Section 10: Frequency-Response Design
SECTION 10: FREQUENCY- RESPONSE DESIGN
ESE 499 – Feedback Control Systems Introduction 2
We have seen how to design feedback control systems using the root locus In this section of the course, we’ll learn how to do the same using the open-loop frequency response
Objectives: Determine static error constants from the open-loop frequency response Determine closed-loop stability from the open-loop frequency response Use the open-loop frequency response for compensator design to: Improve steady-state error Improve transient response
K. Webb ESE 499 3 Steady-State Error from Bode Plots
K. Webb ESE 499 Static Error Constants 4
For unity-feedback systems, open-loop transfer function gives static error constants Use static error constants to calculate steady-state error = lim
𝑝𝑝 𝐾𝐾 = lim𝑠𝑠→0 𝐺𝐺 𝑠𝑠
𝑣𝑣 𝐾𝐾 = 𝑠𝑠lim→0 𝑠𝑠𝐺𝐺 𝑠𝑠 2 𝑎𝑎 𝐾𝐾 𝑠𝑠→0 𝑠𝑠 𝐺𝐺 𝑠𝑠 We can also determine static error constants from a system’s open-loop Bode plot
K. Webb ESE 499 Static Error Constant – Type 0 5
For a type 0 system = lim
𝑝𝑝 At low𝐾𝐾 frequency,𝑠𝑠→0 𝐺𝐺 𝑠𝑠 i.e. below any open-loop poles or zeros
100 + 30 = + 3 + 200 𝑠𝑠 𝑝𝑝 𝐺𝐺 𝑠𝑠 Read𝐺𝐺 𝑠𝑠 directly≈ 𝐾𝐾 from 𝑠𝑠 𝑠𝑠 the open-loop Bode plot 𝐾𝐾𝑝𝑝 Low-frequency gain
K. Webb ESE 499 Static Error Constant – Type 1 6
For a type 1 system = lim
𝑣𝑣 At low frequencies,𝐾𝐾 𝑠𝑠→0 𝑠𝑠i.e.𝑠𝑠 𝑠𝑠below any other open-loop poles or zeros and 𝐾𝐾𝑣𝑣 𝐾𝐾𝑣𝑣 A straight line𝐺𝐺 𝑠𝑠 with≈ 𝑠𝑠 a slope of𝐺𝐺 𝑗𝑗𝑗𝑗20 ≈ /𝜔𝜔 Evaluating this low-frequency asymptote at = 1 yields the velocity constant, − 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑
𝜔𝜔 On the Bode plot, extend the low𝑣𝑣 -frequency asymptote to = 1 𝐾𝐾 Gain of this line at = 1 is 𝜔𝜔 K. Webb 𝜔𝜔 𝐾𝐾𝑣𝑣 ESE 499 Static Error Constant – Type 1 7
85 + 0.1 + 50 = + + 125 𝑠𝑠 𝑠𝑠 𝐺𝐺 𝑠𝑠 2 𝑠𝑠 𝑠𝑠 10𝑠𝑠
K. Webb ESE 499 Static Error Constant – Type 2 8
For a type 2 system = lim 2 𝑎𝑎 At low frequencies,𝐾𝐾 𝑠𝑠→0 𝑠𝑠i.e.𝐺𝐺 below𝑠𝑠 any other open-loop poles or zeros and 𝐾𝐾𝑎𝑎 𝐾𝐾𝑎𝑎 2 2 A straight line𝐺𝐺 𝑠𝑠 with≈ 𝑠𝑠 a slope of𝐺𝐺 𝑗𝑗𝑗𝑗40 ≈/𝜔𝜔 Evaluating this low-frequency asymptote at = 1 yields the acceleration constant,− 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑
𝜔𝜔 On the Bode plot, extend the low-frequency𝑎𝑎 asymptote to = 1 𝐾𝐾 Gain of this line at = 1 is 𝜔𝜔 K. Webb 𝜔𝜔 𝐾𝐾𝑎𝑎 ESE 499 Static Error Constant – Type 2 9
1600 + 0.1 + 5 = + 100 𝑠𝑠 𝑠𝑠 𝐺𝐺 𝑠𝑠 2 𝑠𝑠 𝑠𝑠
K. Webb ESE 499 10 Stability from Open-Loop Bode Plots
K. Webb ESE 499 Stability 11
Consider the following system
We already have a couple of tools for assessing stability as a function of loop gain, Routh Hurwitz 𝐾𝐾 Root locus Root locus: Stable for some values of Unstable for others 𝐾𝐾
K. Webb ESE 499 Stability 12
In this case gain is stable below some value Other systems may be stable for gain above some value Marginal stability point: Closed-loop poles on the imaginary axis at ± For gain = 𝑗𝑗𝜔𝜔1 𝐾𝐾 𝐾𝐾1
K. Webb ESE 499 Open-Loop Frequency Response & Stability
13
Marginal stability point occurs when closed-loop poles are on the imaginary axis Angle criterion satisfied at ±
= 1 and 𝑗𝑗𝜔𝜔1 = 180° Note𝐾𝐾 that𝐾𝐾 𝑗𝑗𝜔𝜔 1180° = 180°∠𝐾𝐾𝐾𝐾 𝑗𝑗𝜔𝜔1 −
