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We now wish to consider new model classes for signals. Towards this end, let { ψ j } j = 1 { lbrace ψ_j rbrace}_{j = 1}^∞ be an orthonormal basis for L 2 ( T , T ) L_2 { \( { - T , T} \)} . Thus for f L 2 f in L_2 we can write f = j = 1 c j ( f ) ψ j f = sum_{j = 1}^∞ c_j { \( f \)} ψ_j where ( c j ( f ) ) 2 { \( {c_j { \( f \)}} \)} in ℓ_2 . We will now build an encoder and decoder and analyze its performance oncompact sets K K . For example, we might want to encode signals in the space X p = { f : ( c j ( f ) ) p }, 0 p 2 X_p ={ lbrace {f : { \( {c_j { \( f \)}} \)} in ℓ_p} rbrace} 0<= p<= 2 with norm f X p : = ( c j ( f ) ) p . However, in this space the unit ball, U ( X p ) U { \( X_p \)} is not compact. To get a compact set we need more structure on the sequence ( c j ) \( c_j \) . Hence we define Y α : = { f : c n ( f ) n α , n = 1 , 2 , } Y^α :={ lbrace {f : \lline c_n { \( f \)} \lline<= n^{ - α} , n = 1 , 2 , dotslow} rbrace} and we define the norm in this space as f Y α : = the smallest c c such that this holds. We now take K = U ( X p ) U ( Y α ) K =U { \( X_p \)} intersection U { \( Y^α \)} to get a compact set. Notice that when α > 0 α>0 is small the requirement for membership in Y α Y^α is very mild.

Next, suppose that we choose a target distortion level ε = 2 m ε =2^{ - m} . Given f f , let Λ k : = Λ k ( f ) = { j { 0 , , N } : 2 k 1 c j ( f ) < 2 k } Λ_k :=Λ_k { \( f \)} ={ lbrace {j in { lbrace {0 , dotslow , N} rbrace} : 2^{ - k - 1}<= \lline c_j { \( f \)} \lline<2^{ - k}} rbrace} for 0 k M 0<= k<= M , where M : = 2 m 2 p M :={⌈ {2 m} over {2 - p} ⌉} . We then choose N N as the smallest integer so that N α 2 M and thus log N C m . It follows from the requirement that f Y α f in Y^α that Λ k { 1 , , N } Λ_k subset { lbrace {1 , dotslow , N} rbrace} for each 0 k M 0<= k<= M .

Recall that # Λ k  2 ( - k - 1 ) p c j Λ k | c j | p f X p p . Since f U ( X p ) U ( Y α ) f in U { \( X_p \)} intersection U { \( Y^α \)} , # Λ k f p 2 ( k + 1 ) p 2 ( k + 1 ) p . Hence, the total number of indices in all of the Λ k Λ_k , 0 k M 0<= k<= M , is O ( 2 M p ) O { \( 2^{M p} \)} .

To encode, for each f f , we can send the following bits:

  • Send log n bits to identify each index in Λ k Λ_k , for 0 k M 0<= k<= M . This will require a total of O ( log N 2 M p ) bits.
  • Send one bit to identify the sign of c j ( f ) c_j { \( f \)} for each j Λ k j in Λ_k , 0 k M 0<= k<= M . This will require O ( 2 M p ) O { \( 2^{M p} \)} bits.
  • Send m m bits to describe each c j ( f ) , j Λ k c_j { \( f \)} ,j in Λ_k , for 0 k M 0<= k<= M . This will require O ( m 2 M p ) O { \( {m 2^{M p}} \)} bits.
Thus the total number of bits used in the encoding is O ( m 2 M p ) O { \( {m 2^{M p}} \)} .

Notice that for each j Λ k j in Λ_k , 0 k M 0<= k<= M , we can recover each c j ( f ) c_j { \( f \)} by c j ¯ = ± i = 0 m b i 2 k i {overline c_j} = +- sum csub {i = 0} csup m b_i 2^{ - k - i} where the sign is given by the sign bit. It follows that c j ( f ) c j ¯ 2 m k \lline c_j { \( f \)} - {overline c_j} \lline<= 2^{ - m - k} for every such coefficient. Here we have used the fact that knowing that j Λ k j in Λ_k means that the first nonzero binary bit of c j ( f ) c_j { \( f \)} is the k k -th bit.

