lecture 3 laplace transform ing. jaroslav jíra, csc. physics for informatics
TRANSCRIPT
What is it good for?
The Laplace transform
• Solving of differential equations
• System modeling
• System response analysis
• Process control application
Solving of differential equations procedure
The Laplace transform
1. Finding differential equations describing the system
2. Obtaining the Laplace transform of these equations
3. Performing simple algebra to solve for output or variable of interest
4. Applying inverse transform to find solution
A system analysis can be done by several simple steps
The definition
The Laplace transform
1. The Laplace transform is an operator that switches a function of real variable f(t) to the function of complex variable F(s).
2. We are transforming a function of time - real argument t to a function of complex angular frequency s.
3. The Laplace transform creates an image F(s) of the original function f(t)
0
)()()( dtetftfLsF st
where s= σ+ iω
Restrictions
The Laplace transform
1. The function f(t) must be at least piecewise continuous for t ≥ 0.
2. |f(t)| ≤ Meαt where M and α are constants. The function f(t) must be bounded, otherwise the Laplace integral will not converge.
3. We assume that the function f(t) = 0 for all t < 0
Inverse Laplace transform
The Laplace transform
1. Inverse transform requires complex analysis to solve
2. If there exists a unique function F(s)=L[f(t)], then there is also a unique function f(t)=L-1[F(s)]
3. Using the previous statement, we can simply create a set of transform pairs and calculate the inverse transform by comparing our image with known results in time scope
j
j
stdsesFj
sFLtf
)(
2
1)()( 1
Basic properties
The Laplace transform
Linearity )()()]()([ 22112211 sFcsFctfctfcL
)(1
)]([a
sFa
atfL Scaling in time
Time shift
Frequency shift
)()]()([ 000 sFettuttfL st
)()]([ asFtfeL at
Another properties
The Laplace transform
Original Image
)(tf
t
dxxf0
)(
)(tf
)(atf
atetf )(
)(tf
)()( tf n
)(ttf
)(tft n
)0()0()(2 fsfsFs
)0(...)0()0()( )1(21 nnnn ffsfssFs
)(sF
)()1( )( sF nn
)(1
sFs
)0()( fssF
)(1
a
sFa
)( asF
)(sF
The most commonly used transform pairs
The Laplace transform
Original Image
t
ate
atte
atet 2
Nnt n ,
2t
Nnet atn ,
a
Original Image
)sinh( t
)cosh( t
)sin( tt
)cos( tt
)sin( teat
)cos( teat
)sin( t
)cos( t
s
a
2
1
s
3
2
s
1
!ns
n
as 1
2)(
1
as
3)(
2
as
1)(
! nas
n
22 s
22 ss
22 s
22 ss
222 )(
2
ss
222
2
)(
s
s
22)( as
22)( as
as
Unit step
The Laplace transformTransform pair deduction
01)(
00)(
tfortu
tfortu
se
sdtetuL stst 11
1)]([00
The unit step u(t)
is defined by
The Laplace image s
tuL1
)]([
Shifted unit step
t
1
0
u(t)
atfortf
atfortf
1)(
0)(
t
1
0
u(t)
aThe Laplace image
s
ee
sdtetfL
as
aa
stst
1
1)]([s
etfL
as
)]([
Unit impulse
The Laplace transformTransform pair deduction
1
11
00)(
0,1
)(
ttandtfortf
ttfort
tf
ss
e
ts
e
tdte
ttfL
sttt stst 1111
)]([1
11
100 11
The unit impulse f(t)
is characterized by unit area under its function
The Laplace image
t
1/t1
0
f(t)
It is a unit impulse for t1 → 0
1lim1
lim)]([1
1
1
1 01
0
s
se
ts
etL
st
t
st
t 1)]([ tL
t1
1
11)]([
ts
etfL
st
Dirac delta
0,0)(
0,)(
tfort
tfort
t0
f(t)
Exponential function
The Laplace transformTransform pair deduction
atetf )(
2
