Root Locus


By Prof. Seungchul Lee
http://iai.postech.ac.kr/
Industrial AI Lab at POSTECH

Table of Contents

1. Motivation for Root LocusΒΆ



  • For example,


$$\text{System} = {s^2 + s + 1 \over s^3 + 4s^2 + Ks + 1}$$

  • Unknown parameter affects poles
  • Poles of system are values of $s$ when


$$s^3 + 4s^2 + \color{red}{K}s + 1= 0$$


  • What value of $K$ should I choose to meet my system performance requirement?



2. Root LocusΒΆ

  • Given the open loop transfer function $G(s)$, the typical closed-loop transfer function is





$${KG(s) \over 1 + KG(s)}$$


  • The root locus of an (open-loop) transfer function $G(s)$ is a plot of the locations (locus) of all possible closed-loop poles with some parameter, often a proportional gain $K$, varied between 0 and $\infty$.
  • The basic form for drawing the root locus


$$1 + KG(s) = 0$$

  • In MATLAB, rlocus(G(s))
  • The same denominator system is



$${G(s) \over 1 + KG(s)}$$
  • But you noticed that in the previous example I used




$$ \begin{align*} s^3 + 4s^2 + Ks + 1 &= 0, \qquad \text{not in the correct form}\\\\ 1 + K{s \over s^3 + 4s^2 + 1} &= 1 + KG(s) = 0 \end{align*}$$


  • Equivalent $G(s)$ and the closed loop system





2.1. Definition: Root LocusΒΆ

  • Root Locus: a graphical representation of closed-loop poles as $K$ varied
  • Based on root-locus graph, we can choose the parameter $K$ for stability and the desired transient response.



  • So why should we care about this?



  • Now that we understand how pole locations affect the system
  • Question: how do we draw root locus ?
    • for more complex system and
    • without calculating poles
  • We will be able to make a rapid sketch of the root locus for higher-order systems without having to factor the denominator of the closed-loop transfer function.
    • You might not use an exact sketch very often in practice, but you will use an approximated one!
    • What does closed loop root locus look like from open loop?
  • The closed-loop system is





$$H(s) = \frac{KG}{1+KG}$$


  • A pole exists when the characteristic polynomial in the denominator becomes zero.


$$1+KG(s) = 0 \implies KG(s) = -1 = 1 \angle (2k+1) \pi, \quad k = 0, \pm1, \pm 2, \cdots$$

  • A value of $s^*$ is a closed loop pole if


$$\begin{cases} \lvert KG(s^*) \rvert = 1 \quad \implies \quad K = \frac{1}{\lvert G(s^*)\rvert}\\\\ \angle KG(s^*) = (2k+1)\pi \end{cases}$$

2.2. Eight RulesΒΆ

  • There will be 8 rules to drawing a root locus
$$ 1 + KG(s) = 1 + K {{Q(s)}\over{P(s)}} = 0 $$
  • Rule 1: There are $n$ lines (loci) where $n$ is the degree of $Q$ or $P$, whichever greater.
  • Rule 2: As $K$ increases from $0$ to $\infty$, the roots move from the pole of $G(s)$ to the zeros of $G(s)$


$$ P(s) + KQ(s) = 0 $$


  • Poles of $G(s)$ are when $P(s) = 0$, $K = 0$

  • Zeros of $G(s)$ are when $Q(s) = 0$, as $K \rightarrow \infty, P(s) + \infty Q(s) = 0$

  • So closed loop poles travel from poles of $G(s)$ to zeros of $G(s)$




  • Poles and zeros at infinity
    • $G(s)$ has a zero at infinity if $G(s \rightarrow \infty) \rightarrow 0$
    • $G(s)$ has a pole at infinity if $G(s \rightarrow \infty) \rightarrow \infty$
  • Example


$$KG(s) = \frac{K}{s(s+1)(s+2)}$$

  • Clearly, this open loop transfer function has three poles $0, -1, -2$. It has no finite zeros.
  • For large $s$, we can see that


$$KG(s) \approx \frac{K}{s^3}$$

  • So this open loop transfer function has three zeros at infinity
  • Rule 3: When roots are complex, they occur in conjugate pairs (= symmetric about real axis)




  • Rule 4: At no time will the same root cross over its path



  • Rule 5: The portion of the real axis to the left of an odd number of open loop poles and zeros are part of the loci
    • which parts of real line will be a part of root locus?




$$G(s) = {Q(s) \over P(s)} = {\prod (s - z_i)\over \prod (s - p_j)}$$


  • For complex conjugate zero and pole pair $\implies \angle G(\cdot) = 0$




  • For real zeros and poles



  • Rule 6: Lines leave (breakout) and enter (breakin) the real axis at $90\,^{\circ}$
  • Rule 7: If there are not enough poles and zeros to make a pair, then the extra lines go to or come from infinity.





  • Rule 8: Lines go to infinity along asymptotes


$n-m$ = # poles - # zeros = number of lines that go to infinity.

