Modern Control - Lec 03 - Feedback Control Systems Performance and Characteristics

LECTURE (3)

Feedback Control Systems

Performance and Characteristics

Assist. Prof. Amr E. Mohamed

Agenda

Introduction

Test Input Signals.

Response of First Order systems.

Response of Second Order Systems.

Higher Order Systems Response.

Steady State Errors of Feedback Control Systems.

Stability Analysis Using Routh-Hurwitz Method

Introduction to PID

2

Introduction

Order of the system:

Consider a system defined by the transfer function:

The order of this system is n which is defined by the highest power for s in

the denominator.

Examples:

3

)()(

)()(

0

1

1

0

1

1

asasa

bsbsb

sR

sCsT

n

n

n

n

m

m

m

m

1st order system 2nd order system

14

5

)(

)(

ssR

sC

44

10

)(

)(2

ss

s

sR

sC

3423

10

)(

)(234

2

ssss

s

sR

sC

4th order system

Introduction

The system type Number:

It is defined as the number of poles at the origin of the open loop transfer function

G(s)H(s).

Consider the open loop transfer function of a system as :

The system of type c and has an order of n+c

Examples:

4

)()()(

0

1

1

0

1

1

asasas

bsbsbsHsG

n

n

n

n

c

m

m

m

m

)4)(1(

50)()(

sssHsG

)3423(

310)()(

2342

2

sssss

ssHsG

System of type 0

System of type 2

Standard Test Signals

Impulse-Function

The impulse signal imitate the sudden shock

characteristic of actual input signal.

Step-function

The step signal imitate the sudden change


5

0 0

0 )()(

t

tAttu

0 t

δ(t)

A

00

0

t

tAtu

)(

0 t

u(t)

A

s

AsU )(

AsU )(

Standard Test Signals

Ramp-function

The ramp signal imitate the constant velocity characteristic of

actual input signal.

Parabolic-function

The parabolic signal imitate the constant acceleration


6

00

0

t

tAttr

)( 0 t

r(t)

00

02

2

t

tAt

tp

)( 0 t

p(t)

3)(

s

AsU

2)(

s

AsU

Relation Between Standard Test Signals

Impulse

Step

Ramp

Parabolic

7

00

0

t

tAt

)(

00

0

t

tAtu

)(

00

0

t

tAttr

)(

00

02

2

t

tAt

tp

)(

dt

d

dt

d

dt

d

Time Response of Control Systems

Time response of a dynamic system is response to an input expressed as

a function of time.

The time response of any system has two components:

Transient response

Steady-state response.

8

System

)()()( tctctc sstr


When the response of the system is changed form rest or equilibrium it takes some

time to settle down.

Transient response is the response of a system from rest or equilibrium to steady state.

The response of the system after the transient response is called steady state

response.

9

0 2 4 6 8 10 12 14 16 18 200

1

2

3

4

5

6x 10

-3

Step Response

Time (sec)

Am

plitu

de Response

Step Input


Transient response dependents upon the system poles only and not on

the type of input.

It is therefore sufficient to analyze the transient response using a step

input.

The steady-state response depends on system dynamics, system type,

and the input quantity.

It is then examined using different test signals by final value theorem.

10

Response of First Order System

The Standard form system transfer function G(s) is given by:

Where K is the D.C gain and T is the time constant of the system.

Time constant is a measure of how quickly a 1st order system responds to a

unit step input.

D.C Gain of the system is ratio between the input signal and the steady

state value of output.

The first order system has only one pole at 1/T

11

1)(

)()(

sT

k

sR

sCsG

Unit-Impulse Response of First Order System

Consider the following 1st order system

12

)(sR

0t

δ(t)

1

1)( sR

Ts

TK

Ts

KsRsGsC

11)()()(

TteT

Ktc /)(

)(sC

1Ts

K

K=1 & T=2s

Unit-Step Response of First Order System

Taking Inverse Laplace of above equation

13

1Ts

K )(sC)(sR

ssUsR

1 )()(

1)()()(

Tss

KsRsGsC

0 t

u(t)

1

TteKtc /1)(

1

1)(

Ts

T

sKsC

K=1 & T= 1.5s

Step Response of 1st order System

System takes five time constants to reach its final value.

