逆小波转换

已知

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

则逆转换为

${\displaystyle x(t)={\frac {1}{C}}_{\psi }\int _{0}^{\infty }\int _{-\infty }^{\infty }{\frac {1}{b^{5/2}}}X_{w}(a,b)\psi ({\frac {t-a}{b}})\,dadb}$ 

其中

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

证明:

由于${\displaystyle X_{w}(a,b)=x(t)*{\frac {1}{\sqrt[{}]{b}}}\psi ({\frac {-t}{b}})}$ 

假设${\displaystyle y(t)={\frac {1}{C}}_{\psi }\int _{0}^{\infty }\int _{-\infty }^{\infty }{\frac {1}{b^{5/2}}}X_{w}(a,b)\psi ({\frac {t-a}{b}})\,dadb}$ 

则${\displaystyle y(t)={\frac {1}{C}}_{\psi }\int _{0}^{\infty }x(t)*\psi ({\frac {-t}{b}})*\psi ({\frac {t}{b}}){\frac {db}{b^{3}}}}$ 

经过傅立叶转换，原本的折积性质变为相乘

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

如果母小波为实函数，则其傅立叶转换有以下性质

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

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

${\displaystyle =X(f){\frac {1}{C}}_{\psi }\int _{0}^{\infty }|{\boldsymbol {\psi }}(f_{1})|^{2}{\frac {df_{1}}{bf}}}$ (使用变数代换${\displaystyle f_{1}=bf}$ )

${\displaystyle =X(f){\frac {1}{C}}_{\psi }\int _{0}^{\infty }|{\boldsymbol {\psi }}(f_{1})|^{2}{\frac {df_{1}}{bf_{1}}}}$ 

${\displaystyle =X(f)}$ 

得证${\displaystyle y(t)=x(t)}$ 

${\displaystyle x(t)=\sum _{m=-\infty }^{\infty }\sum _{n=-\infty }^{\infty }2^{m/2}\psi _{1}(2^{m}-n)X_{w}(n,m)}$ 

${\displaystyle \psi _{1}(t)}$ 为${\displaystyle \psi (t)}$ 的双效函数(dual function)，满足以下正交(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

${\displaystyle \int _{-\infty }^{\infty }2^{m}\psi _{1}(2^{m_{1}}t-n_{1})\psi (2^{m}t-n)\,dt=\delta (m-m_{1})\delta (n-n_{1})}$ 

通常会设计成${\displaystyle \psi _{1}(t)=\psi (t)}$ 

因此离散变数连续小波转换能进行逆转换的条件为:

${\displaystyle \int _{-\infty }^{\infty }2^{m}\psi (2^{m_{1}}t-n_{1})\psi (2^{m}t-n)\,dt=\delta (m-m_{1})\delta (n-n_{1})}$ 

在这里解释的是如何重建(reconstruction)一个经过离散小波转换的函数

以进行一阶离散小波转换，升降频倍率为2为例，可以得到右图的架构

${\displaystyle g_{1}[n],h_{1}[n]}$ 需要满足一些条件才能使${\displaystyle x[n]=x_{0}[n]}$ 

将此流程进行Z转换及化简可得到:

${\displaystyle X_{0}(z)={\frac {1}{2}}[G(z)G_{1}(z)+H(z)H_{1}(z)]X(z)+{\frac {1}{2}}[G(-z)G_{1}(z)+H(-z)H_{1}(z)]X(-z)}$ 

因此为了得到${\displaystyle X_{0}(z)=X(z)}$ ，须满足以下二条件

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

可转换为 ${\displaystyle {\begin{bmatrix}G(z)&H(z)\\G(-z)&H(-z)\end{bmatrix}}{\begin{bmatrix}G_{1}(z)\\H_{1}(z)\end{bmatrix}}={\begin{bmatrix}2\\0\end{bmatrix}}}$ 

化简得到${\displaystyle {\begin{bmatrix}G_{1}(z)\\H_{1}(z)\end{bmatrix}}={\frac {2}{det(H_{m}(z))}}{\begin{bmatrix}H(-z)\\-G(-z)\end{bmatrix}}}$  其中${\displaystyle det(H_{m}(z))=G(z)H(-z)-H(z)G(-z)}$ 

要满足上式须满足以下四个条件，此四条件及上式的关系为当且仅当

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. ${\displaystyle \sum _{p}g_{1}[p]h[2n-p]=0}$ 

证明于参考条目中 因此只要${\displaystyle g_{1}[n],h_{1}[n]}$ 符合上述条件就能将经过离散小波转换的${\displaystyle x_{1,L}[n],x_{1,H}[n]}$ 重建为x[n]

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1. Jian-Jiun Ding (2014) Time-Frequency Analysis and Wavelet Transform （页面存档备份，存于互联网档案馆）