# Drawing a root locus and finding the dominant roots

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

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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 number of roots and zeros. This I have drawn as the green line. The green root loci trends from the finite pole (ie when ) at to the infinite zero (ie when ) at . The orange and purple root loci are the trends of the two finite poles at to zeros at infinity.

As one can see all three roots intersect (ie. have the same value) at so to find the value of at this point we substitute into our characteristic equation {3}

{3}

Thus in summary we have deduced that the value of the gain to obtain all equal roots is

Following from some of the general arguments presented in the earlier Q24. We can identify the numerator and denominator polynomials as

The numerator has finite roots (finite zeros) obtained from

The denominator has finite roots (finite poles) obtained when

two (duplicate) and finite poles

Thus we have root loci branches to our plot

Equating the parameter , thus we can say we have finite poles in the root locus that trend to a zero at as and two poles that trend to the finite zeros and as

We have a root locus that sits between the two poles at and which has zero extension so this cannot be plotted but we need to ear this in mind. We have another root locus on the real axis that extends from and trends to , all from the requirement that the root loci on the real axis must extend to the left of an odd number of poles & zeros.

- 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 on the real axis with one another are determined from

(

As we only have to deduce this for one angle when

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

Multiplying by -1

, and two complex conjugate roots which we don't need to work out as they don't lie on the real axis. As can be seen only lies on any root locus on the real axis so the asymptotes intersect at this point.

In order to plot the shape of the root loci branches we really need to obtain the equations for these branches. This process is quite complex so I have shown it as a separate Appendix A at the end of this work.

The shape of the root loci are derived from {18}

{18}

The root loci plot will then be as shown below (this is a rough plot I have drawn. If one wishes to construct a really accurate plot of the root loci use equation {18} above for different values of to derive the values then plot ). As can be seen we have two finite poles trending to finite zeros of when and one finite pole at trending to when

- Finding the required gain for damping factor conditions:

If given a damping factor of and in order to find the required for this condition we need first to find the point of the intersection of the damping factor with the root loci plot. In order to do this we need to construct lines originating at the origin of the plot for angles of with the negative real axis. This is shown as below.

We then need to find the intersection of either of these lines with the root loci; either will do as they are symmetric about the real axis. I will choose the line at an angle with the negative real axis. Note this represents an angle of to the normal positive real axis so we must use the angle in our analysis for required roots in the subsequent analysis.

Normally if one had graphing paper one could do a graphical analysis with some computer generating tools to generate the root loci plots, then look for the intersection. If one wants to use online root loci graphing tools and to check ones work a good tool can be found here [1] . I do not have such graphing tools so I will mathematically derive the root loci equation and then substitute the condition that the angle of the root loci at the intersection must obey the condition

Derivation of the root loci equation is very complex involving much algebra so I have left it to an Appendix, Appendix A.

As a result I will just write the result here that the intersection of the positive going root loci and the line occurs at a root of

or alternatively because of complex conjugate symmetry at a root of

To find the required gain at this point we just use the condition from the characteristic equation that

And evaluate this at the required root which is

Since our open loop numerator is from before and our denominator is

We evaluate the numerator at this point to be

This has magnitude

We evaluate the denominator at this point to be

This has magnitude

Thus on substitution into

We obtain a value for to meet this condition

- Appendix A: Derivation of the equations for the root loci for the open loop transfer function

We already know the roots (poles and zeros) of the above

Zeros ,

Poles , ,

We can use the angle relation to derive the equation for the loci, that is

{1}

For the specific case in point maximum (since ) then we use the specific angle relation {2}

{2}

- Looking at the contribution to angles due to the zeros

We write out the numerator fully as

And writing the root at some value of as the sum of real and imaginary parts we can substitute for to get

Taking the arguments we get {3}

{3}

- Looking at the contribution to angles due to the poles

And again writing the root at some value of as the sum of real and imaginary parts , we can substitute for to get

Taking the arguments we get {4}

{4}

Thus we can write by back substitution {3} and {4} into {2}

{5}

Re-writing as {6}

{6}

Now we take the of each side of equation {6} above to get

{7}

Now thus {7} simplifies to {8}

{8}

Using the general relation for sum or difference of angles for tangents ie,

where the elements in {8} represent the general angles we can write {9}

{9}

Simplify {9} by cross multiplication of denominators and numerators we get {10} and {11}

{10}

{11}

Collecting like terms we get {12}

{12}

Cancelling common factor of both sides in numerator of {12} we get {13}

{13}

Cross multiplication yields {14} then {15}

{14}

{15}

Simplifying and collecting terms we can write {16}

{16}

If we look at {16} we can see it is a quadratic in thus we write it as the quadratic functional form {17}

{17}

This has solution for given by {18}

{18}

{18} is a general equation that plots the root loci branches for different where we can plot

Now we are in a position to work out the specific real part of the root when the line representing which makes an angle of with the negative axis intersects the root loci plot

As the angle required is at the intersection of lines of we can find

Thus we write in {18}

{19}

{20}

As examining {2} one root is the trivial solution we need to solve for {21}

{21}

Using Excel {21} has a root of

If this is the values of in the root and at the intersection as previously derived we must have a value for given by

Thus the intersection of the damping factor lines at is

- References:

[1] http://users.ece.gatech.edu/bonnie/book/OnlineDemos/InteractiveRootLocus/applet.html

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