1.8 Filtering a signal with something to remove
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Take the off-band perturbation
The constant occupies frequency zero, and the cosine occupies labels three and nine. These labels all lie outside \(U_{12}\). The perturbation has the literal rational vector
Set \(x=c+b\), giving
Theorem QA11: Exact filter recovery and orthogonal energy
The test signal satisfies
and
Proof
QA05 gives the support of \(\widehat c\). Write the cosine as
Thus \(\widehat b\) equals twenty-four at zero, eighteen at three and nine, and zero elsewhere. Multiplication by \(D\) fixes the spectrum of \(c\) and kills the spectrum of \(b\), proving the filter identities. Their supports are disjoint, so Parseval for inner products gives \(\langle c,b\rangle=0\). QA05 gives \(\|c\|^2=12\cdot4=48\), and
QA07 then gives \(\|x\|^2=48+102=150\). The same numbers follow by direct rational squaring of the displayed vectors. \(\square\)
The discarded component is part of a complete explanation of the input. It is called a perturbation because the example declares \(c\) to be the target. It is a perfectly exact signal component rather than an error introduced by the representation. If the task were to recover all of \(x\), deleting \(b\) would lose valid information.