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Stable Signal Representations

Module by: Michael A Lexa, Ronald DeVore, Shriram Sarvotham. E-mail the authors

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

f ^(ω)=0for|ω|π-δ,δ>0,f ^(ω)=0for|ω|π-δ,δ>0,
(1)
and instead of using χ [-π,π] χ [-π,π] , we multiply by another function g ^(ω)g ^(ω) which is very similar in form to the characteristic function, but decays at its boundaries in a smoother fashion (i.e. it has more derivatives). A candidate function g ^g ^ is sketched in Figure 1.

Figure 1: Sketch of g ^g ^.
Figure 1 (g_hat_new.png)

Now, it is a property of the Fourier transform that an increased smoothness in one domain translates into a faster decay in the other. Thus, we can fix our instability problem, by choosing g ^g ^ so that g ^g ^ is smooth and g ^(ω)=1g ^(ω)=1, |ω|π-δ|ω|π-δ and g ^=0g ^=0, |ω|>π|ω|>π. By choosing the smoothness of gg suitably large, we can, for any given m1m1, choose gg to satisfy

|g(t)|C (|t|+1) m |g(t)|C (|t|+1) m
(2)
for some constant C>0C>0.

Using such a g ^g ^, we can rewrite ((Reference)) as

f ^(ω)=F(ω)g ^(ω)=1 2π nZ f(n)e -inω g ^(ω).f ^(ω)=F(ω)g ^(ω)=1 2π nZ f(n)e -inω g ^(ω).
(3)
Thus, we have the new representation
f(t)= nZ f(n)g(t-n),f(t)= nZ f(n)g(t-n),
(4)
where we gain stability from our additional assumption that the signal is bandlimited on [-π-δ,π-δ][-π-δ,π-δ].

Does this assumption really hurt? No, not really because if our signal is really bandlimited to [-π,π][-π,π] and not [-π-δ,π-δ][-π-δ,π-δ], we can always take a slightly larger bandwidth, say [-λπ,λπ][-λπ,λπ] where λλ 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

f(t)= nZ f(n) sinc (π(t-n)),f(t)= nZ fn λg(λt-n).f(t)= nZ f(n) sinc (π(t-n)),f(t)= nZ fn λg(λt-n).
(5)
Shannon's Theorem tells us that if fB A fB A , we should sample ff at the Nyquist rate AA (which is twice the support of f ^f ^) and then take the binary representation of the samples. Our more stable representation says to slightly oversample ff and then convert to a binary representation. Both representations offer perfect reconstruction, although in the more stable representation, one is straddled with the additional task of choosing an appropriate λλ.

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

  1. How many bits do we need to represent ff in B A=1 B A=1 on some interval [-T,T][-T,T] in the norm L [-T,T]L [-T,T]?
  2. Using this methodology, what is the optimal way of encoding?
  3. How is the optimal encoding implemented?

Towards this end, we define

B A :={fL 2 (R):|f ^(ω)|=0,|ω|Aπ}.B A :={fL 2 (R):|f ^(ω)|=0,|ω|Aπ}.
(6)
Then for any fB A fB A , we can write
f= n fn A·sincπ(At-n).f= n fn A·sincπ(At-n).
(7)
Figure 2: Fourier transform of g λ (·)g λ (·).
Figure 2 (figure1.png)
In other words, samples at 0, ±1 A±1 A, ±2 A,±2 A, are sufficient to reconstruct ff. Recall also that sinc(x)=sin(x) xsinc(x)=sin(x) x decays poorly (leading to numerical instability). We can overcome this problem by slight over-sampling. Say we over-sample by a factor λ>1λ>1. Then, we can write

f=fn λAg λ (λAt-n).f=fn λAg λ (λAt-n).
(8)

Hence we need samples at 0, ±1 λA±1 λA, ±2 λA±2 λA, etc. What is the advantage? Sampling more often than necessary buys us stability because we now have a choice for g λ (·)g λ (·). If we choose g λ (·)g λ (·) infinitely differentiable whose Fourier transform looks as shown in Figure 2 we can obtain

|g λ (t)|c λ,k (1+|t|) k ,k=1,2,...|g λ (t)|c λ,k (1+|t|) k ,k=1,2,...
(9)
and therefore g λ (·)g λ (·) decays very fast. In other words, a sample's influence is felt only locally. Note however, that over-sampling generates basis functions that are redundant (linearly dependent), unlike the integer translates of the sinc(·)sinc(·) function.

Figure 3: To reconstruct signals in [-T,T][-T,T], the sampling interval is [-cT,cT][-cT,cT].
Figure 3 (figure2.png)

If we restrict our reconstruction to tt in the interval [-T,T][-T,T], we will only need samples only from [-cT,cT][-cT,cT], for c>1c>1 (see Figure 3), because the distant samples will have little effect on the reconstruction in [-T,T][-T,T].

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