Chapter 7: The Root Locus Method

A self-contained, exam-focused study guide built strictly from the uploaded Ch07 slides and the Desired Outcomes document.

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1. What the Root Locus Is

For a unity-feedback system with controller gain $K$ and plant $G(s)$:

$T(s) = \frac{Y(s)}{R(s)} = \frac{KG(s)}{1 + KG(s)}$

Characteristic equation:

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

Root locus plot: the graphical display of the closed-loop pole locations in the $s$-plane as $K$ varies over $0 \le K \le \infty$.

Why we use it: transient response is closely related to the locations of the closed-loop poles. For many systems, simple gain adjustment moves the closed-loop poles to desired locations, so design reduces to choosing an appropriate $K$.

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2. Angle and Magnitude Conditions

For the more general feedback configuration with feedback path $H(s)$:

$\frac{Y(s)}{R(s)} = \frac{G(s)}{1 + G(s)H(s)}$

Characteristic equation:

$1 + G(s)H(s) = 0 \;\Longrightarrow\; G(s)H(s) = -1$

Because $-1$ has unit magnitude and angle $\pm 180^\circ$:

Angle condition

$\angle G(s)H(s) = \pm 180^\circ(2k+1), \quad k = 0, 1, 2, \ldots$

Magnitude condition

$|G(s)H(s)| = 1$

The values of $s$ that satisfy both conditions are the closed-loop poles.

After rearranging to standard form

$1 + K\frac{(s+z_1)(s+z_2)\cdots(s+z_m)}{(s+p_1)(s+p_2)\cdots(s+p_n)} = 1 + K\frac{Z(s)}{P(s)} = 0$

the angle condition reads

$(\phi_{z_1} + \cdots + \phi_{z_m}) - (\phi_{p_1} + \cdots + \phi_{p_n}) = \pm 180^\circ(2k+1)$

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3. Introductory Example: $G(s) = \dfrac{1}{s(s+2)}$

Closed-loop:

$\frac{Y(s)}{R(s)} = \frac{K}{s^2 + 2s + K}$

Characteristic equation:

$q(s) = s^2 + 2s + K = 0 \;\Longrightarrow\; s_{1,2} = -1 \pm \sqrt{1-K}$

$K$	Poles
0	$0,\ -2$
3/4	$-1/2,\ -3/2$
1	$-1,\ -1$ (breakaway)
5	$-1 \pm 2j$

Observations:

• $K=0$: closed-loop poles equal the open-loop poles.
• $0 < K \le 1$: real distinct poles, no oscillations.
• $K > 1$: complex conjugate poles, oscillatory but stable.
• Loci start at open-loop poles when $K=0$ and travel toward zeros (finite or at infinity) as $K \to \infty$.

Verification at $s_{1,2} = -1 \pm j$

Magnitude: $K = |-1+j|\cdot|-1+j+2| = \sqrt{2}\cdot\sqrt{2} = 2$.

Angle: $-\angle s_1 - \angle(s_1+2) = -135^\circ - 45^\circ = -180^\circ$ ✓

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4. Eight-Step Construction Procedure

Demonstrated on $G(s)H(s) = \dfrac{K}{s(s+1)(s+2)}$ throughout.

Step 1 — Locate open-loop poles and zeros

• Loci start at the poles of $Z(s)/P(s)$ when $K = 0$.
• Loci end at the zeros of $Z(s)/P(s)$ when $K \to \infty$ (finite or at infinity).

For the example: poles $0, -1, -2$; no finite zeros.

Step 2 — Root loci on the real axis (point test)

A point on the real axis lies on a root locus iff the total number of real poles and zeros to its right is odd.

For the example: loci exist on $[-1, 0]$ and on $(-\infty, -2]$.

Step 3 — Asymptotes

$\text{Number of asymptotes} = \#p - \#z$

$\psi = \frac{\pm 180^\circ(2k+1)}{\#p - \#z}, \quad \sigma = \frac{\sum \text{poles} - \sum \text{zeros}}{\#p - \#z}$

For the example: 3 asymptotes; $\psi = \pm 60^\circ, \pm 180^\circ$; $\sigma = -1$.

