Numerical Laplace Transform
inversion
Partial fraction expansion is a popular method to find
inverse Laplace transform. However it requires
finding poles and residues. Finding poles is
expensive and inaccurate. Here, we will discuss a
numerical method for inverse Laplace transform
calculation based on Padé approximation.
From the definition of inverse Laplace transform we
have
(i)
v( t ) 
1
2j
c  j
st
v
(
s
)
e
ds

c  j
If we want to avoid evaluation of poles we mist
approximate the integrand. Approximation of v(s) is
not good since it will have to be repeated for each
st
new function. Instead we approximate e using
Padé approximation (with z = st)
N
e z  R N1M (z) 
PN (z)

Q M (z)
a z
i 0
M
b z
i 1
i
i
i
i
1
Using this result we can replace (i) by
1
v( t ) 
(ii)
2jt
c ' tj
z z
v
 e dz

t
c '  j  
Padé approximation is found by expanding ez and
R N1M in Taylor series and comparing first (M + N + 1)
terms. The solution is known in the closed,
analytical form giving
 N i
 z
(
M

N

i
)
!

i 0
i
 M
M
 (z) i
(
M

N

i
)!

i 0
 i 
N
R N1M
Result for first few N & M values are shown in Table
10.1.1.
Using Padé approximation we approximate v(t) by

v (t )
c '  j
1
z
v( t ) 
v
 R N1M (z)dz

2jt c ' j  t 

if M  N  2
(iii)
z
v
c  t R N1M (z)dz  2j
(residue at poles inside
the closed path)
+
if path C is closed in the left half plane
-
right
(clockwise)
Since
M
Ki
R N1M (z)  
i 1 z  z i
10.1.2
ki, zi known from Table
 zi 
 does
t 
then closing the path C clockwise ( v
have poles in o.h.p)
we have
(iv)
where
M1 
1

1M
z
v( t )    Re  k 1i v i
t i 1 
 t

M
& k 1i  2k i
2

  t  0

for M even
Example
Find v(t) for t - 0.1 and
v(s) 
2s  3
s  1s  2
 exact solution

t
2 t
 v( t )  e  e
Solution
Select M = 2 N = 0
from Table 10.1.2
z1  1  j K i  2 j
M1  1
not
use z1/t = 10 + 10j in (v) to obtain
1
 2 j 23  20 j 
40  j 46
v(t )   Re
 10 Re
 1.72465
t 11  10 j 12  10 j 
32  j 230

the exact value v(t) = 1.72357 (for t = 0.1)
While the method is good for nonperiodic excitations
the error grows with + for periodic excitations. To
improve accuracy we may use stepping algorithm.
Properties of the method
By closing the path C to the left we can prove that
1. Formula (iv) inverts exactly the function
V (s ) 
1
sm
for  M  N  2  m  M  N  1
2. Formula (iv) inverts exactly the first M + N + 1
terms of the Taylor expansion of any time
response.
10.3
Application
Derivatives of time domain responses can be
obtained from the transform
L v
(n)

n
(t )  s v s    v k 1 (0) s n k
n
k 1
and using (v) with v  replaced by
z
t
n
z z
  V 
t t
we have
n
1 M  zi   zi 
v ( t )    k i   V 
t i 1  t   t 
 (n)
t>0
 (n)
[note that the 2nd term doesn't contribute to v (t) t  0 ]
Newton-Raphson iteration can be used to find time
when function raises to a specific value A (important
for event driven simulation).
A
V ' (t k )
k
V (t )
tk
t k 1
a
 zi 
k
  At
i
k
k
v t  A k i 1

t  M
 t k  1
 zi   zi 
k
k
v t

i  k  v k 
t  t 
i 1

t k 1
 
M
 k v t
The method can also be used to find time domain
sensitivities from the frequency domain sensitivities.
Let v(s)  F(s) W (s)
1M
z  z 
v( t )  1  k i F i W i 
t i 1
 t   t 

and
Let F(s) depend on a parameter h. Then time domain
sensitivities can be obtained using M frequency
domain sensitivities as follows
 zi

