Discrete Wavelet Transform

The discrete signal ${\bf x}=[x[0],\cdots,x[N-1]]^T$ is a set of N samples taken from a continuous signal

$$x[m]=x(t_0+m \triangle t),\;\;\;\;(m=0,1,\cdots,N-1)$$

for some initial time $t_0$ and sampling period $\triangle t$. The basis functions ${\bf\varphi}=[\varphi[0],\cdots,\varphi[N-1]]^T$ and ${\bf\psi}=[\psi[0],\cdots, \psi[N-1]]^T$ are also vectors containing $N$ elements. We let $j_0=0$, $N=2^J$, $j=0,1,\cdots,J-1$ and $k=0,1,\cdots,2^j-1$. Correspondingly the wavelet expansion becomes discrete wavelet transform (DWT). The discrete function is represented as a weighted sum in the space spanned by the bases ${\bf\varphi}$ and ${\bf\psi}$:

$$x[m]=\frac{1}{\sqrt{N}}\sum_k W_{\varphi}[j_0,k]\varphi_{j_0... ... \sum_k W_{\psi}[j,k]\psi_{j,k}[m], \;\;\;\;\;(m=0,\cdots,N-1)$$

This is the inverse wavelet transform where the summation over $j$ is for different scale levels and the summation over $k$ is for different translations in each scale level, and the coefficients (weights) are projections of the function onto each of the basis functions:

$$W_{\varphi}[j_0,k]=({\bf x},{\bf\varphi}_{j_0,k})=\frac{1}{\s... ...=0}^{N-1} x[m] \varphi_{j_0,k}[m],\;\;\;\;\mbox{(for all $k$)}$$

$$W_{\psi}[j,k]=({\bf x},{\bf\psi}_{j,k})=\frac{1}{\sqrt{N}}\su... ...m] \psi_{j,k}[m],\;\;\;\;\mbox{(for all $k$ and all $j>j_0$)}$$

where $W_{\varphi}[j_0,k]$ is called the approximation coefficient and $W_{\psi}[j,k]$ is called the detail coefficient. These are the forward wavelet transform.

An Example:

Assume $N=4$-point discrete singal ${\bf x}=[x[0],\cdots,x[N-1]]^T=[1, 4, -3, 0]^T$ and the discrete Haar scaling and wavelet functions are:

$$\left[ \begin{array}{rrrr} 1 & 1 & 1 & 1 1 & 1 & -1 & -1 \... ... \psi_{0,0}[m] \psi_{1,0}[m] \psi_{1,1}[m] \end{array}$$

The coefficient for $V_0$:

$$W_{\varphi}[0,0]=\frac{1}{2}\sum_{m=0}^3 x[m]\varphi_{0,0}[m] =\frac{1}{2}[1\cdot 1+4\cdot 1-3\cdot 1+0\cdot 1]=1$$

The coefficient for $W_0$:

$$W_{\psi}[0,0]=\frac{1}{2}\sum_{m=0}^3 x[m]\psi_{0,0}[m] =\frac{1}{2}[1\cdot 1+4\cdot 1-3\cdot (-1)+0\cdot (-1)]=4$$

The two coefficients for $W_1$:

$$W_{\psi}[1,0]=\frac{1}{2}\sum_{m=0}^3 x[m]\psi_{1,0}[m] =\fr... ...ot \sqrt{2}+4\cdot (-\sqrt{2})-3\cdot 0+0\cdot 0]=-1.5\sqrt{2}$$

$$W_{\psi}[1,1]=\frac{1}{2}\sum_{m=0}^3 x[m]\psi_{1,0}[m] =\fr... ...ot 0+4\cdot 0-3\cdot \sqrt{2}+0\cdot (-\sqrt{2})]=-1.5\sqrt{2}$$

In matrix form, we have

$$\left[ \begin{array}{c} 1 4 -1.5\sqrt{2} -1.5\sqrt{2} ... ...t] \left[ \begin{array}{c} 1 4 -3 0 \end{array} \right]$$

Now the function $x[m]$ can be expressed as a linear combination of these basis functions:

$$x[m]=\frac{1}{2}[W_{\varphi}[0,0]\varphi_{0,0}[m]+W_{\psi}[0,... ...[m]+W_{\varphi}[1,1]\psi_{1,1}[m] \;\;\;\;\;\;(m=0,\cdots,3)$$

or in matrix form:

$$\left[ \begin{array}{r} 1 4 -3 0 \end{array} \right] =... ...y}{c} 1 4 -1.5\sqrt{2} -1.5\sqrt{2} \end{array} \right]$$