chapter 4 block diagrams of control systems

8/15/2019 Chapter 4 Block Diagrams of Control Systems

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EEE 352 Automatic Control Systems

Chapter 4: Block Diagrams of Control Systems

Prof. Dr. Ahmet Uçar

© Dr. Ahmet Uçar EEE 352 Chapter 4 1

Block Diagrams of Control Systems


A block diagram of a system is a pictorial representation of the functions

performed by each component and of the flow of signals. Such a diagram depicts

the interrelationshipsthat exist among the various components.

Differing from a purely abstract mathematical representation, a block diagram

has the advantage of indicating more realistically the signal flows of the actual

system.

The nonlinear and linear systems canbe represented by block diagrams.

The nonlinear systems

F ( x, u)y (t )

Output

signal

u(t )

Input

signal

State initial condition;

x(t =0) 0

The linear systems

Transfer

function

G(s)

R(s)

Input

signal

Y (s)

Output

signal

State initial condition;

x(t =0) 0


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Drawing a Block Diagram of linear systems:

The transfer functions of the components are usually entered in the

corresponding blocks, which are connected by arrows to indicate the direction of

the flow of signals.

Transfer

function

G(s)

R(s)

Input

signal

Y (s)

Output

signal

Branch point

G( s) R( s) Y ( s)

H ( s)

E (s)Gc( s)

Summing point



Drawing a Block Diagram of linear systems with the Laplace transforms.

•Write the equations that describe the dynamic behavior of each component.

•Take the Laplace transforms of these equations, assuming zero initial

conditions.

•Represent each Laplace-transformed equation individually in block form.

•Assemble the elements into a complete block diagram.

ei e0 R

C I

Example 4.1 a): Draw block diagram of the following RC circuit with the

Laplace transforms?

Solution 4.1 a): Let define the error between input and output signal as E (s)

)1()()()( 0 s E s E s E i

E i( s)

E 0( s)

E ( s)


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Example 4.1: a) Write current I(s)

)2()(1

)( s E R

s I

)3()0()(1

1),0(

1

00

00

e s I Cs

E

sdt e Idt

C e

Write output E 0(s)

ei e0 R

C I E (s) I(s)

R

1

Finally, assemble the elements into a complete

block diagram.

E 0(s)I(s)

Cs

1

e0(0)0

E i( s) E 0( s) E ( s) I ( s)

R

1

Cs

1

e0(0)0



Drawing a Block Diagram of linear systems with time domain:


•Represent each equation individually'inblock form.


ei e0, e0(0)0 R

C I

Example 4.1 b): Draw block diagram of the following RC circuit without Laplace

transform ?

Solution 4.1 b): Let define the error between input and output signal as e(t )

)1()()()( 0 t et et e i

ei(t )

e0(t )

e(t )


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Example 4.1 b): Write current i (t )

)2()(1

)( t e R

t i

)3()0()(1

)( 00 t edt t iC t e

Write output e0(t )

ei e0 R

C I e(s) i (t )

R

1

Finally, assemble the elements into a complete block diagram.

e0(t )i(t )

e0(0)0

ei(t ) e0(t )e(t ) i(t )

R

1

e0(0)0

C

1



Block diagram of continuous time control system represented in state space

r mnm

r nnn

R D RC t Dut Cxt y

R B R At But Axt x

,)()()(

,)()()(

s1 B

D

A

C U ( s) Y ( s) X ( s)

In time domain:

In Laplace domain:

dt B

D

A

C u(t ) )(t x x(t ) y(t )

x0(0)0


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Block Diagrams of Control SystemsRules of Block Diagram Algebra


Original Block Diagram Equivalent Block Diagram

1.Combining blocks in

cascade

2. Moving a summing

point behind a block

A complicated block diagram involving many feedback loops can be simplified by

a step-by-step rearrangement. Simplification of the block diagram by

rearrangements considerably reduces the labour needed for subsequent

mathematical analysis.The following rules block diagram algebra can be used.

