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2.19 Lecture 18, Thursday Oct 29, 2015, More root locus

Reminder: No lecture Tuesday Nov 3. Makeup lecture on Wed. Nov 18 at 6 pm.

Will continue root locus. We are using the classical setup

We want to study the behavior of closed loop poles as $K$ changes from zero to infinity. There is a geometric condition which is the driver behind root locus. It is the angle condition. This condition says that a point $s^{\ast }$ is on R.L. iff phase of open loop $G\left ( s\right ) H\left ( s\right )$ evaluated at $s^{\ast }$ is $\pi$

$$\sphericalangle \left . G\left ( s\right ) H\left ( s\right ) \right \vert _{s=s^{\ast }}=\pi$$ When the point $s^{\ast }$ is on the R.L., then the corresponding gain is (notice, this is valid only after we decided the point $s^{\ast }$ is on R.L.) $$k=\frac{1}{\left \vert G\left ( s\right ) H\left ( s\right ) \right \vert _{s=s^{\ast }}}$$ Let us now look at $G\left ( s\right ) H\left ( s\right )$ with two zeros and three poles. $$G\left ( s\right ) H\left ( s\right ) =\frac{\left ( s-z_{1}\right ) \left ( s-z_{2}\right ) }{\left ( s-p_{1}\right ) \left ( s-p_{2}\right ) \left ( s-p_{3}\right ) }$$ We want to know if point $s^{\ast }$ is on R.L.

How to know if $s^{\ast }$ on R.L.? We check the angle $\sphericalangle \left . G\left ( s\right ) H\left ( s\right ) \right \vert _{s=s^{\ast }}$ and see if it $\pi$. We do this graphically, like this

We look at the angle each pole and zero makes with the point $s^{\ast }$, then

$$\sphericalangle \left . G\left ( s\right ) H\left ( s\right ) \right \vert _{s=s^{\ast }}=\left ( \theta _{1}+\theta _{2}\right ) -\left ( \psi _{1}+\psi _{2}+\psi _{3}\right )$$ i.e. we add the zero angles, and subtract the poles angles. Angle is always measured counter clock wise from the horizontal line as shown. If the above adds to $\pi$ then the point $s^{\ast }$ is on R.L. The gain $K$ in this case is $$K=\frac{1}{\left \vert G\left ( s\right ) H\left ( s\right ) \right \vert }=\frac{1}{\frac{\left \vert \left ( s-z_{1}\right ) \left ( s-z_{2}\right ) \right \vert }{\left \vert \left ( s-p_{1}\right ) \left ( s-p_{2}\right ) \left ( s-p_{3}\right ) \right \vert }}=\frac{\left \vert \left ( s-p_{1}\right ) \left ( s-p_{2}\right ) \left ( s-p_{3}\right ) \right \vert }{\left \vert \left ( s-z_{1}\right ) \left ( s-z_{2}\right ) \right \vert }=\frac{D_{1}D_{2}D_{3}}{d_{1}d_{2}}$$ Where $D_{i}$ is the size of the vector from each pole $p_{i}$ to $s^{\ast }$ and $d$ is the length of the vector from each zero $z_{i}$ to $s^{\ast }$

Back to the 9 lemmas. When used together, enable us to draw R.L. for any system.

Lemma 1: R.L. has $n$ branches, where $n$ is number of open loop $GH$ poles.

Lemma 2: We talk about very small $K$ and very large $K$. When $K$ is very small, R.L. is at the open loop poles. This means R.L. always starts from the open loop pole locations. When $K$ is very large, R.L. is at open loop zeros. For example $GH=\frac{\left ( s+1\right ) \left ( s-2\right ) }{\left ( s+4\right ) ^{5}}$, the open loop has poles at $s=-4$, this is where R.L. starts from. Since it has zero at $s=-1$ and zero at $s=2$, then this is where 2 of the branches will end up at. This also means for large gain $K$ the closed loop is not stable, since it ends up at $s=2.$

Since $n=5$ in the above, so we have 5 branches. Two of these end up at $s=-1$ and $s=2$. What about the other three? Those will end up infinity. And we need to check if there will be stable or not.

Lemma 3 (called the number criterion) Which real axis points are on the R.L. ? Since complex poles lead to angle cancellations when point $s^{\ast }$ being checked for is on the real axis, then only  poles and zeros of the open loop decide if a point $s^{\ast }$ is on R.L. or not. A real axis point $s^{\ast }$ is on R.L. iff the number of real axis poles and zeros (taking multiplicity into account) found as we travel towards $s^{\ast }$ from the right side is odd.  

Example: $GH=\frac{\left ( s+2\right ) \left ( s-5\right ) ^{2}\left ( s^{2}+8s+20\right ) }{\left ( s+1\right ) ^{2}\left ( s^{3}-1\right ) \left ( s^{2}+s+1\right ) }$. R.L. has 7 branches since $n=7$ from lemma 1. From lemma 2, we know R.L. starts from open loop poles and end up at open loop zeros. Now we find which part of the real axis are on R.L. We use lemma 3. We just need to mark the open loop pole and zeros that are on the real axis for this. All complex poles and zero have no effect. Hence the diagram is

This system is unstable at high gain $K$ and also unstable to low gain $K$. There are 2 branches that do not end up at zeros since $n=7$ and $m=5$. Next we need to check what happens when $k\rightarrow \infty$ after counting for the open loop zeros, since we know what happens in this case. How does phase behave when $k\rightarrow \infty$? When $K$ is very large, we can approximate $GH=\frac{N\left ( s\right ) }{D\left ( s\right ) }\approx \frac{R_{1}S^{m}}{R_{2}S^{n}}=\frac{R_{1}e^{jm\theta }}{R_{1}e^{jn\theta }}=Ae^{j\left ( m-n\right ) \theta }$. Therefore we need $\left ( m-n\right ) \theta =\pi$ for a point to be on R.L. Hence

$$\theta =\frac{\pi }{m-n}+M\left ( 2\pi \right )$$ From some integer $M$. Lemma 4 will be discussed more next time.