Mathematical Models and Block Diagrams of Systems
Regulation And Control Engineering
CONTENTS Introduction Differential Equations of Physical
Systems The Laplace Transform Transfer Function of Linear Systems Block Diagram
STEP AND PROCEDURE Define the system and its components Formulate the mathematical model and list
the necessary assumptions Write the differential equations describing
the model Solve the equations for the desired output
variables Examine the solutions and the
assumptions If necessary, reanalyze or redesign the
system
INTRODUCTIONS A mathematical model is a set of equations (usually
differential equations) that represents the dynamics of systems.
In practice, the complexity of the system requires some assumptions in the determination model.
The equations of the mathematical model may be solved using mathematical tools such as the Laplace Transform.
Before solving the equations, we usually need to linearize them.
DIFFERENTIAL EQUATIONS
Physical law of the process Differential Equation
Mechanical system (Newton’s laws)Electrical system (Kirchhoff’s laws)
How do we obtain the equations?
Examples:i. ii.
DIFFERENTIAL EQUATIONS Example: Springer-mass-damper system
Assumption: Wall friction is a viscous force.
The time function of r(t) sometimes called forcing function
Linearly proportional to the velocity
)()( tbvtf
DIFFERENTIAL EQUATIONS Example: Springer-mass-damper system
Newton’s 2nd Law:
)()()()( tMatrtkytbv
)()()()(2
2
trtkydt
tdybdt
tydM
DIFFERENTIAL EQUATIONS Example: RLC Circuit
t
tvdiC
tRidt
tdiL0
)()(1)()(
0)( cLR VVVtv
THE LAPLACE TRANSFORM The differential equations are transformed into
algebraic equations, which are easier to solve. The Laplace transformation for a function of time,
f(t) is:
If, , then,
Similarly,
Thus,
0
)}({)()( tfLdtetfsF st
dtdytf )( )0()}({)}({ ytysL
dtdyLtfL
dtdy
dttdysL
dttydL )0()()(2
2
dtdysytyLs
dttydL )0()0()}({)( 22
2
THE LAPLACE TRANSFORM Example: Spring-mass-damper dynamic equation
)()()()(2
2
trtkydt
tdybdt
tydM
)()()]0()([)]0()0()([ 2 sRskYyssYbysysYsM
0)()()( 002 skYbysbsYMsysYMs
Laplace Transform for the equation above:
When r(t)=0, y(0)= y0 and (0)=0:
y
)()()()( 2
0
sqsp
kbsMsybMssY
THE LAPLACE TRANSFORM Example: Spring-mass-damper dynamic equation
Some Definitionsi. q(s) = 0 is called characteristic
equation (C.E.) because the roots of this equation determine the character of the time response.
ii. The roots of C.E are also called the poles of the system.
iii. The roots of numerator polynomial p(s) are called the zeros of the system.
)()()()( 2
0
sqsp
kbsMsybMssY
THE LAPLACE TRANSFORM Transform table:
f(t) F(s)
1. δ(t) 1
2. u(t)
3. t u(t)
4. tn u(t)
5. e-at u(t)
6. sin t u(t)
7. cos t u(t)
s1
2
1s
1
!ns
n
as 1
22 s
22 ss
Impulse functionStep functionRamp function
THE LAPLACE TRANSFORM Transform
Properties
THE LAPLACE TRANSFORM Example: Find the Laplace Transform for the
following.
i. Unit function:
ii. Ramp function:
iii. Step function:
1)( tf
ttf )(
atAetf )(
THE LAPLACE TRANSFORM Transform Theorem
i. Differentiation Theorem
ii. Integration Theorem:
iii. Initial Value Theorem:
iv. Final Value Theorem:
ssFdfL
t )()(0
)(lim)(lim0
ssFtfst
)(lim)0( ssFft
)0()(})({ fssFdt
tdfL
)0()0()(})({ 22
2
fsfsFsdt
tfdL
THE LAPLACE TRANSFORM The inverse Laplace Transform can be obtained using:
Partial fraction method can be used to find the inverse Laplace Transform of a complicated function.
