5 - block diagrams (read-only)bw-drew/cse_lecture_5-uwe.pdfindustrial control ufmf6w-20-2 control...

06/03/2017

© 2017 University of the West of England, Bristol. 1

UWE Bristol

Industrial ControlUFMF6W-20-2

Control Systems EngineeringUFMEUY-20-3

Lecture 5: Block Diagrams and Steady State Errors

Today’s Lecture

• Block diagrams to represent control systems

• Block diagram manipulation• Example• Steady State Errors

Block Diagrams

• Block Diagrams provide a pictorial representation of a system

• Unidirectional operational block representing individual transfer functions

• Three basic elements:– Rectangles - operators– Lines - signals– Circles - additional or subtraction

Block Diagrams: Examples

• y = Ax

• e = r - c

Ax y

r e

c–

+

Block Diagrams: Examples

• y = Ax-Bz

Ax y–

+

zB

• Simple Closed Loop Control System

Closed Loop System

Process

Sensor

+

–Input Error

Feedback

Output

G(s)

H(s)

+

–R(s) E(s)

B(s)

C(s)

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– Transfer function from R(s) to C(s)

Closed Loop System

G(s)

H(s)

+

–R(s) E(s)

B(s)

C(s)

( ) ( ) ( )sBsRsE -=( ) ( ) ( )sCsHsB =

( ) ( ) ( ) ( ) ( )( )sGsCsEsEsGsC =®=

( )( ) ( ) ( ) ( )sCsHsRsGsC

-=®

ïïþ

ïïý

ü



Closed Loop System

G(s)

H(s)

+

–R(s) E(s)

B(s)

C(s)

( )( ) ( ) ( ) ( )sCsHsRsGsC

-= ( ) ( ) ( ) ( )sRsHsG

sC =÷÷ø

öççè

æ+®

1

( )( )

( )( ) ( )sHsGsG

sRsC

+=\1



Closed Loop System

G(s)

H(s)

+

–R(s) E(s)

B(s)

C(s)

( )( )

( )( ) ( )sHsGsG

sRsC

+=1

C(s)R(s)( )( ) ( )sHsGsG

+1


– With unity feedback, H(s) = 1

Closed Loop System

( )( ) ( )

( )( )sGsG

sHsGsG

+®

+ 11

G(s)

1

+

–R(s) E(s)

B(s)

C(s)

• Remove the feedback link from summing junction

Open Loop Transfer Function

G(s)

H(s)

+R(s) E(s)

B(s)

C(s)

( ) ( )sRsE =

( )( ) ( ) ( )sHsGsEsB=

Open Loop Transfer Function given by:

Block Diagram Manipulation

• Diagrams can be manipulated using the following transformations

• Combining Blocks in Series:

06/03/2017



• Moving a summing junction

13


• Moving a pickoff point ahead of a block

14


• Moving a pickoff point behind a block

15


• Moving a summing point ahead of a block

16


• Eliminating a feedback loop

17

Example

G1 G2 G3

H1

H2

1

+ + +

– – –R CE A B J D

HKC

Consider subgroup containing G3 and H1:

413

3

1G

HGG

JC

=+

=

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Example

G1 G2 G4

H2

1

+ +

– –R CE A B J

KC


413

3

1G

HGG

JC

=+

=

Example

G1 G2 G4

H2

1

+ +

– –R CE A B J

KC

Consider subgroup containing G2 and G4:

542 GGGBC

==

Example

G1 G5

H2

1

+ +

– –R C

E A B

KC

Consider subgroup containing G2 and G4:

542 GGGBC

==

Example

G1 G5

H2

1

+ +

– –R C

E A B

KC


625

5

1G

HGG

AC

=+

=

Example

G1 G6

1

+

–R C

E A

C


625

5

1G

HGG

AC

=+

=

Example

G1 G6

1

+

–R C

E

C

Final Closed Loop Transfer Function

( ) 3212213

321

61

61

11 GGGHGHGGGG

GGGG

RC

+++=

+=

06/03/2017


Steady State Errors• Feedback control used to reduce steady-state

errors• Steady-state error is error after the transient

response has decayed• If error is unacceptable, the control system will

need modification• Errors are evaluated using standardised inputs

– Step inputs– Ramp inputs– Sinusoidal inputs

Example – No SS Error

Time

Res

pons

e

Transient

ts

Steady State

Example – No SS Error

Time

Res

pons

e

Transient

ts

Steady State

Error

Causes of Steady State Error

• Errors can be caused by factors including1. Instrumentation of measurement errors2. System non-linearities – deadbands,

hysteresis, saturation etc.3. Form of input signal4. Form of system transfer function5. External disturbances acting on the system,

for example: forces or torques

Error FunctionG(s)

H(s)

+

–R(s) E(s)

B(s)

C(s)

( ) ( ) ( )sBsRsE -=

( ) ( ) ( ) ( ) ( ) ( )sEsGsHsCsHsB ==

( ) ( ) ( ) ( ) ( )sEsHsGsRsE -=

( ) ( ) ( )[ ] ( )sRsHsGsE =+1

( )( ) ( ) ( )sHsGsRsE

+=1

1

Calculating Value

• Use the final value theorem:

