Root Locus

System Modeling and Control · Chapter 12

John N. Chiasson and Aykut C. Satici

Root Locus and the Angle Condition
Root Locus Construction Rules
Asymptotes, Breakaway Points, and Departure Angles
Effects of Open-Loop Poles and Zeros
Satellite Control Design Example

Chapter 12 Root Locus

Let \(G(s)\) be given by \[G(s)=\frac{b(s)}{a(s)}=\frac{b_{m}s^{m}+b_{m-1}s^{m-1}+\cdots+b_{1}s+b_{0}}{s^{n}+a_{n-1}s^{n-1}+\cdots+a_{1}s+a_{0}},\text{ }m<n\] The closed-loop transfer function is \[\frac{C(s)}{R(s)}=\frac{KG(s)}{1+KG(s)}=\frac{Kb(s)\!/\!a(s)}{1+Kb(s)\!/\!a(s)}=\frac{Kb(s)}{a(s)+Kb(s)}.\]
As \(K\) is varied from \(0\) to \(\infty\) we want to know the location of the roots of \[a(s)+Kb(s)=0.\]
\(a(s)+Kb(s)\) is the closed-loop characteristic polynomial.
The roots of \(a(s)+Kb(s)=0\) are the closed-loop poles.
The root locus is a plot of the roots of \(a(s)+Kb(s)=0\) as \(K\) goes from \(0\) to \(\infty.\)

Root Locus for \(\dfrac{1}{s(s+4)}\)

With \(G(s)=\dfrac{1}{s(s+4)}\) we have \[\frac{C(s)}{R(s)}=\frac{KG(s)}{1+KG(s)}=\frac{K\dfrac{1}{s(s+4)}}{1+K\dfrac {1}{s(s+4)}}=\frac{K}{s^{2}+4s+K}.\]

For \(K\geq0\) compute the roots of \[s^{2}+4s+K=0.\]

By the quadratic formula\[s=\dfrac{-4\pm\sqrt{16-4K}}{2}=-2\pm\sqrt{4-K}\] or \[p_{1},p_{2}=\left\{ \begin{array}{ll} -2+\sqrt{4-K},\text{ }-2-\sqrt{4-K}, & 0\leq K\leq4 \\ & \\ -2\pm j\sqrt{K-4}, & 4\leq K.\end{array} \right.\]

Root Locus for \(\dfrac{1}{s(s+4)}\) (continued)

From previous slide: \[p_{1},p_{2}=\left\{ \begin{array}{ll} -2+\sqrt{4-K},\text{ }-2-\sqrt{4-K}, & 0\leq K\leq4 \\ & \\ -2\pm j\sqrt{K-4}, & 4\leq K.\end{array} \right.\]

For \(K=0,2,4,8\) we have\[\begin{array}{|c|c|} \hline K & p_{1},p_{2} \\ \hline 0 & 0,-4 \\ \hline 2 & -2+\sqrt{2},-2-\sqrt{2} \\ \hline 4 & -2,-2 \\ \hline 8 & -2\pm j2 \\ \hline \end{array}\]

Sketch the root locus.

Angle Condition and Root Locus Rules

Sketch the root locus without having to solve for all the roots.

A computer can be used to numerically find the roots.

By sketching the root locus we gain insight into the information it contains.
Sketching the root locus is based on the angle condition.

Define the return difference by \[1+KG(s)=1+K\dfrac{b(s)}{a(s)}=\dfrac{a(s)+Kb(s)}{a(s)}.\]

The zeros of the return difference are the roots of \(a(s)+Kb(s)=0.\)

These are the closed-loop poles.
The poles of the return difference are the roots of \(a(s)=0.\)

These are open-loop poles.
The root locus is a plot of the zeros of \(1+KG(s)=0\) for \(K\geq0.\)
If \(s_{0}\) satisfies \[1+KG(s_{0})=0\] for some \(K>0\) then it satisfies \[G(s_{0})=-\frac{1}{K}.\]

Angle Condition

From previous slide: If \(s_{0}\) satisfies \[1+KG(s_{0})=1+K\dfrac{b(s_{0})}{a(s_{0})}=\dfrac{a(s_{0})+Kb(s_{0})}{a(s_{0})}=0\]

for some \(K>0\) then it satisfies \[G(s_{0})=-\frac{1}{K}.\]

In general, \(s_{0}\) is a complex number.
For \(s_{0}\) to be on the root locus, \(G(s_{0})\) equal to a negative real number.
That is, in polar coordinates, \(s_{0}\) must satisfy \[|G(s_{0})|e^{j\angle G(s_{0})}=-\frac{1}{K}=\frac{1}{K}e^{j\pi}.\]
Thus for \(s_{0}\) to be on the root locus, \(G(s_{0})\) satisfies the angle condition given by \[\angle G(s_{0})=\pi(2\ell+1),\ell=0,\pm1,\pm2,...\]
The corresponding value of \(K\) is given by \[K=\frac{1}{|G(s_{0})|}.\]
Only need the open-loop transfer function \(G(s)\) to see if \(s_{0}\) is a closed-loop pole.

