TRANSPORMASI
60
description
TRANSPORMASI. LAPLACE. Ditemukan oleh Piere Simon Maequis de Laplace tahun (1747-1827) seorang ahli astronomi dan matematika Prancis Menurut; fungsi waktu atau f(t) dapat ditranspormasi menjadi fungsi komplek atau F(s) Dimana s bilangan komplek dari s = s + j 2 p f atau s + j - PowerPoint PPT Presentation
Transcript of TRANSPORMASI
• Ditemukan oleh Piere Simon Maequis de Laplace tahun (1747-1827) seorang ahli astronomi dan matematika Prancis
• Menurut; fungsi waktu atau f(t) dapat ditranspormasi menjadi fungsi komplek atau F(s)– Dimana s bilangan komplek dari s = + j2f
atau + j = frekuensi neper = neper/detik = frekuensi radian = radian/detik
• Hasil TL dari f(t) di beri nama F(s)• Tanda TL diberikan dengan £ atau L, dan
fungsinya di tulis
f(t): nilai komplek dari fungsi sebuah fariabel tF(s): Nilai komplek dari fungsi sebuah fariabel s
)]([)( tfLsF
Inverse Transformasi Laplace
• Inverse (Bilateral) Transform
• NotationF(s) = L{f(t)} variable t tersirat untuk Lf(t) = L-1{F(s)} variable s tersirat
untuk L-1
dtetfsFtfL ts ).()()]([
0
sFtf L
Contoh: Transpormasi Laplace
1. f(t) = A– Jawab
dteAALtfL ts .)()]([
0
0][ ste
sA
sA
sA
eesA
)10(]11[ 0
Contoh 2. f(t) = At Jawab
Dibantu dengan formula integral partsiel yaitu
dteAtAtL st
0
.][
dtetA st
0
.
VUUVUV ''
00
..][ stst dets
AdtetAAtL
2
0
00
)10(10
][1)100(
.1].[
sA
ssA
ess
A
dteets
A
st
stst
• Contoh 3 f(t) = e-at
jawab
0
)(
0
.][
dte
dteeeL
tsa
statat
sa
sa
esa
tsa
1
)10()(
1)(
10
)(
• Contoh 4 : f(t) = t.e-at
dteetetL statat
..].[0
0)(
0)(
0
)(0
)(
0
)(
][)(
1].[)(
1
].[)(
1
.
tsatsa
tsatsa
sa
esa
etsa
dteetsa
dtet
0)( )( tsat etLim
2)(1
)10()(
1)00()(
1].[
sa
sasaetL at
5. f(t) = Sin(t)6. f(t) = Cos (t)7. f(t) = Sin(t+)8. f(t) = e-at. Sin(t)
• Contoh 9;
• f(0+) artinya harga nol untuk fungsi, jika didekati dari arah positif
dtdftf )(
)0()(.)}(.)0(0{
.][
)(][
00
0
0
fsFssFsf
dtefsfe
dfe
dtedtdf
dtdfL
stst
st
st
• Contoh 10; dttftf )()(
00
0
)(])([1
})({)([
dttfedttfes
dtdttfedttfL
st
st
st
)(1])(0[1
)(1}])({[1
0
0
sFs
dttfs
sFs
dttfes
st
s
dttf
ssF
sFs
tfs
)0(
0
])()(
)(1)](1
0intarg])( )0( padafungsiegralahartinyadttf
)0(])( 1
)0( fditulisdapatdttf
f(t) L(f) f(t) L(f)1 1 1/s 7 cos t
2 t 1/s2 8 sin t
3 t2 2!/s3 9 cosh at
4 tn
(n=0, 1,…)
10 sinh at
5 ta
(a positive)
11 eat cos t
6 eat 12 eat sin t
1
!ns
n
1
)1(
as
a
as 1
1
!ns
n
1
)1(
as
a
as 1
22 ss
22 s
22 ass
22 asa
22)(
asas
22)(
as
Some useful Laplace transforms
f(t) F(s)=L[f(t)]
ntate
)t( 1)t(u
t
)atsin()atcos()at(sh)at(ch
)1n(s/!n
2s/1
)as/(1 )as/(a 22 )as/(s 22 )as/(a 22 )as/(s 22
s/1
Some useful Laplace transforms
f(t) F(s)=L[f(t)]
)atsin(ebt ]a)bs/[(a 22
)bs)(as/(1 ]a)bs/[()bs( 22 )atcos(ebt
ba )ab/()ee( atbt
ba )bs)(as/(s )ab/()aebe( atbt
sFt fsFtf LL2211 and
? 22112211 sFasFatfatfa L
sFasFadtetfadtetfa
dtetfatfatfatfaLstst
st
2211
0 220 11
0 22112211
L F(s)f(t)
Laplace Transform Properties• Linear atau Nonlinear?
