Digital Signal Processing Using MATLAB and Wavelets
By Michael Weeks

Chapter 9.10 - Biorthogonal Wavelets

When talking about wavelets, the transform is classified as either orthogonal, or biorthogonal. The previously discussed wavelet transforms (Haar and Daubechies) were both orthogonal. These can be normalized as well, such as by using $1/\sqrt{2}$ and $-1/\sqrt{2}$ instead of $1/2$ and $-1/2$ for the Haar coefficients. If you assume that we just take the square root of $1/2$, you have missed a step. Instead, multiply by $\sqrt{2}$, i.e., $\sqrt{2} (1/2) = \sqrt{2}/2 = 1/\sqrt{2}$. When it is both orthogonal and normal, we call it orthonormal.

 

The following structure (Figure 9.23) shows an example biorthogonal wavelet transform, and its inverse, for a single octave.

Figure 9.23: Biorthogonal wavelet transform.
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Proceeding as before, we start with definitions for $w[n]$ and $z[n]$:

\begin{displaymath}w[n] = a x[n] + b x[n-1] + c x[n-2] + b x[n-3] + a x[n-4] \end{displaymath}

\begin{displaymath}w[n-k] = a x[n-k] + b x[n-k-1] + c x[n-k-2] + b x[n-k-3] + a x[n-k-4] \end{displaymath}

\begin{displaymath}z[n] = d x[n] - f x[n-1] + d x[n-2] \end{displaymath}

\begin{displaymath}z[n-k] = d x[n-k] - f x[n-k-1] + d x[n-k-2] . \end{displaymath}

Now we can define $w_f[n]$ and $z_f[n]$ in terms of $w[n]$ and $z[n]$:

\begin{displaymath}w_f[n] = d w[n] + f w[n-1] + d w[n-2] \end{displaymath}

\begin{displaymath}z_f[n] = -a z[n] + b z[n-1] - c z[n-2] + b z[n-3] - a z[n-4] . \end{displaymath}

Next, we replace the $z[n]$'s and $w[n]$'s with their definitions:

\begin{eqnarray*} w_f[n] &=& ad x[n] + bd x[n-1] + cd x[n-2] + bd x[n-3] + ad x... ...x[n-4] + bd x[n-5] \&-& ad x[n-4] + af x[n-5] - ad x[n-6] . \end{eqnarray*}

These are added together to give us $y[n]$, the output:

\begin{displaymath}y[n] = w_f[n] + z_f[n] . \end{displaymath}

Removing the terms that cancel, and simplifying, gives us:

\begin{displaymath}y[n] = 2(bd+af) x[n-1] + (4bd+2cf) x[n-3] + 2(af+bd) x[n-5] . \end{displaymath}

To make $y[n]$ a delayed version of $x[n]$, we need the above expression to be in terms of only one index for $x$, that is, $x[n-3]$. This can be accomplished if we set $2(bd+af) = 0$, or $bd=-af$. Another condition is that we do not want to have to divide the output by a constant. In this case, that means that the $(4bd+2cf)$ term should be equal to 1. However, we are likely to use this transform with down-samplers and up-samplers, which means that we would want the term $(2bd+cf)$ to be equal to 1.

Here, we repeat the above analysis with the added complexity of down-samplers and up-samplers. Figure  9.24 shows this case.

Figure 9.24: Biorthogonal wavelet transform.
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The definitions for $w[n]$ and $z[n]$ are the same as above. The signals after the down-samplers are $w_d[n]$ and $z_d[n]$. After the up-samplers, the signals are labeled $w_u[n]$ and $z_u[n]$.

\begin{displaymath}w_d[n] = w[n], \;\;\; n\; \mathrm{is\;even} \end{displaymath}

\begin{displaymath}w_d[n], \; \; \mathrm{is \; undefined,} \;\;\; n \; \mathrm{is \; odd} \end{displaymath}

\begin{displaymath}w_u[n] = w_d[n] = w[n], \;\;\; n\; \mathrm{is\;even} \end{displaymath}

\begin{displaymath}w_u[n] = 0, \;\;\; n\; \mathrm{is \; odd} \end{displaymath}

\begin{displaymath}z_d[n] = z[n], \;\;\; n\; \mathrm{is\;even} \end{displaymath}

\begin{displaymath}z_d[n], \; \;\mathrm{is \; undefined,} \;\;\; n \; \mathrm{is \; odd} \end{displaymath}

\begin{displaymath}z_u[n] = z_d[n] = z[n], \;\;\; n\; \mathrm{is\;even} \end{displaymath}

\begin{displaymath}z_u[n] = 0, \;\;\; n\; \mathrm{is \; odd} \end{displaymath}

\begin{displaymath}w_f[n] = d w_u[n] + f w_u[n-1] + d w_u[n-2] \end{displaymath}

\begin{displaymath}z_f[n] = -a z_u[n] + b z_u[n-1] - c z_u[n-2] + b z_u[n-3] - a z_u[n-4] \end{displaymath}

\begin{displaymath}y[n] = w_f[n] + z_f[n] \end{displaymath}

Next, we replace the $z_u[n]$'s and $w_u[n]$'s with their definitions.

\begin{displaymath}w_f[n] = d w_u[n] + d w_u[n-2], \;\; n\; \mathrm{is\;even} \end{displaymath}

\begin{displaymath}z_f[n] = -a z_u[n] - c z_u[n-2] - a z_u[n-4], \;\; n\; \mathrm{is\;even} \end{displaymath}

\begin{displaymath}w_f[n] = f w_u[n-1], \;\; n\; \mathrm{is \; odd} \end{displaymath}

\begin{displaymath}z_f[n] = b z_u[n-1] + b z_u[n-3], \;\; n \; \mathrm{is \; odd} \end{displaymath}

\begin{displaymath}y[n] = d w[n] + d w[n-2] - a z[n] - c z[n-2] - a z[n-4], \;\; n \; \mathrm{is\;even} \end{displaymath}

\begin{displaymath}y[n] = f w[n-1] + b z[n-1] + b z[n-3], \;\; n\; \mathrm{is \; odd} \end{displaymath}

Multiplying out:

\begin{displaymath} \begin{array}{cccccc} y[n] = & d(a x[n] & +\; b x[n-1] & +\... ...x[n-4] & +\; d x[n-5]), & n \; \mathrm{is\;odd.} & \end{array} \end{displaymath}

Simplifying:


\begin{displaymath} \begin{array}{cccccc} y[n] =& ad x[n] &+\; bd x[n-1] &+\; c... ... x[n-4] &+\; bd x[n-5], & n \; \mathrm{is\;odd.} & \end{array} \end{displaymath}

Combining them together, and removing terms that cancel, gives us:

\begin{displaymath} \begin{array}{ccccc} y[n] = & bd x[n-1] &+\; bd x[n-3] & & ... ...bd x[n-3] &+\; bd x[n-5], & n \; \mathrm{is\;odd.} \end{array} \end{displaymath}

Whether $n$ is even or odd,

\begin{displaymath}y[n] = (bd+af) x[n-1] + (2bd + cf) x[n-3] + (bd+af) x[n-5] . \end{displaymath}

We see that this is the same equation as the one for biorthogonal wavelets without down/up-samplers, except without a factor of 2 throughout. This again confirms the notion that the down/up-samplers remove redundant data, making the transform efficient.

© 2026 Infinity Science Press
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