Lecture 12
Inverting Laplace Transforms
Recall: Laplace Transforms
Laplace Transform as a mapping from time-domain to s-domain:
L : f (t) → F (s)
Z∞
y (t) 7→ Y (s) =
e −st y (t)dt
0
How can we go back to the time domain?
This mapping is bijective for most practical cases
ex: time-domain: y (t) = 12
⇒
1
s-domain: Y (s) = 2s
We can also reverse the arrow...
1
s-domain: H(s) = 2s
⇒ time-domain: h(t) = 12
Lecture 12
Inverting Laplace Transforms
Bijectivity of the Laplace Transform
L : F (s) =
R∞
0
(not true for all functions)
e −st f (t)dt
time-domain
s-domain
L−1 : y (t) = L−1 {Y (s)}
f (t)
C
F (s)
C
s
Ce at
C
s−a
tn
n!
s n+1
sin(ωt)
ω
ω 2 +s 2
cos(ωt)
s
ω 2 +s 2
..
.
..
.
Lecture 12
Inverting Laplace Transforms
1
ex: Suppose Y (s) = s+6
, find y (t).
Y (s) =
C
s − a
y (t) = L−1
C = 1,
1
s +6
a = −6
= e −6t
ex: Suppose G (s) = s 212
, find g(t).
+16
G (s) = C
ω
s +ω 2
g (t) = L−1
C = 3,
12
s 2 + 16
= 3 sin(4t)
ω=4
Lecture 12
Inverting Laplace Transforms
s−1
ex: Suppose Y (s) = 9+s
2 , find y (t)
−1
s
+
2
9+s
9 + s2
s
ω
+ 2
=A 2
2
s +ω
s + ω2
| {z } | {z }
Y (s) =
L{sin ωt}
ω=3
L{cos ωt}
3
s
=A 2
+ 2
s +9 s +9
−1 3
s
Y (s) =
+
3 s2 + 9 s2 + 9
1
y (t) = − sin(3t) + cos(3t)
3
A · 3 = −1
A=−
1
3
Lecture 12
Inverting Laplace Transforms
4
ex: Suppose H(s) = s 2 +6s+25
, find h(t).
Kind of looks like previous example, but denominator is not quite right.
Try completing the square.
b
d = 2a
b2
e = c − 4a
as 2 + bs + c = a(s + d)2 + e
4
2
s + 6s + 25
4
=
(s + 3)2 + 16
H(s) =
d = 62 = 3
e = 25 − 36
4 = 16
Looks similar to LT of sin(4t), but s is shifted.
Lecture 12
Inverting Laplace Transforms
First Shift Theorem: Multiplication by e at
L e at f (t) =
Z ∞
Z0 ∞
=
Z ∞
e −st e at f (t)dt
e −(s−a)t f (t)dt
let s̃ = s − a
0
e s̃t f (t)dt = F (s̃)
0
L e at f (t) = F (s − a)
Take home:
If you recognize something in the s-domain but with s → s − a, multiply
by e at in the time domain.
Lecture 12
Inverting Laplace Transforms
ex: Suppose H(s) = (s+3)42 +16 , find h(t).
H(s) =
4
s 2 + 16 s=s+3
h(t) = e
−3t
−1
L
=
L {sin(4t)}
= e −3t sin(4t)
4
2
s + 16
s=s+3
Lecture 12
Inverting Laplace Transforms
3
2
+ s+6
ex: Suppose Y (s) = s(s+6)
, find y (t).
