By gentleness is there an expert who can tell me if the attached inverse Laplace transforms are exact?
Thank you very much.
With best regards
F. M.
Hallo Luc, excuse me but I don't find satisfactory your reply.
The issue concern the response of an active high pass filter, to a sawtooth signal. I find a discrepancy between the calculated response and that obtained by convolution. for this reason I would like to know if there is an error in the calculation of the response. 
With best regards
F. M.
Perhaps, the calculation details can be useful to better understand the problem. Therefore, I attach a picture about the calculations.
Your waveform FMSAW is:
The literature gives the following LaplaceTransform | TimeFunction relationship for a sawtooth LITSAW:
That LITSAW waveform can be converted into yours by:
FMSAW(t)=10*(LITSAW(t)-0.5) = 10*LITSAW(t)-5
Further it is known the following LaplaceTransform | TimeFunction relationships:
and
Then I conclude that the Laplace transform of FMSAW(t) = fmsaw(s)=10*litsaw(s)-5*1/s=
Does this help?
Luc
My signal is a sawtooth (bipolar). To obtain the Laplace transform of it, you have to subtract 1/2 to what you provided, ie subtract 1 / 2s to the transform you provide. Thus obtaining the formula No. 7 of my previous photo. So, your answer, it is not very useful.
in my particular case
... But thank you anyway.
F. M.