  • The angles of the asymptotes
$$ \phi_{A} = {{2k+1}\over{n-m}} 180\,^{\circ} \quad \text{where } k = 0, 1, \cdots , n-m-1$$
  • The centroid of the asymptotes
$$ \sigma_A = {{\sum \text{finite poles} - \sum \text{finite zeros}}\over {n-m}} $$
  • (brief proof) Lines go to infinity along asymtotes


$$ \begin{align*} G(s) &= {1 \over (s+1)(s+2)} \approx {1 \over s}\cdot{1 \over s} \\\\ \angle G(s) &= \pi + 2k\pi \\ &\approx 0 - 2\alpha = \begin{cases} \pi\\ -\pi\\ \end{cases}\\\\ \therefore \alpha &= {\pi \over 2} \;\text{ or}\; -{\pi \over 2} \end{align*} $$$$$$

$$ \phi_A = {(2k+1)\pi \over n - m}$$




  • (brief proof) The centroid of the asymptotes


$$ \sigma_A = {{\sum \text{finite poles} - \sum \text{finite zeros}}\over {n-m}} $$


Given the system transfer function


$$G(s) = \beta\frac{(s-z_1)(s-z_2)\cdots}{(s-p_1)(s-p_2)\cdots} = \beta \frac{s^m - \left(\sum z_i \right)s^{m-1} + \cdots}{s^n - \left(\sum p_i \right)s^{n-1} + \cdots}, \quad \text{assume }\; n>m$$


For a large value of $s$, $G(s)$ approximately looks like having $n-m$ repeated poles at $\sigma_A$ on the real axis


$$ \begin{align*} G(s) = \beta \frac{s^m - \left(\sum z_i \right)s^{m-1} + \cdots}{s^n - \left(\sum p_i \right)s^{n-1} + \cdots} \; &\approx \;\beta \frac{1}{(s-\sigma_A)^{n-m}} = \beta \frac{1}{s^{n-m} - (n-m) \sigma_A s^{n-m-1} + \cdots}\\\\\\ \left(s^m - \left(\sum z_i \right)s^{m-1} + \cdots \right) \left(s^{n-m} - (n-m) \sigma_A s^{n-m-1} + \cdots \right) \; &\approx \; s^n - \left(\sum p_i \right)s^{n-1} + \cdots \\\\ s^n - \left( \sum z_i + (n-m) \sigma_A \right)s^{n-1} + \cdots \; &\approx \; s^n - \left(\sum p_i \right)s^{n-1} + \cdots \\\\\\ \sum z_i + (n-m) \sigma_A &= \sum p_i \\\\ \therefore \; \sigma_A &= \frac{\sum p_i - \sum z_i}{n-m} \end{align*} $$
  • If $n-m = 1$
$$ \phi_{A} = {{2 \centerdot 0 + 1}\over{1}}180 = 180\,^{\circ}$$




  • If $n-m = 2$
$$ \begin{align*} \phi_{A1} &= {{2 \centerdot 0 + 1}\over{2}}180 = 90\,^{\circ}\\ \phi_{A2} &= {{2 \centerdot 1 + 1}\over{2}}180 = 270\,^{\circ} \\\\ \sigma_A &= {{(-2 - 1) - (0)}\over{2}}= -1.5 \end{align*} $$




  • If $n-m = 3$
$$ \begin{align*} \phi_{A1} &= {{2 \centerdot 0 + 1}\over{3}}180 = 60\,^{\circ}\\ \phi_{A2} &= {{2 \centerdot 1 + 1}\over{3}}180 = 180\,^{\circ}\\ \phi_{A3} &= {{2 \centerdot 2 + 1}\over{3}}180 = 300\,^{\circ}\\\\ \sigma_A &= {{(-1 -2 -3) - (0)}\over{3}}= -2 \end{align*} $$




  • If $n-m = 4$
$$ \begin{align*} \phi_{A1} &= {{2 \centerdot 0 + 1}\over{4}}180 = 45\,^{\circ}\\ \phi_{A2} &= {{2 \centerdot 1 + 1}\over{4}}180 = 135\,^{\circ}\\ \phi_{A3} &= {{2 \centerdot 2 + 1}\over{4}}180 = 225\,^{\circ}\\ \phi_{A4} &= {{2 \centerdot 3 + 1}\over{4}}180 = 315\,^{\circ}\\\\ \sigma_A &= {{(1+2-1-2) - (0)}\over{4}}= 0 \end{align*} $$




2.3. Break-away, Break-in PointsΒΆ

  • Break-away is the point where loci leave the real axis.

  • Break-in is the point where loci enter the real axis.

The method is to maximize and minimizes the gain, $K$, using differential calculus.

  • For all points on the root locus,
$$K = -{1\over{G(s)}}$$
  • When ${dK \over ds} = 0$, $K$ is Break-away and Break-in.
$${d \over ds} {1 \over G(s)} = 0$$
  • Determine the breakaway points
$$G(s) = {1 \over (s+2)(s+4)}$$




$$ \begin{align*} 1 + K{1 \over (s+2)(s+4)} = 0 &\Rightarrow s^2 + 6s + 8 + K = 0 \\ &\Rightarrow s = -3 \pm \sqrt{9 - (8 + K)} = -3 \pm \sqrt{1 - K} \end{align*} $$


  • When $K < 1$ : two solutions, overdamped
  • When $K > 1$ : two complex numbers, underdamped
  • With respect to $K$, (as value of $K$ changes)
    • The number of solutions changes $0 \rightarrow 1 \rightarrow 2 \; \text{ or }\; 2 \rightarrow 1 \rightarrow 0$




$$ \begin{align*} 1 + KG(s) &= 0 \implies K = -{1 \over G(s)}\\\\ {dK \over ds} &= 0 \;\text{at a breakaway point}\\\\ K &= -(s+2)(s+4) = -(s^2 + 6s + 8)\\\\ {dK \over ds} &= -(2s + 6) = 0 \\\\ \therefore s &= -3 \end{align*} $$

2.4. Find Angles of Departure/Arrival for Complex Poles/ZerosΒΆ


$$G(s) = {s - 0 \over (s - (-1 + i))(s - (-1 - i))}$$