Step 4 — Breakaway / break-in points

Solve $\dfrac{dK}{ds} = 0$ and keep only roots that lie on root-locus segments.

• Two adjacent open-loop poles on a locus segment ⇒ at least one breakaway.
• Two adjacent open-loop zeros on a locus segment ⇒ at least one break-in.

For the example: $K = -s(s+1)(s+2)$, $\dfrac{dK}{ds} = -(3s^2 + 6s + 2) = 0 \Rightarrow s = -0.4226,\, -1.5774$. Only $s = -0.4226$ is on the locus.

Step 5 — Angle of departure / arrival

$\text{Angle of departure from } p_1 = 180^\circ - \sum_{i\ne 1}\phi_{p_i} + \sum_j\phi_{z_j}$

$\text{Angle of arrival at } z_1 = 180^\circ - \sum_{j\ne 1}\phi_{z_j} + \sum_i\phi_{p_i}$

Slide example: angle of departure from $p_2 = 180^\circ - (120^\circ+90^\circ) + 20^\circ = -10^\circ$.

Step 6 — Imaginary-axis crossings

Apply Routh–Hurwitz; the $K$ that creates a row of zeros, plus the auxiliary polynomial, gives the crossing.

For $q(s) = s^3 + 3s^2 + 2s + K$:

Row	Col 1	Col 2
$s^3$	$1$	$2$
$s^2$	$3$	$K$
$s^1$	$(6-K)/3$	$0$
$s^0$	$K$

$(6-K)/3 = 0 \Rightarrow K = 6$. Auxiliary $U(s) = 3s^2 + 6 = 0 \Rightarrow s = \pm j\sqrt{2}$.

Step 7 — Sketch the loci

The plot is symmetric about the real axis. For $K > 6$ (this example), the system is unstable.

num = [1];
den = [1 3 2 0];
sys = tf(num, den);
rlocus(sys);

Step 8 — Select closed-loop poles and find $K$

Magnitude condition:

$K = \frac{|s+p_1|\,|s+p_2|\cdots}{|s+z_1|\,|s+z_2|\cdots}\bigg|_{s=s_\text{desired}}$

Damping-ratio line: $\zeta = \cos\theta$, with $\theta$ from the negative real axis.

For $\zeta = 0.5$ ($\theta = 60^\circ$): $s_{1,2} = -0.3337 \pm j0.5780$, $K = 1.0383$, third pole $s_3 = -2.3326$.

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5. Worked Example with Real-Axis Zeros

$1 + K\frac{s^2 - 8s + 15}{s^2 + 3s + 2} = 0$

Poles $-1, -2$; zeros $+3, +5$. $\#p - \#z = 0$ → no asymptotes.

Breakaway/break-in: $11s^2 - 26s - 61 = 0 \Rightarrow s = -1.45$ (breakaway), $s = 3.82$ (break-in).

Imaginary-axis crossing: Routh row-of-zeros at $K = 3/8$; auxiliary $\tfrac{11}{8}s^2 + \tfrac{61}{8} = 0 \Rightarrow s = \pm 2.355j$.

Design ($T_p \le 4$ s and $T_s \le 8$ s):

$T_p = \frac{\pi}{\omega_d} \le 4 \Rightarrow |\omega_d| \ge 0.79$

$T_s = \frac{4}{\zeta\omega_n} \le 8 \Rightarrow |\zeta\omega_n| \ge 0.5$

Select locus points satisfying both.

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6. Worked Example — Parameter Other Than Gain

$q(s) = s^3 + (5+\alpha)s^2 + 8.5s + 5 = 0$

Convert to standard form:

$1 + \alpha\frac{s^2}{s^3 + 5s^2 + 8.5s + 5} = 0$

• Zeros: $z_1 = z_2 = 0$ (double at origin).
• Poles: $s_1 = -2$, $s_{2,3} = -1.5 \pm 0.5j$.
• 1 asymptote.

