Control System - Laplase transform, Transfer function

Control Systems
L H E Wijesinghe
BEng (Hons) in Electronic Engineering (SHU-UK)
MSc in Applied Electronics (University of Colombo – SL )

Learning Outcome
i. Laplace transform
ii. Transfer Functions
iii. Block Diagrams and Block Diagram Reduction
iv. Signal Flow graphs

Definition
• Suppose f(t) is a real valued function defined on the interval
(0,Infinity).
• L(f(t)) is called the Laplace Transform of the f(t) and is
denoted by
F(s) = L(f(t))

Theorem 1
L (e at
) =
F(s) = L(f(t)) = 0∫e –st
f(t) .dt
f(t) = e at
F(s) = 0∫e –st
e at
.dt = 0∫e (a-s)t
dt = [e (a-s)t
/ (a-s)] 0
=[e –t(-a+s)
/ (a-s)] 0
=[(0-1)/(a-s)]
=[(-1)/(a-s)]
So for a given constant a
L(e at
) =where s>a
𝑥 ⇒−∝, 𝑦 ⇒ 0 𝑥=0 , 𝑦=1

Time Domain vs. Laplace Domain

Theorem 2
L[C1f1(t) + C2f2(t) ] = C1L[f1(t)] + C2L[f2(t)]
If C1 and C2 are two arbitrary constants
F(s) = L(f(t)) = 0∫e –st
f(t) .dt
f(t) = [C1f1(t) + C2f2(t) ]
F(s) = L(f(t)) = 0∫e –st
[C1f1(t) + C2f2(t) ].dt
=C1[0∫e –st
f1(t).dt] + C2[0∫e –st
f2(t) .dt]
L[f1(t)] L[f2(t)]
L[C1f1(t) + C2f2(t) ] = C1L[f1(t)] + C2L[f2(t)]

Theorem 3
L[a] = if a = Constant
F(s) = L(f(t)) = 0∫e –st
a .dt
= a 0
α
∫e –st
.dt
=a [e –st
/ -s] 0
=a [0-1 /-s]
= a/s

Theorem 4
L[e at
f(t)] = F(s-a) if L(f(t)) = F(s)
L(e at
f(t)) = 0∫e –st
e at
f(t) .dt
= 0∫e –(s-a)t
f(t) .dt
If F(s) = 0∫e –st
f(t) .dt
Substitute s s-a
F(s-a) = 0∫e –(s-a)t
f(t) .dt
= L(e at
f(t))

Theorem 5
L[] = SF(s) – f(0) if L(f(t)) = F(s)
L() = 0∫e –st
dt
= e –st
f(t) |0 + s[0∫e –st
f(t).dt]
F(s)
=(0 f(α) - 1f(0) ) +sF(s)
= sF(s)- f(0)

Theorem 6
L[] = SF(s) –Sf(0) – f’(0) if L(f(t)) = F(s)
() = 0∫e –st
.dt
= e –st
f’(t) |0 + s[0∫e –st
f’(t).dt]
=[0 f’(α) - 1f’(0)] +s[sF(s) –f(0)]
= sF(s)- sf(0) –f’(0)

Note
L(f(t)) = F(s)
L[] = SF(s) – f(0)
L[] = SF(s) –Sf(0) – f’(0)
L[] = SF(s) –Sf(0) – Sf’(0) – f’’(0)
L[] = SF(s) –Sf(0) – Sf’(0) – Sf’’(0) –f’’’(0)
L[] = SF(s) –Sf(0) – Sf’(0) …….– Sf(0) –f(0)

Theorem 7
L[0
t
∫f(t) dt ] = if L(f(t)) = F(s)
g(t) = 0
t
∫f(t).dt
=f(t)
If L{g(t)} =G(s)
L{} =sG(s)-g(0) = sG(s)
L{f(t)} =L{} =sG(s) =s L{g(t)} = sL{0
t
∫f(t).dt}
L{0
t
∫f(t).dt} = =

Partial Fraction
Linear Factors
Repeated Factors
Quadratic Factors

If the highest degree of the numerator is same or
greater than the highest degree of the denominator.
• When highest degree of the numerator - highest degree of
the denominator =+0
= + + C
• When highest degree of the numerator - highest degree of
the denominator =+1
= + + C+ D

Laplace Transform of a
Unit function
F(s) = L(u(t)) = 0∫e –st
u(t) dt
= 0∫e –st
1dt
=[e –st
/ -s] 0
=[0-1 /-s]
=

