ME 3600 Control Systems
Examples – Partial Fraction Expansions
5s  3
( s  1)( s  2)( s  3)
a) the partial fraction expansion of X ( s) , and b) x(t ) .
1. Given: X ( s) 
Find:
Solution: Given the characteristic equation has real, unequal roots, the partial fraction
expansion has the form
K1
K
K
 2  3
s 1 s  2 s  3
where the coefficients Ki (i  1,2,3)
X ( s) 


5s  3
K1   ( s  1) X ( s)s 1  
 1

 ( s  2)( s  3)  s 1
 5s  3 
K 2   ( s  2) X ( s)s 2  
7

(
s

1)(
s

3)

 s 2
 5s  3 
K3   ( s  3) X ( s)s 3  
 6

(
s

1)(
s

2)

 s 3
So, the partial fraction expansion is
 1   7   6 
X ( s)  



 s 1  s  2   s  3 
Using the Laplace transform table gives
x(t )  et  7e2t  6e3t
2. Given:
Find:
5s  15
( s  2)( s 2  3s  9)
a) the partial fraction expansion of X ( s) , and b) x(t ) .
X ( s) 
Solution: Given the characteristic equation has one real root and a pair of complex
conjugate roots, the partial fraction expansion is of the form
 K   As  B 
X ( s)  
 2

 s  2   s  3s  9 
Kamman – ME 3600 – page: 1/2
where the coefficient K may be found as before
5
 5s  15 
K  ( s  2) X ( s)s 2   2


 s  3s  9  s2 7
and the coefficients A and B are found by clearing fractions as follows
5s  15  K ( s 2  3s  9)  ( As  B)( s  2)
 ( K  A) s 2  (2 A  B  3K ) s  (9 K  2 B)
Equating the coefficients of the powers of s on both sides of the equation gives
A K  0
(s 2 )
2 A  B  3K  5
( s1 )
9K  2B  15
(s0 )
Using the first and third equations, we find A   75 and B  30
7 . So, the partial fraction
expansion is
 1  5  s  6  5  1  5  s  (6) 
X ( s)  75 

  2
 
 
2
 s  2  7  s  3s  9  7  s  2  7  ( s  32 )  274 
Using #4 and #18 in the Laplace transform tables gives
0.3335 (rad)
2.5981

 
  tan 1 
or
x(t )  75 e2t  3.055 e1.5t sin(2.5981t   ) 

 6  1.5  
 2.8081 (rad)
The correct choice of  must be made to satisfy the initial conditions of the differential
equation (not given here).
3. Given:
Find:
s5  a1s 4  a2 s 3  a3s 2  a4 s  a5
s 6  11s5  79s 4  427 s3  1510s 2  3800s  4000
What is the form of x(t ) ? Identify the steady-state and transient parts of x(t ) .
X ( s) 
Solution: Using the root solving feature on your calculator, the poles of X ( s) are found to
be 5 j , 2  3.4641 j , 5 , and 2 . So, the partial fraction expansion and x(t ) have the
forms
 K   K   A s  B1   A2 s  B2 
X ( s)   1    2    12
 2

 s  2   s  5   s  25   s  4s  16 
x(t )  K1e2t  K 2e5t  C1e2t sin
transient response


12 t  1  C2 sin(5t  2 )
steady-state response
Kamman – ME 3600 – page: 2/2