is the− open-loop frequency response Marginal stability occurs when: 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 Open-loop gain is: = 0 Open-loop phase is: = 180° 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 K. Webb ∠𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 − ESE 499 Stability from Open-Loop Bode Plots 14
Varying simply shifts gain response up or down Here, stable for smaller gain values𝐾𝐾 < 0 when = 180° 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 Often,∠𝐾𝐾𝐾𝐾 stable𝑗𝑗𝑗𝑗 for− larger gain values > 0 when = 180° 𝐾𝐾𝐾𝐾 𝑗𝑗𝑗𝑗 𝑑𝑑𝑑𝑑 Root∠𝐾𝐾 locus𝐾𝐾 𝑗𝑗𝑗𝑗 provides− this information Bode plot does not
K. Webb ESE 499 Open-Loop Frequency Response & Stability
15
Open-loop Bode plot can be used to assess stability But, we need to know if system is closed-loop stable for low gain or high gain
Here, we’ll assume open-loop-stable systems Closed-loop stable for low gain
Open-loop Bode plot can tell us: Is a system closed-loop stable? If so, how stable? I.e. how close to marginal stability
Two stability metrics: Gain margin Phase margin
K. Webb ESE 499 16 Stability Margins
K. Webb ESE 499 Crossover Frequencies 17
Two important frequencies when assessing stability:
Gain crossover frequency, The frequency at which the 𝜔𝜔open𝑃𝑃𝑃𝑃 -loop gain crosses 0
Phase crossover𝑑𝑑𝑑𝑑 frequency, The frequency at which the 𝜔𝜔open𝐺𝐺𝐺𝐺 -loop phase crosses 180°
− K. Webb ESE 499 Gain Margin 18
An open-loop-stable system will be closed-loop stable as long as its gain is less than unity at the phase crossover frequency
Gain margin, GM The change in open- loop gain at the phase crossover frequency required to make the closed- loop system unstable
K. Webb ESE 499 Phase Margin 19
An open-loop-stable system will be closed-loop stable as long as its phase has not fallen below 180° at the gain crossover frequency − Phase margin, PM The change in open- loop phase at the gain crossover frequency required to make the closed- loop system unstable
K. Webb ESE 499 Gain and Phase Margins from Bode Plots
20
K. Webb ESE 499 Phase Margin and Damping Ratio, 21
PM can be expressed as a function of damping ratio, , as = tan 𝜁𝜁 𝜁𝜁 −1 2𝜁𝜁 𝑃𝑃𝑃𝑃 2 4 For 65° or so, we can− approximate:2𝜁𝜁 + 1+4𝜁𝜁
𝑃𝑃𝑃𝑃 ≤ or 𝑃𝑃𝑃𝑃 𝑃𝑃𝑃𝑃 ≈ 100𝜁𝜁 𝜁𝜁 ≈ 100
K. Webb ESE 499 22 Frequency Response Analysis in MATLAB
K. Webb ESE 499 bode.m 23
[mag,phase] = bode(sys,w) sys: system model – state-space, transfer function, or other w: optional frequency vector – in rad/sec mag: system gain response vector phase: system phase response vector – in degrees If no outputs are specified, bode response is automatically plotted – preferable to plot yourself Frequency vector input is optional If not specified, MATLAB will generate automatically
May need to do: squeeze(mag) and squeeze(phase) to eliminate singleton dimensions of output matrices
K. Webb ESE 499 margin.m 24
[GM,PM,wgm,wpm] = margin(sys)
sys: system model – state-space, transfer function, or other GM: gain margin PM: phase margin – in degrees wgm: frequency at which GM is measured, the phase crossover frequency – in rad/sec wpm: frequency at which PM is measured, the gain crossover frequency
If no outputs are specified, a Bode plot with GM and PM indicated is automatically generated
