Finding the Inverse Laplace Transformation

In summary, the conversation is discussing finding the inverse Laplace transformation of (5s+4)/s^2 e^{-2s}. The attempt at a solution involved using partial fractions and identifying the need for a u(t-2) term due to second shifting. However, the provided answer of 5 u(t-2)(e^{4(t-2)} - e^{-(t-2)}) was deemed incorrect and the correct answer is [5+4(t-2)] u(t-2).
  • #1
izen
51
0

Homework Statement



find the inverse laplace transformation of [itex]\frac{5s+4}{s^2}[/itex] e[itex]^{-2s}[/itex]

Homework Equations

The Attempt at a Solution



I have tried to partial fractions [itex]\frac{5s+4}{s^2}[/itex] and I got [itex]\frac{5}{s}[/itex]+[itex]\frac{4}{s^2}[/itex] and I know that the answer must have u(t-2) because of second shifting ( a=2)

but I looked at the answer from this question is 5 u(t-2)(e[itex]^{4(t-2)}[/itex] - e[itex]^{-(t-2)}[/itex]).

I don't know how to get that answer please help
 
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  • #2
The answer is wrong.
 
  • #3
so my answer is [5+4(t-2)] u(t-2) Is my answer right?
thank you
 
  • #4
izen said:

Homework Statement



find the inverse laplace transformation of [itex]\frac{5s+4}{s^2}[/itex] e[itex]^{-2s}[/itex]

Homework Equations




The Attempt at a Solution



I have tried to partial fractions [itex]\frac{5s+4}{s^2}[/itex] and I got [itex]\frac{5}{s}[/itex]+[itex]\frac{4}{s^2}[/itex] and I know that the answer must have u(t-2) because of second shifting ( a=2)

but I looked at the answer from this question is 5 u(t-2)(e[itex]^{4(t-2)}[/itex] - e[itex]^{-(t-2)}[/itex]).

I don't know how to get that answer please help

First: what is the inverse Laplace transform of ##(5s+4)/s^2##? Next: worry about the effect of the factor ##e^{-2s}##.
 
  • #5
izen said:
so my answer is [5+4(t-2)] u(t-2) Is my answer right?
thank you
Yes, that's correct.
 
  • #6
Thanks you all
 

FAQ: Finding the Inverse Laplace Transformation

What is an inverse Laplace transform?

An inverse Laplace transform is a mathematical operation that takes a function represented in the Laplace domain and converts it back to its original form in the time domain. It is the reverse process of the Laplace transform and is often used in solving differential equations and analyzing systems in control theory.

How is an inverse Laplace transform calculated?

The inverse Laplace transform is calculated using a formula or a table of known Laplace transform pairs. The most common method is to use partial fraction decomposition to break the function into simpler terms, and then use the known transform pairs to convert each term back to the time domain. Other methods include using the residue theorem or using Laplace transform tables.

What is the relationship between the Laplace transform and the inverse Laplace transform?

The Laplace transform and the inverse Laplace transform are mathematical operations that are inverse of each other. The Laplace transform converts a function from the time domain to the Laplace domain, while the inverse Laplace transform converts it back to the time domain. Together, they form a powerful tool for solving differential equations and analyzing systems in engineering and physics.

What are the applications of inverse Laplace transform?

The inverse Laplace transform has many practical applications in engineering and science. It is commonly used in control theory, signal processing, and electrical engineering to solve differential equations and analyze systems. It is also used in physics to study the behavior of physical systems, such as in quantum mechanics and fluid dynamics.

What are the limitations of the inverse Laplace transform?

One limitation of the inverse Laplace transform is that it is not always possible to find an analytical solution, especially for complex functions. In such cases, numerical methods may be used to approximate the inverse transform. Another limitation is that the inverse Laplace transform assumes that the function is Laplace-transformable, which means it must have certain properties, such as being of exponential order and having a finite number of discontinuities.

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