• Linear operator
contoh
• Seperti gambar disamping, muatan awal kapasitor = 0. Tentukan persamaan arusnya;
CR
V
• Transpormasi Laplace
vdtiC
RI
CqRIV
1
sV
sf
sCIRI
vdtiC
RI
SS
0
.
1
1
)0,0(0)0(0
1
qtidtf
CRSR
VI
CRSR
VI
sCRs
VI
sV
sCRI
sV
sCI
RI
s
s
s
s
sS
.1
1.1
.1
.1.
)(
)(
)(
)(
)()(
• Pembalikan transpormasi laplace
CRSR
VCR
SRVI s
.1
1.1
1
1
1)(
1
• Lihat tabel
RCt
t eRVI
)(
Contoh 2
• Gambar RL seperti gambar disamping, jika saklar s di on-kan maka tentukan persamaan arunya
• Persamaan rangkaian
VRIdtdiL
• Transpormasi Laplace
VRIdtdiL
VRIdtdiL
.
)0,0,0)0(,)0( )()( itisVRIisIL ss
• Transpormasi dari cos t
Laplace transformDefinition of function f(t)
0
stdte)t(f)s(F)]t(f[L
Examples
20
st
0
st
0
st
s1dtte)]t(tu[L
s1dte)]t(u[L
1dte)t()]t([L
• f(t)=0 for t<0• defined for t>=0• possibly with discontinuities• f(t)<Mexp(t)[exponential order]• s: real or complex
t
f(t)
Definition of Laplace transform
Laplace transform
0
stdte)t(f)s(F)]t(f[L Examples
1dte)t()]t([L0
st
0st
0
st00 edte)tt()]tt([L
f(t)Diract)t(
t
f(t)
)tt( 0
0t
Laplace transform
0
stdte)t(f)s(F)]t(f[L Examples
s1
sedtedte)t(u)]t(u[L
0
st
0
st
0
st
)]t(u
f(t)Heaviside
t
f(t)
t)]tt(u 0
0ts
es
e)]tt(u[L0
0
st
t
st
0
Laplace transform
0
stdte)t(f)s(F)]t(f[L Examples
20
st
0
st
0
st
sadte
sa
satedtate)]t(r[L
0t,at)t(r f(t)Ramp
t
Laplace transform properties•Linearity
)]t(f[L)s(F 11
)]t(f[L)s(F 22
tstanConsc,c 21
)s(F.c)s(F.c)]t(f[L.c)]t(f[L.c
)]t(f.c)t(f.c[L
2211
2211
2211
Laplace transform properties
• Translation
)as(F)]t(fe[L at a) if F(s)=L[f(t)]
)as(Fdte)t(fdte])t(fe[)]t(fe[L t)as(
0
st
0
atat
Example4s
s)]t2(Cos[L 2
5s2s1s
4)1s(1s)]t2(Cose[L 22
t
Laplace transform properties
• Translation
b) if g(t) = f(t-a) for t>a = 0 for t<a
)s(Fe)]t(g[L as
due)u(fedue)u(fdte])at(f)]t(g[L su
0
as)au(s
0
st
0
at
f(t) g(t)
Example 443
s6
s!3]t[L
2t,0)t(g2t,)2t()t(g 3
4
s2
se6)]t(g[L
Laplace transform properties
•Change of time scale )as(F
a1)]t.a(f[L
)as(F
a1
adue)u(fdte])t.a(f)]t.a(f[L a
su
0
st
0
Example
1s1)]t(Sin[L 2
9s3
13s
131)]t3(Sin[L
22
• Derivatives
)0(f)s(F.s)]t(f[L]dtdf[L)]t('f[L
Laplace transform properties
)
0
st0
st
0
st 0(f)s(sFdt)t(fse)t(fedtfe)]t('f[L
)0(f)s(F.s)]t('f[L
• Derivatives
)0(f)s(F.s)]t(f[L]dtdf[L)]t('f[L
)0('f)0(f.s)s(F.s])t(f[L)]t("f[L 2
Laplace transform properties
