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Mathematical Modeling of Physical Systems: Linear and Non-Linear Systems, Summaries of Engineering

PCI CollegeEngineering

Mathematical models of physical systems, focusing on linear and non-linear systems. Linear systems are those with linear differential equations, while non-linear systems have non-linear equations. Linear systems have the property of superposition, while non-linear systems do not. examples of linear and non-linear systems and their transfer functions.

Typology: Summaries

2017/2018

Uploaded on 12/02/2021

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Mathematical Model of Physical Systems
●Mechanical, electrical, thermal, hydraulic, economic, biological, etc,
systems, may be characterized by differential equations.
●The response of dynamic system to an input may be obtained if these
differential equations are solved.
●The differential equations can be obtained by utilizing physical laws
governing a particular system, for example, Newtons laws for
mechanical systems, Kirchhoffs laws for electrical systems, etc.
e mathematical description of the dynamic Th:
Mathematical models
characteristic of a system.
The first step in the analysis of dynamic system is to derive its model.
Models may assume different forms, depending on the particular system
and the circumstances.
In obtaining a model, we must make a compromise between the
simplicity of the model and the accuracy of results of the analysis.
Linear systems are one in which the equations of the:
Linear systems
model are linear.
A differential equation is linear, if the coefficients are constant or
functions only of the independent variable (time).
(1) t
ey
dt
dy
dt
yd 63
2
2
is linear differential equation.
(2) ty
dt
dy
t
dt
yd
t
dt
yd sin)6( 2
2
2
3
3
is linear differential equation.
or tyytyty sin)6( 2
The most important properties of linear system is that the principle of
superposition is applicable.
In an experimental investigation of dynamic system, if cause and effect
are proportional, thus implying that the principle of superposition holds,
then the system can be considered linear.
Linear time invariant systems: the system which represented by
differential equation whose coefficients are function of time for example
t
eyytyty 22 42)6(
An example of time varying control system is a spacecraft control system.
pf3
pf4
pf5
pf8
pf9
pfa

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Mathematical Model of Physical Systems

● Mechanical, electrical, thermal, hydraulic, economic, biological, etc,

systems, may be characterized by differential equations.

●The response of dynamic system to an input may be obtained if these

differential equations are solved.

●The differential equations can be obtained by utilizing physical laws

governing a particular system, for example, Newtons laws for

mechanical systems, Kirchhoffs laws for electrical systems, etc.

Mathematical models :The mathematical description of the dynamic

characteristic of a system.

The first step in the analysis of dynamic system is to derive its model.

Models may assume different forms, depending on the particular system

and the circumstances.

In obtaining a model, we must make a compromise between the

simplicity of the model and the accuracy of results of the analysis.

Linear systems :Linear systems are one in which the equations of the

model are linear.

A differential equation is linear, if the coefficients are constant or

functions only of the independent variable (time).

(1) y et

dt

dy dt

d y 2 ^3 ^6 

2

is linear differential equation.

(2) y t

dt

dy t dt

d y t dt

d y ( 6 ) 2 2 sin

2 3

3

     is linear differential equation.

or  y   ( 6  t ) y  t^2 y  y sin t

The most important properties of linear system is that the principle of

superposition is applicable.

In an experimental investigation of dynamic system, if cause and effect

are proportional, thus implying that the principle of superposition holds,

then the system can be considered linear.

Linear time invariant systems : the system which represented by

differential equation whose coefficients are function of time for example

 y   ( t  6 ) y  2 t^2 y  4 y  e ^2^ t

An example of time varying control system is a spacecraft control system.

Non linear system: Non linear system are ones which are represented by

non linear equations. Examples are

Y=sin(x) , y=x^2 , z=x 2 +y^2

A differential equation is called non linear if it is not linear. For examples

y A wt dt

dy dt

d y 2 (^ )^2 sin

2   

2 (^21 )^0

2    ydt

dy y dt

d y

2   yydt

dy dt

d y

The most important properties of non linear system is that the principle

of superposition is not applicable.

Because of the mathematical difficulties attached to nonlinear systems,

one often finds it necessary to introduce equivalent linear system which

are valid for only a limited range of operation.

Transfer functions: The transfer function of a linear time-invariant

system is define to be the ratio of the Laplace transform ( z transform for

sampled data systems) of the output to the Laplace transform of the input

(driving function), under the assumption that all initial conditions are

zero.

2- Spring

The elastance, or stiffness k provides a restoring force as represented

by a spring. The reaction force fk on each end of the spring is the same

and is equal to the product of stiffness k and the amount of deformation

of the spring.

End c has a position yc and end d has a position yd measured from

the respective equilibrium positions. The force equation, in accordance

with the Hoks law is

fk = k ( yc - yd )

If the end d is stationary, then yd = 0 and the above equation reduces

to

Fk = k y c

3- dashpot

The reaction damping force f B is approximated by the product of

damping B and the relative velocity of the two ends of the dashpot. The

direction of this force depend on the relative magnitude and direction of

the velocity Dye and Dyf

FB = B (Ve – V f ) = B ( Dye - Dy f )

Mechanical translation quantity and units

Quantity U. S. customary units metric units

Force pounds newtons

Distance feet meters

Velocity feet/second meter/second

Acceleration feet/second

2

meter/second

2

Mass slugs=pound*second^2 /foot kilogram

Stiffness coefficient pounds/foot newtons/meter

Damping coefficient pounds/(foot/second) newtons/(meter/second)

Example : find the T. F. of the figure below

c (^) d

e f

The fundamental law governing mechanical system is Newtons law

for translational system, the law state that

Ma   F

ky x dt

dy f dt

d y M (^) 2   

2

Taking the Laplace transform of each term

[ 2 ] (^2 ( ) ( 0 ) ( 0 ))

2

M sY s sy y

dt

d y

 M    

[ ] f [ sY ( s ) y ( 0 )] dt

dyf  

[ ky ]  kY ( s ) ( x )  X ( s )

Set the initial conditions to zero so that y(0)=0, y ( 0 ) 0 the Laplace

transform can be written

( Ms^2^  fsk ) Y ( s ) X ( s )

Taking the ratio of Y(s) to X(s), we find that the transfer function of the

system is

X s Ms fs k

Y s T F Gs  

Example : find the T.F. of simple mass-spring-damper mechanical system

To draw the mechanical network, the points x a, xb and the reference

are located. The complete mechanical network is drawn in fig. below.