Ramp function
F(s) = L(r(t)) = 0∫e –st
r(t) dt
= 0∫e –st
t dt
If v=t , = e –st
then u= , =1
= x t - 0∫ x 1 dt
=[0-1 /-s]
=

Impulse function
F(s) = L(δ(t) ) = 0∫e –st
δ(t) dt
δ(t)
L(δ(t) )= L{}
=s U(s) – u(0)
= s x - 0
= 1

Inverse Laplace Transform
If output Laplace transform of a system
Y(s) = find the invers Laplace trans form of Y(s).
Step 1 : Initially find the partial fractions of the Y(s).
Y(s) = + = +
Recall B =
y(t) = a + b  Inverse Laplace transform of Y(s)

Example 1
Solve + = 3 where =0
L{ +L{ = L{3}
SF(s) – + F(s) =
F(s) = =
F(s) =
Recall B =
3 U-3 =
When t>0; U=1
3-3 =
0 3=A(s+1) + B(s)
A=3,B=-3

Example 2
The rotational velocity is Ѡ(s) = * , when there is an unit step input
to the system.
Expanding in a partial fraction expansion yields
Ѡ(s) = = + - -
Taking the inverse Laplace transform yields
Recall L{u(t)} = , = , b =
ω(t) = + - -
When t≥0 ; u(t)=1
ω(t) = + - - ; t≥0

Conversion of system equation into
transfer function

Example -:
Derive a model for the mechanical system represented by the
system of mass, spring and dashpot given in the Figure . The
input to the system is the force f and the output is the
displacement y.

According to the free body diagram that indicates forces acting on the mass
Net force  = f – Dashpot reaction force on the mass – Spring reaction force on the mass.
according to the Newton's second law
Net force  =resistance force on the mass due the inertia = ma = m
Dashpot force  = c
Spring force  = k y
f– Dashpot reaction force on the mass – Spring reaction force on the mass = Resistance force on the mass
due the inertia
f - c - k (y) = m
f = m + c + k (y)
The relationship between the input F to the system and the output y is thus described by the
second-order differential equation.

m + c + k (y)}=L{f}
Laplace Transform of the function
= m + c ()+ k
Initial Conditions -:By imposing zero-input zero-output initial conditions
= m + c ()+ k
= m + c ()+ k = [m + c +k]
Resulting Expression -: Transfer function
=
0 0 0

System
• Described by the box outline within the boundary
• Inputs to the system are shown by arrows entering
the box
• Outputs to the system are shown by arrows leaving
the box.

Block
It is most efficient way to describe a process.

Basic Elements
1. Open Loop Control System
It is the one in which the control action is independent of
the output.

Important Features of an Open Loop
System
• Their ability to perform accurately is determined by
their calibration.
• There no problem related to instability.

Basic Elements
2. Close Loop Control System

Comparison Element (Summing Junction)
• Calculate the error or the difference between
desired and the actual outputs.
• It is a part of a Controller.

Actual Output (Controlled Variable)
The quantity or condition that is measured
and controlled.

Desired Output
(Control Signal or Manipulated Variable)
Is the quantity or condition that is varied
by the controller so as to affect the value of
the controlled variable.

Controller
Takes the error (difference between desired and
the actual outputs) into account to create a control
signal in a way to minimize the error.

Process
It Is the process to be controlled.
Feedback
Feed the plant’s actual output to the
controller for the purpose of error calculation.

Important Features of a Close Loop
System
•Reduced effect of nonlinearities and distortions
•Increased accuracy
•Increased bandwidth
•Reduced sensitivity of the ratio of the output to
input to variations in system characteristics
•Tendency toward oscillation or instability .

Block Diagram Application
A Temperature Control System

• Desired output / input -: Desired water temperature.
• Actual output / output -: Actual water temperature
• Control input -: difference between actual and desired water
temperature
• Control output/Process Input -: control signal Burner valve
position according to the error
• Process Output / Output -: Actual water temperature after
changing the burner valve position according to control signal
generated due to previous error.
• Feedback/sensor -: The thermocouple sensor.

Control System - Laplase transform, Transfer function

Transfer Function
• The term gain to relate the input and output of a system
with gain
• When we are working with inputs and outputs described as
functions of we define the transfer function as when all
initial conditions before we apply the input are zero.

Example
Derive the transfer function of the system is equal to

Control System - Laplase transform, Transfer function

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