Y (s) =
3
+2L e −6t
s(s + 6)
| {z }
L{???}
Partial fraction decomposition
A
B
3
= +
s(s + 6)
s
s +6
multiply by denominator
3 = A(s + 6) + B · s
= 6A + (A + B)s
True for all s ⇒ coefficients must match
constant terms:
s terms:
1
2
3 = 6A
A=
0=A+B
B = −A = −
1
2
Lecture 12
Inverting Laplace Transforms
1
1 1
2
−
+
2s
2s +6 s +6
1
3 1
=
+
2s 2s +6
3 1
+ L e −6t
=L
2
2
Y (s) =
y (t) =
1 3 −6t
+ e
2 2
Lecture 12
Inverting Laplace Transforms
s+4
ex: Suppose Y (s) = (s−4)
2 (s+1) , find y (t)
A
C
s +4
B
=
+
+
2
2
(s − 4) (s + 1)
(s − 4)
(s − 4) (s + 1)
s + 4 = A(s + 1) + B (s − 4)(s + 1) +C (s − 4)2
|
{z
}
| {z }
s 2 −3s−4
s 2 −8s+16
2
s + 4 = (B + C )s + (A − 3B − 8C )s + A − 4B + 16C
match coeffiecients
s2 :
s:
B +C =0
⇒ B = −C
B=−
3
25
A − 3B − 8C = 1
A − 5C = 1
⇒ A = 1 + 5C
s 0 :A − 4B + 16C = 4
1 + 25C = 4 ⇒ C =
3
25
A=
40
25
Lecture 12
Y (s) =
Inverting Laplace Transforms
3
1
1
3
1
40
−
+
2
25 (s − 4)
25 (s − 4) 25 (s + 1)
| {z }
| {z }
| {z }
L{e 4t }
L{t}
L{e −t }
s=s−4
40 4t −1 1
3
3
y (t) = e L
− e 4t + e −t
2
25
s
25
25
40 4t
3 4t
3 −t
= te − e + e
|25{z } |25
{z 25 }
yp
yh
Next lecture, we learn how to find Y(s) from an ODE.
Lecture 12
Inverting Laplace Transforms
ex: Suppose Y (s) = ((s+3)2s−4
, find y (t)
+16)(s+4)
s −4
((s + 3)2 + 16)(s + 4)
=
As + B
C
+
2
(s + 3) + 16 s + 4
s − 4 = (As + B)(s + 4) + C (s 2 + 6s + 25)
s − 4 = (A + C )s 2 + (4A + B + 6C )s + 4B + 25C
match coeffiecients
s2 : A + C = 0
⇒ A = −C
A=
8
17
s : 4A + B + 6C = 1
B + 2C = 1
s0 :
⇒ B = 1 − 2C
4B + 25C = −4
4 + 17C = −4 ⇒ C =
−8
17
B=
33
17
Lecture 12
Y (s) =
Inverting Laplace Transforms
8
33
s
1
8
1
+
−
2
2
17 (s + 3) + 16 17 (s + 3) + 16 17 s + 4
| {z }
{z
}
{z
}
|
|
=
L{e −4t }
1
L{e −3t sin(4t)}
4
???
"
#
8
s +3
33
1
3
+
−
2
2
17 (s + 3) + 16 (s + 3) + 16
17 (s + 3)2 + 16
|
{z
} |
{z
}
L{e −3t cos(4t)}
−
y (t) = e
|
1
L{e −3t sin(4t)}
4
8 1
17 s + 4
−3t
8
9
8
cos(4t) +
sin(4t) − e −4t
17
68
17
{z
} | {z }
yh
yp
Next lecture, we learn how to find Y(s) from an ODE.
Lecture 12
Inverting Laplace Transforms
Summary: Inverting the Laplace transform of a function
s+4
ex: Y (s) = (s−4)
2 (s+1)
1. Do some algebra to get a sum of ”easy” terms
partial fraction decomposition
completing the square
1
1 1
1 1
ex: Y (s) = 85 (s−4)
2 − 5 (s−4) + 5 (s+1)
2. Transform back from Y (s) to y (t) using Laplace transform tables
Tackle each term in the sum individually.
Go slowly when applying shift theorems
ex:
8
1
1
1
y (t) = e 4t L−1
− e 4t + e −t
5
s2
5
5
8
1
1
= te 4t − e 4t + e −t
5
5
5