Inverse Laplace Transform
Note that most of
the transforms are
written as fractions
1
Partial-fraction Expansion
P( s) b0 s m  b1s m1  ...  bm1s  bm
F ( s) 

Q( s) a0 s n  a1s n1  ...  an 1s  an
where P(s) and Q(s) are polynomials in the complex frequency
variables s and the coefficients ao, a1, . . . , an, bo, b1, . . . , bm are real
numbers. A rational function is specified completely by the two sets of
real coefficients which define the numerator and denominator
polynomials. On the other hand, the polynomials can also be
expressed in the factored form in terms of their zeros. Thus, an
alternate representation of F(s) is given
m
F ( s)  K
 s  z 
i
i 1
n
 s  p 
j
j 1
2
The first step in the partial-fraction expansion is to put the rational
function into a proper form. We say that a rational function is proper if
the degree of the numerator polynomial is less than the degree of the
denominator polynomial. If the given rational function F(s) is not
proper, i.e.., if the degree of P(s) is greater than or equal to that of
Q(s), we divide (long division) P(s) by Q(s) and obtain
P( s) ˆ
R( s )
F ( s) 
 P( s) 
Q( s )
Q( s )
Pˆ ( s)
The quotient, is a polynomial and R(s) is the remainder; therefore, R(s)
has a degree less than that of Q(s), and the new rational function
R(s)/Q(s) is proper. Since Pˆ ( s)is a polynomial, the corresponding time
function is a linear combination of ,(1),(2), etc., and can be
determined directly using Tables. We therefore go ahead with the new
rational function R(s)/Q(s) which is proper. In the remaining part of this
section we assume that all rational functions are proper.
3
Second Order Polynomial
j
•
•
•
Poles: roots of 𝑄(𝑠) are called and
Zeros: roots of 𝑃(𝑠) are called.
Pole-Zero Plot: plot in the complex frequency
domain zeros as ‘o’ and poles as ‘x’)
•
In general we may write a second order Q(s) as
s 2  2n s  n2
•
The roots are
s1,2      2  1  n



 : damping ratio
n : naturalfrequency
• If the damping ratio >1 (roots are real & distinct ‘simple’)
• If the damping ratio <1 (roots are complex conjugates
• If the damping ratio =1 (roots are real & equal ‘repeated’)
4
Case 1 Simple poles
We start with a simple example as follows:
P( s )
s 2  3s  5
F ( s) 

Q( s) ( s  1)( s  2)( s  3)
We claim that there are constants K1, K2, and K3 such that
K3
K1
K2
s 2  3s  5



( s  1)( s  2)( s  3) s  1 s  2 s  3
s 2  3s  5
K1 
 1.5
( s  2)( s  3) s 1
Heaviside’s
Expansion
s 2  3s  5
K2 
 3
( s  1)( s  3) s 2
s 2  3s  5
K3 
 2.5
( s  1)( s  2) s 3
Compare Heaviside’s Expansion with solving for K or substituting convenient values for s.
5
2


s
 3s  5
-1
f (t )  L 

(
s

1)(
s

2)(
s

3)


3
2.5 
-1  1.5
L 


 s  1 s  2 s  3 
 1.5e  t  3e 2t  2.5e 3t
n
Kj
j 1
s  pj
F (s)  
for t  0
and the residue Kj of the pole pj is given by
K j  s  p j F ( s )
s p j
6
Case 2 Multiple poles
Find the partial-fraction expansion of
1
F ( s) 
( s  1)3 s 2
The function has two multiple poles at s =p1=-1 (third order, n1=3) and
at s = p2 = 0 (second order,n2= 2). Thus, the partial fraction expansion
is of the form
F ( s) 
K13
K11
K12
K 21 K 22



 2
2
3
s  1 s  1 s  1
s
s
To calculate K11, K12, and K13, we first multiply F(s) by (s +1)3 to obtain
( s  1) 3 F ( s ) 
we find
1
s2
1
d ( nm)
n

Km 
(
s

p
)
F ( s ) 
( nm) 
s 
(n  m)! ds
1
K13  2
s
1
s  1
d 1
K12 
ds s 2
2
 3
2
s s 1
s  1
7
K 11  (1 / 2)
d
ds
2
1
2
s

2
s 1
1 6
2 s
4
 3
s 1
Similarly, to calculate K21 and K22, we first multiply F(s) by s2 to obtain
1
s F (s) 
( s  1)3
2
Using, we find
1
K 22 
s  13
1
s 0
d
1
K 21 
ds s  13
 3
s 0
Therefore, the partial-fraction expansion is
1
3
2
1
3 1
F (s) 



  2
2
3
3 2
( s  1) s
s  1 s  1 s  1 s s
The corresponding time function is


1
t
t
1 2 t
f (t )  L 

3
e

2
te

2t e -3 t
3 2
 ( s  1) s 
-1
for t  0
8
Case 3
Complex
poles
The two cases presented above are valid for poles which are
either real or complex. However, if complex poles are present,
the coefficients in the partial-fraction expansion are, in general,
complex, and further simplification is possible.
First, we observe that F(s) is a ratio of polynomials with real
coefficients; hence zeros and poles, if complex, must occur in complex
conjugate pairs. More precisely, if p1 = 1 + j1 is a pole, that is,
*
*
Q1(p1)=0, then p1   1  j 1 is also a pole; that is, Q1(p1 )  0. This is
due to the fact that any polynomial Q(s) with real coefficients has the
property that 𝑄 ∗ 𝑠 = 𝑄(𝑠 ∗ ) for all s.
Let us assume that the rational function has a simple pole at s = p1 =
*
-+ j; then it must have another pole at s = p2 =p1 =- - j. The
partial-fraction expansion of F(s) must contain the following two
terms:
K1
K2

s   j  s   j 
Using formula for simple poles, we obtain
9
K 1   s    j   F (s ) s   j 
K 2   s    j   F (s ) s   j 
Since F(s) is a rational function of s with real coefficients, it follows that
K2 is the complex conjugate of K1.
Once K and K* are determined, the corresponding two complex terms cn be
combined as follows: Suppose K=a+jb. Then K*=a-jb, abd
K
K*
(a  jb )(s    j  )  (a  jb )(s    j  )


s   j  s   j 
(s    j  )(s    j  )
2a (s   )
2b 


2
2
(s   )  
(s   ) 2   2
Once, we find a and b, given,  ,  ,
using the Inverse Laplace transform, we get
 t
 t
2
ae
cos

t

2
be
sin  t u (t )

10
This formula, which gives the corresponding time function for a pair of
terms due to the complex conjugate poles, is extremely useful.
There are good Matlab examples utilizing the command residue
Practice the examples
11