BLOCK DIAGRAM MANIPULATION AND

REDUCTIONBy Schck Tchamna Rodrigue, PhD

October 2014

General Block Diagram of a Control System

( )G s( )R s ( ) ( )Y G s R s=Transfer Function

Reduced Block Diagram of a Control System

( )C s ( )pG s

( )H s

Summing point

E R F=

( )F s

Branchpoint

Forward Path

Feedback Path

Plant

Controller

+-

( )D s

Reference Input

( )R s

Main Feedback Signal

Disturbance

Manipulated Variable

Actuating Signal Systems Error ( )M s

Rodrigue Tchamna

( )Y s

( )C s ( )pG s

( )H s

E R F=

( )F s

+-

( )D s( )R s ( )M s ( )Y s

( )C s ( )pG s

( )H s

E R F=

( )F s

+-( )R s ( )M s ( )Y s

( ) 0if D s =

( ) ( )pG s C s

( )H s

E R F=

( )F s

+-( )R s ( )Y s

( ) ( )pG s C s

( )H s

E R F=

( )F s

+-( )R s ( )Y s

( )G s

( )H s

( )E s

( )F s

+-( )R s ( )Y s

( )G s Is called Direct or Forward transfer function( )H s Is called Feedback transfer function

( ) ( )G s H s Is called Open-loop transfer function

( )( )

Y sR s

Is called Closed-loop transfer function or Control ratio

( )( )

Y sE s

Is called Feedforward transfer function

Closed Loop Transfer FunctionY GE=

E R F R HY= = ( )G s

( )H s

( )E s

( )F s

+-( )R s ( )Y s

( )Y G R HY GR GHY= = Y GHY GR+ =

( )I GH Y GR+ =( ) 1Y I GH G

R= +

For SISO 1Y GR GH=

+

( )G s

( )H s

( )E s

( )F s

++( )R s ( )Y s

For SISO 1Y GR GH=

Positive Feedback

The Characteristic equation of the system is defined as an equation obtained by setting thedenominator polynomial of the transfer function to zero.

For SISO 1Y GR GH=

+

1 0GH+ =The Characteristic equation of the above system is

Block Diagram Reduction RulesFor many real systems, the block diagram of a Single Input-Single Output (SISO),feedback control system may involve several feedback loops and summing points.In principle, the block diagram of (SISO) closed loop system, no matter howcomplicated it is, it can be reduced to the standard single loop form.

( )G s( )R s ( ) ( )Y G s R s=Transfer Function

Block Diagram Reduction Rules

1 Combine all cascade blocks

2 Combine all parallel Blocks. Eliminate all minor (interior) feedback loops

3 Moving a pickoff point behind a block

4 Moving a pickoff point ahead a block

5 Moving a summing point behind a block

6 Moving a summing point ahead a block

Manipulation Original Block Diagram Equivalent Block Diagram Equation

1Combine all cascade blocks

2 Combine all parallel Blocks. Eliminate all minor (interior) feedback loops

3 Moving a pickoff pointbehind a block

4 Moving a pickoff pointahead a block

5 Moving a summing pointbehind a block

6Moving a summing pointahead a block

1G +R Y

2G

1GR Y

2G

GR Y

R 1/ GG

R YR

GR Y GR Y

G

1R YG+2R

YG +

G1R2R

YG

+1

R

2R+1

R

1/ G2R

YG

1 2G GR Y

1 2G GR Y

1 2Y G G R=

( )1 2Y G G R=

YY GR RG

= =

Y GR=

( )1 2Y G G R=

1 2

21

Y GR RRG RG

=

=

( )G s

( )H s

( )E s

( )F s

+-( )R s ( )Y s

( )G s

( )H s

( )E s

( )F s

++( )R s ( )Y s

++ +- ++1R

2R 3R

-Y Y3R

2R

1R1 2 3Y R R R= +

1Y GHR GH=

+

1Y GHR GH=

( )1G s

( )D s

( )2G s( )E s

( )H s( )F s

( )R s ( )Y s-+ ++

Using the block diagrams rules, find the correct block Diagram among the block diagram a) b) or c)

Example 1

( )1G s ( )2G s( )E s

( )H s( )F s

( )R s ( )Y s-+

( )D s

++

( )21

G s

a)

( )1G s ( )2G s( )E s

( )H s( )F s

( )R s ( )Y s-+

( )D s

++ b)

( )1G s ( )2G s( )E s

( )H s( )F s

( )R s ( )Y s-+

( )D s

++

( )2G s

c)

c)

