Inverse Laplace Transform Involving a Step (Heaviside) Function

[cswall]

Hello,

I have a fairly straight forward question regarding an inverse laplace transform. The function follows along with the solution. The solution (provided in my text) has been confirmed by Maple; however, i cannot account for the (1/3) in the solution. Any help would be greatly appreciated; thanks in advance!

function to transform:
$$F(s)=\frac{e^{-2s}}{(s+2)*(s-1)}$$
for Maple: exp(-2*s)/((s+2)*(s-1))

solution:
$$f(t)=\frac{1}{3}*Heaviside(t-2)*(-e^{-2t+4}+e^{t-2})$$
for Maple: (1/3)*Heaviside(t-2)*(-exp(-2*t+4)+exp(t-2))

On a side note, this is my first post, and any suggestions regarding my formatting, etiquette, or topic selection would also be appreciated.

 

[rbj]

$$F(s) = \frac{e^{-2s}}{(s+2)(s-1)}$$

you need to think of it as

$$F(s) = e^{-2s} G(s)$$

where

$$G(s)=\frac{1}{(s+2)(s-1)} = \frac{-1/3}{s+2}+\frac{1/3}{s-1}$$

(i did the partial fraction expansion for you since it was so easy.)

now, find out what the inverse L.T. of G(s) is (gonna be unstable or unbounded, but who cares?), then, knowing that intermediate result, tell us what multiplying e^-2s does to it.

 

[tacman]

Hello!

The inverse Laplace transform involving a step (Heaviside) function can be solved using the convolution theorem. In this problem, the function is F(s)=\frac{e^{-2s}}{(s+2)*(s-1)}, which can be rewritten as F(s)=\frac{1}{s+2}-\frac{1}{s-1}. Using the inverse Laplace transform, we get f(t)=e^{-2t}-e^{t}.

However, in order to incorporate the Heaviside function into the solution, we need to take into account the shifting property of the Laplace transform. This means that for every term in the Laplace transform, we need to add a step function with the corresponding shift. In this case, we have a shift of 2 for the first term and a shift of -1 for the second term.

Therefore, the solution should be f(t)=Heaviside(t-2)*e^{-2t}-Heaviside(t+1)*e^{t}. Simplifying this expression, we get f(t)=Heaviside(t-2)*(-e^{-2t+4}+e^{t-2}), which is the same as the solution provided in your text. The (1/3) coefficient is just a factor that comes from simplifying the expression.

I hope this helps clarify why the (1/3) is included in the solution. As for your formatting and etiquette, it looks great to me! Just keep asking questions and participating in discussions, and you'll do great. Happy learning!