New SECTION 9: ROOT-LOCUS DESIGN - College of...

ESE 499 – Feedback Control Systems

SECTION 9: ROOT-LOCUS DESIGN

K. Webb ESE 499

Introduction2

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Introduction

Consider the following unity-feedback system

Assume 𝐷𝐷 𝑠𝑠 = 𝐾𝐾 A proportional controller

Design for 8% overshoot Use root locus to determine 𝐾𝐾 to

yield required 𝜁𝜁

𝜁𝜁 = −ln 0.08

𝜋𝜋2 + ln2 0.08= 0.63

Desired poles and gain: 𝑠𝑠1,2 = −2 ± 𝑗𝑗𝑗.5 𝐾𝐾 = 2.4

𝑇𝑇 𝑠𝑠 =3𝐾𝐾

𝑠𝑠2 + 4𝑠𝑠 + 3 + 3𝐾𝐾

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Introduction

Overshoot is 8%, as desired, but steady-state error is large: 𝑒𝑒𝑠𝑠𝑠𝑠 = 29.4%

Position constant:𝐾𝐾𝑝𝑝 = lim

𝑠𝑠→0𝐺𝐺 𝑠𝑠

𝐾𝐾𝑝𝑝 = lim𝑠𝑠→0

3𝐾𝐾𝑠𝑠 + 1 𝑠𝑠 + 3

= 𝐾𝐾

𝐾𝐾𝑝𝑝 = 2.4

Steady-state error:

𝑒𝑒𝑠𝑠𝑠𝑠 =1

1 + 𝐾𝐾𝑝𝑝=

11 + 𝐾𝐾


1 + 2.4= 0.294

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Introduction

Let’s say we want to reduce steady-state error to 2% Determine required gain


1 + 𝐾𝐾𝑝𝑝=

11 + 𝐾𝐾

= 0.02

𝐾𝐾 =1

0.02− 1 = 49

Transient response is degraded 𝑂𝑂𝑂𝑂 = 59.2%

Can set overshoot orsteady-state error via gain adjustment Not both simultaneously

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Introduction

Now say we want OS = 8% and 𝑡𝑡𝑠𝑠 ≈ 1 𝑠𝑠𝑒𝑒𝑠𝑠, we’d need:𝜁𝜁 = 0.63 and 𝜎𝜎 = 4.6

Desired poles are not on the root locus

Closed-loop poles can exist only on the locus If we want poles elsewhere,

we must move the locus Modify the locus by adding

dynamics (poles and zeros) to the controller A compensator

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Introduction

We’ll learn how to use root-locus techniques to design compensators to do the following: Improve steady-state error Proportional-integral (PI) compensator Lag compensator

Improve dynamic response Proportional-derivative (PD) compensator Lead compensator

Improve dynamic response and steady-state error Proportional-integral-derivative (PID) compensator Lead-lag compensator

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Compensation Configurations

Two basic compensation configurations: Cascade compensation

Feedback compensation

We will focus on cascade compensation

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Improving Steady-State Error9

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Improving Steady-State Error

We’ve seen that we can improve steady-state error by adding a pole at the origin An integrator System type increased by one for unity-feedback

For example, consider the previous example Let’s say we are happy with

8% overshoot and the corresponding pole locations

But, want to reduce steady-state error to 2% or less

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System is type 0 Adding an integrator to 𝐷𝐷 𝑠𝑠 will increase it to type 1 Zero steady-state error for constant reference

Let’s first try a very simple approach:

Plot the root locus for this system How does the added pole at the origin affect the locus?

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Now have 𝑛𝑛 −𝑚𝑚 = 3asymptotes to 𝐶𝐶∞

𝜃𝜃𝑎𝑎 = 60°, 180°, 300°𝜎𝜎𝑎𝑎 = −1.33

Locus now crosses into the RHP Integrator has had a

destabilizing effect on the closed-loop system

System is now type 1, but Desired poles are no longer

on the root locus

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Desired poles no longer satisfy the angle criterion:∠𝐷𝐷 𝑠𝑠1 𝐺𝐺 𝑠𝑠1 = − 𝜙𝜙1 + 𝜙𝜙2 + 𝜙𝜙3∠𝐷𝐷 𝑠𝑠1 𝐺𝐺 𝑠𝑠1 = − 128.8° + 111.9° + 68.1°∠𝐷𝐷 𝑠𝑠1 𝐺𝐺 𝑠𝑠1 = −308.8° ≠ 180°

Excess angle from the additional pole at the origin, 𝜙𝜙1

How could we modify 𝐷𝐷 𝑠𝑠 to satisfy the angle criterion at 𝑠𝑠1? A zero at the origin would do it, of course But, that would cancel the desired pole at the origin

How about a zero very close to the origin?

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Now, 𝜓𝜓1 ≈ 𝜙𝜙1 Angle contributions nearly cancel 𝑠𝑠1 is not on the locus, but very close

The closer the zero is to the origin, the closer 𝑠𝑠1 will be to the root locus

Let 𝑧𝑧𝑐𝑐 = −0.1 Controller transfer function:

𝐷𝐷 𝑠𝑠 = 𝐾𝐾𝑠𝑠 + 0.1

𝑠𝑠 Plot new root locus to see how

close it comes to 𝑠𝑠1

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Now only two asymptotes to 𝐶𝐶∞

𝜃𝜃𝑎𝑎 = 90°, 270° 𝜎𝜎𝑎𝑎 = −1.95

Real-axis breakaway point: 𝑠𝑠 = −1.99

𝑠𝑠1 not on locus, but close Closed-loop poles with 𝜁𝜁 = 0.63:

𝑠𝑠1,2 = −1.96 ± 𝑗𝑗𝑗.44 Gain: 𝐾𝐾 = 2.37

Determined from the MATLAB root locus plot

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Initial transient relatively unchanged Pole/zero pair near the origin nearly

cancel 2nd-order poles close to desired

location

Zero steady-state error Pole at origin increases system type

to type 1 Slow transient as error is integrated

out

2nd-order approximation is valid Poles: 𝑠𝑠 = −0.07,

𝑠𝑠 = −1.96 ± 𝑗𝑗𝑗.44 Zeros: 𝑠𝑠 = −0.1

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Ideal Integral Compensation17

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Proportional-Integral Compensation

The compensator we just designed is an ideal integral or proportional-integral (PI) compensator

Control input to plant, 𝑈𝑈 𝑠𝑠 , has two components: One proportional to the error, plus One proportional to the integral of the error

𝑈𝑈 𝑠𝑠 = 𝐸𝐸 𝑠𝑠 𝐾𝐾𝑠𝑠 + 𝑎𝑎𝑠𝑠

= 𝐾𝐾𝐸𝐸 𝑠𝑠 +𝐾𝐾𝑎𝑎𝑠𝑠𝐸𝐸 𝑠𝑠

Equivalent to:

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PI Compensation – Summary

PI compensation

𝐷𝐷 𝑠𝑠 = 𝐾𝐾𝑠𝑠 + 𝑎𝑎𝑠𝑠

= 𝐾𝐾𝑝𝑝 +𝐾𝐾𝑖𝑖𝑠𝑠

Controller adds a pole at the origin and a zero nearby

Pole at origin (integrator) increases system type, improves steady-state error

Zero near the origin nearly cancels the added pole, leaving transient response nearly unchanged

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PI Compensation – Zero Location

Compensator zero very close to the origin: Closed-loop poles moved very little from uncompensated

location Relatively low integral gain, 𝐾𝐾𝑖𝑖 Closed-loop pole close to origin – slow Slow transient as error is integrated out

Compensator zero farther from the origin: Closed-loop poles moved farther from uncompensated

location Relatively higher integral gain, 𝐾𝐾𝑖𝑖 Closed-loop pole farther from the origin – faster Error is integrated out more quickly

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PI Compensation – Zero Location

Root locus and step response variation with 𝑧𝑧𝑐𝑐:

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Lag Compensation22

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Lag Compensation

PI compensation requires an ideal integrator Active circuitry (opamp) required for analog implementation Susceptible to integrator windup

An alternative to PI compensation is lag compensation Pole placed near the origin, not at the origin Analog implementation realizable with passive components

(resistors and capacitors)

Like PI compensation, lag compensation uses a closely-spaced pole/zero pair Angular contributions nearly cancel Transient response nearly unaffected

System type not increased Error is improved, not eliminated

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Lag Compensation – Error Reduction

Consider the following generic feedback system

A type 0 system, assuming 𝑝𝑝𝑖𝑖 ≠ 0, ∀𝑖𝑖 Position constant:

𝐾𝐾𝑝𝑝𝑝𝑝 = lim𝑠𝑠→0

𝐺𝐺 𝑠𝑠 = 𝐾𝐾𝑧𝑧1𝑧𝑧2 ⋯𝑝𝑝1𝑝𝑝2 ⋯

Now, add lag compensation

The compensated position constant :

𝐾𝐾𝑝𝑝𝑐𝑐 = 𝐾𝐾𝑧𝑧1𝑧𝑧2 ⋯𝑝𝑝1𝑝𝑝2 ⋯

𝑧𝑧𝑐𝑐𝑝𝑝𝑐𝑐

= 𝐾𝐾𝑝𝑝𝑝𝑝𝑧𝑧𝑐𝑐𝑝𝑝𝑐𝑐

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Lag Compensation – Error Reduction

Compensator pole is closer to the origin than the compensator zero, so

𝑧𝑧𝑐𝑐 > 𝑝𝑝𝑐𝑐 and 𝐾𝐾𝑝𝑝𝑐𝑐 > 𝐾𝐾𝑝𝑝𝑝𝑝 For large improvements in 𝑒𝑒𝑠𝑠𝑠𝑠, make 𝑧𝑧𝑐𝑐 ≫ 𝑝𝑝𝑐𝑐

But, to avoid affecting the transient response, we need 𝑧𝑧𝑐𝑐 ≈ 𝑝𝑝𝑐𝑐

As long as both 𝑧𝑧𝑐𝑐 and 𝑝𝑝𝑐𝑐 are very small, we can satisfy both requirements: 𝑧𝑧𝑐𝑐 ≫ 𝑝𝑝𝑐𝑐 and 𝑧𝑧𝑐𝑐 ≈ 𝑝𝑝𝑐𝑐

𝐾𝐾𝑝𝑝𝑐𝑐 = 𝐾𝐾𝑧𝑧1𝑧𝑧2 ⋯𝑝𝑝1𝑝𝑝2 ⋯

𝑧𝑧𝑐𝑐𝑝𝑝𝑐𝑐

= 𝐾𝐾𝑝𝑝𝑝𝑝𝑧𝑧𝑐𝑐𝑝𝑝𝑐𝑐

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Lag Compensation – Example

Apply lag compensation to our previous example Design for a 10x improvement of the position constant

Want 𝑝𝑝𝑐𝑐 ≈ 0 (relative to other poles) Let 𝑝𝑝𝑐𝑐 = 0.01

Want a 10x improvement in 𝐾𝐾𝑝𝑝 𝑧𝑧𝑐𝑐 = 10𝑝𝑝𝑐𝑐 = 0.1

Lag pole and zero differ by a factor of 10 Static error constant improved by a factor of 10

