Chapter 12: Root Locus

Closed-Loop Transfer Function

The closed-loop transfer function is:

\[ G(s) = \frac{C(s) P(s)}{1 + C(s) P(s)} = \frac{K}{s + 1 + K} \]

This is a first-order system with a single pole located at:

\[ \boxed{s = -1 - K} \]

As \( K \to \infty \), the pole moves toward \( s = -\infty \). The system becomes increasingly fast and remains stable for all positive values of \( K \) as shown in the figure:

We can also visualize the motion of this pole as \( K \) changes using Mathematica:

Pole Behavior for Real Poles

For this simple case, the root locus consists of a single branch entirely along the real axis. The key behavior is:

- At \( K = 0 \), the pole is at \( s = -1 \) (the open-loop pole)
- As \( K \) increases, the pole moves leftward on the real axis
- The pole never becomes complex
- As \( K \to \infty \), \( s \to -\infty \)

This reflects the typical behavior for systems with a single real pole: increasing gain improves speed and reduces rise/settling time without introducing oscillation.

Characteristic Equation

\[
1 + \frac{K}{(s + 1)(s + 2)} = 0 \Rightarrow (s + 1)(s + 2) + K = 0
\Rightarrow s^2 + 3s + (2 + K) = 0
\]

The closed-loop poles are:

\[
\boxed{s = \frac{-3 \pm \sqrt{1 - 4K}}{2}}
\]

Qualitative Behavior

- When \( K < \frac{1}{4} \), the poles are real and distinct
- When \( K = \frac{1}{4} \), the poles are repeated (critically damped)
- When \( K > \frac{1}{4} \), the poles become complex conjugates
- As \( K \to \infty \), the poles approach:
\[
s \approx \frac{-3 \pm j\sqrt{4K - 1}}{2}
\]
The real part stays fixed at \( -\frac{3}{2} \), but the imaginary part grows.

Thus, for second-order systems, increasing \( K \) can cause poles to become complex and oscillatory — but they still remain in the left-half plane because the real part remains negative.

Characteristic Equation

\[
1 + K \cdot \frac{s - 1}{(s + 1)(s + 2)} = 0
\]

<h3>Closed-Loop Poles</h3>

\[
\boxed{s = \frac{-(3 + K) \pm \sqrt{(3 + K)^2 - 4(2 - K)}}{2}}
\]

As \( K \) increases:
- The constant term \( 2 - K \) becomes negative when \( K > 2 \)
- This implies the product of the roots is negative → one pole must be positive → the system becomes <strong>unstable</strong>

Key Insight

- For simple systems, increasing \( K \) moves poles to \( -\infty \)
- For higher-order systems with RHP zeros, increasing \( K \) can <strong>destabilize</strong> the system
- Root locus allows us to track how poles move and determine safe values of \( K \)

Summary

- In first-order systems like \( P(s) = \frac{1}{s + 1} \), increasing \( K \) moves the pole leftward — faster and more stable
- In second-order systems, poles can become complex as \( K \) increases — leading to oscillations
- If the plant has right-half-plane zeros, increasing \( K \) can push poles into the RHP — causing instability
- Root locus gives a graphical way to understand this behavior and guide controller design

Root locus plots are useful because they allow us to:

- Predict closed-loop pole locations without solving for roots numerically
- Design controllers by adding poles/zeros
- Assess how gain \( K \) influences stability, damping, and response time

Rule 0: Symmetry About the Real Axis

The root locus is always symmetric with respect to the real axis. This is because complex poles always appear in conjugate pairs, so their paths mirror each other about the real axis.

Rule 1: Number of Branches and Endpoints

Let \( L(s) = \frac{N(s)}{D(s)} \) with:

- \( n = \) number of poles of \( L(s) \)
- \( m = \) number of zeros of \( L(s) \)

Then:

- The root locus has \( n \) branches, one for each pole
- \( m \) branches terminate at the finite zeros of \( L(s) \)
- \( n - m \) branches go to infinity, approaching asymptotes

Note: \( n \geq m \) must hold to preserve causality.

Rule 2: Real-Axis Segments

A point \( s \) on the real axis belongs to the root locus if the total number of poles and zeros to the right of \( s \) on the real axis is odd.

This rule can be applied visually using a vertical line test.

