Stability and Steady State Error Notes

Stability and Steady State Error

Introduction

• The lecture focuses on stability and steady-state error in control systems, crucial for controller design.

• Modeling of mechanical and electrical systems has been covered; now, the focus shifts to controller design.

Consultation Hours and Schedule

• Consultation hours: Wednesdays, 4:30-5:30 PM in P638 (Gardens Point).

• Today's consultation will be in the atrium due to room booking.

• Week 12 consultation in 0 Block, room 702 (check announcements).

• PST1 marking is being finalized; results will be released early next week.

• No tutorial this week due to public holiday.

• PST release planned for the week after the mid-semester break, due in week 15.

Lecture Overview

• Topics: Stability (definition, importance, identification) and Steady State Error.

• Goal: Control system performance (transient/frequency response) by controlling stability and minimizing steady-state error.

Control Engineer Aims

• Stability: Prevent system blow-out.

• Steady State Error: Achieve desired output (target).

• Transient Response: Ensure smooth and quick response.

Stability

• Most important aspect: Without stability, other characteristics are irrelevant.

• Definition: Property of the system itself, independent of the input.

Stability Cases:

• Purely Stable: System approaches its natural response as t \rightarrow \infty.

• Unstable: System oscillates with increasing amplitude or experiences exponential growth.

• Marginally Stable: Continuous oscillations with constant amplitude (a limiting case).

• Consideration: Systems with bounded inputs; unbounded inputs make it difficult to have bounded outputs.

Bounded Input, Bounded Output (BIBO) Stability

• A linear system is stable if every bounded input results in a bounded output.

• A system is unstable if any bounded input results in an unbounded output.

S-Plane and Pole Location

• Poles on the S-plane (with real and imaginary parts) determine system stability.

• Poles on the left-hand side of the S-plane: Stable system.

Second-Order System Example

• Second-order system with poles at -\sigma \pm j\omega_d.

• Response involves an exponential term e^{-\sigma t}.

  • If -\sigma is positive, the solution grows exponentially (unstable).

  • If -\sigma is negative, the solution decays exponentially (stable).

• Oscillatory components may be present, leading to damped oscillations that fade out over time.

Poles on the Right-Hand Side

• Only one pole on the right-hand side is sufficient to cause instability.

• The effect of poles on the left-hand side decays over time, while the right-hand side pole causes exponential growth.

Poles on the Imaginary Axis

• Marginal stability may occur if poles are purely imaginary with multiplicity of one.

• Multiplicity greater than one leads to instability.

Unity Feedback System Example

• Unity Feedback System: Feedback loop transfer function is 1.

• Error (E) represents the difference between output and input.

• Transfer function for the system: C/R = G / (1 + G)

Stability Check

• Check the poles of the closed-loop transfer function.

• Compute the roots of the denominator.

• If poles are on the right-hand side of the plane, the system is unstable.

Controller Impact

• Controllers (e.g., gain factor/proportional controller) affect pole locations.

• Free parameters in the denominator may impact stability depending on their values.

• Goal: Determine the range of parameters to guarantee system stability.

Routh-Hurwitz Criterion

• Used to determine the number of poles on the left and right-hand sides of the S-plane without finding their exact values.

• It indicates the stability of the system.

Routh-Hurwitz Table Construction

• Start with the denominator of the transfer function in polynomial form.

• Create a table with one row for each power of s (from highest to lowest, including s^0).

Filling the Table

• First row: Coefficients of the odd powers of s.

• Second row: Coefficients of the even powers of s.

• Subsequent Rows: Computed from the rows above using determinants.

Determinant Calculation

• Calculate determinants of entries above the current cell in the table.

• The general form -\frac{det\begin{bmatrix} a & b\ c & d \end{bmatrix}}{c}, where a and c are from the column above, and b and d are to the right.

Sign Changes

• Examine the first column of the Routh-Hurwitz table for sign changes.

• Each sign change indicates the presence of a pole on the right-hand side of the S-plane.

Example

• The example has a transfer function with a denominator of 2s^4 + 5s^3 + s^2 + 2s + 1. The Routh table is constructed to determine stability.

  • Constructed the Routh Table, with rows corresponding to powers of s from 4 down to 0.

  • Calculated the entries in rows 3, 4, and 5 using the determinant method. For example, the first entry in row 3 (s^2^) is computed as follows:
    a = -\frac{\begin{vmatrix} 2 & 1 \ 5 & 2 \end{vmatrix}}{5} = -\frac{(2 \cdot 2 - 1 \cdot 5)}{5} = -\frac{-1}{5} = \frac{1}{5}

  • Sign Changes Identification: Upon completing the table, inspect the first column for sign changes. If changes occur, it indicates right-half plane (RHP) poles, confirming instability.

Simplifications

• Multiplying a row by a constant can simplify calculations without affecting the sign changes.

Example Result

• Two sign changes indicate two poles on the right-hand side, confirming instability.

Routh-Hurwitz Criterion Example

• The transfer function is T(s) = \frac{1}{2s^4 + 5s^3 + s^2 + 2s + 1}.

• The system is unstable due to sign changes in the first column of the Ruth-Hurwitz table.

Special Cases

• Case 1: Zero in the First Column: Substitute a small positive constant \epsilon for the zero and proceed with calculations.

• Case 2: Entire Row of Zeros: Compute the derivative of the polynomial from the row above and use the resulting coefficients to complete the table.