14

Unit-Ramp Response of First Order System

The ramp response is given as

15

1Ts

K)(sC)(sR

2

1)(

ssR

1)(

2

Tss

KsC

TtTeTtKtc /)(

0 5 10 150

2

4

6

8

10

Time

c(t

)

Unit Ramp Response

Unit Ramp

Ramp Response

K=1 & T= 1s

error

Table 3.1 Summary of response of a LTI first order system

16

/1)( tetc

/1)( tetc

/1)( tettc

input: output:

unite ramp

unite step

unite impulse )()( 0 tuttr

)()( 1 tututr

)(2 tutr

dt

d

dt

d

dt

d

dt

d

Second Order Systems

A general second-order system (without zeros) is characterized by the

following transfer function.

1722

2

2 nn

n

sssR

sC

)(

)(

)2()(

2

n

n

sssG

Open-Loop Transfer Function

Closed-Loop Transfer Function

Second Order Systems

damping ratio of the second order system, which is a measure of

the degree of resistance to change in the system output.

un-damped natural frequency of the second order system, which

is the frequency of oscillation of the system without damping

18

22

2

2)(

)(

nn

n

sssR

sC

n

Example#2

Determine the un-damped natural frequency and damping ratio of the

following second order system.

Compare the numerator and denominator of the given transfer function

with the general 2nd order transfer function.

19

42

42

sssR

sC

)(

)(

22

2

2)(

)(

nn

n

sssR

sC

42 n sec/radn 2 ssn 22

422 222 ssss nn 50.

1 n

Example#3

Example 3: For the second order system described by the closed loop

transfer function T(s), n and ξ .

Compare with the standard equation we have:

20

643

6

256124

24

)(

)()(

22

sssssR

sCsT

632642 nnn kandandSince

75.08.016

3,8 kandn

Second Order Systems - Poles

The second order system Transfer function is

The characteristic polynomial of a second order system is:

The closed-loop poles of the system are

21

22

2

2)(

)(

nn

n

sssR

sC

1

1

2

2

2

1

nn

nn

s

s

0))((2 21

22 ssssss nn

12

442,

2222

11

nn

nnnss

Response of Second Order Systems

According the value of , a second-order system can be set into one

of the four categories:

Case 1: Over damped response (ξ > 1)

Case 2: Critically damped response (ξ = 1)

Case 3: Under damped response (0 < ξ <1)

Case 4: No damped response (ξ = 0)

22

Unit-Step Response of Second Order Systems

Case 1: Over damped response (ξ > 1)

The two roots of the characteristic equation s1 and s2 are real and distinct.

Example#4: Calculate and plot the output of the system with the following

transfer function:

Solution: With unit step input, its response is:

23

δ

jω

23

2)(

2

sssT

2

1

1

21

)23(

1)()()(

2

sssssssTsRsC

1

1

2

2

2

1

nn

nn

s

s


The corresponding time domain output is given by:

24

)(]21[)}({)( 21 tueesCLtc tt

time [sec]

outp

ut

signal

1


Case 2: Critically damped response (ξ = 1)

The two roots of the characteristic equation s1 and s2 are real and equal.


transfer function:


25

δ

jω

44

45)(

2

ss

ssT

22 )2(

3

2

11

)44(

45)()()(

ssssss

ssTsRsC

ns 2,1



26

)(]31[)}({)( 221 tuetesCLtc tt

time [sec]

outp

ut

signal

1


Case 3: Under damped response (ξ < 1)

The two roots of the characteristic equation s1 and s2 are complex conjugates of on another.


transfer function:


27

δ

jω2

2,1 1 nn js

dn js 2,1

42

4)(

2

sssT

42

21

)42(

4)()()(

22

ss

s

sssssTsRsC

Unit-Step Response of Second Order Systems The corresponding time domain output is given by:

28time [sec]

outp

ut

signal

1

)()sin(

1

11)}({)(

2

1 tutsCLtc d

t

en

𝑤ℎ𝑒𝑟𝑒 𝜃 =

1 − 𝜉2

𝜉


Case 4: Undamped response (ξ = 0)

The two roots of the characteristic equation s1 and s2 are imaginary poles.


transfer function:


29

δ

jω

njs 2,1

4

4)(

2

ssT

222 2

1

)4(

4)()()(

s

s

ssssTsRsC

)()sin(1)}({)( 1 tutsCLtc n

The transient response as a function of the damping ratio ξ

30time [s]

outp

ut

signal

01.0

4.0

2.0

5.0

3.07.0

6.0

8.0

2

Time-Domain Specification

For 0< ξ <1 and ωn > 0, the 2nd order system’s response due to a unit

step input looks like

31

Time-Domain Specification – Delay Time

Delay-Time (Td): The delay (Td) time is the time required for the

response to reach half the final value the very first time.