Lag pole/zero are very close together relative to poles at 𝑠𝑠 = −1,−3 Angular contributions nearly cancel Transient response nearly unaffected

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Root locus and step response of lag-compensated system

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Now, let 𝑧𝑧𝑐𝑐 = 0.4 and 𝑝𝑝𝑐𝑐 = 0.04 2nd-order poles moved more Faster low-frequency closed-loop pole Faster overall response

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Lag Compensation – Summary

Lag compensation

𝐷𝐷 𝑠𝑠 = 𝐾𝐾 𝑠𝑠+𝑧𝑧𝑐𝑐𝑠𝑠+𝑝𝑝𝑐𝑐

, where 𝑝𝑝𝑐𝑐 < 𝑧𝑧𝑐𝑐

Controller adds a pole near the origin and a slightly-higher-frequency zero nearby

Static error constant improved by 𝑧𝑧𝑐𝑐/𝑝𝑝𝑐𝑐 Angle contributions from closely-spaced pole/zero

nearly cancel Transient response is nearly unchanged

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Improving Transient Response30

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Improving Transient Response

Consider the following system

Root locus: Three asymptotes to 𝐶𝐶∞

at 60°, 180°, and 300° Real-axis breakaway

point: 𝑠𝑠 = −1.88 Locus crosses into RHP

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Design proportional controller for 10% overshoot 𝐾𝐾 = 1.72

Overshoot < 10% due to third pole

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Now, decrease settling time to 𝑡𝑡𝑠𝑠 ≈ 1.5 𝑠𝑠𝑒𝑒𝑠𝑠 Maintain same overshoot (𝜁𝜁 = 0.59)

𝜎𝜎 ≈4.6𝑡𝑡𝑠𝑠

= 3.1

Desired poles: 𝑠𝑠1,2 = −3.1 ± 𝑗𝑗4.23 Not on the locus

Must add compensation to move the locus where we want it Derivative compensation

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Ideal Derivative Compensation34

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Proportional-Derivative Compensation

One way to improve transient response is to add the derivative of the error to the control input to the plant

This is ideal derivative or proportional-derivative (PD) compensation

𝑈𝑈 𝑠𝑠 = 𝐸𝐸 𝑠𝑠 𝐾𝐾𝑝𝑝 + 𝐾𝐾𝑑𝑑𝑠𝑠 = 𝐾𝐾 𝑠𝑠 + 𝑧𝑧𝑐𝑐 𝐸𝐸 𝑠𝑠 Compensator transfer function:

𝐷𝐷 𝑠𝑠 = 𝐾𝐾 𝑠𝑠 + 𝑧𝑧𝑐𝑐 Compensator adds a single zero at 𝑠𝑠 = −𝑧𝑧𝑐𝑐

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PD Compensation

Compensator zero will change the root locus Placement of the zero allows us to move the locus to place

closed-loop poles where we want them

One less asymptote to 𝐶𝐶∞ 𝑛𝑛 −𝑚𝑚 decreased by one

Asymptote origin changes

𝜎𝜎𝑎𝑎 =Σ𝑝𝑝𝑖𝑖 − Σ𝑧𝑧𝑖𝑖𝑛𝑛 − 𝑚𝑚

As 𝑧𝑧𝑐𝑐 increases (moves left), 𝜎𝜎𝑎𝑎 moves right, toward the origin

As 𝑧𝑧𝑐𝑐 decreases (moves right), 𝜎𝜎𝑎𝑎 moves further into the LHP

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PD Compensation

Derivative compensation allows us to speed up the closed-loop response Control signal proportional to (in part) the derivative of

the error When the reference, 𝑟𝑟(𝑡𝑡), changes quickly:

Error, 𝑒𝑒(𝑡𝑡), changes quickly Derivative of the error, �̇�𝑒 𝑡𝑡 , is large Control input, 𝑢𝑢 𝑡𝑡 , may be large

Derivative compensation anticipates future error and compensates for it

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PD Compensation – Example 1

Now add PD compensation to our example system Root locus depends on 𝑧𝑧𝑐𝑐

Let’s first assume 𝑧𝑧𝑐𝑐 < 3

Two real-axis segments −6 ≤ 𝑠𝑠 ≤ −3 Between pole at −1 and 𝑧𝑧𝑐𝑐

Two asymptotes to 𝐶𝐶∞

𝜃𝜃𝑎𝑎 = 90°, 270°

𝜎𝜎𝑎𝑎 = 𝑧𝑧𝑐𝑐−102

As 𝑧𝑧𝑐𝑐 varies from 0 … 3, 𝜎𝜎𝑎𝑎 varies from −5 …− 3.5

Breakaway point between −6 …− 3

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As 𝑧𝑧𝑐𝑐 moves to the left, 𝜎𝜎𝑎𝑎 moves to the right Moving 𝑧𝑧𝑐𝑐 allows us to move the locus

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Now move the zero further to the left: 𝑧𝑧𝑐𝑐 > 3 Still two real-axis segments

−6 ≤ 𝑠𝑠 ≤ −𝑧𝑧𝑐𝑐 −3 ≤ 𝑠𝑠 ≤ −1

Two asymptotes to 𝐶𝐶∞ 𝜃𝜃𝑎𝑎 = 90°, 270°

𝜎𝜎𝑎𝑎 = 𝑧𝑧𝑐𝑐−102

As 𝑧𝑧𝑐𝑐 varies from 3 …∞, 𝜎𝜎𝑎𝑎varies from −3.5 …∞

Breakaway point between − 3 …− 1

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Asymptote origin continues to move to the right

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PD Compensation – Calculating 𝑧𝑧𝑐𝑐

