- #1

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The book is almost no help on this and I unfortunatley missed the class lecture on it.

Thanks.

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- Thread starter supersix2
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- #1

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The book is almost no help on this and I unfortunatley missed the class lecture on it.

Thanks.

- #2

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supersix2,

Please try to ask specific questions. In the meantime, here are some good sites (found via google.com) on the topic.

http://en.wikipedia.org/wiki/Dirac_delta_function

http://tutorial.math.lamar.edu/AllBrowsers/3401/DiracDeltaFunction.asp [Broken]

- Warren

Please try to ask specific questions. In the meantime, here are some good sites (found via google.com) on the topic.

http://en.wikipedia.org/wiki/Dirac_delta_function

http://tutorial.math.lamar.edu/AllBrowsers/3401/DiracDeltaFunction.asp [Broken]

- Warren

Last edited by a moderator:

- #3

Science Advisor

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- #4

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- #5

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It's the simplest one of them all, once you see it.

The Dirac delta is really what is called a linear functional. It is very straightforward. Essentially, it assigns the value of a function via the standard scalar product for functions. In other words, for a function [itex]f[/itex] of a real argument, and [itex]\mathbb{R}\ni a \geq 0[/itex],

[tex]\delta_a(f) = \langle \delta(x-a), \ f(x) \rangle = \int_{0}^\infty \delta(x-a)f(x) \ dx = f(a).[/tex]

This may all look daunting, but it's about the simplest thing you'll ever learn once you understand it.

Let's look at this again:

[tex]\int_{0}^{\infty} \delta(x-a) f(x) \ dx = f(a)[/tex]

and try to apply it to finding Laplace transforms. We want the Laplace transform of [itex]\delta (t-a)[/itex], [itex] a \geq 0 [/itex] (which is really a misnomer, because [itex]\delta[/itex] isn't really a function at all - you actually need to define a completely new transformation to deal with generalized functions, and it has different properties! I won't worry about that in this post, though).

From the definition of the Laplace transform (as you already have it), this is

[tex]{\cal L}\{ \delta(t-a) \} = \int_{0}^\infty \delta(t-a)e^{-st} \ dt.[/tex]

But look at what I wrote above! From the definition of the delta disribution, this integral is trivial. Let [itex]f(t) = e^{-st}[/itex], so our integral is just

[tex]\int_{0}^\infty \delta(t-a)f(t) \ dt[/tex]

which from the definition of the delta is just the*value* of [itex]f(x)[/itex] when [itex]x=a[/itex], or in other words, it's just [itex]f(a)[/itex]. In this case,

[tex]f(a) = e^{-as},[/tex]

so we just get

[tex]{\cal L}\{\delta(t-a)\} = e^{-as}.[/tex]

Now, of course, the question is how to find things like

[tex]{\cal L}\{ \delta(t-a) f(t) \},[/tex]

but I'm not going to tell you how to do that (at least not yet). I want you to look at what I've posted above (specifically the identities concerning the delta distribution itself), and see if you can figure it out on your own. Post your work here and we'll comment!

The Dirac delta is really what is called a linear functional. It is very straightforward. Essentially, it assigns the value of a function via the standard scalar product for functions. In other words, for a function [itex]f[/itex] of a real argument, and [itex]\mathbb{R}\ni a \geq 0[/itex],

[tex]\delta_a(f) = \langle \delta(x-a), \ f(x) \rangle = \int_{0}^\infty \delta(x-a)f(x) \ dx = f(a).[/tex]

This may all look daunting, but it's about the simplest thing you'll ever learn once you understand it.

Let's look at this again:

[tex]\int_{0}^{\infty} \delta(x-a) f(x) \ dx = f(a)[/tex]

and try to apply it to finding Laplace transforms. We want the Laplace transform of [itex]\delta (t-a)[/itex], [itex] a \geq 0 [/itex] (which is really a misnomer, because [itex]\delta[/itex] isn't really a function at all - you actually need to define a completely new transformation to deal with generalized functions, and it has different properties! I won't worry about that in this post, though).