K. Webb ESE 499 25 Frequency-Response Design
K. Webb ESE 499 Frequency-Response Design 26
In a previous section of notes, we saw how we can use root-locus techniques to design compensators Two primary objectives of compensation Improve steady-state error Proportional-integral (PI) compensation Lag compensation Improve dynamic response Proportional-derivative (PD) compensation Lead compensation
Now, we’ll learn to design compensators using a system’s open-loop frequency response We’ll focus on lag and lead compensation
K. Webb ESE 499 27 Improving Steady-State Error
K. Webb ESE 499 Improving Steady-State Error 28
Consider the system above with a desired phase margin of 50°
According to the Bode plot: 𝑃𝑃𝑃𝑃 ≈ = 130° at = 3.46 / 𝜙𝜙 − = 12.1 Gain𝜔𝜔𝑃𝑃𝑃𝑃 is 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 at 𝐾𝐾𝑃𝑃𝑃𝑃 − 𝑑𝑑𝑑𝑑 = = 12. = 4 Set𝜔𝜔 𝑃𝑃𝑃𝑃 for desired phase margin 𝐾𝐾 −𝐾𝐾𝑃𝑃𝑃𝑃 1𝑑𝑑𝑑𝑑
K. Webb ESE 499 Improving Steady-State Error 29
Can read the position constant directly from the Bode plot: = 14.8 5.5
𝑝𝑝 Note𝐾𝐾 that 𝑑𝑑𝑑𝑑 → 50°, as desired 𝑃𝑃𝑃𝑃 ≈ Gain margin is = 17.9
𝐺𝐺𝐺𝐺 𝑑𝑑𝑑𝑑
K. Webb ESE 499 Improving Steady-State Error 30
Steady-state error to a constant reference is 1 = = 0.154 15.4% 1 + 𝑒𝑒𝑠𝑠𝑠𝑠 → 𝐾𝐾𝑝𝑝
K. Webb ESE 499 Improving Steady-State Error 31
Let’s say we want to reduce steady-state error to < 5% 𝑠𝑠𝑠𝑠 Required position 𝑒𝑒 constant 1 > 1 = 19 0.05 𝑝𝑝 Increase𝐾𝐾 gain− by 4x Bode plot shows desired position constant But, phase margin has been degraded significantly
K. Webb ESE 499 Improving Steady-State Error 32
Step response shows that error goal has been met But, reduced phase margin results in significant overshoot and ringing Error improvement came at the cost of degraded phase margin Would like to be able to improve steady-state error without affecting phase margin Integral compensation Lag compensation
K. Webb ESE 499 33 Integral Compensation
K. Webb ESE 499 PI Compensation 34
Proportional-integral (PI) compensator: 1 + 1 = 𝑇𝑇𝐷𝐷𝑠𝑠 𝐷𝐷 𝑠𝑠 Low-frequency𝑇𝑇𝐷𝐷 gain𝑠𝑠 increase Infinite at DC System type increase
For 1/ Gain unaffected 𝐷𝐷 Phase𝜔𝜔 ≫ affected𝑇𝑇 little PM unaffected
Susceptible to integrator overflow Lag compensation is often preferable
K. Webb ESE 499 35 Lag Compensation
K. Webb ESE 499 Lag Compensation 36
Lag compensator + 1 = , > 1 + 1 𝑇𝑇𝑇𝑇 Objective: add a gain𝐷𝐷 𝑠𝑠 of 𝛼𝛼at low frequencies𝛼𝛼 without affecting phase margin 𝛼𝛼𝛼𝛼𝛼𝛼 Lower-frequency pole: 𝛼𝛼= 1/ Higher-frequency zero: = 1/ 𝑠𝑠 − 𝛼𝛼𝛼𝛼 Pole/zero spacing determined by 𝑠𝑠 − 𝑇𝑇 For 1/ 𝛼𝛼 Gain: ~20 log Phase:𝜔𝜔 ≪ ~0°𝛼𝛼𝛼𝛼 𝛼𝛼 𝑑𝑑𝑑𝑑 For 1/ Gain: ~0 Phase:𝜔𝜔 ≫ ~0°𝑇𝑇 𝑑𝑑𝑑𝑑 K. Webb ESE 499 Lag Compensation vs. 37
Gain increased at low frequency only 𝛼𝛼 Dependent on + 1 DC gain: 20log = + 1 𝛼𝛼 𝑇𝑇𝑇𝑇 𝐷𝐷 𝑠𝑠 𝛼𝛼 Phase lag added 𝛼𝛼between𝑑𝑑𝑑𝑑 𝛼𝛼𝛼𝛼𝛼𝛼 compensator pole and zero 0° 90° Dependent on ≤ 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 ≤ Lag pole/zero well𝛼𝛼 below crossover frequency Phase margin unaffected
K. Webb ESE 499 Lag Compensator Design Procedure 38
Lag compensator adds gain at low frequencies without affecting phase margin
