Product of two dirac impulses

[Question (Gast)]

Hallo, what is the product of two identical Dirac impulses? I mean 
delta(t) times delta(t). It is delta(t)? Why delta(t) times f(t) equals 
f(0). In this case f(t) is delta(t)? Any solution or hint by you? 
Thanks!

[Florian P. (db1pf)]

Hello,

look at the definition of the delta Impulse. You will find
d(x)=1 for x=0, else 0
So d(x) * d(x) gives 2*d(x)

Using the definition, you can give an answer onto your second question 
by yourself.
Btw: I guess the question is wrong. I think you mean d(t)*f(t) = 
d(t)*f(0) or Integral over d(t)*f(t) results to f(0)

I hope this was useful,
best regards,
Florian

[sqrt (Gast²) = +- (Gast)]

Florian Pfanner schrieb:
> Hello,
>
> look at the definition of the delta Impulse. You will find
> d(x)=1 for x=0, else 0
> So d(x) * d(x) gives 2*d(x)
>
> Using the definition, you can give an answer onto your second question
> by yourself.
> Btw: I guess the question is wrong. I think you mean d(t)*f(t) =
> d(t)*f(0) or Integral over d(t)*f(t) results to f(0)
>
> I hope this was useful,
> best regards,
> Florian

I think d(0) is infinity.

[Question (Gast)]

Florian Pfanner schrieb:
> Hello,
>
> look at the definition of the delta Impulse. You will find
> d(x)=1 for x=0, else 0
> So d(x) * d(x) gives 2*d(x)
>
> Using the definition, you can give an answer onto your second question
> by yourself.
> Btw: I guess the question is wrong. I think you mean d(t)*f(t) =
> d(t)*f(0) or Integral over d(t)*f(t) results to f(0)
>
> I hope this was useful,
> best regards,
> Florian
Okay. Thanks!
And ist than d(x-x0) * d(x) = d(x0) =0?

[Nase (Gast)]

Florian Pfanner schrieb:
> d(x)=1 for x=0, else 0
> So d(x) * d(x) gives 2*d(x)

Why that?

d(x) * d(x) with x = 0 gives 1*1 = 1 ..?

[A. S. (rava)]

first of all you should clarify whether we're talking about discrete 
time or continuous time.

In the first case delta[t] is 0 exept for t=0 where it is 1. This is an 
easy answer to your question. And you've already got it.

The continuous time case is more elaborate as delta(t) is 0 for all t 
exept for t=0 where it converges to positive infinity. This is due to 
the fact that the integral over delta(t) needs to be 1 in both domains.

So in theory delta(t)*delta(t) should be equivalent to delta(t) as the 
integral of both is 1=1².

[Question (Gast)]

A. S. schrieb:
> first of all you should clarify whether we're talking about
> discrete
> time or continuous time.
>
> In the first case delta[t] is 0 exept for t=0 where it is 1. This is an
> easy answer to your question. And you've already got it.
>
> The continuous time case is more elaborate as delta(t) is 0 for all t
> exept for t=0 where it converges to positive infinity. This is due to
> the fact that the integral over delta(t) needs to be 1 in both domains.
>
> So in theory delta(t)*delta(t) should be equivalent to delta(t) as the
> integral of both is 1=1².

I mean the continuous time case.

[Werner (Gast)]

Florian Pfanner schrieb:
> look at the definition of the delta Impulse. You will find
> d(x)=1 for x=0, else 0

You missed the integration
http://en.wikipedia.org/wiki/Dirac_delta_function#As_a_measure

[Florian (Gast)]

Nase schrieb:
> Why that?
>
> d(x) * d(x) with x = 0 gives 1*1 = 1 ..?
You are right. I should not write forum posts while I put my child into 
bed ;)


Question schrieb:
>> Florian
> Okay. Thanks!
> And ist than d(x-x0) * d(x) = d(x0) =0?
Yes, if x0 != 0.