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I have a continuous function \begin{equation}G_p(s) = {1\over s^2} \end{equation} which I am trying to combine with a zero order hold (with a sampling time of 1 second) to produce a discrete function. I started by combining the plant with the hold to obtain \begin{equation} {(1-e^{-sT})\over s^3} \end{equation} I know \begin{equation} {(1-e^{-sT})} \end{equation} is equal to

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

\begin{equation} {1/s^3} \end{equation} is equal to \begin{equation} {T{^2}z^{-1}(1+z^{-1})\over (1-z^{-1})^3} \end{equation} where T = 1. Next, I multiplied these together to obtain, \begin{equation} {z^{-1}(z-1)(1+z^{-1})\over z(1-z^{-1})^3} \end{equation} which cancels down to \begin{equation} {z^{-1}+z^{-2}\over z^3-2z^{2}+z} \end{equation}

When I convert the continuous function to a discrete function in matlab, I obtain \begin{equation} {0.5z+0.5\over z^{2}-2z+1} \end{equation}

Why am I getting a different result than matlab? Is there something wrong with my method?

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  • \$\begingroup\$ That is the correct pulse TF. How do the results differ? \$\endgroup\$
    – Chu
    May 19, 2017 at 7:41

1 Answer 1

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My Z table doesn't have 1/s3. It has 2/s3. The Z transform for 2/s3 is;

\begin{equation} {T{^2}z(z+1)\over (z-1)^3} \end{equation}

To make this work with 1/s3, you have to take multiply this transform by 1/2. Using this and your equation;

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

gives the Matlab result.

Here is a link to a second table that confirms the transform I have; Laplace to Z transform table with 1/s^3

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  • \$\begingroup\$ The OP has divided by 2, and has the correct pulse TF. \$\endgroup\$
    – Chu
    May 19, 2017 at 22:21
  • \$\begingroup\$ Chu, I've edited my answer with a link to a online table that confirms 1/S^3 is z(z+1)/(z-1)^3. This table has 1/s^3 instead of the one in a my book that has 2/s^3. It's on the 4th row of the second page. Are you saying the OP's z-transform is equivalent to the ones in the tables? I'm missing something because I can't get them to be equal. \$\endgroup\$
    – owg60
    May 19, 2017 at 23:48
  • \$\begingroup\$ The s and z transforms of \$t^2\$ are \$\frac{2}{s^3}\$ and \$\frac{Tz(z+1)}{(z-1)^3}\$ \$\endgroup\$
    – Chu
    May 20, 2017 at 6:57

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