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Drawing a root locus and finding the dominant roots

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Please see my attached questions. I actually only need no. 24 and 25.

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We can write this function as the ratio of two polynomials in , a numerator polynomial and a denominator polynomial , thus

On examination we can equate

In a negative feedback system the closed loop transfer function is given in general by

As for the feedback function we have (for a unity feedback system) and thus we get a closed loop transfer function as given by {1} below

{1}

Putting in values for the specific case and we obtain {2}

{2}

The denominator of which we define as the "characteristic equation", call it , for . Thus we can write as {3}

{3}

- Finding the "finite" Poles and zeros of the open loop transfer function

Hence looking at these polynomials separately when either the numerator or denominator equate to zero to find the finite (when ) poles and zeros of the open loop transfer function.

- From the numerator:

has one finite zero at obtained by equating the numerator to zero, thus

We thus denote the number of zeros by a parameter

- From the denominator:

has roots (finite poles) when (duplicate pole) and when

We thus denote the number of zeros by a parameter

Using Evans rules we have

Asymptotes to the root loci and thus poles that go to infinity as

Therefore for the particular case we have

asymptotes

- The intersection of the asymptotes on the real axis in the plane can further be deduced from another of Evans rules, namely

For the particular case the summation of poles is

The summation of zeros is

Thus we get an intersect point for asymptotes of

- The angles that the asymptotes make with one another are determined from

(

When we get

When we get

- We get the intersection of the root locus plot with the real axis in the plane when

Using the quotient rule for differentiation and the notation ,

Thus we can determine the intersect of the root locus with the real axis in the plane when we have the condition below

Looking at out numerator and denominator polynomials again we have

So

So

Thus on substitution we solve for

or

(double root)

We thus get an asymptotic intersection at

- Root Loci plots

The root loci plots are shown on the plot below. We have one root loci on the real axis to the left of an odd number of poles and zeros between the two poles at , this has zero extension as one cannot draw an extended line between 0 and 0 of zero length. We have one root locus on the real axis extending between and As defined by the rules the root locus must extend away from an odd ...

Solution Summary

The expert draws a root locus and finding the dominant roots/

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See Also This Related BrainMass Solution

Root Locus Analysis & Design

Root Locus Analysis & Design. See attached file for full problem description.

1) Draw the root locus of L (attached). of for positive and negative gain on graph paper.

Find (graphically) the gain, k, to get l = 0.5 for the dominant poles and specify the corresponding corner frequency.

2) Draw the root locus of L (attached) of for positive and negative gain on graph paper. Calculate breakpoints, jw axis crossings and departure angles.

3) Draw the root locus of L (attached) of for positive and negative gain on graph paper.

Find (graphically) the gain k, to get l = 0.5 for the dominant poles and specify the corresponding frequency.

4) Draw the root locus of L (attached) of for positive and negative gain on graph paper. Calculate breakpoints, jw axis crossings and departure angles.

5) Draw the root locus of L (attached) of for positive and negative gain on graph paper.

Find (graphically) the gain k, to get l = 0.5 for the dominant poles and specify the corresponding frequency.

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