# Approximation of z-transform transfer function

It is common in control theory to approximate a transfer function neglecting the high order terms, in example, a transfer function with two poles: $$P=\frac{1}{(as+1)(bs+1)}=\frac{1}{abs^2+(a+b)s+1}$$ approximated by this $$P_1=\frac{1}{(a+b)s+1}$$

with one pole in $$-1/(a+b)$$ .

Now consider the corresponding discrete z-tranform of P applying Zero-order-hold:

$$P_z=(1-z^{-1}) \mathcal{Z}\left(\mathcal{L}^{-1}\left(\frac{P}{s}\right)_{t=kT}\right)= \frac{(e^{-b_i T} b_i - e^{-a_i T}a_i + e^{-a_i T}e^{-b_i T}(a_i - b_i))z^{-2}+ (a_i - b_i + e^{-a_i T} b_i - a_ie^{-b_i T})z^{-1}}{e^{-a_i T}e^{-b_i T}(a_i - b_i)z^{-2}+(e^{-a_i T}+e^{-b_i T})(b_i-a_i)z^{-1} +a_i - b_i}$$

with $$a_i=\frac{1}{a} \ \ b_i=\frac{1}{b}$$. The result is again a fractional polinomial. My question is, could there be a method to neglect the high order terms in a similar way of that in continuos to obtain a lower order z-transform tf like this:

$$P_{z1}=\frac {z^{-1}}{d_0+d_1z^{-1} }$$

(a z-trasform of a one pole process). Thanks.

• Why not discretize $P_1$ instead of $P$? Also note that $P_1$ is only a reasonable approximation of $P$ if $a$ and $b$ are real. For computing $P_{z1}$ you could try balanced truncation method. Commented Jun 11 at 18:20