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0
)()()( dttyetyLsY st
0
)()()( dttyetyLsY st
La transformada de Laplace
2
Sea f(t) una función definida para t ≥ 0, su transformada de Laplace se define como:
donde s es una variable compleja
Se dice que la transformada de Laplace de f(t) existe si la integral converge.
dtetfsFtfL st
0
)()()}({
.iws
La transformada de Laplace
3
Pierre-Simon Laplace (1749 - 1827)
"Podemos mirar el estado presente del universo como el efecto del pasado y la causa de su futuro.Se podría condensar un intelecto que en cualquier momento dado sabría todas las fuerzas que animan la naturaleza y las posiciones de los seres que la componen, si este intelecto fuera lo suficientemente vasto para someter los datos al análisis, podría condensar en una simple fórmula el movimiento de los grandes cuerpos del universo y del átomo más ligero; para tal intelecto nada podría ser incierto y el futuro así como el pasado estarían frente sus ojos."
4
Observa que la transformada de Laplace es una
integral impropia, uno de sus límites es infinito:
0 0
( ) lim ( )h
s t s t
he f t dt e f t dt
( ) ( ),f t F sL
( ) ( ),
( ) ( ), etc.
y t Y s
x t X s
L
L
Notación:
5
Condiciones suficientes de existencia de la TL
Si f(t) es continua a trozos en [0, ∞) y
),0[,|)(| tMetf at
Es decir, f(t) es de orden exponencial en el infinito:
0|)(|lim
bt
tetftqb
Entonces:
L{f(t)} = F(s) existe s > a.
dtetfsFtfL st
0
)()()}({
6
Unicidad de la TL
Si f1(t) y f2(t) poseen la misma TL:
)()()(
:por definida nulafunción lay0
0)(
21
0
tftftN
N(t)a
dttNa
L{f1(t) } = L{f2(t) }= F(s),
Entonces el teorema de Lerch garantiza que
7
se
sdtesFL tsst 11
1)(1
0
1
0
Calcula la transformada de f(t) = 1:
ssFtf
1)(1)(
Nota: Obviamente L{a} = a/s y L{0} = 0.
8
1
0
1
0
1
0
0 )(
nstn
stn
stnstnn
tLs
ndtet
s
n
dts
ent
s
etdtetsFtL
Calcula la transformada de f(t) = tn:
1
!)(1)(
ns
nsFtf
10
1
!
1
nn
nn
s
ntL
stL
tLs
ntL
9
1
1
1
1
)(
0
1
0
1
0
se
s
dtedteesFeL
ts
tssttt
Calcula la transformada de f(t) = e-t:
1
1)()(
ssFetf t
10
asas
Ae
as
A
dtAedteAesFAeL
tas
tasstatat
,)(
)(
0
0
0
Calcula la transformada de f(t) = Aeat:
asas
AsFAetf at
,)()(
11
dteatsens
a
s
adt
s
eatsena
s
eat
s
a
dts
eata
s
eatsendteatsensFatsenL
ststst
ststst
0 22
0
0
0
0
0
)()()cos(
)cos()()()()(
Calcula la transformada de f(t) = sen(at):
22)()()(
as
asFatsentf
222
2
2
2
;1as
aI
s
aI
s
a
Ejercicio: calcula F(s) para f(t) = cos(at)
12
)()cos(
11
)(
)()cos(
2222
22
0
0
0
atseniLatLas
ai
as
s
as
ias
ias
ias
iase
ias
dtedteesFeL
atseniate
tias
tiasstiatiat
iat
Calculemos la transformada de f(t) = eiat:
13
c
1
t
0 if ( )
1 if
t cu t c
t c
La función Heaviside o escalón unidad:
c0
1
0
1 1
( ) ( ) lim
lim lim ( )
hs t s t
hc
h s cs t s h s cs sch h
u t c e u t c dt e dt
ee e e s
L
14
Función delta de Dirac
/1
a a
área = 1Sea la función parametrizada:
t
)(lim)( 0 tfat
s
ee
s
e
s
etfL
sas
saas
11
)()(
ass
ass
as es
see
s
eetfL
000 lim1
lim)(lim
)(tf
)()(1
)( atuatutf
Observemos que
15
ta
1)(
)(
tL
eatL as
)( at )(t
Así la transformada de la función delta de Dirac es:
16
Step function and delta function
There are two common functions which are used to represent very rapidly changing quantities. The first of these is the step function, u(t), defined by:
00
01)(
t
ttu
t
u(t)
17
Step function and delta function
If the step function ‘switches on’ at t = a it is defined by:
at
atatu
0
1)(
u(t-a)
t =a t
18
Step function and delta function
The step function can be considered as the limiting case of a very steep “ramp” function:
2b
19
Step function and delta function
The 2nd function is the delta function which is used to represent point loads and other large inputs applied over very small areas. It is defined by:
0
00)(
t
tt
But it also satisfies:
1)(
dtt
20
Step function and delta function
The delta function can be considered as a limiting case as shown below:
0 lim b
1/2b
2b
(t)
t
21
Step function and delta function
The delta function for a point load at t=a is given by:
1)(
a
a
dtat
It has the properties:
at
atat
0)(
)()()( afdttfata
a
22
Laplace transform of the step function
For the step function
The Laplace transform is:
00
01)(
t
ttu
0
0 0
10
11)}({ e
se
sdtetuL stst
stuL
1)}({
23
Laplace transform of the step function
For the step function centred at t=a
The Laplace transform is:
s
eatuL
as
)}({
at
atatu
0
1)(
a
st
a
sta
st es
dtedteatuL1
010)}({0
24
Laplace transform of the delta function
For the delta function
The Laplace transform is:
1)}({ tL
1)()}({0
0
t
stst edttetL
0
00)(
t
tt
25
Laplace transform of the delta function
For the delta function centred at t=a
The Laplace transform is:
aseatL )}({
as
at
stst eedtateatL
0
)()}({
at
atat
0)(
26
Funciones periódicas
Supongamos que f (t) es una función periódica de periodo T. Entonces:
)(1
1)()( 1 sF
etfLsF
sT
donde F1(s) es la tranformada de Laplace de la función f(t) sobre el primer periodo y cero fuera.
