What is the inverse laplace transform of (1+sA)/(c+sB) ?

dirac(t)+inverse laplace transform+matlab

what is the inverse laplace transform of

(1 + sA)/(c + sB)

Laplace Transform

(1/b)(e^(-ct/b)) + (a/b) d(e^(-ct/b))/dt

i made it i'm not really sure

Added after 1 minutes:

the first term its correct but im not really sure about the second one.

Laplace Transform

i am not sure but there may be a delta function added to the 2nd term >> L-1[s*f(t)]= f'(t)+L-1[f(0)]
where f(0) is a constant so it will give a delta at t=0
i am not sure too

Laplace Transform

MATLAB says:

a/b*dirac(t)+(b-c*a)/b^2*exp(-c*t/b)

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

Laplace Transform

hello v_c,
can u provid a good and simple refrence for LATEX.
thnx.

Re: Laplace Transform

I don't know if there is anything simple -- I learned mostly from looking at the input code and seeing what happens when something is changed. It is a markup language similar to HTML if you are familiar with that.

There is a book by Leslie Lamport (who wrote the initial LaTeX) which is pretty good
Amazon.com: LaTeX: A Document Preparation System (2nd Edition) (0785342529838): Leslie Lamport: Books

There is also a book by Mittelbach, et. al.
Amazon.com: The LaTeX Companion (Tools and Techniques for Computer Typesetting) (0785342362992): Frank Mittelbach, Michel Goossens, Johannes Braams, David Carlisle, Chris Rowley: Books

I have the hardcopies of the first edition to both of these books. Other than these you should be able to find lots of good resources on the web. If you do a Google search on "LaTeX tutorial" you get lots of hits. Browse through these to see what you like

LaTeX tutorial - Google Search

Best regards and happy \[\LaTeX\\]ing,
v_c

Re: Laplace Transform

here u can use this file to help ur study