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Inverse Laplace Transform
So far, we have dealt with the problem of finding the Laplace transform for a given function f(t), t > 0,
L {f(t)} = F(s) = # 0 "e!stf(t)dt
Now, we want to consider the inverse problem, given a function F(s), we want to find the function f(t) whose Laplace transform is F(s). We introduce the notation L -^1 {F(s)} = f(t) to denote such a function f(t), and it is called the inverse Laplace transform of F.
Remark : The inverse Laplace transform is not unique: If
g(t) =
1 !if! 0 < t < 3 ! 8 !if!t = 3 1 !if!t > 3
#^ $
%^ $
then L {g(t)} = 1/s and L {1} = 1/s So, both functions have the same Lapalce transform, therefore 1/s has two inverse transforms. But, the only continuous function with Laplace transform 1/s is f(t) =1.
A crude, but sometimes effective method for finding inverse Laplace transform is to construct the table of Laplace transforms and then use it in reverse to find the inverse transform.
Example :
Since L {1} = 1/s, then L -^1 {1/s} = 1
Since L {t} = 1/s^2 , then L -^1 {1/s^2 } = t
Since L {cos at} = (^) s (^2) +s a 2 , then L -^1 { (^) s (^2) +s a 2 } = cos at
The following properties will simplify our calculations:
- Linear Property If c 1 and c 2 are constants, L -^1 {c 1 F 1 (s) + c 2 F 2 (s)}= c 1 L -^1 {F 1 (s)} + c 2 L -^1 {F 2 (s)}
Example : L -^1 { (^4) s! (^) s^32 } = 4 L -^1 {1/s } - 3 L -^1 {1/s^2 } = 4 – 3t.
- Inverse Translation Property Given F(s) = L {f(t)}, since L {eatf(t)} = F(s – a), L -^1 {F(s – a)} = eatf(t) = eat^ L -^1 {F(s)}
Example :
- Find L -^1 { (^) s (^2) + 154 s + 13 }
L -^1 { (^) s (^2) + 154 s + 13 } = L -^1 { (^) s (^2) + 415 s + 4 + 9 } = L -^1 { s (^2) + 415 s + 4
( ) +^9
} = L -^1 {^3 (^5 )
( s + 2 )^2 + 33
= 5 L -^1 {^3
( s + 2 )^2 + 33
} = 5 L -^1 {^3
( s! (! 2 ))^2 + 33
} = 5 e-2t^ L -^1 { (^) s (^2) +^3 32 } = 5 e-2t^ sin 3t
- Find L -^1 { (^) s (^2) +s 6 +s^1 + 25 }
L -^1 { (^) s (^2) +s 6 +s^1 + 25 } = L -^1 { (^) s (^2) + 6 ss^ + +^1 9 + 16 } = L -^1 { s^ +^1
(^ s 2 +^6 s^ +^9 ) +^16
L -^1 { (s^ +^3 )^!^2
(^ s +^3 )^2 +^42
} = L -^1 { s^ +^3
(^ s +^3 )^2 +^42
} - L -^1 {^2
(^ s +^3 )^2 +^42
} = e-3t^ L -^1 { (^) s (^2) +s 42 } –
e-3t^ ½ L -^1 { (^) s (^22) +(^2 4 ) 2 } = e-3t[cos 4t – ½ sin 4t]
- Find L -^1 { (^) s (^31) + s}.
L -^1 { (^) s (^3 1) + s} = L -^1 {^1
s (s 2 + 1 )
Let’s use partial fractions 1
s ( s^2 + 1 )
= A s + Bs s 2 ++^ C 1 = As
(^2) + A + Bs (^2) + Cs
s ( s^2 + 1 )
then, 1 = (A + B)s^2 + Cs + A
A = 1, C = 0 , A + B = 0, B = - 1 L -^1 {^1
s (s 2 + 1 )
}= L -^1 { (^1) s! (^) s 2 s+ 1 } = L -^1 { (^1) s} - L -^1 { (^) s 2 s+ 1 } = 1 – cos t
Properties :
- Commutative : f ∗g = g∗f Proof:
Since f! g(t) = $ 0 tf(")g(t # ")d",
Make the substitution u = t - τ, then τ = t – u. Since 0 < τ < t, then t < u < 0 and dτ = - du, so
f! g(t) = $ 0 tf(")g(t # ")d" = $ t^0 f(t # u)g(u) (# du) = # $t^0 f(t # u)g(u)du = $ 0 tf(t # u)g(u)du =
g! f(t)
Distributive : f ∗(g + k) = f∗g + f∗k
Associative : (f ∗g)∗k = f∗(g∗k)
f ∗0 = 0∗f = 0
Remark : f ∗ 1 ≠ f If f(t) = cos t, then
f! 1 (t) = $ 0 tcos(t " #)! 1 !d# =!" sin(t " #) 0 t^ = " sin 0 + sin t = sin t % f(t)
Theorem : In the above conditions, we have L {f∗g} = L {f}⋅ L {g} = F(s)⋅G(s) then
L -^1 {F(s)⋅G(s)} = f! g(t) = $ 0 tf(")g(t # ")d"= " 0 tg(!)f ( t # !)d!
Example :
L -^1 {^1
s (s 2 + 1 )
}= L -^1 { (^1) s! (^) s 2 s+ 1 } = f∗g = g∗f
Since L -^1 { (^1) s} = 1 and L -^1 { (^) s (^2 1) + 1 } = sin t
f! g(t) = $ 01 sin!"!d" = # cos " 0 t^ = 1 # cos t
- L -^1 { (^) s (^2)! 1 s! 12 }= L -^1 { (^) s!^1 4 " (^) s +^1 3 } = f∗g = g∗f
Since L -^1 { (^) s!^1 4 } = e4t^ L -^1 { (^1) s} = e4t1 and L -^1 { (^) s +^1 3 } = e-3t^ L -^1 { (^1) s} = e-3t 1
f! g(t) = $ 0 te^4 (t"#)e"^3 #d# = $ 0 te^4 te"^4 #e"^3 #d# = e^4 t^ $ 0 te"^7 #d# = e^4 t^ e
" 7 # " 7 0
t (^) = e 4 t e"^7 t " 7 +^
e^4 t 7 "^
e"^3 t 7