T
st dttfesF0
1 )()(
t t
T
27
)()(
)()(
,)()(
)()(
)()(
0
00
0
)(
0
0
0
sFedttfe
dfeedttfe
TtdTfedttfe
dttfedttfe
dttfesF
sTT
st
ssTT
st
TsT
st
T
stT
st
st
Demostración
28
Ejemplo: onda cuadrada
a 2a
aT 2
)(1
1)( 12
sFe
sFas
asasa
a
sta
st ees
dtedttfesF 222
0
1
1)()(
)1(
1
)1()(
2
2
asas
asas
eses
eesF
29
Tabla de transformadas de Laplace
2 2
2 2
2 2
2 2
1
sen
cos
sen
cos
!
at
at
n atn
ts
st
s
e ts a
s ae t
s a
nt e
s a
ase
s
nt
t
s
t
at
nn
1
!
s
1
1 1
1
1
2
30
31
32
33
34
35
La TF es un caso particular de la TL
dtetffFtfF ti )()(ˆ)()]([
Supongamos que es complejo: = + i
dteetfdtetfif tittii )()()(ˆ )(
Antitransformando tendríamos:
deifetf tit )(ˆ2
1)(
36
Recordemos que = + i:
deife
tf tit
)(ˆ2
)(
deiftf tii )()(ˆ2
1)(
)Im(
)(ˆ2
1)( deftf ti
Re ()
Im()
-γ
tief )(ˆ es analítica para todo perteneciente a la región en rojo.Haciendo s = i( + i) llegamos a la transformada de Laplace.
37
Al proceso inverso de encontrar f(t) a partir de F(s) se le conoce como transformada inversa de Laplace y se obtiene mediante:
conocida también como integral de Bromwich o integral de Fourier-Mellin.
i
i
st tdsesFi
tfsFL
0,)(
2
1)()}({1
Transformada inversa de Laplace
38
Re(s)
Im(s)
γ
i
i
st tdsesFi
tfsFL
0,)(
2
1)()}({1
γ determina un contorno vertical en el plano complejo,tomado de tal manera que todas lassingularidades de F(s) queden a su izquierda.
Con condiciones de existencia:
)(lim)2(
0)(lim)1(
ssF
sF
s
s
39
Por ejemplo, determinemos:
Puesto que la función a invertir tiene un polo en s = -1, entonces basta con tomar γ > -1. Tomemos γ = 0 y el contorno de integración C de la figura.
21
)1(
1
sL
Re(s)
Im(s)
γ=0-1
C1R
-R
ds
s
e
idsesF
i C
sti
i
st2)1(2
1)(
2
1
iR
iRC
stst
s
e
ids
s
e
i1
22 )1(2
1
)1(2
1
0 por la desigualdad ML cuando R→∞ con t≥0.
21
121 )1(
1lim
)1(Res
2
2
sLtee
ds
d
s
e
i
i tst
s
st
s
Haciendo R→∞ y utilizando teoría de residuos:
40
Sea F(s) una función analítica, salvo en un número finito de polos que se encuentran a la izquierda de cierta vertical Re(s) = γ. Y supongamos que existen m, R, k > 0 tq. para todo s del semiplano Re(s) γ y |s| > R, tenemos que
ks
msF |)(|
).( de polos losson s,...,s,s donde
)(Res)}({
n21
1
1
sF
sFesFLn
k
st
ss k
Entonces si t > 0:
En particular, sea F(s) = N(s)/D(s), con N(s) y D(s) polinomios de grado n y d respectivamente, d > n; entonces podemos usar la igualdad anterior.
41
Ejemplo, determinar:
21
)1)(2(
1)(
ssLtf
.1sy 2s :doble otroy simple uno polos, dos posee
)1)(2()(
21
2
ss
esFe
stst
9
3
2lim
)1(lim
)1)(2(Res
)1)(2(Res)(
2
122
2122
tttst
s
st
s
st
s
st
s
etee
s
e
ds
d
s
e
ss
e
ss
etf
42
1. Linealidad: Si c1 y c2 son constantes, f1(x) y f2(x) son funciones cuyas transformadas de Laplace son F1(x) y F2(x), respectivamente;
entonces:
).()()}()({ 22112211 sFcsFctfctfcL
La transformada de Laplace es un operador lineal.
Propiedades
43
)()(
)()(
)()(
)()(
2211
0 22
0 11
0 2211
2211
tfLctfLc
dtetfcdtetfc
dtetfctfc
tfctfcL
stst
st
Demostración:
44
2. Desplazamiento temporal
)(
)(
)(
)()()(
)()(
0
0
0
0
0
0
0
00
0
sFe
tt
dfee
dtttfe
dtttuttfesX
dttfesF
st
sst
t
st
st
st
0
000 ,0
),()()()(
tt
ttttfttutftg
)()}()({
)()}({0
0 sFettutfL
sFtfLst
45
Ejemplo:
3
31
s
eL
s
3
2 2
stL
332 2
)3()3(s
etutL s
)3()3(2
1 23
31
tuts
eL
s
3t
46
Shift in t
Remember the definition of the Laplace transform:
Answer: The Laplace transform assumes all functions are zero for t<0.
