Laplace Transforms
1. Standard notation in dynamics and control
(shorthand notation)
2. Converts mathematics to algebraic operations
3. Advantageous for block diagram analysis
Chapter 3
Docsity.com
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Guidelines and tips
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Laplace Transforms - Process Control - Lecture Slides, Slides of Process Control
This lecture is from Process Control course. Some key points for this lecture are: Laplace Transforms, Standard Notation, Dynamics, Shorthand Notation, Converts Mathematics, Algebraic Operations, Advantageous, Block Diagram Analysis, Unit Step Function, Difference
Typology: Slides
2012/2013
1 / 25
This page cannot be seen from the preview
Don't miss anything!
Related documents
Partial preview of the text
Download Laplace Transforms - Process Control - Lecture Slides and more Slides Process Control in PDF only on Docsity!
Laplace Transforms
- Standard notation in dynamics and control
(shorthand notation)
- Converts mathematics to algebraic operations
- Advantageous for block diagram analysis
Chapter 3
∫
∞
0
f ( t ) = f (t)e dt
-st L
Laplace Transform
Example 1:
0 0
- -( ) ( )
0 0 0
0
( ) 0
1 1 ( ) -
( ) 0
st st
bt bt st b s t b s t
st
a a a a ae dt e s s s
e e e dt e dt e b s s b
df df f e dt s (f) f( ) dt dt
∞ ∞ −
∞ ∞ (^) ∞
- − +
∞
= = − ^ = − − =
= = = ^ =
′ = (^) = = −
∫
∫ ∫
∫
L
L
L L L
Usually define f(0) = 0 (e.g., the error)
Chapter 3
{ }
2 2
2 2 2 2
cos
s ω
s
s ω
s jω
s ω
s jω
s jω s jω
e e
t =
-j t j t
ω + ω
L ω L
Note that
cos sin
cos sin
j t
jwt
e t j t
e t j t
j
ω ω ω
ω ω
−
{ }
2 2
sin
2
j t j t e e t
j
s
ω ω
ω
ω
ω
- − (^) −
= (^)
=
L L
Chapter 3
Difference of two step inputs S(t) – S(t-1)
Note that S(t-1) is the step starting at t = 1.
By Laplace transform
1 ( )
s e F s s s
−
= −
Can be generalized to steps of different magnitudes
(a 1 , a 2 ).
Chapter 3
0
1 1 ( ) (1 )
h st hs F s e dt e h hs
− − = = − ∫
Let h→0, f(t) = δ(t) (Dirac delta)
Laplace transforms can be used
in process control for:
- Solution of differential equations (linear)
- Analysis of linear control systems
( frequency response )
- Prediction of transient response for
different inputs
Chapter 3
- Chapter
Example 3.
Solve the ODE,
dy y y dt
- = =
First, take L of both sides of (3-26),
( ( ) ) ( )^
2 5 sY s 1 4 Y s s
− + =
Rearrange,
5 2 (3-34) 5 4
s Y s s s
=
Take L-1^ ,
1 5 2
5 4
s y t s s
−
(^) + = (^)
L
From Table 3.1,
0.5 0.5 (3-37)
t y t e
− = +
Chapter 3
Example 2:
system at rest (s.s.)
Step 1 Take L.T. (note zero initial conditions)
0 0
0 0 0 0
6 2 11 6 4 2
2
3
3
at t= dt
du
y( )=y( )=y ( )=
u dt
du y dt
dy
dt
d y
dt
d y
=
′ ′′
- = +
3 2 s Y(s)+ 6 s Y(s)+ 11 sY(s) + 6 Y s = ( ) 4 sU(s) + U(s) 2
Chapter 3
To find transient response for u(t) = unit step at t > 0
- Take Laplace Transform (L.T.)
- Factor, use partial fraction decomposition
- Take inverse L.T.
For α 2 , multiply by (s+1), set s=-1 (same procedure
for α3, α 4 )
3
5 α 2 = 1 , α 3 =− 3 , α 4 =
3
1
3
5 3 3
(^1 )
→∞ →
- − +
− − −
t y(t )
y(t)= e e e
t t t
Step 3. Take inverse of L.T.
You can use this method on any order of ODE,
limited only by factoring of denominator polynomial
(characteristic equation)
Must use modified procedure for repeated roots,
imaginary roots
Chapter 3
) 3
5 / 3
2
3
1
1
3
1 (
−
s s s s
Y(s)=
One other useful feature of the Laplace transform
is that one can analyze the denominator of the transform
to determine its dynamic behavior. For example, if
the denominator can be factored into (s+2)(s+1).
Using the partial fraction technique
The step response of the process will have exponential terms
e-2t^ and e-t^ , which indicates y(t) approaches zero. However, if
We know that the system is unstable and has a transient
response involving e 2t^ and e -t^. e 2t^ is unbounded for
large time. We shall use this concept later in the analysis
of feedback system stability.
3 2
1
2 s + s +
Y(s)=
2 1
1 2
s
α
s
α Y(s)=
s s (s )(s )
Y(s)= 1 2
1
2
1 2
- −
= − −
Chapter 3
Linearity
{ ( )^ ( )}
where and are constants.
af t bg t
a f t g t
aF s bG s
a b
L
L + bL
Sifting Theorems
( ) ( ) ( )
{ ( )} { ( )}
{ ( )} (^ )
0 0
0 0
(a) Dead time
( )
(b) Multiplication by
t s t s
bt
bt
g t f t t S t t
g t e f t e F s
e
e f t F s b
− −
−
= − −
= =
= +
L L
L