Solving LTI ODE Using Laplace and Fourier Transforms
Laplace transform
Define auxiliary function
g
(
,
t
):
Pb. 1:
if lim
t
f
(
t
)
0:
FT is not defined
Pb. 2:
FT does
not
capture well sinusoidal with damping/divergence
Fourier transform (FT) has 2 problems:
Its FT is:
Laplace transform:
Inverse Laplace
Laplace transform properties
Independent
variable:
t
Dependent
variable:
y
Function is a
mapping:
y
=
f
(
t
)
Time domain
t
y
f
(
t
)
F
(
s
)
e
–
at
1/(
s+a
)
sin(
at
)
a
/(
s
2
+a
2
)
cos(
at
)
s
/(
s
2
+a
2
)
(
t
)
1 /
s
t
k
/k!
(
k
0)
1
/
s
k+
1
f
(
t
)
F
(
s
)
g
(
t
)=
df
(
t
)/
dt
G
(
s
)
=s F
(
s
) –
f
(0)
g
(
t
)=
f
(
t
)
dt
G
(
s
)
=
(1/
s
)
F
(
s
)
g
(
t
)=
(
t–a
)
f
(
t–a
)
G
(
s
)
=
(e
–
at
)
F
(
s
)
g
(
t
) = e
–
at
f
(
t
)
G
(
s
)
=F
(
s+a
)
Application to solve LTI – ODE
Take
: m
=2,
c
=4,
k
=20, and
f
1
=60
Inverse transform
To get the inverse transform: Partial fractions
Required to inverse:
1.
Get factors
of
D
n
(
s
):
First get solutions of D
n
(
s
)
=
0: say
p
1
,
p
2
, …,
p
n
D
n
(
s
) = (
s
–
p
1
) (
s
–
p
2
) … (
s
–
p
n
)
2.
If complex
conjugate solutions:
p
1
=
+
j
,
p
2
=
–
j
,
group: (
s
–
p
1
) (
s
–
p
2
)
= (
s
2
+ 2
s
+
2
+
2
)
= ((
s
+
)
2
+
2
)
G
(
s
) =
N
m
(
s
) /
D
n
(
s
)
3.
Rewrite
G
(
s
) as:
4.
Get
A
i
by recombining the RHS
and comparing with LHS
N
&
D
are polynomials in
s
of degree
m
,
n
:
n
m
f
(
t
)
F
(
s
)
(
t
)
1 /
s
t
k
/k!
(
k
0)
1
/
s
k+
1
e
–
at
1/(
s+a
)
e
–
bt
f
(
t
)
F
(
s+b
)
sin(
at
)
a
/(
s
2
+a
2
)
cos(
at
)
s
/(
s
2
+a
2
)
Get the inverse from:
To get the inverse transform:
To get the inverse of:
1. Get factors of the denominator:
s
(
s+a
) (
s+b
)
[where:
a
= – 1 + 3
j
,
b
= – 1 – 3
j
]; [Note that (
s+a
)(
s+b
) = (
s
+1)
2
+ 3
2
]
2. Expand original fraction to partial fractions:
Use partial fractions
3. Get numerators by equating both sides
4. Use tables
f
(
t
)
F
(
s
)
(
t
)
1 /
s
t
k
/k!
(
k
0)
1
/
s
k+
1
e
–
at
1/(
s+a
)
e
–
bt
f
(
t
)
F
(
s+b
)
sin(
at
)
a
/(
s
2
+a
2
)
cos(
at
)
s
/(
s
2
+a
2
)
Application of Laplace and Fourier transforms to solve Linear Time-Invariant Ordinary Differential Equations (LTI ODEs) with specific functions and parameters. Learn how to find the inverse transform using partial fractions and handle complex solutions.