Mostly we do not need to know this.
Question: What happens to f(t) for t<0 ?
0
)()( dttfesF st
47
Shift in t
Define a shifted function by:
atatf
attg
)(
0)(
t
f(t-a) f(t)
t = a
48
Shift in t
The shifted function can also be defined by:
)()()( atfatutg
The Laplace transform of the shifted function is given by:
00
)()()()( dtatfatuedttgesG stst
a
sta
st dtatfedtatfesG )(1)(0)(0
a
st dtatfesG )(0)(
49
Shift in t
Substitute =t-a:
)()()(0
sFedfeesG sassa
)()()( sFetgLsG sa
aa
as dfesG
)()( )(
50
Example - Shift in t
Calculate the Laplace transform of a square wave shown by the diagram below
1
t=a t=2a t=4a t=0
51
Example - Shift in t
Note that the 1st pulse can be constructed as superposition of two step functions:
Step Function: u(t)
Step Function: -u(t-a)
t=0
t=a
52
Example - Shift in t
Then, the Laplace transform of the first pulse is :
)()(pulse 1st atuLtuLL
asas
ess
e
sL
1
1
1pulse 1st
53
Example - Shift in t
To obtain the Laplace transform of the 2nd pulse, we note that it is the 1st pulse shifted in time by 2a:
1
t=a t=2a t=4a t=0
1
t=a t=2a t=4a t=0
54
Example - Shift in t
Thus the Laplace transform is given by:
pulse 1pulse 2 st2nd LeL as
Similarly the Laplace transform of the 3rd pulse is (it is shifted by 4a):
pulse 1pulse 3 st4rd LeL as
55
Example - Shift in t
Thus the Laplace transform of the whole square wave is given by:
pulse 3pulse 2pulse 1
wavesquarerdndst LLL
L
pulse 11 st42 Lee asas
asasas es
ee 11
1 42
56
Example - Shift in t
In this case we can sum the series (it is a geometric series): 1242 11
asasas eee
Thus:
asas
ese
L
1
1
1
1 wavesquare
2
)1(
11
1
)1)(1(
1as
asasas es
esee
57
)(
)()()(
)()(
0
)(
0
0
asF
dttfedttfeesX
dttfesF
tasatst
st
22 )(
11
asteL
stL at
3. Desplazamiento en frecuencias
Ejemplo:
)()}({
)()}({
asFtfeL
sFtfLat
58
4. Cambio de escala en tiempo
)/()/1(
)(1
)()(
)()(
0
)/(
0
0
asFa
atdfea
dtatfesX
dttfesF
as
st
st
a
sF
aatfL
sFtfL
1)}({
)()}({
59
5. Derivada de la transformada de Laplace
)(
)(
)()(
)()(
0
0
0
ttfL
dtttfe
dttfeds
dsF
ds
d
dttfesF
st
st
st
)()(
)}({)(
ttfLsF
tfLsF
60
6. Transformada de Laplace de las derivadas de una función
La transformada de Laplace de la derivada de una función está dada por:
donde f(0) es el valor de f(t) en t = 0.
La transformada de Laplace de la segunda derivada de una función está dada por:
)0()()}('{ fssFtfL
)0(')0()()}(''{ 2 fsfsFstfL
61
En forma similar:
Demostración:
)0()0(')0()()}({ )1(21)( nnnnn ffsfssFstfL
)0()()()0(
)()()(')('
0
00
0
fssFdttfesf
dttfsetfedttfetfL
st
ststst
0)(lim
tfe st
t
62
Supongamos que:
)0()0(')0()()}({ )2(321)1( nnnnn ffsfssFstfL
)0()0(')0()(
)0()()()0(
)()()()(
)1(21
)1()(
0
)1()1(
0
)1(
0
)1(
0
)()(
nnnn
nnnstn
nstnstnstn
ffsfssFs
ftfsLdttfesf
dttfsetfedttfetfL
Entonces: 0)(lim )1(
tfe nst
t
63
64
Gracias a esta propiedad y a la linealidad de la TL podemos convertir una ec. diferencial como
" 3 ' 4 ( 1)
(0) 1, '(0) 2
y y y t u t
y y
en una ec. algebraica
2
2 1( )*( 3 4) ( 1) ss
s eY s s s s
Resolver paray(t)
Resolver para Y(s)
Ec. Diferencial
Transformada de Laplace
Ec. Algebraica
Si resolvemos la ec. algebraica:
2
2 2
( 1) ( 1)( )
( 3 4)
s ss s e eY s
s s s
y encontramos la transformada inversa de Laplace de la solución, Y(s), encontraremos la solución de la ec. diferencial.
Ec. Algebraica
Solución de la Ec. Diferencial
Inversa de la Transformada
de Laplace
La transformada inversa de Laplace de:
es
4 43 32 15 80 4 16
4325 5
( ) ( 1)( + ( ) )
( )( ( ) )
t tee
t t
y t u t e e t
u t e e
2
2 2
( 1) ( 1)( )
( 3 4)
s ss s e eY s
s s s
4 43 32 15 80 4 16
4325 5
( ) ( 1)( + ( ) )
( )( ( ) )
t tee
t t
y t u t e e t
u t e e
es la solución de la ec. diferencial:
" 3 ' 4 ( 1)
(0) 1, '(0) 2
y y y t u t
y y
De modo que:
Para conseguirlo hemos aplicado:
Primero, que la TL y su inversa son lineales:
1 1 -1
( ) ( ) ( ) ( ) ,
( ) ( ) ( ) ( )
cf t g t c f t g t
cF s G s c F s G s
L = L +L
L = L +L
2
'( ) ( ) (0),
''( ) ( ) (0) '(0)
f t s f t f
f t s f t s f f
L = L
L = L
and
etc...
