Class 2– Topics of discussion
 Mathematical modelling
 Laplace transform, its inverse and few problems
 Transfer function
 Impulse response
Mathematical modelling of control systems
 Differential equation- Describe the input and output relationship of a system difficult to be modelled as a block
diagram (SISO system)
 Objective - Develop the mathematical model
Step 1: Apply fundamental physical laws of science and engineering
Example- Electrical networks  Ohm’s laws and Kirchhoff’s current and voltage laws
Mechanical systems Newton’s laws
Step 2: Obtain the mathematical model
Mathematical model – a) transfer function in frequency domain
b) state space equations in time domain
Mathematical modelling of control systems
 Mathematical representation as shown in Fig 6(a) where input, output, and system are distinct
and separate parts
 System and subsystems in each block will be represented by transfer function blocks
Fig 6: Block diagram representation
Mathematical modelling of control systems
 Mathematical model is linear  obeys principle of superposition and homogeneity
r1(t)
G
r2(t)
G
c1(t)
r1(t)
a1
c2(t)
r2(t)
a2
a1r1 (t)
a2r2 (t)
a1r1(t)+a2r2(t)
+
+
c3 (t)
G
If c3 (t)=a1c1(t)+a2c2(t)G linear
Mathematical modelling of control systems
 Coefficients of the differential equation constant  Model is linear time invariant
 Coefficients of the differential equation functions of time  Model is linear time varying
Laplace transform of a function f(t)
If f(t) has impulse function then lower limit starts from 0- otherwise it starts from 0
Inverse Laplace transform
Laplace Transform Table
Laplace Transform Theorems
Problems
Problem 1: Find the Laplace transform of f(t)= Ae-atu(t)
Solution:
Problem 2: Find the inverse Laplace transform of F1(s)= 1/(s+3)2
ILT
tu(t)
ILT
Therefore, f1(t)=e-3ttu(t)
e-attu(t)
Inverse Laplace transform by partial fraction expansion
Step 1: Write F1(s) = N(s)/D(s)
Step 2: Check the order of N(s) and D(s)
a) Order of N(s) < order of D(s)  partial fraction expansion
b) Order of N(s) ≥ order of D(s) First divide N(s) by D(s) till result has numerator
order < denominator order  perform partial fraction expansion
Step 3: Perform Inverse Laplace transform on each term
 Example:
Solution:
Step 2: Here order of N(s) >order of D(s) divide N(s) by D(s) till result has numerator
order < denominator order
 Perform long division
Final form is
 Step 3: Find Inverse Laplace transform
Differentiation theorem
Using differentiation theorem and linearity theorem
Linearity theorem
LT
LT
1
s-f(0-)
Homework
Using partial fraction, expand the functions like F(s) = 2/(s2+s+5) into sum of terms and find the ILT for each term
Three cases: Case 1- Roots of the Denominator of F (s) are Real and Distinct
Case 2- Roots of the Denominator of F (s) are Real and Repeated
Case 3 - Roots of the Denominator of F (s) are Complex or Imaginary
Case 1- Roots of the Denominator of F (s) Are Real and Distinct
 Problem1 :
Solution:
Step 1: Writing the above in partial fraction
Roots s= -1, -2
 (1)
Step 2: Find K1 and K2
a) Find K1  Multiply Eq. (1) on both sides by (s+1) and let s=-1 and finally, K1 =2
b) K2  Multiply (1) by (s+2) and let s=-2 and finally, K2 =-2
Step 3: ILT of F(s) = f(t)= (2e-t -2e-2t)u(t)
Case 1- Roots of the Denominator of F (s) Are Real and Distinct
Problem 2 : Solve for y(t) if all the initial conditions are zero
--- (1)
Solution : Step 1: Find Y(s)
Step 2: Write in partial fraction expansion
Step 3: Find K1, K2 and K3
Step 4: Find ILT of Y(s)
Case 2- Roots of the Denominator of F (s) Are Real and Repeated
Example:
Solution:
(s+2)2 Root at s=-2 is a multiple root of multiplicity 2
Step 1: Write in partial fraction expansion
Step 2: Find K1, K2 and K3
Step 3: Find ILT of F(s)
Case 3- Roots of the Denominator of F (s) Are Complex or Imaginary
Example:
Solution:
Step 1: Write in partial fraction expansion
---- (18)
(s2+2s+5) = (s+1+2j)(s+1-2j)
Roots s= -1-2j, -1+2j --> complex
-----(19)
Step 4 :Simplify the second term in Eq. (22)
Denominator
s2+2s+5= (s+1)2+22  a=1, ω=2
Numerator
-3/5(s+2)  -3/5 [(s+1)+(1/2)2]
 A= 1, B=1/2
Step 5: Find the ILT of F(s)
Transfer function
 A function which allows separation of input, system and output to three distinct parts
 nth order linear time invariant differential equation
◦
-----------
(34)
Where c(t) is the output, r(t) is the input, ai’s and bi’s are the system parameters
 Taking Laplace transform on both sides of Eq. (34)
-----------
(35)
Transfer function
 Assuming all the initial conditions = 0, then Eq. (35)
Transfer function = Output transform, C(s)/ Input transform, R(s) with zero initial conditions
◦
 Block diagram of transfer function
-----------
(36)
Transfer function
 Output C(s) can be found by
Problem1: Find the transfer function represented by
 Solution:- Step 1 : Taking LT on both sides
sC(s)+2C(s)=R(s)
Step 2: Find Transfer function G(s)
Transfer function
Problem2: Find the response c(t) of the system G(s)= 1/(s+2) to an input r(t)=u(t), assuming zero initial
conditions
Solution: Step 1: Find the C(s)
C(s)= G(s)R(s) =
Step 2: Use partial fraction expansion
= K1/s + K2 /(s+2)
Step 3: Find K1 and K2
Transfer function
Step 4: Find C(s) from partial fraction expansion
Step 5: Take the inverse Laplace transform of each term of C(s)
Homework problems
Problem 3: Find transfer function G(s)=C(s)/R(s) corresponding to the differential equation
Problem 4: Find the differential equation corresponding to the transfer function
Summary
 Mathematical model of a system
Laplace transform and its inverse
Concept of partial fraction expansion applied to solution of differential equation
 Definition of Transfer function and few problems