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

Started by 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
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
> 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).

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