Week 7

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

Suppose there are two signals $x_1(t)$ and $x_2(t)$ that pass through the same sampling points generated by sampling $x_1(t)$. Also, assume that $x_2(t)$ has a higher frequency. This implies that $X_2(\omega)$ has a larger bandwidth than $X_1(\omega)$. Even though the signal is bandlimited, it is not bandlimited enough, and we would need to increase $\omega_s$ (decrease $T_s$) to recover the signal $x_2(t)$.

%\img{images/sampling/sincs/func/graph.ps}

%In this case we are looking at two trigonometric functions, which happen to be $\cos(\omega_0 t)$ and $\cos((\omega_0 + 2\pi)t)$. This demonstrates why in discrete time we have $2\pi$-periodicity of trigonometric functions. \\ \\

We saw previously what happens when you sample below the Nyquist rate to a bandlimited signal $x$, such that $\omega_s \lt 2D$:

%\img{images/sampling/aliasing/overlap/overlap.ps}

The higher frequencies fold into the lower frequencies, and the area where the triangles collide adds up and we get echos that cause the higher frequency content of the signal to be unrecoverable.

%\img{images/sampling/impulse/triangles/aliased/impulse.ps}

After we sample below the Nyquist rate, we pass the signal through a low pass filter.

%\img{images/sampling/aliasing/aliasing/antialiassys/orig.ps}

If we had an ideal low pass filter that only kept values from $-\frac{\omega_s}{2}$ to $\frac{\omega_s}{2}$, then we would bandlimit the signal, in this case, directly where the overlap occurs:

%\img{images/sampling/aliasing/overlap/lpf.ps}

We would end up with ``garbled'' information in the intervals $(-\omega_s/2,-\omega_s + D)$ and $(\omega_s - D, \omega_s/2)$.

%\img{images/sampling/aliasing/aliasing/triangles/filtered.ps}

How can use solve this? Using \emph{Anti-alias Filtering}. This consists of bandlimiting the signal $x$ in a pre-processing stage, and then sampling the bandlimited version:

%\img{images/sampling/aliasing/aliasing/antialiassys/anti.ps}

If we set the cutoff frequency $\omega_c \leq \frac{\omega_s}{2}$ in the low pass filter, then we have the Nyquist rate for this bandlimited signal. So then we would have our anti-aliased signal $x_{aa}(t)$ and its corresponding frequency spectra:

%\img{images/sampling/aliasing/aliasing/antialiased/filtered.ps}

Now we sample this frequency spectra, resulting with the following spectra:

%\img{images/sampling/impulse/triangles/aliased/anti/freq.ps}

Now if we use an ideal low pass filter with a cutoff frequency of $\omega_s/2$, we can fully recover the anti-aliased version of our signal $x$. This is preferable, because there is more reliable data than before when we had ``garbled'' information in the intervals $(-\omega_s/2,-\omega_s + D)$ and $(\omega_s - D, \omega_s/2)$. That is the difference between sampling first versus bandlimited and then sampling first.


Anti-Aliasing

Lets revisit $x(t) = \sin(\omega_0t)$. Now, that signal, if we sample it at a particular rate and create a discrete time signal, which is essentially a sampled version of the continuous time signal: we can get a discrete time signal out of continuous time signal.

%\img{images/sampling/sin/graph.ps}

and let $x_d(n) =x(nT_s) = \sin(n\omega_0 T_S)$.

%\img{images/sampling/sin/justk.ps}

What if we let our sampling frequency be $\omega_s = \frac{3}{2}\omega_0$? We know we will have aliasing based on the Nyquist rate since we would need $2\omega_0$ to recover the signal safely. Our sampling period would be $T_s = \frac{4\pi}{3\omega_0}$. This gives us $x_d(n) = \sin(4\pi n/3)$.

\begin{align*}
\hat{x}(t) &= \sin(-\omega_0t/2) \\
\hat{x}_d(n) &= \hat{x}(nT_s) \\
\hat{x}_d(n) &= \sin(-\omega_0nT_s/2) \\
\hat{x}_d(n) &= \sin(-2\pi n/3) \\
\hat{x}_d(n) &= \sin(-2\pi n/3 + 2\pi n) = \sin(4\pi n/3) = x_d(n) \\
\hat{x}_d(n) &= x_d(n) \\
\end{align*}

The result is that the two signals when sampled, produce the same discrete time signal. 

%\img{images/sampling/ratevsresolution/sin/graph.ps}



7.2 Aliasing