taking inverse laplace

[mark_neil2]

Hi,

I want to take inverse laplace of an transfer fucntion that I have generated by analysis of a circuit and I am having difficulty with it. Please see the attached file. It is saved in MC13. The parameters x1, x2, x3 are initial conditions of caps and they are real and positive numbers.

The second equation takes the inverse laplace once numbers assigned to x1, x2, x3. But I want to calculate the inverse laplace in equation 3 as a function of x1, x2, x3 as parameters. Then I will replace x1, x2, x3 with some equations once the inverse laplace is found.

How can I calculate the inverse laplace of equation (1) having x1, x2, x3 as parameters in the solution?

Thanks.

Mark.

[MikeArmstrong]

Like the attached?

Mike

[mark_neil2]

Hi Mike,

Thanks for your reply. However, I use MC13 and I couldn't open the file you attached. I asked a friend to open it in MC15 and save it in MC13 for me. After I open it in MC13, it didn't work. I am looking for solution that works with Maple symbolic engine in MC13. I understand you replaced s with s^-1. For some reason it still doesn't take the inverse laplace. MC15 takes the inverse laplace of the same function. That means the solution exists.

By the way, can we replace s with 1/s in taking inverse laplace anytime I want?

Thanks.

Mark.

[MikeArmstrong]

Re-saved in M11 format.

Mike

[mark_neil2]

Did this work in MC11? I opened it up in MC13 and it is still not working: it doesn't take the inverse laplace of Equation (1). It just returns the original expression with its parameters rearranged.

Mark.

[MikeArmstrong]

By the way, can we replace s with 1/s in taking inverse laplace anytime I want?

I think that would be a solution. Have you tried it?

Mike

[mark_neil2]

it doesn't seem to work. I have tried it on (1/ (1+s^2)), for which the solution is sin(t). If I replace s with 1/s and take the inverse laplace again, it displays the result as -sin(t) + delta(t).

Mark.