Can someone explain Laplace transformations

[brandy]

Can someone explain Laplace transformations! i don't understand ittttt.


[edit] sorry. i hadnt even heard of laplace transformations until i found it in my assignment.
basically i want to know how to use them
so, some simple, well explained examples perhaps?

 

[blade123]

Uhh...try giving a little more information next time, what don't you get?

 

[rock.freak667]

Essentially it transforms a function f(t) into a usually simpler form F(s).

 

[brandy]

sorry. i hadnt even heard of laplace transformations until i found it in my assignment.
basically i want to know how to use them
so, some simple, well explained examples perhaps?

 

[HallsofIvy]

The Laplace transform of the function f(x) is
[tex]L\{f\}(s)= \int_0^\infty f(x)e^{-sx}dx[/tex].

The nice thing about the Laplace transform is that it changes the derivative of f into an algebraic function of L\{f\}(s).

For example, the Laplace transform of df/dx is
[tex]\int_0^\infty \frac{df}{dx}e^{-sx}dx[/tex]

Integrate that using "integration by parts" with [itex]u= e^{-sx}[/itex] and [itex]dv= (df/dx)dx[/itex]. Then [itex]du= -se^{-sx}dx[/itex] and [itex]v= f(x)[/itex].
[tex]\int udv= uv\right|_0^\infty- \int_0^\infty v du= -se^{-sx}f(x)\right|_0^\infty+ s\int_0^\infty f(x)e^{-sx}dx[/tex]
[tex]= -sf(0)+ sL\{f\}(s)[/tex]

So the differential equation problem
[tex]\frac{dy}{dx}= f(x,y)[/tex]
with initial condition [itex]y(0)= y_0[/itex] becomes
[tex]-sy_0+ sL{y}(s)= L\{f(x,y)\}[/itex]

You could solve that, algebraically, for L{y}(s), then reverse the transform.

Unfortunately, while there is a straightforward (if not always easy) integral for the transform, the inverse transform tends to be less clear. Typically, what one does is break L{f}(s) into parts, the look up the inverse transform of each part in a "table of Laplace transforms".

I am of the opinion that there are always better ways of solving differential equations. Laplace transforms have two uses- allowing engineers a "standardized" way of looking up solutions to differential equations and giving an abstract way of writing solutions to use in theoretical discussions even if it is impossible to find the inverse transform.

 

[brandy]

1 why does the vdu part not have an interval??

2 and what happened to the infinite interval thing i can only see the 0 subed in. shouldn't there be an infinity subbed in or like a limit or something...?

3 with sL{y}(s)= L\{f(x,y)\}
how does that work? i don't quite think i understand the jump (sorry)

4 if someone wrote L{f} is L{f}(s) implied? or do they mean something different.

 

[rock.freak667]

1 why does the vdu part not have an interval??

It does, HallsofIvy just left it out is all.

2 and what happened to the infinite interval thing i can only see the 0 subed in. shouldn't there be an infinity subbed in or like a limit or something...?

for 'uv' you will have e^-sxf(x), for the upper limit of ∞, the e^-sx as x goes to ∞, e^-sx goes to 0. So as x goes to ∞, e^-sxf(x) goes to zero.

for the lower limit of x=0, that you should be able to follow.

3 with sL{y}(s)= L\{f(x,y)\}
how does that work? i don't quite think i understand the jump (sorry)

HallsofIvy showed that the laplace transform of dy/dx (or in his example df/dx) is -sf(0)+sL{y(x)}(s).

For dy/dx = f(x,y), if you take the laplace transform on both sides, you will get

L{dy/dx} = L{f(x,y)}

to get -sy(0)+sL{y(x)}(s) = L{f(x,y)}

and y(0) = y₀ to get -sy₀ + sL{y(x)}(s) = L{f(x,y)}.

4 if someone wrote L{f} is L{f}(s) implied? or do they mean something different.

Usually when you apply a Laplace transform, you transform whatever domain it was in, into the 's' domain.

so if you had L{f(x)}, you would get some function F(s). So it would be the same thing, but it would be better to write L{f(x)} since you can have f in terms of x and y and so on.

 

[HallsofIvy]

Thanks, I've fixed the limits on the integral. (A little late!)

 

[Skrew]

An important part of the Laplace Transform is the almost one to one nature of the transfer function to the original differential equation.

If this were not the case, the Laplace transform would be useless.

 

[brandy]

wow thanks heaps! lol some facepalmable mistakes there.
with 4, so it does mean the same thing or the domain is s in the case of L{f}(s)
i don't think the domain is s because its outside of the {}. am i right in assuming that L{f}(s) is the same as L{f} and merely just different notation?

does sL{f} mean L{f}(s) multiplied by s??

 

[rock.freak667]

does sL{f} mean L{f}(s) multiplied by s??

Yes.

Usually when you have to do the transforms by hand it is useful to put something like let F(s) = L{f(t)} so that you don't continuously have to write out L{f(t)}.

so 'sL{f(t)}' can be rewritten as 'sF(s)'.

 

[paulfr]

The Laplace Transform reduces a differential equation to an algebraic equation

Like the Fourier Transform, it gives a different perspective on a time domain function.
The FT yields the spectral content of the time domain function; that is its frequency components
and thus the steady state response.
The LT does this too [using complex frequency, e^ -st] but also contains the transient response.

In electrical engineering one can not do without it.
It is a tool for both synthesis and analysis.