CLASS 10
LAPLACE TRANSFORM
L f (t ) = e − st f (t ) dt = F ( s )
0
L y (t ) = Y ( s )
L g (t ) = G ( s )
Example:
1
2
−3
(a ) L 1 = ; L 2 = ; L −3 =
s
s
s
1!
1
2! 2 1
3! 3 2 1 6
4! 24
(b) L t 1 = 1+1 = 2 ; L t 2 = 2+1 = 3 ; L t 3 = 3+1 =
= 4 ; L t 4 = 4 +1 = 5
4
s
s
s
s
s
s
s
s
s
1
1
1
1
1
(c) L e at =
; L e 4t =
; L e3t =
; L e −2t =
=
s−a
s−4
s −3
s − ( −2 ) s + 2
(d ) L sin kt =
k
2
2
3
3
;ex : L sin 2t = 2
= 2
; L sin 3t = 2
= 2
2
2
2
s +k
s +4
s +9
s + ( 2)
s + ( 3)
(e) L cos kt =
s
s
s
s
s
;ex : L cos 2t = 2
= 2
; L cos 3t = 2
= 2
2
2
2
s +k
s +4
s +9
s + ( 2)
s + ( 3)
2
2
Example:
1
2
−3
(a ) L−1 = 1; L−1 = 2; L−1 = −3
s
s
s
1
2
6
24
(b) L−1 2 = t ; L−1 3 = t 2 ; L−1 4 = t 3 ; L−1 5 = t 4
s
s
s
s
1 at −1 1 4t −1 1 3t −1 1 −2t
(c) L−1
= e ;L
= e ;L
= e ;L
=e
s − a
s − 4
s − 3
s + 2
3
k
2
−1
(d ) L−1 2
= sin kt ; L−1 2
= sin 2t ; L 2
= sin 3t
2
s + k
s + 4
s + 9
s
s
s
s
s
−1
−1
−1
−1
(e) L−1 2
=
cos
kt
;
L
=
L
=
cos
2
t;
=
L
=
L
= cos 3t
2
2
2
2
2
2
2
s + k
s + 4
s + 2
s + 9
s + ( 3)
LAPLACE TRANSFORM FOR DERIVATIVE
dy
i ) L = sY ( s ) − y (0)
dt
d2 y
ii ) L 2 = s 2Y ( s ) − sy (0) − y (0)
dt
d3y 3
iii ) L 3 = s Y ( s ) − s 2 y (0) − sy (0) − y(0)
dt
dy
+ 6 y = e 4t , y (0) = 2.
dt
Step1: Laplace Transform each term
example past sem final exam Qn :
dy
L + 6 L y = L e 4 t
dt
1
s−4
Step 2 : Substitute y (0) = 2
sY ( s) − y (0) + 6Y ( s) =
sY ( s) − 2 + 6Y ( s) =
1
s−4
Step3 : Simplify Y ( s )
1 + 2 ( s − 4 ) 1 + 2s − 8 2s − 7
1
+2=
=
=
s−4
s−4
s−4
s−4
2s − 7
Y (s) ( s + 6 ) =
s−4
A ( s + 6) + B ( s − 4)
2s − 7
A
B
Y (s) =
=
+
=
( s − 4 )( s + 6 ) s − 4 s + 6
( s − 4 )( s + 6 )
sY ( s ) + 6Y ( s ) =
*partial fraction
2s − 7 = A ( s + 6 ) + B ( s − 4 )
A : Let s − 4 = 0 s = 4
2 ( 4 ) − 7 = A ( 4 + 6 ) + B ( 4 − 4 ) 1 = 10 A A =
1
10
B : Let s + 6 = 0 s = −6
2 ( −6 ) − 7 = A ( −6 + 6 ) + B ( −6 − 4 ) −19 = −10 B B =
1
19
2s − 7
Y ( s) =
= 10 + 10
( s − 4 )( s + 6 ) s − 4 s + 6
Step 4 : Inverse Laplace Transform
L−1 Y ( s ) =
y (t ) =
1 −1 1 19 −1 1
L
+ L
10
s − 4 10
s + 6
1 4t 19 −1 1 1 4t 19 −6t
e + L
= e + e .
10
10
10
s − ( −6 ) 10
19
10
dy
− y = 1, y (0) = 0.
dt
Step1: Laplace Transform each term
Easy Example :
dy
L − L y = L 1
dt
1
s
Step 2 : Substitute y (0) = 0
sY (s) − y (0) − Y (s) =
1
s
Step3 : Simplify Y ( s )
sY (s) − 0 − Y ( s) =
Y ( s ) ( s − 1) =
Y (s) =
1
s
1
A
B
A( s − 1) + B ( s )
= +
=
s ( s − 1) s s − 1
s ( s − 1)
*partial fraction
1 = A( s − 1) + B( s )
A : Let s = 0 − A = 1 A = −1
B : Let s − 1 = 0 s = 1 1 = B
Y ( s) =
1
−1
1
=
+
s ( s − 1) s s − 1
Step 4 : Inverse Laplace Transform
1
1
L−1 Y ( s ) = − L−1 + L−1
s
s − 1
y (t ) = −1 + et .#
0
