Electronic – Calculating the pulse transfer function

controltransfer functionz transform

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?

Best Answer

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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