Laplace Transforms

Laplace Transform

Luma Naji Mohammed Tawfiq

Why use Laplace Transforms?

Find solution to differential equation using algebra  Relationship to Fourier Transform allows easy way to characterize systems  No need for convolution of input and differential equation solution  Useful with multiple processes in system 

How to use Laplace Find differential equations that describe system  Obtain Laplace transform  Perform algebra to solve for output or variable of interest  Apply inverse transform to find solution 

What are Laplace transforms? 

F(s)  L{f ( t )}   f ( t )e st dt 0   j

1 1 st f ( t )  L {F(s)}  F ( s ) e ds  2j   j     

t is real, s is complex!

Inverse requires complex analysis to solve Note “transform”: f(t)  F(s), where t is integrated and s is variable Conversely F(s)  f(t), t is variable and s is integrated Assumes f(t) = 0 for all t < 0

Evaluating F(s) = L{f(t)}  let

Hard Way – do the integral f (t)  1 

1 1 F(s)   e st dt   (0  1)  s s 0

let

f ( t )  e at 



0

0

F(s)   e at e st dt   e ( s  a ) t dt 

let

1 sa

f ( t )  sin t 

F(s)   e st sin( t )dt 0

Integrate by parts

Evaluating F(s)=L{f(t)}- Hard Way  udv  uv   vdu

remember let

u  e  st , du  se st dt

dv  sin( t )dt , v   cos( t )   st st   e sin( t )dt  [e cos( t ) ]  s  e st cos( t )dt  

0

0

0 

 e (1)  s  e st cos( t )dt st



 se



sin( t )dt  1  s

2

st e  sin( t )dt  0



u  e  st , du  se st dt

(1  s 2 )  e st sin( t )dt 1

dv  cos( t )dt , v  sin( t )



0 st e  sin( t )dt 



  e st cos( t )dt 

0

0

[ e

st

0

0

let

Substituting, we get:

st





sin( t ) ]  s  e 0

0

 st



sin( t )dt  e (0)  s  e st

0

st

sin( t )dt

1 1  s2

It only gets worse…

Evaluating F(s) = L{f(t)} This is the easy way ...  Recognize a few different transforms  See table 2.3 on page 42 in textbook  Learn a few different properties  Do a little math 

Table of selected Laplace Transforms 1 f ( t )  u ( t )  F(s)  s 1 f ( t )  e u ( t )  F(s)  sa  at

s f ( t )  cos( t )u ( t )  F(s)  2 s 1 1 f ( t )  sin( t )u ( t )  F(s)  2 s 1

More transforms f ( t )  t u ( t )  F(s) 

n!

n

s

n 1

0! 1 n  0, f ( t )  u ( t )  F(s)  1  s s 1! n  1, f ( t )  tu ( t )  F(s)  2 s 5! 120 n  5, f ( t )  t 5 u ( t )  F(s)  6  6 s s

f ( t )  ( t )  F(s)  1

Note on step functions in Laplace 

Unit step function definition: u ( t )  1, t  0 u ( t )  0, t  0



Used in conjunction with f(t)  f(t)u(t) because of Laplace integral limits: 

L{f ( t )}   f ( t )e dt 0

st

Properties of Laplace Transforms Linearity  Scaling in time  Time shift  “frequency” or s-plane shift  Multiplication by tn  Integration  Differentiation 

Properties: Linearity

L{c1f1 ( t )  c 2f 2 ( t )}  c1F1 (s)  c 2 F2 (s) Example : L{sinh( t )} 

Proof :

1 t 1 t y{ e  e }  2 2 1 1 L{e t }  L{e  t }  2 2 1 1 1 (  ) 2 s 1 s 1 1 (s  1)  (s  1) 1 ( )  2 s2 1 s2 1

L{c1f1 ( t )  c 2 f 2 ( t )}  

st [ c f ( t )  c f ( t )] e dt  2 2  11 0 



0

0

c1  f1 ( t )e st dt  c 2  f 2 ( t )e st dt  c1F1 (s)  c 2 F2 (s)

Properties: Scaling in Time 1 s L{f (at )}  F( ) a a Example : L{sin( t )} 1 1 (  1)  2 s  ( )  1 2 ( 2 ) 2  s   s 2  2

Proof :

L{f (at )}  

st f ( at ) e dt   0

let

u  at , t   a

u 1 , dt  du a a s

( ) u 1 f (u )e a du   a0

1 s F( ) a a

Properties: Time Shift

L{f ( t  t 0 ) u ( t  t 0 )}  e  a ( t 10) u ( t  10)}  Example : L{e e 10s sa

Proof :

 st 0

F(s)

