逆小波轉換

逆小波轉換(inverse wavelet transform)為小波轉換的反函數，小波轉換大致分為三類 連續小波轉換 離散變數連續小波轉換 離散小波轉換 分別介紹此三種的反函數 已知 X w ( a , b ) = 1 | ( b ) | ∫ − ∞ ∞ x ( t ) ψ ( t − a b ) d t {\displaystyle X_{w}(a,b)={\frac {1}{\sqrt {|(b)|}}}\int _{-\infty }^{\infty }x(t)\psi ({\frac {t-a}{b}})\,dt} 則逆轉換為 x ( t ) = 1 C ψ ∫ 0 ∞ ∫ − ∞ ∞ 1 b 5 / 2 X w ( a , b ) ψ ( t − a b ) d a d b {\displaystyle x(t)={\frac {1}{C}}_{\psi }\int _{0}^{\infty }\int _{-\infty }^{\infty }{\frac {1}{b^{5/2}}}X_{w}(a,b)\psi ({\frac {t-a}{b}})\,dadb} 其中 C ψ = ∫ 0 ∞ | ψ ( f ) | 2 | f | d f < ∞ {\displaystyle C_{\psi }=\int _{0}^{\infty }{\frac {|{\boldsymbol {\psi }}(f)|^{2}}{|f|}}df<\infty } 證明: 由於 X w ( a , b ) = x ( t ) ∗ 1 b ψ ( − t b ) {\displaystyle X_{w}(a,b)=x(t)*{\frac {1}{\sqrt[{}]{b}}}\psi ({\frac {-t}{b}})} 假設 y ( t ) = 1 C ψ ∫ 0 ∞ ∫ − ∞ ∞ 1 b 5 / 2 X w ( a , b ) ψ ( t − a b ) d a d b {\displaystyle y(t)={\frac {1}{C}}_{\psi }\int _{0}^{\infty }\int _{-\infty }^{\infty }{\frac {1}{b^{5/2}}}X_{w}(a,b)\psi ({\frac {t-a}{b}})\,dadb} 則 y ( t ) = 1 C ψ ∫ 0 ∞ x ( t ) ∗ ψ ( − t b ) ∗ ψ ( t b ) d b b 3 {\displaystyle y(t)={\frac {1}{C}}_{\psi }\int _{0}^{\infty }x(t)*\psi ({\frac {-t}{b}})*\psi ({\frac {t}{b}}){\frac {db}{b^{3}}}} 經過傅立葉轉換，原本的摺積性質變為相乘 Y ( f ) = 1 C ψ ∫ 0 ∞ X ( f ) ψ ( − b f ) ψ ( b f ) d b b {\displaystyle Y(f)={\frac {1}{C}}_{\psi }\int _{0}^{\infty }X(f){\boldsymbol {\psi }}(-bf){\boldsymbol {\psi }}(bf){\frac {db}{b}}} 如果母小波為實函數，則其傅立葉轉換有以下性質 ψ ( − f ) = ψ ∗ ( f ) , ψ ( − b f ) ψ ( b f ) = ψ ∗ ( f ) ψ ( b f ) = | ψ ( b f ) | 2 {\displaystyle {\boldsymbol {\psi }}(-f)={\boldsymbol {\psi }}^{*}(f),{\boldsymbol {\psi }}(-bf){\boldsymbol {\psi }}(bf)={\boldsymbol {\psi }}^{*}(f){\boldsymbol {\psi }}(bf)=|{\boldsymbol {\psi }}(bf)|^{2}} Y ( f ) = X ( f ) 1 C ψ ∫ 0 ∞ | ψ ( b f ) | 2 d b b {\displaystyle Y(f)=X(f){\frac {1}{C}}_{\psi }\int _{0}^{\infty }|{\boldsymbol {\psi }}(bf)|^{2}{\frac {db}{b}}} = X ( f ) 1 C ψ ∫ 0 ∞ | ψ ( f 1 ) | 2 d f 1 b f {\displaystyle =X(f){\frac {1}{C}}_{\psi }\int _{0}^{\infty }|{\boldsymbol {\psi }}(f_{1})|^{2}{\frac {df_{1}}{bf}}} (使用變數代換 f 1 = b f {\displaystyle f_{1}=bf} ) = X ( f ) 1 C ψ ∫ 0 ∞ | ψ ( f 1 ) | 2 d f 1 b f 1 {\displaystyle =X(f){\frac {1}{C}}_{\psi }\int _{0}^{\infty }|{\boldsymbol {\psi }}(f_{1})|^{2}{\frac {df_{1}}{bf_{1}}}} = X ( f ) {\displaystyle =X(f)} 得證 y ( t ) = x ( t ) {\displaystyle y(t)=x(t)} x ( t ) = ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ 2 m / 2 ψ 1 ( 2 m − n ) X w ( n , m ) {\displaystyle x(t)=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }2^{m/2}\psi _{1}(2^{m}-n)X_{w}(n,m)} ψ 1 ( t ) {\displaystyle \psi _{1}(t)} 為 ψ ( t ) {\displaystyle \psi (t)} 