From what I have learnt from control systems classes, the dominant poles of a system are those that can be used to analyze a higher order system in terms of the formulas of a 2nd order system. I took the following system as an example:

$$G(s) = \frac{s^2 + 2s + 3}{10s^3 + 0.1s^2 + 23.3s + 30}$$

and tried verifying some results using MATLAB for a damping ratio of 0.48:

G = tf([1 2 3], [10 0.1 23.3 30]); % open loop TF

As seen from the root locus plot, I get the gain for the damping ratio as K = 58.3 and an expected peak overshoot of 17.9%.

enter image description here

K = 58.3;
H = 1;
T = feedback(K*G, H); % closed loop TF

enter image description here

From the step response I tried getting the peak overshoot as:

$$ \frac{1.13 - 0.852}{0.852} \times 100 \approx 32.63 \% $$

which is nowhere close to the overshoot predicted from the second order system (= 17.9%). I tried comparing other parameters like peak time, rise time, etc of this system with the predicted 2nd order values and none of them matched. The only parameter that did match (exactly) was the settling time, with a value of 3.45s.

I understand that I am dealing with a higher order system here so the characteristics must be different so how good is this approximation really? It has been used throughout my book especially in the design part. Any help is appreciated, thanks!


1 Answer 1


It depends entirely on how close the less dominant poles are to the dominant poles. A simple way to understand what is happening is consider poles on the real negative axis for continuous time systems: each pole at location $x$ has an impulse response given by $e^{-xt}$ and with that we see how poles further away from the $j\omega$ axis will decay to insignificant amounts much more quickly. This also applies generally to poles anywhere in the negative half plane in the s-plane (or equivalently inner circle for discrete time systems in the z-plane): the further the pole is from the $j\omega$ axis the faster it decays. This is why poles closest to the $j\omega$ axis are dominant— relative to other poles they take a long time to decay and no longer contribute to the overall impulse response. So for example while poles 100x further from the $j\omega$ axis can be neglected in observation of the final impulse response, poles only 2x to 3x away will still have a significant contribution. Specific examples of the effect of a single pole are shown in the graphic below.

example poles

With two poles the effect is the same; if the poles are complex (and conjugate symmetric as is the case with real systems) we get an additional oscillation but the rate of decay is similarly based solely on the distance from the $j\omega$ axis, as demonstrated in the graphic below showing different examples where all have a distance of one from the $j\omega$ axis and we see the same rate of decay for all cases:

2nd order all pole

Further and specific to the OP's case, there are two dominant zeros closer to the $j\omega$ axis than the two dominant poles, and with the third pole at approximately -3.5 which is not significantly larger than the real portion of the dominant poles which is at approximately -1.16. See the pole zero map below for the closed loop system given by the OP with K = 58.3; the presence of the two zeros would be the most significant factor causing the step response to deviate from a second order system with just two poles which the overshoot figures are providing:

pole zero map

The presence of the zeros will significantly alter the resulting impulse response, and zeros closer to the $j\omega$ axis than the poles will impose a sharp leading edge to the response as we see formed with such a structure in lead networks (zero and pole with zero closer to the $j\omega$ axis) in contrast to lag networks (zero and pole with pole closer to the $j\omega$ axis). Below are example graphics I have showing the effect of a single zero on the two real pole example given above:

zeros in transfer function

If we eliminate the zeros and the weaker pole, we do get the step response with the predicted 17.9% overshoot as shown in the plot below:

only dominant poles

If we only eliminate the two zeros and leave the third weaker pole, we get a significantly less overshoot of approximately 11.5%:

3 poles

The lesson is that the dominant pole approximations only hold when we truly have two dominant poles, with any other poles (and zeros!) significantly (as in 10x or more) further away from the $j\omega$ axis. The more accurate we desire the response to be, the further away the weaker poles and zeros need to be.

Related to this is a chart I have showing the general considerations for pole placement when working with a root locus plot such as what the OP has presented. With the root locus we see the trajectory of the poles, starting with the open loop poles and zeros, as we vary the loop gain in the closed loop system. I find it interesting how the math works out that the trajectories will all start with the open loop poles and end at the open loop zeros which never move (each pole has a "companion" zero, if we don't see the zero, it's not that it doesn't exist, that particular zero is at infinity, and sure enough in the OP's case we see one of the root locus paths going off to infinity, and in my example below, all three paths go off to infinity in different directions):

root locus


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