32

Time-Domain Specification – Rise Time

Rise-Time (TR): The rise time is the time required for the response to

rise from

10% to 90% of its final value, over damped systems

5% to 95% of its final value, Critical damped systems

or 0% to 100% of its final value. under damped systems

33

d

rt

n

n

2

1 1tan

Time-Domain Specification – Peak Time

Peak Time (Tp): The peak time is the time required for the response to

reach the first (maximum) peak of the overshoot.

34

d

pt

Time-Domain Specification – Maximum Overshoot

Maximum Overshoot (MP): is the maximum peak value of the response

curve measured from unity.

Maximum percent overshoot (P.O): is defined as follows:

35

eMP

)

1

(2

%100.21xeOP

%100. xvaluefinal

OPMp

Time-Domain Specification – Rise Time

The settling time (Ts): is the time required for the response curve to

reach and stay within a range about the final value of size specified by

absolute percentage of the final value (usually 2% or 5%).

36

nst

4

Settling Time (2%)

nst

3

Settling Time (5%)

Example#7

For the control system shown in Figure, determine k and a that satisfies

the following requirements:

a) Maximum percentage overshoot P.O =10%.

b) The 5% settling time ts = 1 sec.

Solution: The closed loop transfer function is given by

37

)(sR

-+

)(sC

2

1

sas

k

)(sT

)2()2()2)((

2

11

2

1

)(

)(2 kaass

k

ksas

k

sas

k

sas

k

sR

sC

Example#7

The maximum percent overshoot (P.O )is given by:

For 5%, the settling time ts is given by:

From these two equations we get a + 2 = 6 then a = 4 and k = 17

38

6.0

100

10.

21

eOP

133

ne

st 5n

305.02&22 2 nn kaa

Higher Order Systems Response

The natural response of higher-order systems consists of a sum of

terms, one term for each characteristic root:

For each distinct real characteristic root, there is a real exponential term in

the system natural response.

For each pair of complex conjugate roots, there is a an exponential

sinusoidal term in the system natural response.

Repeated roots give additional terms involving power of time times the

exponential.

39


Example#8: Analyze the system with the following transfer function:

Solution: With unit step input, apply the partial fraction, its response is

given by:


40

)134)(4)(1(

58

5281379

58)(

2

2

234

2

ssss

s

ssss

ssT

22

54321

)3()2(41)(

1)(

s

ksk

s

k

s

k

s

ksT

ssC

)3cos()(24

321

tkkktc eAeettt


Example#9: Analyze the system with the following T.F

Solution: With unit step input, apply the partial fraction, its response is

given by:


Where the natural frequencies and damping ratios are given by:

41

)178)(134)(4)(1(

23967)(

22

23

ssssss

ssssT

17813431)(

1)(

2

76

2

54321

ss

ksk

ss

ksk

s

k

s

k

s

ksT

ssC

)cos(2

2

2)cos(2

1

1)( 2211

3

321

1

1

1

12

11

1

ttkkktc d

t

d

t

tt eAeAee

nn

97.02

8sec,/12.41755.0

2

4sec,/6.313

22

2

11

1 n

n

n

n radandrad

The s-Plane Root Location and The Transient Response

42

Example#10

Consider the system shown in following figure, where damping ratio is

0.6 and natural undamped frequency is 5 rad/sec. Obtain the rise time tr,

peak time tp, maximum overshoot Mp, and settling time 2% and 5%

criterion ts when the system is subjected to a unit-step input.

Solution:

ξ = 0.6 and ωn = 5 rad/sec

43

Example#10

Rise Time:

Peak Time:

44

drt

21

1413

n

rt.

rad 9301 2

1 .)(tan

n

n

str 550

6015

9301413

2.

.

..

dpt

st p 7850

4

1413.

.

Example#10 Settling Time (2%):

Settling Time (5%):

Maximum Overshoot:

45

nst

4 sts 331

560

4.

.

nst

3 sts 1

560

3

.

