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

Introduction

When using the Laplace-transform

H s t h t s t
it is often useful to be able to find h t given H s . There are at least 4 different methods to do this:
  1. Inspection
  2. Partial-Fraction Expansion
  3. Power Series Expansion
  4. Contour Integration

Inspection method

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

When given H s s s α with an ROC of s α we could determine "by inspection" that h t α t u t

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Partial-fraction expansion method

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

H s B s A s k 0 M b k s k k 0 N a k s k
This can also expressed as
H s a 0 b 0 k 1 M 1 c k s k 1 N 1 d k s
where c k represents the nonzero zeros of H s and d k represents the nonzero poles.

If M N then H s can be represented as

H s k 1 N A k 1 d k s
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 H s in the above form. If M N then H s can be expressed as
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

Find the inverse z-transform of H s 1 2 s s -2 1 -3 s 2 s -2 where the ROC is s 2 . In this case M N 2 , so we have to use long division to get H s 1 2 1 2 7 2 s 1 -3 s 2 s -2 Next factor the denominator. H s 2 -1 5 s 1 2 s 1 s Now do partial-fraction expansion. 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 , h t 1 2 δ t 9 2 2 t u t -4 u t

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Demonstration of partial fraction expansion

A demonstration involving Partial Fraction Expansion
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 athttp://www.wolfram.com/products/player/download.cgi)

Khan lecture on partial fraction expansion

video from Khan Academy

Power series expansion method

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

H s t h t s t
then each term of the sequence h t can be determined by looking at the coefficients of the respective power of s t .

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

H s s 2 1 2 s 1 1 2 s 1 s s 2 5 2 s 1 2 s
In this case, since there were no poles, we multiplied thefactors of H s . Now, by inspection, it is clear that h t δ t 2 5 2 δ t 1 1 2 δ t δ t 1 .

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

Suppose H s t logbase --> 1 α s Noting that t logbase --> 1 x t 1 -1 t 1 x t t Then H s t 1 -1 t 1 α t s t t Therefore H s -1 t 1 α t t t 1 0 t 0

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Contour integration method

Without going in to much detail

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

Demonstration of contour integration

A demonstration involving Contour Integration
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 athttp://www.wolfram.com/products/player/download.cgi)

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.

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
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Joan Reply
J, combine like terms 7x-4y
Bridget Reply
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Rachael Reply
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Asali
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Asali
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linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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China
Cied
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Cesar
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AMJAD
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
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Azam
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Damian
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Azam
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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Source:  OpenStax, Signals and systems. OpenStax CNX. Aug 14, 2014 Download for free at http://legacy.cnx.org/content/col10064/1.15
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