的雙效函數(dual function)，滿足以下正交(orthogonal)特性 ∑ m = − ∞ ∞ ∑ n = − ∞ ∞ 2 m ψ 1 ( 2 m − n ) ψ ( 2 m t 1 − n ) = δ ( t − t 1 ) {\displaystyle \sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }2^{m}\psi _{1}(2^{m}-n)\psi (2^{m}t_{1}-n)=\delta (t-t_{1})} 或是 ∫ − ∞ ∞ 2 m ψ 1 ( 2 m 1 t − n 1 ) ψ ( 2 m t − n ) d t = δ ( m − m 1 ) δ ( n − n 1 ) {\displaystyle \int _{-\infty }^{\infty }2^{m}\psi _{1}(2^{m_{1}}t-n_{1})\psi (2^{m}t-n)\,dt=\delta (m-m_{1})\delta (n-n_{1})} 通常會設計成 ψ 1 ( t ) = ψ ( t ) {\displaystyle \psi _{1}(t)=\psi (t)} 因此離散變數連續小波轉換能進行逆轉換的條件為: ∫ − ∞ ∞ 2 m ψ ( 2 m 1 t − n 1 ) ψ ( 2 m t − n ) d t = δ ( m − m 1 ) δ ( n − n 1 ) {\displaystyle \int _{-\infty }^{\infty }2^{m}\psi (2^{m_{1}}t-n_{1})\psi (2^{m}t-n)\,dt=\delta (m-m_{1})\delta (n-n_{1})} 在這裡解釋的是如何重建(reconstruction)一個經過離散小波轉換的函數 以進行一階離散小波轉換，升降頻倍率為2為例，可以得到右圖的架構 g 1 [ n ] , h 1 [ n ] {\displaystyle g_{1}[n],h_{1}[n]} 需要滿足一些條件才能使 x [ n ] = x 0 [ n ] {\displaystyle x[n]=x_{0}[n]} 將此流程進行Z轉換及化簡可得到: X 0 ( z ) = 1 2 [ G ( z ) G 1 ( z ) + H ( z ) H 1 ( z ) ] X ( z ) + 1 2 [ G ( − z ) G 1 ( z ) + H ( − z ) H 1 ( z ) ] X ( − z ) {\displaystyle X_{0}(z)={\frac {1}{2}}[G(z)G_{1}(z)+H(z)H_{1}(z)]X(z)+{\frac {1}{2}}[G(-z)G_{1}(z)+H(-z)H_{1}(z)]X(-z)} 因此為了得到 X 0 ( z ) = X ( z ) {\displaystyle X_{0}(z)=X(z)} ，須滿足以下二條件 G ( z ) G 1 ( z ) + H ( z ) H 1 ( z ) = 2 {\displaystyle G(z)G_{1}(z)+H(z)H_{1}(z)=2} G ( − z ) G 1 ( z ) + H ( − z ) H 1 ( z ) = 0 {\displaystyle G(-z)G_{1}(z)+H(-z)H_{1}(z)=0} 可轉換為 [ G ( z ) H ( z ) G ( − z ) H ( − z ) ] [ G 1 ( z ) H 1 ( z ) ] = [ 2 0 ] {\displaystyle {\begin{bmatrix}G(z)&H(z)\\G(-z)&H(-z)\end{bmatrix}}{\begin{bmatrix}G_{1}(z)\\H_{1}(z)\end{bmatrix}}={\begin{bmatrix}2\\0\end{bmatrix}}} 化簡得到 [ G 1 ( z ) H 1 ( z ) ] = 2 d e t ( H m ( z ) ) [ H ( − z ) − G ( − z ) ] {\displaystyle {\begin{bmatrix}G_{1}(z)\\H_{1}(z)\end{bmatrix}}={\frac {2}{det(H_{m}(z))}}{\begin{bmatrix}H(-z)\\-G(-z)\end{bmatrix}}} 其中 d e t ( H m ( z ) ) = G ( z ) H ( − z ) − H ( z ) G ( − z ) {\displaystyle det(H_{m}(z))=G(z)H(-z)-H(z)G(-z)} 要滿足上式須滿足以下四個條件，此四條件及上式的關係為若且唯若 ∑ p g [ p ] g 1 [ 2 n − p ] = δ [ n ] {\displaystyle \sum _{p}g[p]g_{1}[2n-p]=\delta [n]} ∑ p h [ p ] h 1 [ 2 n − p ] = δ [ n ] {\displaystyle \sum _{p}h[p]h_{1}[2n-p]=\delta [n]} ∑ p g [ p ] h 1 [ 2 n − p ] = 0 {\displaystyle \sum _{p}g[p]h_{1}[2n-p]=0} ∑ p g 1 [ p ] h [ 2 n − p ] = 0 {\displaystyle \sum _{p}g_{1}[p]h[2n-p]=0} 證明於參考條目中 因此只要 g 1 [ n ] , h 1 [ n ] {\displaystyle g_{1}[n],h_{1}[n]} 符合上述條件就能將經過離散小波轉換的 x 1 , L [ n ] , x 1 , H [ n ] {\displaystyle x_{1,L}[n],x_{1,H}[n]} 重建為x[n] 數值分析 訊號處理 Jian-Jiun Ding (2014) Time-Frequency Analysis and Wavelet Transform （页面存档备份，存于互联网档案馆）