Thread: finding inverse Laplace transforms

Find the inverse Laplace transform of: (9s+6)/((s^2)-49)

I separate this into two fractions: 9s/(s^2-49) + 6/(s^2-49)

For the first fraction, I think I am correct in using the rule that for a transform of s/(s^2-a^2) ...the inverse transform is coshat. And for the second fraction, the inverse transform of a/(s^2-a^2) is sinhat. I'm not sure what the "h" means or how to use it.

Thanks =)

Originally Posted by plopony

Find the inverse Laplace transform of: (9s+6)/((s^2)-49)

I separate this into two fractions: 9s/(s^2-49) + 6/(s^2-49)

For the first fraction, I think I am correct in using the rule that for a transform of s/(s^2-a^2) ...the inverse transform is coshat. And for the second fraction, the inverse transform of a/(s^2-a^2) is sinhat. I'm not sure what the "h" means or how to use it.

Thanks =)

Clearly you are not familiar with hyperbolic functions, so you shouldn't be using them. I suggest you solve your question using partial fractions - you ought to know what the inverse Laplace transform of 1/(s - 7) and 1/(s + 7) are.