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
Example 1:


a
 a a
L(a)=  ae-st dt   e st   0     
0
s
 s s
0


L(e )=  e e dt   e-(b+s)t dt 
-bt
-bt -st
0
 df
L(f )  L 
 dt
0

1
1
-e  ( b s)t  
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
 d2 f 
df
 dφ 
L 2  = L
 where φ =
dt
 dt 
 dt 
 sφ(s) - φ(0)
= s  sF(s) - f(0)  f (0)
= s2F(s)  sf(0)  f (0)
etc. for
d nf
dt n
 e-jt  e  jt 
L(cos ωt) = L 

2


=
1 1
1 



2  s  jω s  jω 
1  s  jω
s  jω 
 2
 2

2
2 s ω
s  ω2 
s
= 2
s  ω2
=
 e  j t - e  jt 
L(sin ωt) = L 

2j


ω
= 2
s  ω2
Let h→0, f(t) = δ(t) (Dirac delta)
F ( s)  
h
0
1  st
1
e dt  (1  e  hs )
h
hs
If h = 1, rectangular pulse input
Difference of two step inputs S(t) – S(t-1)
(S(t-1) is step starting at t = h = 1)
By Laplace transform
note :
e jt  cos t  j sin t
e  jt  cos t  j sin t
j  1
L(δ) = 1
1 e s
F ( s)  
s s
Can be generalized to steps of different magnitudes (a1, a2).
Laplace transforms can be used in process control for:
1. Solution of differential equations (linear)
2. Analysis of linear control systems
(frequency response)
3. Prediction of transient response for
different inputs
Please see Table 3.1 in Text
Example 3.1
Solve the ODE,
5
dy
 4y  2
dt
y 0  1
(3-26)
First, take L of both sides of (3-26),
5  sY  s   1  4Y  s  
Rearrange,
Y s 
2
s
5s  2
s  5s  4 
(3-34)
Take L-1,
 5s  2 
y  t   L1 

 s  5s  4  
From Table 3.1,
y  t   0.5  0.5e0.8t
(3-37)
Example 2:
d3y
d2y
dy
du
 6 2  11  6 y  4  2u
3
dt
dt
dt
dt
y( 0 )=y( 0 )=y( 0 )=0
du
 0 at t=0
dt
To find transient response for u(t) = unit step at t > 0
1. Take Laplace Transform (L.T.)
2. Factor, use partial fraction decomposition
3. Take inverse L.T.
Step 1
Take L.T. (note zero initial conditions)
s3Y(s)+ 6s 2Y(s)+11sY(s)  6Y ( s) = 4sU(s) + 2U(s)
Rearranging,
4 s+2
1

2
s  6 s  11s  6 s
1
U(s)  (unit step input)
s
Y(s)=
3
Step 2a. Factor denominator of Y(s)
s(s 3+6s 2+11s+6 )=s(s+1 )(s+2 )(s+3 )
Step 2b. Use partial fraction decomposition
α
4s+2
α
α
α
 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
4 s+2
α 
 α
 α1  s  2  3  4 
(s+1 )(s+2 )(s+3 ) s 0
 s  1 s  2 s  3  s 0
2
1
 α1 
1 2  3
3
For 2, multiply by (s+1), set s=-1 (same procedure for 3, 4)
α2  1, α3  3, α4 
5
3
Step 3. Take inverse of L.T.
(Y(s)=
1
1
3
5/3



)
3s s  1 s  2 s  3
1
5
y(t)=  e t  3e  2t  e 3t
3
3
1
t 
y(t ) 
3
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
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
1
s 2  3s  2
Y(s)=
the denominator can be factored into (s+2)(s+1).
Using the partial fraction technique
Y(s)=
α1
α
 2
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  s  2 (s  1 )(s  2 )
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)
“offset”
s 0
Example 3: step response
1 a
τs  1 s
a
sY(s) 
τs  1
Y(s) 
lim
s 0
a
a
τs  1
B. Time-shift theorem
y(t)=0 t < to
L yt- =e-s Y(s)
C. Initial value theorem
lim y(t)= lim sY(s)
t 0
s 
For Y(s) =
y( 0 )=0
y()=
1
3
4s+2
s(s+1 )(s+2 )(s+3 )
By initial value theorem
By final value theorem