Week 10 Notes: Root Locus and Controller Design

Week 10: Root Locus and Controller Design

Introduction

• Review of root locus concepts from the previous week.
• Using root locus for system response optimization.
  • Improving transient response.
  • Guaranteeing stability.
  • Reducing steady-state error.
• Focus on gain adjustment and its impact on root locus.

Gain Adjustment and Transient Response

• Changing the gain $K$ moves the system's closed-loop poles along the root locus.
• Poles move from open-loop poles towards open-loop zeros (or infinity).
• Objective: Improve transient response by selecting an appropriate gain $K$.
• Process:
  • Sketch the root locus.
  • Identify a desired pole location on the root locus that satisfies the angular criterion.
  • Determine the gain $K$required to place the closed-loop pole at the desired location.
• Gain Computation: $K = \frac{\text{Product of finite pole lengths}}{\text{Product of finite zero lengths}}$
  • Geometrically, this involves measuring the lengths of vectors from open-loop poles and zeros to the desired closed-loop pole location.

Example: Gain Adjustment for Settling Time

• Starting with a third-order transfer function in the forward path.
• Addressing the second-order approximation:
  • Ensuring the third pole (or zero) is far enough to the left to validate the approximation.
• Objective: Achieve a settling time of 2 seconds.
• Relating settling time to $\zeta$ (damping ratio) and $\omega<em>n$ (natural frequency): $T</em>s = \frac{4}{\zeta \omega_n}$
• If $T<em>s = 2$ seconds, then $\zeta \omega</em>n = 2$.
• On the s-plane, this corresponds to a vertical line at $\sigma = -2$.
• Locate the intersection of this vertical line with the root locus to find the desired pole location.

Procedure

1. Identify the Sigma data needed to achieve the desired settling time.
2. Intersect the vertical line (representing the sigma value) with the root locus.
3. Select the pole, typically focusing on the one above the x-axis for a positive imaginary component.
4. Measure the distance (or length) of each pole and zero from the selected pole.
5. Compute K using the formula. If there are no zeros, treat the denominator as 1.

Calculation and Validation

• Example pole location: $-2 + j2.8$.
• Compute distances from poles to the desired pole using the Pythagorean theorem.
  • Example: Distance from pole at 0 to $-2 + j2.8$ is $\sqrt{2^2 + 2.8^2}$. Distance from pole at $-6$ to $-2+j2.8$ is $\sqrt{4^2 + 2.8^2}$
• Calculate $K$ as the product of pole lengths divided by the product of zero lengths.
• In the example, K is approximately 47 (or 48 using MATLAB).
• Verify the second-order dominance after closing the loop with the computed gain $K$.

Last Exercise: Guaranteeing a Damping Ratio

• Objective: Guarantee a damping ratio $\zeta$ of 0.707 (which corresponds to an angle of 45 degrees).
• First Step: Sketch the root locus.
• Second Step: Find the gain $K$.
• Transfer function: $G(s) = K \frac{s^2 + 12s + 100}{s(s+1)}$

Steps

1. Identify the zeros of the transfer function.
2. Represent the zeros and the poles

Zeros Identification

• Zeros are complex conjugates in this case.
• Roots of the numerator $s^2 + 12s + 100$ are $-6 \pm 8j$.

Poles Identification

• Poles are at $s = 0$ and $s = -1$.
• Represent poles and zeros on the s-plane.

Root Locus Sketching

• On the real axis, the root locus exists to the left of an odd number of poles and zeros.
• In this case, the root locus exists between 0 and -1.
• The number of zeros matches the number of poles, so there are no asymptotes.
• Since the root locus starts from poles and goes to zeros, there must be a breakout point between 0 and -1.
• The root locus is symmetric about the real axis.

Breakout Point Calculation

• Use the equation: $\sum<em>{i=1} \frac{1}{\sigma + z</em>i} = \sum<em>{j=1} \frac{1}{\sigma + p</em>j}$
• Where ${z<em>i}$ are the zero locations, and ${p</em>j}$ are the pole locations.
• Substituting the zeros $-6 \pm 8j$ and poles $0, -1$ yields:
  • $\frac{1}{\sigma -6 + 8j} + \frac{1}{\sigma -6 - 8j} = \frac{1}{\sigma} + \frac{1}{\sigma + 1}$
• Simplifies to: $\frac{2\sigma + 12}{\sigma^2 + 12\sigma + 100} = \frac{2\sigma + 1}{\sigma^2 + \sigma}$
• Further simplification results in: $\sigma^3 + 13\sigma^2 + 112\sigma + 100 = 0$
• Solving this cubic equation gives three roots:
  • One real root and two complex conjugate roots.
• The real root is approximately $\sigma = -0.51$, which is the breakout point.
• The complex roots are not acceptable because they are not on the real axis.

Arrival Angle Computation

• Compute the angle of arrival to the zeros.
• Use the formula: $\sum \theta<em>{poles} - \sum \theta</em>{zeros} = 180(2k+1)$
• Where the angles are measured from the horizontal axis.
• Calculate angles from each pole and zero to a point close to the zero of interest.

Damping Ratio Consideration

• Desired damping ratio $\zeta = 0.707$ corresponds to an angle of 45 degrees.
• Draw a line from the origin at 45 degrees.
• The intersection of this line with the root locus gives the desired pole location.
• Since the breakout point is at $\sigma = -0.51$, the desired pole is approximately at $-0.53 + 0.53j$.

