# Calculating pulse transfer function

I've been tasked with finding the pulse transfer function G(z) of the combination of the following equation $$G_p(s) = {10\over (s+1)(s+2)(s+3)}$$ and a zero order hold.

I started by combining the hold with the equation to obtain

$${10(1-e^{-sT})\over s(s+1)(s+2)(s+3)}$$

and then splitting this up to the equations

\begin{align} {1\over s(s+1)} \tag1\\ {1\over (s+2)(s+3)}\tag2 \\ {10(1-e^{-sT})}\tag3 \end{align}

finding the z transform of each one, and finally combining them together. I'm not sure if I've went about this in the right way, the answer I obtained seems very complicated and I can't find any similar examples. Can anyone who knows how to do this tell me if I'm going about this correctly and point me in the right direction?

• hm, what's your plan on combining the z-transforms? You can't just multiply them; $\mathcal Z\left\{a\cdot b\right\}\ne \mathcal Z\left\{a\right\} \cdot \mathcal Z\left\{b\right\}$! Apr 24, 2017 at 13:03
• To follow up on what @MarcusMüller is saying. There are properties of the Z-Transform that you have to consider.
– user103380
Apr 24, 2017 at 13:36

First z-transform $\small(1-e^{-sT})$ $\rightarrow$ $\small(1-z^{-1})=\large\frac{z-1}{z}$. Then write the remaining Laplace expression in partial fractions: $$\small 10\left(\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+2}+\frac{D}{s+3 }\right)$$
Z-transform the bracket term-by-term using standard Laplace/z transform tables, and combine with $\large\frac{z-1}{z}$.
• By combine, do you just mean multiply? Would I be correct in thinking the final answer looks something like this? $$\small {10z-1 \over z}\left(\frac{1}{6(1-z^{-1})}-\frac{1}{2(1-e^{-T}z^{-1})}+ \frac{1}{2(1-e^{-2T}z^{-1})}-\frac{1}{6(1-‌​e^{-3T}z^{-1}) }\right)$$ Apr 24, 2017 at 20:23
• Yes, it's fine. But there's a typo: should be $\large \frac{10(z-1)}{z}$. Also it's better to remove negative powers of $z$.