Basic design procedure: Adjust gain to achieve the desired phase margin Add compensation, increasing low-frequency gain to achieve desired error performance
Same as adjusting gain to place poles at the desired damping on the root locus, then adding compensation Root locus is not changed Here, the frequency response near the crossover frequency is not changed
K. Webb ESE 499 Lag Compensator Design Procedure 39 1. Adjust gain, , of the uncompensated system to provide the desired phase margin plus 5° … 10° (to account for small phase lag added𝐾𝐾 by compensator) 2. Use the open-loop Bode plot for the uncompensated system with the value of gain set in the previous step to determine the static error constant 3. Calculate as the low-frequency gain increase required to provide the desired error performance 4. Set the upper𝜶𝜶 corner frequency (the zero) to be one decade below the crossover frequency: 1/ = /10 Minimizes the added phase lag at the crossover frequency 𝑃𝑃𝑃𝑃 5. Calculate the lag pole: 1/ 𝑇𝑇 𝜔𝜔 6. Simulate and iterate, if necessary 𝛼𝛼𝛼𝛼 K. Webb ESE 499 Lag Example – Step 1 40
Design a lag compensator for the above system to satisfy the following requirements < 2% for a step input 12% 𝑒𝑒𝑠𝑠𝑠𝑠 First,%𝑂𝑂𝑂𝑂 determine≈ the required phase margin to satisfy the overshoot requirement ln = = 0.559 + ln 𝑂𝑂𝑂𝑂 𝜁𝜁 − 2 2 𝜋𝜋 = 55𝑂𝑂𝑂𝑂.9°
Add ~10° to account for compensator phase at 𝑃𝑃𝑃𝑃 ≈ 100𝜁𝜁 = 65.9° 𝜔𝜔𝑃𝑃𝑃𝑃 K. Webb 𝑃𝑃𝑃𝑃 ESE 499 Lag Example – Step 1 41
Plot the open-loop Bode plot of the uncompensated system for = 1 Locate frequency where phase𝐾𝐾 is 180° + = 114.1° This is , the desired − crossover𝑃𝑃𝑃𝑃 frequency− 𝑃𝑃𝑃𝑃 =𝜔𝜔2.5 /
Gain𝜔𝜔𝑃𝑃𝑃𝑃 at 𝑟𝑟𝑟𝑟is 𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 = 8.4 0.38 𝜔𝜔𝑃𝑃𝑃𝑃 𝐾𝐾𝑃𝑃𝑃𝑃 Increase𝐾𝐾𝑃𝑃𝑃𝑃 the− gain𝑑𝑑𝑑𝑑 by→ 1/ = 8.4 2.63 𝐾𝐾𝑃𝑃𝑃𝑃 K. Webb 𝐾𝐾 𝑑𝑑𝑑𝑑 → ESE 499 Lag Example – Step 2 42
Gain has now been set to yield the desired phase margin of = 65.9°
Use the new open-loop 𝑃𝑃𝑃𝑃 bode plot to determine the static error constant
Position constant of the uncompensated system given by the DC gain: = 11.14 3.6
𝐾𝐾𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑 →
K. Webb ESE 499 Lag Example – Step 3 43
Calculate to yield desired steady-state error improvement Steady-state error: 𝛼𝛼1 = < 0.02 1 + 𝑒𝑒𝑠𝑠𝑠𝑠 The required𝐾𝐾 position𝑝𝑝 constant: 1 > 1 = 49 = 50
𝐾𝐾𝑝𝑝 − → 𝐾𝐾𝑝𝑝 Calculate𝑒𝑒𝑠𝑠𝑠𝑠 as the required position constant improvement𝛼𝛼
= = 13.9 = 14 𝐾𝐾𝑝𝑝 K. Webb𝛼𝛼 → 𝛼𝛼 ESE 499 𝐾𝐾𝑝𝑝𝑝𝑝 Lag Example – Steps 4 & 5 44
Place the compensator zero one decade below the crossover frequency, = 2.5 / 1 = 0.25 / 𝜔𝜔𝑃𝑃𝑃𝑃 / 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 = 4 𝑇𝑇 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 The𝑇𝑇 compensator𝑠𝑠𝑠𝑠𝑠𝑠 pole: . 1/ = 0 25 1/ = 0.018 / 𝛼𝛼𝑇𝑇 14 Lag compensator𝛼𝛼𝛼𝛼 𝑟𝑟𝑟𝑟 transfer𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 function + 1 = + 1 𝑇𝑇𝑠𝑠 𝐷𝐷 𝑠𝑠 𝛼𝛼 + 1 = 14𝛼𝛼𝛼𝛼𝛼𝛼 + 1 4𝑠𝑠 𝐷𝐷 𝑠𝑠 K. Webb 56𝑠𝑠 ESE 499 Lag Example – Step 6 45
Bode plot of compensated system shows:
= 60.5° = 50.5 𝑃𝑃𝑃𝑃 𝐾𝐾𝑝𝑝
K. Webb ESE 499 Lag Example – Step 6 46
Lag compensator adds gain at low frequencies only
Phase near the crossover frequency is nearly unchanged
K. Webb ESE 499 Lag Example – Step 6 47
