IPC - Lectures 16-18 (Laplace Transform).pdf

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1. 1
Instrumentation and Process
Control (CHE 323)
Lectures 16-18: Laplace Transforms

2. •A number of mathematical modelsthat describe the
dynamic operation of selected processeswere
developed in previous lectures.
•Solving such models—that is, finding the output
variables as functions of timefor some change in the
input variable(s)—requires either analyticalor
numerical integrationof the differential equations.
Sometimes considerable effort is involved in obtaining the
solutions.
•One important class of models includes systems
described by linear ordinary differential equations (ODEs).
•Such linear systems represent the starting point for many
analysis techniques in process control.

3. Laplace Transforms
•Laplace transformcan significantly reduce the effort
required to solve and analyzethe linear differential
equation models.
•Transformationconverts ordinary differential
equationsto algebraic equations, which can simplify
the mathematical manipulations required to obtain a
solution or perform an analysis.

4. Laplace Transforms of Representative
Functions
The Laplace transform of a function f(t) is defined as
•where F(s) is the symbol for the Laplace transform, s is a complex
independent variable, f(t) is some function of time to be transformed,
andis an operator, defined by the integral.
•The function f(t) must satisfy mild conditions that include being
piecewise continuous for 0 < t <∞; this requirement almost always
holds for functions that are useful in process modeling and control.
•When the integration is performed, the transform becomes a function
of the Laplace transform variable s. The inverse Laplace transform
operates on the function F(s) and converts it to f(t).

5. Laplace Transforms of Representative
Functions
Laplace transform and the inverse Laplace transform are
linear operators:
Wheredenotes a particular operation to be performed,
such as differentiation or integration with respect to time. If
, then Eq. 3-2 becomes:
Therefore, the Laplace transform of a sum of functions x(t)
and y(t) is the sum of the individual Laplace transforms,
X(s) and Y(s)

6. 6
The unit step functionis an important input that is used
frequently in process dynamics and control.
The Laplace transform of the unit step function is:
If the step magnitude is a, the Laplace transform is a/s.

7. 7
Derivatives.The transform of a first derivative of f is
important because such derivatives appear in dynamic
models:
where F(s) is the Laplace transform of f(t).
Model solutions are most easily obtained assuming that
time starts (i.e., t = 0) at the moment the process model is
first perturbed. For example, if the process initially is
assumed to be at steady stateand an input undergoes a
unit step change, zero time is taken to be the moment at
which the input changes in magnitude.

8. 8
•In many process modeling applications, functions
are defined so that they are zero at initial time—that
is, f(0) = 0.
•In these cases, Eq. 3-9 simplifies to
•The Laplace transform for higher-order derivatives can
be found by repeated application of Eq. 3-9:

9. 9
Exponential Functions.The Laplace transform of an
exponential function is important because exponential
functions appear in the solution to most linear differential
equations. For an exponential, e
−bt
, with b > 0,

10. Example 1:-st st
0
0
-bt -bt -st -(b+s)t ( s)t
0
00
-st
0
a a a
L(a)= ae dt e 0
s s s
11
L(e )= e e dt e dt -e
b+s s+b
df df
L(f ) L e dt
dt dt
b








     
 

  


  





Usually define f(0) = 0 (e.g., the error)f(0)sL(f)

11. Example 3.1
Solve the ODE,5 4 2 0 1 (3-26)
dy
yy
dt
  
First, take L of both sides of (3-26), 
2
5 1 4sY s Y s
s
  
Rearrange,
 
52
(3-34)
54
s
Ys
ss



Take L
-1
,
 
152
54
s
yt
ss





L
From Table 3.1 (line 11),
0.8
0.5 0.5 (3-37)
t
y t e



12. Other Transforms
etc. for 
(0)fsf(0)F(s)s =
(0)ff(0)-sF(s)s =
φ(0) - φ(s)
dt
df
=φ where
dt
dφ
L=
dt
fd
L
2
2
2

















s 22
2222
ωs
s
=
ωs
jωs
ωs
jωs
2
1
=
jωs
1
jωs
1
2
1
=






















 




j t j t
22
e - e
L(sin ωt) =L
2j
ω
=
sω n
n
dt
fd 








2
ee
L=ωt) L(cos
tjt-j
Note:

13. Example 3.2
system at rest (s.s.)
Step 1Take L.T. (note zero initial conditions)32
32
6 11 6 4
0 0 0 0
d y d y dy
y
dt dt dt
y( )= y ( )= y ( )=
   
  32 4
6 11 6 ( )s Y(s)+ s Y(s)+ sY(s) Y s =
s


14. Rearranging,
Step 2a. Factor denominator of Y(s)
Step 2b. Use partial fraction decomposition
Multiply by s, set s = 032
4
( 6 11 6)
Y(s)=
s s s s   ))(s+)(s+)=s(s+s++s+s(s 3216116
23 31 2 4
4
1 2 3 1 2 3
αα α α
s(s+ )(s+ )(s+ ) s s s s
   
   324
1
00
1
4
1 2 3 1 2 3
42
1 2 3 3
ss
ααα
αs
(s+ )(s+ )(s+ ) s s s
α


   

  



15. For a
2, multiply by (s+1), set s=-1 (same procedure
for a
3, a
4)2 3 4
2
22
3
α , α , α    
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)2 2 2 2/3
( + )
3 1 2 3
Y(s)=
s s s s

  

17. Table 3.1 Laplace Transforms for Various Time-Domain Functions
a
f(t) F(s)

18. Table 3.1 Laplace Transforms for Various Time-Domain Functions
a
f(t) F(s)

19. Table 3.1 Laplace Transforms for Various Time-Domain
Functions
a
(continued)
f(t) F(s)