G1( s) G2( s)Y ( s) R( s)

G1( s) R( s)

G2( s) Y ( s) X ( s)

R( s)G1( s)

Y ( s)

B( s)

R( s) Y ( s)

G1( s)

G1( s)

B( s)




3. Moving a branch point

ahead of a block R( s)G1( s)

Y ( s)

Y ( s)

R( s) Y ( s)

G1( s)

G1( s)

Y ( s)

4. Moving a branch

point behind a block R( s)

G1( s) Y ( s)

R( s)

R( s) Y ( s)G1( s)

R( s)

)(

1

1 sG

R( s)G1( s)

Y ( s)

B( s)

R( s)G1( s)

B( s)

)(

1

1 sG

Y ( s)

5. Moving a summing point

ahead of a block


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6. Eliminating a feedback

loop Y ( s)G( s)

R( s) E ( s)+

H ( s)

Y ( s) R( s)

)()(1

)(

s H sG

sG

Example 4.2: Consider the system shown in the following Figure.

Simplify this diagram?

G1 R

H 2

G2 G3Y

H 1


Rules of Block Diagram Algebra


Example 4.2:(cont.)

G1 R

H 2

G2 G3Y

H 1

G1 R

H 2/G1

G2 G3Y

H 1

By moving the summing point of the negative feedback loop containing H2 outside the

positivefeedback loop containing H1:

Eliminating the positive feedback loop:

R

H 2/G1

G3Y

121

21

1 H GG

GG


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Example 4.2 : (cont.)

The elimination of the loop containing H2/G1:

R Y

232121

321

1 H GG H GG

GGG

R

H 2/G1

G3Y

121

21

1 H GG

GG

Finally, eliminating the feedback loop results overall system transfer function:

R Y

321232121

321

1 GGG H GG H GG

GGG

Remark. The numerator of the closed-loop transfer function Y (s)/R(s) is the product of

the transfer functions of the feedforward path: G1(s)G2(s)G3(s)

The denominator of Y (s)/R(s) is equal to

1 - (product of the transfer functions around each loop)




Example 4.2 :(cont.)

The system transfer function: R Y

321232121

321

1 GGG H GG H GG

GGG

Remark.

The numerator of the closed-loop transfer function Y (s)/R(s) is the product of

the transfer functions of the feedforward path: G1(s)G2(s)G3(s)

The denominator of Y (s)/R(s) is equal to

1 -

(product of the transfer functions around each loop)

G1 R

H 2

G2 G3Y

H 1

321232121

321232121

1

)(1

GGG H GG H GG

GGG H GG H GG

The positive feedback loop yields a negative term in the denominator.


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Block Diagrams of Control Systems in MATLAB


Example 4.3: Find the transfer function of the following system with given blocks

transfer functions.

Y ( s)G2

R( s) H 2

G3 G4

H 1

G1

H 3

10

1)(1

s

sGMATLAB Command

%%%%%%%%%%%%%%%%%%%%%

% Modelling blocks

ng1=[1]; dg1=[1 10];sysg1=tf(ng1,dg1)

ng2=[1]; dg2=[1 1];sysg2=tf(ng2,dg2)

ng3=[1 0 1]; dg3=[1 4 4];sysg3=tf(ng3,dg3)ng4=[1 1]; dg4=[1 6];sysg4=tf(ng4,dg4)

nh1=[1 1]; dh1=[1 2];sysh1=tf(nh1,dh1)

nh2=[2]; dh2=[1];sysh2=tf(nh2,dh2)

nh3=[1]; dh3=[1];sysh3=tf(nh3,dh3)

1

1)(2

s

sG

44

1

)( 2

2

3

s s

s

sG

6

1)(4

s

s sG

2

1)(1

s

s s H

2)(2 s H

1)(3 s H



Example 4.3:

Y ( s)G2

R( s)

H 2/G4

G3 G4

H 1

G1

H 3

sys2

sys4 sys3

sys1

sys6

sys

sys5

MATLAB Command

% Block Diagram Reduction

sys1=sysh2/sysg4

sys2=series(sysg3,sysg4)

sys3=feedback(sys2,sysh1,+1)

712219631282517106620512

25664

)(

)(23456

2345

s s s s s s

s s s s s sys

s R

sY

sys4=series(sysg2,sys3)

sys5=feedback(sys4,sys1,-1)

sys6=series(sysg1,sys5)

sys=feedback(sys6,sysh3)


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MATLAB Command

sysmr = minreal(sys)

Eliminates cancels pole-zero pairs in transfer functions or zero-pole-gain

models.

The output sysmr has minimal order and the same response characteristics

as the original model sys.

Ones many feedback loops simplified it is difficult to observe pole zero pairs toreduce system transfer function futher. However this can readly achived with

Matlab.

712219631282517106620512

25664

)(

)(23456

2345

s s s s s s

s s s s s sys

s R

sY

Example 4.14: (cont.)



MATLAB Commandsysmr = minreal(sys)

712219631282517106620512

25664

)(

)(23456

2345

s s s s s s

s s s s s sys

s R

sY

33.597.12313775.7208.16

1667.025.025.025.008333.0

)(

)(2345

234

s s s s s

s s s s sysmr

s R

sY

Both transfer functions have the same response characteristics

Example 4.3:


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Homework 4.1: Simplify the block diagram shown in Figure and obtain theclosed loop transfer function Y (s)/R(s).

Homework 4.2: Simplify the block diagram shown in Figure and obtain the closed

loop transfer function Y (s)/R(s).

R( s) Y ( s)

G1( s) G2( s) G3( s)

H 1( s)

H 2( s)

H 3( s)

R( s) Y ( s) E 1(s)G1( s) G2( s)

H 1( s) H 2( s)

H 3( s)

G3( s) G4( s)

E 2(s)



Drawing a Block Diagram of nolinear systems:


•Represent each equation individually'inblock form.


Example 4.4: Draw block diagram of the following nonlinear system for non

zero state initial conditions? The system output is y = x ?

x y

x x x x

036.0 2


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Block Diagram of Nonlinear Systems


x y

x x x x

036.0 2

Solution 4.4: We start to draw the block diagram from the system output y

1 x(t ) y(t )

and considering the integration process as; x(t )

x0(0)0

)(t x

1 y(t ) x(t )

x0(0)0

)(t x

)(t x

0)0(0 x

0.6

3

( )2



Example 4.5: Draw block diagram of the following nonlinear system for non

zero state initial conditions? The system output is y = x 1 ?

1

21

2

12

21

)1(

x y

u x x x x

x x

m u

x1 xe= 022 )( x xc

3

111)( x x xk

Physical

System

f ( x )

Output

y s(t )

Input

u(t ) 0

State initial

condition ;

x(t =0) 0


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1

21

2

12

21

)1(

x y

u x x x x

x x

m u

x1 xe= 022 )( x xc

3

111)( x x xk

Solution 4.5: We start to draw the block diagram from the system output y andconsidering the integration process.

1 y(t ) x1(t )

x10(0)0

)(2 t x

)(2 t x

1

( )3

u(t )

x20(0)0



Homework 4.3: Draw block diagram of the following nonlinear system for non

zero state initial conditions? The system output is y = x .

Homework 4.4: Draw block diagram of the following nonlinear system for non

zero state initial conditions? The system output is y = x 1 .

u x x x x )1( 2

2

3

112

21

x x x x

x x


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MATLAB/Simulink


Simulink provides a graphical editor, customizable block libraries, and solvers for

modeling and simulating dynamic systems. It is integrated with MATLAB, enabling

you to incorporate MATLAB algorithms into models and export simulation results

to MATLAB for further analysis.