095.022

6.01

6.0141.3

1

eeM p

%5.9100.21

eOP

Example#11

For the system shown in Figure, determine the values of gain K and

velocity-feedback constant Kh so that the maximum overshoot in the

unit-step response is 0.2 and the peak time is 1 sec. With these values of

K and Kh, obtain the rise time and settling time. Assume that J=1 kg-m2

and B=1 N-m/rad/sec.

46

Example#11

47

Example#11

Comparing above T.F with general 2nd order T.F

48

Nm/rad/sec and Since 11 2 BkgmJ

KsKKs

K

sR

sC

h

)()(

)(

12

22

2

2 nn

n

sssR

sC

)(

)(

Kn K

KKh

2

1 )(

Example#11

49

Maximum overshoot is 0.2.

Kn K

KKh

2

)1(

2021

.ln)ln(

e

The peak time is 1 sec

dpt

245601

1413

.

.

n

21

14131

n

.

533.n

Example#11

50

Kn K

KKh

2

1 )(

96.3n

K533.

512

533 2

.

.

K

K

).(.. hK512151224560

1780.hK

Example#11

51

963.n

nst

4

nst

3

21

n

rt

str 65.0 sts 48.2sts 86.1

Example#12

When the system shown in Figure(a) is subjected to a unit-step input,

the system output responds as shown in Figure(b). Determine the values

of a and c from the response curve.

52

)( 1css

a

Example#13

For the RLC shown in Fig. 3.17, determine the natural frequency n and

the damping ratio ξ for the transfer function T(s)= X(s) / F(s), if M = 1

kg, b= 4 Ns/m and k = 10 N/m.

Solution:

The D.E for this system is given by:

The closed loop transfer function is given by

Then

53

)()()()(

2

2

tftKxdt

tdxf

dt

txdM v

m

kS

m

fs

m

ksfsMsF

sX

vv

22

/11

)(

)(

10

22sec/10

/

1

m

fandrad

mk

vnn

K

vf

M)(tf

)(tx

Fig. 3.15 RLC network

Example#14

For the RLC shown in Fig. 3.16, determine the natural frequency n and

the damping ratio ξ for the transfer function T(s)= I(s) / E(s).

Solution:

The closed loop transfer function is given by

Then

54

11

1

)(

)(2

RCSLC

Cs

CSLsR

sE

sI

s

L

CR

L

Rand

LCnn

22

1

Fig. 3.15 RLC network

LCs

L

R

sL

sE

sI

s1

1

)(

)(

2

Example#15

Figure (a) shows a mechanical vibratory system. When 2 lb of force

(step input) is applied to the system, the mass oscillates, as shown in

Figure (b). Determine m, b, and k of the system from this response

curve.

55

Example#16

Given the system shown in following figure, find J and D to yield 20%

overshoot and a settling time of 2 seconds for a step input of torque T(t).

56

Example#16

57

Example#16

58

Steady-State Error

59

Steady-state error

Steady-state error

Consider the feedback control system shown in Fig. 3.16.

The closed loop transfer function is given by:

The system error is equal to:

By application of final value theorem the steady state error is:

Fig. 3.16

)(sR )(sC)(sE)(sG

)(sH

)()(1

)(

)(

)()(

sHsG

sG

sR

sCsT

)()()()()(

)()()()()()()()( sR

sHsGsHsG

sHsGsRsRsHsCsRsE

1

1

1

)()(1

)(lim)(lim)(lim

00 sHsG

ssRssEtee

sstss

Static Error Constants

Static Position Error Constant Kp [Step-error]

Static Velocity Error Constant Kv [Ramp-error]

Static Acceleration Error Constant Ka [Parabolic-error]

)()(lim0

sHsGKs

p

p

ssK

e

1

1

)()(lim0

sHssGKs

v

v

ssK

e1

)()(lim 2

0sHsGsK

sa

a

ssK

e1

Summary of steady-state errors

Input

Type #

step input ramp input acc. input

type 0

system

type 1

system

type 2

system

2)(

2ttr 1)( tr ttr )(

pK1

1

vK

1

aK

1

0

0 0

Example

Find the steady state error of the system shown in Fig. 3.17, when the

reference input r(t) is:

a) δ(t) b) u(t) c) t u(t)

Solution:

The steady state error is given by

)(sR )(sC)(sE

4

5

s

1

2

s

Fig. 3.17

)()()(1

1lim)(lim

00sR

sHsGssE

sssse

Example (Cont.)