For nodes a & b

f  fk  k ( xa  xb ) (1)

f (^) kfmfBMDxbBDx b

2 (2) where D=d/dt

Example :

Automobile suspension some system of one wheel

M 1 = mass of the automobile

B = the shock absorber

k 1 =the spring

k 2 = elestance of the tire

M 2 = mass of the wheel

Two independent displacement exist, so we must write

Two equations

21 (^12 )^1 (^12 )

2 1 k x x dt

dx dt

dx B dt

d x M    

1 2 1 2 2 2 1 2

2

2 1 (^ ) ( dt ) k ( x x ) kx

dx dt

dx f t B dt

d x M      

M 1 s^2 X 1 (s) + B(sX 1 (s)-sX 2 (s))+k 1 (X 1 (s)-X 2 (s)) =

M 2 s^2 X 2 (s) + B(sX 2 (s)-sX 1 (s))+k 1 (X 2 (s)-X 1 (s))+k 2 X 2 (s) =F(s)

2 1 2

2 1 2 2 1

3 1 2

4 1 2

1 1 ( ) ( ) ( )

MMs BM M s KM kM s kBs kk

B k F s

X s T s s      

Mechanical Rotational System

Consider the system shown below. The system consists of a load inertia

and a viscous – friction damper.

J = moment of inertia of

The load kg. m^2

f = Viscous – Friction coefficient Newton / (rad /sec)

ω = angular velocity, red / sec

T = torque applied to the system, Newton.m

For mechanical rotational systems Newtons low states that

J   T Where α = angular acceleration, rad /sec^2

Applying Newtons law to the present system, We obtain

J  f  T

Then the transfer function of this system is found to be

T s Js f

s

 1 ( )

( )

where ( s ){ ( t )}, T ( s ){ T ( t )}

Electrical Systems

LRC circuit. Applying kirchhoffs voltage law to the system shown. We

obtain the following equation;

dt Ri C idt ei

di L    

Cidteo

Equation (1)&(2) give a mathematical model of the circuit. Taking the

L.T. of equations (1)&(2), assuming zero initial conditions, we obtain

( ) ( ) I s E s Cs

LsI sRI s   i

I s E s C s o

2

0

E s LCs RCs

E s

thetransfer function

i

T = k (^) f i (^) f k 1 i (^) a For a constant field current T = k ia where k is a motor-torque constant {(eb =k 2 ψω)} For constant flux

dt

d eb kb

 where kb is a back emf constant………(1)

The differential equation for the armature circuit is

a a Raia eb ea ........(^2 ) dt

di L   

The armature current produces the torque which is applied to the inertia

and friction; hence

2 ..........(^3 )

2 T Ki a dt

d f dt

Jd   

Assuming that all initial conditions are condition are zero/and taking the L.T. of equations (1) ,(2)&(3), we obtain Kp sθ(s) = Eb (s) (La s+Ra ) Ia (s)+Eb (s) =Ea (s) (Js^2 +fs) θ(s) = T(s) = KIa (s) The T.F can be obtained is

.........? ( ) ( ( ) )

sLJs^2 Lf RJ s R f KK chek

K

E s

s a aaaab

Example:

Field-Controlled dc motor

Find the T.F ( )

E s

s f

For the field-contnalled dc motor shown in figure below

The torque T developed by the motor is proportional to the product of the air gap flux ψ and armature current ia so that T = k 1 ψ ia Where k 1 is constant T = k 2 i (^) f Where k 2 is constant

f f f

f f Ri e dt

di L   ……….(1)

2 2.

2 T ki f dt

d f dt

d J   

By taking the L.T. of eqs. (1)&(2) & assuming zero initial conditions we get (Lf s +Rf) If (s) = Ef (s) (Js^2 + fs) θ(s) = k 2 If (s) The T.F. of the this system is obtained as

...........? () ( )( )

chek S Ls R Js f

K

E s

s f ff

H.W

For the positional servomechanism obtain the closed-loop T.F. for the positional servomechanism shown below, Assume that the in-put and out put of the system are input shaft position and the output shaft r = reference input shaft, radian c = out put shaft, radian θ = motor shaft, radian k1 = gain of potentiometer error detector = 24/π volt/rad kp = amplifier gain = 10 kb = back emf const.= 5.510-2^ volts-sec/rad K = motor torque constant = 610-5^ Ib-ft-sec^2 R (^) a = 0.2 Ω La = negligible J (^) m = 110-3^ Ib-ft-sec^2 fm = negligble J (^) l = 4.410-3^ Ib-ft-sec^2 f (^) L = 4*10 -2^ Ib-ft/rad/sec n = gear ratio N 1 /N2 =1/ Hint J = J (^) m+n^2 J (^) l ,f = fm+n^2 f (^) L

Answer ( 0. 13 1 )

E s s s

s a