ECE307-2
The Inverse Laplace Transform
Electrical and Computer Engineering Department
Cal Poly Pomona
The Laplace Transform
Inverse Transform
Techniques
In general, we define Laplace function as
F (s ) =
a s n + an −1s n −1 + ... + a1s + a0
N (s )
= n m
D(s ) bms + bm −1s m −1 + ... + b1s + b0
F (S ) =
N (s )
D(s )
is called proper rational function if n<m
F (S) =
N (s )
D (s )
is called improper rational function if n>=m
Only proper rational function can be expressed as a sum of partial
fraction
•The roots of N(s)=0 are called the zeros of F(s)
•The roots of D(s)=0 are called the poles of F(S)
•Use partial fraction expression to break F(s) down into simple terms
•Find the inverse transform from the table for each simple term.
ECE 307-2 # 2
1
Inverse Transform Techniques
Distinct Real Roots of D(s) (simple poles)
F (s ) =
N (s )
N (s )
=
D(s )
(s + p1)(s + p2 )...(s + pm )
where are simple poles and for all . N(s) has less than the degree of D(s).
Then, the partial fraction expansion
F (s ) =
N (s )
K1
K2
Km
=
+
+ ... +
D(s ) (s + p1) (s + p2 )
(s + pm )
The expansion of coefficients are known as the residues of F(s).
Ki = (s + pi )F(s)
s =− pi
We can find the inverse transform of each term since
L−1 
K 
− at
 = Ke
s + a 
ECE 307-2 # 3
Inverse Transform Techniques
Distinct Real Roots of D(s) (simple poles)
Example:
F (s ) =
K3
96(s + 5)(s + 12) K1
K2
≡
+
+
s(s + 8)(s + 6)
s (s + 8) (s + 6)
To find K1, we multiply both sides by s and the evaluate both sides at s=0.
K1 = sF (s ) =
96(s + 5)(s + 12)
(s + 8)(s + 6)
=
s =0
96(0 + 5)(0 + 12)
= 120
(0 + 8)(0 + 6)
To find K2, we multiply both sides by (s+8) and the evaluate both sides at s=-8.
K 2 = (s + 8)F (s ) =
96(s + 5)(s + 12)
s(s + 6)
=
s =−8
96( −8 + 5)( −8 + 12)
= −72
( −8)( −8 + 6)
To find K3, we multiply both sides by (s+6) and the evaluate both sides at s=-6.
K 3 = (s + 6)F (s ) =
96(s + 5)(s + 12)
s(s + 8)
=
s =−6
96( −6 + 5)( −6 + 12)
= 48
( −6)( −6 + 8)
ECE 307-2 # 4
2
Inverse Transform Techniques
Distinct Real Roots of D(s) (simple poles)
Example:
F (s ) =
96(s + 5)(s + 12) 120
72
48
≡
−
+
s(s + 8)(s + 6)
s
(s + 8) (s + 6)
Take the inverse transform
 96(s + 5)(s + 12) 
72
48 
−1  120
−
+
f (t ) = L−1 
≡L 

+
+
+
+ 6) 
s
(
s
8)(
s
6)
s
(
s
8)
(
s



The result
f (t ) = (120 − 72e −8t + 48e −6t ) u(t )
ECE 307-2 # 5
Inverse Transform Techniques
Repeated Real Roots (Repeated poles)
Let say F(s) has 3 repeated poles
F (s ) =
N (s )
N (s )
=
D(s ) (s + p1)3 (s + p2 )...(s + pm )
F (s ) =
K13
K11
K12
K2
Km
+
+
+
... +
(s + p1)3 (s + p1)2 (s + p1) (s + p2 )
(s + pm )
We can find K with
K11 = (s + p1)3 F(s )
s =− p1
K12 =
d
(s + p1)3 F (s )
ds 
K13 =
1 d2
(s + p1)3 F (s )
2! ds 2 
s =− p
s =− p1
ECE 307-2 # 6
3
Inverse Transform Techniques
Repeated Real Roots (Repeated poles)
In general form, if you have n repeated poles at
K1i =
1 di
(s + p1)n F (s )
i ! ds i 
s =− p1
for i = 1,2,..., n
The inverse transform of
1 
t n −1
e − at
=

n
 (s + a)  (n − 1)!

