- #1
Terocamo
- 47
- 0
I have recently digged up a post in the forum about a confusion arise from definition of Dirac Delta function and I am actually really bothered by it (link to the thread).
When people talk about sampling some function f(x) with Dirac Comb, or impulse train, they would be talking about the product of f(x) and the Dirac Comb:
$$f'(x)=\sum_{n=-\infty}^{\infty}f(x)\cdot\delta(x-nT)$$
And as we all know δ(0)=infinite instead of 1, this doesn't make sense to me since III(x) would be a train of infinite instead of 1.
However, Wikipedia and some textbooks about Signal Processing almost always leave out the integration as in the definition of Dirac Delta:
$$f(X)=\int^\infty_{-\infty}\delta(x-X)\cdot f(x)dx$$
I understand that Dirac Delta is not well defined in mathematics, but when I think deeply, I don't see the why Dirac Delta have to be infinite at x=0. Why Dirac use +infinite instead of 1 or any other values? And can it really sample a function without the integration?
When people talk about sampling some function f(x) with Dirac Comb, or impulse train, they would be talking about the product of f(x) and the Dirac Comb:
$$f'(x)=\sum_{n=-\infty}^{\infty}f(x)\cdot\delta(x-nT)$$
And as we all know δ(0)=infinite instead of 1, this doesn't make sense to me since III(x) would be a train of infinite instead of 1.
However, Wikipedia and some textbooks about Signal Processing almost always leave out the integration as in the definition of Dirac Delta:
$$f(X)=\int^\infty_{-\infty}\delta(x-X)\cdot f(x)dx$$
I understand that Dirac Delta is not well defined in mathematics, but when I think deeply, I don't see the why Dirac Delta have to be infinite at x=0. Why Dirac use +infinite instead of 1 or any other values? And can it really sample a function without the integration?