Rule 3: Asymptotes of Branches Going to Infinity

For \( n - m \) branches that go to infinity, the asymptotes intersect the real axis at:

\[
\alpha = \frac{\sum \text{(poles)} - \sum \text{(zeros)}}{n - m}
\]

and depart at angles:

\[
\theta_k = \frac{(2k + 1)\pi}{n - m}, \quad k = 0, 1, 2, \dots
\]

Rule 4: Angle of Departure and Arrival

For complex poles and zeros not on the real axis, the angle of departure from a pole and angle of arrival at a zero can be calculated using:

\[
\phi_{\text{departure}} = 180^\circ - \sum \angle(\text{vectors from other poles and zeros})
\]

This rule is tedious and typically used only for precision analysis but can be necessary to perform.

Given System

Consider the open-loop transfer function:

\[
L(s) = \frac{s + 4}{(s + 2)(s^2 + 4s + 20)}
\]

This system has:

- One finite zero at \( z_1 = -4 \)
- Three poles:
- One real pole at \( p_1 = -2 \)
- Two complex poles at \( p_{2,3} = -2 \pm j4 \)

Step 1: Number of Branches and Asymptotes

- Number of poles: \( n = 3 \)
- Number of zeros: \( m = 1 \)
- Number of asymptotes: \( n - m = 2 \)

Centroid of Asymptotes:

\[
\alpha = \frac{p_1 + p_2 + p_3 - z_1}{n - m}
= \frac{-2 + (-2 + j4) + (-2 - j4) - (-4)}{2}
= \frac{-6 + 4}{2} = -1
\]

Asymptote Angles:

\[
\theta_k = \frac{(2k + 1)\pi}{n - m}, \quad k = 0, 1
\Rightarrow \theta_0 = 90^\circ, \quad \theta_1 = 270^\circ
\]

These asymptotes indicate that two branches go to infinity vertically, one upward and one downward from \( s = -1 \).

Step 2: Angle of Departure from a Complex Pole

Compute the angle of departure from the pole at \( p_2 = -2 + j4 \):

\[
\phi_d = \pi + \sum_j \angle(p - z_j) - \sum_{i \neq p} \angle(p - p_i)
\]

Where:

- \( p = -2 + j4 \)
- \( z_1 = -4 \)
- \( p_1 = -2 \), \( p_3 = -2 - j4 \)

Compute angles:

\[
\angle(p - z_1) = \arg(2 + j4) = \tan^{-1}\left(\frac{4}{2}\right) = 63.4^\circ
\]

\[
\angle(p - p_1) = \arg(j4) = 90^\circ, \quad \angle(p - p_3) = \arg(j8) = 90^\circ
\]

Final calculation:

\[
\phi_d = 180^\circ + 63.4^\circ - (90^\circ + 90^\circ)
= 180 + 63.4 - 180 = \boxed{63.4^\circ}
\]

So the root locus departs from the complex pole at \( -2 + j4 \) at an angle of approximately \( 63.4^\circ \).

Summary of This Example

- The system has 3 poles and 1 zero
- The root locus has 2 asymptotes at \( \pm 90^\circ \), centered at \( \alpha = -1 \)
- The root locus leaves the complex pole at \( -2 + j4 \) at an angle of \( 63.4^\circ \)
- Symmetry rule guarantees the conjugate pole departs at \( 63.4^\circ \) below the real axis

Rule 5: Breakaway and Break-in Points

When two or more branches of the root locus come together or separate on the real axis, they do so at breakaway (diverging) or break-in (converging) points. These points are always located on the real axis and must lie within segments of the root locus (as determined by Rule 2).

To calculate them, write:

\[
L(s) = \frac{n(s)}{d(s)}
\]

Then apply the quotient rule to the open-loop transfer function (excluding the gain \( k \)), and solve:

\[
\frac{d}{ds} \left( \frac{n(s)}{d(s)} \right) = 0
\]

That is, compute:

\[
\frac{n'(s) d(s) - n(s) d'(s)}{d(s)^2} = 0
\]

This simplifies to:

\[
n'(s) d(s) - n(s) d'(s) = 0
\]

Only real-valued roots of this equation that lie on valid segments of the real axis are valid breakaway or break-in points.

