Short review on Laplace Transform I ...
After the discussions on bilateral/unilateral Laplace Transforms, I made a brief enquiry on how this topic is treated in the literature , observing how different authors handle the matter.
Do they consider " by default " the unilateral or bilateral transform ?
Furtherly : how non mathematicians approach the Laplace transform and how explain its applications in technical fields such as, for instance, the Telecom world ?
Herebelow you can find how some authors deal with the subject.
A) F.G.TRICOMI -ISTITUZIONI DI ANALISI SUPERIORE
The Laplace transform is the linear functional operator which makes correspondence between:
* a (real or complex) function $F(t)$, specified in the points of a given fixed line "l " of the complex plane
and
* the $f(x)$ function given by : $f(x)=int_(l) e^(-xt)F(t)dt$ .
Generally the "l " curve is the positive real axis and the corresponding Laplace transform $ccL^(I) $ is defined (unilateral or single side) $f(x)=ccL_x^I[F(t)]=int_0^(oo)e^(-xt)F(t)dt$.
In other cases the "l " curve is the whole real axis and the corresponding transform $ccL^(II)$ ( bilateral or double side) is defined as :
$f(x)=ccL_x^(II)[F(t)]=int_(-oo)^(+oo)e^(-xt)F(t)dt$.
By default we will consider the unilateral(single side) transform, assuming as usual that the integral between $0$ and $+oo$ will be intended as the limit,for $ omega rarr +oo$ of the corresponding integral between $0$ and $ omega$.
For example, in the case $ F(t)=1 $ we get:
$f(x)=int_0^(+oo)e^(-xt)dt=-(1/x)[e^(-xt)]_0^(+oo)=1/x-(1/x) *lim_(t rarr +oo)e^(-xt)$;being $|e^(-xt)|=e^(-t*Re(x))$ , we can conclude that in the whole half plane $Re(x)>0 $ , we have $ccL_x[1]=1/x $, while for $Re(x)<=0$ the corresponding Laplace integral diverges or is undetermine.
Do they consider " by default " the unilateral or bilateral transform ?
Furtherly : how non mathematicians approach the Laplace transform and how explain its applications in technical fields such as, for instance, the Telecom world ?
Herebelow you can find how some authors deal with the subject.
A) F.G.TRICOMI -ISTITUZIONI DI ANALISI SUPERIORE
The Laplace transform is the linear functional operator which makes correspondence between:
* a (real or complex) function $F(t)$, specified in the points of a given fixed line "l " of the complex plane
and
* the $f(x)$ function given by : $f(x)=int_(l) e^(-xt)F(t)dt$ .
Generally the "l " curve is the positive real axis and the corresponding Laplace transform $ccL^(I) $ is defined (unilateral or single side) $f(x)=ccL_x^I[F(t)]=int_0^(oo)e^(-xt)F(t)dt$.
In other cases the "l " curve is the whole real axis and the corresponding transform $ccL^(II)$ ( bilateral or double side) is defined as :
$f(x)=ccL_x^(II)[F(t)]=int_(-oo)^(+oo)e^(-xt)F(t)dt$.
By default we will consider the unilateral(single side) transform, assuming as usual that the integral between $0$ and $+oo$ will be intended as the limit,for $ omega rarr +oo$ of the corresponding integral between $0$ and $ omega$.
For example, in the case $ F(t)=1 $ we get:
$f(x)=int_0^(+oo)e^(-xt)dt=-(1/x)[e^(-xt)]_0^(+oo)=1/x-(1/x) *lim_(t rarr +oo)e^(-xt)$;being $|e^(-xt)|=e^(-t*Re(x))$ , we can conclude that in the whole half plane $Re(x)>0 $ , we have $ccL_x[1]=1/x $, while for $Re(x)<=0$ the corresponding Laplace integral diverges or is undetermine.
Risposte
C) G.C. BAROZZI -MATEMATICA PER L'INGEGNERIA DELL'INFORMAZIONE
Foreword
Let $f :I rarr CC$ be a real or complex values function, defined on an interval $I$ containing the real positive axis :$R_(+)=[0,+oo) sube I$.
Definition 1 - We will say that $f $ is L-transformable if it exists $ s in CC$ such that the function $t rarr e^(-st)f(t) $ is summable on $RR_+$; in such case we will call Laplace integral of $ f $ , the integral:
$ int_0^(+oo) e^(-st)f(t)dt $.
Definition 2- We will call Laplace transform of $f $, the function $F(s)=int_0^(+oo)e^(-st)f(t)dt $, where the integral converges.
Let $f : I rarr CC$ be a function defined on an interval $I$, being $R_+ sube I $ ; we will indicate with the symbol $f_+$
the function:
$ f_+(t) =f(t) $ for $t >=0 $ ; $=0 $ elsewhere.
If $f $ is defined on $RR$, we have $f_+(t) =H(t)f(t) $. The functiona as $f_+(t) $ , which are null for $t< 0 $, are often called SIGNALS.
Formally : a function $f:RR rarr CC $null for negative values of its argument and Laplace transformable, is called a SIGNAL.