( )1G s( )R s

-+ ( )2G s

( )H s

-+( )Y s

( )1G s( )R s

-+( )Y s2

21GG H+

( )R s1 2

2

1 2

2

1

11

G GG HG G

G H

+

++

( )Y s ( )R s 1 22 1 21G G

G H G G+ +

( )Y s

Example 2

( )R s1 2 3

2 3 2

1 2 3 1

2 3 2 3

1

11

G G GG G H

G G G HG G H G

+

++

( )Y s

( )R s-+

( )Y s

1

3

HG

1 2 3

2 3 21G G GG G H+

( )R s1 2 3

2 3 2 1 2 11G G G

G G H G G H+ +( )Y s

( )R s-+ ( )1G s ( )2G s-+

( )Y s( )3G s

2H

1H

( )R s-+ ( )1G s ( )2G s-+

( )Y s( )3G s

2H

1

3

HG

( )R s-+ ( )1G s

( )Y s

1

3

HG

2 3

2 3 21G GG G H+

Example 3: Simplify the following block diagram First Method

Move pickoff point

Behind block

Negative feedback loop

Negative feedback loop

( )R s 1 2 32 3 2 1 2 11

G G GG G H G G H+ +

( )Y s

( )R s-+ ( )1G s ( )2G s-+

( )Y s( )3G s

2H

1H

( )R s-+

( )Y s

1

3

HG

1 2 3

3 2 21G G GG G H+

( )R s-+ ( )1G s ( )2G s -+

( )Y s( )3G s

2 2G H1H

( )R s-+ ( )1G s ( )2G s -+

( )Y s( )3G s

2 2G H1

3

HG

( )R s-+ 1 2G G

( )Y s

1

3

HG

3

3 2 21G

G G H+

Second Method (more complicated)

Move summing point

Behind block

Move pickoff point

Behind block

Negative feedback loop

Negative feedback loop

Example 3: Simplify the following block diagram

Example 4 : Simplify the following block diagram

R-+ ++ 1G 4G

1H

3G

2G ++

2H

Y R-+ ++ 1 4G G 2 3G G+

2H

Y

1H

R-+ 2 3G G+

2H

Y1 41 4 11

G GG G H

Parallel and Cascade blocks

reduction

Positivefeedback loop

Cascade blocks reduction

R-+

2H

Y( )1 4 2 31 4 11

G G G GG G H

+

Negativefeedback loop R

Y( )

( )

1 4 2 3

1 4 1

1 4 2 32

1 4 1

1

11

G G G GG G H

G G G GH

G G H

+

++

R ( )( )

1 4 2 3

1 4 1 1 4 2 3 21G G G G

G G H G G G G H+

+ +

Y

R-+ -+ 1G 4G

1H

3G

2G ++

2H

Y

Example 5 : assume the following model is for a given plant

1 2sG

s=

+2

23

Gs

=+

332

Gs

=+ 4

35

sGs+

=+

1 1H = 226

sHs+

=+

R ( )( )

1 4 2 3

1 4 1 1 4 2 3 21G G G G

G G H G G G G H+

+ +

Y

R ( )( )

1 4 2 3

1 4 1 1 4 2 3 21G G G G

G G H G G G G H+

+ + +

Y

Using the original and the simplified block diagram. Simulate the system using Simulink :

Multi inputs Single Output (MISO)

( )1G s ( )2G s( )E s

( )H s( )F s

( )R s ( )Y s-+ ++

1) Set one input to zero, 2) find the equivalent block Diagram, 3) then set the otherinput to zero, 4) find its equivalent block Diagram, 5) then sub up the two results

( )1G s

( )D s

( )2G s( )E s

( )H s( )F s

( )R s ( )Y s-+ ++

1 2G GE

HF

R Y-+

Step 1

Step 2

1 2

1 21G GY

R G G H=

+

( )1G s

( )D s

( )2G s( )E s

( )H s( )F s

( )R s ( )Y s-+ ++

2

1 21GY

D G G H=

+

Step3

1) Set one input to zero, 2) find the equivalent block Diagram, 3) then set the otherinput to zero, 4) find its equivalent block Diagram, 5) then sub up the two results

( )1G s

( )D s

( )2G s( )E s

( )H s

( )Y s-+ ++

Step 4

( )2G s

1G H

( )Y s( )D s

++ ( )2G s

1G H

( )Y s( )D s ++ ( )2G s

1G H

( )Y s( )D s +-

1 2

1 21G GY

R G G H=

+2

1 21GY

D G G H=

+

( )

1 2 2

1 2 1 2

21

1 2

1 1

1

G G GY R DG G H G G HG G R D

G G H

= ++ +

= ++

( )1G s

( )D s

( )2G s( )E s

( )H s

( )R s ( )Y s-+ ++

Homework 1

( )R s-+ ( )1G s ( )2G s-+

( )Y s( )3G s

-+

1H

2H

( )4G s

3H

( )R s-+

( )Y s

1H

1 2 3 4

2 3 2 3 4 31G G G G

G G H G G H+ +

1) Prove that Block Diagram 1 = Block Diagram 2

Block Diagram 1

Block Diagram 2

2) Determine the final Transfer function of the system3) Simulate using Simulink

1 2sG

s=

+2

23

Gs

=+

332

Gs

=+ 4

35

sGs+

=+

1 1H = 226

sHs+

=+

3 2

16

Hs

=+

Homework 2

Using the superposition principle, Compute the transfer function Y in terms of R and D

Simulate the system for a step input disturbance, for R = 0;

LECTURE 6

Systems Time Domain Analysis

BLOCK DIAGRAM MANIPULATION AND REDUCTIONSlide Number 2Slide Number 3Slide Number 4Closed Loop Transfer FunctionSlide Number 6Slide Number 7Block Diagram Reduction RulesSlide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Multi inputs Single Output (MISO)Slide Number 20Slide Number 21Homework 1Homework 2LECTURE 6Systems Time Domain Analysis