Root Locus

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

# 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.

• 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

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