**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

1. 連續小波轉換
2. 離散變數連續小波轉換
3. 離散小波轉換

分別介紹此三種的反函數

連續小波轉換反函數
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已知

X\_{w}(a,b)={\frac {1}{{\sqrt {|(b)|}}}}\int \_{{-\infty }}^{{\infty }}x(t)\psi ({\frac {t-a}{b}})\,dt

則逆轉換為

{\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

{\displaystyle C\_{\psi }=\int \_{0}^{\infty }{\frac {|{\boldsymbol {\psi }}(f)|^{2}}{|f|}}df<\infty }

證明:

由於{\displaystyle X\_{w}(a,b)=x(t)\*{\frac {1}{\sqrt[{}]{b}}}\psi ({\frac {-t}{b}})}

假設{\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{\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

{\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

{\displaystyle {\boldsymbol {\psi }}(-f)={\boldsymbol {\psi }}^{\*}(f),{\boldsymbol {\psi }}(-bf){\boldsymbol {\psi }}(bf)={\boldsymbol {\psi }}^{\*}(f){\boldsymbol {\psi }}(bf)=|{\boldsymbol {\psi }}(bf)|^{2}}

{\displaystyle Y(f)=X(f){\frac {1}{C}}\_{\psi }\int \_{0}^{\infty }|{\boldsymbol {\psi }}(bf)|^{2}{\frac {db}{b}}}

{\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{\displaystyle f\_{1}=bf})

{\displaystyle =X(f){\frac {1}{C}}\_{\psi }\int \_{0}^{\infty }|{\boldsymbol {\psi }}(f\_{1})|^{2}{\frac {df\_{1}}{bf\_{1}}}}

{\displaystyle =X(f)}

得證{\displaystyle y(t)=x(t)}

離散變數連續小波轉換反函數
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{\displaystyle x(t)=\sum \_{m=-\infty }^{\infty }\sum \_{n=-\infty }^{\infty }2^{m/2}\psi \_{1}(2^{m}-n)X\_{w}(n,m)}

\psi \_{1}(t)為\psi (t)的雙效函數(dual function)，滿足以下正交(orthogonal)特性

{\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

{\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{\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:

{\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})}

離散小波轉換反函數
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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

DWT reconstruction
{\displaystyle g\_{1}[n],h\_{1}[n]}需要滿足一些條件才能使{\displaystyle x[n]=x\_{0}[n]}

將此流程進行Z轉換及化簡可得到:

{\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{\displaystyle X\_{0}(z)=X(z)}，須滿足以下二條件

1. {\displaystyle G(z)G\_{1}(z)+H(z)H\_{1}(z)=2}
2. {\displaystyle G(-z)G\_{1}(z)+H(-z)H\_{1}(z)=0}

可轉換為
{\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{\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{\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

1. {\displaystyle \sum \_{p}g[p]g\_{1}[2n-p]=\delta [n]}
2. {\displaystyle \sum \_{p}h[p]h\_{1}[2n-p]=\delta [n]}
3. {\displaystyle \sum \_{p}g[p]h\_{1}[2n-p]=0}
4. {\displaystyle \sum \_{p}g\_{1}[p]h[2n-p]=0}

證明於參考條目中
因此只要{\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{\displaystyle x\_{1,L}[n],x\_{1,H}[n]}重建為x[n]

相關條目
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* 數值分析
* 訊號處理

參考
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1. Jian-Jiun Ding (2014) Time-Frequency Analysis and Wavelet Transform （页面存档备份，存于）