Gain Calculation

• Evaluate the lengths of vectors from each pole and zero to the desired pole.
• Use the formula: $K = \frac{\text{Product of pole lengths}}{\text{Product of zero lengths}}$
• Desired pole location: $s = -0.53 + 0.53j$
• Calculate the lengths from the poles at 0 and -1 and the zeros at $-6 \pm 8j$.
• $K = \frac{\sqrt{0.53^2 + 0.53^2} \cdot \sqrt{0.47^2 + 0.53^2}}{\sqrt{5.47^2 + 7.47^2} \cdot \sqrt{5.47^2 + 8.53^2}} = 0.0057$

Controller Design Using Root Locus

Controller Design Goals

1. Stability: Ensure the closed-loop system is stable.
2. Desired Output: Achieve the desired output with minimal steady-state error.
3. Transient Response: Obtain a fast and smooth transient response.

Achieving Design Goals

• Modify the Gain, but it's limited because:
  • The root locus shape is fixed.
  • Conflicting requirements might not be simultaneously satisfied (e.g., settling time and overshoot).

Compensators

• A state compensator adds zeros and poles to the open-loop transfer function.
• Modifies the root locus shape, affecting stability, transient response, and steady-state error.

Controller Categories

• PI (Proportional Integral) or Lag Compensators:
  • Improve steady-state performance.
  • Integrator reduces steady-state error.
  • Potential stability issues.
• PD (Proportional Derivative) or Lead Compensators:
  • Improve transient response and stability.
  • Derivative component in the transfer function.
  • Considerations similar to PI versus Lag.
• PID (Proportional Integral Derivative) Compensators or Lead-Lag Compensators:
  • Combine both PI and PD characteristics.
  • Increased complexity.
  • Trade-offs in performance and cost.

PI and Lag Compensators for Steady-State Performance Improvement

• Increasing system type reduces or eliminates steady-state error.
• Increasing $K$ to meet transient requirements can compromise stability.
• PI compensators add a pole at the origin (integrator) to eliminate steady-state error for a step input.

Challenges with Pure Integrators

• Adding an integrator changes the shape of the root locus.
• May result in a higher settling time and potential instability.

PI Compensator

• Compensator form: $G<em>c(s) = K</em>p + \frac{K<em>i}{s} = \frac{K</em>p s + K_i}{s}$
• Adds a pole at the origin and a zero nearby to mitigate the effect of the integrator on transient response.

PI Compensator Transfer Function

• $G<em>c(s) = K</em>p \frac{s + \frac{K<em>i}{K</em>p}}{s}$
• $K_p$ is the proportional gain.
• $-\frac{K<em>i}{K</em>p}$ is the location of the zero.

Example: PI Compensator Design

• Plant transfer function: $G(s) = \frac{K}{(s+1)(s+2)}$
• Initial Design (Gain Controller Only):
  • Damping ratio $\zeta = 0.707$.
  • Desired pole: $-1.5 \pm j1.5$.
  • Kp = 1.25
  • Settling time = $2.67$ seconds
  • Steady-state error = $0.444$
• Adding a Pure Integrator to Remove Steady-State Error
• New transfer function: $G(s) = \frac{K}{s(s+1)(s+2)}$
• The system is now type 1, so the steady-state error for step input is 0.
• However, the root locus is bent, resulting in
  • Higher settling time.
  • Potential for instability.

PI Compensator Design Example

• Place a zero at $-0.5$.
• Using PI compensator: $G<em>c(s) = K</em>p \frac{s+0.5}{s}$
• The new transfer function: $G(s) = K_p \frac{s+0.5}{s(s+1)(s+2)}$
• Choose the desired poles
• Compute Kp and Ki.

Lag Compensator

• Use passive components.
• $G<em>c(s) = \frac{s+z</em>{lag}}{s+p_{lag}}$
• Requires no external power source.
• Does not eliminate steady-state error entirely but reduces it.

Compensator Design Guideline

• Place PLAG is smaller than zLAG.
• Effect
  The error constants increase by $\frac{Z<em>{lag}}{P</em>{lag}}$

Compensator Design Example

• Goal: design a Lag Compensator, so the constant ratio grows 5 times.
• Assume z = 0.5
• Calculate $p = \frac{0.5}{5} = 0.1$

Transient and Stability Improvement: PD and Lead Compensators

PD Compensator

• Based on choosing the pole location and a 0,
• $G<em>c(s) = K</em>p + K<em>d s = K</em>p (1 + \frac{K<em>d}{K</em>p} s)$
• Proportional derivative compensation: a gain that is added and derivative given from adding an S.

Lead Compensator

• If S is a derivative part.

Example

• Start with the transfer function, same than beginning of lecture: $G(s) = \frac{K}{(s+1)(s+2)}$
• You want settling time to be half on what it was initially
• The overshoot has to remain the same.

Steps to follow

1. Choose the pole position. For the Example -\$3+3i
2. Apply Angle-criterion to compute where to place the zero.

• Angle condition: Zero minus two poles needs to equal $180(2k+1)$
• Applying Trigonometry- Compute Tangents to deduct angles theta.
• $tan(\theta) = \frac{opposite}{adjacent}$

Compensator expression using calculated angle

• Compute constant: K=3
• Final Root-Locus can be seen
• Expressions for Constants: $\frac{K<em>d}{K</em>p} = 5.33$

Notes

• PD controllers can cause an increased overshoot on the system
• Placing 0 too far increases the risk of oscillations & vice versa.
• Derivatives increase noise which could be a risk

PD Implementation

• Can be realized with Active components.
• Is not as simple and it requires external sources to supply Energy.

PD Implementation can be expressed from constants

• Remember resistance/capacitor values will compute constants

Advantages and disadvantages of Active and Passive component implementations

Lead compensator

• Advantages:
• Uses passive components. No external power source.
• Great advantage to reduce noise.
• Disadvantages:
  • Isn't so easy to produce robust stability design

Notes

• When building Lead Compensators, it's good to set the Zero for the Pole