Steady-state error requirement has been satisfied
Overshoot spec has been met Though slow tail makes overshoot assessment unclear
K. Webb ESE 499 Lag Compensator – Summary 48 + 1 = + 1 𝑇𝑇𝑇𝑇 𝐷𝐷 𝑠𝑠 𝛼𝛼 Higher-frequency zero: =𝛼𝛼𝛼𝛼𝛼𝛼1/ Place one decade below crossover frequency, 𝑠𝑠 − 𝑇𝑇 Lower-frequency pole: = 1/ 𝜔𝜔𝑃𝑃𝑃𝑃 sets pole/zero spacing 𝑠𝑠 − 𝛼𝛼𝛼𝛼 DC𝛼𝛼 gain: 20 log
10 Compensator𝛼𝛼 → adds low-𝛼𝛼frequency𝑑𝑑𝑑𝑑 gain Static error constant improvement Phase margin unchanged
K. Webb ESE 499 49 Improving Dynamic Response
K. Webb ESE 499 Improving Dynamic Response 50
We’ve already seen two types of compensators to improve dynamic response Proportional derivative (PD) compensation Lead compensation
Unlike with the lag compensator we just looked at, here, the objective is to alter the open-loop phase
We’ll look briefly at PD compensation, but will focus on lead compensation
K. Webb ESE 499 51 Derivative Compensation
K. Webb ESE 499 PD Compensation 52
Proportional-Derivative (PD) compensator: = + 1
Phase𝐷𝐷 𝑠𝑠 added𝑇𝑇𝐷𝐷 𝑠𝑠near (and above) the crossover frequency Increased phase margin Stabilizing effect
Gain continues to rise at high frequencies Sensor noise is amplified Lead compensation is usually preferable
K. Webb ESE 499 53 Lead Compensation
K. Webb ESE 499 Lead Compensation 54
With lead compensation, we have three design parameters: Crossover frequency, Determines closed-loop bandwidth, ; risetime, ; peak time, ; and settling time, 𝜔𝜔𝑃𝑃𝑃𝑃 𝐵𝐵𝐵𝐵 𝑟𝑟 𝜔𝜔 𝑡𝑡 Phase𝑡𝑡𝑝𝑝 margin, PM 𝑡𝑡𝑠𝑠 Determines damping, , and overshoot Low-frequency gain 𝜁𝜁 Determines steady-state error performance
We’ll look at the design of lead compensators for two common scenarios, either Designing for steady-state error and phase margin, or Designing for closed-loop bandwidth and phase margin
K. Webb ESE 499 Lead Compensation 55
Lead compensator + 1 = , < 1 + 1 𝑇𝑇𝑇𝑇 𝐷𝐷 𝑠𝑠 𝛽𝛽 Objectives: add phase lead near𝛽𝛽𝛽𝛽𝛽𝛽 the crossover frequency and/or alter the crossover frequency Lower-frequency zero: = 1/ Higher-frequency pole: = 1/ 𝑠𝑠 − 𝑇𝑇 Zero/pole spacing determined by 𝑠𝑠 − 𝛽𝛽𝛽𝛽 For 1/ 𝛽𝛽 Gain: ~0 Phase:𝜔𝜔 ≪ ~0°𝑇𝑇 𝑑𝑑𝑑𝑑 For 1/ Gain: ~20 log 1/ Phase:𝜔𝜔 ≫ ~0°𝛽𝛽𝛽𝛽 𝛽𝛽 𝑑𝑑𝑑𝑑 K. Webb ESE 499 Lead Compensation vs. 56 + 1 = , < 1 + 1 𝛽𝛽 𝑇𝑇𝑇𝑇 𝐷𝐷 𝑠𝑠 𝛽𝛽 𝛽𝛽𝛽𝛽𝛽𝛽 determines: Zero/pole spacing 𝛽𝛽 Maximum compensator phase lead, High-frequency 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 compensator gain
K. Webb ESE 499 Lead Compensation – 57 , zero/pole spacing, determines maximum𝜙𝜙𝑚𝑚𝑚𝑚 phase𝑚𝑚 lead 1 𝛽𝛽 = sin 1 + −1 − 𝛽𝛽 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 Can use a desired 𝛽𝛽to determine 𝑚𝑚𝑚𝑚𝑚𝑚 1 sin 𝜙𝜙 = 𝛽𝛽 1 + sin − 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽 occurs at 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 1 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 = 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚
𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 1 = 𝑇𝑇 𝛽𝛽 𝑇𝑇 𝑚𝑚𝑚𝑚𝑚𝑚 K. Webb 𝜔𝜔 𝛽𝛽 ESE 499 Lead Compensation – Design Procedure