Y segundo, la TF de las derivadas de una función son:
A este método se le conoce como cálculo de Heaviside.
Por ejemplo:
012
1
012
01
)0()0(')0()(
0)()}0()({)}0(')0()({
0)()(')(''
asas
fafsfsF
sFafssFafsfsFs
tfatfatf
Y antitransformando obtendremos la solución.
Veamos un ejemplo concreto: Resolver la ec. diferencial
)4)0(y0()(2)(' 3 ftetftf t
tt
t
tt
eetfss
sF
ssFssF
ssFfssF
eLtfLtfL
etftfLetftf
32
3
33
5)(3
1
2
5)(
03
1)(24)(
03
1)(2))0()((
0}{)}({2)}('{
0})(2)('{;0)(2)('
74
Ejemplo
Resolver
2222 )1()1(
1)(
s
e
ssY
s
0)0()0(,0
0sin
yyt
ttyy
11
1
)sin()(sin
sin)(sin)()(
22
2
s
e
s
ttutL
ttutLsYsYs
s
tt
tttt
ttttutttty
cos
0cossin
)cos()()sin()(cossin)(
21
21
21
21
75
Ejemplo:
2
1
1
1
23
1)(
)(2)(3)(
2
2
sse
ssesY
esYssYsYs
ss
s
)1(2)1()1()( tt eetuty
Resolver 0)0()0(),1(23 yytyyy
76
7. Transformada de Laplace de la integral de una función
s
sFtfL
sduufL
t )()}({
1)(
0
)(1
)(11
)(
)()(
)()(
000
00
0
sFs
dttfes
es
df
dtdfesX
dttfesF
ststt
tst
st
Si existe la TL de f(t) cuando Re(s) > p ≥ 0, entonces:
para Re(s) > p.
77
s
sFduufL
t )()(
0
)(2
)(1
1
1
1}{;
2
20
sarctguarctgduut
tsenL
stsenLdte
t
tsen
t
tsenL
ss
st
sduuF
t
tfL )(
)(
)()(con tfLsF Ejemplo:
8. Transformada de Laplace de f(t)/t
78
Useful theorems - Integration
Integration. Given:
Then
0
)()()( dttfetfLsF st
t
dfsFs
L0
1 )()(1
79
Example - Integration
Use the integration result to find the inverse transform of:
Put
)1(
1
ss
tt
tt
ee
edess
L
1)1(
)1
110
0
1
tetfs
sF
)()1(
1)(
Then
80
)cos()()(Si attftg
24
222
2222
0
4
2
222
2
)()()()(
)(1
)(
as
aaisa
aisa
i
aiasa
aiasa
i
dtet
atsensG
atsent
tg
st
2
)()()(
iasFiasFsG
acon
Ejemplo:
)()()(Si atsentftg
2
)()()(
iasFiasFisG
acon
9. TF de f(t)cos(at) y f(t)sen(at)
81
10. Teorema del valor final
Si existe, entonces:
11. Teorema del valor inicial
El valor inicial f(0) de la función f(t) cuya transformada de Laplace es F(s), es:
)(lim tft
)(lim)(lim 0 ssFtf st
)(lim)(lim)0(0
ssFtff st
82
Recordemos que
la operación se conoce
como la convolución de y y se denota como
La transformada de Laplace de esta operación está dada por:
dtff )()( 21
)(1 tf ),(2 tf
)}({)}({)}(*)({
)()()}(*)({
2121
2121
tfLtfLtftfL
sFsFtftfL
).(*)( 21 tftf
12. Integral de convolución
83
0,0
0,)()()(*)( 0
t
tdtgftgtft
Si trabajamos con funciones que son cero para para t < 0, entonces la convolución queda:
Así que para estas funciones podemos definirla convolución como:
t
tdtgftgtf0
)0(,)()()(*)(
84
44
1
2
)()()(*)(
2
0
22
0
)(2
ttt
t t
etdee
dedtgftgtf
)}({)}({)}(*)({ 2121 tfLtfLtftfL Ejemplo: Verificar que funciona para f(t) = t y g(t) = e-2t
con valores 0 para t < 0.
)2(
1
)2(
1
4
11
4
11
2
1
}{4
1}1{
4
1}{
2
1
44
1
2
2
2
2
2
ss
sss
eLLtL
etL
t
t
)2(
1
)2(
11
)2(
1}{;
1}{
22
22
ssss
seL
stL t
85
De hecho, podemos utilizar la convolución para encontrar transformadas inversas de Laplace:
1
1
11
)1(
1
0
21
21
tede
etss
Lss
L
tt
t
t
86
Convolution theorem
For any functions f(t) and g(t) with:
0
)()()( dttfetfLsF st
0
)()()( dttgetgLsG st
Then
)()()()(0
sGsFdgtfLt
convolution integral
87
Convolution theorem and inversion
For any functions F(s) and G(s) with:
)()(1 tfsFL
Then
)()(1 tgsGL
t
dgtfsGsFL0
1 )()()()(
Do not use unless really necessary!
