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