2 Questions on inverse Laplace Transformation

In summary, we have discussed finding the inverse Laplace transforms of two given functions. For the first function, we factored the denominator and used partial fractions to simplify the expression. For the second function, we used the convolution theorem and found the integral of the function. However, there was a minor mistake in the final answer, as the variable should have been t instead of tau. It is also recommended to use partial fractions for this function. The last date for submitting the application form for IIT is usually in April, and it is advised to use a textbook specifically for geology for preparation.
  • #1
Reshma
749
6
1] Find the inverse Laplace transform of the given function, i. e.
[tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

And this is how I proceeded about:
[tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

[tex]= L^{-1}\left[\frac{1}{(s+1)(s^2 - s +1)}\right][/tex] (By factorising the denominator)

I am stuck at this step and can't proceed furthur. Someone help.

2]Find the inverse Laplace Transform of the given function, i. e.
[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

I did solve this one, just want to know if its correct.
[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

[tex]= 3L^{-1}\left[3\over {s(s^2 + 9)}}\right][/tex]

We know:
[tex]L^{-1}\left[3\over {s^2 + 9}}\right] = \sin 3t = F(t)[/tex]
[tex]L^{-1}\left[1\over s}\right] = 1 = G(t)[/tex]

By the convolution theorem;
[tex]L^{-1}{f(s)} = F(t) * G(t) = \int_0^t G(\tau)F(t - \tau)d\tau[/tex]

So,
[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]
[tex]=3\int_0^t sin3(t - \tau)d\tau = (\cos 3\tau - 1)[/tex]

Please care to see if this is right. Thanks in advance.
 
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  • #2
Reshma said:
1] Find the inverse Laplace transform of the given function, i. e.
[tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

And this is how I proceeded about:
[tex]L^{-1}\left[1\over {s^3 + 1}}\right][/tex]

[tex]= L^{-1}\left[\frac{1}{(s+1)(s^2 - s +1)}\right][/tex] (By factorising the denominator)

I am stuck at this step and can't proceed furthur. Someone help.
Partial fractions, of course!
[tex]\frac{1}{(s+1)(s^2-s+1)}= \frac{A}{s+1}+ \frac{Bs+ C}{s^2- s+ 1}[/tex]
s2- s+ 1 cannot be factored using real coefficients but you can complete the square: s2- s+ 1= (s- 1/2)2+ 3/4.

2]Find the inverse Laplace Transform of the given function, i. e.
[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

I did solve this one, just want to know if its correct.
[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]

[tex]= 3L^{-1}\left[3\over {s(s^2 + 9)}}\right][/tex]

We know:
[tex]L^{-1}\left[3\over {s^2 + 9}}\right] = \sin 3t = F(t)[/tex]
[tex]L^{-1}\left[1\over s}\right] = 1 = G(t)[/tex]

By the convolution theorem;
[tex]L^{-1}{f(s)} = F(t) * G(t) = \int_0^t G(\tau)F(t - \tau)d\tau[/tex]

So,
[tex]L^{-1}\left[9\over {s(s^2 + 9)}}\right][/tex]
[tex]=3\int_0^t sin3(t - \tau)d\tau = (\cos 3\tau - 1)[/tex]

Please care to see if this is right. Thanks in advance.

Well, it is [itex] cos (3t)- 1[/itex], not [itex]cos(3\tau )-1[/itex]!

Again, I would have done that by partial fractions:
[tex]\frac{9}{s(s^2+ 9)}= \frac{1}{s}- \frac{s}{s^2+ 9}[/itex]
which then gives what you have.
 
  • #3
Cool, that makes sense. I will follow the technique you've suggested. Thank you so very much for your time.
 
  • #4
hi

hi... reshma thanks a lot for ur updates of iit which is very usefull to me.ho w is ur preparation going.please tell me when is the last date if submitting the application form. which book book ru using for geology.eagerly waiting fror ur reply :smile:

from
prakash chennai
 
  • #5
prakashlava said:
hi... reshma thanks a lot for ur updates of iit which is very usefull to me.ho w is ur preparation going.please tell me when is the last date if submitting the application form. which book book ru using for geology.eagerly waiting fror ur reply :smile:

from
prakash chennai
Welcome to PF. Please do not hijack other threads to post your questions. Choose a thread relevant to your topic. Please read the Forum Guidelines: https://www.physicsforums.com/showthread.php?t=5374

Good luck!
 

1. What is an inverse Laplace transform?

The inverse Laplace transform is a mathematical operation that is used to convert a function from the frequency domain to the time domain. It is the reverse process of the Laplace transform, which is used to convert a function from the time domain to the frequency domain. Inverse Laplace transforms are commonly used in engineering and physics to solve differential equations and analyze systems in the time domain.

2. How do you perform an inverse Laplace transform?

To perform an inverse Laplace transform, you first need to have the Laplace transform of the function you want to convert. Then, you can use a table of Laplace transforms or a mathematical formula to find the inverse transform. The process involves finding the poles and residues of the function and using them to determine the inverse transform. In some cases, partial fraction decomposition and integration may also be necessary. Advanced techniques such as contour integration can also be used for more complex functions.

3. What is the difference between a Laplace transform and an inverse Laplace transform?

The Laplace transform and inverse Laplace transform are two mathematical operations that are used to convert a function between the time and frequency domains. The Laplace transform is used to convert a function from the time domain to the frequency domain, while the inverse Laplace transform is used to convert a function from the frequency domain to the time domain. They are inverse operations of each other, meaning that applying the Laplace transform to a function and then applying the inverse Laplace transform will result in the original function.

4. What are some real-world applications of inverse Laplace transforms?

Inverse Laplace transforms have many practical applications in engineering, physics, and mathematics. They are commonly used to solve differential equations in the time domain, analyze control systems and circuits, and study the behavior of physical systems. Inverse Laplace transforms are also used in signal processing, image processing, and communication systems to convert signals from the frequency domain to the time domain for analysis and manipulation.

5. Are there any limitations to using inverse Laplace transforms?

While inverse Laplace transforms are a powerful tool for converting functions between the time and frequency domains, there are some limitations to their use. They may not be applicable to functions that do not have a Laplace transform or have a very complex one. In addition, inverse Laplace transforms can be challenging to calculate for functions with multiple poles or complex poles. In these cases, alternative methods such as numerical approximations or advanced techniques like the method of steepest descent may be used.

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