t
dtgfsGsFL0
1 )()()()( Or
88
Convolution theorem – example
Use the convolution theorem to invert: 22 )1( s
s
Put
Thus
)1()(
2
s
ssF
)1(
1)(
2
ssG
ttf cos)( ttg sin)(
89
Convolution theorem – example
Now use
t
dgtfsGsFL0
1 )()()()(
Thus
t
dts
sL
022
1 )sin()cos()1(
90
Convolution theorem – example
Expand out
After much work (see notes)
tts
sL sin
2
1
)1( 221
tt
t
dtdt
dts
sL
00
022
1
sinsinsin sincoscos
)sin()cos()1(
91
1
1)(
41)(;
1
10)}(*{41)(
1
1)0()}({4)0()(
}{)(4)(;)()(4)(
)(1
)}({}{
)()(*
0
2
ssX
sssX
stxtLsssX
shthsLxssX
eLthdt
dLtx
dt
dLedssxst
dt
dtx
dt
d
sXs
txLtL
tt
thtxt
t
1)0(;)()(4)(0
xedssxsttx
dt
d tt
Resolver la ec.integro-diferencial:
92
ttt eeetx
ssssX
sss
ssX
ssX
sssX
22
2
3
1
3
1)(
2
1
3
1
2
1
1
1
3
1)(
)3)(2)(1()(
1
1)(
41)(
Antitransformando:
93
Raíces del denominador D(s) o polos de F(s):
Caso I – Polos reales simples
Caso II – Polos reales múltiples
Caso III – Polos complejos conjugados
Caso IV – Polos complejos conjugados múltiples
)( as 2)( as
))(( *asas
01
1
01
1
)(
)()(
bsbs
asasa
sD
sNsF
mm
m
nn
nn
Desarrollo en fracciones parciales: Se utiliza para facilitar el cálculo de la transformada inversa, descomponiendo la función en componentes más sencillos.
2*))(( asas
94
Caso I – Polos reales simples )( as
32
)3)(2(
1
6
1
)(
)()(
23
s
C
s
B
s
A
sss
s
sss
s
sD
sNsF
Ejemplo
as
A
95
15
2
)2(
1
3
10
3
)3(
1
2
6
1
)3)(2(
1
3
2
0
s
s
s
ss
s
s
C
ss
s
s
B
ss
s
s
A
32)3)(2(
1)(
s
C
s
B
s
A
sss
ssF
assD
sNasA
)(
)()(
96
)3)(2(
)2()3()3)(2(326
123
sss
sCssBsssAs
C
s
B
s
A
sss
s
)2()3()3)(2(1 sCssBsssAs
Ass
s
s
0)3)(2(
1
)6()23()(
)2()3()6(12
222
ACBAsCBAs
ssCssBssAs
16;123;0 ACBACBA
métodoalternativo
y resolver...
97
3
1
15
2
2
1
10
31
6
132
6
1)(
23
sss
s
C
s
B
s
Asss
ssF
La transformada inversa de Laplace es:
tt eetf 32
15
2
10
3
6
1)(
98
Otro ejemplo
2
1
1
2
1
1
211
)2)(1)(1(
372
)2)(1(
372)(
2
2
2
ssss
C
s
B
s
A
sss
ss
ss
sssF
1)1)(3(
3148
)1)(1(
372
2)3)(2(
372
)2)(1(
372
1)1)(2(
372
)2)(1(
372
2
2
1
2
1
2
s
s
s
ss
ssC
ss
ssB
ss
ssA Transformada inversa de Laplace:
ttt eeetf 22)(
99
Caso II – Polos reales múltiples 2)( as
12)1)(2(
44
)(
)()(
22
23
s
D
s
C
s
B
s
A
sss
ss
sD
sNsF
Ejemplo
)()( 2 as
B
as
A
Polos realessimples
Polos realesmúltiples
100
3)1)(2(
44
2)1)(2(
44
0
230
23
2
s
s
ss
ss
ds
d
s
B
ss
ss
s
A
assD
sNasA
)(
)()( 2
assD
sNas
ds
dB
)(
)()( 2
)1)(2(
44)(
2
23
sss
sssF
101
Transformada inversa de Laplace:
tt eettf 232)(
1
1
2
113
12
12
)1)(2(
44)(
2
2
2
23
ssss
s
D
s
C
s
B
s
A
sss
sssF
102
En general, para polos reales múltiples:
sN
sDsF n
r pspspssD 21
n
nr
rr
r
ps
a
ps
a
ps
a
ps
b
ps
b
ps
bsF
3
3
2
2
1
11
1
1
1
1
1!
1
ps
r
j
j
jr pssFds
d
jb
ipsii pssFa
1
1
1
1
]))(([)!1(
1
]))(([!
1
]))(([
]))(([
11
1
1
1
11
1
ps
rr
r
ps
rj
j
jr
ps
rr
psr
r
pssFds
d
rb
pssFds
d
jb
pssFds
db
pssFb
103
Caso III – Polos complejos conjugados
ejemplo
))(( *asas
iaas
B
as
B
s
A
ss2,
)4(
4*
*
2
2
1
)2(
4
2
1
)2(
4
14
4
2
*
2
02
is
is
s
issB
issB
sA
conjugados complejos
*
11
2
11
asass
Transformada inversa de Laplace:
)2cos(1)( ttx
104
ejemplo
iaas
B
as
B
ss
s43,
256
4*
*
2
)4(8
1
43
4
)4(8
1
43
4
43
*
43
iis
sB
iis
sB
is
is
Transformada inversa de Laplace:
)cos(2)( teBtf t
245.0,4,3
,8
17),4(
8
1
BiB
)245.04cos(4
17)( 3 tetf t
donde
105
Se trata de repetir los métodos usados en los casos II y III,teniendo en cuenta que trabajamos con complejos.