For this particular system, we’ve seen: Additional zero decreased the number of asymptotes to 𝐶𝐶∞ by one

A stabilizing effect – locus does not cross into the RHP Adjusting 𝑧𝑧𝑐𝑐 allows us to move the asymptote origin

left or right

Next, we’ll determine exactly where to place 𝑧𝑧𝑐𝑐 to place the closed-loop poles where we want them

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Desired 2nd-order poles: 𝑠𝑠1,2 = −3.1 ± 𝑗𝑗4.23 Calculate required value for 𝑧𝑧𝑐𝑐 such that these points are on the

locus

Must satisfy the angle criterion∠𝐷𝐷 𝑠𝑠1 𝐺𝐺 𝑠𝑠1 = 180°∠𝐷𝐷 𝑠𝑠1 𝐺𝐺 𝑠𝑠1 = 𝜓𝜓𝑐𝑐 − 𝜙𝜙1 − 𝜙𝜙2 − 𝜙𝜙3𝜓𝜓𝑐𝑐 = 180° + 𝜙𝜙1 + 𝜙𝜙2 + 𝜙𝜙3

𝜙𝜙1 = 116.4°𝜙𝜙2 = 91.35°𝜙𝜙3 = 55.57°

The required angle from 𝑧𝑧𝑐𝑐:𝜓𝜓𝑐𝑐 = 83.3°

Next, determine 𝑧𝑧𝑐𝑐

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Compensator zero, 𝑧𝑧𝑐𝑐, must contribute 𝜓𝜓𝑐𝑐 = 83.3° at 𝑠𝑠1,2 = −3.1 ± 𝑗𝑗4.23

Calculate the required value of 𝑧𝑧𝑐𝑐𝜓𝜓𝑐𝑐 = ∠ 𝑠𝑠1 + 𝑧𝑧𝑐𝑐 = ∠ −3.1 + 𝑗𝑗4.23 + 𝑧𝑧𝑐𝑐 = 83.3°

𝜓𝜓𝑐𝑐 = tan−14.23

𝑧𝑧𝑐𝑐 − 3.1

tan 𝜓𝜓𝑐𝑐 =4.23

𝑧𝑧𝑐𝑐 − 3.1

𝑧𝑧𝑐𝑐 =4.23

tan 𝜓𝜓𝑐𝑐+ 3.1 =

4.23tan 83.3°

+ 3.1

The required compensator zero:𝑧𝑧𝑐𝑐 = 3.6

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Locus passes through desired points

Closed-loop poles at 𝑠𝑠 = −3.1 ± 𝑗𝑗4.23 for 𝐾𝐾 = 1.6

Third closed-loop pole at 𝑠𝑠 = −3.8 Close to zero at 𝑠𝑠 =− 3.6

2nd-order approximation likely justified

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Settling time reduced, as desired

Overshoot is a little higher than 10% Higher order pole and

zero do not entirely cancel

Iterate to further refine performance, if desired

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PD Compensation – Summary

PD compensation𝐷𝐷 𝑠𝑠 = 𝐾𝐾 𝑠𝑠 + 𝑧𝑧𝑐𝑐

Controller adds a single zero

Angular contribution from the compensator zero allows the root locus to be modified

Calculate 𝑧𝑧𝑐𝑐 to satisfy the angle criterion at desired closed-loop pole locations Use magnitude criterion or plot root locus to determine

required gain

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Lead Compensation48

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Sensor Noise

Feedback control requires measurement of a system’s output with some type of sensor Inherently noisy Measurement noise tends to be broadband in nature I.e., includes energy at high frequencies

High-frequency signal components change rapidly Large time derivatives

Derivative (PD) compensation amplifies measurement noise

An alternative is lead compensation Amplification of sensor noise is reduced

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Lead Compensation

PD compensation utilizes an ideal differentiator Amplifies sensor noise Active circuitry (opamp) required for analog implementation

An alternative to PD compensation is lead compensation Compensator adds one zero and a higher-frequency pole

𝐷𝐷 𝑠𝑠 = 𝐾𝐾 𝑠𝑠+𝑧𝑧𝑐𝑐𝑠𝑠+𝑝𝑝𝑐𝑐

, where 𝑝𝑝𝑐𝑐 > 𝑧𝑧𝑐𝑐

Pole can be far enough removed to have little impact on 2nd-order dynamics

Additional high-frequency pole reduces amplification of noise Analog implementation realizable with passive components

(resistors and capacitors)

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Lead Compensation – Example

Apply lead compensation to our previous example system Desired closed-loop poles:

𝑠𝑠1,2 = −3.1 ± 𝑗𝑗4.23

Angle criterion must be satisfied at 𝑠𝑠1∠𝐷𝐷 𝑠𝑠1 𝐺𝐺 𝑠𝑠1 = 180°∠𝐷𝐷 𝑠𝑠1 + ∠𝐺𝐺 𝑠𝑠1 = 180°∠𝐷𝐷 𝑠𝑠1 = 180° − ∠𝐺𝐺 𝑠𝑠1∠𝐺𝐺 𝑠𝑠1 = − 𝜙𝜙1 + 𝜙𝜙2 + 𝜙𝜙3∠𝐺𝐺 𝑠𝑠1 = −263.3°

Required net angle contribution from the compensator:

∠𝐷𝐷 𝑠𝑠1 = 443.3° = 83.3°

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For 𝑠𝑠1 to be on the locus, we need ∠𝐷𝐷 𝑠𝑠1 = 83.3° Zero contributes a positive angle Higher-frequency pole contributes a smaller negative angle Net angular contribution will be positive, as required:

∠𝐷𝐷 𝑠𝑠1 = ∠ 𝑠𝑠1 + 𝑧𝑧𝑐𝑐 − ∠ 𝑠𝑠1 + 𝑝𝑝𝑐𝑐 = 83.3°

Compensator angle is the angle of the ray from 𝑠𝑠1 through 𝑧𝑧𝑐𝑐and 𝑝𝑝𝑐𝑐

∠𝐷𝐷 𝑠𝑠1 = 𝜃𝜃𝑐𝑐 = 88.3°

Infinite combinations of 𝑧𝑧𝑐𝑐 and 𝑝𝑝𝑐𝑐 will provide the required 𝜃𝜃𝑐𝑐

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An infinite number of possible 𝑧𝑧𝑐𝑐/𝑝𝑝𝑐𝑐 combinations All provide 𝜃𝜃𝑐𝑐 = 83.3° Different static error

constants Different required gains Different location of other

closed-loop poles

No real rule for how to select 𝑧𝑧𝑐𝑐 and 𝑝𝑝𝑐𝑐

Some options: Set 𝑝𝑝𝑐𝑐 as high as acceptable

given noise requirements Place 𝑧𝑧𝑐𝑐 below or slightly left

of the desired poles

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Root locus and Step response for 𝑧𝑧𝑐𝑐 = 0.5, 𝑝𝑝𝑐𝑐 = 8.5 Lower-frequency pole/zero do not adequately cancel

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Root locus and Step response for 𝑧𝑧𝑐𝑐 = 1.5, 𝑝𝑝𝑐𝑐 = 11.3 Effect of lower-frequency pole/zero reduced

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Root locus and Step response for 𝑧𝑧𝑐𝑐 = 2.5, 𝑝𝑝𝑐𝑐 = 19.2 Lower-frequency pole/zero very nearly cancel

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Root locus and Step response for 𝑧𝑧𝑐𝑐 = 3.2, 𝑝𝑝𝑐𝑐 = 48.5 Higher-frequency pole/zero almost completely cancel

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Here, 𝑧𝑧𝑐𝑐 = 2.5 or 𝑧𝑧𝑐𝑐 = 3.2 are good choices

Steady-state error varies Error depends on

gain required for each lead implementation

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Lead Compensation – Summary

Lead compensation

𝐷𝐷 𝑠𝑠 = 𝐾𝐾𝑠𝑠 + 𝑧𝑧𝑐𝑐𝑠𝑠 + 𝑝𝑝𝑐𝑐

, where 𝑝𝑝𝑐𝑐 > 𝑧𝑧𝑐𝑐

Controller adds a lower-frequency zero and a higher-frequency pole

Net angular contribution from the compensator zero and pole allows the root locus to be modified Allows for transient response improvement

Infinite number of possible 𝑧𝑧𝑐𝑐/𝑝𝑝𝑐𝑐 combinations to satisfy the angle criterion at the design point

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Improving Error and Transient Response60

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Improving Error and Transient Response

PI (or lag) control improves steady-state error PD (or lead) control can improve transient response

Using both together can improve both error and dynamic performance PD or lead compensation to achieve desired transient

response PI or lag compensation to achieve desired steady-state error

Next, we’ll look at two types of compensators: Proportional-integral-derivative (PID) compensator Lead-lag compensator

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Improving Error and Transient Response

Two possible approaches to the design procedure:1. First design for transient response, then design for

steady-state error Response may be slowed slightly in the process of

improving steady-state error

2. First design for steady-state error, then design for transient response Steady-state error may be affected

In either case, iteration is typically necessary We’ll follow the first approach, as does the text

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PID Compensation63

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Proportional-Integral-Derivative Compensation

Proportional-integral-derivative (PID) compensation Combines PI and PD

compensation PD compensation adjusts

transient response PI compensation improves

steady-state error

Controller transfer function:

𝐷𝐷 𝑠𝑠 = 𝐾𝐾𝑝𝑝 +𝐾𝐾𝑖𝑖𝑠𝑠

+ 𝐾𝐾𝑑𝑑𝑠𝑠 =𝐾𝐾𝑑𝑑𝑠𝑠2 + 𝐾𝐾𝑝𝑝𝑠𝑠 + 𝐾𝐾𝑖𝑖

𝑠𝑠 Two zeros and a pole at the origin

Pole/zero at/near the origin determined through PI compensator design

Second zero location determined through PD compensator design

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PID Design Procedure

PID compensator design procedure:

1. Determine closed-loop pole location to provide desired transient response

2. Design PD controller (zero location and gain) to place closed-loop poles as desired

3. Simulate the PD-compensated system, iterate if necessary4. Design a PI controller, add to the PD-compensated system,

and determine the gain required to maintain desired dominant pole locations

5. Determine PID parameters: 𝐾𝐾𝑝𝑝, 𝐾𝐾𝑖𝑖, and 𝐾𝐾𝑑𝑑6. Simulate the PID-compensated system and iterate, if

necessary

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PID Compensation – Example

Design PID compensation to satisfy the following specifications: 𝑡𝑡𝑠𝑠 ≈ 2 𝑠𝑠𝑒𝑒𝑠𝑠%𝑂𝑂𝑂𝑂 ≈ 20% Zero steady-state error to a constant reference

First, design PD compensator to satisfy dynamic specifications

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Calculate desired closed-loop pole locations𝜎𝜎 ≈

4.6𝑡𝑡𝑠𝑠

= 2.3

𝜁𝜁 = −ln 0.2

𝜋𝜋2 + ln2 0.2= 0.46

𝜔𝜔𝑑𝑑 =𝜎𝜎𝜁𝜁

1 − 𝜁𝜁2 =2.3

0.461 − 0.462

𝜔𝜔𝑑𝑑 = 4.49

Desired 2nd-order poles:𝑠𝑠1,2 = −2.3 ± 𝑗𝑗4.49

Uncompensated root locus does not pass through the desired poles Gain adjustment not sufficient Compensation required

