Hi all, Sorry for my rustiness on Laplace transforms. I used the Laplace transform to solve the following equation, and got some results. Could you please comment on the correctness of my solution? The equation is: f(t)=a+b*Integrate( exp(-c*(t-s)) * f(s), ds, s from 0 to t) Here f(t) is a function that only takes values on [0, +inf). Here "*" denotes the multiplication. a, b, and c are constants. I argue that the integration above is a convolution and should be handled with Unilateral Laplace transform, since the range of integration is limited to within [0, t]. Taking Laplace transform on both sides, I obtained: F(s)= a/s + b/(s+c)*F(s), where F(s) is the Unilateral Laplace transform of f(t). Thus I got: F(s)= a*(s+c)/s/(s+c-b) =a/(s+c-b) + ac/s/(s+c-b) =a*exp(-(c-b)*t)*U(t) +ac/(c-b) * (1-exp(-(c-b)*t))*U(t) =a/(c-b)*(c-b*exp(-(c-b)*t))*U(t) where U(t) is the step function such that U(t)=1, when t>=0 and U(t)=0, for t<0. --------------- I based on my Unilateral Laplace transforms on the following table: http://en.wikipedia.org/wiki/Laplace_transform I hope I was right in the above calculations. But given my rustiness, I might not have realized it if I did something wrong... Please help me comment on the correctness of above calculations. Thank you very much! -M

# Please comment on my solution of this equation using Laplace transform

On May 15, 12:22 am, "Mike" <michael.monkey.in.the.jun...@gmail.com> wrote:> Hi all, > > Sorry for my rustiness on Laplace transforms. I used the Laplace transform > to solve the following equation, and got some results. Could you please > comment on the correctness of my solution? > > The equation is: > > f(t)=a+b*Integrate( exp(-c*(t-s)) * f(s), ds, s from 0 to t) > > Here f(t) is a function that only takes values on [0, +inf). Here "*" > denotes the multiplication. a, b, and c are constants. > > I argue that the integration above is a convolution and should be handled > with Unilateral Laplace transform, since the range of integration is limited > to within [0, t]. > > Taking Laplace transform on both sides, I obtained: > > F(s)= a/s + b/(s+c)*F(s), > > where F(s) is the Unilateral Laplace transform of f(t). > > Thus I got: > > F(s)= a*(s+c)/s/(s+c-b) > > =a/(s+c-b) + ac/s/(s+c-b) > > =a*exp(-(c-b)*t)*U(t) +ac/(c-b) * (1-exp(-(c-b)*t))*U(t) > > =a/(c-b)*(c-b*exp(-(c-b)*t))*U(t) > > where U(t) is the step function such that U(t)=1, when t>=0 and U(t)=0, for > t<0. > > --------------- > > I based on my Unilateral Laplace transforms on the following table: > > http://en.wikipedia.org/wiki/Laplace_transform > > I hope I was right in the above calculations. But given my rustiness, I > might not have realized it if I did something wrong... Please help me > comment on the correctness of above calculations.The inverse Laplace transform of A/(A+s) is A*exp(-A*t), and the inverse LT of A/s/(A+s) is int(A*exp(-A*x), x=0..t) = 1-exp(-A*t) (assuming t > 0). So, your answer does not look OK to me; the second term is correct, but the first one is not. R.G. Vickson> Thank you very much! > > -M

On Tue, 15 May 2007 00:22:37 -0700, "Mike" <michael.monkey.in.the.jungle@gmail.com> wrote:>Hi all, > >Sorry for my rustiness on Laplace transforms. I used the Laplace =transform=20>to solve the following equation, and got some results. Could you please=20 >comment on the correctness of my solution? > >The equation is: > >f(t)=3Da+b*Integrate( exp(-c*(t-s)) * f(s), ds, s from 0 to t) > >Here f(t) is a function that only takes values on [0, +inf). Here "*"=20 >denotes the multiplication. a, b, and c are constants. > >I argue that the integration above is a convolution and should be =handled=20>with Unilateral Laplace transform, since the range of integration is =limited=20>to within [0, t]. > >Taking Laplace transform on both sides, I obtained: > >F(s)=3D a/s + b/(s+c)*F(s), > >where F(s) is the Unilateral Laplace transform of f(t). > >Thus I got: > >F(s)=3D a*(s+c)/s/(s+c-b) > >=3Da/(s+c-b) + ac/s/(s+c-b) > >=3Da*exp(-(c-b)*t)*U(t) +ac/(c-b) * (1-exp(-(c-b)*t))*U(t) > >=3Da/(c-b)*(c-b*exp(-(c-b)*t))*U(t) > >where U(t) is the step function such that U(t)=3D1, when t>=3D0 and =U(t)=3D0, for=20>t<0.[snip] I got the same result ... f(t) =3D a/(c-b) ( c - b exp[-(c-b) t] ) =20

