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Product of two dirac impulses
Gast
#3873002
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! 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
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
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?
Gast
#3873065
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 ..? 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
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.
Gast
#3873103
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
Gast
#3888010
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. Antwort schreiben |
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