Angle Condition for \(\dfrac{1}{s(s+4)}\)

\[\begin{aligned} G(s) & =\dfrac{1}{s(s+4)}=\dfrac{1}{\underset{s}{\underbrace{|s|e^{j\angle s}}}\text{ }\underset{s+4}{\underbrace{|s+4|e^{j\angle(s+4)}}}}=\dfrac {1}{|s||s+4|}e^{-j\angle s}e^{-j\angle(s+4)} \\ & \\ \Longrightarrow\text{ }\angle G(s) & =-\angle s-\angle(s+4). \end{aligned}\]

Let the point “\(s\)” be on the root locus.
\(s+4=s-(-4)\) is a vector from \(-4\) to \(s.\)
\(s\) is a vector from \(0\) to \(s.\)
The figure shows \(\angle s+\angle(s+4)=\pi,\) i.e., \[\angle G(s)=-\angle s-\angle(s+4)=-\pi.\]

Angle Condition for \(\dfrac{1}{s(s+4)}\) (continued)

For example, \(s=-2+2j\) is on the root locus and \[\begin{aligned} G(-2+2j) & =\dfrac{1}{(-2+j2)(-2+2j+4)}=\frac{1}{(-2+2j)(2+2j)}=-\frac{1}{8} \\ & \\ \Longrightarrow\angle G(-2+2j) & =\angle(-1/8)=\pi. \end{aligned}\]
On the other hand \(s=-4+2j\) is not on the root locus and \[\begin{aligned} G(-4+2j) & =\dfrac{1}{(-4+j2)(-4+2j+4)}=\frac{1}{(-4+2j)(2j)}=-\frac {1}{4+8j} \\ & \\ \Longrightarrow\angle G(-4+2j) & \neq\pi(2\ell+1),\text{ }\ell=0,\pm 1,\pm2,..... \end{aligned}\]

Angle Condition for \(\dfrac{1}{s(s+4)}\) (continued)

Let \(s=-1\) which is on on the root locus with \[\angle s=\pi \text{and }\angle(s-(-1))=0.\]

Thus \[\angle G(s)=-\angle s-\angle(s+4)=-\pi\] so that \(s=-1\) satisfies the angle condition.
On the other hand, let \(s=-5\), so \[\angle s=\pi \quad \text{and} \quad \angle(s-(-5))=\pi.\]

Thus \[\angle G(s)=-\angle s-\angle(s-(-5))=-2\pi\] showing \(s=-5\) does not satisfy the angle condition.

EXAMPLE Root Locus for \(\dfrac{1}{s(s+4)}\)

The return difference is \[1+KG(s)=1+K\dfrac{b(s)}{a(s)}=\dfrac{a(s)+Kb(s)}{a(s)}=\frac{s(s+4)+K}{s(s+4)}.\]

Its zeros are the closed-loop poles.

(1) The root locus starts on the open-loop poles.

I.e., for \(K=0\) the closed-loop poles are the roots of \[a(s)+Kb(s)=a(s)=s(s+4)=0.\]

(2) The number of branches of the root locus is 2.

This is because \(a(s)+Kb(s)=s^{2}+4s+K=0\) has two roots.

(3) The root locus is symmetric with respect to the real axis.

This is simply because the complex roots of \[a(s)+Kb(s)=s^{2}+4s+K=0\]

come in complex conjugate pairs.

Root Locus for \(\dfrac{1}{s(s+4)}\) (continued)

(4) For any \(s_{0}\) on the root locus, \[\angle G(s_{0})=\angle\dfrac{1}{s_{0}(s_{0}+4)}=\pi(2\ell+1),\ell=0,\pm 1,\pm2,....\]

Root Locus for \(\dfrac{1}{s(s+4)}\) (continued)

(5) A breakaway point occurs at \(s=-2\) for \(K=4.\)