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PD compensator design Determine the required angular contribution of the

compensator zero to satisfy the angle criterion at 𝑠𝑠1∠𝐷𝐷𝑝𝑝𝑑𝑑 𝑠𝑠1 𝐺𝐺 𝑠𝑠1 = 180°∠𝐷𝐷𝑝𝑝𝑑𝑑 𝑠𝑠1 = 180° − ∠𝐺𝐺 𝑠𝑠1∠𝐺𝐺 𝑠𝑠1 = −𝜙𝜙1 − 𝜙𝜙2 − 𝜙𝜙3𝜙𝜙1 = ∠ 𝑠𝑠1 + 1 = 106.15°𝜙𝜙2 = ∠ 𝑠𝑠1 + 3 = 81.14°𝜙𝜙3 = ∠ 𝑠𝑠1 + 6 = 50.51°∠𝐺𝐺 𝑠𝑠1 = −237.8°∠𝐷𝐷𝑝𝑝𝑑𝑑 𝑠𝑠1 = 180° + 237.8° = 417.8°

Required angle from PD zero𝜓𝜓𝑝𝑝𝑑𝑑 = 57.8°

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Use required compensator angle to place the PD zero, 𝑧𝑧𝑝𝑝𝑑𝑑∠𝐷𝐷𝑝𝑝𝑑𝑑 𝑠𝑠1 = ∠ 𝑠𝑠1 + 𝑧𝑧𝑝𝑝𝑑𝑑

∠𝐷𝐷𝑝𝑝𝑑𝑑 𝑠𝑠1 = ∠ −2.3 + 𝑧𝑧𝑝𝑝𝑑𝑑 + 𝑗𝑗4.49

tan 𝜓𝜓𝑝𝑝𝑑𝑑 =4.49

𝑧𝑧𝑝𝑝𝑑𝑑 − 2.3

𝑧𝑧𝑝𝑝𝑑𝑑 =4.49

tan 57.8°+ 2.3 = 5.13

PD compensator transfer function:

𝐷𝐷𝑝𝑝𝑑𝑑 𝑠𝑠 = 𝐾𝐾 𝑠𝑠 + 5.13

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PD-compensated root locus Determine required gain

from MATLAB plot, or Apply the magnitude

criterion:

𝐾𝐾 =1

𝐷𝐷𝑝𝑝𝑑𝑑 𝑠𝑠1 𝐺𝐺 𝑠𝑠1

𝐾𝐾 =𝑠𝑠1 + 1 𝑠𝑠1 + 3 𝑠𝑠1 + 6

15 𝑠𝑠1 + 5.13𝐾𝐾 = 1.55

PD compensator:𝐷𝐷𝑝𝑝𝑑𝑑 𝑠𝑠 = 1.55 𝑠𝑠 + 5.13

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Performance specifications not met exactly Higher-frequency

pole/zero do not entirely cancel

Close enough for now – may need to iterate when PI compensation is added

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Next, add PI compensation to the PD-compensated system Add a pole at the origin and a zero close by

𝐷𝐷𝑝𝑝𝑖𝑖 𝑠𝑠 =𝑠𝑠 + 𝑧𝑧𝑝𝑝𝑖𝑖𝑠𝑠

Where should we put the zero, 𝑧𝑧𝑝𝑝𝑖𝑖? In this case, open-loop pole at the origin will become a closed-

loop pole near −𝑧𝑧𝑝𝑝𝑖𝑖 Very small 𝑧𝑧𝑝𝑝𝑖𝑖 yields very slow closed-loop pole Error integrates out very slowly

Small 𝑧𝑧𝑝𝑝𝑖𝑖 means PI compensator will have less effect on the PD-compensated root locus

Simulate and iterate

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Step response for various 𝑧𝑧𝑝𝑝𝑖𝑖 values:

Here, 𝑧𝑧𝑝𝑝𝑖𝑖 = 0.8 works well Moving 𝑧𝑧𝑝𝑝𝑖𝑖 away from the

open-loop pole at the origin moves the 2nd-order poles significantly:

𝑠𝑠1,2 = −1.86 ± 𝑗𝑗3.63

Faster low-frequency closed-loop pole means error is integrated out more quickly

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The resulting PID compensator:

𝐷𝐷 𝑠𝑠 = 𝐾𝐾𝑠𝑠 + 0.8

𝑠𝑠𝑠𝑠 + 5.13

Required gain: 𝐾𝐾 = 1.15

𝐷𝐷 𝑠𝑠 =1.15𝑠𝑠2 + 6.817𝑠𝑠 + 3.718

𝑠𝑠

𝐷𝐷 𝑠𝑠 =𝐾𝐾𝑑𝑑𝑠𝑠2 + 𝐾𝐾𝑝𝑝𝑠𝑠 + 𝐾𝐾𝑖𝑖

𝑠𝑠

The PID gains:𝐾𝐾𝑝𝑝 = 6.817, 𝐾𝐾𝑖𝑖 = 3.718, 𝐾𝐾𝑑𝑑 = 1.15

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Step response of the PID-compensated system: Settling time is a little

slow A bit of margin on the

overshoot Iterate

First, try adjusting gain alone

If necessary, revisit the PD compensator

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Increasing gain to 𝐾𝐾 = 1.25 speed things up a bit, while increasing overshoot

Recall, however that root locus asymptotes are vertical Increasing gain will have

little effect on settling time

If further refinement is required, must revisit the PD compensator

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How valid was the second-order approximation we used for design of this PID-compensated system?