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Presentation Transcript
Laplace transform 1 Fourier transform (FT) has 2 problems: t Pb. 1: if limt f(t) 0: Pb. 2: FT does not capture well sinusoidal with damping/divergence FT is not defined 3 Define auxiliary function g( , t): l=0.2 =0.2 l=0 =0 f(t) l=-0.2 = 0.2 1 0 0 0 t t 2 ( ) t ( ) ( ) t e = ( ) = t , g t f t 1 ( ) ( ) t e ( ) j t = t F s f t e dt 0 Its FT is: 0 1 2 3 4 5 ( ) ( ) + j t = e f t dt -1 0 Laplace transform: ( ) f t ( ) -2 = t cos e t ( + ( ) ( ) = 0 st F s e f t dt ) + + 0 s ( ) = F s 1 + j ( ) ( ) ( ) 2 = Inverse Laplace st 2 0 f t e F s ds s 2 j j Automatic Control 211 ASU MN Sabry 2021 II.2.b 1
Laplace transform properties Independent variable: t Independent variable: s Im Dependent variable: y Dependent variable: Y Im C Re C Re Function is a mapping: y=f(t) Function is a mapping: Y=F(s) y t Time domain s domain f (t) g(t)=df (t)/dt g(t)= f (t) dt g(t)= (t a) f(t a) G(s)=(e at) F (s) g(t) = e atf(t) F (s) G(s)=s F (s) f(0) G(s)=(1/s) F (s) f (t) e at sin(at) cos(at) (t) tk/k! (k 0)1/ sk+1 F (s) 1/(s+a) a/(s 2 +a 2) s/(s 2 +a 2) 1 / s G(s)=F(s+a) Automatic Control 211 ASU MN Sabry 2021 II.2.b 2
Application to solve LTI ODE x(t) ( ) 0; ( ) 0 = ( ) f ( ) + + = 2 2 md x t dt cdx t dt ( ) f t k x t = f t ( ) t k m = x dx dt f(t) 1 = = c 0 0 t t Take: m=2, c=4, k=20, and f1=60 ( ) ( ) ( ) 5 + + = 2 2 4 20 60 s X s s X s X s s x (t) 4 30 2 + ( ) 3 = X s ( ) + 2 10 s s s 2 Inverse transform 1 0 ( ) ( ) t ( ) ( ) = t t 3 cos 3 sin 3 3 x t e t e t time t 0 2 4 6 Automatic Control 211 ASU MN Sabry 2021 II.2.b 3
To get the inverse transform: Partial fractions N & D are polynomials in s of degree m, n: n m Required to inverse: G(s) = Nm(s) / Dn(s) 1. Get factors of Dn(s): First get solutions of Dn(s) = 0: say p1, p2, , pn Dn(s) = (s p1) (s p2) (s pn) 2. If complex conjugate solutions: p1 = +j , p2 = j , group: (s p1) (s p2) = (s2 + 2 s + 2+ 2) = ((s+ )2 + 2) 3. RewriteG(s) as: Get the inverse from: f (t) (t) tk/k! (k 0)1/ sk+1 e at e btf(t) sin(at) cos(at) F (s) 1 / s 1/(s+a) F(s+b) a/(s 2 +a 2) s/(s 2 +a 2) + A A As + A + ( ) = + + ... 3 n G s 1 2 ( ) 2 s p s p s 2 3 n A A ( ) = + G s 1 2 p N.B. if p1 = p2: use: ( ) 2 s p s 1 1 4. Get Aiby recombining the RHS and comparing with LHS Automatic Control 211 ASU MN Sabry 2021 II.2.b 4
To get the inverse transform: 30 2 + ( ) = Use partial fractions X s To get the inverse of: ( ) + 2 10 s s s 1. Get factors of the denominator: s (s+a) (s+b) [where: a = 1 + 3j , b = 1 3j]; [Note that (s+a)(s+b) = (s+1)2 + 32] 2. Expand original fraction to partial fractions: A Bs C X s s s + f (t) (t) tk/k! (k 0)1/ sk+1 e at e btf(t) sin(at) cos(at) F (s) 1 / s + ( ) = + ( ) 2 + 2 1 3 3. Get numerators by equating both sides 1/(s+a) F(s+b) a/(s 2 +a 2) s/(s 2 +a 2) ( s ) 2 + + + 1 1 3 s + 1 s ( ) = 3 X s ( ) 2 1 ( + ) + 1 s 1 s 1 3 3 ( ) = 3 X s ( ) ( ) 2 2 + + + 2 2 1 3 1 3 s s 4. Use tables ( ) ( ) t ( ) ( ) = t t 3 cos 3 sin 3 3 x t e t e t Automatic Control 211 ASU MN Sabry 2021 II.2.b 5
