1.4 Stable signal representations

To fix the instability of the Shannon representation, we assume that the signal is slightly more bandlimited than before

Now, it is a property of the Fourier transform that an increased smoothness in one domain translates into a faster decay in theother. Thus, we can fix our instability problem, by choosing $\widehat{g}$ so that $\widehat{g}$ is smooth and $\widehat{g}\left(\omega \right)=1$ , $\left|\omega \right|\le \pi -\delta$ and $\widehat{g}=0$ , $\left|\omega \right|>\pi$ . By choosing the smoothness of $g$ suitably large, we can, for any given $m\ge 1$ , choose $g$ to satisfy

Using such a $\widehat{g}$ , we can rewrite ( ) as

Does this assumption really hurt? No, not really because if our signal is really bandlimited to $[-\pi ,\pi ]$ and not $[-\pi -\delta ,\pi -\delta ]$ , we can always take a slightly larger bandwidth, say $[-\lambda \pi ,\lambda \pi ]$ where $\lambda$ is a little larger than one, and carry out the same analysis as above.Doing so, would only mean slightly oversampling the signal (small cost).

Recall that in the end we want to convert analog signals into bit streams. Thus far, we have the two representations

In practical situations, we shall be interested in approximating $f$ on an interval $[-T,T]$ for some $T>0$ and not for all time. Questions we still want to answer include

• How many bits do we need to represent $f$ in $B_{A=1}$ on some interval $[-T,T]$ in the norm $L_{\infty }[-T,T]$ ?
• Using this methodology, what is the optimal way of encoding?
• How is the optimal encoding implemented?

Towards this end, we define

Hence we need samples at 0, $\pm \frac{1}{\lambda A}$ , $\pm \frac{2}{\lambda A}$ , etc. What is the advantage? Sampling more often than necessary buys us stability because we now have a choicefor $g_{\lambda }(·)$ . If we choose $g_{\lambda }(·)$ infinitely differentiable whose Fourier transform looks as shown in we can obtain

If we restrict our reconstruction to $t$ in the interval $[-T,T]$ , we will only need samples only from $[-cT,cT]$ , for $c>1$ (see ), because the distant samples will have little effect on the reconstruction in $[-T,T]$ .