This is where \(a(s)+Kb(s)=s^{2}+4s+K=0\) has a double root, that is, \[\left. 1+K\frac{1}{s(s+4)}\right\vert _{K=4}=\left. \dfrac{s^{2}+4s+K}{s(s+4)}\right\vert _{K=4}=\dfrac{(s+2)^{2}}{s(s+4)}\]

\[\begin{aligned} \!\!\!\!\!\!\!\left. \frac{d}{ds}\left( 1+4\frac{1}{s(s+4)}\right) \right\vert _{s=-2}\!\!\!\!\!\!\! & =\!\!\!\!\left. \frac{d}{ds}\left( \dfrac{(s+2)^{2}}{s(s+4)}\right) \right\vert _{s=-2} \\ \!\!\!\!\!\!\! & =\!\!\!\!\left. -\dfrac{1}{s(s+4)}\left( \dfrac{d}{ds}(s^{2}+4s)\right) (s+2)^{2}\right\vert _{s=-2}\!\!\!\!+\left. \left( \dfrac{1}{s(s+4)}\dfrac{d}{ds}(s+2)^{2}\right) \right\vert _{s=-2} \\ \!\!\!\!\!\!\! & =\!\!\!\!0. \end{aligned}\]

Example Root Locus for \(\dfrac{1}{s(s+2)(s+4)}\)

\(G(s)=\dfrac{1}{s(s+2)(s+4)}=\dfrac{1}{|s||s+2||s+4|}e^{-j\left( \angle s+\angle(s+2)+\angle(s+4)\right) }.\)
\(1+KG(s)=1+K\dfrac{1}{s(s+2)(s+4)}=\dfrac{s(s+2)(s+4)+K}{s(s+2)(s+4)}=0.\)
\(G(s)=-\dfrac{1}{K}\Longrightarrow\angle G(s)=\pi(2\ell+1),\ell =0,\pm1,\pm2,....\)

Sketch the root locus.

(1) With \(K=0\) we have \(s(s+2)(s+4)+K=s(s+2)(s+4)=0\) or \(s=0,-2,-4.\)

The root locus starts on the open-loop poles.

(2) \(1+KG(s)=0\) or \(G(s)=\dfrac{1}{s(s+2)(s+4)}=-\dfrac{1}{K}.\)

As \(K\rightarrow\infty,\) the 3 roots go to infinity, i.e., \(|s|\rightarrow\infty.\)

Root Locus for \(\dfrac{1}{s(s+2)(s+4)}=\dfrac{1}{|s||s+2||s+4|}e^{-j\left( \angle s+\angle(s+2)+\angle (s+4)\right) }=-\dfrac{1}{K}\)

(3) Real Axis Root Locus

(i) Consider \(s\) real with \(-2<s<0\) so \(\angle s=\pi,\angle (s+2)=0,\angle(s+4)=0.\)

Then \[\angle G(s)=-\angle s-\angle(s+2)-\angle(s+4)=-\pi.\]

These values of \(s\) are on the root locus.

(ii) Consider \(s\) real with \(-4<s<-2\) so \(\angle s=\pi,\angle (s+2)=\pi,\angle(s+4)=0.\)

Then \[\angle G(s)=-\angle s-\angle(s+2)-\angle(s+4)=-2\pi.\]

These values of \(s\) are not on the root locus.

Root Locus for \(\dfrac{1}{s(s+2)(s+4)}=\dfrac{1}{|s||s+2||s+4|}e^{-j\left( \angle s+\angle(s+2)+\angle (s+4)\right) }=-\dfrac{1}{K}\)

(3) Real Axis Root Locus (continued)

(iii) Consider \(s\) real with \(s<-4\) so \(\angle s=\pi,\angle (s+2)=\pi,\angle(s+4)=\pi.\)

Then \[\angle G(s)=-\angle s-\angle(s+2)-\angle(s+4)=-3\pi.\]

These values of \(s\) are on the root locus.

(iv) Finally, consider \(s\) real with \(s>0\) so \(\angle s=0,\angle (s+2)=0,\angle(s+4)=0.\)

Then \[\angle G(s)=-\angle s-\angle(s+2)-\angle(s+4)=0.\]

These values of \(s\) are not on the root locus.

Root Locus for \(\dfrac{1}{s(s+2)(s+4)}\) (continued)

(4) Breakaway Point

A breakaway point occurs at a value of \(K\) for which \[1+K\dfrac{1}{s(s+2)(s+4)}=\dfrac{s(s+2)(s+4)+K}{s(s+2)(s+4)}=0\] has multiple zeros.

That is, \(s_{b}\) is a breakaway point for a value of \(K_{b}\) such that \[1+K_{b}\dfrac{1}{s(s+2)(s+4)}=\dfrac{s(s+2)(s+4)+K_{b}}{s(s+2)(s+4)}=\frac{(s-s_{b})^{2}(s-s_{1})}{s(s+2)(s+4)}.\]
Two branches breaking away from the real axis come together as a double root.

