简单的小波压缩变换工作的原理如下:

信号和求导信号,形成的求同信号序列和求导信号序列的长度均为输入信号的长度的一半.如果输入信号序列是A={a(1),...,a(n)}求同信号S={s(1),...,s(n/2)}和求异信号D={d(1),...,d(n/2)}的计算方式为:

for i=1,...,n/2

s(i)=a(2*i-1)+a(2*i)

d(i)=a(2*i-1)-a(2*i)

例如输入信号是:

5,2,3,2,5,7,9,6

求出的求同信号和求异信号和求同信号分别是:

s={7,5,12,15}

d={3,1,-2,3}

将求同信号和求异组合之后，原信号变换为：

A ={7,5,12,15,3,1,-2,3}

同样过程再对当前的信号A 的前半部分操作，直到输入信号的长度为1，变换过程结束。

还使用上面的例子，输入的原始信号为5，2，3，2，5，7，9，6，该信号经过一系列变换后得到小波压缩变换结果为39，-15，2，-3，3，1，-2，3，变换过程如下

5，2，3，2，5，7，9，6

7，5，12，15，3，1，-2，3

12，27，2，-3，3，1，-2，3

39，-15，2，-3，3，1，-2，3

原始信号的输入长度是2的整数次幂，并且输入信号中的每个元素为0和255之间的整数 -Simple wavelet compression transform the principle of work is as follows:

Signals and signal derivation, formed for the same signal sequence and signal sequence derivation of the length of the input signal are half the length. If the input signal sequence is A = (a (1 ),..., a (n)) for the same signal S = (s (1 ),..., s (n/2)) and the difference signal D = (d (1 ),..., d (n/2)) method for calculating as follows:

for i = 1 ,..., n/2　

s (i) = a (2* i-1)+ a (2* i)

d (i) = a (2* i-1)-a (2* i)

For example, the input signal is:

5,2,3,2,5,7,9,6

Seek to derive the signal and difference signal and for the same signals are:

s = (7,5,12,15)

d = (3,1,-2,3)

Will seek the same combination of signals and difference, the original signal is transformed into:

A　= (7,5,12,15,3,1,-2,3)

The same process again on the current signal A　of the first half of the operation, until the length of the input signal is 1, change the course of The End.

Also used the above exampl