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For the time delay $e^{-sT}$ I shall find the Pade Approximation for $M = 0$ and $N = 1$.

$f(s) = \sum_i^{\infty} a_is^i \approx \dfrac{\sum_{n=0}^{N} b_is^i}{\sum_{m=0}^{M} c_is^i}$

$e^{-sT} = \sum_{i=0}^{\infty}\dfrac{(-sT)^i}{i!}$

Using the Taylor approximation for $e^{-sT}$ I yield

$1 -sT \approx \frac{P_N(s)}{Q_M(s)}$

$1 -sT \approx \frac{b_0 + b_1 s}{c_0}$

with $c_0 = 1$ per definition according to my materials.

This leads me to

$1 -sT \approx b_0 + b_1 s$

$b_0 = 1$

$b_1 = -T$

If I now want to check the step response of $e^{-sT}$ and my approximation MATLAB complains about my function having more zeroes than poles, which is not surprising as I thought I would get my transfer function from $G(s) = \frac{P_N(s)}{Q_M(s)}$.

I'm sure there is a basic misunderstanding of what I am doing here on my side.

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There is no misunderstanding at all. The Padé approximant you found is correct.

The "problem" is that you chose $M$ and $N$ such that the transfer function you get to approximate the delay is improper. Namely, the order of the numerator exceeds the order of the denominator.

In Control Theory, improper systems are not too useful because they cannot be realized in real life. A transfer function with more zeros than poles contains pure differentiators, so the transfer function represents a system that is non-causal (assuming stability, as Matt pointed out in the comments below), and non-causal systems cannot be physically realized.

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  • $\begingroup$ Just a remark: systems with an improper transfer function can be causal, but not causal and stable at the same time. See my answer for a possible causal step response. $\endgroup$ – Matt L. Jan 16 '18 at 13:55
  • $\begingroup$ @MattL. I tend to (mistakenly) ignore unstable systems, so thanks for pointing that out. $\endgroup$ – Tendero Jan 16 '18 at 14:06
  • $\begingroup$ It's all a bit (?) theoretical, I admit. $\endgroup$ – Matt L. Jan 16 '18 at 14:30
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As explained in Tendero's answer everything you've calculated is correct. Matlab is supposed to complain because the system with transfer function

$$H(s)=1-sT\tag{1}$$

has (assuming causality) a step response

$$a(t)=\mathcal{L}^{-1}\left\{\frac{1-sT}{s}\right\}=u(t)-T\delta(t)\tag{2}$$

where $\delta(t)$ is the Dirac delta impulse. The system is unstable and its step response is unbounded/ill-defined at $t=0$.

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  • $\begingroup$ This system is unstable for both causal and noncausal cased, right? That impulse would appear no matter what. Are there any examples where a system can be either causal and unstable OR noncausal and stable, depending on the ROC? $\endgroup$ – Tendero Mar 13 '18 at 14:10
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    $\begingroup$ @Tendero: Yes (to your first question). Examples of causal and unstable, or non-causal and stable systems are easy to find: just put a pole in the right half-plane and you'll have a causal unstable system or a non-causal stable system, e.g. $H(s)=1/(s-1)$ $\endgroup$ – Matt L. Mar 13 '18 at 19:50
  • $\begingroup$ There's something that is not clear to me. Sorry if this question is kind of silly. But given for example the transfer function $H(s)=s^2/(s+1)$, one can find a ROC such that the system is both causal and stable. However, the transfer function is improper. As far as I understand from your comment in my answer above, this is not possible. So where am I making the mistake? $\endgroup$ – Tendero Mar 14 '18 at 2:35

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