Laplace transform of dirac delta function

[Kambiz_Veshgini]

let S be the Unit Step function

for a function with a finite jump at t0 we have:

(*) L{F'(t)}=s f(s)-F(0)-[F(t0+0)-F(t0-0)]*exp(-s t0)]

so:

L{S'(t-k)}=s exp(-s k)/s-0-[1-0]*exp(-s k) = 0 & k>0

but S'(t-k)=deltadirac(t-k) and we know that L{deltadirac(t-k)}=exp(-s k)

so why do I get ZERO when using the formula (*)

 

[uart]

let S be the Unit Step function

for a function with a finite jump at t0 we have:

(*) L{F'(t)}=s f(s)-F(0)-[F(t0+0)-F(t0-0)]*exp(-s t0)]

Ok, I'm not familiar with that one, but I think you're mis-applying it.

The expression I'm familar with is just the L{F'(t)}=s f(s)-F(0) part.

My suspicion is that the [F(t0+0)-F(t0-0)]*exp(-s t0)] is just inserted to manually take care of the dirac impulse that results from the finite discontinuity and that in this case the f(s) you should be using is that of the original function without the discontinuity. Note that the unit step without the step is a pretty simple function. :)

 

[tacman]

The reason you are getting zero when using the formula (*) is because the Laplace transform of the Dirac delta function is not defined using the same formula as a function with a finite jump at t0. The Dirac delta function is a distribution, not a function, and therefore has a different definition for its Laplace transform.

The Laplace transform of the Dirac delta function is defined as:

L{δ(t-k)} = exp(-sk)

This means that the Laplace transform of the Dirac delta function is simply the exponential function with a negative argument.

In your calculation, you are using the formula for a function with a finite jump, which does not apply to the Dirac delta function. This is why you are getting zero as the result.

In conclusion, the Laplace transform of the Dirac delta function is not calculated using the same formula as a function with a finite jump, and instead has its own unique definition. So, it is important to use the correct formula when calculating Laplace transforms.