Factoring Higher Order Polynomials: A Practical Method for Control System Design

In summary, the conversation is discussing how to determine the parameters of a PI controller in order to achieve specific specifications for a second order LTI system. The method being considered is decomposing the transfer function into a first order system and a second order system and comparing it to a generic analysis of a second order system. Another approach involves calculating the parameters for a second order system without zeros and then relating it to the transfer function denominator. The speaker also mentions a helpful method for factoring polynomials.
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Homework Statement


Determine the parameters of the PI controller such that two of the closed-loop poles of the transfer function Gclr(s) correspond to the poles of a second order LTI system with the following specifications: i) overshoot S% = 25%; and ii) settling time ts(5%) = 120 s.

Homework Equations


3. The Attempt at a Solution [/B]

I have the following system:

$$G(s)=\frac{2K(s+z)}{1000s^3+110s^2+(1+2K)s+2Kz}$$

Now I need to have this system meeting certain specifications (not relevant to what I'm asking now) such as a certain value of overshoot and of settling time. And I have to determine z and K for that.

Anyway for that my guess is that I must decompose my transfer function into the sum of a first order system and a second order system (without zeros I suppose) and compare the second order parcel to the generic analysis of second order system. In theory I know that's what I have to do, but I'm a bit stuck. How do I decompose my denominator?

My first attempt consisted of making z=0. There I would have a transfer functios of

$$G(s)=\frac{2Ks}{1000s^3+110s^2+(1+2K)s}$$

And by cancelling the zero at the origin (it's valid to do that right?)

$$G(s)=\frac{2K}{1000s^2+110s+(1+2K)}$$

And since the static gain of this system is $$\frac{2K}{1+2K}$$ we must have, after some algebraic manipulation:

$$ G(s)=\frac{2K}{(1+2K)} \frac{0.001}{\frac{s^2}{1+2K}+ \frac{0.11s}{1+2K} + 0.001} $$

Now the second term of this product corresponds to the standard second order system without zeros. I should now apply know formulas and obtain the desired overshoot and settling time.

Is this correct?
But now can't z be different than zero? How do I deal with that in that case?

Then I made another try where I calculate the parameters of a second order system without zeros has to be to meet those specifications. I came up with a denominator of the system $$ s^2 + 0.04999 s + 0.00383$$. and two complex conjugate poles $$ s = - 0.025 \pm j0.05663$$. Now how can I relate this denominator to the transfer function denominator and therefore determine my parameters?

Thanks in advance!
 
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Related to Factoring Higher Order Polynomials: A Practical Method for Control System Design

What is a third order system?

A third order system refers to a mathematical model that describes the behavior and response of a system with three independent state variables.

How is a third order system different from a first or second order system?

A first order system has only one independent state variable, while a second order system has two. A third order system has three, making it more complex and capable of modeling more complex systems.

What are some real-world examples of third order systems?

Examples of third order systems include electrical circuits with three energy storage elements, such as inductors, capacitors, and resistors, as well as mechanical systems with three degrees of freedom, such as a three-mass-spring system.

How is a third order system typically modeled?

A third order system can be modeled using differential equations, transfer functions, or state-space representations, depending on the specific system and the desired level of detail and accuracy.

What are the advantages and limitations of modeling a third order system?

The advantages of modeling a third order system include being able to accurately represent complex systems and predict their behavior. However, the limitations include the increased complexity and difficulty in analyzing and solving the resulting equations or functions.

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