Product of two dirac impulses

#3873034
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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
Gast #3873035
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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.
Gast #3873041
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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?
#3873079
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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.
Gast #3873097
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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.

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