L−1 
Therefore f(t) can be obtain
t 2 − p1t
e + fr (t )
2!
f (t ) = K11e − p1t + K12te − p1t + K13
ECE 307-2 # 7
Inverse Transform Techniques
Repeated Real Roots (Repeated poles)
Example:
K 23
100(s + 25) K1
K 21
K 22
≡
+
+
+
s(s + 5)3
s (s + 5)3 (s + 5)2 (s + 5)
K1 = sF (s )
s =0
=
100(s + 25)
(s + 5)3
K 21 = (s + 5)3 F (s )
s =−5
K 22 =
d
(s + 5)3 F (s )
ds
K 23 =
1 d2
(s + 5)3 F (s )
2! ds 2
s =−5
=
=
s =−5
s =0
= 20
100(s + 25)
s
d 100(s + 25)
ds
s
=
s =−5
s =−5
1 d 2 100(s + 25)
2 ds 2
s
=
= −400
100s − 100(s + 25)
s2
s =−5
=
1 d −2500
2 ds s 2
s =−5
=
s =−5
=
−2500
= −100
s2
1 5000s
2 s4
s =−5
= 20
ECE 307-2 # 8
4
Inverse Transform Techniques
Repeated Real Roots (Repeated pole)
Example:
100(s + 25) 20
400
100
20
≡
−
−
+
s(s + 5)3
s (s + 5)3 (s + 5)2 (s + 5)
The inverse transform of
400 2 −5t


f (t ) = L−1{F (s )} =  20 −
t e − 100te −5t + 20e −5t  u(t )
2


Therefore f(t) can be obtain
f (t ) =  20 − (200t 2 + 100t − 20)e −5t  u(t )
ECE 307-2 # 9
Inverse Transform Techniques
Distinct Complex Root (Complex poles)
Let’s have F(s) that the roots of D(s) are complex that are s = −α ± j β
F (s ) =
a1s + a0
K
K*
=
+
s + b1s + b0 s + α − j β s + α + j β
2
We can find K with
K=
a1s + a0
s +α + jβ
K* =
a1s + a0
s +α − jβ
s =−α + j β
s =−α − j β
= K eθ = K ∠θ
= K e −θ = K ∠ − θ
Result
f (t ) = L−1 {F (s )} =  2 K e −αt cos( β t + θ ) u(t )
ECE 307-2 # 10
5
Inverse Transform Techniques
Distinct Complex Root (Complex poles)
Example:
F (s ) =
K1 =
K2 =
10(s 2 + 119)
K1
K2
K 2*
=
+
+
(s + 5)(s 2 + 10s + 169) s + 5 s + 5 − j 12 s + 5 + j 12
10(s 2 + 119)
(s 2 + 10s + 169)
s =−5
= 10
10(s 2 + 119)
(s + 5)(s + 5 + j 12)
s =−5 + j 12
=
10{( −5 + j 12)2 + 119}
= 0 + j 4.166
( −5 + j 12 + 5)( −5 + j 12 + 5 + j 12)
K 2 = 0 + j 4.166 = 4.166e j 90
K 2* = 0 − j 4.166 = 4.166e − j 90
Result
f (t ) = L−1 {F (s )} = 10e −5t − 8.33e −5t cos(12t + 90) u(t )
ECE 307-2 # 11
Inverse Transform Techniques
Repeated Complex Roots (Complex poles)
Let’s have F(s) that the roots of D(s) are complex that are s = −α ± j β
F (s ) =
a1s + a0
K1
K2
K1*
K 2*
=
+
+
2
2
2
(s + b1s + b0 )
(s + α − j β ) s + α − j β (s + α + j β )
s + α + jβ
2
We can find K1 with
K1 =
a1s + a0
(s + α + j β )2
s =−α + j β
= K1 eθ = K1 ∠θ1
and find K2
K2 =
d  a1s + a0 
ds  (s + α − j β )2 
s =−α + j β
= K 2 e −θ = K 2 ∠ − θ 2
The result
f (t ) = L−1 {F (s )} =  2 K1 te −αt cos( β t + θ1) + 2 K 2 e −αt cos( β t + θ 2 ) u(t )
ECE 307-2 # 12
6
Inverse Transform Techniques
Repeated Complex Roots (Complex poles)
Let’s have F(s) that the roots of D(s) are complex that are s = −α ± j β
F (s ) =
a1s + a0
K1
K2
K1*
K 2*
=
+
+
2
2
2
(s + b1s + b0 )
(s + α − j β ) s + α − j β (s + α + j β )
s + α + jβ
2
ECE 307-2 # 13
7