0
20 00
10
)(1)]()([
ss
edt
s
e
s
etdttetutfL
stststst
Linear function
asas
tasedtedteetutfL tasstat
1
)(
)()]()([
00
)(
0
aseat
1
ˆ
ttf )(
vuuvvu
0
)]()([ dttetutfL st
Per partes integration
s
evev
utust
st
;'
1';
2
1ˆs
t
The Laplace transformTransform pair deduction
200 00
22 122
02
)]()([ss
dttes
dts
te
s
etdtettutfL st
ststst
Square function 2)( ttf
vuuvvu
0
2)]()([ dtettutfL st
Per partes integration
s
evev
tutust
st
;'
2';2
32 2ˆs
t
The n-th power function nttf )(1
!ˆ
nn
s
nt
The Laplace transformTransform pair deduction
Cosine function )cos()( ttf
isisdte
eedtettutfL st
titist 11
2
1
2)cos()]()([
0 0
22ˆ)cos(
s
st2222
)()(
2
1)]()([
s
s
s
isistutfL
Sine function )sin()( ttf
isisidte
i
eedtewttutfL st
titist 11
2
1
2)sin()]()([
0 0
22ˆ)sin(
s
t2222
2
2
1)()(
2
1)]()([
s
i
is
isis
itutfL
The Laplace transformTransform pair deduction
Time shift
a
stdteatfatuatfL )()]()([
)(ˆ)( sFeatf as
t0
f(t)
a
The original function f(t) is shifted in time to f(t-a)f(t-a)
.,0,
,
xtasandxatAs
axtanddtdxthenatxLet
0 0
)( )()()]([ dxexfedxexfatfL sxasaxs
Frequency shift
0
)(
0
)()()]([)]([ asFdtetfdtetfetfeL tasstatat
)()]([ asFtfeL at
The Laplace transformProperty deduction
00
0 )()0(0])[()]([])(
[ dtetfsfdtetftfedt
tdfL ststst
Time differentiation ])(
[dt
tdfL
vuuvvu
0
)(])(
[ dtedt
tdf
dt
tdfL st
Per partes integration
)(;)(
'
';
tfvdt
tdfv
seueu stst
)0()(])(
[ fssFdt
tdfL
The Laplace transformProperty deduction
00
0
00
)(1
)(]1
[])(1
[])([ dtetfs
dttfes
dxxfes
dxxfL ststt
stt
Time integration ])([0t
dxxfL vuuvvu
0 00
)(])([ dtedxxfdxxfL sttt
Per partes integration
stst
t
es
vev
tfudxxfu
1
;'
)(';)(0
)(1
])([0
sFs
dxxfLt
The algorithm of inverse Laplace transform
Inverse Laplace transform
Since the F(s) is mostly fractional function, then the most important step is to perform partial fraction decomposition of it.
Depending on roots in denominator, we are looking for the following functions, where A and B are real numbers:
for a single real root s= a
for a double real root s= a
for a triple real root s= a
for a pair of pure imaginary roots s= ± iω
for a pair of complex conjugated roots s= a ± iω
as
A
as
B
as
A
2)(
as
C
as
B
as
A
23 )()(
22
s
BAs
22)( as
BAs
Two distinct real roots
Inverse Laplace transformBasic examples of partial fraction decomposition to find the original f(t)
34
32)(
2
ss
ssF
1334
32)(
2
s
B
s
A
ss
ssF
)3()1(32 sBsAs
2
121:1
2
323:3
BBs
AAs
1
1
2
1
3
1
2
3)(
sssF
0,2
1
2
3)( 3 teetf tt
The equation s2 + 4s + 3= 0 has two distinct real roots s1= -3 and s2= -1
We have to find coefficients
A and B for
Multiplying the equation
by its denominator
Now we can substitute
Decomposed F(s)
so the original function
One real root
and one real double root
Inverse Laplace transform
2
2
)3)(1(
53)(
ss
ssF
3)3(1)3)(1(
53)(
22
2
s
C
s
B
s
A
ss
ssF
)3)(1()1()3(53 22 ssCsBsAs
133:
16232:3
248:1
2
ACCAs
BBs
AAs
3
1
)3(
16
1
2)(
2
ssssF
0,162)( 33 teetetf ttt
The denominator has a single root s1= -1 and a double root s23= -3
We are looking for coefficients
A, B and C
Multiplying the equation
by its denominator
Now we can substitute to get A,B;
the C coefficient can be obtained
by comparison of s2 factors
Decomposed F(s)
so the original function
Two pure imaginary roots
