dr ian r. manchester - pws.yazd.ac.ir · pdf file3 transient performance and the s-plane ... 6...

Dr Ian R. Manchester

Dr. Ian R. Manchester Amme 3500 : Introduction

Week Content Notes

1 Introduction

2 Frequency Domain Modelling

3 Transient Performance and the s-plane

4 Block Diagrams

5 Feedback System Characteristics Assign 1 Due

6 Root Locus

7 Root Locus 2 Assign 2 Due

8 Bode Plots

9 Bode Plots 2

10 State Space Modeling Assign 3 Due

11 State Space Design Techniques

12 Advanced Control Topics

13 Review Assign 4 Due

•  Review mathematical justification for

feedback over open loop control

•  Introduce the PID controller

•  Performance specification: steady-state error

•  Introduction to parameter sensitivity aka

“robustness”


Desired

Value Output +

-

Feedback

Signal

error

Controller Plant

Control

Signal

r(t)

e(t) = r(t) − y(t)Error:


(Assuming perfect measurement)

•  Perfect feed-forward controller: Gn

•  Output and error:

•  Only zero when:

(Perfect Knowledge)

(No load or disturbance)


r

d

unity feedback

controller plant

reference output

A sensor

disturbance

y =GK

1+KGr −

G

1+GKd

Reference to Output

Transfer Function

Disturbance to Output

Transfer Function


rd

Equivalent Open Loop Block Diagram


y =GK

1+KGr −

G

1+GKd

Reference to Output

Transfer Function

Disturbance to Output

Transfer Function

•  Simplest possible feedback control

•  useful for thermostats, similarly simple problems

•  Often results in oscillations

•  Rapid changes in control input are often infeasible

•  Nonlinear: can be very difficult to analyse Dr Ian R. Manchester

•  A very large number (at least 95%) of feedback controllers used in practice have the following simple form:

•  This is called a “PID”

–  Proportional (P) term feeds back on the current error

–  Integral (I) feedback can eliminate steady state error

–  derivative (D) term can improve dynamic response


u(t) = KPe +K

Ie(τ)dτ

0

t

∫ +KD

˙ e

•  Suppose we just use the “P” term, i.e. proportional control

•  We have seen how this form of feedback is able to minimize the effect of disturbances by increasing K

•  This may adversely affect other performance, in particular overshoot

K

-

+

R(s) E(s) C(s) G(s)


•  Let’s compute an example of a torsional mass-spring-damper system with proportional control

•  J = 1 N m, k = 5 N m/rad and b = 6 N m s/rad

b = 6 N m s/rad

* N.S. Nise (2004) �

Control Systems Engineering

�

Wiley & Sons

k = 5 N m s/rad


•  Recall the differential equation

•  The transfer function is

•  The closed loop response is

K

- +

R(s) E(s) C(s) G(s)

J ˙ ̇ θ + b ˙ θ + kθ = τ


K=5

K=10

K=100

K=1000


•  Proportional (P) control acts like a spring

pulling the output towards the reference.

•  A very stiff spring (large gain K) can result in

highly oscillatory response.

•  Why not add damping? This is PD control


u(t) = KPe +K

D˙ e Kp+Kds

- +

R(s) E(s) C(s) G(s)

U(s)


•  D term adds damping and differentiates the reference


P control:

K=100

PD control:

K=100+20s

•  Note the anticipatory “kick” at the beginning,

and the damping of oscillations

Same torsional setup as

before

•  We have looked at transient response for second order systems

•  This can be defined by

–  Rise time, Tr

–  Settling time, Ts

–  Peak time, Tp

–  Percentage overshoot.

•  These can all be specified by PD control

cfinal=steady state response

* N.S. Nise (2004)

�

Control Systems Engineering

�

Wiley & Sons


•  Why use integral control? (The “I” part

in PID)

•  Another important characteristic of

systems in general is their steady state

performance

•  Steady-state error is the difference

between the reference input and output

for a particular input as t→∞ time

output

input

Steady state

error


•  Intuitively: if the controller continues to see an

error signal over a long time, then the integral

term “builds up”, and increases control signal

•  Slowly it builds up to the level required to

completely kill the error


u(t) = KPe +K

Ie(τ)dτ

0

t

∫ +KD

˙ e

•  For a PD controller, if the error and its

derivative are both zero, then

•  So, a zero-steady-state-error response to a step

must result in zero control input.

•  This is not possible if there is some force (e.g.

a spring, or gravity, or wind) pulling the

system away from the reference Dr Ian R. Manchester

u = KPe +K

D˙ e = 0

dr ian r. manchester - pws.yazd.ac.ir · pdf file3 transient performance and the s-plane ... 6...

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