laplace trasform

Ratch

Joined Mar 20, 2007
1,070
Mark44,

Ratch, why are you calling this a false equation? It has three real solutions in s, with approximate values -7.4, -0.8, and 0.2.
Because the OP asked for the inverse Laplace transform of the left side of the equation. The inverse Laplace transform changes the s-domain to the time-domain. So it would be a time-domain expression equaling a s-domain term of 1/s. That's a no-no.

Ratch
 

Ratch

Joined Mar 20, 2007
1,070
Mark44,

Maybe that's what he meant, but he didn't say anything about applying it only to the left side.
Yes he did. Otherwise he would have asked for the inverse Laplace on the right side too. What if he asked for the integral on the left side. Would you assume he meant the integral on the right side also?

Ratch
 

silvrstring

Joined Mar 27, 2008
159
Hey guys, the whole thing needs to be solved.

If the OP had written X(s)*(s^2+8s+4)=1/s, then it would have been clear that the problem is already in the s-domain. Then, X(s) = 1/[s*(s^2+8s+4)].

Solving L^(-1)[X(s)] would give the answer the OP is looking for, but he hasn't posted any work...at all.

I think he thinks this is a magic answer box. Not even a "Hello." Just type in the problem and wait. Rude.
 

Mark44

Joined Nov 26, 2007
628
Mark44,
Yes he did. Otherwise he would have asked for the inverse Laplace on the right side too. What if he asked for the integral on the left side. Would you assume he meant the integral on the right side also?
Ratch
It's not clear to me what "Yes he did" is agreeing with, since there are two statements in the thread just before your reply: "Maybe that's what he meant" and "he didn't say anything about applying it (Laplace transform) only to the left side."

If you meant "Yes he did" to apply to the second statement above, I disagree. His exact words were, "inverse laplace transform of s^2+8s+4=1/s". You are apparently interpreting what shem wrote as if there were parentheses on the left side, which there clearly aren't, nor which it would be reasonable to assume their presence. I agree that it would be meaningless to apply a Laplace transform (or any other operation that results in a change of value) to only one side.
 
Top