and you see the half of the reversed speed. The negative indicates time reversal, and the division by 2 indicates a slower speed.

The Carriage Wheel Effect