From the definition of the Laplace transform (as you already have it), this is

[tex]{\cal L}\{ \delta(t-a) \} = \int_{0}^\infty \delta(t-a)e^{-st} \ dt.[/tex]

But look at what I wrote above! From the definition of the delta disribution, this integral is trivial. Let [itex]f(t) = e^{-st}[/itex], so our integral is just

[tex]\int_{0}^\infty \delta(t-a)f(t) \ dt[/tex]

which from the definition of the delta is just the

[tex]f(a) = e^{-as},[/tex]

so we just get

[tex]{\cal L}\{\delta(t-a)\} = e^{-as}.[/tex]

Now, of course, the question is how to find things like

[tex]{\cal L}\{ \delta(t-a) f(t) \},[/tex]

but I'm not going to tell you how to do that (at least not yet). I want you to look at what I've posted above (specifically the identities concerning the delta distribution itself), and see if you can figure it out on your own. Post your work here and we'll comment!

Last edited:

- #6

- 26

- 0

Here's one I was working on

Code:

`[tex]\int_{\infty}^\infty\exp(2t)\cos(t) \delta(t)f(t) \ dt[/tex]`

I got the answer of 1.

My reasoning...

The way I understand is when evaluating the integral of a delta function of

Code:

`[tex]\delta(t-a)[/tex]`

If not where did I go wrong?

BTW that integral is from -infinity to infinity

I couldn't figure how to make a -infinity using the code.

- #7

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You know guys, I'm really learning a lot about the Dirac Delta function from you all. I've come to the following conclusion:

[tex]\mathop \lim\limits_{k\to 0}\int_0^k f(t)\frac{1}{k}[H(t)-H(t-k)]dt=f(0)[/tex]

where I'm representing the Dirac function as the limit of a narrow square-wave pulse (difference of two Heaviside functions) at the origin that always has an area of 1.

I interpret the integral as saying: take a function and multiply it by a "narrow" square wave pulse at the origin and integrate it through the width of the pulse but take the limit as the width of the square wave goes to zero but the height goes to infinity. You'd think the integral is 0 but it's not! Amazing . . . Or am I wrong Data?

Don't wish to interfere with your help here. I'm learning too!

Edit: I should point out to SuperSix that the expression:

[tex]\frac{1}{k}(H(x)-H(x-k))[/tex]

is only the Dirac Delta function at the limit as k goes to zero Ok? Any similiarly "limiting" process could substitute for the Dirac function. Suppose I could have just used parabolas but not sure how to express the limiting process with those. Maybe I could ask you Supersix: we have a parabola opening downward, centered at the y-axis, but the x-intercepts keep getting closer to 0, yet the area is always kept to 1 which means the y-intercept is getting larger. Would that work?

[tex]\mathop \lim\limits_{k\to 0}\int_0^k f(t)\frac{1}{k}[H(t)-H(t-k)]dt=f(0)[/tex]

where I'm representing the Dirac function as the limit of a narrow square-wave pulse (difference of two Heaviside functions) at the origin that always has an area of 1.

I interpret the integral as saying: take a function and multiply it by a "narrow" square wave pulse at the origin and integrate it through the width of the pulse but take the limit as the width of the square wave goes to zero but the height goes to infinity. You'd think the integral is 0 but it's not! Amazing . . . Or am I wrong Data?

Don't wish to interfere with your help here. I'm learning too!

Edit: I should point out to SuperSix that the expression:

[tex]\frac{1}{k}(H(x)-H(x-k))[/tex]

is only the Dirac Delta function at the limit as k goes to zero Ok? Any similiarly "limiting" process could substitute for the Dirac function. Suppose I could have just used parabolas but not sure how to express the limiting process with those. Maybe I could ask you Supersix: we have a parabola opening downward, centered at the y-axis, but the x-intercepts keep getting closer to 0, yet the area is always kept to 1 which means the y-intercept is getting larger. Would that work?

Last edited:

- #8

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- #9

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On a side note my test today was completely void of the delta function. In fact it only had 4 questions. One was a second-order linear non-homogenous ODE with a discontinuous unit step function that you needed to use a Laplace transform to answer so that was easy. Then the other 3 questions involved using matrices to find solutions to a system of first order equations. Those were also easy. I blazed through the exam in a half hour it was nuts.

- #10

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Put the "-" (by pressing the "-" lock from the keyboard) before the "\infty" code sequence.

Daniel.

Daniel.

- #11

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[tex]f(x)=\frac{1}{a\sqrt{2\pi}}e^{-\frac{x^2}{2a^2}}[/tex]

Where a is the width of the gaussian. Think of this function's behavour in the limit a->0.

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