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).
d2y
dy
(1)
 3  6 y  et
is linear differential equation.
2
dt
dt
d3y
d 2 y 2 dy

(
6

t
)
t
 y  sin t
is linear differential equation.
(2)
dt 3
dt 2
dt
y  (6  t ) y  t 2 y  y  sin t
or
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  2t 2 y  4 y  e 2t
An example of time varying control system is a spacecraft control system.
١
Non linear system: Non linear system are ones which are represented by
non linear equations. Examples are
Y=sin(x) , y=x2 , z=x2+y2
A differential equation is called non linear if it is not linear. For examples
d2y
dy
 ( ) 2  y  A sin wt
2
dt
dt
2
d y
dy
 ( y 2  1)  y  0
2
dt
dt
2
d y dy

 y  y3  0
2
dt
dt
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.
٢
Example: Consider the linear time-invariant system
a0 y ( n )  a1y ( n 1)  ...  an 1 y  a n y  b0 x ( m )  b1 x ( m 1)  ...  bm 1 x b m x
nm
(1)
take the Laplace transform (ℓ) of y(t)
ℓ(a0y(n)) = aoℓ(y(n)) = ao[snY(s)-sn-1y(0)-s(n-2)y/(o)-…-y(n-1)(0)
ℓ(a1y(n-1)) = a1ℓ(y(n-1)) = a1[sn-1Y(s)-sn-2y(0)-s(n-3)y/(o)-…-y(n-2)(0)

ℓ(any)

= anℓ(y) = anY(s)
same thing is applied to obtain the L.T. of x(t).by substitute all initial
condition to zero. The transfer function of the system become.
Transfer Function  G ( s ) 
b0 s m  b1s m 1  ...  bm 1s  bm
a0 s n  a1s n 1  ...  an 1s  an
● Transfer function is not provide any information concerning the
physical structure of the system (the T.F. of many physically different
system can be identical).
● The highest power of s in the denominator of T. F. is equal to the
order of the highest derivative term of the output. If the highest power of
s is equal to n the system is called an nth order system.
How you can obtain the transfer function (T. F.)?
1- Write the differential equation of the system
2- Take the L. T. of the differential equation, assuming all initial
condition to be zero.
3- Take the ratio of the output to the input. This ratio is the T. F. .
Mechanical translation system
Consider the mass – spring – dashpot system
1- Mass
a
b
M
A force applied to the mass produces an acceleration of the mass.
The reaction force fm is equal to the product of mass and acceleration and
is opposite in direction to the applied force in term of displacement y, a
velocity v, and acceleration a, the force equation is
where D = d/dt
Fm = M a = M D v = M D2 y
٣
2- Spring
c
d
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 )
to
If the end d is stationary, then yd = 0 and the above equation reduces
Fk = k yc
3- dashpot
e
f
The reaction damping force fB 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 – Vf ) = B ( Dye - Dyf )
Mechanical translation quantity and units
U. S. customary units
metric units
Quantity
Force
pounds
newtons
Distance
feet
meters
Velocity
feet/second
meter/second
2
meter/second2
Acceleration
feet/second
Mass
slugs=pound*second2/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
٤
The fundamental law governing mechanical system is Newtons law
for translational system, the law state that
Ma 
M

F
d2y
dy
f
 ky  x
2
dt
dt
Taking the Laplace transform of each term
[M
d2y
]  M (s2Y (s)  sy(0)  y(0))
2
dt
dy
]  f [ sY ( s )  y (0)]
dt
[ky ]  kY ( s )
( x)  X ( s )
[ f
Set the initial conditions to zero so that y(0)=0, y (0)  0 the Laplace
transform can be written
( Ms 2  fs  k )Y ( s )  X ( s )
Taking the ratio of Y(s) to X(s), we find that the transfer function of the
system is
T .F .  G ( s ) 
Y (s)
1

