Re: Laplace Transform
To do this by hand, you must first simplify things a little bit. First rewrite the expression so that the coefficients multiplying \[s\] are 1. This might not be necessary, but it makes the inverse Laplace transform a bit easier. So we can write the quotient as
\[ \frac{1+sA}{c+sB} = \frac{A}{B} \frac{s+\frac{1}{A}}{s+\frac{c}{B}}\]
Now let's just look at the quotient \[(s+\frac{1}{A})/(s+\frac{c}{B})\]. We can use long division to rewrite this as a constant plus another term. Doing the long division yields
\[\frac{s+\frac{1}{A}}{s+\frac{c}{B}} = 1 + \frac{\frac{1}{A}-\frac{c}{B}}{s+\frac{c}{B}}\]
Now since the quotient has a term of \[\frac{A}{B}\] in front, our expression becomes
\[\frac{A}{B} + \frac{A}{B} \frac{\frac{1}{A}-\frac{c}{B}}{s+\frac{c}{B}}\]
So this new expression should be identical to the original one that you are trying to find the inverse Laplace transform for. You can check to see by getting a common denominator, adding the two term to make sure they match your original quotient.
The nice thing about this new form is that you can use a table of Laplace transform quite easily now. The transform of the first term is the delta function multiplied by a constant \[\frac{A}{B}\] and the second term has a transform of an exponential of form \[e^{-\frac{c}{B} t}\] multiplied by a constant of
\[\frac{A}{B} (\frac{1}{A}-\frac{c}{B})\]. So we get the following
\[\frac{A}{B} \rightarrow \frac{A}{B} \delta(t) \]
\[\frac{A}{B} \frac{\frac{1}{A}-\frac{c}{B}}{s+\frac{c}{B}} \rightarrow \frac{A}{B} (\frac{1}{A}-\frac{c}{B}) e^{-\frac{c}{B} t} u(t)\]
So the tranform of the total expression is the sum of the tranforms. Simplifying the constant in front of the second term and adding yields
\[\frac{A}{B} \delta(t) + (\frac{1}{B}-\frac{Ac}{B^2}) e^{-\frac{c}{B} t} u(t)\]
I think this matches what grobar got from MATLAB. I hope you find the information useful. Note that \[\delta(t)\] is the Dirac delta function and \[u(t)\] is the Heaviside step function.
Best regards,
v_c