# me132 interconnections lec3

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1 ME 132 Dynamic Systems and Feedback Lecture 3 Feedback Interconnections Closed-Loop VS Open-Loop Control Static and Linear Analysis Summary The structure of control systems Basic control system (open-loop vs closed-loop) Basic block-diagram algebra – static and linear systems – Open-loop control – Closed-loop control Example: Cruise Control for a Car – Open-loop control – Closed-loop control Advantages and disadvantages of feedback 2 The Structure of Control Systems 3 Basic Control Systems Open-Loop System 4 Closed-Loop System

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ME 132 Dynamic Systems and Feedback

Lecture 3

Feedback Interconnections

Closed-Loop VS Open-Loop Control

Static and Linear Analysis

Summary•  The structure of control systems•  Basic control system (open-loop vs closed-loop)•  Basic block-diagram algebra – static and linear

systems– Open-loop control– Closed-loop control

•  Example: Cruise Control for a Car – Open-loop control– Closed-loop control

2

The Structure of Control Systems3

Basic Control Systems•  Open-Loop System

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•  Closed-Loop System

Open-loop system – static blocks5

controller plant

•  r - reference input•  d - disturbance input•  y - output•  e = r – y - error

Open-loop system – linear static blocks6

controller plant

y = Gm

m = u+ dy = G(u+ d)

u = Kr

y = G(u+ d) y = G(Kr + d)

Open-loop system – linear static blocks7

controller plant

y = G(Kr + d)

e = (1�GK)r �Gd e = r � y

Closed-loop system – static blocks8

controller plant

•  Inputs–  r - reference input–  d - disturbance input

•  Outputs–  y - plant output–  u - controller output–  e = r – y - error

Closed-loop system – linear static blocks9

controller plant

Plant: y = G(u+ d)

Controller:u = Ke

e = r � yu = K(r � y)

Closed-loop system – static blocks10

controller plant

y = G(u+ d) u = K(r � y)

y = �GKy +GKr +Gd

y =GK

1 +GKr +

G

1 +GKd

Closed-loop system – static blocks11

controller plant

y =GK

1 +GKr +

G

1 +GKd

e =1

1 +GKr � G

1 +GKd e = r � y

We want to make small

Closed-loop VS Open-loop12

e =1

1 +GKr � G

1 +GKd

e = r � y

e = (1�GK)r �Gd

Making K large, results in e being small

Difficult to select K

K ⇡ r �Gd

Gr

Feedback system’s algebra13

controller plant

•  Inputs–  r - reference input–  d - disturbance input

•  Outputs–  y - plant output–  u - controller output–  e = r – y - error

y =GK

1 +GKr +

G

1 +GKd

e =1

1 +GKr � G

1 +GKd

Closed-loop transfer functions

| {z }Gc(r!e)

| {z }Gc(d!e)

| {z }Gc(r!y)

| {z }Gc(d!y)

Feedback system’s algebra14

•  Open-loop transfer function

Ge!f

= Go

= GK

Negative feedback system’s algebra15

•  Closed-loop transfer function

Gclosed loop

=

Gforward path

1 +Gopen loop

Gc(r!y) =

Gfp(r!y)

1 +Go

=GK

1 +GK

Gc(d!y) =

Gfp(d!y)

1 +Go

=G

1 +GK

Negative feedback system’s algebra16

•  Closed-loop transfer function

Gclosed loop

=

Gforward path

1 +Gopen loop

Gc(r!e) =

Gfp(r!e)

1 +Go

=1

1 +GK

Gc(d!e) =

Gfp(d!e)

1 +Go

=�G

1 +GK

Example: Cruise Control for a Car We want to regulate the linear speed of a car, y, to a

desired value r

Static car model:

G

H

Nominal car parameters

speed

control input

disturbance input

Cruise Open-Loop Control

G

HOpen-loop control:

u = Kol

r

Assume nominal system:

y = Go

Kol

r

G = Go, H = Ho, d = 0

Cruise Closed-Loop Control

Closed-loop control:

Closed-loop response:

| {z }Gc(r!y)

| {z }Gc(d!y)

y =GK↵

1 +GKfbr +

�H

1 +GKfbd

Cruise Closed-Loop Control Objective:

y =GK↵

1 +GKfbr +

�H

1 +GKfbd

Closed-loop response :

y ⇡ r

Assume nominal system:

| {z }⇡1

G = Go, H = Ho, d = 0

Cruise Closed-Loop Control Objective:

y =GK↵

1 +GKfbr +

�H

1 +GKfbd

Closed-loop response :

y ⇡ r

Assume nominal system: G = Go, H = Ho, d = 0

| {z }⇡�0.01

Ho

1 +GoKfb= 0.01 Kfb & Ho

0.01Go

Closed-Loop VS Open-Loop - Disturbance Rejection

Kfb & Ho

0.01Go

G

H

yol

= r �Hod ycl = r � 0.01d

Ho = 5 >> 0.01Closed-loop system exhibits superior external disturbance rejection

Open-Loop – Sensitivity to Plant Variations

G

H

Assume no disturbances (d = 0).

Question: •  what is the variation of the output y around the nominal

output yo when there is a variation of the plant G around the nominal plant Go?

Open-Loop – Sensitivity to Plant Variations

G

H

S =

✓Go

yo

◆@y

@G

����G = G

o

y = y

o

=

✓Go

yo

◆@�

G

G

o

r�

@G

�����G = G

o

y = y

o

yo = r

S = 1

y =G

Go

r

1 percent deviation of the plant G from its nominal value Go will approximately result in a 1 percent deviation in the output y

Closed-Loop – Sensitivity to Plant Variations

S =

✓Go

yo

◆@y

@G

����G = G

o

y = y

o

y =GK↵

1 +GKfbr

yo =GoK↵

1 +GoKfbr

=

✓Go

yo

◆ @⇣

GKff1+GKfb

r⌘

@G

������G = G

o

y = y

o

Closed-Loop – Sensitivity to Plant Variations

y =GK↵

1 +GKfbr

yo =GoK↵

1 +GoKfbr

Ho

1 +GoKfb= 0.01

remember that we chose …

and Ho = 5

Scl =1

1 +GoKfb⇡ 0.002 << S

ol

Closed-Loop VS Open-Loop - Sensitivity

G

H

Closed-loop system is much less sensitive to plant variations

Scl =1

1 +GoKfb⇡ 0.002S

ol

= 1

Closed-Loop – Sensitivity to Noise

noise

| {z }�Gc(n!y)

Closed-Loop – Sensitivity to Noise

Gc(n!y) =GKfb

1 +GKfb

Ideally we require��Gc(n!y)

�� << 1

But this is in conflict with “high-gain” feedback,

which is necessary for disturbance rejection and robustness to plan variations. More on this later …

|GoKfb| >> 1

Summary•  The structure of control systems

– Open-loop control– Closed-loop control

•  Advantages of feedback :– Closed-loop systems exhibit superior external

disturbance rejection– Closed loop systems are significantly less

sensitive to plan parameter variations•  Disadvantages of high-gain feedback

– High sensitivity to measurement noise– May destabilize feedback system (dynamic

models)

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