To decode we simply set f ¯ = k = 0 M j Λ k c j ¯ ψ j . {overline f} = sum csub {k = 0} csup M sum csub {j in Λ_k} {overline c_j} ψ_j

We now analyze the error we have incurred in such an encoding. The square of the error will consist of two parts. The first corresponds to the j Λ k j in Λ_k , 0 k M 0<= k<= M . For each such j j we have c j ( f ) c j ¯ 2 m k \lline c_j { \( f \)} - {overline c_j} \lline<= 2^{ - m - k} and so the total square error for this is C k = 1 M 2 k p 2 2 m 2 2 k c 2 2 m because p 2 p<= 2 . The second part of the error corresponds to all the coefficients which have magnitude 2 M . We have that this sum does not exceed c j > 2 M c j 2 2 M ( 2 p ) j = 1 c j p 2 2 m . sum csub { \lline c_j \lline>2^{ - M}} \lline c_j \lline^2<= 2^{ - M { \( {2 - p} \)}} sum csub {j = 1} csup ∞ \lline c_j \lline^p<= 2^{ - 2 m} "." Thus the total error we incur is O ( 2 m ) O { \( 2^{ - m} \)} .

In summary, by allocating O ( m 2 m 1 p 1 2 ) bits we achieve distortion C 2 m C 2^{ - m} . Equivalently, by allocating n log n bits, we achieve distortion C n ( 1 p 1 2 ) C n^{ - { \( {1 ∕ p - 1 ∕ 2} \)}} .

This is within a logarithmic factor of the optimal encoding given by Kolmogorov entropy of the class U ( X p ) Y α U { \( X_p \)} intersection Y^α . A slightly more careful argument can remove this logarithm.

The wavelet basis

In the method above we failed to achieve the optimal performance because of the cost involved in identifying which indices were in each Λ k Λ_k . We will now describe a method that can do better, using the Haar basis for L 2 [ 0 , 1 ] L_2 { \[ {0 , 1} \]} . Thus, we first define the scaling function φ : = χ [ 0 , 1 ] . φ :=χ_{ \[ {0 , 1} \]} "." Next, we define the mother wavelet ψ : = χ [ 0 , 1 2 ] χ [ 1 2 , 1 ] . ψ :=χ_{ \[ {0 , 1 over 2} \]} - χ_{ \[ {1 over 2 , 1} \]} "." We then define the remaining wavelets recursively. They are obtained by dilations and shifts of the mother wavelet on dyadic intervals: ψ J : = 2 k 2 ψ [ 0 , 1 ] ( 2 k x j ) ψ_J :=2^k over 2 ψ_{ \[ {0 , 1} \]} { \( {2^k x - j} \)} where J = [ j 2 k , ( j + 1 ) 2 k ] J ={ \[ {j 2^{ - k} , { \( {j + 1} \)} 2^{ - k}} \]} are dyadic intervals. We denote by D + D_+ the collection of all dyadic intervals contained in [ 0 , 1 ] \[ {0 , 1} \] . Then, the collection of functions { φ } { ψ J } J D + forms an orthonormal basis for L 2 [ 0 , 1 ] L_2 { \[ {0 , 1} \]} .

A key property of wavelets is that a tree structure can be placed on the coefficients due to the use of dyadic intervals in their construction. Thus, let T k : = { j : c j 2 k } T_k :={ lbrace {j : \lline c_j \lline>= 2^{ - k}} rbrace} and T k + 1 T k = Λ k . T_{k + 1} - T_k =Λ_k "."

We define T ¯ k {overline T}_k as the smallest tree containing T k T_k . Given any binary tree of size n n , we can encode the tree with at most O ( n ) O { \( n \)} bits, in the process outperforming the encoder described in above.

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Source:  OpenStax, Compressive sensing. OpenStax CNX. Sep 21, 2007 Download for free at http://cnx.org/content/col10458/1.1
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