c2: basic of mechanical and electrical control system

Lecture 2

• Basic of mechanical and electrical control system

• By

• Oladokun Sulaiman

Note

Revise :

• Differential equation

• Partial fraction

2.0 Objective

• Free body and block diagram• Block diagram• Obtain the differential equation• Obtain Laplace transform of the differential

equations• Solve the resulting algebraic transform• Mathematical modeling of physical system

RECAP

implified block diagram of closed-loop control system

• Process or plant:– System to be controlled including actuators and power amplifiers

• Sensor:– Instrumentation that measures output and converts it to a signal

• Compensator or controller:– System added to enhance performance of control loop– Output or controlled variable is y(t) is the variable we wish to control– Input is a measure of (but not always equal to) the desired system

output u(t)– Error is desired output minus actual output: e(t)= y(t) - u(t)

Gc(s) Gp(s)Output

H(s)

Input

Y(s)U(s)

Compensatoror controller

Processor plant

Sensor

test waveform

•Refer to control engineering system page 19

Types of input

0 t1 t2 t3t4

Time, second

50

100

Forc

e,

N

0 t1 t2 t3t4

Time, second

50

100

Forc

e,

N

0 t1 t2 t3t4

Time, second

50

100

Forc

e,

N

0 t1 t2 t3

Time, second

50

100

Forc

e,

N

Two step input response

Ramp Input Response

Time Constant

• It is defined as time taken by a control plant to achieve output responseequal to 63% of its desired value.

Control System Response

• Time response

• Frequency response

• Steady state response

• Transient response

• Undershoot

• Overshoot

• Settling time

Control system stability

• Damping factor

• Damping ratio

• Routh’s stability criteria

• Nyquist stability criteria

fig_01_11

Control System Design Process

•Control engineering system page 16

2.1 Introduction : Mathematical modeling

• In order to analyze and design a control system knowledge of its behavior through mathematical terms is essential

• The mathematical equations are derived from law of physics i.e Newton

• Analysis can be done in two operating condition: Steady and Transient

• Change as a result of input or disturbance• Output will depend on the system variables and how they

interact• Description of dynamic system is obtain from differential

equation’• models of the physical system• Solutions of these mathematical equations simulate the

response of the physical system which they represent

2.2 Differential equation• Input and output relationship of a linear measurement

system - ordinary differential equation (ODE):

• u = input, y = output; u and y varies with t• n > m and a, b = constant coefficients

• DE contain variable and rate of change of or derivative of the variable in control system

• Ordinary differential equation (ODE) are main concern in control system , they contain singe dependent and independent variable which is usually time

• The order DE relate to the index of the highest derivative

yadt

dya

dt

da

dt

yda

n

n

nn

n

n 011

1

1

Example 2.1 : Spring mass balance damper

xo

xi

mk

c

=

Fs

FD

Fm

2

2

)()(dt

xdm

dt

dx

dt

dxcxxk ooi

oi

ii

ooo x

m

k

dt

dx

m

cx

m

k

dt

dx

m

c

dt

xd

2

2

2.3 Physical system modeling

• To obtain linear approximation of physical system Time response solution is obtain:

• Obtain the differential equation

• Obtain Laplace transform of the differential equations

• Solve the resulting algebraic transform

2.3a. Laplace transformDefinition of Laplace transformation of

f(t):

where s = + j = a complex variable– Inverse Laplace transformation– f(t) = L-1[F(s)]– L[Af(t)] = AL[f(t)]– L[f1(t) + f2(t)] = L[f1(t)] + L[f2(t)]

For step input f(t) = 0 t < 0– = A t > 0 – Laplace transform:

0 0

)()]([)()]([ dtetftfdtesFtfL stst

A

f(t)

t

dtAeALtfLsF st

0

][)()(

s

Ae

s

A st

0

Example 2• Find the time response xo(t) for this system if step input xi(t)=1

and initial condition xo(0)=0

• Differential equation: • For k =1,c = 1;

• Laplace transform

• Partial fraction

• Inverse Laplace transform:

0)( 00

dt

dxcxxk i

xi

k

c

xo

k = 1c = 1 dt

dxc 0

k(xi-x0)

ioo kxkx

dt

dxc

ioo xx

dt

dx

ssXssX

1)()( 00

)1(

11)(

sssX o

tetx 1)(0

Transient response Steady-state response

xo(t)

t

1

table_02_01


table_02_02•Control engineering system page 34

2.3b. Transfer Functions

• Defined as the ratio of the Laplace Transform of the output to the Laplace Transform of the input to the system

• G(s) = Y(s)/X(s)

• X(s) Y(s)G(s)

Transfer Function

• An assembly of linked components within a boundary.