They then leave the real axis as a complex conjugate pair.
Thus at a breakaway point we have \[\frac{d}{ds}\underset{\frac{(s-s_{b})^{2}(s-s_{1})}{s(s+2)(s+4)}}{\underbrace{\left( \right. 1+K_{b}\dfrac{1}{s(s+2)(s+4)}\left. \right) }}\left. \right\vert _{s=s_{b}}=K_{b}\dfrac{d}{ds}\underset{G(s)}{\underbrace{\left( \dfrac{1}{s(s+2)(s+4)}\right) }}=0.\] Summarizing, a breakaway point \(s_{b}\) must be a solution to \(\left. \dfrac{d}{ds}G(s)\right\vert _{s=s_{b}}=0.\)

Root Locus for \(\dfrac{1}{s(s+2)(s+4)}\) (continued)

(4) Breakaway Point (continued) \[\begin{aligned} \frac{d}{ds}G(s) & =-\left( \dfrac{1}{s(s+2)(s+4)}\right) ^{2}\frac{d}{ds}\left( s(s+2)(s+4)\right) \\ & =-\left( \dfrac{1}{s(s+2)(s+4)}\right) ^{2}\frac{d}{ds}\left( s^{3}+6s^{2}+8s\right) . \end{aligned}\]

\(\dfrac{d}{ds}G(s)=0\) requires solving \(3s^{2}+12s+8=0\) \(\Longrightarrow\) \[s=\frac{-12\pm\sqrt{(12)^{2}-4(3)(8)}}{6}=-2\pm\frac{2}{\sqrt{3}}=-0.845,-3.146\]

Only \(s=-0.845\) is on the real axis root locus and is therefore the only breakaway point.

Root Locus for \(\dfrac{1}{s(s+2)(s+4)}\) (continued)

(5) \(j\omega\) axis intercepts

Apply the Routh-Hurwitz test to \[s(s+2)(s+4)+K=s^{3}+6s^{2}+8s+K.\] The Routh table is\[\begin{array}{cccccc} s^{3} & & 1 & & 8 & \\ s^{2} & & 6 & & K & \\ s & & (48-K)/6 & & 0 & \\ 1 & & K & & & \end{array}\]

Stable for \(0<K<48.\)
Two right half-plane poles for \(K>48.\)
\(\Longrightarrow\) At \(K=48\) there are two poles on the \(j\omega\) axis.

To find these poles solve the auxiliary equation \(6s^{2}+48=0\) to obtain \(s=\pm j2\sqrt{2}=\pm j2.83.\)

Root Locus for \(\dfrac{1}{s(s+2)(s+4)}\) (continued)

Using the above results we (Matlab!) can sketch the root locus as shown below.

Root Locus for \(\dfrac{1}{s(s+2)(s+4)}\) (continued)

% Root Locus of G(s) = 1/(s(s+2)(s+4)) = 1/(s3+6s2+8s)

den = [1 6 8 0];

num = [1];

rlocus(tf(num,den))

% Pretty up the plot by making the linewidth thicker,

%`` ``the marker size and`` font size bigger.

h = findobj(gca, ’Type’, ’line’);

set(h, ’LineWidth’, 6)

set(h, ’MarkerSize’, 20)

set(gca,’FontSize’,24)

% Set range of x-axis [-10,0] and y-axis [-2,2]

v = [-10 10 -10 10];

axis(v)

axis square

% Title the plot and label the axes

str_G = ’G(s) = 1/(s(s+2)(s+4))’;

title(str_G,’FontSize’,20)

xlabel(’Re(s)’,’FontSize’,20)

ylabel (’Im(s)’,’FontSize’,20)

Root Locus Rules

The previous examples used the following root locus rules.

(1) The root locus starts on the open-loop poles (\(K=0\)).

(2) The root locus ends on the open-loop zeros (\(K=\infty\)).

(3) The root locus is symmetric with respect to the real axis.

(4) Any \(s_{0}\) on the root locus satisfies the angle condition, i.e., \[\angle G(s_{0})=\pm(2\ell+1)\pi\text{ for }\ell=0,1,2,...\]

(used to find the real-axis root locus)

(5) A breakaway point from the real axis is a solution to \(dG(s)/ds=0.\)

(6) The \(j\omega\) axis intercepts may be found using the Routh-Hurwitz test.

