Two-Channel Critically Sampled Filter Banks

Let's begin with a simple two-channel case, with lowpass analysis filter $H_0(z)$ , highpass analysis filter $H_1(z)$ , lowpass synthesis filter $F_0(z)$ , and highpass synthesis filter $F_1(z)$ :

The outputs of the two analysis filters are then

$$X_k(z) = H_k(z)X(z), \quad k=0,1$$

After downsampling, the signals become

$$V_k(z) = \frac{1}{2}\left[X_k(z^{1/2}) + X_k(-z^{1/2})\right], \; k=0,1$$

After upsampling, the signals become

$$Y_k(z) = V_k(z^2) = \frac{1}{2}[X_k(z) + X_k(-z)]$$

$$= \frac{1}{2}[H_k(z)X(z) + H_k(-z)X(-z)],\; k=0,1$$

After substitutions and rearranging, the output $\hat{x}$ is a filtered replica plus an aliasing term:

$$\hat{X}(z) = \frac{1}{2}[H_0(z)F_0(z) + H_1(z)F_1(z)]X(z)$$

$$+ \frac{1}{2}[H_0(-z)F_0(z) + H_1(-z)F_1(z)]X(-z)$$

$$\hbox{(Filter Bank Reconstruction)}$$ (1)

We require the second term (the aliasing term) to be zero for perfect reconstruction. This is arranged if we set

$$F_0(z) = \quad\! H_1(-z)$$

$$F_1(z) = -H_0(-z)$$

$$\hbox{(Aliasing Cancellation Constraints)} \protect$$ (2)

Thus,

• The synthesis lowpass filter $F_0(z)$ is the rotation by $\pi$ of the analysis highpass filter $H_1(z)$ on the unit circle. If $H_1(z)$ is highpass, cutting off at $\omega=\pi/2$ , then $F_0(z)$ will be lowpass, cutting off at $\pi/2$ .
• The synthesis highpass filter $F_1(z)$ is the negative of the $\pi$ -rotation of the analysis lowpass filter $H_0(z)$ .

Note that aliasing is completely canceled by this choice of synthesis filters $F_0,F_1$ , for any choice of analysis filters $H_0,H_1$ .

For perfect reconstruction, we additionally need

$$c = H_0(z)F_0(z) + H_1(z)F_1(z)$$

$$\hbox{(Filtering Cancellation Constraint)} \protect$$ (3)

where $c=Ae^{-j\omega D}$ is any constant $A>0$ times a linear-phase term corresponding to $D$ samples of delay.

Choosing $F_0$ and $F_1$ to cancel aliasing,

$$c = H_0(z)H_1(-z) - H_1(z)H_0(-z)$$

$$\hbox{(Filtering and Aliasing Cancellation)} \protect$$ (4)

Perfect reconstruction thus also imposes a constraint on the analysis filters, which is of course true for any band-splitting filter bank.

Let ${\tilde H}$ denote $H(-z)$ . Then both constraints can be expressed in matrix form as

$$\left[\begin{array}{cc} H_0 & H_1 \\ [2pt] {\tilde H}_0 & {\tilde H}_1 \end{array}\right]\left[\begin{array}{c} F_0 \\ [2pt] F_1 \end{array}\right]=\left[\begin{array}{c} c \\ [2pt] 0 \end{array}\right]$$