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i have this open loop system , and i've been asked to find out the response $C(kT)$ due to a unit step input. enter image description here I am able to find the transfer function without the delay unit i.e $$\frac{C(z)}{R(z)}=\left(1-z^{-1}\right)\cdot \mathcal{Z}\left\{\frac{1}{s(s+a)}\right\}$$ i can find this ZT easily and find $C(kT)$

But how to deal with the exponential term i.e $e^{-s}$ ? then my pulse transfer function becomes $$\frac{C(z)}{R(z)}=\left(1-z^{-1}\right)\cdot \mathcal{Z}\left\{\frac{e^{-s}}{s(s+a)}\right\}$$ Can i also consider it as a delay ? and take it out of the braces ? i.e $$\frac{C(z)}{R(z)}=\left(1-z^{-1}\right)\cdot z^{-1} \cdot \mathcal{Z}\left\{\frac{1}{s(s+a)}\right\}$$

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  • $\begingroup$ The exponential term is just a delay of 1 sample. However, there are a few things in your question that are unclear. What is $R(z)$? When you refer to "transfer function", the transfer function of which (part of the) system do you mean? And finally, what do you mean by $\mathcal{Z}\{\cdot\}$? This symbol usually refers to the Z-transform, but you can't take the Z-transform of a function (a Laplace transform in your case). You can only take the Z-transform of a sequence, like $C(z)$ is the Z-transform of the sequence $c[k]$. And finally, where does the term $1-z^{-1}$ come from? $\endgroup$ – Matt L. Sep 10 '17 at 11:23
  • $\begingroup$ yeah $R(z)$ means nothing but Z transform of the input signal , by transfer function i meant the usual pulse transfer function which is ratio of $C*(kT)$ and $R*(kT)$ , yes Z means Z transform and the argument inside Z will be first converted to time domain then i'll take ZT ( i've skipped this detail , you can find this approach in any book , i follow ogata discrete control), and the term $(1-z^{-1})$ came from the ZT of the ZOH's numerator i.e $\mathcal{Z}\left(1-e^{-sT}\right)=1-z^{-1}$ @MattL. $\endgroup$ – Zeno San Sep 10 '17 at 12:07
  • $\begingroup$ This answer to a related question actually has everything you need. $\endgroup$ – Matt L. Sep 10 '17 at 15:57
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    $\begingroup$ This video should also give some answers. $\endgroup$ – fibonatic Sep 11 '17 at 5:11
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In my opinion, the most straightforward way to solve this problem is to consider the time domain signal as it passes through the signal chain. A unit step at the input when sampled and put through a ZOH becomes just another continuous-time unit step. Consequently, the Laplace transform of the continuous-time signal just before the sampler at the output is given by

$$C(s)=\frac{e^{-s}}{s(s+a)}=\frac{e^{-s}}{a}\left[\frac{1}{s}-\frac{1}{s+a}\right]\tag{1}$$

which corresponds to the time domain signal

$$c(t)=\frac{1}{a}\left[1-e^{-a(t-1)}\right]u(t-1)\tag{2}$$

Sampling $c(t)$ at $t=kT$ gives

$$c(kT)=\frac{1}{a}\left[1-e^{-a(kT-1)}\right]u(kT-1)\tag{3}$$

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