Using convolution for Laplace transform

In summary: You are correct, using integration by parts you can get the result e^-st*(integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt.
  • #1
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Homework Statement


Using convolution theorem for Laplace theorem,, show that


Homework Equations


inverse Laplace transform (1/(S^3/2*(s-1)) = (2*e^t)/Pi^1/2 intregral (from 0 to t) e^-x*x^1/2dx.


The Attempt at a Solution



The inverse Laplace above is a product of 1/s^3/2 and 1/(s-1)

and both terms are the Laplace transform of 2/Pi^1/2*t^1/2 and e^t respectively.

changing variable t = u gives:2/Pi^1/2*u^1/2 and e^u

using convolution: integral (from 0 to infinity) e^-st integral (from 0 to t)
(2/Pi^1/2*u^1/2 )(e^u)du dt.

I don't know how to go about with the integral from here onwards..Help!
 
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  • #2
The "convolution theorem" you are told use says that the Laplace transform of a convolution of functions is the product of the Laplace transforms of the functions. Isn't that exactly what you need here?
 
  • #3
I am sorry but I don't really get you. Laplace transform of a convolution of functions is the product of the Laplace transforms of the functions.

Isnt that what I am trying to do. finding the product of the inverse Laplace (which as given) and converting the products to its functions.

Is there a rule or formula for convolution theorem of Laplace theorem?
 
  • #4
sensitive said:

Homework Statement


Using convolution theorem for Laplace theorem,, show that


Homework Equations


inverse Laplace transform (1/(S^3/2*(s-1)) = (2*e^t)/Pi^1/2 intregral (from 0 to t) e^-x*x^1/2dx.


The Attempt at a Solution



The inverse Laplace above is a product of 1/s^3/2 and 1/(s-1)

and both terms are the Laplace transform of 2/Pi^1/2*t^1/2 and e^t respectively.
Good. You are saying that [itex]\frac{1}{s^{3/2}}\frac{1}{s-1}[/itex] is the product of the Laplace transforms of [itex]\frac{2}{\sqrt{\pi t}}[/itex] and [itex]e^t[/itex]. You also know that the Laplace transform of a convolution is the product of the individual functions. So [itex]\frac{1}{s^{3/2}}\frac{1}{s-1}[/itex] is the Laplace transform of the convolution of [itex]\frac{2}{\sqrt{\pi t}}[/itex] and [itex]e^t[/itex]. If you can show that the convolution of those function is equal to [itex]\frac{2e^t}{\sqrt{\pi}}\int_0^t e^{-x\sqrt{x}}dx[/itex], you are done!

By the way, what "Precalculus" class deals with the Laplace transform?
 
  • #5
My problem now is to prove it because now the function has integral form 0 to t where does that come from?

Is there a convolution rule that i need to know to be able to prove it?

And I am so sorry if I post it in the wrong section. Under which section should post this Laplace problems? thx.
 
  • #6
First what is the DEFINITION of the convolution of two functions? I think that is the formula you need.

Second, this should be in "Calculus and Beyond". I will go ahead and move it.
 
  • #7
Oh I just found an appropriate section for this problem. Sorry for posting the problem at the wrong area. I'll make sure Ill post any problem/s in the appropraite sectionnext time.
 
  • #8
There was some typing error, in the original equation.

inverse Laplace transform (1/(S^3/2*(s-1)) = (2*e^t)/Pi^1/2 intregral (from 0 to t) (e^-x)*(x^1/2)dx.

The inverse Laplace above is a product of 1/s^3/2 and 1/(s-1)

and both terms are the Laplace transform of (2/Pi^1/2)*t^1/2 and e^t respectively.

Using convolution theorem,
integral (from 0 to infinity) e^-st (integral (from 0 to t) f (t - x)*g(x)dx)dt

putting the Laplace transform it becomes
integral (from 0 to infinity) e^-st (integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt

But can I do this,
(integral (from 0 to infinity) e^-st dt)*(integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt

If i can do that then it is already proven
(2*e^t)/Pi^1/2 intregral (from 0 to t) (e^-x)*(x^1/2)dx.
 
  • #9
Well, actually i don't think I can do the following:
(integral (from 0 to infinity) e^-st dt)*(integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt

I think the integration
integral(from 0 to infinity) of e^-st*(integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)dt using integration by parts

integration of e^-st (from 0 to infinity), will give 1.

hence and leaving (integral (from 0 to t) (e^ (t - x))*(2/Pi^1/2)*x^1/2dx)

am I right? I hope I am making sense.
 

1. What is the purpose of using convolution for Laplace transform?

The purpose of using convolution for Laplace transform is to simplify the mathematical process of calculating the Laplace transform of a function. Convolution allows us to break down a complex function into simpler components, making it easier to apply the Laplace transform.

2. How does convolution work for Laplace transform?

Convolution works for Laplace transform by taking two functions, f(t) and g(t), and multiplying them together. The result is then integrated over all values of t. This process is repeated for every value of t, resulting in a new function that is the convolution of the two original functions.

3. Is convolution necessary for calculating the Laplace transform?

No, convolution is not necessary for calculating the Laplace transform. It is only used as a mathematical tool to simplify the process. The Laplace transform can also be calculated using algebraic manipulations, but convolution is often a more efficient method.

4. Can convolution be used for any type of function?

Yes, convolution can be used for any type of function as long as it satisfies certain mathematical criteria. This includes functions that are continuous, bounded, and have finite integrals. However, the process may become more complex for more complicated functions.

5. Are there any limitations to using convolution for Laplace transform?

The main limitation of using convolution for Laplace transform is that it can only be used for functions that are defined for all real numbers. This means that functions with singularities or discontinuities may not be suitable for this method. In these cases, other techniques may need to be used to calculate the Laplace transform.

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