The Block Diagram

Dr.-Ing. Erwin SitompulPresident University

Lecture 2Feedback Control Systems

http://zitompul.wordpress.com

President University Erwin Sitompul FCS 2/1


The Block DiagramBlock diagram is a graphical tool to visualize the model of a

system and evaluate the mathematical relationships between its components, using their transfer functions.

In many control systems, the system equations can be written so that their components do not interact except by having the input of one part be the output of another part.

The transfer function of each components is placed in a box, and the input-output relationships between components are indicated by lines and arrows.

Chapter 3 Dynamic Response


The Block Diagram

1 1 1( ) ( ) ( )U s G s Y s

Using block diagram, the system equations can be simplified graphically, which is often easier and more informative than algebraic manipulation.



Elementary Block DiagramsBlocks in series Blocks in parallel with their

outputs added

21 2

1

( )( )

Y s GGU s

1 2( )( )Y s G GU s

Summingpoint

Pickoffpoint



Single loopnegative feedback

1

1 2

( )( ) 1

GY sR s GG

2 1( ) ( ) ( )Y s R s Y s G G

1 1 2( ) ( ) ( )Y s G R s GG Y s

1 2 1( ) 1 ( )Y s GG G R s

1

1 2

( )( ) 1

GY sR s GG

What about single loop with positive feedback?

?Negativefeedback

Elementary Block DiagramsChapter 3 Dynamic Response


Block Diagram AlgebraChapter 3 Dynamic Response


Transfer Function from Block DiagramExample:Find the transfer function of the system shown below.

( )( )( )Y sT sR s

2

2

2 4

2 41

ssss

2

2 42 4s

s s



Example:Find the transfer function of the system shown below.

Transfer Function from Block DiagramChapter 3 Dynamic Response


Transfer Function from Block Diagram

1 2 5 1 6

1 3 1 2 41GG G GGGG GG G

1 2

1 3

1 2 4

1 3

1 65

21

( )( )( ) 1

G GG GG G GG G

GY sT s GR s G



?

Transfer Function from Block Diagram

1 2

1 2

( )( ) 1RY s GGR s GG H

When D(s) 0,

When R(s) 0,2 2

1 2 1 2

( )( ) 1 ( ) 1DY s G GD s G G H GG H

Example:Find the response of the system Y(s) to simultaneous application of the reference input R(s) and disturbance D(s).



Transfer Function from Block Diagram( ) ( ) ( )R DY s Y s Y s

1 2 2

1 2 1 2

( ) ( ) ( )1 1GG GY s R s D sGG H GG H

21

1 2

( ) ( ) ( )1GY s G R s D sGG H



Definition of Pole and ZeroConsider the transfer function F(s):

The system response is given by:

The poles are the values of s for which the denominator A(s) = 0.

The zeros are the values of s for which the numerator B(s) = 0.

( ) ( )( )Y s F sU s

( )( ) ( ) ( ) ( )( )B sY s F s U s U sA s

( )( )( )B sF sA s

Numerator polynomial

Denominator polynomial



Effect of Pole Locations

( ) 1( )( )Y sH sU s s

The impulse response will be an exponential function:( ) 1( )ty t e t

1( ) ( )Y s U ss

When σ > 0, the pole is located at s < 0, The exponential expression y(t) decays. Impulse response is stable.

When σ < 0, the pole is located at s > 0, The exponential expression y(t) grows with time. Impulse response is referred to as unstable.

A form of first-order transfer function

•How?


Consider the transfer function F(s):


•PFE


2

2 1( )3 2sH s

s s

1 11 1( ) 31 2

h ts s

L L

2( ) ( 3 ) 1( )t th t e e t

●The terms e–t and e–2t, which are stable, are determined by the poles at s = –1 and –2.

●This is true for more complicated cases as well.

●In general, the response of a transfer function is determined by the locations of its poles.