The construction of a model is simplified with click-and-drag mouse operations.

Simulink includes a comprehensive block library of toolboxes to analyze and

design the linear and nonlinear systems.

Open Simulink Library Browser

Start MATLAB, and then in the MATLAB Command Window,

a) enter

» simulink or

b) by clicking the Simulink icon in the MATLAB toolbar or

Simulink Block Diagram of Control Systems


c ) by clicking the MATLAB Start button, then selecting Simulink > Library Browser

The Simulink Library Browser in Figure 2 opens.


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Create a New Simulink Model from Simulink Library Browser1. From the Simulink Library Browser menu, select File> New> Model.

An empty model opens in the Simulink Editor;

2. Use Simulink library and build the system block diagram.

3. In the Simulink Editor, select File > Save.

4. In the Save As dialog box, enter a name for your model, and then click Save.

Simulink saves your model.



Example 4.6: Obtain SIMULINK model the following RC circuit.

ei e0, e0(0)0 R

C I

The complete block diagram of the system is;

ei(t ) e0(t )e(t ) i(t )

R

1

e0(0)0

C

1


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The block diagram of the circuit/system can be directly implemented in

SIMULINK, as shown in the following Figure:

Solution 4.6:Assume ei(t ) is a DC input.

ei(t ) e0(t )e(t ) i(t )

R

1

e0(0)0

C

1

yo

ei; e0

ei; e0

ei

-1

e0(0)

1

sxo

Integrator

1

Gain1

1/2

1/R

1

1/C



Example 4.7: A Damped Pendulum System is depicted Figure shows it can

rotate freely around its fixed point P. It consists of a rod with angle

represented by from the vertical position. Using the equation of motion for

a damped pendulum given by

Obtain SIMULINK model the following nonlinear systems.

2122

21

sinml

T x

l

g x

ml

c x

x x

c

v

h

P

2sin

ml

T

l

g

ml

c c

here is the angle of the pendulum from the vertical (in radians),c= 0.15 is the velocity damping term (in 1/sec),

m= 1 is the mass of the pendulum (in kilograms),

l =2.5 is the length (in meters),

g= 9.81 is the acceleration due to gravity (in m/s2) and

T c is the force input (in N).

Changing state variable as leads to),(),( 21 x x


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The SIMULINK model is given in the following Figure:

Solution 4.7:

2122

21

sinml

T x

l

g x

ml

c x

x x

c

x2

x2x1

x1

r

r

1

s

int x2

1

s

int x1

-K-c/ml1

K

c/ml

XY Graph

Tc

sin

Sin(x1)

Scope

K

1/ml



Open an Existing Model

Open an existing Simulink model from the Simulink Library Browser.

1. From the Simulink Library Browser menu, select File > Open.

2. In the Open dialog box, select the model file that you want to open, and then

click Open.


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Block Diagram of Control Systems


Homework 4.5: Model the following system in SIMULINK and depict the timeresponse of the system output y = x for;

a) u=0 and state initial condition; x0=(1,2).

b) u=1 and state initial condition; x0=(0,0).

u x x x x )1( 2

2

2

112

21

)1( x x x x

x x

Homework 4.6: Model the following system in SIMULINK and depict x1 versus

x2 for

a) state initial condition; x0=(0.2,0.2).

b) state initial condition; x0=(4,4).



2

3

112

21

x x x x

x x

Homework 4.7: Model the following system in SIMULINK.

a) depict the time response of the system output y = x 1 for state initial condition;

x0=(-1,1).

and depict x1 versus x2 for

b) state initial condition; x0=(-0.5,0).

c) state initial condition; x0=(0.5,0).

d) state initial condition; x0=(1.5,0).


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EEE 352 Automatic Control Systems

Chapter 4: Block Diagrams of Control Systems

Remarks and Questions?

chapter 4 block diagrams of control systems

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