• The motor car is a good example; mechanical, electrical, control and suspension sub-systems within a body-chassis boundary.

• A system may have one input and a related output dependent on the effect of that system (transfer function G).

•0 = G I

•The boundary, represented as a "black box", may include a complex system which need not be analysed if G is provided. •More complex systems have interconnecting links to related systems. •A system must have input, process, output, and in most systems a source of power and a means of control.

Transfer Function Expression

Characteristics Equation

0012

2 aSaSa

•Denominator of the transfer function equated to zero is the characteristics equation of the system

•Characteristics equation of the system determines the response of the control system

012

2

01)(aSaSa

bSbsG

Characteristics Equation

Order of control systems

• Zero Order System

• First Order System

• 2nd Order System

)( 2121

01

asass

bsb

)( 2120

01

asass

bsb

)( 2122

01

asass

bsb

(Source: Instrumentation and Control Systems by Leslie Jackson)

Poles and Zeros

• Roots of the Characteristics equation are called poles of the system

Roots of numerator of the TF are called zeros of the system

;3,2,1;5;1)2(

;2,1,2)1(

6116

56)2

23

42)1

23

2

2

poleszeros

poleszerosAns

sss

ss

ss

s

Example

2.4 System modeling

Step for drawing block diagram:

• Step 1: Free body diagram

• Step 2: Mathematical equations

• Step 3: Block diagram

2.4a Mechanical system: Spring

• Spring• Where k = stiffness, x= displacemnet

• Fs= Fx or

• Transfer function = k or 1/k• Output variable = transfer function input variable• Spring with free at both ends

• Block diagram Fs = k(x1 – x2)

Fs xx

Fsk

kx1 x2

kx1(t)

x2(t)

+-

x1- x2 Fs(t)

4. Mechanical system: Mass

• Fm = ma

• Force F acting on mass m

• Use D-operator where: D = d/dt and D2 =d2/dt2

• Fm = mD2x

m

Fm(t) x(t)

2

1

mD

2

2

dx

xdmFm

Fm(t)x(t)

Spring mass system

•Equations:

• Fs = k(xi – x0)• Since Fm = ma and Fs = Fm• mD2x0(t) = Fs = k(xi – x0)kxi(t)xi(t)-xo(t)-+

• Block diagram for spring-mass system

m

xi(t)xo(t)

kxi(t) xi(t)-xo(t)

-+

Damper

• c = damping coeff.

• x = displacement

• dx(t)/dt = velocity

• FD(t) = cdx(t)/dt

• FD(t) = cDx

x(t) FD(t)

cD

Spring-damper system

• Force on spring: Fs = k(xi – xo)• Force on damper: FD = cDxo

xi(t)

xo(t)

kxi(t) Fs=FD

-+

Block diagram for spring-damper system

1/cD

Spring-mass-damper system

F = ma• Fs1 - FD = Fm

Fs1 FD

Fm

kxi(t) Fs

-+

FD

-+

xo(t)Fm

cD

2

1

mD

Spring Mass System

)(2

2

2

2

ForceDisturbingFkydt

dyf

dt

ydm

kyForceSpringdt

dyfForceDamping

dt

ydmForcengAccelerati

ksfmssF

sYsGTF

sFsYkfssm

2

2

1

)(

)()(

)()()(

fig_02_15

Example 2

•Draw free body diagram

•Determine forces and direction i.e applied force to the right and impeding forces to the left spring, viscous damper, acceleration

•Write the differential equation

Contd

2 ( ) ( ) ( ) ( )vMS s f sX s KX s F s

2( )vMS f s K

•Take la place transform

•Or

•Solve for transfer function

•G(s)=X(s)/F(s) = 1/

2( ) ( ) ( )vMS f s K X s F s

fig_02_17


fig_02_18

fig_02_20


fig_02_21

fig_02_22

table_02_05

2.4b. Electrical system modeling Electrical components: resistance R, capacitance C and inductance LVariables: voltage V and current i