Worked Example

Let the open-loop transfer function be:

\[
L(s) = \frac{k(s+3)}{s^2 + 3s + 2}
\]

Zeros: \( s = -3 \)
Poles: \( s = -1, -2 \)
This system has two poles and one zero, so one branch will go to the zero and one will go to infinity.

Step 1: Use the Quotient Rule

Let:

\[
n(s) = s + 3, \quad d(s) = s^2 + 3s + 2
\]

Then:

\[
n'(s) d(s) - n(s) d'(s) = 1 \cdot (s^2 + 3s + 2) - (s + 3)(2s + 3)
\]

Expand both terms:

\[
(s^2 + 3s + 2) - (2s^2 + 9s + 9) = -s^2 - 6s - 7
\]

Solve:

\[
-s^2 - 6s - 7 = 0 \Rightarrow s^2 + 6s + 7 = 0
\]

\[
s = \frac{-6 \pm \sqrt{36 - 28}}{2} = \frac{-6 \pm \sqrt{8}}{2} = \frac{-6 \pm 2\sqrt{2}}{2} = -3 \pm \sqrt{2}
\]

\[
\Rightarrow s \approx -4.414,\quad s \approx -1.586
\]

Step 2: Check Real-Axis Segments (Rule 2)

From Rule 2: A point on the real axis is part of the root locus if the number of real poles and zeros to the right of that point is odd.

Real poles: \( s = -1,\ -2 \)
Real zero: \( s = -3 \)

Segments of the real axis:

- \( (-\infty, -3) \): 3 elements to the right → on locus
- \( (-3, -2) \): 2 to the right → not on locus
- \( (-2, -1) \): 1 to the right → on locus
- \( (-1, \infty) \): 0 to the right → not on locus

So the breakaway and break-in points:

\[
\boxed{-4.414 \text{ and } -1.586}
\]

are both valid — they lie on segments of the real axis where the root locus exists.

Step 3: Asymptote Calculation

Number of poles: \( n = 2 \), number of zeros: \( m = 1 \)
So there is one branch going to infinity.

Asymptote:

\[
\alpha = \frac{\sum \text{poles} - \sum \text{zeros}}{n - m} = \frac{-1 + (-2) - (-3)}{1} = 0
\]

Angle:

\[
\theta_k = \frac{(2k+1)\pi}{n - m} = \pi \Rightarrow 180^\circ
\]

Conclusion

- Root locus starts at poles \( s = -1 \) and \( s = -2 \)
- One branch terminates at the zero at \( s = -3 \)
- One branch goes to infinity along the real axis at angle \( 180^\circ \)
- Breakaway at \( s \approx -1.586 \), break-in at \( s \approx -4.414 \)

Step 1: Try Proportional Control

\[
C(s) = K
\]

If the root locus of \( K P(s) \) does not intersect the desired region for any value of \( K \), then P control is insufficient.

Let us consider the open-loop transfer function:

\[
L(s) = \frac{1}{s(s + 1)}
\]

This is a Type 1, second-order system with:

- Poles at \( s = 0 \) and \( s = -1 \)
- No finite zeros
- System order: \( n = 2 \), number of zeros \( m = 0 \)
- \( n - m = 2 \) branches of the root locus

Closed-loop characteristic equation (with P controller):

\[
1 + K \cdot \frac{1}{s(s + 1)} = 0
\Rightarrow s(s + 1) + K = 0
\Rightarrow s^2 + s + K = 0
\]

Closed-loop poles:

\[
s = \frac{-1 \pm \sqrt{1 - 4K}}{2}
\]

Behavior of the poles as \( K \) varies:

- For \( K < \frac{1}{4} \), poles are real and negative
- For \( K = \frac{1}{4} \), repeated real pole (critical damping)
- For \( K > \frac{1}{4} \), poles become complex conjugates
- Real part of poles approaches \( -\frac{1}{2} \) as \( K \to \infty \)

Desired Closed-Loop Pole Region

The design specifications require:

- Settling time \( t_s < 2 \) seconds
- Maximum overshoot \( < 10\% \)

These constraints define the following pole region in the complex \( s \)-plane:

- Minimum damping ratio: \( \zeta > 0.528 \)
- Minimum exponential decay rate: \( \sigma > 2.3 \)
- Minimum natural frequency: \( \omega_n > 4.35 \)

This defines a shaded “safe” region in the left-half plane where:

- The poles lie to the left of \( \operatorname{Re}(s) = -2.3 \)
- The poles lie within a cone defined by \( \theta = \cos^{-1}(0.528) \approx 58.1^\circ \)
- The poles lie outside a circle of radius \( \omega_n = 4.35 \)

Compare with Root Locus of P-Controlled System

\[
L(s) = \frac{K}{s(s + 1)} \Rightarrow \text{Characteristic equation: } s^2 + s + K = 0
\]

This gives the closed-loop poles:

\[
s = \frac{-1 \pm \sqrt{1 - 4K}}{2}
\]

As \( K \to \infty \), the real part of the poles approaches:

\[
\operatorname{Re}(s) \to -\frac{1}{2}
\]

This does not satisfy the requirement \( \operatorname{Re}(s) < -2.3 \). The root locus stays too close to the imaginary axis, and overshoot and settling time requirements are violated.

Conclusion: P Control is Insufficient

\[
\boxed{\text{Proportional control alone cannot enter the desired region.}}
\]

The closed-loop poles never reach the region defined by \( \zeta > 0.528 \), \( \sigma > 2.3 \), and \( \omega_n > 4.35 \). Therefore, additional control (e.g., \( K_d \), \( K_i \)) is needed to meet specs.

Therefore, we must increase the complexity of the controller.

In the next step, we can add a zero via lead control:

\[
C(s) = K \cdot (s + z)
\]

This will bend the root locus toward the desired pole region by pulling branches leftward, enabling faster and more stable response.

Step 2: Use Lead Compensation

To move the root locus into the desired region, add a zero via lead compensation:

\[
C(s) = K \cdot \frac{s + z}{s + p}, \quad z < p
\]

This zero attracts the locus toward the desired region. For effective placement:

- Place the zero near the desired pole location
- Position the pole further left to ensure lead (phase advancement)

Example System with Lead Compensation

Suppose we have the open-loop transfer function:

\[
L(s) = \frac{k (s + 1.5)}{s(s + 1)(s + 8)(s + 6)}
\]

This corresponds to a unity-feedback system with:

- One compensator zero at \( s = -1.5 \)
- Four finite poles:
- \( s = 0 \) (integrator)
- \( s = -1 \)
- \( s = -6 \) (lead compensator pole)
- \( s = -8 \)
- System order \( n = 4 \), number of zeros \( m = 1 \)
- \( n - m = 3 \) root locus branches go to infinity

Root Locus Geometry

Centroid of asymptotes:

\[
\alpha = \frac{0 + (-1) + (-6) + (-8) - (-1.5)}{3} = \frac{-13.5}{3} = -4.5
\]

Asymptote angles:

\[
\theta_k = \frac{(2k + 1)\pi}{n - m}, \quad k = 0, 1, 2
\Rightarrow \theta = 60^\circ, 180^\circ, 300^\circ
\]

The branches to infinity depart from \( s = -4.5 \) at those angles.

Controller Purpose and Interpretation

The compensator adds a zero at \( s = -1.5 \), which attracts the root locus into the desired pole region, while the pole at \( s = -6 \) keeps the phase lead localized.

This lead compensator bends the locus so that for some \( k > 0 \), we can achieve:

- Damping ratio \( \zeta > 0.6 \)
- Settling time \( t_s < 1.5 \, \text{seconds} \)

The corresponding pole region has:

\[
\operatorname{Re}(s) < -2, \quad \angle(\theta) > \cos^{-1}(0.6) \approx 53.1^\circ
\]

Result:

By applying this compensator, we confirmed via root locus and pole calculations that the system achieves poles such as:

\[
s = -2.67 \pm 4.47j
\]

Which gives:

- \( \zeta = \cos(\theta) \approx 0.51 \)
- \( \omega_n = \sqrt{(-2.67)^2 + (4.47)^2} \approx 5.22 \)
- \( t_s \approx \frac{4.6}{\sigma} = \frac{4.6}{2.67} \approx 1.72 \, \text{seconds} \)

So this compensator brings the poles close to the desired region. With further tuning, we can fully satisfy the damping and settling time requirements.