Time (sec)0 1 2 3 4

-0.5

0

0.5

1

1.5

2

( )h t

Example:Find the impulse response of H(s),

2 1( 1)( 2)

ss s

1 31 2s s



Effect of Pole LocationsTime function of impulse response assosiated with

the pole location in s-plane

LHP RHP

LHP : left half-plane RHP : right half-plane



Representation of a Pole in s-Domain The position of a pole (or a zero)

in s-domain is defined by its real and imaginary parts, Re(s) and Im(s).

In rectangular coordinates, the complex poles are defined as(–s ± jωd).

Complex poles always come in conjugate pairs.

A pair of complex

poles



Representation of a Pole in s-Domain The denominator corresponding to a complex pair will be:

2 2( ) ( )( )

( )d d

d

A s s j s js

On the other hand, the typical polynomial form of a second-order transfer function is:

2

2 2( )2

n

n n

H ss s

Comparing A(s) and denominator of H(s), the correspondence between the parameters can be found:

2 1n d n andζ : damping ratio ωn : undamped natural frequencyωd : damped frequency



Representation of a Pole in s-Domain Previously, in rectangular coordinates,

the complex poles are at (– ± jωd). In polar coordinates, the poles are at

(ωn, sin–1ζ), as can be examined from the figure.

d

2 2n d

n

21d n



Unit Step Resonses of Second-Order System

( )y t

nt



Example:Find the correlation between the poles and the impulse response of the following system, and further find the exact impulse response.


2

2 1( )2 5sH s

s s

Since2

2 2( ) ,2

n

n n

H ss s

2 2 0.447n

2 5 5 2.24 rad secn n

The exact response can be otained from:

2

2 1( )2 5sH s

s s

2 2

2 1( 1) 2

ss

1 2s j poles at


President University Erwin Sitompul FCS 2/22Time (sec)

( )h t

0 1 2 3 4 5 6-1

-0.5

0

0.5

1

1.5

2

Effect of Pole LocationsTo find the inverse Laplace transform, the righthand side of the last equation is broken into two parts:

2 2

2 1( )( 1) 2

sH ss

2 2 2 2

1 1 22( 1) 2 2 ( 1) 2s

s s

1( ) ( )h t H sL12 cos 2 sin 2 1( )2

t te t e t t

Damped sinusoidal oscillation



Time Domain SpecificationsSpecification for a control system design often involve certain requirements associated with the step response of the system:1. Delay time, td, is the time required for the response to reach

half the final value for the very first time.2. Rise time, tr, is the time needed by the system to reach the

vicinity of its new set point.3. 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.

4. Overshoot, Mp, is the maximum peak value of the response measured from the final steady-state value of the response (often expressed as a percentage).

5. Peak time, tp, is the time required for the response to reach the first peak of the overshoot.



Time Domain Specifications

,10% 90% ,5% 95% ,0% 100%, , r r rt t t

, 1% , 2% , 5%, , s s st t t

( ) ( )% 100%

( )p

p

y t yM

y



1( )1

H ss

The step response of first-order system in typical form:

is given by:1 1( )

1Y s

s s

1 1

(1 )s s

1 1 1( )(1 )

y ts s

L

( ) (1 ) 1( )ty t e t • : time constant• For first order system, Mp

and tp do not apply

First-Order SystemChapter 3 Dynamic Response


Second-Order System2

2 2( )2

n

n n

H ss s

is given by:2

2 2

1( )2

n

n n

Y ss s s

2 2 2 2

1( ) ( )

n n

n d n d

ss s s

1( ) ( )y t Y sL

2( ) 1 cos sin

1nt

d dy t e t t

21 cos sin

1n nt t

d de t e t


The step response of second-order system in typical form:


Second-Order SystemTime domain specification

parameters apply for most second-order systems.

Exception: overdamped systems, where ζ > 1 (system response similar to first-order system).

Desirable response of a second-order system is usually acquired with 0.4 < ζ < 0.8.