Caso IV – factores complejos conjugados múltiples
2*))(( asas
106
Partial Fractions – General Case
In the solution of ODEs by the Laplace Transform
method, expressions of the form often occur.
Here P(s) and Q(s) are both polynomials.)(
)(
sP
sQ
These are easy to invert if they are written in partial
fraction form:
)()()()(
)(
2
2
1
1
n
n
s
a
s
a
s
a
sP
sQ
Here )())(()( 21 nssssP
107
Partial Fractions – Complex Roots
If 1 and 2 are a complex conjugate pair, then we can avoid complex numbers by combining these factors into a quadratic expression. Say 1 = -a+ib, 2= -a-ib, then:
Then
2221 )())(( basss
)()()()(
)(
3
322
n
n
s
a
s
a
bas
BsA
sP
sQ
108
Partial Fractions – Repeated Roots
If 1 = 2 then the partial fraction form is:
Etc, etc
)()()()()(
)(
3
32
1
2
1
1
n
n
s
a
s
a
s
a
s
a
sP
sQ
If 1 = 2 = 3 then the partial fraction form is:
)()()()()(
)(3
1
32
1
2
1
1
n
n
s
a
s
a
s
a
s
a
sP
sQ
109
Method 1 - Cover-up Rule
Basic idea:
• Multiply by a factor (s - i)
• Put s = i
• Evaluate ai
)()()()(
)(
2
2
1
1
n
n
s
a
s
a
s
a
sP
sQ
110
Cover-up Rule
Find the partial fraction form for:
The partial fraction form is:
)1(
1
ss
)1()1(
1
s
B
s
A
ss
111
Cover-up Rule
To calculate A, multipy by s and put s=0:
)1()1(
1 :
s
BsA
ss
101
1 :0Put AAs
112
Cover-up Rule
To calculate B, multipy by s+1 and put s=-1:
BAss
s )1(1
:)1(
101
1 :1Put
BBs
Thus the partial fraction form is:
)1(
11
)1(
1
ssss
113
Cover-up Rule
The above procedure can be carried out by “covering up”. For A use:
Now put s = 0, ignoring the covered up items:
)1()1(
1
s
B
s
A
ss
110
1
AA
114
Cover-up Rule
Similarly for B, “covering up” gives:
Now put s = -1, ignoring the covered up items:
11
1
BB
)1()1(
1
s
B
s
A
ss
115
Complex Cover-up Rule
Find the coefficients A, B, C in:
Find A using the standard cover-up idea:
)1()1()1)(1(
122
s
CBs
s
A
ss
)1()1()1)(1(
122
s
CBs
s
A
ss
116
Complex Cover-up Rule
Put s+1 = 0, that is s = -1:
2
1
)1)1((
12
AA
The coefficients B and C are more difficult to calculate. A modified version of the cover-up rule involves using complex factors
117
Complex Cover-up Rule
Multiply the whole equation by (s+i):
i)()1(
i)()1(
i)()1)(1(
122
ss
CBss
s
As
ss
i)(i)i)((
i)()1(
i)(i)i)()(1(
1
sss
CBss
s
A
ssss
118
Complex Cover-up Rule
Now put (s+i)=0, that is s = -i:
i)(i)(
)1(i))(1(
1
s
CBss
s
A
ss
i)2(
i)()0(
)1(i)2)(1i(
1
CB
i
A
After tidying up:
CB
i2
)1i(
119
Complex Cover-up Rule
Equate real and imaginary parts:
Final partial fraction:
2
1 ,
2
1 CB
)1()1()1)(1(
12
21
21
21
2
s
s
sss
120
Method 2 - Substitution of values
Basic idea
• Put s = bi for i=1,2,…,n. Here bi are convenient, easy to work with, numbers
• Obtain n equations in the n unknown ai
• Solve for ai
)()()()(
)(
2
2
1
1
ni
n
iii
i
b
a
b
a
b
a
bP
bQ
121
Substitution of values
Find the coefficients A, B, C in:
)1()1(
122
s
C
s
B
s
A
ss
)1()1(
122
s
C
s
B
s
A
ss
Put s=0:1
)10(
1
BB
Use cover-up for B :)( 2s
122
Substitution of values
Use cover-up for C:
)1()1(
122
s
C
s
B
s
A
ss
Put s=-1:1
)1(
12
CC
123
Substitution of values
We cannot use cover-up for A. So far we have:
)1(
11
)1(
122
sss
A
ss
One good way to calculate A is to substitute a convenient value for s. Say s =1:
)11(
1
1
1
1)11(1
122
A
12
11
2
1 AA
124
Substitution of values
Substitute back:
)1(
111
)1(
122
sssss
125
Example - substitution of values
Substitute values to work out A and B in:
)1()1(
1
s
B
s
A
ss
Put s=1: BABA
2
1
2
1
212
1
Put s=2: BABA
3
2
3
1
326
1
126
Example - substitution of values
Subtract two equations
1166
1 AB
B
Substitute back
)1(
11
)1(
1
ssss
127
Method 3 - Equate coefficients
Basic idea:
• Multiply whole equation by P(s)
• Equate coefficients of each power of s
• Solve resulting equations for ai
ii
nn
s
sPsP
sPasPasPasQ
)(
)( where
),(...)()()( 2211
128
Example - Equate coefficients
Calculate A and B in:
)1()1(
1
s
B
s
A
ss
Multiply by s(s+1): BssA )1(1
Equate coefficients:
BA0 :sFor
A1 :1For 1 ,1 BA
129
Example - Equate coefficients
Calculate A and B in:
Multiply by (s+1)(s2 +1):
)1)(()1(1 2 sCBssA
)1()1()1)(1(
122
s
CBs
s
A
ss
BA0 :sFor 2
Equate coefficients:
)()()(1 2 CAsCBsBA
0 :For CBs
2
1 ,
2
1 ,
2
1 CBA 1 :1For CA
130
Preferred method for partial fractions
Calculate as many coefficients as possible using the simple cover-up rule
Calculate the remaining coefficients by:
• Substituting values for s
• Using the complex cover-up method