Special Case 1: Zero in the First Column

• When the first element in a row is zero, replace it with a small positive constant \epsilon.

• Continue with calculations, assessing the sign changes by considering the limit as \epsilon approaches zero.

Special Case 2: Entire Row of Zeros

• If an entire row consists of zeros, take the derivative of the polynomial from the row above.

• Use the coefficients of the derivative polynomial to complete the Routh-Hurwitz table.

Example

• Given polynomial: s^4 + 6s^2 + 8.

• Derivative: 4s^3 + 12s.

• Use coefficients 4 and 12 to complete the table.

Stability Analysis with Variable Gain K

• The closed-loop system transfer function is T(s) = \frac{K}{s^3 + 2s^2 + 4s + K}.

Routh-Hurwitz Table

• Construct the Routh-Hurwitz table with the coefficients of the characteristic equation.

• Express the entries in terms of K.

Stability Condition

• For stability, all elements in the first column must be positive.

• This leads to the condition 0 < K < 8.

Steady State Error

• Assumes the system is stable.

• Quantifies the difference between the desired output (set point) and the actual output as t \rightarrow \infty.

Error Definition

• Error is defined as E(s) = R(s) - C(s), where R is reference and C is output.

• For a unity feedback configuration, E(s) = \frac{1}{1 + G(s)} R(s)

Final Value Theorem

• Used to compute the steady-state error: e{ss} = \lim{s \to 0} sE(s)

Input Types:

• Step, Ramp, and Parabola.

Steady State Error Characteristics

• Finite Steady State Error

• Zero Steady State Error

• Infinite Steady State Error

Steady State Error and Input Types - Step Input

• For a step input (R(s) = \frac{1}{s}), the steady-state error is:
  e{ss} = \lim{s \to 0} \frac{s}{1 + G(s)} \cdot \frac{1}{s} = \frac{1}{1 + \lim_{s \to 0} G(s)}

Steady State Error and Input Types - Ramp Input

• For a ramp input (R(s) = \frac{1}{s^2}), the steady-state error is:
  e{ss} = \lim{s \to 0} \frac{s}{1 + G(s)} \cdot \frac{1}{s^2} = \lim_{s \to 0} \frac{1}{sG(s)}

Steady State Error and Input Types - Parabola Input

• For a parabolic input (R(s) = \frac{1}{s^3}), the steady-state error is:
  e{ss} = \lim{s \to 0} \frac{s}{1 + G(s)} \cdot \frac{1}{s^3} = \lim_{s \to 0} \frac{1}{s^2G(s)}

Error Constants

• Position Constant Defined as: Kp = \lim{s \to 0} G(s)

• Velocity Constant Defined as: Kv = \lim{s \to 0} sG(s)

• Acceleration Constant Defined as: Ka = \lim{s \to 0} s^2G(s)

Steady-State Error Formulas (Unity Feedback)

• For a Step Input: e{ss} = \frac{1}{1 + Kp}

• For a Ramp Input: e{ss} = \frac{1}{Kv}

• For a Parabolic Input: e{ss} = \frac{1}{Ka}

Steady State Error Example

• Given G(s) = \frac{10(s + 20)}{(s + 50)}{s(s + 25)(s + 40)}

Stability Check First

• Find transfer function: T(s) = \frac{G(s)}{1 + G(s)}

• Created the Routh Table, with rows corresponding to powers of s from 3 down to 0.
  G(s) = \frac{10000 + 700s + 10s^{2}}{s^{3} + 75s^{2} + 1700s}
  \frac{sG(s)}{1 + G(s)} = \frac{10(s + 20)(s + 50)}{s(s + 25)(s + 40)}

• The system is stable because there are no sign changes. With value of (1566.7) calculated in the table at (S1a).

• With no sign change in the table, there are no right-hand poles and thus it is stable.

Steady State Error Calculation

• Step Input (r(s) = 1/s)
  Kp = \lim{s \to 0} G(s) = \infty
  e{ss} = \frac{1}{1 + Kp} = 0

• Ramp Input (r(s) = 1/s^2)
  Kv = \lim{s \to 0} sG(s) = 10
  e{ss} = \frac{1}{Kv} = 0.1

• Parabola Input (r(s) = 1/s^3)
  Ka = \lim{s \to 0} s^2G(s) = 0
  e{ss} = \frac{1}{Ka} = \infty

• A Routh table was constructed and the result that resulted in an infinite steady state error was unstable and discounted.

Non-unity Step Error

 A non unity step error with a value of (5) was calculated from the equation:
\frac{5}{sG(s)}

Type Zero, Type One and Type Two System

• The integrators, with the general term of \frac{1}{s}, also seen in the first part of the unit, are the poles at zero.

Impact of Integrators

• Adding an integrator in the forward path can eliminate steady-state error for a step input.

• A Proportional-Integral (PI) controller contains an integrator.

Steady State Error with Integrators

• No integrators(m=0, Type zero system) = Finite Kb

• One integrator (m=1, Type one system) = One finite Kv

• Two Integrators (m=2, Type two system) = Two finite Ka.

  Important that if one of the following transfer functions were unstable, then one of the following conditions should be present:

Summary Table of System and Input Types

Kp = \lim{s \to 0} G(s),
Kv = \lim{s \to 0} sG(s),
Ka = \lim{s \to 0} s^2G(s)

System Stability Considerations

The next lecture will start with equivalent unity figment and it may address the system stability considerations.