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

I started by combining the hold with the equation to obtain

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

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?

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    \$\begingroup\$ 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\}\$! \$\endgroup\$ Apr 24, 2017 at 13:03
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    \$\begingroup\$ To follow up on what @MarcusMüller is saying. There are properties of the Z-Transform that you have to consider. \$\endgroup\$
    – user103380
    Apr 24, 2017 at 13:36

1 Answer 1

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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}\$.

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  • \$\begingroup\$ By combine, do you just mean multiply? Would I be correct in thinking the final answer looks something like this? \begin{equation} \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) \end{equation} \$\endgroup\$
    – Ca01an
    Apr 24, 2017 at 20:23
  • \$\begingroup\$ 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\$. \$\endgroup\$
    – Chu
    Apr 26, 2017 at 11:01

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