2( ) 1 cos sin

1nt

d dy t e t t



Rise Time The step response expression of the second order system is

now used to calculate the rise time, tr,0%–100%:

Since , this condition will be fulfilled if:

2( ) 1 1 cos sin

1n rt

r d r d ry t e t t

0n rte

2cos sin 0

1d r d rt t

or, 21

tan d rt

d

21n

d n

11 tan dr

d

t

( )

d

φ

tan d

tr is a function of ωd



Settling TimeUsing the following rule:

sin cos cos( ),A B C

The step response expression can be rewritten as:

2 2 1, tan AC A BB

with:

2

( ) 1 cos( )1

nt

dey t t

1

2tan

1

where: ts is a function of ζ

1

n



Settling Time The time constant of the envelope curves shown previously is

1/ζωn , so that the settling time corresponding to a certain tolerance band may be measured in term of this time constant.

, 2%44sn

t , 1%

55sn

t , 5%

33sn

t



Peak Time

2

2

( ) cos sin1

sin cos1

n

n

tn d d

t dd d d

y t e t t

e t t

When the step response y(t) reaches its maximum value (maximum overshoot), its derivative will be zero:

2( ) 1 cos sin

1nt

d dy t e t t

2

2( ) sin

1nt n

d dy t e t



Peak Time

2

2( ) sin

1n pt n

p d d py t e t

0

≡00, , 2 ,3 ,d pt

At the peak time,

Since the peak time corresponds to the first peak overshoot,

pd

t

tp is a function of ωd



Maximum OvershootSubstituting the value of tp into the expression for y(t),

2( ) 1 cos sin

1n pt

p d p d py t e t t

2( ) 1 cos sin

1n d

py t e

211 e

( ) ( )% 100%

( )p

p

y t yM

y

21

( ) ( )

(1 ) 1p pM y t y

e

21pM e

21% 100%pM e if y(∞) = 1



Example 1: Time Domain SpecificationsExample:Consider a system shown below with ζ 0.6 and ωn 5 rad/s. Obtain the rise time, peak time, maximum overshoot, and settling time of the system when it is subjected to a unit step input.

2

2 2

( )( ) 2

n

n n

Y sR s s s

After block diagram simplification,

Standard form of second-order system



11 tan dr

d

t

11 4tan4 3

pd

t

0.6, 5 rad/sn 2 21 1 0.6 5 4 rad/sd n

0.6 5 3 rad/sn

In second quadrant

Example 1: Time Domain Specifications

1 ( 0.927)4 0.554 s

4

0.785 s



, 2%4 4 1.333 s

0.6 5sn

t

, 5%3 3 1 s

0.6 5sn

t

21% 100% 9.48%pM e

21( ) ( ) (1 ) 1p pM y t y e

2 0.6 0.81 0.0948pM e e

2

2 2

( )( ) 2

n

n n

Y sR s s s

Check y(∞) for unit step input, if

Example 1: Time Domain SpecificationsChapter 3 Dynamic Response


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

Time (sec)

Ampl

itude


0.554 s, 0.785 s

9.48%, 1.333 sr p

p s

t t

M t



Example:For the unity feedback system shown below, specify the gain K of the proportional controller so that the output y(t) has an overshoot of no more than 10% in response to a unit step.

( 2)

( 2)

( )( ) 1

Ks s

Ks s

Y sR s

2 2

Ks s K

2 2n

0.592

2n K

21% 10% 0.1pM e 1 1 1.689

0.592n

2 21.689 2.853nK 0 2.853K

Example 2: Time Domain SpecificationsChapter 3 Dynamic Response



0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

Time (sec)

Ampl

itude

: K = 2: K = 2.8: K = 3



Homework 2No.1

Obtain the overall transfer function of the system given below.

Deadline: 18.05.2012, at 07:30.

No.2, FPE (5th Ed.), 3.20.Note: Verify your design using MATLAB and submit also the printout of the unit-step response.

++


The Block Diagram

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