Conclusion:

This example shows how lead compensation:

- Attracts the root locus into more desirable locations in the complex plane
- Enables design for speed and damping not achievable by P control alone
- Can be implemented by adding a zero near the origin and a pole further left

Step 3: Use Integral Control to Improve Steady-State

If tracking or steady-state error is unsatisfactory, add an integrator:

\[
C(s) = \frac{K(s + z)}{s}
\]

This makes the system Type 1 and improves steady-state accuracy, but introduces a pole at the origin and a new third-order dynamic, which must be carefully analyzed (e.g., using Routh-Hurwitz).

Example System with Integral Control

Let the plant be:

\[
P(s) = \frac{s + 2}{s^2 + s + 3}
\]

We define the controller as pure integral control:

\[
C(s) = \frac{K}{s}
\]

So the open-loop transfer function becomes:

\[
L(s) = C(s) P(s) = \frac{K(s + 2)}{s(s^2 + s + 3)}
\]

This system has:

- Poles at:
- \( s = 0 \) (integrator)
- \( s = -\frac{1}{2} \pm j \frac{\sqrt{11}}{2} \)
- One zero at \( s = -2 \)
- System order \( n = 3 \), number of finite zeros \( m = 1 \)
- Number of asymptotes: \( n - m = 2 \)

Asymptotes and Geometry

Centroid of asymptotes:

\[
\alpha = \frac{0 + (-\frac{1}{2} + j \frac{\sqrt{11}}{2}) + (-\frac{1}{2} - j \frac{\sqrt{11}}{2}) - (-2)}{2} = \frac{-1 + 2}{2} = \boxed{\frac{1}{2}}
\]

Asymptote angles:

\[
\theta_k = \frac{(2k + 1)\pi}{2}, \quad k = 0, 1
\Rightarrow \theta_0 = 90^\circ, \quad \theta_1 = 270^\circ
\]

So the root locus will have branches going vertically to infinity from \( \alpha = 0.5 \).

Closed-Loop Characteristic Equation

We form the unity-feedback closed-loop system:

\[
1 + L(s) = 0
\Rightarrow s(s^2 + s + 3) + K(s + 2) = 0
\Rightarrow s^3 + s^2 + 3s + K(s + 2) = 0
\Rightarrow \boxed{s^3 + s^2 + (3 + K)s + 2K = 0}
\]

This is a cubic polynomial. To determine stability, we apply the Routh-Hurwitz criterion.

Routh Table

Let:

\[
a_3 = 1, \quad a_2 = 1, \quad a_1 = 3 + K, \quad a_0 = 2K
\]

Stability Conditions (All First Column Elements > 0)

- \( 1 > 0 \)
- \( 1 > 0 \)
- \( 3 - K > 0 \Rightarrow K < 3 \)
- \( 2K > 0 \Rightarrow K > 0 \)

So for stability:

\[
\boxed{0 < K < 3}
\]

Interpretation

- Adding the integrator improved steady-state tracking (Type 1 system)
- The new pole at \( s = 0 \) increased the system order to 3
- Stability is limited to a bounded range of \( K \)
- Beyond \( K = 3 \), the system becomes unstable due to a sign change in the Routh table

Why the Integrator Improves Tracking

When we add an integrator via \( C(s) = \frac{K}{s} \), the open-loop system becomes Type 1:

\[
L(s) = \frac{K}{s(s + 1)(s + 2)}
\]

This guarantees:

- Zero steady-state error to step inputs:
\[
e_{ss} = \lim_{s \to 0} \frac{1}{1 + L(s)} = 0
\]
- Perfect tracking of step references
- Automatic error correction over time due to integral accumulation

However, adding the integrator increases system order and can introduce instability if \( K \) is too large. Stability must be ensured via Routh-Hurwitz or root locus analysis.

Critical Gain for Stability

Suppose \( s = j \omega \) lies on the imaginary axis — this is the boundary of marginal stability. Then we can find the critical gain \( K_{\text{crit}} \) such that:

\[
1 + K_{\text{crit}} L(j\omega) = 0
\Rightarrow K_{\text{crit}} = \frac{1}{|L(j\omega)|}
\]

This tells us the maximum gain before the system becomes unstable.

Example:

Let the critical frequency be \( \omega_{\text{crit}} \), found by solving:

\[
\angle L(j \omega) = \pm 180^\circ
\]

Then compute \( |L(j \omega_{\text{crit}})| \), and solve:

\[
K_{\text{crit}} = \frac{1}{|L(j \omega_{\text{crit}})|}
\]

This can be visualized and calculated on a Bode plot as well.