Subband Coding

Consider a two-channel filter bank for 1D discrete signal $x[n]$ as shown. The idea is to separate the signal in frequency domain into two subbands (e.g., high-band and low-band) so that each band can be down-sampled for feature extraction, compression, and transmission and then up-sampled and combined for recontruction of the original signal. Perfect reconstruction $x'[n]=x[n]$ is possible if no information is lost during compression and transmision.

Applying the down and up sampling expressions to the subband system, we have

$$X'(z) = \frac{1}{2}G_0(z)[H_0(z)X(z)+H_0(-z)X(-z)] +\frac{1}{2}G_1(z)[H_1(z)X(z)+H_1(-z)X(-z)]$$

$$= \frac{1}{2}[G_0(z)H_0(z)+G_1(z)H_1(z)]X(z) +\frac{1}{2}[G_0(z)H_0(-z)+G_1(z)H_1(-z)]X(-z)$$

For perfect reconstruction, the output signal must be identical to the original signal $x[n]=x'[n]$ or $X(z)=X'(z)$, i.e.,

$$\left\{ \begin{array}{l} G_0(z)H_0(-z)+G_1(z)H_1(-z)=0 \\ G_0(z)H_0(z)+G_1(z)H_1(z)=2 \end{array} \right.$$ (1)

which can be written in matrix form:

$$\left[ \begin{array}{rr} H_0(-z) & H_1(-z) H_0(z) & H_1(z... ...} \right] =\left[ \begin{array}{r} 0 2 \end{array} \right]$$

where

$${\bf H}(z)=\left[ \begin{array}{rr} H_0(-z) & H_1(-z) H_0(z) & H_1(z) \end{array} \right]$$

Solving this for $G_0(z)$ and $G_1(z)$, we get:

$$\left[ \begin{array}{r} G_0(z) G_1(z) \end{array} \right] = {\bf H}(z)^{-1}\left[ \begin{array}{r} 0 2 \end{array} \right... ...H_0(-z) \end{array} \right] \left[ \begin{array}{r} 0 2 \end{array} \right]$$

$$= \frac{2}{det{\bf H}(z)}\;\left[ \begin{array}{r} -H_1(-z) H_0... ...t{\bf H}(-z)}\;\left[ \begin{array}{r} H_1(-z) -H_0(-z) \end{array} \right]$$

where

$$det{\bf H}(z)=H_0(-z)H_1(z)-H_0(z)H_1(-z)=-det{\bf H}(-z)$$

is the determinant of matrix ${\bf H}(z)$. From the above, we get

$$G_0(z)=\frac{2}{det{\bf H}(-z)}H_1(-z),\;\;\;\;\; G_1(z)=-\frac{2}{det{\bf H}(-z)}H_0(-z)$$

Replacing $z$ by $-z$ in the second equation above, we get

$$G_1(-z)=-\frac{2}{det{\bf H}(z)}H_0(z)=\frac{2}{det{\bf H}(-z)}H_0(z)$$

From the above two equations we can also get

$$G_1(-z)H_1(-z)=G_0(z)H_0(z),\;\;\;\mbox{or}\;\;\;G_1(z)H_1(z)=G_0(-z)H_0(-z)$$ (2)

Substituting Equaitons (2) into Equatioon (1), we get

$$G_0(z)H_0(z)+G_0(-z)H_0(-z)=2$$

Due to the following Z-transform: property

$$X(-z)=\sum_n x[n](-z)^{-n}=\sum_n x[n](-1)^n z^{-n}={\cal Z}(x[n](-1)^n)$$

and convolution theorem, the time domain expression of the above equation is

$$\sum_k g_0[k]h_0[n-k]+\sum_k (-1)^k g_0[k](-1)^{n-k}h_0[n-k] =\sum_k g_0[k]h_0[n-k]+(-1)^n \sum_k g_0[k] h_0[n-k]=2\delta[n]$$

The two terms on the left side of the equation cancel each other whenever $n$ is odd, and the summation only has the even terms:

$$\sum_k g_0[k]h_0[2n-k]=(g_0[n],h_0[2n-k])=\delta[n]$$

where $(a[n],b[n])=\sum_n a[n]^* b[n]$ is the inner product of the two signals $a[n]$ and $b[n]$.

In addition to this relation between $G_0$ and $H_0$, from Eqautions (1) and (2), we can also get three additional relations for $G_1$ and $H_1$, $G_1$ and $H_1$, $G_1$ and $H_1$. These four relations can be summarized as:

$$\left\{ \begin{array}{l} G_0(z)H_0(z)+G_0(-z)H_0(-z)=2 \\ ... ...0(-z)=0 \\ G_0(z)H_1(z)+G_0(-z)H_1(-z)=0 \end{array} \right.$$

and equaivalently in time domain:

$$\left\{ \begin{array}{l} \sum_k g_0[k]h_0[2n-k]=(g_0[n],h_0[... ...sum_k g_0[k]h_1[2n-k]=(g_0[n],h_1[2n-k])=0 \end{array} \right.$$

These four relations can be further summarized as

$$(h_i[2n-k],g_j[k])=\delta[i-j]\delta[n]$$

Filter banks satisfying this condition are called biorthogonal.