)1n()1(2n1nn
)n()0(f.....)0(fs)0(fs)s(Fs)]t(f[L
)1i(n
1i
inn)n(
)0(f.s)s(Fs])t(f[L
•If discontinuity in a
)]a(f)a(f[e)0(f)s(F.s)]t('f[L as
)a(f)a(f
• Derivatives examplesLaplace transform properties
22s)]t(Sin[L
22ss)]t(Cos[L
dt)]t(Sin[d1)t(Cos
2222 ss
)s(s)0(Sin)]t(Sin[Ls)]t(Cos[L
)t(Cosdt
)]t[sin(d
)t(Sindt
)]t(Cos[d
dt
)]t(Cos[d1)t(Sin
)s()0(Cos)]t(Cos[Ls)]t(Sin[L 22
Remarques sur la dérivation
Deux cas à prévoirdt
)t(df)t(u dt)]t(f)t(u[d
0
st dtdt
)t(dfe]dt
)t(df)t(u[L
0
st0
st dt)t(fse]e)t(f[]dt
)t(df)t(u[L
En intégrant par parties
)0(f)s(sF]dt
)t(df)t(u[L
dt)t(df)t(u
dt)t(du)t(f
dt)]t(f)t(u[d
a)
b)
)0(f)]0(f[L)]t()t(f[L]dt
)t(du)t(f[L
)s(sF]dt
)]t(f)t(u[d[L Si f(t) et toutes ses dérivées sont nulles pour t<0, alors on
peut ne pas tenir compte des valeurs initiales pour étudier le comportement
t
0s
)s(F]du)u(f[L
)s(F)0(g)]t(g[sL)]t(g[L
)t(f)t(g
t
0
]du)u(f)t(g
Laplace transform properties
• Integral
)]t(f[L)s(F
Laplace transform properties
Multiplication by t
dt)t(fe[dsd)s(F
ds)s(dF
0
st'
Leibnitz’s rule
)]t(tf[Ldt])t(tf[e]dt)t(fe[sds
)s(dF
0
stst
0
)s(F)]t(tf[L '
More general
ds)s(Fd)1()]t(ft[L
nnn
Laplace transform properties
Division by t
t)t(f)t(g )t(tg)t(f
)s(Fds
)s(dGds
)]t(g[dL)]t(f[L
s
s
du)u(fdu)u(f)s(G
s
du)u(f]t
)t(f[L
• Periodic function )t(f)Tkt(f k,t
sT
T
0
st
e1
dte)t(f)s(F)]t(f[L
Laplace transform properties
.......dt)t(fedt)t(fedt)t(fe)s(F)]t(f[LT3
T2
stT2
T
stT
0
st
.......du)T2u(fedu)Tu(fedt)t(fe)s(F)]t(f[LT
0
)T2u(sT
0
)Tu(sT
0
st
.......du)u(feedu)u(feedt)t(fe)s(F)]t(f[LT
0
susT2T
0
susTT
0
st
]dt)t(fe[e)s(F)]t(f[LT
0
st
0n
nsT
sT0n
nsT
e11e
Hint
sT0n
nsT
e11e
p1
p1ps1nn
0q
qn
1nn
1nn432n
n432n
p1)p1(s
pp...........ppppps
p...........pppp1s
p1p1ps
1nn
0q
qn
p1
1ps0q
q
1p
Laplace transform properties
Sine and cosine are periodic functions )t(jSin)t(Cose tj
dtedtee)]t(Sin[jL)]t(Cos[L]e[L0
t)sj(
0
sttjtj
sT
T
0
t)sj(
tj
e1
dte]e[L
]1e[sj
1]1ee[sj
1esj
1dte sTsTTjT0
t)sj(T
0
t)sj(
22tj
sjs
)js)(js(js
js1]e[L
Laplace transform propertiesExample
t1
-10 1 2 3
f(t)
)2s(th
s1)s(F
Laplace transform properties
Periodic function
dtedtedte)t(f1
0
st1
0
stT
0
st
s)e1(
s)1e(e1e]e
s1[e
s1[dte)t(f
2ssss2
1
st1
0
stT
0
st
)e1(s)e1(
)e1)(e1(s)e1)(e1(
)e1(s)e1()s(F s
s
ss
ss
s2
2s
)2s(th
s1
)ee(se
)ee(e)s(F2s
2s
2s