Asymptotes and Their Intercept

For \(|s|\) large the branches of the root locus are asymptotic to straight lines.
These straight lines are called asymptotes.
These asymptotes intersect the real axis at angles given by \[\theta_{\ell}=\frac{2\ell+1}{n-m}\pi,\text{ }\ell=0,1,2,...,n-m-1.\]
Further these asymptotes intersect on the real line at \[\sigma_{1}=-\frac{a_{n-1}-b_{m-1}/b_{m}}{n-m}.\]

Asymptotes and Their Intercept

\[\begin{aligned} G(s)=\frac{b(s)}{a(s)} & =\frac{b_{m}s^{m}+b_{m-1}s^{m-1}+\cdots +b_{1}s+b_{0}}{s^{n}+a_{n-1}s^{n-1}+\cdots+a_{1}s+a_{0}} \\ & =\frac{b_{m}\!\left( s^{m}+\dfrac{b_{m-1}}{b_{m}}s^{m-1}+\cdots +\dfrac{b_{1}}{b_{m}}s+\dfrac{b_{0}}{b_{m}}\right) }{s^{n}+a_{n-1}s^{n-1}+\cdots+a_{1}s+a_{0}}. \end{aligned}\] In the special case where \(m=0,\) that is, \[G(s)=\frac{b_{0}}{s^{n}+a_{n-1}s^{n-1}+\cdots+a_{1}s+a_{0}}\] we take \(b_{-1}=0\) so that \[\sigma_{1}=-\frac{a_{n-1}-b_{-1}/b_{0}}{n-m}=-\frac{a_{n-1}-0}{n-m}.\]

This rule is best explained by an example.

Asymptotes and their Intersection for \(G(s)=\dfrac {1}{s(s+2)(s+4)}\)

\(G(s)=\dfrac{1}{s(s+2)(s+4)}=\dfrac{1}{s^{3}+6s^{2}+8s}\)

Then the angles of the asymptotes with respect to the real axis are (\(n=3,m=0\)) \[\begin{aligned} \theta_{\ell} & =\frac{(2\ell+1)\pi}{3-0},\ell=0,1,2 \\ & =\frac{\pi}{3},\frac{3\pi}{3},\frac{5\pi}{3} \\ & =\frac{\pi}{3},\pi,-\frac{\pi}{3} \end{aligned}\] and these asymptotes intersect at\[\begin{aligned} \sigma_{1} & =-\frac{a_{n-1}-0}{n-m},\text{ }n=3,m=0 \\ & =-\frac{6-0}{3-0},\text{ }a_{2}=6 \\ & =-2. \end{aligned}\]

Asymptotes and their Intersection for \(G(s)=\dfrac {1}{s(s+2)(s+4)}\)

\(G(s)=\dfrac{1}{s(s+2)(s+4)}=\dfrac{1}{s^{3}+6s^{2}+8s}.\)

\(\theta_{\ell}=\dfrac{\pi}{3},\pi,-\dfrac{\pi}{3}\) and \(\sigma_{1}=-2.\)

The asymptotes intersect the \(j\omega\) axis at \(\pm j2\tan(\pi/3)=\pm2\sqrt {3}=\pm j3.46.\)

Asymptotes and their Intersection for \(G(s)=\dfrac {1}{s(s+2)(s+4)}\)

Matlab plot of the root locus (See slide ).

Root Locus of \(G(s)=\dfrac{s+4}{s(s+2)}=\dfrac{|s+4|}{|s||s+2|}e^{j\left( \angle(s+4)-\angle s-\angle(s+2)\right) }.\)

Return Difference

\[1+KG(s)=1+K\dfrac{s+4}{s(s+2)}=\dfrac{s(s+2)+K(s+4)}{s(s+2)}\] Angle Condition \(G(s)=-\dfrac{1}{K}\) \(\Longrightarrow\) \[\angle(s+4)-\angle s-\angle(s+2)=\pi(2\ell+1),\ell=0,\pm 1,\pm2,....\]

(1) \(K=0.\) Then \(s(s+2)+K(s+4)=s(s+2)=0\) \(\Longrightarrow\) \(\ s=0,-2.\)

(2) \(K=\infty.\) From \(G(s)=\dfrac{s+4}{s(s+2)}=-\dfrac{1}{K}\) \(\ \Longrightarrow\) \(s\rightarrow-4\) or \(|s|\rightarrow\infty\) as \(K\rightarrow\infty\).

\(G(s)\) has one zero at \(-4\) and one zero at \(|s|=\infty\).

The root locus ends on the open-loop zeros.

Root Locus of \(G(s)=\dfrac{s+4}{s(s+2)}=\dfrac{|s+4|}{|s||s+2|}e^{j\left( \angle(s+4)-\angle s-\angle(s+2)\right) }.\)

(3) Real Axis Root Locus

(i) Consider \(s\) real with \(-2<s<0\) so \(\angle s=\pi,\angle (s+2)=0,\angle(s+4)=0.\)

Then \[\angle G(s)=\angle(s+4)-\angle s-\angle(s+2)=-\pi.\]

These values of \(s\) are on the root locus.

