Laplace Transform of 5{\delta (t-3)}+4{u(t-2)}+3{e^{-5(t-3)}u(t-4)}

In summary, the Laplace Transform of the given expression is 5e^{-3t}+4\frac{1}{s}e^{-2t} + ?. The first and second terms can be solved using the equations L\{ f(t-t_0)u(t-t_0) \} = e^{-st_0}F(s) and L\{ e^{-\alpha t}u(t) \} = \frac{1}{s+\alpha}, respectively. However, t_0 in f and u are not the same, so the first equation cannot be used. Instead, try breaking up the exponential using algebra first.
  • #1
EugP
107
0

Homework Statement


Find the Laplace Transform of the following:

[tex]5{\delta (t-3)}+4{u(t-2)}+3{e^{-5(t-3)}u(t-4)}[/tex]


The Attempt at a Solution


I know how to do the first and second terms, but I'm having trouble with the third.

[tex]5e^{-3t}+4\frac{1}{s}e^{-2t} + ?[/tex]

I think I should be using either

[tex]L\{ f(t-t_0)u(t-t_0) \} = e^{-st_0}F(s)[/tex]

or

[tex]L\{ e^{-\alpha t}u(t) \} = \frac{1}{s+\alpha}[/tex]

or both.

The problem is, in the first one, I don't know what to put for [tex]F(s)[/tex].

Any help would be appreciated.
 
Last edited:
Physics news on Phys.org
  • #2
Unfortunately, you can't use your first equation because [tex]t_0[/tex] in f and u are not the same.

Try breaking up the exponential using algebra first.
 

FAQ: Laplace Transform of 5{\delta (t-3)}+4{u(t-2)}+3{e^{-5(t-3)}u(t-4)}

What is the Laplace Transform of 5{\delta (t-3)}?

The Laplace Transform of 5{\delta (t-3)} is 5e^{-3s}, where s is the variable in the Laplace domain. This is because the Dirac delta function, represented by \delta (t), has a Laplace Transform of 1.

How is the Laplace Transform of a unit step function represented?

The Laplace Transform of a unit step function, also known as the Heaviside function, is represented by \frac{1}{s}, where s is the variable in the Laplace domain. This is because the unit step function, represented by u(t), has a Laplace Transform of \frac{1}{s}.

What is the Laplace Transform of 3{e^{-5(t-3)}u(t-4)}?

The Laplace Transform of 3{e^{-5(t-3)}u(t-4)} is \frac{3}{s+5}e^{-8s}, where s is the variable in the Laplace domain. This is because the Laplace Transform of the exponential function e^{-at}u(t-b) is \frac{1}{s+a}e^{-bs}, where a and b are constants.

How do you find the Laplace Transform of a sum of functions?

To find the Laplace Transform of a sum of functions, you can use the linearity property of the Laplace Transform. This property states that the Laplace Transform of a sum of functions is equal to the sum of the individual Laplace Transforms of each function. In other words, you can find the Laplace Transform of each function separately and then add them together.

Can the Laplace Transform be used to solve differential equations?

Yes, the Laplace Transform can be used to solve differential equations. This is because the Laplace Transform converts a differential equation into an algebraic equation, which is often easier to solve. After finding the solution in the Laplace domain, you can use the inverse Laplace Transform to convert it back to the time domain.

Back
Top