Laplace transform of the dirac delta function

In summary: I used the wrong property when integrating the Dirac delta term. In summary, the correct solution for L[t^2 - t^2 δ(t-1))] is \frac{2}{s^3} - e^-s.
  • #1
november1992
120
0

Homework Statement



L[[itex]t^{2}[/itex] - [itex]t^{2}[/itex]δ(t-1)]

Homework Equations



L[ [itex]t^{n}[/itex]f(t)] = ([itex]-1^{n}[/itex]) [itex]\frac{d^{n}}{ds^{n}}[/itex] L[f(t)]

L[δ-t] = e^-ts

The Attempt at a Solution



My teacher wrote [itex]\frac{2}{s^{3}}[/itex] [itex]-e^{s}[/itex] as the answer.

I got [itex]\frac{2}{s^{3}}[/itex] + [itex]\frac{e^-s}{s}[/itex] + 2 [itex]\frac{e^-s}{s^2}[/itex] + [itex]\frac{2e^-s}{s^3}[/itex]
 
Last edited:
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  • #2
You should show us how you derived your answer. That way we can best help you see where you went wrong. Right now we have to guess.

My guess is that you tried to use the identity under "Relevant equations" on the Dirac delta term, which likely made your problem more complicated than it needed to be.

Are you aware of the rule ##f(t)\delta(t-t_0) = f(t_0)\delta(t-t_0)##? This will simplify your calculation.
 
  • #3
I don't see that formula in my textbook.

When I plug in the numbers I get: [itex]\frac{2}{s^3}[/itex] + [itex]\frac{e^-s}{s}[/itex]

This is what I did to get my first answer.
L[t^2 - t^2 δ(t-1))]


[itex]\frac{2}{s^3}[/itex] + [itex]\frac{d}{ds}[/itex] ([itex]\frac{d}{ds}[/itex] ([itex]\frac{e^(-s)}{s}[/itex])




[itex]\frac{2}{s^3}[/itex] +[itex]\frac{d}{ds}[/itex] ([itex]\frac{e^-s * s + e^-s}{s^2}[/itex] + [itex]\frac{e^-s*s^2 + 2s*e^-s}{s^2}[/itex])

[itex]\frac{2}{s^{3}}[/itex] + [itex]\frac{e^-s}{s}[/itex] + 2 [itex]\frac{e^-s}{s^2}[/itex] + [itex]\frac{2e^-s}{s^3}[/itex]


Okay I just checked the answer on wolframalpha and what my teacher had was the correct answer. Why is it that L[t^n*d(t-a)] = e^-as

Edit:
Nevermind, I've found my mistake.
 
Last edited:

1. What is the Laplace transform of the Dirac delta function?

The Laplace transform of the Dirac delta function is equal to 1. This means that the Laplace transform of the Dirac delta function is a constant value of 1 for all values of s in the complex plane.

2. What is the significance of the Laplace transform of the Dirac delta function?

The Dirac delta function is often used to model impulse functions or point sources of energy in physical systems. The Laplace transform of the Dirac delta function allows us to analyze and solve problems involving these types of functions in the frequency domain.

3. How is the Laplace transform of the Dirac delta function calculated?

The Laplace transform of the Dirac delta function is calculated using the Laplace transform integral formula, which involves integrating the Dirac delta function from 0 to infinity. This integral evaluates to 1, giving us the result of 1 for the Laplace transform of the Dirac delta function.

4. Is the Laplace transform of the Dirac delta function a continuous function?

No, the Laplace transform of the Dirac delta function is not a continuous function. It is a discontinuous function with a value of 1 for all values of s in the complex plane.

5. How is the Laplace transform of the Dirac delta function used in solving differential equations?

The Laplace transform of the Dirac delta function is often used in solving initial value problems involving differential equations. By transforming the differential equation into the frequency domain using the Laplace transform, we can solve it using algebraic methods and then inverse transform the solution back to the time domain.

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