Laplace Transforms
Important analytical method for solving linear ordinary
diff
differential
ti l equations.
ti
- Application to nonlinear ODEs? Must linearize first.
Laplace transforms play a key role in important process
control concepts and techniques.
- Examples:
Transfer functions
Frequency response
Control system design
Stability analysis
Definition
The Laplace transform of a function, f(t), is defined as
F ( s ) L f (t ) f t e st dt (3-1)
0
where F(s) is the symbol for the Laplace transform, L is the
Laplace transform operator, and f(t) is some function of time, t.
Note: The L operator transforms a time domain function f(t)
into an s domain function, F(s). s is a complex variable:
s = a + bj, j 1
Inverse Laplace Transform, L-1:
By definition, the inverse Laplace transform operator, L-1,
converts an s-domain
s domain function back to the corresponding time
domain function:
f t L1 F s
Important Properties:
Both L and L-1 are linear operators. Thus,
L ax t by t aL x t bL y t
aX s bY s (3-3)
(3 3)
where:
- x(t) and y(t) are arbitrary functions
- a andd b are constants
t t
- X ( s ) L[ x (t )] and Y ( s ) L[ y (t )]
Similarly,
L1 aX s bY s ax t b y t
Laplace Transforms of Common
Functions
i
1. Constant Function
Let f(t) = a (a constant).
constant) Then from the definition of the
Laplace transform in (3-1),
a a a
L a ae st dt e st 0 (3-4)
0 s s s
0
2. Step Function
The unit step function is widely used in the analysis of process
control problems. It is defined as:
0 for t 0
S (t ) (3 5)
1 for t 0
Because the step function is a special case of a constant, it
follows from (3-4) that
1
L S t (3-6)
s
3. Derivatives
This is a very important transform because derivatives appear
in the ODEs we wish to solve. In the text (p.53), it is shown
that
df
L sF s f 0 (3-9)
dt
i i i l condition
initial di i at t = 0
Similarly,
y for higher
g order derivatives:
dn f n n 2 1
s F s s f 0 s f 0
n 1
L n
dt
... sf 0 f 0
n2 n 1
(3-14)
(3 14)
where:
- n is
i an arbitrary
bi positive
i i integer
i
dk f
- f
(k )
(0) k
dt t 0
S i l Case:
Special C All Initial
I i i l Conditions
C di i are Zero
Z
f 0 f 0 ... f 0 . Then
1 n 1
S
Suppose Th
dn f n
L n s F s
dt
In process control problems, we usually assume zero initial
conditions Reason:
conditions. R This corresponds to the nominal steady state
when deviation variables are used, as shown in Ch. 4.
4. Exponential Functions
Consider f t e bt where b > 0.
0 Then
Then,
b s t
L e bt ebt e st dt e dt
0 0
1 b s t 1
e (3-16)
(3 16)
bs 0 sb
5 Rectangular
5. R t l Pulse
P l Function
F ti
It is defined by:
0 for t 0
f t h for
f 0 t tw (3
(3-20)
20)
0 for t t
w
f t
tw
Time, t
The Laplace transform of the rectangular pulse is given by
F s
h
s
1 e t w s (3-22)
6. Impulse Function (or Dirac Delta Function)
The impulse function is obtained by taking the limit of the
rectangular pulse as its width, tw, goes to zero but holding
1
the area under the pulse constant at one. (i.e., let h )
tw
Let, (t ) impulse function
Then, L t 1
S l ti off ODE
Solution ODEs by
b Laplace
L l Transforms
T f
Procedure:
1. Take the L of both sides of the ODE.
2. Rearrange
g the resultingg algebraic
g equation
q in the s domain to
solve for the L of the output variable, e.g., Y(s).
3 Perform a partial fraction expansion.
3. expansion
4. Use the L-1 to find y(t) from the expression for Y(s).
Table 3
3.1.
1 Laplace Transforms
See page 54 of the text.
Example 3.1
S l the
Solve th ODE,
ODE
dy
5 4y 2 y 0 1 (3-26)
(3 26)
dt
First, take L of both sides of (3-26),
2
5 sY s 1 4Y s
s
Rearrange,
Rearrange
5s 2
Y s (3-34)
s 5s 4
Take L-1,
5s 2
y t L1
s 5 s 4
From Table 3.1,
y t 0.5 0.5e0.8t (3-37)
Partial Fraction Expansions
Basic idea: Expand a complex expression for Y(s) into
simpler terms, each of which appears in the Laplace
table Then you can take the L-1 of both sides of
Transform table.
the equation to obtain y(t).