(ii) Consider \(s\) real with \(-4<s<-2\) so \(\angle s=\pi,\angle (s+2)=\pi,\angle(s+4)=0.\)

Then \[\angle G(s)=\angle(s+4)-\angle s-\angle(s+2)=0.\]

These values of \(s\) are not on the root locus.

Root Locus of \(G(s)=\dfrac{s+4}{s(s+2)}=\dfrac{|s+4|}{|s||s+2|}e^{j\left( \angle(s+4)-\angle s-\angle(s+2)\right) }.\)

(3) Real Axis Root Locus (continued)

(iii) Consider \(s\) real with \(s<-4\) so \(\angle s=\pi,\angle (s+2)=\pi,\angle(s+4)=\pi.\)

Then \[\angle G(s)=\angle(s+4)-\angle s-\angle(s+2)=-\pi.\]

These values of \(s\) are on the root locus.

(iv) Finally, consider \(s\) real with \(s>0\) so \(\angle s=0,\angle (s+2)=0,\angle(s+4)=0.\)

Then \[\angle G(s)=-\angle s-\angle(s+2)-\angle(s+4)=0.\]

These values of \(s\) are not on the root locus.

Root Locus of \(G(s)=\dfrac{s+4}{s(s+2)}\)

Breakaway Point

Solve \[\frac{d}{ds}\underset{G(s)}{\underbrace{\left( \dfrac{s+4}{s(s+2)}\right) }}=0.\]

\[\frac{d}{ds}G(s)=\dfrac{1}{s(s+2)}-\dfrac{s+4}{\left[ s(s+2)\right] ^{2}}\frac{d}{ds}\left( s^{2}+2s\right) =\frac{s(s+2)-(s+4)(2s+2)}{\left[ s(s+2)\right] ^{2}}=-\frac{s^{2}+8s+8}{\left[ s(s+2)\right] ^{2}}\]

\(\dfrac{d}{ds}G(s)=0\) requires \(s^{2}+8s+8=0\), or \[s=\frac{-8\pm\sqrt{64-32}}{2}=-4\pm2\sqrt{2}=-6.828,-1.172\]

Both roots of \(dG/ds=0\) are on the real axis root locus and so are breakaway pts.

Root Locus of \(G(s)=\dfrac{s+4}{s(s+2)}\)

\(j\omega\) axis intercepts\[1+K\dfrac{s+4}{s(s+2)}=\dfrac{s(s+2)+K(s+4)}{s(s+2)}=0\]

The closed-loop poles are the roots of \[s(s+2)+K(s+4)=s^{2}+(K+2)s+4K=0.\]

Are there values of \(K\geq0\) for which the roots are on the \(j\omega\) axis?
This is a second-order polynomial which is stable for \(K>-2.\)

\(\Longrightarrow\) There is no value of \(K\geq0\) for which the root locus intercepts the \(j\omega\) axis.

Root Locus of \(G(s)=\dfrac{s+4}{s(s+2)}\)

Asymptotes \(n=2,m=1\)

Asymptote angles: \[\begin{aligned} \theta_{\ell} & =\frac{\pi(2\ell+1)}{n-m},\text{ }\ell=0,1,...,n-m-1 \\ & =\frac{\pi(2\ell+1)}{2-1},\text{ }\ell=0. \end{aligned}\]

That is, \(\theta_{0}=\pi.\)

Asymptote intersection with the real axis \[\begin{aligned} \sigma_{1} & =-\frac{a_{n-1}-b_{m-1}/b_{m}}{n-m},\text{ }n=3,m=1 \\ & =-\frac{2-4/1}{2-1}\text{ as }a_{2}=2 \\ & =2. \end{aligned}\]

That is, \(\sigma_{1}=2.\)

There is only one asymptote whose angle is \(\pi.\)

Root Locus of \(G(s)=\dfrac{s+4}{s(s+2)}\)

Sketch Previously

The full sketch is then

Using the angle condition we show that the root locus that is off the real axis is a circle!

