EGR 261 – Inverse Laplace Transforms using MATLAB
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Residue method using MATLAB
[R,P,K] = residue(N, D) finds the residues, poles and direct term of a partial fraction expansion of
the ratio of two polynomials N(s)/D(s).
If there are no multiple roots,
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Vectors B and A specify the coefficients of the numerator and denominator polynomials in
descending powers of s.
R = column vector of residues
P = column vector of pole locations
K = row vector of direct terms
The number of poles is n = length(D)-1 = length(R) = length(P).
The direct term coefficient vector is empty if length(N) < length(D), otherwise length(K) =
length(N) - length(D)+1.
If P(j) = ... = P(j+m-1) is a pole of multiplicity m, then the expansion includes terms of the
form
[N, D] = residue(R,P,K), with 3 input arguments and 2 output arguments, converts the partial
fraction expansion back to the polynomials with coefficients in N and D.
Reference: type help residue in MATLAB
EGR 261 – Inverse Laplace Transforms using MATLAB
Example – Residue method using MATLAB
V(s) 
4s
s  3s  2
2

4s
s  1s  2 

8
s2

-4
s 1
Note that the
largest pole is
listed first
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EGR 261 – Inverse Laplace Transforms using MATLAB
Example – Residue method using MATLAB
2s  23s  79s
4
V(s) 
3
2
 96s  88
s  4 s  6 
 2s  3s  1 
2
Could be found
by hand using
long division
3
10
s6

4
s4
EGR 261 – Inverse Laplace Transforms using MATLAB
Example – Residue method using MATLAB
V(s) 
s3
s  1  s  2 
2

1
s2

-1
s 1
Note that the
largest pole is
listed first

4
2
s  12
Note that repeated roots
are from lowest power to
highest power
EGR 261 – Inverse Laplace Transforms using MATLAB
Example – Residue method using MATLAB
5s  64s  99
2
V(s) 
s  1 s 2
 6s  25


1.5 - j5
s  3  j4

1.5  j5
s  3  j4
Answer expressed
using complex roots

5
2
s  1 

s
3s  49
2
 6s  25

2
 s  1 
Answer expressed
using quadratic
factor
EGR 261 – Inverse Laplace Transforms using MATLAB
Finding inverse Laplace transforms using MATLAB
Although we just saw how we could use MATLAB to perform partial fraction
expansion using the residue method, an even more powerful function is available for
finding inverse Laplace transforms: ilaplace( ).
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ilaplace(F) - the inverse Laplace transform of symbolic F with default independent
variable s. The default return is a function of t. The inverse Laplace transform is
applied to a function of s and returns a function of t.
ilaplace(F, y) - returns a function of y instead of a function of t
- ilaplace(F, t) is the same as ilaplace(F)
ilaplace(F, y, x) - finds the inverse transform as a function of x (instead of s) and
returns a function of y (instead of t)
- ilaplace(F, t, s) is the same as ilaplace(F)
Note: Since we are using one-sided Laplace transforms, there is an implied u(t)
associated with functions. So if ilaplace (2/s) returns the value 2, then it is implied that
f(t) = 2u(t).
Examples: See the following pages
EGR 261 – Inverse Laplace Transforms using MATLAB
Examples – using MATLAB
Verify the following relationships:
-1  1 
L  2   tu(t)
s 
10
20 
-1  2
L  
  2 - 10e
s
s

2
s

3




-1  10s
L  2
  10cos(2t)u (t)
s

4


 10e - 4s 
L 
  10u(t - 4)
s


-1
- 2t
 20e
- 3t
u(t)
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EGR 261 – Inverse Laplace Transforms using MATLAB
Examples
Find inverse Laplace transforms for the some of the following functions and use
MATLAB to verify the results. (Optional: Also find the poles and residues.)
A) V(s) 
C) F(s) 
E) F(s) 
2s
B) I(s) 
s4
2e
10s
s  12s  100
2
-4s
s  4 
D) V(s) 
2
2s  10
s  2 s  3 
2s
s  2  2 s  3 
F) V(s) 
4s
s  3s  2
2
s  1 s  2 
2s  23s  79s
4
G) V(s) 

4s
3
2

8
s2
 96s  88
s  4 s  6 

-4
s 1
 2s  3s  1 
2
10
s6

4
s4
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