• Resistor -> VR = iR

• Inductance -> VL = Ldi/dt = LDi•

• Capacitance->i = CdV/dt = CDV

Circuit theory• Series

V = V1 + V2• Parallel• V = V1 = V2

– i = i1 + i2

+ -VR

+ -

VL

+ -

VCi

+ V

V1 V2

V1 V2

+

-

V

Example 3• Vi = VL + VC + VR; VR = Vo • VL = Ldi/dt = LDi

• CdVC/dt = i; VC = i/CD• Vo = iR

VL

Vo

+

-

Vi

VC

L1

Vo

+

-

Vi

VA

L2

C

+

-

R

i1

i2 i3

Equations:i1 = i2 + i3Vi – VA = L1Di1VA = L2Di2i3 = (VA – Vo)CDVo = i3R

Block diagram

•Use D-operator where D = d/dt and D2 =d2/dt2

Vi i1

-

+

-

+

VA

DL1

1L2D

+R

-

VA

i3

i2 CDi3

Vo

Vo

Example

• Subtitute i(t)=dq(t)

• Subtitute capacitor voltage charge relationship Q(T)=CvC(t)

• Take la place transform, rearrange terms and simplify

• Solve TF Vc(s)/V(s)

Summing the voltage around the loop ->assuming zero initial condition

2( 1) ( )CLCs RCs v V s

2 1

( ) 1/

( )C

CLC

v s LCRv s s SL

R-L-C Circuit

)()(11

)()(11

)()(

1

1

0

0

sEsIsC

sEsIsC

sIRsIsL

givestransformLaplaceTaking

eidtC

eidtC

Ridt

diL

i

i

•Combining two equations from previous slide gives

1

1)(

2

sRCsCLsGTF

TF of spring mass system and RLC circuit are mathematically similar and will give identical response

•Solution for electrical can also be done through KVL, KCL,

• voltage divider, current divider


Steps for electrical modeling

• Replace passive element value wit their impedance

• Replace all source and time value with their Laplace transform

• Assume transform current and current direction is each loop

• Write KVL around each loop• Solve simultaneous equation for the output• Form the transfer function

fig_02_05

fig_02_11

2.3 Block diagram manipulation

• Block diagram – can be simplified to fewer blocks

• Block diagram transformation and reduction –refer to table

• Output input relationship – transfer function

Block diagram manipulation

2 = F1(D)1 ; 3 = F2(D)2 3 = F1(D) F2(D)1 • Transfer function = 3 / 1 = F1(D) F2(D)

F1(D)1

F2(D)3

2

F1(D)F2(D)1

3

F1(s)

F2(s)

++

3

41 2

F1(D)+F2(D)1

4

Series block diagram reduction

Parallel block diagram reduction

4 = 2 + 3 4 = F1(D)1 + F2(D)1 4 = [F1(D) + F2(D)]1

Transfer function = 4 / 1 = F1(D) + F2(D)

Closed loop block diagram

• The negative feedback:• E(s) = U(s) – B(s) = U(s) – H(s)Y(s)• Y(s) = G(s)E(s) = G(s)[U(s) - H(s)Y(s)]• Y(s)[1 + G(s)H(s)] = G(s)U(s)

G(s)

H(s)

E(s)

-+

Input

Output

B(s)

)()(1

)(

)(

)(

sHsG

sG

sU

sY

U

(s)

Y(s)

)()(1

)(

sHsG

sG

Close-loop transfer function = (Forward transfer function)/(1+ Open-loop transfer function)

Example

• Derive transfer function for spring-mass system

200 mD

kxxx i

-+

xixo

2mD

k

kmD

k

x

x

i

o

2

x

i

x

o

kmD

k

2

Multiple-loop feedback control systemTo eliminate G3(s)G4(s)H(s), move H2 behind G4(s)Rule 4:

G4

H2 /G4

G4

H2

G3G4H1 = positive feedback control systemRule 6 – eliminate feedback loops

•Eliminate inner loop containing H2/G4 and G2 and G3G4/(1-G3G4H1)

Reduce the loop containing H3

Revise your digital last semester digital electronics

Tutorial Exercises • Define transfer function of a control

system.

• Define what is time constant.

• Describe the advantages of control and automation

Home work

Chapter 2 : NISE Control ->• Answer all short questions• Problem 7,9,10,16,17,20,28

• Due date January 7

Summary

• Free body and block diagram

• Block diagram

• Obtain the differential equation

• Obtain Laplace transform of the differential equations

• Solve the resulting algebraic transform

• Mathematical modeling of physical system

c2: basic of mechanical and electrical control system

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