Transcript Chapter 3 Laplace Transforms 1. Standard notation in dynamics and control (shorthand notation) 2.

Chapter 3
Laplace Transforms
1. Standard notation in dynamics and control
(shorthand notation)
2. Converts mathematics to algebraic operations
3. Advantageous for block diagram analysis
Laplace Transform

L(f (t)) =  f (t)e -st dt
0
Chapter 3
Example 1:

a st 
 a a
L(a)=  ae dt   e   0     
0
s
 s s
0
-st


L(e )=  e e dt   e
-bt
-bt -st
0
 df
L(f )  L 
 dt
0
-(b+s)t

1
1
 ( b s)t
-e
 
dt 
0
b+s
s+b

df -st


  e dt  sL(f) f(0)
 0 dt
Usually define
f(0) = 0
(e.g., the error)
Other Transforms
Chapter 3
 d2 f 
df
 dφ 


L 2  = L
 w he re φ =
dt
 dt 
 dt 
 sφ (s )- φ (0)
= ss F(s )- f(0)  f (0)
= s2F(s )  s f(0) f (0)
d nf
etc. for
dt n
 e-jt  e  jt 

L(cosωt) = L
2


=
1 1
1 



2  s  jω s  jω 
1  s  jω
s  jω 



2  s2  ω2 s2  ω2 
s
= 2
s  ω2
=
 e  jt - e  jt 
L(sin ωt) = L 

2j


ω
= 2
s  ω2
Note:
Chapter 3
Table 3.1 Laplace Transforms for Various Time-Domain Functionsa
Chapter 3
f(t)
F(s)
Table 3.1 Laplace Transforms for Various Time-Domain Functionsa
Chapter 3
f(t)
F(s)
Table 3.1 Laplace Transforms for Various Time-Domain
Functionsa (continued)
f(t)
F(s)
Example 3.1
Chapter 3
Solve the ODE,
dy
5  4y  2
y 0  1
dt
First, take L of both sides of (3-26),
2
5  sY  s   1  4Y  s  
s
Rearrange,
5s  2
Y s 
s  5s  4 
Take L-1,
(3-26)
(3-34)
1 
5s  2 
y t   L 

 s  5s  4  
From Table 3.1 (line 11),
y  t   0.5  0.5e0.8t
(3-37)
Example:
d3y
d2y
dy
 6 2  11  6 y  4
3
dt
dt
dt
y( 0 )= y( 0 )= y( 0 )= 0
Chapter 3
system at rest (s.s.)
Step 1
Take L.T. (note zero initial conditions)
4
3
2
s Y(s)+ 6s Y(s)+11sY(s)  6Y ( s ) =
s
Rearranging,
Y(s)=
4
( s 3  6s 2  11s  6) s
Chapter 3
Step 2a. Factor denominator of Y(s)
s(s3+6s 2+11s+6 )=s(s+1 )(s+2 )(s+3 )
Step 2b. Use partial fraction decomposition
α
α
α
α
4
 1 2  3  4
s(s +1 )(s + 2 )(s + 3 ) s s  1 s  2 s  3
Multiply by s, set s = 0
α
α 
4
 α
 α1  s  2  3  4 
(s +1 )(s + 2 )(s + 3 ) s 0
 s  1 s  2 s  3  s 0
4
2
 α1 
1 2  3
3
Chapter 3
For a2, multiply by (s+1), set s=-1 (same procedure
for a3, a4)
2
α2  2, α3  2, α4  
3
Step 3. Take inverse of L.T.
(Y(s)=
2
2
2
2/3

+

)
3s
s 1 s  2
s3
2
2
 2e  t  2e 2t  e 3t
3
3
2
t 
y(t) 
t  0 y (0)  0. (check original ODE)
3
y(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
Laplace transforms can be used in process
control for:
Chapter 3
1. Solution of differential equations (linear)
2. Analysis of linear control systems
(frequency response)
3. Prediction of transient response for
different inputs
Factoring the denominator polynomial
Chapter 3
1.
2
3s 2  4 s  1
1
3s  4 s  1  (3s  1)( s  1)  3( s  )( s  1)
3
2
Transforms to e-t/3, e-t
Real roots = no oscillation
2.
2s
s2  s  1
Chapter 3
1
3
1
3
1
3
s 2  s  1  (s  
j )( s  
j)  (s  )2  ( )2
2 2
2 2
2
2
3
3
t , e0.5t cos
t
2
2
Complex roots = oscillation
0.5t
Transforms to e sin
L1[ 3(
3
2
1
3
(s  )2  ( )2
2
2
)(
1
s
2
1
3
(s  )2  ( )2
2
2
From Table 3.1, line 17 and 18

t
2
3
3
y(t )  e (sin( t )  3 cos( t ))
2
2
)]
Chapter 3
Let h→0, f(t) = δ(t) (Dirac delta)
L(δ) = 1
1  st
1
F ( s )   e dt  (1  e  hs ) Use L’Hopital’s theorem
0 h
hs
(h→0)
If h = 1, rectangular pulse input
h
Difference of two step inputs S(t) – S(t-1)
(S(t-1) is step starting at t = h = 1)
Chapter 3
By Laplace transform
1 e s
F ( s)  
s s
Can be generalized to steps of different magnitudes
(a1, a2).
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
Chapter 3
1
Y(s)= 2
s  3s  2
the denominator can be factored into (s+2)(s+1).
Using the partial fraction technique
α1
α2
Y(s)=

s  2 s 1
The step response of the process will have exponential terms
e-2t and e-t, which indicates y(t) approaches zero. However, if
Y(s)=
1
1

s 2  s  2 (s  1 )(s  2 )
We know that the system is unstable and has a transient
response involving e2t and e-t. e2t is unbounded for
large time. We shall use this concept later in the analysis
of feedback system stability.
Other applications of L( ):
A.
Final value theorem
y( )= lim sY(s)
Chapter 3
s 0
“offset”
Example 3: step response
1 a
τ s1 s
a
sY(s) 
τ s1
Y(s) 
lim
s 0
a
a
τ s1
offset (steady state error) is a.
B. Time-shift theorem
y(t)=0 t < θ
L yt- =e
-s
Y(s)
C. Initial value theorem
lim y(t)= lim sY(s)
Chapter 3
t 0
s 
4s+2
For Y(s)=
s(s+1 )(s+2 )(s+3 )
y( 0 )=0
y()=
1
3
by initial value theorem
by final value theorem
Chapter 3
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