58 1. Determine loop gain, , to satisfy either steady-state error requirements or bandwidth requirements: a) Set to provide the 𝐾𝐾required static error constant, or b) Set to place the crossover frequency an octave below the desired closed𝐾𝐾 -loop bandwidth 2. Evaluate𝐾𝐾 the phase margin of the uncompensated system, using the value of just determined 3. If necessary, determine the required PM from or overshoot specifications.𝐾𝐾 Evaluate the PM of the uncompensated system and determine the required phase lead at the crossover𝜁𝜁 frequency to achieve this PM. Add ~10° additional phase – this is 4. Calculate from 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 5. Set = . Calculate from and 𝛽𝛽 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 6. Simulate and iterate, if necessary 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝜔𝜔𝑃𝑃𝑃𝑃 𝑇𝑇 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽
K. Webb ESE 499 Closed-Loop Bandwidth and Transient Response
59
Closed-loop bandwidth, , is one possible design criterion How is it related to transient response? 𝜔𝜔𝐵𝐵𝐵𝐵 For a second-order system (or approximate second-order system): Closed-loop bandwidth and damping ratio and natural frequency, and
𝜁𝜁 𝜔𝜔𝑛𝑛 = 1 2 + 4 4 + 2 2 4 2 𝜔𝜔𝐵𝐵𝐵𝐵 𝜔𝜔𝑛𝑛 − 𝜁𝜁 𝜁𝜁 − 𝜁𝜁 Closed-loop bandwidth and ±1% settling time,
4.6 𝑡𝑡𝑠𝑠 1 2 + 4 4 + 2 2 4 2 𝐵𝐵𝐵𝐵 𝜔𝜔 ≈ 𝑠𝑠 − 𝜁𝜁 𝜁𝜁 − 𝜁𝜁 Closed-loop bandwidth𝑡𝑡 𝜁𝜁 and peak time,
4 𝑡𝑡𝑝𝑝 = 1 2 + 4 4 + 2 1 2 4 2 𝐵𝐵𝐵𝐵 𝜔𝜔 2 − 𝜁𝜁 𝜁𝜁 − 𝜁𝜁 𝑝𝑝 K. Webb 𝑡𝑡 − 𝜁𝜁 ESE 499 Double-Lead Compensation 60
A lead compensator can add, at most, 90° of phase lead If more phase is required, use a double-lead compensator + 1 = + 1 2 𝑇𝑇𝑇𝑇 𝐷𝐷 𝑠𝑠 For phase lead over ~𝛽𝛽60°𝛽𝛽𝛽𝛽 … 70°, 1/ must be very large, so typically use double-lead compensation 𝛽𝛽
K. Webb ESE 499 Lead Compensation – Example 1 61
Consider the following system
Design a compensator to satisfy the following < 0.1 for a ramp input < 15% 𝑒𝑒𝑠𝑠𝑠𝑠 Here,%𝑂𝑂𝑂𝑂 we’ll design a lead compensator to simultaneously adjust low-frequency gain and phase margin
K. Webb ESE 499 Lead Example 1 – Steps 1 & 2 62
The velocity constant for the uncompensated system is = lim
𝑣𝑣 𝐾𝐾 = 𝑠𝑠lim→0 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠= + 1 𝑣𝑣 𝐾𝐾 Steady-𝐾𝐾state error𝑠𝑠→0 is 𝐾𝐾 1 𝑠𝑠 = < 0.1
𝑒𝑒𝑠𝑠𝑠𝑠 = > 10 𝐾𝐾𝑣𝑣 Adding a bit of margin 𝐾𝐾𝑣𝑣 𝐾𝐾 = 12
Bode plot shows the resulting phase margin𝐾𝐾 is = 16.4°
K. Webb 𝑃𝑃𝑃𝑃 ESE 499 Lead Example 1 – Step 3 63
Approximate required phase margin for < 15% Design for 13% %𝑂𝑂𝑂𝑂 First calculate the required damping ratio ln = = 0.545 + ln 𝑂𝑂𝑂𝑂 𝜁𝜁 − 2 2 Approximate corresponding𝜋𝜋 𝑂𝑂𝑂𝑂 PM, and add 10° correction factor + 10° = 64.5°
Calculate the𝑃𝑃𝑃𝑃 required≈ 100𝜁𝜁 phase lead = 64.5° 16.4° = 48°
K. Webb ESE 499 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 − Lead Example 1 – Steps 4 & 5 64
Calculate from 1 sin 𝛽𝛽= 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 = 0.147 1 + sin − 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽 Set = , as𝜙𝜙 determined𝑚𝑚𝑚𝑚𝑚𝑚 from Bode plot, and calculate 𝑚𝑚𝑚𝑚𝑚𝑚 𝑃𝑃𝑃𝑃 𝜔𝜔 𝜔𝜔= = 3.4 / 𝑇𝑇 𝜔𝜔𝑚𝑚=𝑚𝑚𝑚𝑚 𝜔𝜔𝑃𝑃𝑃𝑃= 𝑟𝑟𝑟𝑟𝑟𝑟=𝑠𝑠0𝑠𝑠𝑠𝑠.76 . . 6 1 1 𝑚𝑚𝑚𝑚𝑚𝑚 3 4 0 147 The resulting𝑇𝑇 𝜔𝜔 lead 𝛽𝛽compensator transfer function is + 1 0.766 + 1 = = 12 + 1 0.113 + 1 𝑇𝑇𝑇𝑇 𝑠𝑠 K. Webb 𝐾𝐾𝐷𝐷 𝑠𝑠 𝐾𝐾 ESE 499 𝛽𝛽𝛽𝛽𝛽𝛽 𝑠𝑠 Lead Example 1 – Step 6 65 0.766 + 1 = 12 0.113 + 1 𝑠𝑠 𝐾𝐾𝐷𝐷 𝑠𝑠 The lead compensator Bode plot 𝑠𝑠
