In the canonical series cascade form, the transfer
function *H*(*z*) is
written as a product of first-order and second-order transfer functions:

$${H}_{i}\left(z\right)=\frac{u\left(z\right)}{e\left(z\right)}={H}_{1}\left(z\right)\cdot {H}_{2}\left(z\right)\cdot {H}_{3}\left(z\right)\dots {H}_{p}\left(z\right).$$

This equation yields the canonical series cascade form.

Factoring *H*(*z*) into *H _{i}*(

For example, one factorization of *H*(*z*) might
be

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

You must also take into consideration that the ordering of the
individual *H _{i}*(

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

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

The series cascade form example transfer function is given by

$${H}_{ex}\left(z\right)=\frac{\left(1+0.5{z}^{-1}\right)\left(1+1.7{z}^{-1}+{z}^{-2}\right)}{\left(1+0.1{z}^{-1}\right)\left(1-0.6{z}^{-1}+0.9{z}^{-2}\right)}.$$

The realization of *H _{ex}*(

fxpdemo_series_cascade_form

at the MATLAB^{®} command line.