PID Controller (Tutorial)SUBJECT: CONTROL SYSTEMS

INSTRUCTOR: DR MOHSIN JAMIL

Example#1(2nd Order Mass Damper System)

Given is the 2nd order mass damper System. Design a PD Controller for the given system such that

Settling time

Solution

Approach to Solve the problem

Get the Mathematical Model of System(Modelling)

Open Loop Response of the System.

Transfer Function of Proposed Controller.

Closed-Loop Transfer Function.

Comparison With Standard 2nd Order System to get Controller Gains.

Closed-Loop Response of the System.

Mathematical Modelling of Mass Damper System

Using Newtons 2nd Law of Motion F ma

input force f t

spring force kx t

damping force bx t

2

2

( )

1

mx t f t kx t bx t

Taking LaplaceTransform

ms X s F s kX s bsX s

X s

F s ms bs k

2

1

10 1

X sTF

F s s s

Open Loop Response of SystemStep Response

Time (sec)

Am

plit

ude

0 20 40 60 80 100 1200

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

System: sys

Final Value: 1

System: sys

Time (sec): 72.6

Amplitude: 1.02

System: sys

Peak amplitude: 1.6

Overshoot (%): 60.4

At time (sec): 10.2

System: sys

Rise Time (sec): 3.69

Closed-Loop Transfer Function

2

2

1

10 1

1 10 1 1

c p d

p dc

c d p

ControllerTF K s k k s

PlantTF G ss s

k k sK s G sClosed LoopTF H s

K s G s s k s k

Gain Calculation

Compare with Standard 2nd Order System we will get

22 2

2

0.1

12 1

10 10

p dn

pn n d

k k s

ks s ks s

2

12

10

1

10

dn

p

n

k

k

8

4.6s

n

t

4.611.5

0.8*0.5n

1321

183

p

d

k

k

21325 183

10 184 1322

sH s

s s

Closed-Loop ResponseStep Response

Time (sec)

Am

plit

ude

0 0.1 0.2 0.3 0.4 0.5 0.60

0.2

0.4

0.6

0.8

1

1.2

1.4

System: sys

Peak amplitude: 1.18

Overshoot (%): 17.8

At time (sec): 0.186

System: sys

Rise Time (sec): 0.0707

System: sys

Settling Time (sec): 0.44

System: sys

Final Value: 0.999

Analysis

Resulting Overshoot is 17% and the Settling time is 0.4 sec. so fromresult it can be concluded that PD controller deals very well with thesettling time but a bit problem with overshoot but still it reduced fromthe 60 % to 17 %

Common MATLAB Commands

Defining Transfer Function >>num=[1 0] %Nominator of Transfer Function with increasing power of s

form right to left

>>dnum=[10 2 1] %Denominator of Transfer Function

>>sys=tf(num,dnum) %Command to define transfer function

>>sys1=feedback(sys,1) %System with unity feedback closed-loop transfer function

System Analysis >>step(sys) %Gives the response of system for step input

>>impulse(sys) %Gives the response of system for impulse input

>>pzmap(sys) %Gives Pole-Zero Map of system

>>rlocus(sys) %Gives Root locus Plot for system

>>bode(sys) %Gives Bode Plots of system

>>nyquist(sys) %Gives Nyquist plot of system

Practice Exercise

Change the Value of Gains and see the effect on the output.

See the effect of using proportional, derivative and integral controllersindividually and with different combinations.

See the effect of P, I and D controller effect on the different perimeterslike settling time, overshoot, rise time and steady state errors.

Get Familiar with MATLAB and SIMULINK to perform simulations.

Practice Problem(DC Motor Speed Control)

Design a PD Controller for the Given DC Motor Model such that Settling time is less than 2 sec and overshoot less than 5 %.

0.01

0.1

0.01

1

0.5

J

b

K

R

L