2
X ( s ) Ms  fs  k
Example: find the T.F. of simple mass-spring-damper mechanical system
To draw the mechanical network, the points xa, xb and the reference
are located. The complete mechanical network is drawn in fig. below.
For nodes a & b
(1)
(2) where D=d/dt
f  f k  k ( xa  xb )
f k  f m  f B  MD xb  BDxb
2
٥
f k  f m  f B  M xb  BD x b
or
It is possible to obtain one equation relating xa to f, xb to xa, or xb to f
by combining equation (1) & (2)
k ( MD 2  BD) xa  ( MD 2  BD  k ) f
(3) check
(4)
(5)
( MD 2  BD  k ) xb  kxa
( MD 2  BD) xb  f
xa MD 2  BD  k
G1 

f
k ( MD 2  BD)
Let
(6)
xb
k
(7)

2
xa MD  BD  k
1
x
(8)
G b 
2
f
MD  BD
Or by taking L.T. to equation 3,4,5 with all initial condition equal to zero. The
transfer functions are
G2 
xa ( s ) Ms 2  Bs  k

f ( s ) k ( Ms 2  Bs)
x ( s)
k
G2 ( s )  b

2
xa ( s ) Ms  Bs  k
(9)
G1 ( s ) 
G
(10)
1
xb ( s)

2
f ( s) Ms  Bs
(11)
Note that the equations 8 & 11 are equal to the product of the (6) * (7) & (9)*(10)
&
G(s)=G1(s)G2(s)
G=G1G2
The block diagram representing the mathematic operation
f
f
G1(D)
G1(s)
xa
xa
G2(D)
G2(s)
٦
xb
f
xb
f
G (D)
G (s)
xb
xb
Example:
Automobile suspension some system of one wheel
M1= mass of the automobile
B = the shock absorber
k1 =the spring
k2= elestance of the tire
M2= mass of the wheel
Two independent displacement exist, so we must write
Two equations
d 2 x1
dx dx
M 1 2   B( 1  2 )  k1 ( x1  x2 )
dt
dt
dt
M1
d 2 x2
dx dx
 f (t )  B( 2  1 )  k1 ( x2  x1 )  k2 x2
2
dt
dt
dt
M1s2X1(s) + B(sX1(s)-sX2(s))+k1(X1(s)-X2(s)) =0
M2s2X2(s) + B(sX2(s)-sX1(s))+k1(X2(s)-X1(s))+k2X2(s) =F(s)
T (s) 
X 1 (s)
Bs  k1

4
3
F ( s ) M 1M 2 s  B( M 1  M 2 ) s  ( K1M 2  k2 M 1 ) s 2  k 2 Bs  k1k 2
٧
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. m2
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 /sec2
Applying Newtons law to the present system, We obtain
J  f  T
Then the transfer function of this system is found to be
( s )
1

where ( s)  { (t )} ,
T ( s )  {T (t )}
T ( s) Js  f
Electrical Systems
LRC circuit. Applying kirchhoffs voltage law to the system shown. We
obtain the following equation;
L
di
1
 Ri   idt  ei .......(1)
dt
C
1
 idt  eo .......( 2)
C
Equation (1)&(2) give a mathematical model of the circuit. Taking the
L.T. of equations (1)&(2), assuming zero initial conditions, we obtain
LsI ( s )  RI ( s ) 
11
I ( s )  Ei ( s )
Cs
11
I ( s )  Eo ( s )
Cs
the transfer function
E0 ( s )
1

2
E i ( s ) LCs  RCs  1
١
Complex impedances.
Consider the circuit shown below then the T.F of this
circuit is
Eo ( s )
Z 2 ( s)

Ei ( s ) Z1 ( s )  Z 2 ( s )
Where
Z1= Ls+R
1
Z2 
Cs
Z(s) = E(s)/I(s)
Hence the T.F. Eo(s)/Ei(s) can be found as follows;
1
Eo ( s )
1
Cs