Breakaway: $s^4 - 8.5s^2 - 10s = 0 \Rightarrow s(s^3 - 8.5s - 10) = 0$; keep $s_1 = 0$.

Angle of departure from $p_1 = -1.5 + 0.5j$:

$180^\circ - (90^\circ + 45^\circ) + (161.6^\circ + 161.6^\circ) = 368.2^\circ \equiv 8.2^\circ$

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7. PID Controller

$G_c(s) = K_P + \frac{K_I}{s} + K_D s = \frac{K_D(s+z_1)(s+z_2)}{s}$

$u(t) = K_P\, e(t) + K_I\!\int e(t)\,dt + K_D\,\dot e(t)$

Gain	Meaning
$K_P$	Output proportional to error
$K_I$	Output proportional to time error is present
$K_D$	Output proportional to rate of change of error

Structural insight: PID adds one pole at the origin and two LHP-placeable zeros.

Effect of increasing PID gains on step response

Gain ↑	%Overshoot	Settling Time	Steady-State Error
$K_P$	Increases	Minimal impact	Decreases
$K_I$	Increases	Increases	Zero (eliminates)
$K_D$	Decreases	Decreases	No impact

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8. PID + Root Locus Example

Plant: $\dfrac{s+6}{(s+3)(s^2+9)}$ — poles $-3, \pm 3j$; zero $-6$.

(a) P-controller

$\#p - \#z = 2$, $\psi = \pm 90^\circ$, $\sigma = \dfrac{(-3-3j+3j)-(-6)}{2} = 1.5$ (RHP). Cannot stabilize.

(b) PD-controller (zero at $-9$)

$\#p - \#z = 1$, $\psi = \pm 180^\circ$ (heads left). Can be stabilized.

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9. Mass-Spring-Damper PID Example

Plant from $m\ddot x + b\dot x + kx = F$ with $m=1, b=10, k=20$:

$\frac{X(s)}{F(s)} = \frac{1}{s^2 + 10s + 20}$

Unit step: open-loop $x_{ss} = 0.05$, $e_{ss} = 0.95$.

Closed-loop TFs:

P: $T(s) = \dfrac{K_p}{s^2 + 10s + (20+K_p)}$

PD: $T(s) = \dfrac{K_d s + K_p}{s^2 + (10+K_d)s + (20+K_p)}$

PID: $T(s) = \dfrac{K_d s^2 + K_p s + K_i}{s^3 + (10+K_d)s^2 + (20+K_p)s + K_i}$

Controller	$K_P$	$K_I$	$K_D$	$x_{ss}$	$e_{ss}$	Notes
Open loop	—	—	—	0.05	0.95	very large error
P	300	—	—	0.9375	0.0625	oscillatory
PD	300	—	10	0.9375	0.0625	same error, fewer oscillations, faster $T_s$
PID	350	300	50	1	0	fast rise, small OS, no error

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10. How to Choose PID Gains

With a TF model: MATLAB / Simulink PID Tuner.

Without a model — Ziegler–Nichols:

1. Set $K_I = K_D = 0$.
2. Slowly increase $K_P$ until output has stable, consistent oscillations at ultimate gain $K_u$ with period $T_u$.

Controller	$K_P$	$T_i$	$T_d$	$K_I$	$K_D$
P	$0.5\,K_u$	—	—	—	—
PI	$0.45\,K_u$	$0.80\,T_u$	—	$0.54\,K_u/T_u$	—
PD	$0.8\,K_u$	—	$0.125\,T_u$	—	$0.10\,K_u T_u$
Classic PID	$0.6\,K_u$	$0.5\,T_u$	$0.125\,T_u$	$1.2\,K_u/T_u$	$0.075\,K_u T_u$
Pessen Integral	$0.7\,K_u$	$0.4\,T_u$	$0.15\,T_u$	$1.75\,K_u/T_u$	$0.105\,K_u T_u$
Some overshoot	$0.33\,K_u$	$0.50\,T_u$	$0.33\,T_u$	$0.66\,K_u/T_u$	$0.11\,K_u T_u$
No overshoot	$0.20\,K_u$	$0.50\,T_u$	$0.33\,T_u$	$0.40\,K_u/T_u$	$0.066\,K_u T_u$

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11. Full Worked Study Question (from Desired Outcomes)

$GH(s) = \frac{K(s+5)}{s^2 + 4s + 3}$

Poles $-1, -3$; finite zero $-5$; one zero at infinity.

