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I've been tasked with showing that the pulse transfer function G(z) of the following plant \begin{equation}G_p(s) = {12\over (s+1)(s+4)} \end{equation} being sampled and held by a zero order hold with a sampling period T = 0.2s is equal to \begin{equation}G(z) = {0.1745z^{-1}+0.1249z^{-2}\over1-1.268z^{-1}+0.3678z^{-2} } \end{equation}

I started by combining the plant with the hold to obtain \begin{equation} {12(1-e^{-sT})\over s(s+4)(s+1)} \end{equation}

I know \begin{equation} {(1-e^{-sT})} \end{equation} is equal to

\begin{equation} {(z-1)/z} \end{equation}

and I split the remaining equation into partial fractions, found the z transform of each, and recombined them to obtain,

\begin{equation}{0.173z^{-1}+0.126z^{-2}\over1-2.27z^{-1}+1.64z^{-2}-0.368z^{-3} } \end{equation} combined with the z transform of the hold, this equals

\begin{equation}G(z) = {0.173-0.047z^{-1}-0.126z^{-2}\over z-2.27+1.64z^{-1}-0.368z^{-2} } \end{equation}

This is not equal to the transfer function I am supposed to obtain- Does anyone see what I am doing wrong?

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2 Answers 2

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You have used the correct expression for the zero-order hold, and the transfer function is indeed: \begin{equation} H(s) = \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)} \end{equation} However, it's unclear how you proceeded from here. It looks like you've simply replaced s by z (z is actually equal to esT) and then split the function into partial fractions.

Alternatively, you can first find h(t) and then find H(z):

\begin{eqnarray} H(s) &=& \dfrac{12(1-e^{-sT})}{s(s+1)(s+4)} \\ &=& (1-e^{-sT}) \left \{ \dfrac{3}{s} - \dfrac{4}{s+1} + \dfrac{1}{s+4} \right \} \end{eqnarray} Taking the Inverse Laplace Transform: \begin{eqnarray} h(t) &=& f(t)u(t) - f(t-T)u(t-T) \end{eqnarray} where \begin{equation} f(t) = 3 - 4e^{-t} + e^{-4t} \end{equation} H(z) can now be found from h(t). You can refer to this link for a table of formulae: http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html

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  • \$\begingroup\$ After splitting the function up into partial fractions and obtaining \begin{eqnarray} H(s) &=& \ (1-e^{-sT}) \left \{ \dfrac{3}{s} - \dfrac{4}{s+1} + \dfrac{1}{s+4} \right \} \end{eqnarray} , I found the z-transform of each part of the equation using a table of z transforms, so \begin{equation} {\dfrac{3}{s} = \dfrac{3}{1-z^{-1}} } \end{equation} \begin{equation} {\dfrac{4}{s+1} = \dfrac{4}{1-e^{-T}z^{-1}} } \end{equation} \begin{equation} {\dfrac{1}{s+4} = \dfrac{1}{1-e^{-4T}z^{-1}} } \end{equation} \begin{equation} {(1-e^{-sT}) = \dfrac{z-1}{z}} \end{equation} \$\endgroup\$
    – Ca01an
    May 14, 2017 at 20:02
  • \$\begingroup\$ I then combined them to find H(z) \$\endgroup\$
    – Ca01an
    May 14, 2017 at 20:03
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There's a common factor of \$\small (z-1)\$ in the numerator and denominator of your final expression. Cancel these and you're left with:

$$\small G(z)=0.17\frac{z(z+0.73)}{z^2-1.29z+0.38}$$

compare with your required target expression: \begin{equation}\small G(z) = {0.1745z^{-1}+0.1249z^{-2}\over1-1.268z^{-1}+0.3678z^{-2} } =0.1745\frac{z(z+0.7158)}{z^2-1.268z+0.3678}\end{equation}

Note: I've worked with positive powers of z (easier to type into root solver!) and rounded all calculations to 2 d.p.

\$Addendum\$

To illustrate, consider the simple process TF: \$\small G_p(s)=\large\frac{1}{s+1}\$, and sample/hold: \$\small G_H(s)=\large\frac{1-e^{-sT}}{s}\$.

The combined TF is: $$\small G(s)=\frac{(1-e^{-sT})}{s}\times \frac{1}{(s+1)}\small=(1-e^{-sT})\frac{1}{s(s+1)}=(1-e^{-sT})\left( \frac{1}{s}-\frac{1}{s+1} \right)$$

Taking z-transforms: $$\small G(z)=\frac{z-1}{z}\left( \frac{z}{z-1}-\frac{z}{z-a} \right)$$ where \$\small a=e^{-T}\$

If the bracket is evaluated first, we have:

$$\small G(z)= \frac{(z-1)}{z}\times\frac{z(1-a)}{(z^2-(1+a)z+a)}=\frac{(1-a)(z-1)}{(z^2-(1+a)z+a)} $$ ...and the \$\small (z-1)\$ factor in the denominator is not apparent.

However, multiplying through by \$\frac{(z-1)}{z}\$ first, gives the simplified form: $$\small G(z)=1-\frac{z-1}{z-a}=\frac{1-a}{z-a} $$

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  • \$\begingroup\$ Do you have any idea how the common factor of (z-1) got in there? Is it just a problem with my math? \$\endgroup\$
    – Ca01an
    May 14, 2017 at 20:05
  • \$\begingroup\$ The ZOH z-TF is \$\frac{z-1}{z}\$, and the z-transform of \$\frac{1}{s}\$ is \$\frac{z}{z-1}\$, so the 's' in the denominator of Gp(s) puts a (z-1) factor in the denominator of G(z) \$\endgroup\$
    – Chu
    May 14, 2017 at 20:21
  • \$\begingroup\$ ...I'll add a note to my answer to illustrate what's happening. \$\endgroup\$
    – Chu
    May 14, 2017 at 20:42
  • \$\begingroup\$ Thank you for illustrating it, this makes things clearer. \$\endgroup\$
    – Ca01an
    May 15, 2017 at 19:00

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