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 2j 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 sa
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) sa 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 sa
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 01f ( t )
L{sin( t )} g ( t )e st dt
Example : L{D 01 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