K. Webb ESE 499 Lead Example 1 – Step 6 66
Lead-compensated system: = 48.5° = 7.2 / 𝑃𝑃𝑃𝑃 High𝜔𝜔-𝑃𝑃𝑃𝑃frequency𝑟𝑟𝑟𝑟 compensator𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 gain increased the crossover frequency Phase was added at the previous crossover frequency PM is below target
Move lead zero/pole to higher frequencies Reduce the crossover frequency increase Improve phase margin
K. Webb ESE 499 Lead Example 1 – Step 6 67
As predicted by the insufficient phase margin, overshoot exceeds the target = 20.9% > 15%
Redesign%𝑂𝑂𝑂𝑂 compensator for higher Improve phase margin 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 Reduce overshoot
K. Webb ESE 499 Lead Example 1 – Step 6 68
The steady-state error requirement has been satisfied = 0.08 < 0.1
𝑠𝑠𝑠𝑠 Will𝑒𝑒 not change with compensator redesign Low-frequency gain will not be changed
K. Webb ESE 499 Lead Example 1 – Step 6 69
Iteration yields acceptable value for = 5.5 rad/sec 𝑚𝑚𝑚𝑚𝑚𝑚 Maintain same zero/pole spacing, , and,𝜔𝜔 therefore, same 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 Recalculate zero/pole time constants:𝛽𝛽 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 1 1 = = = 0.4742 5.5 0.147 𝑇𝑇 =𝜔𝜔0𝑚𝑚.𝑚𝑚147𝑚𝑚 𝛽𝛽0.4742 = 0.0697
The updated𝛽𝛽𝛽𝛽 lead ⋅compensator transfer function: 0.4742 + 1 = 12 0.0697 + 1 𝑠𝑠 K. Webb 𝐷𝐷 𝑠𝑠 ESE 499 𝑠𝑠 Lead Example 1 – Step 6 70
Crossover frequency has been reduced = 5.58 /
Phase𝜔𝜔𝑃𝑃𝑃𝑃 margin 𝑟𝑟𝑟𝑟is close𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 to the target = 58.2°
Dip𝑃𝑃𝑃𝑃 in phase is apparent, because is now placed at point of lower 𝑚𝑚𝑚𝑚𝑚𝑚 open-loop𝜔𝜔 phase
K. Webb ESE 499 Lead Example 1 – Step 6 71
Overshoot requirement now satisfied
= 14.7% < 15%
Low%𝑂𝑂𝑂𝑂-frequency gain has not been changed, so error requirement is still satisfied
Design is complete
K. Webb ESE 499 Lead Compensation – Example 2 72
Again, consider the same system
Design a compensator to satisfy the following 1.2 (±1%) 10% 𝑡𝑡𝑠𝑠 ≈ 𝑠𝑠𝑠𝑠𝑠𝑠 Now,%𝑂𝑂𝑂𝑂 we’ll≈ design a lead compensator to simultaneously adjust closed-loop bandwidth and phase margin
K. Webb ESE 499 Lead Example 2 – Step 1 73
The required damping ratio for 10% overshoot is
= = 0.5912 ln 𝑂𝑂𝑂𝑂 2 2 Given the required𝜁𝜁 − damping𝜋𝜋 +ln 𝑂𝑂𝑂𝑂 ratio, calculate the required closed-loop bandwidth to yield the desired settling time
. = 1 2 + 4 4 + 2 4 6 2 4 2 𝐵𝐵𝐵𝐵 𝑠𝑠 𝜔𝜔 =𝑡𝑡7𝜁𝜁.52 − /𝜁𝜁 𝜁𝜁 − 𝜁𝜁
We’ll initially 𝜔𝜔set𝐵𝐵𝐵𝐵 the gain, 𝑟𝑟𝑟𝑟, 𝑟𝑟to 𝑠𝑠place𝑠𝑠𝑠𝑠 the crossover frequency, , one octave below the desired closed-loop bandwidth 𝐾𝐾 𝜔𝜔𝑃𝑃𝑃𝑃 = /2 = 3.8 /
𝜔𝜔𝑃𝑃𝑃𝑃 𝜔𝜔𝐵𝐵𝐵𝐵 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠
K. Webb ESE 499 Lead Example 2 – Step 1 74
Plot the Bode plot for = 1 Determine the loop gain at the desired crossover 𝐾𝐾 frequency = 23.3
Adjust𝐾𝐾 𝑃𝑃𝑃𝑃so that− the𝑑𝑑𝑑𝑑 loop gain at the desired crossover𝐾𝐾 frequency is 0 1 = = 23.3 = 14.7𝑑𝑑𝑑𝑑
𝐾𝐾 𝑑𝑑𝑑𝑑 𝐾𝐾𝑃𝑃𝑃𝑃 K. Webb ESE 499 Lead Example 2 – Steps 2 & 3 75
Generate a Bode plot using the gain value just determined Phase margin for the uncompensated system: = 14.9°
Required𝑃𝑃𝑀𝑀𝑢𝑢 phase margin to satisfy overshoot requirement: = 59.1°