(a) For

(b) For

(c) For

01.

1

2

4

51

1lim

0

ss

sssse

)()( ttr

14

41.

10)1)(4(

)1)(4(lim

0

sss

sss

ssse

)()( tutr

)()( tuttr

20

1.

10)1)(4(

)1)(4(lim

sss

sss

ssse

Example

Find the steady state error of the system shown in Fig. 3.18, when the

reference input r(t) is:

a) δ(t) b) u(t) c) t u(t)

Solution:

The steady state error is given by

)()()(1

1lim)(lim

00sR

sHsGssE

sssse

)(sR )(sC)(sE

)4(

10

ss

Fig. 3.18

Example (Cont.)

(a) For

(b) For

(c) For

)()( ttr

)()( tutr

)()( tuttr

01.10)4(

)4(lim

0

ss

sss

ssse

01

.10)4(

)4(lim

0

sss

sss

ssse

10

41.

10)4(

)4(lim

20

sss

sss

ssse

Example

For the system shown in Fig. 3.19, determine:

a) The system type number

b) Static position error constant Kp

c) Static velocity error constant Kv

d) ess, when r(t) is: u(t) and t u(t)

Solution:

Fig. 3.19

)(sR )(sC)(sE)5(

10

ss

3

2

s

Example (Cont.)

a) The open loop transfer function is given by

b) static position error constant Kp

c) Static velocity error constant Kv

d) ess, when r(t) is: u(t) and t u(t)

))3)(5(

20)()(

ssssHsG Then the system is type number =1

)3)(5(

20lim)()(lim

00 sssssHsGKp

ss

15

20

)5)(5(

20lim)()(lim

00

sss

ssHssGKv

ss

20

151)(0

1

1)(

ke

ke

v

ss

p

ssrampstep

Routh-Hurwitz criterion

70

The Concept of Stability

A stable system is a dynamic system with a bounded response to a bounded

input.

Consider the closed loop transfer function of a system as :

The characteristic equation or polynomial of the system which is given by:

For the system described by T(s) to be stable, the root of the characteristic equation

must lie in the left half plane.

The Routh-Hurwitz criteria or test is a numerical procedure for determining the

number of right half-plane (RHP) and imaginary axis roots of the characteristic

polynomial.

)(

)()(

0

1

1

0

1

1

s

sN

asasa

bsbsbsT

n

n

n

n

m

m

m

m

0

1

1)()( asasasQs n

n

n

n

The Routh-Hurwitz Method Stability Criteria

The method requires two steps:

Step #1: Generate a date table called a Routh table as follows:

Consider the characteristic equation which is given by:

01

2

2

3

3

4

4)()( asasasasasQs

Table 3.3 Initial layout for Routh table

Rules for Routh table creation

Any row of the Routh table can be

multiplied by a positive constant without

changing the values of the rows below.

To avoid the division by zero, an epsilon is

assigned to replace the zero in the first

column.


Further rows of the schedule are then completed as follows:

Table 3.4 Completed Routh table


Step #2: Interpret the Routh table to tell how many closed-loop system poles are inthe:

left half-plane

right half-plane.

The number of roots of the polynomial that are in the right half-plane is equal to thenumber of sign changes in the first column.

Notes: 1- If the coefficients of the characteristic equation have differing algebraic, there is at

least one RHP root. For Example:

• Has definitely one or more RHP roots.

2- If one or more of the coefficients of the characteristic equation have zero value, there areimaginary or RHP roots or both. For Example:

• has imaginary axis roots indicated by missing s3 term.

102357)()( 2345 ssssssQs

173823)()( 2456 ssssssQs

Example

For the system shown in Fig. 3.20, determine T(s) and then apply the

Routh-Hurwitz Method to check the its Stability .

Solution: :

Step #1: Find the system closed loop transfer function T(s)

)(sR )(sC)(sE

)5)(3)(2(

1000

sss

Fig. 3.20

10303110

1000

)5)(3)(2(

10001

)5)(3)(2(

1000

)()(1

)()(

23

ssssss

sss

sHsG

sGsT

Example (Cont.) Therefore, the characteristic equation is given by:

Step #2: Generate the Routh table as follows:

Step #3: Since. There are two sign changes in the left column.

• therefore, the system is unstable and has two roots in the right hand side.

