# 0.4 Bases, orthogonal bases, biorthogonal bases, frames, tight  (Page 3/5)

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$g\left(t\right)=\sum _{k}\phantom{\rule{0.166667em}{0ex}}{a}_{k}\phantom{\rule{0.277778em}{0ex}}cos\left(kt\right)$

where the basis vectors (functions) are

${f}_{k}\left(t\right)=cos\left(kt\right)$

and the expansion coefficients are obtained as

${a}_{k}=⟨g\left(t\right),\phantom{\rule{0.166667em}{0ex}}{f}_{k}\left(t\right)⟩=\frac{2}{\pi }{\int }_{0}^{\pi }g\left(t\right)\phantom{\rule{0.166667em}{0ex}}cos\left(kt\right)\phantom{\rule{0.166667em}{0ex}}dx.$

The basis vector set is easily seen to be orthonormal by verifying

$⟨{f}_{\ell }\left(t\right),\phantom{\rule{0.166667em}{0ex}}{f}_{k}\left(t\right)⟩=\delta \left(k-\ell \right).$

These basis functions span an infinite dimensional vector space and the convergence of [link] must be examined. Indeed, it is the robustness of that convergence that is discussed in this section under the topic ofunconditional bases.

## Sinc expansion example

Another example of an infinite dimensional orthogonal basis is Shannon's sampling expansion [link] . If $f\left(t\right)$ is band limited, then

$f\left(t\right)\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\sum _{k}f\left(Tk\right)\phantom{\rule{0.166667em}{0ex}}\frac{sin\left(\frac{\pi }{T}t-\pi k\right)}{\frac{\pi }{T}t-\pi k}$

for a sampling interval $T<\frac{\pi }{W}$ if the spectrum of $f\left(t\right)$ is zero for $|\omega |>W$ . In this case the basis functions are the sinc functions with coefficients which are simply samples of the originalfunction. This means the inner product of a sinc basis function with a bandlimited function will give a sample of that function. It is easy tosee that the sinc basis functions are orthogonal by taking the inner product of two sinc functions which will sample one of them at the pointsof value one or zero.

## Frames and tight frames

While the conditions for a set of functions being an orthonormal basis are sufficient for the representation in [link] and the requirement of the set being a basis is sufficient for [link] , they are not necessary. To be a basis requires uniqueness of the coefficients. Inother words it requires that the set be independent , meaning no element can be written as a linear combination of the others.

If the set of functions or vectors is dependent and yet does allow the expansion described in [link] , then the set is called a frame [link] . Thus, a frame is a spanning set . The term frame comes from a definition that requires finite limits on an inequality bound [link] , [link] of inner products.

If we want the coefficients in an expansion of a signal to represent the signal well, these coefficients should have certain properties. They arestated best in terms of energy and energy bounds. For an orthogonal basis, this takes the form of Parseval's theorem. To be a frame in asignal space, an expansion set ${\varphi }_{k}\left(t\right)$ must satisfy

${A\parallel g\parallel }^{2}\le \sum _{k}|⟨{\phi }_{k},g⟩{|}^{2}\le B{\parallel g\parallel }^{2}$

for some $0 and $B<\infty$ and for all signals $g\left(t\right)$ in the space. Dividing [link] by ${\parallel g\parallel }^{2}$ shows that $A$ and $B$ are bounds on the normalized energy of the inner products. They “frame" thenormalized coefficient energy. If

$A\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}B$

then the expansion set is called a tight frame . This case gives

${A\parallel g\parallel }^{2}\phantom{\rule{4pt}{0ex}}=\phantom{\rule{4pt}{0ex}}\sum _{k}{|⟨{\phi }_{k},g⟩|}^{2}$

which is a generalized Parseval's theorem for tight frames. If $A=B=1$ , the tight frame becomes an orthogonal basis. From this, it can be shown thatfor a tight frame [link]

$g\left(t\right)={A}^{-1}\sum _{k}⟨{\phi }_{k}\left(t\right),g\left(t\right)⟩\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}{\phi }_{k}\left(t\right)$

which is the same as the expansion using an orthonormal basis except for the ${A}^{-1}$ term which is a measure of the redundancy in the expansion set.

If an expansion set is a non tight frame, there is no strict Parseval's theorem and the energy in the transform domain cannot be exactly partitioned. However,the closer $A$ and $B$ are, the better an approximate partitioning can be done. If $A=B$ , we have a tight frame and the partitioning can be done exactly with [link] . Daubechies [link] shows that the tighter the frame bounds in [link] are, the better the analysis and synthesis system is conditioned. In other words, if $A$ is near or zero and/or $B$ is very large compared to $A$ , there will be numerical problems in the analysis–synthesis calculations.

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