# Meaning of the “pole” of a transfer function with time delay

I learned in basic control theory that we can determine the stability of an LTI system by the signs of the real part of the poles of its transfer function $H(s)=\frac{N(s)}{D(s)}$. For a rational system, where both $N(s)$ and $D(s)$ are polynomials is s, it is easily understandable because a positive real part of the pole will cause an exponential increase in time domain, thus being unstable. My question is, if the system has a delay of T seconds somewhere, then $H(s)$ will contain a factor $\exp(-sT)$, sometimes in the denominator. In this case, why can we still determine the stability by looking at the real part of the "poles" of $H(s)$ (points in s plane where $D(s)=0$)?

An example:

$$N(s)=s$$

$$D(s)=as^2+bs+c+d \cdot \exp(-sT)$$

where a, b, c, d, and T are constants.

In this case, we can not do partial fraction expansion as we do for rational transfer functions. Then, how to link the time domain behavior to the roots of $D(s)$?

I believe the property you're referring to is

$$\mathcal{L}\left\{ f(t-\tau) \right\} = e^{-s\tau}\mathcal{L}\left\{f(t)\right\}$$

This means that your transfer function does not really follow that rule, and does not really have a time delay:

$$H(s) = \frac{s}{as^2 + bs + c + de^{-sT}}$$

Solving the inverse Laplace transform for this function will most likely involve numerical methods. What is possible, is to have something like this:

$$H(s) = \frac{s + de^{-sT}}{as^2 + bs + c}$$

Because it can be separated in:

$$H(s) = \frac{s}{as^2 + bs + c} + e^{-sT}\frac{d}{as^2 + bs + c}$$

You can see that the second term also can be separated in partial fractions, like the first term. The poles of the denominator will then say something about the stability of the transfer function.

Stability in general

The formula for the inverse Laplace transform is

$$f(t) = \frac{1}{2\pi j}\lim_{T\to \infty}\int_{\sigma-jT}^{\sigma+jT} F(s)e^{st}ds$$

where $\sigma$ is chosen to be greater than all singularities of $F(s)$ on the complex plane (in our case such that it includes all poles).

This integral is equivalently solved by the Cauchy residue theorem:

$$f(t) = \mathcal{L}^{-1}\left\{ F(s) \right\} = \sum_{all\ poles\ of\ F(s)} Res\left[ F(s)e^{st} \right]$$ Remember, this is general, ie. it always works out for a transfer function with whatever singularities!

So it doesn't matter whether or not the poles come from a polynomial or from a transcendental function. As long as the singularities are all in the left-half plane, their residue will always contain an exponential that decays to 0 rather than infinity. Any singularities in the RHP will always lead to an exponential that explodes.

Appendix

It can be noted that $e^{-sT}$ can be expanded in its Taylor series:

$$e^{z} = \sum_{n=0}^{+\infty} \frac{z^n}{n!}$$

So this means your example transfer function can be written as

\begin{align} H(s) &= \frac{s}{as^2 + bs + c + d\cdot (\sum_{n=0}^{+\infty} \frac{(-sT)^n}{n!})} \\ &= \frac{s}{\sum_{n=0}^{+\infty} A_ns^n} \end{align}

Where $A_0 = c + d$, $A_1 = b - d\cdot T$, $A_2 = a + d\frac{T^2}{2}$, $A_n = d\frac{(-T)^n}{n!}, \forall n>2$.

So this kind of transfer function has "poles" all over the place. This is to illustrate that you definitely can't deal with this like a regular second-order transfer function.

Appendix

The residue of a (simple) pole is

$$Res_{s=a}\left[ H(s) \right] = \lim_{s\to a} (s-a)F(s)$$

For poles with multiplicity $n$:

$$Res_{s=a}\left[ H(s) \right] = \frac{1}{(n-1)!} \lim_{s\to a} \frac{d^{n-1}}{ds^{n-1}}\left( (s-a)^n H(s) \right)$$

• The Taylor expansion in your Appendix is a good point I think, in that it reveals that there will be plenty of poles over the space as you suggested. In my case, the exponential is indeed in the denominator due to the delay being occurring in a feedback loop. Numerical methods are generally required, I agree, but I would like to know why conceptually we can determine the stability by looking at the roots of the denominator. – George C Jun 22 '18 at 11:07
• @GeorgeC I edited my reponse to what I think you meant. – Sven B Jun 22 '18 at 12:45
• George, the roots of the denominator are the poles of the transfer function. And none of these poles must not appear in the right half of the s-plane (RHP). Hence, we watch the locations of all the poles while varying the gain. Usung this method, we can see at which gain the poles move to the RHP (instability). This indicates where the stability limit is. – LvW Jun 22 '18 at 14:01

The expression for a time delay (dead time) in the frequency domain exp(-sT) is a transcendental function. It cannot be modelled with lumped elements and, hence, cannot be directly combined with a transfer function which consists of polynomes in "s". However, for this purpose, it can be approximated by the following expression (Pade approximation of second order):

$$H_d(s)=\frac{1-\frac{sT}{2}+\frac{s^2 T^2}{12}}{1+\frac{sT}{2}+\frac{s^2 T^2}{12}}$$

This is a rational function (allpass) and appears as a factor in the loop gain and appears, of course, in the closed-loop transfer function. Here it can be treated like each other block in the closed loop.

• Yes, I have learned that the Pade is a common technique for approximating delays. However, even if we don't do Pade, we still determine the stability by finding the "poles" of the original transfer function. As an example, the open-loop transfer function is H(s)=KG(s), where G(s)=exp(-sT)/(s(s+1)). The textbook just draws the root locus of the closed-loop pole versus the gain K (no Pade approximation). Why can we do this? – George C Jun 22 '18 at 11:16
• OK - I know what you mean. In my answer I have given the second-order Pade approximation. Of course, the third-order or even the 15th order is more exact. A simulator is able to handle higher orders than n=2 - but I do not know exctly what the simulator does. I suppose the order of approximation is limited to any large value and, finally, the simulator connects all the calculated poles and, thus, is able to produce a root locus. – LvW Jun 22 '18 at 12:34