eee3420 lecture01 rev2011

1Week © Vocational Training Council, Hong Kong.

│ Lecture 1 │

Introduction to Control Systems

EEC3420 Industrial ControlDepartment of Electrical Engineering

2Week© Vocational Training Council, Hong Kong.

EEE3420 Industrial Control

Learning objectives

Understand the basic concept in control systems.

Know what is a Transfer Function.

Appreciate the PID control process.

Know what criteria leading to a stable control system.



Open-loop control system

An open-loop control system is one in which the control signal of the process is independent of the process output

control accuracy is determined by the calibration of the plant



Advantages & disadvantage of open-loop control system

Advantages

simple and inexpensive

no stability problem

Disadvantages

cannot compensate for any disturbances that add to the controller’s driving signal



Closed-loop control system

A closed-loop control system depends on the output of the process to adjust the signal controlling the closed loop

process output is compared to the user command and an output from the plant



Advantages & disadvantages of closed-loop control system

Advantages

less sensitive to noise, disturbances and changes in the environment

transient response and steady-state error can be controlled more conveniently and with greater flexibility

Disadvantages

relatively expansive

may be unstable if not properly designed



Architecture of a closed-loop control system

Controlled Variable (CV).

Set point.

Error = set point – current value of CV.

Manipulated Variable.

Feedback Loop.



Feedback control real-time scheduling

Choices for control variables, manipulated variables, set points

Choice of appropriate control functions

Stability problem of feedback control in the context of real-time scheduling?

How to tune control parameters?

How significant is the overhead and how to minimize it?

How to integrate a runtime analysis of time constraints with scheduling algorithms?



Using negative feedback control system

typically more stable

less sensitive to variation in component values

more immune to noise



Transfer function of a control system

The transfer function of a control system is defined as the ratio of the output to the input

predict how the system will perform if the transfer function is known

output depends on both the present input and the past history of the input, so the output c(t) is a convolution product of the input r(t) and the system g(t)




c(t) = r(t) * g(t)

∫∞

⋅−=0

)()( τττ dgtr

with the use of Laplace transform, we get

∫ ∫∫∞ −∞∞ − ⋅

⋅−=⋅==

0 00)()()()]([)( dtedgtrdtetctcLsC stst τττ




Let t’ = t – τ then t = t’ + τ

the convolution product of r(t)*g(t) is now transformed into an algebraic product of R(s).G(s)

)()()(')'(

')()'()()()(

0 0

'

0

)'(

00 0

sGsRdegdtetr

dtedgtrdtedgtrsC

sst

tsst

⋅=

⋅⋅⋅=

⋅

⋅=⋅

⋅−=

∫ ∫

∫ ∫∫ ∫∞ ∞ −−

∞ +−∞∞ −∞

ττ

τττττ

τ

τ




time-domain output c(t) may be obtained by the inverse Laplace operation

open-loop control system can be represented in frequency domain as shown above

the output C(s) is given by G(s) x R(s).

∫∞+

∞−

− ⋅==jk

jk

st dsesCj

sCLtc )(2

1)]([)( 1

π



PID Control System PID – Controller is the most widely used control strategy

in industry used for various control problems such as automated

systems or plants consists of three different elements

P Proportional control I Integral control D Derivative control



PID Control System

for control loop to work properly, the PID loop must be properly tuned



The PID transfer function

)(teKP P ×=

∫ ⋅×= dtteKI I )(

dt

tedKD D

))((×=

P Proportional control,

D Derivative control,

I Integral control,




The total controller output,

We get,

Use the Laplace transform,

∫ ×+⋅×+=dt

tedKdtteKteKtu DIP

))(()()()(

∫ ∫ →→→→s

xdtx

sdtsx

dt

dxs

dt

d;1

;;

)1()()(

)(sK

s

KKsG

sE

sUD

IPC ⋅++==




Re-arrange to get, sT

sTsTTKsG

I

IDIPC ⋅

+⋅+⋅⋅= )1()(

2

Where

KP is the proportional gain

TI is the integral time constant

TD is the derivative time constant




the three different adjustments (KP, TI, TD) interact with each other

it can be very difficult and time consuming to tune these three values in order to get the best performance according to the design specifications of the system.



A Thermal Control System

• electrical heater of heat capacity Ch & thermal resistance Rho

• oven of heat capacity Co & thermal resistance Ro

• environment temperature Te, set-point temperature Ts

• temperature controller adjusts the power W by comparing To with Ts



On-Off Control of the Thermal System• the simplest form of control

• when the oven is cooler than the set-point temperature the heater is turned on at maximum power M

• once the oven is hotter than the set-point temperature the heater is switched off completely

Red line: set-point temperature

Green line: actual temperature

Blue line: Delivered Power



Proportional Control of the Thermal System

• proportional controller attempts to perform better than the On-Off type by applying power W to the heater in proportion to the difference in temperature between the oven and the set-point

• KP is known as the proportional gain of the controller

( )OSP TTKW −×=



PD Control of the Thermal System

• add D-Control (proportional to the time-derivative of the error signal) to mitigate the stability and overshoot problems that arise from a high gain proportional controller

• adjust KD (the damping constant) to achieve a critically damped response

( ) ( )dt

TTdKTTKW OSDOSP

−×+−×=



PID Control of the Thermal System• add I-Cotnrol (proportional to

the time-integral of the error signal) to change the heater power until the time-averaged value of the temperature error is zero