Inverse Laplace transform
16
74)(
2
s
spF
16
4
4
7
164
16
74222
ss
s
s
s
0,4sin4
74cos4)( ttttf
,sinˆcosˆ 2222t
sandt
s
s
Since we know, that
it will be helpful to rearrange
the original formula
Now we can directly
write the result
One real root
and two pure imaginary roots
Inverse Laplace transform
)4)(6(
72)(
2
ss
ssF
46)4)(6(
7222
s
CBs
s
A
ss
s
)6)(()4(72 2 sCBssAs
60
51)
10
197(
6
1647:
40
190:
40
194019:6
0
2
CCAs
ABBAs
AAs
4
2
120
51
440
19
6
1
40
19)(
22
ss
s
ssF
0,2sin120
512cos
40
19
40
19)( 6 tttetf t
We are looking for coefficients
A, B and C
Multiplying the equation
by its denominator
Now we can substitute to get A;
B,C coefficients can be obtained
by comparison of s0,s2 factors
Decomposed F(s)
so the original function
Two complex conjugated roots
Inverse Laplace transform
134
54)(
2
ss
ssF
9)2(944134 222 sssss
9)2(
3
9)2(
)2(4
9)2(
3)2(4
134
542222
ss
s
s
s
ss
s
tes
ste
s
ts
st
s
etfsF
tt
t
3cosˆ9)2(
2;3sinˆ
9)2(
3
3cosˆ9
;3sinˆ9
3
)(ˆ)2(
22
22
22
2
0),3sin3cos4()( 2 tttetf t
We have to rearrange the denominator in the first step
Decomposed F(s)
Now we have to assemble all necessary relations
so the original function
Example 1
Find the x(t) on the interval <0,∞)
Solving of differential equations by the Laplace transform
0)0(,32 xxx
)(ˆ)( txsX
)0()(ˆ)( xsXstx
)2(
3)(
3)(20)(
sssX
ssXsXs
0),1(2
3)( 2 tetx t
The image of desired function is
From the former definitions we know, that
Then we can write
2)2(
3
s
B
s
A
ssBssA )2(3
2
323:2
2
323:0
BBs
AAs
)2(2
3
2
3)(
sssX
The original function
Example 2
Function on the right side
Solving of differential equations by the Laplace transform
3)0(,2sin4 xtxx
22ˆsin
s
t)0()(ˆ)( xsXstx
4
2)(43)(
2 s
sXsXs
Necessary relations
Equation in the Laplace form
44)4)(4(
222
s
CBs
s
A
ss)4)(()4(2 2 sCBssA
10
10:
10
1202:4
2
ABBAs
AAs5
2
4
42442:0
ACCAs
)4)(4(
2
4
3)(
4
23)4)((
22
ssssX
sssX
Example 2 - continued
Solving of differential equations by the Laplace transform
452
101
4101
4
3)(
2
s
s
sssX
0,2sin5
12cos
10
1
10
31)( 4 tttetx t
A formula for the X(s) after the partial fraction decomposition
after some small arrangements
The original function
452
101
41031
)(2
s
s
ssX
4
2
5
1
410
1
41031
)(22
ss
s
ssX
Example 3
Homogeneous second order LDR
Solving of differential equations by the Laplace transform
4)0(;2)0(,096 xxxxx
0)(9)2)((642)(2 sXsXsssXs
)0()(ˆ)( xsXstx
0,)102()( 3 tettx t
Equation in the Laplace form
The original function
Necessary relations
)0()0()(ˆ)( 2 xxssXstx
96
162)(162)96)((
22
ss
ssXssssX
222 )3(
10
3
2
)3(
10)3(2
)3(
162)(
sss
s
s
ssX
Example 4
Inhomogeneous second order LDR
Solving of differential equations by the Laplace transform
4)0(;0)0(,2cos24 xxtxx
4
2)(4402)(
22
s
ssXsXs
)0()(ˆ)( xsXstx
0,2sin)4(2
1)( ttttx
Equation in the Laplace form
The original function
Necessary relations
)0()0()(ˆ)( 2 xxssXstx
22222
)4(
2
4
4)(
4
24)4)((
s
s
ssX
s
sssX
ttttx 2sin2
12sin2)(
222 )(
2ˆsin
s
sttknowing that
Example 5
Integro-differential equation
Solving of differential equations by the Laplace transform
1)0(,1)(0
xdxxt
ssssX
ssX
sssX
11)
1)((
1)(
11)(
)0()(ˆ)( xsXstx
0,sincos)( ttttx
Equation in the Laplace form
The original function
Necessary relations );(1ˆ)(
0
sXs
dxt
1
1
11
1)(1)1)((
2222
ss
s
s
ssXsssX