We can convert the function to a sum of simpler terms for which we know the inverse Laplace Transform.
j
j
stdsesFj
tf )(21)(
)()()()( 21 sFsFsFsF n
)()()()( 12
11
1 sFLsFLsFLtf n
)()()( 21 tftftf n
THE LAPLACE TRANSFORM We will consider three cases and show that F(s) can
be expanded into partial fraction:i. Case 1:
Roots of denominator A(s) are real and distinct.ii. Case 2:
Roots of denominator A(s) are real and repeated.iii. Case 3:
Roots of denominator A(s) are complex conjugate.
THE LAPLACE TRANSFORM Case 1: Roots of denominator A(s) are real and
distinct.Example:
Solution:
)2)(1(2)(
sssF
21)(
sB
sAsF It is found that:
A = 2 and B = -2
22
12
sstt eetf 222)(
THE LAPLACE TRANSFORM Case 1: Roots of denominator A(s) are real and
distinct.
Problem: Find the Inverse Laplace Transform for the following.
)2)(1(3)(
ssssF
THE LAPLACE TRANSFORM Case 2: Roots of denominator A(s) are real and
repeated.Example:
Solution:
2)2)(1(2)(
sssF
2)2(21)(
sC
sB
sAsF It is found that:
A = 2, B = -2 and C = -2
2)2(2
22
12
sss
ttt teeetf 22 222)(
THE LAPLACE TRANSFORM Case 3: Roots of denominator A(s) are complex
conjugate.Example:
Solution:
)52(3)( 2
sss
sF
52)( 2
ssCBs
sAsF
It is found that:A = 3/5, B = -3/5and C = -6/5
52
25353
2 sss
s
22 2)1(
)2)(21()1(5353
ss
s
THE LAPLACE TRANSFORM Case 3: Roots of denominator A(s) are complex
conjugate.Example:
Solution:
)52(3)( 2
sss
sF
)2sin212(cos
53
53)( ttetf t
THE LAPLACE TRANSFORM Problem: Find the solution x(t) for the following
differential equations.
i.
ii.
,023 xxx
,352 xxx
bxax )0(,)0(
bxax )0(,)0(
THE TRANSFER FUNCTION The transfer function of a linear system is the ratio of the
Laplace Transform of the output to the Laplace Transform of the input variable.
Consider a spring-mass-damper dynamic equation with initial zero condition.
)()()(
sInputsOutputsG
)()()()(2 sRskYsbsYsYMs
THE TRANSFER FUNCTION
The transfer function is given by the following.
kbsMssRsYsG
2
1)()()(
Y(s)R(s)kbsMs 2
1
THE TRANSFER FUNCTION Electrical Network Transfer Function
Component V-I I-V V-Q Impedance
Admittance
THE TRANSFER FUNCTION Problem: Obtain the transfer function for the following RC
network.
THE TRANSFER FUNCTION Problem: Obtain the transfer function for the following
RLC network.
Answer:
THE TRANSFER FUNCTION Mechanical System Transfer Function
Problem: Find the transfer function for the mechanical system below.
)(tukyybym
The external force u(t) is the input to the system, and the displacement y(t) of the mass is the output.
The displacement y(t) is measured from the equilibrium position.
The transfer function of the system.
BLOCK DIAGRAM A block diagram of a system is a practical representation
of the functions performed by each component and of the flow of signals.
Cascaded sub-systems:
Transfer Function G(s) Outpu
tInpu
t
BLOCK DIAGRAM Feedback Control System
BLOCK DIAGRAM Feedback Control System
Therefore,
The negative feedback of the control system is given by:Ea(s) = R(s) – H(s)Y(s)Y(s) = G(s)Ea(s)
)]()()()[()( sYsHsRsGsY
)()(1)(
)()(
sHsGsG
sRsY
BLOCK DIAGRAM Reduction Rules
BLOCK DIAGRAM Reduction Rules
BLOCK DIAGRAM Problem:
BLOCK DIAGRAM Problem:
FURTHER READING… Chapter 2
i. Dorf R.C., Bishop R.H. (2001). Modern Control Systems (9th Ed), Prentice Hall.
ii. Nise N.S. (2004). Control System Engineering (4th Ed), John Wiley & Sons.
THE END…
“The whole of science is nothing more than a refinement of everyday thinking…”