• KI is the integral gain parameter

( ) ( ) ( )dt

TTdKdtTTKTTKW OSDDSIOSP

−×+−×+−×= ∫



Analysis of a Control System using Transfer Function

given that,6116

10)(:

23 +++=

ssssGprocessThe P

and the feedback path: H(s) = 1




for P-control only, if KP = 3, then GC(s) = KP =3, and

36116

30

)()(1

)()(23 +++

=⋅+

⋅=ssssGsG

sGsG

PC

PC

i

o

θθ




for PI-control, if KP = 2.7 and TI = 1.5, then

and the transfer function is:

275.495.1695.1

275.40

)()(1

)()(234 ++++

+=⋅+

⋅=ssss

s

sGsG

sGsG

PC

PC

i

o

θθ

s

s

sT

sTK

sTKsG

I

IP

IPC 5.1

)15.1(7.2)1(11)(

+=⋅

+⋅=

⋅

+=




for PID-control, if KP = 2,

TI = 0.9 and TD = 0.6, then

and the transfer function is:

204.237.204.59.0

20188.10

)()(1

)()(234

2

++++++=

⋅+⋅=

ssss

ss

sGsG

sGsG

PC

PC

i

o

θθ

s

ss

sT

sTsTTKsG

I

IDIPC 9.0

)19.054.0(2)1()(

22 ++=⋅

+⋅+⋅⋅=




The transfer functions give the step responses as shown on the right

• for P-control (the red curve) – a steady state error occurs

• for PI-control (the blue curve) – the response becomes more oscillatory and needs longer to settle, the error disappears

• for PID-control (the green curve) – the overshoot and the number of oscillatory cycles are much reduced



Concluding remarks of the PID Effect

• In general, we may observe that

• P term is used to adjust the speed of response.

• I term provides zero error.

• D term introduces damping.



3. Stability of Control System• a control system responds to an input by undergoing a

transient response before reaching a steady-state

• the total response of a system consists of two parts, namely, the natural response and the forced response

• natural response describes the way the system dissipates or acquires energy, the nature of this response is dependent only on the system

• the nature of the forced response is dependent on the input



3. Stability of Control System• for a linear system, we can write

Total response = Natural response + Forced response

• for a control system to be useful, the natural response must eventually approach zero, thus leaving only the forced response

• a stable control system will always return to a stable operating state

• in an unstable system, any disturbance will result in oscillations building up until some parts fails



3. Stability of Control System

• Oscillatory System

• between the stable state and the unstable state lies the conditionally stable system in which oscillations neither increase nor decrease

• each cycle being identical to the previous one and results in sustainable oscillation.



Stability of First Order SystemConsider the response of a control

system with transfer function (s+2)/(s+5) under a step input with R(s) = 1/s

• a pole on the real axis generates an exponential response of the form e-αt, where –α is the pole location on the real axis.

• if α is positive, the transient response will decay to zero

• if α is negative, then the transient response will grow and the system will be unstable.



Stability of Second Order System

As long as the poles of the output function lies on the left hand side of the complex plane

• the system output will not grow without bound

• it will be stable



Stability of Second Order System



Stability of Higher Order System



Stability of Higher Order System - Routh table

The Routh table method will yield the stability information of a system without the need to solve for the system poles.

The method requires two steps:

(1) generate a Routh table

For a fourth order system given on the right,

Construct the table for the denominator by using the formulae shown on the right

Similar formulae are used for system with order higher than 4.

s4 a4 a2 a0

s3 a3 a1 0

s2 b1=(a2*a3-a1*a4)/a3 b2=(a0*a3- 0*a4)/a3 0

s1 c1=(a1*b1-a3*b2)/b1 0 0

s0 d1=(b2*c1-0*b1)/c1 0 0




(2) interpret the Routh table The number of poles that are

in the right half plane is equal to the number of sign change in the first column of the Routh table.

Note: For the special cases such as the element in the first column is equal to zero or the elements in the entire row are equal to zero will not be treated here, please refer to descriptions in books dealing with control theory.

s4 a4 a2 a0

s3 a3 a1 0

s2 b1=(a2*a3-a1*a4)/a3 b2=(a0*a3- 0*a4)/a3 0

s1 c1=(a1*b1-a3*b2)/b1 0 0

s0 d1=(b2*c1-0*b1)/c1 0 0




Example

Determine the stability of the system on the right by Routh table method.

Solution

As there are two sign changes in the first column of the Routh table, so there are two poles lying in the right half plane and hence the system is not stable.

s3 a3 = 1 a1 = 31 0

s2 a2 = 10 a0 = 1030 0

s1 b1=(31*10-1*1030)/10

= -72

0 0

s0 c1 = (1030*(-72)-0*10)/(-72)

= 1030

0 0



Summary of Introduction to Control System

• Closed-loop control system is less sensitive to noise, disturbances and changes in the environment.

• The transfer function of a control system is the ratio of the output to the input.

• Proportional control is used to adjust the speed of response.

• Integral control provides zero error.

• Differential control introduces damping.

• If the poles of the output function lies on the left hand side of the complex plane, the system will be stable.



Introduction to Control System

End of Lecture 1

RevisionNorman S. Nise, Control Systems Engineering, Fourth Edition, Johne Wiley & Sons, Inc., page 177 to page 183.