• Equating coefficients
131
More on differential equations
ODE problem
Apply Laplace transform
)(011
1
1 tfyadt
dya
dt
yda
dt
ydn
n
nn
n
nn
n
bdt
ydb
dt
dyby
)0( ..., ,)0( ,)0(1
1
21
)()()()( sFsQsYsP
132
More on differential equations
Rearrange
)(
)(
)(
)()(
sP
sF
sP
sQsY
)(
)(
sP
sQ This term comes from the initial conditions. To invert, convert into partial fraction form, then use tables and useful rules
)(
)(
sP
sF This term comes from the right hand side of the ODE. To invert, convert into partial fraction form (if possible) then use tables. Otherwise use partial fractions on 1/P(s) , invert, and then apply the convolution theorem
133
Transfer function (optional extra)
System Input: f(t) Output: y(t)
If we forget about the initial transient in )(
)(
)(
)()(
sP
sF
sP
sQsY
Then
)(
1
)(
)(
)(
)()(
sPsF
sY
sP
sFsY
134
Transfer function (optional extra)
Thus the transfer function Can be written as:
inputL
outputL
sF
sY
sP
)(
)(
)(
1
)(
1
sP
If f(t)=(t) then F(s)=1 and
Thus the transfer function is the Laplace transform of the response of the system to an impulse (delta function)
)(
1)(
sPsY
135
Example (Laplace transform solution of an ODE)
Solve the following problem using the Laplace transform method:
Step 1. Define your transform
,sin10232
2
tydt
dy
dt
yd 0)0(,1)0(
dt
dyy
0
)()()( dttyetyLsY st
136
Example (Laplace transform solution of an ODE)
Step 2. Transform the ODE:
tLydt
dy
dt
ydL sin1023
2
2
tLyLdt
dyL
dt
ydL sin10}2{3
2
2
tLyLdt
dyL
dt
ydL sin10}{23
2
2
137
Example (Laplace transform solution of an ODE)
Use formulae from the tables
1
110)(2)0()(3
)0()0()(
2
2
ssYyssY
sydt
dysYs
Tidy up
1
10)0()3()0()(]23[
22
sys
dt
dysYss
138
Example (Laplace transform solution of an
ODE) Step 3. Use the initial conditions and solve for Y(s):
1
10)0()3()0()(]23[
22
sys
dt
dysYss
1
10)3()(]23[
22
sssYss
)1)(23(
10
23
3)(
222
sssss
ssY
139
Example (Laplace transform solution of an ODE)
Step 4. Find the partial fraction forms. First
21)2)(1(
3
23
32
s
B
s
A
ss
s
ss
s
Cover-up for A:
21)2)(1(
3
s
B
s
A
ss
s
Put s = -1:2
2)1(
3)1(
AA
140
Example (Laplace transform solution of an ODE)
Cover-up for B:
21)2)(1(
3
s
B
s
A
ss
s
Put s = -2:1
1)2(
3)2(
BB
Substitute back:
2
1
1
2
)2)(1(
3
ssss
s
141
Example (Laplace transform solution of an ODE)
Now work out the partial fraction form for:
Use cover-up for C
211)2)(1)(1(
1022
s
D
s
C
s
BAs
sss
211)2)(1)(1(
1022
s
D
s
C
s
BAs
sss
Put s = -1: 5)21(1)1(
102
CC
142
Example (Laplace transform solution of an ODE)
Use cover-up for D
211)2)(1)(1(
1022
s
D
s
C
s
BAs
sss
Put s = -2: 2)12(1)2(
102
DD
Result so far:
2
2
1
5
1)2)(1)(1(
1022
sss
BAs
sss
143
Example (Laplace transform solution of an ODE)
To find B, put s = 0 :
20
2
10
5
10
0
)20)(10)(10(
10
BA
1152
10 BB
144
Example (Laplace transform solution of an ODE)
To find A, put s = 1 :
21
2
11
5
11
11
)21)(11)(11(
10
A
3
2
2
5
2
1
322
10
A
33
415
3
52 AA
145
Example (Laplace transform solution of an
ODE) Substitute back:
2
2
1
5
1
13
)2)(1)(1(
1022
sss
s
sss
Combine both partial fractions:
)1)(23(
10
23
3)(
222
sssss
ssY
2
2
1
5
1
13
2
1
1
2)(
2 sss
s
sssY
146
Example (Laplace transform solution of an ODE)
Tidy up:
Step 5. Invert using tables. Look at each term separately:
2
3
1
7
1
1
13)(
22
ssss
ssY
ts
sL cos
121
ts
L sin1
12
1
tes
L
1
11 tes
L 21
2
1
147
Example (Laplace transform solution of an ODE)
Combine to invert Y(s):
2
13
1
17
1
1
13
2
3
1
7
1
1
13)}({)(
112
12
1
2211
sL
sL
sL
s
sL
ssss
sLsYLty
tt eettty 237sincos3)(
148
Inversion of typical terms in partial fractions
(I) Inversion of)( as
A
From tables: atAeas
AL
)(1
(II) Inversion ofnas
A
)(
)!1(
1
)(
1111
n
tAe
sLAe
s
ALe
as
AL
nat
nat
nat
n
149
Inversion of typical terms in partial fractions
(III) Inversion of22)( bas
BsA
221
221
)(
)()(
)( bas
asBBaAL
bas
BsAL
2222
122
1
)( bs
Bs
bs
BaALe
bas
BsAL at
221
221)(
bs
sBL
bs
bL
b
BaAe at
150
Inversion of typical terms in partial fractions
From tables:
btBbt
b
BaAe
bas
BsAL at cossin
)(
)( 221
151
Even more on Laplace transforms and ODE’s
The Laplace transform method gives the result:
)(
)(
)(
)()(
sP
sF
sP
sQsY
Here:
)())((
1
)(
1
21 nssssP
152
Even more on Laplace transforms and ODE’s
In the special case when f(t) is a polynomial, exponential, sine or cosine or a sum of such terms, then F(s) can be written in the partial fraction form:
Thus the partial fraction form for F(s)/P(s) will be:
kk
s
C
s
C
s
C
s
B
s
B
s
BsF
221
1
1
1
1
1
1
)()()()(
)()()(
)()()()(
)(
1
1
1
1
1
1
221
1
1
1
1
1
1
s
a
s
a
s
a
s
c
s
c
s
c
s
b
s
b
s
b
sP
sFkk
153
Even more on Laplace transforms and ODE’s
Thus the part of y(t) arising from the inverse of F(s)/P(s) will contain the following parts:
• The same kinds of functions as in the rhs f(t). (This part corresponds to the particular integral of Module 3)
• A sum of several exponentials of the form below(This part corresponds to the complementary function of Module 3)
ti
iea
154
Even more on Laplace transforms and ODE’s
The real part of all the i determine whether or not the system is stable:
• If any Re{i} are positive the solution will grow exponentially with time. That means the system is unstable.
• If all Re{i} are negative the solution will decay exponentially with time. That means the system is stable.
155
Even more on Laplace transforms and ODE’s
• If all Re{i} are zero the solution will oscillate with time. That means the system is stable – except for the case of resonance.
• The case of resonance occurs when one of the i is the same as one of the i. In this case terms like that below will occur: 2)( i
i
s
a
This leads to terms in the solution of the form below which correspond to weak instability. tite
156
Final example (summary of all methods)
Solve the following problem using the integrating factor method, the guessing method and the Laplace transform method:
tydt
dy with 1)0( y
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Final example (summary of all methods)
(a) Integrating factor method (Module 1).
Compare with the standard form:
)()( tfytgdt
dy
In our case g(t)=1, thus the integrating factor is given by:
tedttg )(exp
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Final example (summary of all methods)
Multiply the ODE by the integrating factor:
teyedt
dye ttt
Integrate:
teyedt
d tt
Use the product rule in reverse:
Cdtteye tt
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Final example (summary of all methods)
Use integration by parts on the rhs:
Use the initial condition y(0)=1:
Divide by the integrating factor:
Cete
CdteteCdtteye
tt
tttt
tCety 1
2101 0 CCe
Substitute back: tety 21
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(b) Guessing method (see Module 3).
Final example (summary of all methods)
Thus:
Step 1. Find the complementary function. Solve:
101
tCF Cey
Try y=Cet. This gives the characteristic equation:
0 ydt
dy
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Final example (summary of all methods)
Step 2. Find the particular integral. Try
Equate coefficients:
Substitute in the ODE:
btayPI
bdt
dyPI
tbtab
1: term bt
10: term1 baab
tyPI 1
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Final example (summary of all methods)
Step 3. Combine the particular integral and the complementary function. The general solution is:
CFPI yyy tCety 1
Step 4. Use the initial condition y(0)=1:
2011 0 CCe
Substitute back:tety 21
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Final example (summary of all
methods) (c) Laplace Tranform method (see Module 5)
Step1. Define your transform:
Step 2. Transform the ODE:
0
)()()( dttyetyLsY st
2
1)()]0()([
ssYyssYtLy
dt
dyL
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Final example (summary of all methods)
Tidy up:
Step 3. Use the initial condition y(0)=1 and solve for Y(s):
2
1)0()()1(
sysYs
2
11)()1(
ssYs
)1(
1
1
1)(
2
ssssY
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Final example (summary of all methods)
Step 4. Put in partial fraction form. Find A,B,C for:
Use cover-up to find B:
)1()1(
122
s
C
s
B
s
A
ss
)1()1(
122
s
C
s
B
s
A
ss 1 0 Bs
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Final example (summary of all methods)
Use cover-up to find C:
)1()1(
122
s
C
s
B
s
A
ss1 101 Css
At this stage we have:
)1(
11
)1(
122
sss
A
ss
Put s = 1 : 1)11(
1
1
1
1)11(1
122
AA
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Final example(summary of all methods) Substitute back:
)1(
111
)1(
122
sssss
Combine all terms in Y(s):
)1(
1
1
1)(
2
ssssY
1
111
1
1)(
2
sssssY
1
211)(
2
ssssY
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Final example (summary of all methods) Step 5. Invert using tables:
)1(
211)}({)(
211
sssLsYLty
tetty 21)(
)1(
12
11)( 1
211
sL
sL
sLty