Real-axis loci: between $-1$ and $-3$; between $-5$ and $-\infty$.

Asymptotes: $\#p-\#z = 1$, $\psi = \pm 180^\circ$ (single asymptote — no centroid needed).

Breakaway/break-in:

$K = -\frac{s^2 + 4s + 3}{s+5}, \quad \frac{dK}{ds} = -\frac{(2s+4)(s+5) - (s^2+4s+3)}{(s+5)^2} = 0$

$s^2 + 10s + 17 = 0 \Rightarrow s = -5 \pm 2\sqrt{2}$

$s_1 = -7.83,\quad s_2 = -2.17$

• $s = -2.17$: breakaway between poles $-1$ and $-3$, at $K = 0.343$.
• $s = -7.83$: break-in between zero $-5$ and $-\infty$, at $K = 11.66$.

MATLAB code

clc; clear; close all;
num = [1 5];
den = [1 4 3];
GH = tf(num, den);
figure;
rlocus(GH);
grid on;
title('Root Locus of GH(s) = K(s+5) / (s^2 + 4s + 3)');
xlabel('Real Axis');
ylabel('Imaginary Axis');

Range of $K$ for overdamped response (real poles)

$0 < K \le 0.343 \quad\text{or}\quad 11.66 < K < \infty$

Range of $K$ for underdamped response (complex poles)

$0.343 < K < 11.66$

Find $K$ for $T_s = 0.8$ s

$T_s \approx \dfrac{4}{\zeta\omega_n} \Rightarrow \zeta\omega_n = 5$. Vertical line at $\sigma = -5$.

Grid intersection: $s = -5 \pm 2.8j$.

$K = \left|\frac{s^2+4s+3}{s+5}\right|_{s=-5+2.8j} = \frac{16.8}{2.8} = 6$

Verification: $T(s) = \dfrac{K(s+5)}{s^2 + (4+K)s + (3+5K)}$; $2\zeta\omega_n = 4+K \Rightarrow 10 = 4+K \Rightarrow K = 6$. With $K=6$: $q(s) = s^2+10s+33$, roots $-5 \pm 2\sqrt{2}j \approx -5 \pm 2.8j$ ✓

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12. Useful Derivations

Asymptote-angle formula

At a far test point, all pole/zero angles ≈ $\psi$. The angle condition becomes

$\#z\,\psi - \#p\,\psi = \pm 180^\circ(2k+1) \Rightarrow \psi = \frac{\pm 180^\circ(2k+1)}{\#p - \#z}$

Breakaway condition $dK/ds = 0$

From $P(s) + KZ(s) = 0$, incrementing $K$ by $\Delta K$ and dividing by the original characteristic polynomial yields, at a root of multiplicity $n$:

$\frac{\Delta K}{\Delta s} = -\frac{(\Delta s_i)^{n-1}}{c_i} \;\xrightarrow{\Delta s \to 0}\; \frac{dK}{ds} = 0$

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13. Key Facts to Remember

1. Loci start at open-loop poles ($K=0$) and end at open-loop zeros ($K\to\infty$, finite or at infinity).
2. Number of branches = number of open-loop poles.
3. Plot is symmetric about the real axis.
4. Real-axis test: odd count of real poles + zeros to the right.
5. Number of asymptotes = $\#p - \#z$.
6. $j\omega$-axis crossings: Routh–Hurwitz + auxiliary polynomial.
7. PID contributes one pole at the origin and two LHP-placeable zeros.
8. $T_s \approx 4/(\zeta\omega_n)$ → vertical line in $s$-plane.
9. $T_p = \pi/\omega_d$ → horizontal line in $s$-plane.
10. $\zeta = \cos\theta$, $\theta$ measured from the negative real axis.