• Inputs can be– Step

– Ramp

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

sHsGsE

sHsGsRsE

´+

=®+

=1

11

1

System dependent

Input dependent

( ) ( )ssEtEEstss 0limlim®¥®

==

( ) 2sAsR =

( )sAsR =

A is step amplitude

A is step velocity

06/03/2017


Example

• System and Feedback Transfer functions:

• Error

( ) ( ) ( ) 1 and 1

1=

+= sH

sssG

t

( )( ) ( ) ( )

( )

( )( ) 111

111

11

1++

+=

++

=+

=ssss

sssHsGsR

sEtt

t

Example

• Step input

• Steady State Error:

( )( )

( )( ) ( ) ( )

( ) 111

111

+++

=®++

+=

ssss

sAsE

ssss

sRsE

tt

tt

( ) ( )( )

( )( ) 0

1010010

111lim

0=

+++

=++

+==

® tt

tt Assss

sAsssEE

sss

• as Ess = 0 and H(s) = 1, steady state output will be same as input

Rh=1

C

Example

• Ramp input


( )( )

( )( ) ( ) ( )

( ) 111

111

2 +++

=®++

+=

ssss

sAsE

ssss

sRsE

tt

tt

( ) ( )( )

( )( ) AA

ssss

sAsssEE

sss =++

+=

+++

==® 1010

0111

1lim 20 tt

tt

• in this case Ess is not zero and during application of ramp input C will lag R by A

Time - s

Response

A

R

C

Example

• System and Feedback Transfer functions:

• Error

( ) ( ) 2 and 1

1=

+= sH

ssG

t

( )( ) ( ) ( ) 3

1

121

11

1++

=

++

=+

=ss

ssHsGsR

sEtt

t

06/03/2017


Example

• Step input


( )( ) ( )

31

31

++

=®++

=ss

sAsE

ss

sRsE

tt

tt

( ) ( )330

1031lim

0

AAss

sAsssEE

sss =++

=++

==® t

ttt

• in this case, error is not zero and the output will not be equal to the input

• steady state output Css can also be found using the Final Value Theorem as follows :

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

( )23

11 sRss

sHsEsRsCsCsHsRsE ÷

øö

çèæ

++

-=-

=®-=tt

32311

2311lim)(lim

00

AAsA

sssssCC

ssss =÷øö

çèæ -=÷

øö

çèæ

++

-==®® t

t

• resulting response for A = 1:

R

C

(b) ramp input:)(lim

0ssEE

sss ®= ÷÷

ø

öççè

æ++

=® 20 )3(

)1(limssAss

s tt

¥=´+´+

=0)03()01( A

tt

error continues to increase which is not acceptable

Time - s

Response R

C

Today’s lecture• Block Diagrams pictorial representation of

control system• Rectangles represent operations• Lines are signals• Summing junctions enable addition/subtraction• Manipulation Techniques to reduce block

diagrams to transfer function• Steady State Errors help to determine what

happens to signal in steady state

An electrical motor is used in a closed loop system to control the angular position of an inertial load. The position of the load, which is directly connected to the motor, is measured by a simple rotary potentiometer. The output signal from the transducer is compared with the input demand and the resulting error signal is passed to a voltage/current amplifier. The input demand is converted from angular displacement to voltage before being connected to the summing junction.

Example Block Diagram

06/03/2017


• Closed Loop

Example

DCAmplifier

Battery

Anglesetting

DC motor

Angle

Turntable

+–

Tachometer

Amplifier DCmotor Turntable

ControlDevice Actuator Process

Desired angle(voltage)

Actual angle+–

Potentiometer

Sensor

Error

Measured angle(voltage)

Gain

Gain

system block diagram:

An electrical motor is used in a closed loop system to control the angular position of an inertial load.


The position of the load, which is directly connected to the motor, is measured by a simple rotary potentiometer.


The output signal from the transducer is compared with the input demand and the resulting error signal is passed to a voltage/current amplifier.

The input demand is converted from angular displacement to voltage before being connected to the summing junction.

system block diagram: • system equations:

IkT mm =-torqueMotor(a)

eAVI =-Amplifier(b)

ooL csJsT qq +=- 2Load(c)

ipi kV q=-demandInput(d)

oTo kV q=-Feedback(e)

oie VVV -=-Error(f)

06/03/2017


• system block diagram

reduced block diagram

csJs +21Tm oq

mkI

APk

Tk

VeVi

Vo

iq +

-

+G(s)

H(s)-

R C csJskAksG Pm

+= 2)(

P

T

kksH =)(

• Closed Loop Transfer function for system:

Tm

Pm

i

o

kAkcsJskAks++

= 2)(qq

5 - block diagrams (read-only)bw-drew/cse_lecture_5-uwe.pdfindustrial control ufmf6w-20-2 control...

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