Pole at 𝑠𝑠 = −0.78 Nearly canceled by the zero at 𝑠𝑠 =− 0.8

Pole at s = −5.5 Not high enough in frequency to be

negligible, but Partially canceled by zero at 𝑠𝑠 =− 5.13

But, validity of the assumption is not really important Used as starting point to locate poles Iteration typically required anyway

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PID Compensation – Summary

PID compensation

𝐷𝐷 𝑠𝑠 = 𝐾𝐾𝑝𝑝 +𝐾𝐾𝑖𝑖𝑠𝑠 + 𝐾𝐾𝑑𝑑𝑠𝑠 =

𝐾𝐾𝑑𝑑𝑠𝑠2 + 𝐾𝐾𝑝𝑝𝑠𝑠 + 𝐾𝐾𝑖𝑖𝑠𝑠

Two zeros and a pole at the origin Cascade of PI and PD compensators

PD compensator Added zero allows for transient response improvement

PI compensator Pole at the origin increases system type Nearby zero nearly cancels angular contribution of the pole,

limiting its effect on the root locus

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Lead-Lag Compensation79

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Lead-Lag Compensation

Just as we combined derivative and integral compensation, we can combine lead and lag as well Lead-lag compensation Lead compensator improves transient response Lag compensator improves steady-state error

Compensator transfer function:

𝐷𝐷 𝑠𝑠 = 𝐾𝐾𝑠𝑠 + 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑𝑠𝑠 + 𝑝𝑝𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑

𝑠𝑠 + 𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙𝑠𝑠 + 𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙

Lead compensator adds a pole and zero - 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 < 𝑝𝑝𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 Lag pole/zero close to the origin - 𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙 > 𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙 ≈ 0

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Lead-Lag Design Procedure

Lead-lag compensator design procedure:

1. Determine closed-loop pole location to provide desired transient response

2. Design the lead compensator (zero, pole, and gain) to place closed-loop poles as desired

3. Simulate the lead-compensated system, iterate if necessary4. Evaluate the steady-state error performance of the lead-

compensated system to determine how much of an improvement is required to meet the error specification

5. Design the lag compensator to yield the required steady-state error performance

6. Simulate the lead-lag-compensated system and iterate, if necessary

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Lead-Lag Compensation – Example

Design lead-lag compensation to satisfy the following specifications:

𝑡𝑡𝑠𝑠 ≈ 2 𝑠𝑠𝑒𝑒𝑠𝑠 %𝑂𝑂𝑂𝑂 ≈ 20% 2% steady-state error to a constant reference

First, design the lead compensator to satisfy the dynamic specifications

Then, design the lag compensator to meet the steady-state error requirement

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Design the lead compensator to achieve the same desired dominant 2nd-order pole locations:

𝑠𝑠1,2 = −2.3 ± 𝑗𝑗4.49

Again, an infinite number of possibilities

Let’s assume we want to limit the lead pole to 𝑠𝑠 =− 100 due to noise considerations Lower pole frequency results in amplification of less noise

𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠 = 𝐾𝐾𝑠𝑠 + 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑𝑠𝑠 + 100

Apply the angle criterion to determine 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑

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∠𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠1 𝐺𝐺 𝑠𝑠1 = 180° → ∠𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠1 = 180° − ∠𝐺𝐺 𝑠𝑠1∠𝐺𝐺 𝑠𝑠1 = −𝜙𝜙1 − 𝜙𝜙2 − 𝜙𝜙3

𝜙𝜙1 = ∠ 𝑠𝑠1 + 1 = 106.15°𝜙𝜙2 = ∠ 𝑠𝑠1 + 3 = 81.14°𝜙𝜙3 = ∠ 𝑠𝑠1 + 6 = 50.51°

∠𝐺𝐺 𝑠𝑠1 = −237.8°∠𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠1 = 180° + 237.8°∠𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠1 = 417.8° = 57.8°∠𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠1 = 𝜓𝜓𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 − 𝜙𝜙𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑

𝜙𝜙𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 = ∠ 𝑠𝑠1 + 100 = 2.63°𝜓𝜓𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 = ∠𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠1 + 𝜙𝜙𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑

𝜓𝜓𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 = 60.43°

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Next, calculate 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 from 𝜓𝜓𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑𝜓𝜓𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 = ∠ 𝑠𝑠1 + 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑

𝜓𝜓𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 = tan−1𝐼𝐼𝑚𝑚 𝑠𝑠1

𝑅𝑅𝑒𝑒 𝑠𝑠1 + 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑

tan 𝜓𝜓𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 =𝐼𝐼𝑚𝑚 𝑠𝑠1

𝑅𝑅𝑒𝑒 𝑠𝑠1 + 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑

𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 =𝐼𝐼𝑚𝑚 𝑠𝑠1

tan 𝜓𝜓𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑− 𝑅𝑅𝑒𝑒 𝑠𝑠1

𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 =4.49

tan 60.43°+ 2.3

𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 = 4.85

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Lead compensator:

𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠 = 𝐾𝐾𝑠𝑠 + 4.85𝑠𝑠 + 100

From magnitude criterion or MATLAB plot, K = 156

Lead compensator transfer function:

𝐷𝐷𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 𝑠𝑠 = 156𝑠𝑠 + 4.85𝑠𝑠 + 100

Next, simulate the lead-compensated system to verify dynamic performance and to evaluate steady-state error

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Performance specifications not met exactly Higher-frequency

pole/zero do not entirely cancel

Close enough for now –may need to iterate when lag compensation is added, anyway

Steady-state error is 13.8%

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Desired position constant:


1 + 𝐾𝐾𝑝𝑝= 0.02

𝐾𝐾𝑝𝑝 =1𝑒𝑒𝑠𝑠𝑠𝑠

− 1 = 49

𝐾𝐾𝑝𝑝 for lead-compensated system

𝐾𝐾𝑝𝑝,𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 =1

0.138− 1 = 6.26

Required error constant improvement𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙

=𝐾𝐾𝑝𝑝

𝐾𝐾𝑝𝑝,𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑= 7.83

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Arbitrarily set 𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙 = 0.01 To achieve desired error, we need

𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙 = 8 ⋅ 𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙 = 0.08 The lag compensator transfer function:

𝐷𝐷𝑙𝑙𝑎𝑎𝑙𝑙 𝑠𝑠 =𝑠𝑠 + 0.08𝑠𝑠 + 0.01

Magnitudes of lag pole/zero effectively cancel, so required gain is unchanged:

𝐾𝐾 = 156 Lead-lag compensator transfer function:

𝐷𝐷 𝑠𝑠 = 156𝑠𝑠 + 4.85𝑠𝑠 + 100

𝑠𝑠 + 0.08𝑠𝑠 + 0.01

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Root locus and closed-loop poles/zeros for the lead-lag-compensated system:

Second-order poles:𝑠𝑠1,2 = −2.27 ± 𝑗𝑗4.46

Other closed-loop poles:𝑠𝑠 = −100.2𝑠𝑠 = −5.2𝑠𝑠 = −0.07

Closed-loop zeros:𝑠𝑠 = −0.08𝑠𝑠 = −4.85

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Step response for the lead-lag-compensated system: Steady-state error

requirement is satisfied Slow closed-loop pole at 𝑠𝑠 = −0.07 results in very slow tail as error is eliminated

Can speed this up by moving the lag pole/zero away from the origin Dominant poles will move

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Increase the lag pole/zero frequency by 8x𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙 = 0.08 and 𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙 = 0.64

Lag pole/zero now affect the root locus significantly Dominant poles move:

𝑠𝑠1,2 = −1.9 ± 𝑗𝑗3.8 Required gain for 𝜁𝜁 = 0.46

changes:𝐾𝐾 = 123

Reduced gain will reduce 𝐾𝐾𝑝𝑝 𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙/𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙 ratio must increase

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Some iteration shows reasonable transient and error performance for:

𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙 = 0.08

𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙 = 0.8

𝐾𝐾 = 130

Lead-lag compensator:

𝐷𝐷 𝑠𝑠 = 130𝑠𝑠 + 4.85𝑠𝑠 + 100

𝑠𝑠 + 0.8𝑠𝑠 + 0.08

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Lead-Lag Compensation – Summary

Lead-lag compensation

𝐷𝐷 𝑠𝑠 = 𝐾𝐾 𝑠𝑠+𝑧𝑧𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑠𝑠+𝑝𝑝𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙

𝑠𝑠+𝑧𝑧𝑙𝑙𝑙𝑙𝑙𝑙𝑠𝑠+𝑝𝑝𝑙𝑙𝑙𝑙𝑙𝑙

, 𝑝𝑝𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 > 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑 and 𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙 < 𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙

Two zeros and two poles Cascade of lead and lag compensators

Lead compensator Added pole/zero improves transient response

Lag compensator Steady-state error improved by 𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙/𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙 Nearby zero partially cancels angular contribution of the pole,

limiting its effect on the root locus May introduce a slow transient

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Compensator Summary

Type Transfer function Improves Comments

PI𝐾𝐾

𝑠𝑠 + 𝑧𝑧𝑐𝑐𝑠𝑠

Error • Pole at origin• Zero near origin• Increases system type• May introduce a slow transient• Active circuitry required• Susceptible to integrator windup

Lag𝐾𝐾

𝑠𝑠 + 𝑧𝑧𝑐𝑐𝑠𝑠 + 𝑝𝑝𝑐𝑐

Error • Pole near the origin• Small negative zero• 𝑧𝑧𝑐𝑐 > 𝑝𝑝𝑐𝑐• Error constant improved by 𝑧𝑧𝑐𝑐/𝑝𝑝𝑐𝑐• May introduce a slow transient• Passive circuitry implementation

possible

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Compensator Summary


PD 𝐾𝐾 𝑠𝑠 + 𝑧𝑧𝑐𝑐 Transient response • Zero at −𝑧𝑧𝑐𝑐 contributes angle to satisfy angle criterion at desired closed-loop pole location

• Active circuitry required• Amplifies sensor noise

Lead𝐾𝐾

𝑠𝑠 + 𝑧𝑧𝑐𝑐𝑠𝑠 + 𝑝𝑝𝑐𝑐

Transient response • Lower-frequency zero• Higher-frequency pole• Net angle contribution satisfies

angle criterion at design point• Added pole helps reduce

amplification of higher-frequency sensor noise

• Passive circuitry implementation possible

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Compensator Summary


PID 𝐾𝐾𝑝𝑝 +𝐾𝐾𝑖𝑖𝑠𝑠

+ 𝐾𝐾𝑑𝑑𝑠𝑠Error & transient response

• PD compensation improves transient response

• PI compensation improves steady-state error

• Active circuitry• Amplifies noise

Lead-lag𝐾𝐾

𝑠𝑠 + 𝑧𝑧𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑𝑠𝑠 + 𝑝𝑝𝑙𝑙𝑒𝑒𝑎𝑎𝑑𝑑

𝑠𝑠 + 𝑧𝑧𝑙𝑙𝑎𝑎𝑙𝑙𝑠𝑠 + 𝑝𝑝𝑙𝑙𝑎𝑎𝑙𝑙

Error & transient response

• Lead compensation improves transient response

• Lag compensation improves steady-state error

• Passive circuitry implementation possible

• Amplification of high-frequency noise reduced

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