2s
2s
2s
Laplace transform propertiesExample
1
t
0 1 2 3
)e1(se
s1)s(F s
s
2
Laplace transform properties
)e1(s1
se
s1
se
sedtte s
2
s
22
ss1
0
st
1
02
sts1
0
st1
0
st1
0
st
se
sedte
s1
stedtte
)e1(se
s1
e1
dtte)s(F s
s
2s
1
0
st
•Limit behaviourInitial value
Laplace transform properties
)0(f)s(sF)]t(f[L
s
0]dt)t(fe[Lim0
stExponential order
}s)]{s(sFlim[}0t)]{t(f[Lim
0t)0(f)]t(f[Lim
s)0(f)]s(sF[Lim
•Limit behaviourFinal value
}0s)]{s(sFlim[}t)]{t(f[Lim
Laplace transform properties
)0(f)s(sF)]t(f[L
)]0(f)p(f][plim[dt)t(fdt)t(fe]0s[Lim00
st
Laplace transform applications
C
R
e0.(t) v(t)
RC circuit
0)0(v
)t(vdtdvRC)t(e
Equation describing the circuit
Laplace transform ]RCs1)[s(V)s(V)s(RCsV)s(E
RCs1)s(E)s(V
Laplace transform applicationsRCs1
)s(E)s(V
)t(e)t(e 0
Impulse function
0e)s(E RCs1e)s(V 0
RCt
Impulse response
CRt
0 eRCe)t(v
RCe0
CRt
e
RCe
)RC1s(
1)s(V 0
)1RCs(se)s(sV 0
s
0s
Laplace transform applications
RCs1)s(E)s(V
)t(ue)t(e 0
Step function
se)s(E 0 )RCs1(s
e)s(V 0
)RC1s(
ese)s(V 00
]e1[eeee)t(v CRt
0CR
t
00
e0
RC
0e63,0
]e1[e)t(v crt
0
Laplace transform applications
Step function and initial conditions v(0) 0
RCs1)0(RCv
)RCs1(se)s(V 0
)RC1s(
e)0(vse)s(V 00
CRt
00 e]e)0(v[e)t(v
)0(RCv]RCs1)[s(V)s(V)]0(v)s(sV[RC)s(E
crt
00 e]e)0(v[e)t(v
0e
)0(v
1RCs]e)0(v[RCse)s(sV 0
0
Laplace transform applicationsRCs1
)s(E)s(V
t)t(r)t(e
Ramp function
2s1)s(E
)RCs1(s1)s(V 2
)RC1s(
RCs
RCs1)s(V 2
CRt
RCeRCt)t(v
RC
CRt
)t(v
CRt
e1dtdv
)1RCs(s)RC(RC
s1)s(sV
2
RC1s
v
)RC1s(RC
)s(E)s(V 0
seE)s(E
as0
(Heaviside)
crt
0cr
at
0 ev]e1[E)at(u)t(v
)at(uE)t(e 0
Laplace transform properties
a
0E
t
RC1s
v
)RC1s(sRC
eE)s(V 0as
0
RC1s
v]
RC1s
1s1[eE)s(V 0as
0
crt
0cr
at
0 ev]e1[E)t(v
crt
0ev)t(v
at
a t
0E
0v
Laplace transform properties
crt
0cr
at
0 ev]e1[E)at(u)t(v
at
RC1s
sv
)RC1s(RC
eE)s(sV 0as
0
Laplace transform properties
Limits
s )0(vv)s(sV 0
Initial value
0s )(vE)s(sV 0
Final value
e(t)E(s)
v(t)V(s)
R
CRC1a,
as)s(aE
)RC1s(RC
)s(E)s(V
Harmonic analysis
)tsin(e)t(e 0 220 s
e)s(E
)22
0
s)(as(ae)s(V
)s
CBsas
A(ae)s(V 220
22
22
22
aaC
a1B
a1A
)s
ss
aas
1(a
ae)s(V 2222220
Laplace transform properties
)s
ss
aas
1(a
ae)s(V 2222220
)]tcos()tsin(RC
1e[a
ae)t(v CRt
220
Laplace transform properties
RC)(tg2)RC(1
1)(Cos
]e)sin()t)[sin((Cose)t(v CRt
0
Forced Transient