F

 v( t )
1 M
t
   ki 
h
t i 1
h



 W z i 
 
 t 
Example 10.3.1
Calculate time domain sensitivities of v2 wrt network
elements. Input J(t) = 1(t), skip size h = 0.1. Use M
= 2 N = 0 with z = 1 + j , k1 = -2j
C
=
1
G1
=1
J=1
/S
G2 V
=1 2
t
1 2
exact solution v 2 ( t )  2 e
v 2 ( t ) 1 
t 
 1  e 2
G 1
4 2
t  0.1
 0.4756147123
t
t  0.1
 0.249697723
v 2 ( t ) t  2
 e  0.0237807356
C
4
t
From analysis
v1 (s) 
s 1
1
& v 2 (s) 
 1
 1
2s s  
2 s  
 2
 2
Adjoint system
 sC   v1a   0 
G1  sC
 a     yields
  sC

G 2  sC v 2   1

v1a (s) 
s
 1
2 s  
 2
 (s  1)
v a2 (s) 
1

2 s  
2

so the frequency domain sensit (Ch. 6)
1
v 2
 (s  1) L
1
t  2
a
 1  v1 v1 


1


e
2
G 1
4
2


1

n s  
2

L1
v 2
1
t
a
a
 s v1  v 2  v1  v 2  
 e t / 2
2
C
4
1

4 s  
2

Using

 

the numerical evaluation of
t
v 2 ( t )
v (t )
& 2
G1
C
are
as follows




v 2 ( t )
 g  j10 2 j  0.2496885962
1  1  (s  1) 
  Re  k


10
Re
2  s  z i 10 j10
2
G 1
t


4
10
.
5

j
10
1


t
 4 s   


 2 





v 2 ( t )
1 
1

  Re  k 1
2 
C
t
1


 4 s   


 2 

s 10 j10
 10 Re
 2j
 0.0237575355
4(10.5  j10) 2
Stepping algorithm
To increase accuracy of numerical inversed Laplace
transform we "reset" the problem after each step in
time.
Consider system equs
TC (s)X C  WC (s)  I
(1)
where TC (s)  G  sC , c denotes complex vectors and
matrices
I vector of initial conditions Civi0 or -LiIi0
substituting
x(t) for t = h
s
zi
h
we can solve (1) to obtain solution
1

1M
x    Re k 1i X Ci
h i 1

x Ci
 zi 
 xC 
h
for the next step the vector I is obtained from
I = Cx
If step h does not change then LU factorization of the
TC(s) can be used for all time steps( as Tc (s) does not
change).
For the rational transfer function
M
v(s)  F(s) W (s) 
N(s)
W (s) 
D(s)
a s
a 1
i
i 1
n
s   b i s i 1
i 1
define an auxiliary system
x 1 (s) 
1
W(s)
D(s)
so D(s)x1 (s)  W(s) (2)
then
V(s)  N(s)X1 (s)
Equation (2) can be solved for X1 as follows
s
n

 b n s n  ...  b1 X1 (s)  W(s)  introduce x1, x2, . . . xn
x 11  x 2
x 12  x 3
.
.
.
x 1n  W  b n x u  b n 1 x n 1  ...  b1 x 1
In the matrix form (for n = 4)
 x 1'   0
 ' 
x 2    0
 x 3'   0
 ' 
 x 4   b1
1
0
0
1
0
 b2
0
b3
0
0 
1

b4 
 x 1  0 
 x  0 
 2  W
 x 3  0 
   
 x 4  1
or
x 1  Ax  Bw
is s-domain
sX - I0 = AX + BW  (s1 - A)X = BW + IC
(3)
initially IC = 0, and (3) can be solved using the
stepping algorithm similarly like (1).
Algorithm implementing solution of (3) is shown in
Fig. 10.4.1
Stability
1
is tested on a differntial equation x  x
using Laplace transform sX  x 0  x
X
x0
s
from (*) (inverse Lapl. tr.)
M
x0
ki
1 M
x1 
k

x
 i
0 
h i 1 z i
i 1 h  z i

h
M
for z = utiv= h
x1  x 0 
i 1
.
.
.
ki
z  zi
 M ki
x n  x 0  
 i 1 z  z i
M
ki
1
stable if 
z

z
i 1
i
or

 n

R N1M (z) 1