2
Ei ( s ) Ls  R  1
LCs  RCs  1
Cs
Example
Armature-Controlled dc motors
The dc motors have separately excited fields. They are either armature-controlled with
fixed field or field-controlled with fixed armature current. For example, dc motors
used in instruments employ a fixed permanent-magnet field, and the controlled signal
is applied to the armature terminals.
Consider the armature-controlled dc motor shown in the following figure.
Ra = armature-winding resistance, ohms
La = armature-winding inductance, henrys
ia = armature-winding current, amperes
if = field current, a-pares
ea = applied armature voltage, volt
eb = back emf, volts
θ = angular displacement of the motor shaft, radians
T = torque delivered by the motor, Newton*meter
J = equivalent moment of inertia of the motor and load referred to the motor shaft
kg.m2
f = equivalent viscous-friction coefficient of the motor and load referred to the motor
shaft. Newton*m/rad/s
T = k1 ia ψ where ψ is the air gap flux, ψ = kf if ,
k1 is constant
٢
T = kf if k1 ia
For a constant field current
where k is a motor-torque constant
T = k ia
For constant flux
eb  kb
d
dt
{(eb=k2ψω)}
where kb is a back emf constant………(1)
The differential equation for the armature circuit is
La
dia
 Raia  eb  ea ........(2)
dt
The armature current produces the torque which is applied to the inertia
and friction; hence
Jd 2
d
 f
 T  Kia ..........(3)
2
dt
dt
Assuming that all initial conditions are condition are zero/and taking the L.T. of
equations (1) ,(2)&(3), we obtain
Kpsθ(s) = Eb(s)
(Las+Ra) Ia(s)+Eb(s) =Ea(s)
(Js2+fs) θ(s) = T(s) = KIa(s)
The T.F can be obtained is
 (s)
K
.........chek ?

2
Ea ( s ) s ( La Js  ( La f  Ra J ) s  Ra f  KK b )
Example:
Field-Controlled dc motor
Find the T.F
 ( s)
For the field-contnalled dc motor shown in figure below
E f ( s)
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 = k1 ψ ia
Where k1 is constant
T = k2 if
Where k2 is constant
di f
Lf
 R f i f  e f ……….(1)
dt
d 2
d
J 2  f
 T  k2i f . ………(2)
dt
dt
٣
By taking the L.T. of eqs. (1)&(2) & assuming zero initial conditions we get
(Lf s +Rf) If (s) = Ef (s)
(Js2 + fs) θ(s) = k2 If (s)
The T.F. of the this system is obtained as
K2
 (s)

...........chek ?
E f ( s ) S ( L f s  R f )( Js  f )
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.= 5.5*10-2 volts-sec/rad
K = motor torque constant = 6*10-5 Ib-ft-sec2
Ra = 0.2 Ω
La = negligible
Jm = 1*10-3 Ib-ft-sec2
fm = negligble
Jl = 4.4*10-3 Ib-ft-sec2
fL = 4*10-2 Ib-ft/rad/sec
n = gear ratio N1/N2=1/10
Hint J = Jm+n2Jl ,f = fm+n2fL
 ( s)
0.72
Answer

Ea ( s ) s (0.13s  1)
٤
UNIVERSITY OF TECHNOLOGY
DEPARTMENT OF ELECTRICAL & ELECTRONIC ENGINEERING
FIRST TERM EXAMINATION
Jan. ,13th, 2009
Subject: Control Engineering
Year: Third
time allowed 75 minute
Attempt All Questions
Q1: (A) Define four of the following statements, (1) state vector, (2)
transducer, (3) open loop control system, (4) transfer function, (5) linear
time invariant system.
[2 Marks]
(B) For the electrical circuit shown in Fig. (1)
(1) Derive the transfer function Y(s)/U(s).
[2 Marks]
(2) Find the state space equations and then draw the block diagram of
state space equations.
[2 Marks]
R
L
u
Fig. (1)
C
R
y
Q2: For the block diagram of feedback control system shown in Fig.(2)
the output is
Y(s)=M(s)R(s)+Mw(s)W(s)
Find the transfer functions M(s) and Mw(s).
[4 Marks]
W(s)
R(s) +
_
G1
+
_
H3
G2
+
_
+
+
G3
H2
H1
Fig. (2)
Y(s)