Root Locus of \(G(s)=\dfrac{s+4}{s(s+2)}\)

Show \(s=-4+\sqrt{8}e^{j\theta}\) is on the root locus for \(0\leq\theta \leq2\pi.\) \[\begin{aligned} \left. G(s)\right\vert _{s=-4+\sqrt{8}e^{j\theta}}=\left. \dfrac {s+4}{s(s+2)}\right\vert _{s=-4+\sqrt{8}e^{j\theta}} & =\frac{\sqrt {8}e^{j\theta}}{(-4+\sqrt{8}e^{j\theta})(-2+\sqrt{8}e^{j\theta})} \\ & =\frac{\sqrt{8}e^{j\theta}}{8-6\sqrt{8}e^{j\theta}+8e^{2j\theta}} \\ & =\frac{\sqrt{8}}{8e^{-j\theta}-6\sqrt{8}+8e^{j\theta}} \\ & =\frac{\sqrt{8}}{16\cos(\theta)-6\sqrt{8}} \\ & =-\frac{\sqrt{8}}{\underset{16.97}{\underbrace{6\sqrt{8}}}-16\cos(\theta)}. \end{aligned}\] That is, \(G(-4+\sqrt{8}e^{j\theta})=-\) \(\underset{>\text{ }0\text{ for all }\theta}{\underbrace{\dfrac{\sqrt{8}}{16.97-16\cos(\theta)}}}\).

\(\angle G(-4+\sqrt{8}e^{j\theta})=\pi\) for all \(0\leq\theta\leq2\pi\).

Angles of Departure

\(G(s)=\dfrac{1}{s(s^{2}+2s+2)}=\dfrac{|s+4|}{|s||s-(-1+j)||s-(-1-j)|}e^{-j\left( \angle s+\angle(s-(-1+j))+\angle s-(-1-j)\right) }.\)

The denominator \(G(s)\) factors as \(s(s^{2}+2s+2)=s[s-(-1+j)][s-(-1-j)].\)

The pair of complex conjugate open-loop poles leads to the angles of departure.

The closed-loop poles are the zeros of \[1+KG(s)=1+K\dfrac{1}{s(s^{2}+2s+2)}=\dfrac{s^{3}+2s^{2}+2s+K}{s(s^{2}+2s+2)}.\] By the angle condition CL poles satisfy \[G(s)=-\frac{1}{K}\text{ }\Longrightarrow\angle G(s)=\pi(2\ell+1),\ell =0,\pm1,\pm2,....\] or \[-\angle s-\angle\lbrack(s-(-1+j)]-\angle\lbrack s-(-1-j)]=\pi (2\ell+1),\ell=0,\pm1,\pm2,....\]

Root Locus of s³ + 2s² + 2s + K = 0

\(G(s)=\dfrac{1}{s(s^{2}+2s+2)}=\dfrac{|s+4|}{|s||s-(-1+j)||s-(-1-j)|}e^{-j\left( \angle s+\angle(s-(-1+j))+\angle s-(-1-j)\right) }\)

(1) \(K=0.\) Then \[s^{3}+2s^{2}+2s+K=s(s^{2}+2s+2)=s[s-(-1+j)][s-(-1-j)]=0\]

The root locus starts on the open-loop poles \(s=0,-1\pm j.\)

(2) \(K=\infty.\) From \(G(s)=\dfrac{1}{s(s^{2}+2s+2)}=-\dfrac{1}{K}\) we have \(|s|\rightarrow\infty\) as \(K\rightarrow\infty.\)

\(G(s)\) has three zeros at \(|s|=\infty\) since \[G(s)=\dfrac{1}{s(s^{2}+2s+2)}\approx\frac{1}{|s|^{3}}\text{ for }|s|\rightarrow\infty.\]

The root locus ends on the open-loop zeros.

Root Locus of s³ + 2s² + 2s + K = 0

(3) Real Axis Root Locus

(i) \(s\) real with \(-\infty<s<0\) so \(\angle s=\pi,\angle \lbrack(s-(-1+j)]=-\angle\lbrack(s-(-1-j)]\). \[\angle G(s)=-\angle s\underset{0}{\underbrace{-\angle \lbrack(s-(-1+j)]-\angle\lbrack s-(-1-j)]}}\text{ }=-\angle s=-\pi\]

These values of \(s\) are on the root locus.

(ii) \(s\) real with \(s>0\) so \(\angle s=0,\angle\lbrack (s-(-1+j)]=-\angle\lbrack(s-(-1-j)].\) \[\angle G(s)=-\angle s\underset{0}{\underbrace{-\angle \lbrack(s-(-1+j)]-\angle\lbrack s-(-1-j)]}}\text{ }=-\angle s=0\]

These values of \(s\) are not on the root locus.

