# Significance of “virtual” zeroes/poles introduced by the Pade expansion of $e^{sT}$

Let $s$ be a complex parameter and $T$ a real. The function $e^{sT}$ is entire. It can also be expanded in a Pade approximant via

$$e^{sT} = \frac{e^{sT/2}}{e^{-sT/2}} = \frac{1+ sT/2 + \frac{(sT/2)^2}{2!}+ \cdots}{1 - sT/2 + \frac{(sT/2)^2}{2!} - \cdots},$$ which upon truncation to first order yields the bilinear mapping

$$\frac{2+sT}{2-sT}.$$

Approximations of the exponential function by bilinear mappings are found frequently in digital signal processing, linearized analysis of delay systems, and other applications in systems theory, where they are known as "Tustin's method" or sometimes "Tustin's approximation".

In all of these applications, poles and zeros introduced into a system transfer function are important factors which determine system stability, transient characteristics, and frequency response. The character of a system, say $e^{sT}H(s)$, where $H(s)$ is an arbitrary rational transfer function, may in fact be changed considerably if the exponential factor is approximated by Tustin's method, despite the fact that numerically the two are close when the latter is defined and nonzero, since the introduction of the pole-zero combination affects relative stability margins and the shape of the system's frequency response.

Clearly the pole-zero combination is "virtual" in the sense that they do not arise from physical considerations, but it is unclear to me what their exact significance is.

• Say a rational function has order $n$ if it has at most $m$ zeros and $n-m$ poles, then $\frac{1+ sT/2 + \cdots+\frac{(sT/2)^{n}}{n!}}{1- sT/2 + \cdots+\frac{(-sT/2)^{n}}{n!}}$ is the rational function of order $2n$ approximating the best $e^{sT}$ around the origin. – reuns Apr 17 '17 at 21:34
• Actually, the closed loop system has an infinite number of poles and zeros when there is delay in the open loop system, so the pole-zero combination is not "virtual". Tustin's approximation can be used to deal with the infinite number of poles/zeros with a finite approximation. Clearly, this cannot work sometimes. – obareey Apr 18 '17 at 5:27
• @obareey I am concerned only insofar as the system is regarded as a meromorphic function. In this case, as outlined above, no poles can arise from the multiplication by $e^{sT}$, which is entire. – SZN Apr 18 '17 at 5:33
• @ALB I see your point now. See my answer. – obareey Apr 18 '17 at 6:33
• @claude leibovici Have you noticed this question ? – Jean Marie Jun 6 '17 at 23:27

In my comment I mistakenly assumed that your system is like $$\dot{x}(t) = a x(t-T) + u(t)$$ which has an infinite number of poles that can be described with Lambert-W function. But instead your system is like $$\dot{x}(t) = a x(t) + b u(t-T)$$ which becomes like $$\dot{x}(t) = a x(t) + b k x(t-T)$$ for the closed loop system, which also has an infinite number of poles.
But you are right, for the open loop system, there cannot be additional zeros/poles when there is input or output delay. Assuming a causal transfer function (hence $e^{-sT}$ multiplier) the delay effect can be obtained with a non-minimum phase system, because it oscillates around zero (depending on the order) before it fades out and non-delay part takes over. See the figure in page 31 of http://engineering.nyu.edu/mechatronics/Control_Lab/Criag/Craig_RPI/2002/Week2/First-Order_Process_Time_Delay_2002.pdf to see what I mean.
• @Pait I've been thinking about this for some time and it is genuinly hard to grasp. I think "virtual" zeros may not be virtual after all. Because $e^{-sT}$ has infinitely many zeros at $\Re(s)=\infty$ and infinitely many poles at $\Re(s)=-\infty$, whatever that means. The approximated system's zeros and poles move towards infinity as the order increases. – obareey Apr 30 '17 at 18:15
• Also, the frequency response of $e^{-sT}$ and its approximation gives some interesting results. They have both very similar frequency responses, as expected. Their gain is 0dB at all frequencies. Interestengly, phase equals to $T$ at $\omega=1$ for both. But at other frequencies phase is different, yet $e^{-sT}$ delays the system exactly $T$ seconds for all frequency inputs. But its approximation fails to do so at high frequencies. – obareey Apr 30 '17 at 18:28