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Quick question about Dirac delta functions 
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#1
Feb1913, 08:54 AM

P: 75

What does the square of a Dirac delta function look like? Is the approximate graph the same as that of the delta function?



#4
Feb1913, 11:22 AM

Mentor
P: 18,061

Quick question about Dirac delta functions
George is right. There is no way that we can make sense of squares of the Dirac Delta function. Distributions are very strange things and their properties are subtle. Nice operations like squaring don't always make sense.
So the square doesn't exist and so the approximate graph also doesn't exist. 


#5
Feb1913, 07:46 PM

P: 2,251

often this question comes up in an electrical engineering class when one is faced with convolving an impulse [itex] \delta(t) [/itex] with another impulse. i.e. what would happen if you had a linear, timeinvariant system with impulse response [itex] h(t) = \delta(t) [/itex] and you input to that system [itex] x(t) = \delta(t) [/itex]. obviously, the output should be [itex] y(t) = \delta(t) [/itex], but how do you get that from the convolution integral?



#6
Feb1913, 09:27 PM

HW Helper
P: 1,391

Although it looks like the convolution integral will generate the squared delta function integral when t = 0, one has to remember that 'functions' like the delta function are meant to exist under integrals, so t has to remain a free variable. Fixing its value doesn't really make much sense. Of course, this is a physicist's way of looking at the issue, so there are some gaps in the formality and rigor, and mathematicians should feel free to shore it up (or tear it down, as the case may be) with the appropriate rigor. 


#7
Feb1913, 09:52 PM

Sci Advisor
P: 820

First, let ##f(x)## and ##g(y)## be distributions on ##\Omega \subset \mathbb{R}##. We define the two variable distribution ##f \oplus g## as ##(f \oplus g, \, \phi(x,y)) := (f(x), \, (g(y), \, \phi(x,y)))##. We now define ##(f * g, \phi) := (f(x) \oplus g(y), \phi(x+y))##. Now this is welldefined in any of the following cases:
Dirac has compact support therefore the convolution exists. Edit: Now that I think about it, the simples method would be a approximation to identity argument. You would still need to show that the identity holds irrespective of your choice of representative though. 


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