Root Locus of s³ + 2s² + 2s + K = 0

(4) Breakaway Point

Check where the return difference has multiple zeros. \[1+K\dfrac{1}{s(s^{2}+2s+2)}=\dfrac{s^{3}+2s^{2}+2s+K}{s(s^{2}+2s+2)}=0.\]

A breakaway point must satisfy \[\frac{d}{ds}\underset{G(s)}{\underbrace{\left( \dfrac{1}{s(s^{2}+2s+2)}\right) }}\text{ }=0\]

or \[\frac{d}{ds}G(s)=-\dfrac{1}{[s(s^{2}+2s+2)]^{2}}\frac{d}{ds}(s^{3}+2s^{2}+2s)=-\dfrac{1}{[s(s^{2}+2s+2)]^{2}}(3s^{2}+4s+2)=0.\]

Solving \(3s^{2}+4s+2=0\) gives \[s=\frac{-4\pm\sqrt{16-24}}{6}=-\frac{2}{3}-\pm j\frac{\sqrt{2}}{3}\]

These roots are not on the real axis. Thus there are no breakaway points.

Root Locus of s³ + 2s² + 2s + K = 0

(5) \(j\omega\) axis intercepts The Routh table for \(s^{3}+2s^{2}+2s+K=0\) is \[\begin{array}{cccccc} s^{3} & & 1 & & 2 & \\ s^{2} & & 2 & & K & \\ s & & (4-K)/2 & & 0 & \\ 1 & & K & & & \end{array}\]

Closed-loop system is stable for \(0<K<4.\)
There are two right half-plane roots for \(K>4.\)

\(\Longrightarrow\) For \(K=4\) there are two roots on the \(j\omega\) axis.
The auxiliary equation \(2s^{2}+K=2s^{2}+4=0\) then gives \(s=\pm j\sqrt {2}.\)

Root Locus of \(G(s)=\dfrac{1}{s(s^{2}+2s+2)}\)

(6) Asymptotes The angles of the asymptotes are (\(n=3,m=0\)) \[\theta_{\ell}=\frac{\pi(2\ell+1)}{n-m},\text{ }\ell=0,1,...,n-m-1=\frac {\pi(2\ell+1)}{3-0},\text{ }\ell=0,1,2.\]

I.e., \(\theta_{0}=\pi/3,\theta_{1}=\pi,\theta_{2}=5\pi/3\) \((-\pi /3).\)

Their intercept on the real axis is\[\begin{aligned} \sigma_{1}=-\frac{a_{n-1}-b_{m-1}/b_{m}}{n-m},\text{ }n=3,m=0 & =-\frac {2-0}{3-0}\text{ as }a_{2}=2,b_{0}=1 \\ & =-2/3. \end{aligned}\]

Two of these asymptotes intersect the \(j\omega\) axis at \(\pm j(2/3)\tan (\pi/3)=\pm j1.155.\)

Root Locus of \(G(s)=\dfrac{1}{s(s^{2}+2s+2)}\)

(7) Angles of Departure

For \(K=0\) the roots are at \(s=0,s=-1+j,\) and \(s=-1-j.\)
As \(K\) is increased slightly, the closed-loop poles will leave these roots.
The branch that starts at \(-1+j\) is slightly away from this point.

Then, under these conditions, \[\begin{aligned} \angle s & \approx\pi/2+\pi/2=3\pi/4 \\ \angle\lbrack s-(-1-j)] & \approx\pi/2. \end{aligned}\]

Root Locus of \(G(s)=\dfrac{1}{s(s^{2}+2s+2)}\)

(7) Angles of Departure For \(K\) slightly positive we have\[\angle s\approx3\pi/4\text{ and }\angle\lbrack s-(-1-j)]\approx\pi/2.\]

For\(s\) close to \(-1+j\) and on the root locus it must satisfy the angle condition:\[-\angle s-\underset{\text{unknown}}{\underbrace{\angle\lbrack(s-(-1+j)]}}-\angle\lbrack s-(-1-j)]=\pi(2\ell+1),\ell=0,\pm1,\pm2,....\]

Or, with \(\ell=0\), we have \(-\pi/2-\angle\lbrack(s-(-1+j)]-3\pi/4=\pi\). \[\Longrightarrow\text{ }\angle\lbrack(s-(-1+j)]=-5\pi/4-\pi=-\pi/4-2\pi =-\pi/4\text{ }\operatorname{mod}2\pi.\]

By symmetry we have \(\angle\lbrack(s-(-1-j)]=\pi/4.\)

Root Locus of

\(G(s)=\dfrac{1}{s(s^{2}+2s+2)}\)

The root locus starts on the open-loop poles.
The branch starting at \(s=0\) continues out along the negative real axis to \(-\infty.\)
The branch that starts at \(-1+j\) leaves that point at an angle of \(-\pi/4.\)

As \(K\) increases this branch intercepts the \(j\omega\) axis at \(j\sqrt{2}\).

As \(K\rightarrow\infty\) this branch approaches its asymptote.

Effect of Open-Loop Poles on the Root Locus

Effect of Open-Loop Zeros on the Root Locus