Foreword
Let $f :I rarr CC$ be a real or complex values function, defined on an interval $I$ containing the real positive axis :$R_(+)=[0,+oo) sube I$.
Definition 1 - We will say that $f $ is L-transformable if it exists $ s in CC$ such that the function $t rarr e^(-st)f(t) $ is summable on $RR_+$; in such case we will call Laplace integral of $ f $ , the integral:
$ int_0^(+oo) e^(-st)f(t)dt $.
Definition 2- We will call Laplace transform of $f $, the function $F(s)=int_0^(+oo)e^(-st)f(t)dt $, where the integral converges.
Let $f : I rarr CC$ be a function defined on an interval $I$, being $R_+ sube I $ ; we will indicate with the symbol $f_+$
the function:
$ f_+(t) =f(t) $ for $t >=0 $ ; $=0 $ elsewhere.
If $f $ is defined on $RR$, we have $f_+(t) =H(t)f(t) $. The functiona as $f_+(t) $ , which are null for $t< 0 $, are often called SIGNALS.
Formally : a function $f:RR rarr CC $null for negative values of its argument and Laplace transformable, is called a SIGNAL.
B) GILARDI -ANALISI TRE
-Foreword
The Fourier transform allows to pass from ODE ( Ordinary Differential Equations) to algebraic equations and from some kinds of PDE (Partail Differential Equations) to ODE.
However the Fourier transform is not the best tool for solving a Cauchy problem on the half real axis in the case of an ODE : its use infact restricts the search for a solution to tempered distributions and, for instance the exp function is not tempered .
Now it becomes clear the usefulness of introducing another type of transform , similar to Fourier one as far as formal properties are concerned, but which allows transformation of exp functions .
The Laplace transform is exactly in this condition.
1.
The Laplace transform
The classical definiton, which makes reference to functions defined on the half axis ($t >=0)$ is the following :
$(ccLu)(s) =int_0^(+oo) e^(-st)u(t)dt $ (1.1)
which provides the value $(ccLu)(s) $ if the integral converges.
In view of the extension of the definition to the distributions ( called also generalized functions
see here) ,it is more suitable to assume that the functions we consider are defined in $RR$ and are null for $ t<0 $.
In this case the above formula becomes:
$(ccLu)(s) =int_(-oo)^(+oo) e^(-st)u(t)dt $ (1.2)
We will always consider the formula (1.2).
As a consequence a formula like :
$ccL(sint) =1/(1+s^2) $ which is correct with reference to formula (1.1) , must be substituted by the following :
$ccL(sint H(t))=1/(1+s^2) $ , where H is the Heaviside function : $H=0 $ if $t<0 ; H = 1$ if $ t>0 $.
Continue
-Foreword
The Fourier transform allows to pass from ODE ( Ordinary Differential Equations) to algebraic equations and from some kinds of PDE (Partail Differential Equations) to ODE.
However the Fourier transform is not the best tool for solving a Cauchy problem on the half real axis in the case of an ODE : its use infact restricts the search for a solution to tempered distributions and, for instance the exp function is not tempered .
Now it becomes clear the usefulness of introducing another type of transform , similar to Fourier one as far as formal properties are concerned, but which allows transformation of exp functions .
The Laplace transform is exactly in this condition.
1.
The Laplace transform
The classical definiton, which makes reference to functions defined on the half axis ($t >=0)$ is the following :
$(ccLu)(s) =int_0^(+oo) e^(-st)u(t)dt $ (1.1)
which provides the value $(ccLu)(s) $ if the integral converges.
In view of the extension of the definition to the distributions ( called also generalized functions
see here) ,it is more suitable to assume that the functions we consider are defined in $RR$ and are null for $ t<0 $.
In this case the above formula becomes:
$(ccLu)(s) =int_(-oo)^(+oo) e^(-st)u(t)dt $ (1.2)
We will always consider the formula (1.2).
As a consequence a formula like :
$ccL(sint) =1/(1+s^2) $ which is correct with reference to formula (1.1) , must be substituted by the following :
$ccL(sint H(t))=1/(1+s^2) $ , where H is the Heaviside function : $H=0 $ if $t<0 ; H = 1$ if $ t>0 $.
Continue
"Paolo90":
I do thank you for this thread, Camillo. Really, this is one of the most useful and interesting things I've ever read. Thanks a lot. I'll save this page - when it is finished. Thanks again.
Paolo
It will take some time to be finished
I see that the two approach for Laplace transform that Camillo showed us are different. I think that Gilardi's one is very strange... very different from what I'm used to see.
Laplace (and Fourier) transform is a very usefull stuff...
... one of the most interesting mathematical stuff ever!
Laplace (and Fourier) transform is a very usefull stuff...
"Paolo90":
Really, this is one of the most useful and interesting things I've ever read.
... one of the most interesting mathematical stuff ever!
I do thank you for this thread, Camillo. Really, this is one of the most useful and interesting things I've ever read. Thanks a lot. I'll save this page - when it is finished. Thanks again.
Paolo
Paolo
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