2.21 Lecture 20, Nov. 5, 2015. Roots Locus completed

Reminder: Exam 2 on Nov 12 Thursday. Learn relation between damping ratio \(\zeta \) and damping cone. Exam will cover up to root locus. Will finish root locus today. Quick reminder of lemmas covered so far

1.

Mark open pole loops (where R.L. starts) and open loop zeros.

2.

Tells us that R.L. starts at open loop poles and ends up with open loop zeros. Some branches left will end up at \(\pm \infty \)

3.

Tells us what happens on the real axis. The segments that R.L. will travel over on the real axis. Called the ”odd” lemma

4.

Tell us how to generate the asymptotes of R.L. using \(\theta =\frac{180^{0}\pm k360}{n-m}\). Example, for \(n=6,m=3\), (where \(n,m\) are number of open loop poles and zeros respectively) then \(\theta =60^{0}\pm k120^{0}\) for \(k=0,1,\cdots \)

Reader: With \(n-m>2\) argue that closed loop is unstable. (answer: with \(n-m>2\), an asymptote will be moving to the RHS. Hence one of the branches will eventually move to the RHS for large gain, which means unstable).

Lemma 5: This lemma for finding from which point on the real axis the asymptotes will start. The centroid is given by\[ \sigma _{c}=\frac{{\displaystyle \sum } poles-{\displaystyle \sum } zeros}{n-m}\] The above is for poles and zeros of the open loop, not the closed loop!. For example, for \(GH=\frac{\left ( s+2\right ) ^{2}}{\left ( s+1\right ) \left ( s+6\right ) \left ( s+8\right ) ^{4}}\) then \(\sigma _{c}=-8.75\) and the angles at \(\theta =\frac{180^{0}\pm k360}{4}=45^{0}\pm 90^{0}\). So the asymptotes are

Example: \(GH=\frac{1}{s\left ( s+1\right ) \left ( s+2\right ) }\). Lemma 1: \(n=3,m=0\). Lemma 2: Start from open loop poles. Lemma 3: Real axis, use the odd criteria. Lemma 4: Find the asymptotes and the centroid. \(\sigma _{c}=\frac{0-1-2}{3}=-1\). \(\theta =\frac{180\pm k360}{3}=60^{0}\pm k120\). The result is this root locus

We still need to find where the break away points are and where root locus intersects the imaginary axis at.

Reader: Without using root locus, find the gain \(K\) what makes the closed loop unstable. (answer: use Routh table).

Lemma 6: This lemma tells us where the break points are on the real axis. Solve \(\frac{dK}{ds}=0\). Note not all solution points will be valid. Using the above example \(GH=\frac{1}{s\left ( s+1\right ) \left ( s+2\right ) }\), then the characteristic polynomial is \begin{align*} K+GH & =0\\ \left ( s+1\right ) \left ( s+2\right ) +K & =0\\ K & =-s^{3}-3s^{2}-2s\\ \frac{dK}{ds} & =-3s^{2}-6s-2 \end{align*}

Hence we solve \(-3s^{2}-6s-2=0\), and find \(s=-1.\,\allowbreak 577\) and \(s=-0.423\). But \(s=-1.\,\allowbreak 577\) is not on root locus (from above) so only \(s=-0.423\) is on root locus, and that is the breakaway point.

Lemma 7 Departure angle. R.L. depart each pole. We want to find the departure angles. Use\[{\displaystyle \sum } \sphericalangle z_{i}-{\displaystyle \sum } \sphericalangle p_{i}=180^{0}\pm k360^{0}\text{ \ \ \ \ }k=0,1,\cdots \] Where \({\displaystyle \sum } \sphericalangle z_{i}\) is sum of all angles from all zeros to the pole in question (the one we want to find the departure angle from) and \({\displaystyle \sum } \sphericalangle p_{i}\) is the sum of all the angles from each pole to the pole in question. In the above, the left side will contain only one unknown, which is \(\theta \), the angle of departure of that one pole. Do the above for each pole at a time.

Lemma 8 Apply the same method as lemma 7, but now do it for each zero at a time to find the arrival angles at each zero.

Lemma 9 Find where root locus crosses the imaginary axis. Use Routh table for this.

Example: \(GH=\frac{s^{2}+4s+8}{s^{2}\left ( s+4\right ) }\)Reader: reproduce the solution below

Reader: Do this problem sent to us by email also. Find root locus for \(G=\frac{K}{s\left ( s+4\right ) \left ( s^{2}+4s+20\right ) }\)

See my HW6 for more detailed root locus steps.

2.21.1 Summary of root locus

1.

Mark on plot all open loop poles and open zero locations.

2.

R.L. starts from open loop poles and ends up at open loop zeros. Some will end up at \(\pm \infty \) when \(n-m>0\) (which is almost always the case)

3.

Mark on real line where R.L. exist. Use the odd criteria.

4.

Find centroid of asymptotes \(\sigma =\frac{{\displaystyle \sum } p_{i}-{\displaystyle \sum } z_{i}}{n-m}\). i.e. sum of all poles minus sum of all zeros. \(n\) is number of poles, and \(m\) is number of zeros (these are for the open loop, not the closed loop!)

5.

Find asymptotes angles. \(\theta =\frac{180^{0}\pm 360^{0}}{n-m}\)

6.

Find breakaway points. From \(1+KG=0\) find \(k=f\left ( s\right ) \) and solve \(\frac{dK}{ds}=0\) for \(s\). These are locations where the breakaway and break-in points on root locus will be.

7.

Find departure angles for each pole (the complex ones, the ones on real line will have \(180^{0}\)) and arrival angles for the zeros (also complex ones). This can be done using \({\displaystyle \sum } \sphericalangle z_{i}-{\displaystyle \sum } \sphericalangle p_{i}-=180^{0}\pm 360^{0}k\), where in the LHS, we have one unknown angle each time. This is the angle of either the departure of arrival. Then solve for it. The above is sum of angles that all other poles and zeros make with the point in question. See HW 6 for details.

8.

Find where root locus crosses the imaginary line. Find \(K\) which makes the closed loop unstable. Then solve for the polynomial above that row in Routh table for this \(K\) and solve for \(s\). This will be where it crosses the imaginary line. See HW6 for details.