Heaviside function and dirac delta

In summary, the equation is telling us that the integral between ±∞ of δ(t)F(t)dt is equal to F(0).
  • #1
KateyLou
17
0

Homework Statement



Hi there, i am trying to do a proof that H'(t)= δ(t)

Homework Equations



We have been given the following:

F is a smooth function such that lim (t-->±∞)F(t)=0
Therefore the integral between ±∞ of [H(t)F(t)]'=[H(t)F(t)]-∞=0

I understand it up until this point; however next it says:
"Integration by parts:
(1) = Integral between ±∞ of [H(f)'F(t)]dt
(2) = -the integral between ±∞ of H(t)F'(t)dt
(3) = -the integral between ∞ and 0 of F'(t)dt
(4) = [-F(t)]0
(5) = F(0)
(6) = Integral between ±∞ of δ(t)F(t)dt

The Attempt at a Solution



I don't know where they have got theequation from in (1) or (2) or (3)! I get 4 though and 5! Although i don't then get 6!

I think if i knew where (1) came from i maybe could get through the rest but i just don't know where it has come from?
 
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  • #2
KateyLou said:

Homework Statement



Hi there, i am trying to do a proof that H'(t)= δ(t)

Homework Equations



We have been given the following:

F is a smooth function such that lim (t-->±∞)F(t)=0
Therefore the integral between ±∞ of [H(t)F(t)]'=[H(t)F(t)]-∞=0

Do you understand that this says [tex]\int_{-\infty}^{\infty} \left(\frac{d}{dt}(H(t)F(t))\right)dt=0[/tex]?

Since this is true for all functions F(t) such that lim (t-->±∞)F(t)=0, what must then be true of [tex]\frac{d}{dt}(H(t)F(t))[/tex]?

Equation (1) follows from using that along with the product rule :[tex]\frac{d}{dt}(H(t)F(t))=H'(t)F(t)+H(t)F'(t)[/tex]
 
  • #3
Yes i think I get the =0 part (by the way how do you write equations like that?)
And thank you re: the product rule - not sure how i managed to miss that! will go back to it and see how far i get this time!
 
  • #4
Right i have now completed this up to the last stage - i do not understand how it has gone from F(0) to [tex]\int_{-\infty}^{\infty} \left(\delta (t)F(t)\right)dt[/tex]
 
  • #5
And that is what defines the delta "function" (I have put "function" in quotes because it is not, of course, a "function", it is a "distribution" or "generalized function").
 
  • #6
What definition are you using for the derivative here? Note that the heaviside function is discontinuous, so it isn't differentiable using the derivative you used in Calc I. (similarly, H(x)F(x) is not differentiable)

The generalizations I know of the derivative take (1) = (2) to be the definition. (With the requirement that F be a sufficiently good function; here we need at least differentiable and vanishing at infinity)

As for the last question, what do you know about the Dirac delta distribution? Even if you know only one thing about it, it's probably the thing you need to use...
 
  • #7
i don't know what definition i am using... I have taken that straight from the maths notes we were given... (sorry!)

What do i know about the dirac delta function...well, the intergral of it is one, and that happens at t=0?
 
  • #8
KateyLou said:
What do i know about the dirac delta function...well, the intergral of it is one, and that happens at t=0?
D'oh, you got me. :frown: If that's the only thing you've been taught about the dirac delta distribution, then you are incapable of finishing this problem. (And so shame on your teacher for assigning it!) The step you are stuck upon is often used as the definition of the dirac delta.
 
  • #9
ha! ok!
So what is this definition then? I think i shall go an google it...
 
  • #10
kateylou said:
ha! Ok!
So what is this definition then? I think i shall go an google it...

(5) = (6)
 
  • #11
oh! lol!
Thank you guys :-)
 

FAQ: Heaviside function and dirac delta

What is the Heaviside function?

The Heaviside function, also known as the unit step function, is a mathematical function that is equal to 0 for negative input values and 1 for positive input values. It is often used to model systems that have a sudden change in behavior at a specific point.

What is the Dirac delta function?

The Dirac delta function, also known as the delta function, is a mathematical function that is equal to 0 for all input values except for 0, where it is equal to infinity. It is often used to model impulses or point masses in physical systems.

What is the relationship between the Heaviside function and the Dirac delta function?

The Heaviside function and the Dirac delta function are closely related. The Heaviside function can be defined as the integral of the Dirac delta function, and the Dirac delta function can be expressed as the derivative of the Heaviside function. In other words, the Heaviside function is the cumulative distribution function of the Dirac delta function.

How are the Heaviside function and the Dirac delta function used in applications?

The Heaviside function and the Dirac delta function are used in a variety of applications, including signal processing, control systems, and differential equations. They are particularly useful in modeling systems with discontinuities or sudden changes in behavior.

Are the Heaviside function and the Dirac delta function the same thing?

No, the Heaviside function and the Dirac delta function are not the same thing. While they are closely related and often used together, they are distinct mathematical functions with different properties and uses.

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