L{f ( t  t 0 )u ( t  t 0 )}  

st f ( t  t ) u ( t  t ) e dt  0 0  0 

st f ( t  t ) e dt  0 

t0

let

u  t  t0, t  u  t0 t0

s ( u  t 0 ) f ( u ) e du   0

e

st 0



st 0 su f ( u ) e du  e F(s)  0

Properties: S-plane (frequency) shift  at

L{e f ( t )}  F(s  a ) Example : L{e  at sin( t )}   (s  a ) 2  2

Proof :

L{e  at f ( t )}  

 at st e f ( t ) e dt   0



(s  a ) t f ( t ) e dt   0

F(s  a )

Properties: Multiplication by tn n n n d L{t f ( t )}  (1) F ( s ) n ds Example :

Proof :

L{t n u ( t )}  (1) n n! s n 1



L{t n f ( t )}   t n f ( t )e st dt  0

n

d 1 ( ) n ds s



n st f ( t ) t e dt   0 

n  (1) n  f ( t ) n e st dt  s 0 

n n st n  (1) f ( t )e dt (1) F(s) n  n s 0 s n

The “D” Operator 1.

2.

Differentiation shorthand

Integration shorthand t

if

g ( t )   f ( t )dt 

then

Dg ( t )  f ( t )

df ( t ) Df ( t )  dt d2 2 D f (t)  2 f (t) dt

t

if g( t )   f ( t )dt a 1 then g( t )  D a f ( t )

Properties: Integrals

F(s) L{D f ( t )}  s 1 0

Proof :

g ( t )  D 01f ( t ) 

L{sin( t )}   g ( t )e st dt

Example : L{D 01 cos( t )} 

0

let u  g( t ), du  f ( t )dt

1 s 1 ( )( 2 )  2 s s 1 s 1 L{sin( t )}

1 dv  e st dt , v   e st s 1 1 F(s) st  st  [ g ( t )e ]0   f ( t )e dt  s s s t

g ( t )   f ( t )dt If t=0, g(t)=0 0



 0



for (t  )   f (t )e  st dt   so 0 f (t )dt  g (t )   slower than e st  0

Properties: Derivatives (this is the big one) 

L{Df ( t )}  sF(s)  f (0 ) Example : L{D cos( t )}  s2   f ( 0 ) 2 s 1 s2 1  2 s 1 s 2  (s 2  1) s2  1 1  L{ sin( t )} 2 s 1

Proof :



d f ( t )e st dt dt 0

L{Df ( t )}  

u  e st , du  se st

let

d dv  f ( t )dt , v  f ( t ) dt 

[e st f ( t )]0  s  f ( t )e st dt  0

 f (0  )  sF(s)





Difference in f (0 ), f (0 ) & f (0) The values are only different if f(t) is not continuous @ t=0  Example of discontinuous function: u(t) 

f (0  )  lim u ( t )  0 t 0

f (0  )  lim u ( t )  1 t 0

f (0)  u (0)  1

Properties: Nth order derivatives

L{D f ( t )}  ?

2

let

g ( t )  Df ( t ), g (0)  Df (0)  f ' (0)  L{D 2 g ( t )}  sG (s)  g (0)  sL{Df ( t )}  f ' (0)  s(sF(s)  f (0))  f ' (0)  s 2 F(s)  sF(0)  f ' (0)

L{D n f ( t )}  s n F(s)  s ( n 1) f (0)  s ( n  2) f ' (0)    sf ( n  2)' (0)  f ( n 1)' (0)

NOTE: to take L{D n f ( t )} you need the value @ t=0 for D n 1f ( t ), D n  2 f ( t ),...Df ( t ), f ( t )  called initial conditions!

We will use this to solve differential equations!

Properties: Nth order derivatives Start with L{Df ( t )}  sF(s)  f (0) Now apply again L{D 2f ( t )} let g( t )  Df ( t ) and Dg ( t )  D 2 f ( t ) then L{Dg ( t )}  sG (s)  g(0) remember g( t )  Df ( t )  g (0)  f ' (0) G (s)  L{g ( t )}  L{Df ( t )}  sF(s)  f (0)

 L{Dg ( t )}  sG (s)  g (0)  s[sF(s)  f (0)]  f ' (0)  s 2 F(s)  sf (0)  f ' (0)

Can repeat for D3f ( t ), D 4 f ( t ), etc.

L{D n f ( t )}  s n F(s)  s ( n 1) f (0)  s ( n  2) f ' (0)    sf ( n  2)' (0)  f ( n 1)' (0)

Relevant Book Sections Modeling - 2.2  Linear Systems - 2.3, page 38 only  Laplace - 2.4  Transfer functions – 2.5 thru ex 2.4 