Mike wrote: (snip) I don't understand the step from:> F(s)= a/s + b/(s+c)*F(s),to> F(s)= a*(s+c)/s/(s+c-b)-- glen

On Tue, 15 May 2007 15:39:47 -0800, glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:>Mike wrote: > >(snip) > >I don't understand the step from: > >> F(s)=3D a/s + b/(s+c)*F(s),Solve for F(s) F(s) [ 1 - b/(s+c) ] =3D a/s F(s) =3D a/s (s+c)/(s+c-b)> >to > >> F(s)=3D a*(s+c)/s/(s+c-b) > >-- glen

On May 15, 7:48 pm, Passerby <inva...@invalid.invalid> wrote:> On Tue, 15 May 2007 15:39:47 -0800, glen herrmannsfeldt > > <g...@ugcs.caltech.edu> wrote: > >Mike wrote: > > >(snip) > > >I don't understand the step from: > > >> F(s)= a/s + b/(s+c)*F(s), > > Solve for F(s) > > F(s) [ 1 - b/(s+c) ] = a/s > > F(s) = a/s (s+c)/(s+c-b) > > > > >to > > >> F(s)= a*(s+c)/s/(s+c-b) > > >-- glenI was told that there would be no mathematics required in this group.

On 16 Mai, 02:53, Don Stockbauer <don.stockba...@gmail.com> wrote:> On May 15, 7:48 pm, Passerby <inva...@invalid.invalid> wrote: > > > > > > > On Tue, 15 May 2007 15:39:47 -0800, glen herrmannsfeldt > > > <g...@ugcs.caltech.edu> wrote: > > >Mike wrote: > > > >(snip) > > > >I don't understand the step from: > > > >> F(s)= a/s + b/(s+c)*F(s), > > > Solve for F(s) > > > F(s) [ 1 - b/(s+c) ] = a/s > > > F(s) = a/s (s+c)/(s+c-b) > > > >to > > > >> F(s)= a*(s+c)/s/(s+c-b) > > > >-- glen > > I was told that there would be no mathematics required in this group.They just don't make them like they used to.

Passerby wrote:> On Tue, 15 May 2007 15:39:47 -0800, glen herrmannsfeldt > <gah@ugcs.caltech.edu> wrote: > >> Mike wrote: >> >> (snip) >> >> I don't understand the step from: >> >>> F(s)= a/s + b/(s+c)*F(s), > > Solve for F(s) > > F(s) [ 1 - b/(s+c) ] = a/s > > F(s) = a/s (s+c)/(s+c-b)I think you mean F(s) = (a/s)*(s+c)/(s+c-b). ... Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Don Stockbauer wrote: ...> I was told that there would be no mathematics required in this group.There isn't. A little high-school algebra is helpful, but no one will send you away if you can't do it. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

On Wed, 16 May 2007 14:28:11 -0400, Jerry Avins <jya@ieee.org> wrote:>Passerby wrote: >> On Tue, 15 May 2007 15:39:47 -0800, glen herrmannsfeldt >> <gah@ugcs.caltech.edu> wrote: >>=20 >>> Mike wrote: >>> >>> (snip) >>> >>> I don't understand the step from: >>> >>>> F(s)=3D a/s + b/(s+c)*F(s), >>=20 >> Solve for F(s) >>=20 >> F(s) [ 1 - b/(s+c) ] =3D a/s >>=20 >> F(s) =3D a/s (s+c)/(s+c-b) > >I think you mean F(s) =3D (a/s)*(s+c)/(s+c-b).In math newsgroups, like sci.math, the two expressions are understood to be equivalent. The 'space' implies multiplication (as in a textbook), and unnecessary operators can make math newsgroup postings much less readable. It is generally understood that the "standard order of operations" rules also apply (unless stated otherwise).> > ... > >Jerry