Please comment on my solution of this equation using Laplace transform

Started by Mike May 15, 2007
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 


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) > > >-- glen
I 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. &macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
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. &macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
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