Example:
s5
Y s (3-41)
s 1 s 4
Perform a partial fraction expansion (PFE)
s5
1 2 (3-42)
s 1 s 4 s 1 s 4
where coefficients 1 and 2 have to be determined.
determined
To find 1 : Multiply both sides by s + 1 and let s = -1
s5 4
1
s4 s 1 3
To find 2 : Multiply both sides by s + 4 and let s = -4
s5 1
2
s 1 s 4 3
A General PFE
C id a generall expression,
Consider i
N s N s
Y s (3-46a)
Ds n
s bi
i1
Here D(s) is an n-th order polynomial with the roots s bi
all being real numbers which are distinct so there are no repeated
roots.
The PFE is:
N s n
i
Y s (3-46b)
i 1 s bi
n
s bi
i1
Note: D(s) is called the characteristic polynomial.
Special Situations:
Two other types of situations commonly occur when D(s) has:
i) Complex roots: e.g., bi 3 4 j ( j 1)
ii) Repeated
R t d roots
t (e.g.,
( b1 b2 3 )
For these situations, the PFE has a different form. See SEM
text (pp. 61-64) for details.
Example 3.2
For the ODE, y 6 y 11 y 6 y 1 , with zero initial conditions
resulted in the expression
1
Y s (3-40)
3 2
s s 6 s 11s 6
The denominator can be factored as
s s 3 6 s 2 11s 6 s s 1 s 2 s 3 (3-50)
Note: Normally, numerical techniques are required in order to
calculate the roots.
The PFE for (3-40) is
1
Y s 1 2 3 4 ((3-51))
s s 1 s 2 s 3 s s 1 s 2 s 3
Solve for coefficients to get
1 1 1 1
1 , 2 , 3 , 4
6 2 2 6
(For example,
(F findd , by
l fi b multiplying
lti l i both
b th sides
id byb s and
d then
th
setting s = 0.)
S b i
Substitute numerical
i l values
l into
i (3-51):
( )
1/ 6 1/ 2 1/ 2 1/ 6
Y (s)
s s 1 s 2 s 3
Take L-1 of both sides:
1/ 6 1 1/ 2 1 1/ 2 1 1/ 6
L1 Y s L1 L L L
s s 1 s 2 s 3
From Table 3.1,
1 1 t 1 2t 1 3t
y t e e e (3-52)
6 2 2 6
Important Properties of Laplace Transforms
1. Final Value Theorem
It ccan be used too findd thee steady-state
s e dy s e value
v ue of
o a closed
c osed loop
oop
system (providing that a steady-state value exists.
Statement of FVT:
lim y t lim sY s
t s 0
providing that the limit exists (is finite) for all
Re s 0, where Re (s) denotes the real part of complex
variable, s.
Example:
Suppose,
5s 2
Y s (3 34)
(3-34)
s 5s 4
Then,
Then
5s 2
y lim y t lim 0.5
t s 0 5 s 4
2. Transform
f off an Integral
g
L f (t*)dt * 1s F (s)
0
t
3. Time Delayy (Translation
( in Time))
Time delays occur due to fluid flow, time required to do an
analysis (e.g.,
(e g gas chromatograph).
chromatograph) The delayed signal fd(t)
can be represented as
C pter 3
f d (t ) f (t t 0 ) S (t t 0 ) t 0 time delay
Note: f d (t ) 0 for t t 0
Chap
Also,
L ( f d (t)) e st 0 F ( s )
Example:
Suppose, 1 e 2 s
Y ( s)
(4s 1)(3s 1)
21
Y ( s ) Y1 ( s ) Y2 ( s )
1 e 2 s
(4s 1)(3s 1) (4 s 1)(3s 1)
Therefore
C pter 3
y (t ) y1 (t ) y 2 (t )
e t 4 e t 3 [e (t 2) 4 e ( t 2 ) 3 ]S (t 2)
Chap
or
e t 4 e t 3 0t 2
y (t ) t 4 t 3
2.6487e 2.9477e t2
22
