In the canonical parallel form, the transfer function *H*(*z*) is
expanded into partial fractions. *H*(*z*) is
then realized as a sum of a constant, first-order, and second-order
transfer functions, as shown:

$${H}_{i}\left(z\right)=\frac{u\left(z\right)}{e\left(z\right)}=K+{H}_{1}\left(z\right)+{H}_{2}\left(z\right)+\dots +{H}_{p}\left(z\right).$$

This expansion, where *K* is
a constant and the *H _{i}*(

As in the series canonical form, there is no unique description
for the first-order and second-order transfer function. Because of
the nature of the Sum block, the ordering of the individual filters
doesn't matter. However, because of the constant *K*,
you can choose the first-order and second-order transfer functions
such that their forms are simpler than those for the series cascade
form described in the preceding section. This is done by expanding *H*(*z*) as

$$\begin{array}{c}H\left(z\right)=K+{\displaystyle \sum _{i=1}^{j}{H}_{i}}\left(z\right)+{\displaystyle \sum _{i=j+1}^{p}{H}_{i}}\left(z\right)\\ =K+{\displaystyle \sum _{i=1}^{j}\text{\hspace{0.17em}}\frac{{b}_{i}}{1+{a}_{i}{z}^{-1}}}+{\displaystyle \sum _{i=j+1}^{p}\text{\hspace{0.17em}}\frac{{e}_{i}+{f}_{i}{z}^{-1}}{1+{c}_{i}{z}^{-1}+{d}_{i}{z}^{-2}}}.\end{array}$$

The first-order diagram for *H*(*z*) follows.

The second-order diagram for *H*(*z*) follows.

The parallel form example transfer function is given by

$${H}_{ex}\left(z\right)=5.5556-\frac{3.4639}{1+0.1{z}^{-1}}+\frac{-1.0916+3.0086{z}^{-1}}{1-0.6{z}^{-1}+0.9{z}^{-2}}.$$

The realization of *H _{ex}*(

fxpdemo_parallel_form

at the MATLAB^{®} command line.