Add𝑃𝑃𝑃𝑃 10°≈to100𝜁𝜁 account for crossover frequency increase = 69.1°
Required𝑃𝑃𝑃𝑃 phase lead from the compensator = = 54.2°
K. Webb 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 𝑃𝑃𝑃𝑃 − 𝑃𝑃𝑀𝑀𝑢𝑢 ESE 499 Lead Example 2 – Steps 4 & 5 76
Calculate zero/pole spacing, , from required phase lead, 1 sin = 𝛽𝛽 = 0.1040 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 1 + sin − 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 𝛽𝛽 Calculate zero and pole time constants𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚 = = 0.8228 1 𝑇𝑇 =𝜔𝜔𝑚𝑚0.𝑚𝑚0855𝑚𝑚 𝛽𝛽 𝑠𝑠𝑠𝑠𝑠𝑠
The resulting lead compensator transfer function:𝛽𝛽𝛽𝛽 𝑠𝑠𝑠𝑠𝑠𝑠 + 1 = + 1 𝑇𝑇𝑇𝑇 𝐾𝐾𝐾𝐾 𝑠𝑠 𝐾𝐾 0. + 1 = 14.𝛽𝛽7𝛽𝛽𝛽𝛽 0. + 1 8228𝑠𝑠 𝐾𝐾𝐾𝐾 𝑠𝑠 K. Webb 0855𝑠𝑠 ESE 499 Lead Example 2 – Step 6 77
Bode plot of the compensated system = 49.8° Substantially below target𝑃𝑃𝑃𝑃
Crossover frequency is well above the desired value = 9.44 / Iteration𝑃𝑃𝑃𝑃 will likely be required𝜔𝜔 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠
K. Webb ESE 499 Lead Example 2 – Step 6 78
Overshoot exceeds the specified limit = 19.1° > 10%
Settling%𝑂𝑂𝑂𝑂 time is faster than required = 0.98 < 1.2
𝑠𝑠 Iteration𝑡𝑡 is𝑠𝑠 required𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠 Start by reducing the target
𝑃𝑃𝑃𝑃 K. Webb 𝜔𝜔 ESE 499 Lead Example 2 – Step 6 79
Must redesign the compensator to meet specifications Must increase PM to reduce overshoot Can afford to reduce crossover, , to improve PM
Try various combinations of the𝜔𝜔 following𝑃𝑃𝑃𝑃 Reduce crossover frequency, Increase compensator zero/pole frequencies, 𝑃𝑃𝑃𝑃 Increase added phase lead, 𝜔𝜔 , by reducing 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 Iteration shows acceptable 𝜙𝜙results𝑚𝑚𝑚𝑚𝑚𝑚 for: 𝛽𝛽 = 2.4 / = 3.4 / 𝜔𝜔𝑃𝑃𝑃𝑃 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 = 52° 𝜔𝜔𝑚𝑚𝑚𝑚𝑚𝑚 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 K. Webb 𝜙𝜙 ESE 499 Lead Example 2 – Step 6 80
Redesigned lead compensator: 0. + 1 = 6.27 0. + 1 8542𝑠𝑠 𝐾𝐾𝐷𝐷 𝑠𝑠 1013𝑠𝑠 Phase margin: = 62°
Crossover𝑃𝑃𝑃𝑃 frequency: = 4.84 /
𝜔𝜔𝑃𝑃𝑃𝑃 𝑟𝑟𝑟𝑟𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠
K. Webb ESE 499 Lead Example 2 – Step 6 81
Dynamic response requirements are now satisfied
Overshoot: = 8%
Settling%𝑂𝑂𝑂𝑂 time: = 1.09
𝑡𝑡𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠
K. Webb ESE 499 Lead Compensation – Example 2 82
Lead compensator adds gain at higher frequencies Increased crossover frequency Faster response time
Phase added near the crossover frequency Improved phase margin Reduced overshoot
K. Webb ESE 499 Lead Compensation – Example 2 83
Step response improvements: Faster settling time Faster risetime Significantly less overshoot and ringing
K. Webb ESE 499 Lead-Lag Compensation 84
If performance specifications require adjustment of: Bandwidth Phase margin Steady-state error Lead-lag compensation may be used + 1 + 1 = + 1 + 1 𝑇𝑇𝑙𝑙𝑙𝑙𝑙𝑙𝑠𝑠 𝑇𝑇𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑠𝑠 𝐷𝐷 𝑠𝑠 𝛼𝛼 Many possible design𝛼𝛼 𝑇𝑇procedures𝑙𝑙𝑙𝑙𝑙𝑙𝑠𝑠 𝛽𝛽𝑇𝑇–𝑙𝑙𝑙𝑙𝑙𝑙one𝑙𝑙𝑠𝑠 possibility: 1. Design lag compensation to satisfy steady-state error and phase margin 2. Add lead compensation to increase bandwidth, while maintaining phase margin
K. Webb ESE 499