10303110)( 23 ssss

Divide by 10

Example

Apply the Routh-Hurwitz Method to determine the stability of a the closed

loop system whose transfer function is given by:

Solution: Generate the Routh table as follows:

123653

10)(

2345

ssssssT

There are two sign changes in the left

column, therefore, the system has two

RHP roots and hence it is unstable

Example

Apply the Routh-Hurwitz Method to determine the values of K that make the system

stable.

Solution: Generate the Routh table as follows:

03)60(825413 2345 KsKssss

1

13

5s

4s

3s

2s

1s

0s

7.47

K212.06.65

K212.06.65

K163.0K10539402

K3

54

82

K60

K3

K769.060

K3

0

0

0

0

0 0

0K212.06.65 309K

0K163.0K10539402

35K 0K

Then the values of k that makes the system stable are 350 K

Example

Apply the Routh-Hurwitz Method to check the stability of a system whose

characteristic equation is given by:

Step #1: Generate the Routh table as follows:

35632)( 2345 ssssss

Note:To avoid the division by

zero, an epsilon is

assigned to replace the

zero in the first column.

Table 3.6The Routh table

Example (Cont.)

Step #2: Interpret the Routh table to tell how many closed-loop system poles are in

the right half-plane. To begin the interpretation, we must assume a sign, positive or

negative, for the quantity ε as illustrated in Routh table.

Table 3.6 The Routh table

Step #3: Interpret the Routh table to tell how many closed-loop system poles are in

the right half-plane There are two sign changes in the left column, therefore, Q(s)

has two RHP roots and hence it is unstable.

Introduction to PID Control

81

Introduction This introduction will show you the characteristics of the each of proportional

(P), the integral (I), and the derivative (D) controls, and how to use them to

obtain a desired response.

In this tutorial, we will consider the following unity feedback system:

Plant: A system to be controlled

Controller: Provides the excitation for the plant; Designed to control the

overall system behavior

The PID controller

The transfer function of the PID controller looks like the following:

𝑮𝒄 𝒔 = 𝑲𝒑 +𝑲𝑫 𝒔 +𝑲𝑰𝒔

Kp = Proportional gain

KI = Integral gain

Kd = Derivative gain

The PID controller

First, let's take a look at how the PID controller works in a closed-loop system using the

schematic shown above. The variable [E(s)] represents the tracking error, the difference

between the desired input value [R(s)] and the actual output [C(s)].

This error signal (e) will be sent to the PID controller, and the controller computes both the

derivative and the integral of this error signal.

The signal [U(s)] just past the controller is now equal to the proportional gain (Kp) times

the magnitude of the error plus the integral gain (Ki) times the integral of the error plus

the derivative gain (Kd) times the derivative of the error.

𝒖 𝒕 = 𝑲𝒑 𝒆(𝒕) + 𝑲𝑰 𝒆 𝒕 . 𝒅𝒕 + 𝑲𝑫𝒅 𝒆(𝒕)

𝒅𝒕

This signal (u) will be sent to the plant, and the new output will be obtained. This new

output will be sent back to the sensor again to find the new error signal (e). The controller

takes this new error signal and computes its derivative and its integral again.

This process goes on and on.

The characteristics of P, I, and D controllers

A proportional controller (Kp) will have the effect of reducing the rise time and will

reduce ,but never eliminate, the steady-state error.

An integral control (Ki) will have the effect of eliminating the steady-state error, but it

may make the transient response worse.

A derivative control (Kd) will have the effect of increasing the stability of the system,

reducing the overshoot, and improving the transient response.

Effects of each of controllers Kp, Kd, and Ki on a closed-loop system are summarized

in the table shown below.

RISE TIME OVERSHOOT SETTLING TIME Steady-State

Response

Kp Decrease Increase Small Change Decrease

Ki Decrease Increase Increase Eliminate

Kd Small Change Decrease Decrease Small Change

Example Problem

Suppose we have a simple mass, spring, and damper problem.

The modeling equation of this system is

Let

M = 1kg

fv = 10 N.s/m

k = 20 N/m

F(s) = 1 = unit step

𝑋(𝑠)

𝐹(𝑠)=

1

𝑠2 + 10 𝑠 + 20

K

vf

M)(tf

)(tx

)()()()(

2

2

tftKxdt

tdxf

dt

txdM v

𝑋(𝑠)

𝐹(𝑠)=

1

𝑀 𝑠2 + 𝑓𝑣 𝑠 + 𝐾

Open-loop step response

Let's first view the open-loop step response.

