## Parallel Form

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 Hi(z) are the first- and second-order transfer functions, follows.

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

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

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 Hex(z) using fixed-point Simulink® blocks is shown in the following figure. You can display this model by typing

`fxpdemo_parallel_form`

at the MATLAB® command line.