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Inverse Laplace Transform

Module by: Dan Calderon. E-mail the author

Based on: Inverse Z-Transform by Benjamin Fite

Summary: This module describes the inverse Laplace transform (based on Inverse Z-transform module by Benjamin Fite, notation changed).

Introduction

When using the Laplace-transform

Hs= t =htst H s t h t s t
(1)
it is often useful to be able to find ht h t given Hs H s . There are at least 4 different methods to do this:

Inspection Method

This "method" is to basically become familiar with the Laplace-transform pair tables and then "reverse engineer".

Example 1

When given Hs=ssα H s s s α with an ROC of |s|>α s α we could determine "by inspection" that ht=αtut h t α t u t

Partial-Fraction Expansion Method

When dealing with linear time-invariant systems the z-transform is often of the form

Hs=BsAs= k =0M b k sk k =0N a k sk H s B s A s k 0 M b k s k k 0 N a k s k
(2)
This can also expressed as
Hs= a 0 b 0 k =1M1 c k s-1 k =1N1 d k s-1 H s a 0 b 0 k 1 M 1 c k s k 1 N 1 d k s
(3)
where c k c k represents the nonzero zeros of Hs H s and d k d k represents the nonzero poles.

If M<N M N then Hs H s can be represented as

Hs= k =1N A k 1 d k s-1 H s k 1 N A k 1 d k s
(4)
This form allows for easy inversions of each term of the sum using the inspection method and the transform table. If the numerator is a polynomial, however, then it becomes necessary to use partial-fraction expansion to put Hs H s in the above form. If MN M N then Hs H s can be expressed as
Hs= r =0MN B r sr+ k =0N1 b k ' sk k =0N a k sk H s r 0 M N B r s r k 0 N 1 b k ' s k k 0 N a k s k
(5)

Example 2

Find the inverse z-transform of Hs=1+2s-1+s-213s-1+2s-2 H s 1 2 s s -2 1 -3 s 2 s -2 where the ROC is |s|>2 s 2 . In this case M=N=2 M N 2 , so we have to use long division to get Hs=12+12+72s-113s-1+2s-2 H s 1 2 1 2 7 2 s 1 -3 s 2 s -2 Next factor the denominator. Hs=2+-1+5s-1(12s-1)(1s-1) H s 2 -1 5 s 1 2 s 1 s Now do partial-fraction expansion. Hs=12+ A 1 12s-1+ A 2 1s-1=12+9212s-1+-41s-1 H s 1 2 A 1 1 2 s A 2 1 s 1 2 9 2 1 2 s -4 1 s Now each term can be inverted using the inspection method and the Laplace-transform table. Thus, since the ROC is |s|>2 s 2 , ht=12δt+922tut4ut h t 1 2 δ t 9 2 2 t u t -4 u t

Demonstration of Partial Fraction Expansion

Figure 1: Interactive experiment illustrating how the Partial Fraction Expansion method is used to solve a variety of numerator and denominator problems. (To view and interact with the simulation, download the free Mathematica player at http://www.wolfram.com/products/player/download.cgi)
A demonstration involving Partial Fraction Expansion

Figure 2: video from Khan Academy
Khan Lecture on Partial Fraction Expansion

Power Series Expansion Method

When the z-transform is defined as a power series in the form

Hs= t =htst Hs t h t s t
(6)
then each term of the sequence ht h t can be determined by looking at the coefficients of the respective power of st s t .

Example 3

Now look at the Laplace-transform of a finite-length sequence.

Hs=s2(1+2s-1)(112s-1)(1+s-1)=s2+52s+12+s-1 H s s 2 1 2 s 1 1 2 s 1 s s 2 5 2 s 1 2 s
(7)
In this case, since there were no poles, we multiplied the factors of Hs H s . Now, by inspection, it is clear that ht=δt+2+52δt+1+12δt+δt1 h t δ t 2 5 2 δ t 1 1 2 δ t δ t 1 .

One of the advantages of the power series expansion method is that many functions encountered in engineering problems have their power series' tabulated. Thus functions such as log, sin, exponent, sinh, etc, can be easily inverted.

Example 4

Suppose Hs=log t (1+αs-1) H s t 1 α s Noting that log t (1+x)= t =1-1t+1xtt t 1 x t 1 -1 t 1 x t t Then Hs= t =1-1t+1αtstt H s t 1 -1 t 1 α t s t t Therefore Hs={-1t+1αtt  if  t10  if  t0 H s -1 t 1 α t t t 1 0 t 0

Contour Integration Method

Without going in to much detail

ht=12πjrHsst1d s h t 1 2 s r H s s t 1
(8)
where r r is a counter-clockwise contour in the ROC of Hs H s encircling the origin of the s-plane. To further expand on this method of finding the inverse requires the knowledge of complex variable theory and thus will not be addressed in this module.

Demonstration of Contour Integration

Figure 3: Interactive experiment illustrating how the contour integral is applied on a simple example. For a more in-depth discussion of this method, some background in complex analysis is required. (To view and interact with the simulation, download the free Mathematica player at http://www.wolfram.com/products/player/download.cgi)
A demonstration involving Contour Integration

Conclusion

The Inverse Laplace-transform is very useful to know for the purposes of designing a filter, and there are many ways in which to calculate it, drawing from many disparate areas of mathematics. All nevertheless assist the user in reaching the desired time-domain signal that can then be synthesized in hardware(or software) for implementation in a real-world filter.

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