Create a new m-file and add in the following code:

>> num=1;

>> den=[1 10 20];

>> step (num,den)

Running this m-file in the Matlab command window should give you the

plot shown below.

Open-loop step response The DC gain of the plant transfer function is

1/20, so 0.05 is the final value of the output to

an unit step input. This corresponds to the

steady-state error of 0.95, quite large indeed.

Furthermore, the rise time is about one second,

and the settling time is about 1.5 seconds.

Let's design a controller that will reduce the rise

time, reduce the settling time, and eliminates

the steady-state error.

Closed Loop with P-Controller

The closed-loop T.F is𝐶(𝑠)

𝑅(𝑠)=

𝐾𝑝

𝑠2 + 10 𝑠 + (20 + 𝐾𝑝)

Let the proportional gain (Kp) equals 300 and change the m-file to thefollowing:>> Kp = 300;

>> num = [Kp];

>> den = [1 10 20+Kp];

>> t = 0 : 0.01 : 2 ;

>> step (num,den,t)

Closed Loop with P-Controller

Running this m-file in the Matlab command window should gives you the

following plot.

The plot shows that the proportional

controller reduced both the rise time

and the steady-state error, increased

the overshoot, and decreased the

settling time by small amount.

Closed Loop with PD-Controller

The closed-loop T.F is

𝐶(𝑠)

𝑅(𝑠)=

𝐾𝑝 + 𝐾𝐷 𝑆

𝑠2 + 10 + 𝐾𝐷 𝑠 + (20 + 𝐾𝑝)

Let Kp equals 300 and Kd equals 10, then change the m-file to the following:

>> Kp = 300; KD = 10;

>> num = [KD Kp];

>> den = [1 10+KD 20+Kp];

>> t = 0 : 0.01 : 2 ;

>> step (num,den,t)

Closed Loop with PD-Controller


following plot.

The plot shows that the derivative

controller reduced both the overshoot

and the settling time, and had small

effect on the rise time and the steady-

state error.

Closed Loop with PI-Controller

The closed-loop T.F is𝐶(𝑠)

𝑅(𝑠)=

𝐾𝑝 𝑠 + 𝐾𝐼

𝑠3 + 10 𝑠2 + 20 + 𝐾𝑝 𝑠 + 𝐾𝐼

Let Kp equals 300 and Ki equals 70, then change the m-file to the following:

>> Kp = 300; KI = 70;

>> num = [Kp KI];

>> den = [1 10 20+Kp KI];

>> t = 0 : 0.01 : 2 ;

>> step (num,den,t)

Closed Loop with PI-Controller


following plot.

We have reduced the proportional gain

(Kp) because the integral controller also

reduces the rise time and increases the

overshoot as the proportional controller

does (double effect). The above response

shows that the integral controller

eliminated the steady-state error.

Closed Loop with PID-Controller

The closed-loop T.F is

𝐶(𝑠)

𝑅(𝑠)=

𝐾𝐷 𝑠2 + 𝐾𝑝 𝑠 + 𝐾𝐼

𝑠3 + (10 + 𝐾𝐷)𝑠2 + 20 + 𝐾𝑝 𝑠 + 𝐾𝐼

Let Kp equals 350, Kd equals 50 and Ki equals 300, then change the m-file tothe following:>> Kp = 350; KI = 300; KD = 50;

>> num = [KD Kp KI];

>> den = [1 10+ KD 20+Kp KI];

>> t = 0 : 0.01 : 2 ;

>> step (num,den,t)

Closed Loop with PID-Controller


following plot.

Now, we have obtained the system with no

overshoot, fast rise time, and no steady-

state error.

General tips for designing a PID controller

When you are designing a PID controller for a given system, follow the

steps shown below to obtain a desired response.

1) Obtain an open-loop response and determine what needs to be improved

2) Add a proportional control to improve the rise time

3) Add a derivative control to improve the overshoot

4) Add an integral control to eliminate the steady-state error

5) Adjust each of Kp, Ki, and Kd until you obtain a desired overall response.

Modern Control - Lec 03 - Feedback Control Systems Performance and Characteristics

Engineering

Transcript of Modern Control - Lec 03 - Feedback Control Systems Performance and Characteristics