Confusion Surrounding Dirac Delta Comb Sampling: Why is δ(0) Infinite?

[Terocamo]

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?

 

[BvU]

The Dirac function needs to be normalized. And 'infinitely' narrow. Hence the function value -- which is 'never' used. All we use is multiplication in one domain and convolution in the other. I'm not a mathematician, but I don't think there is any ambiguity about the Dirac function.

Some of the confusion in the thread you mention is because they do Osgood injustice, perhaps because upon casual reading $\partial * \rho$ is misread as a multiplication instead of a convolution.

Depending on your interests (fundamental math or application) I wouldn't worry too much about the distribution aspects and follow the mainline arguments. But then, I'm a physicist...

 

[jasonRF]

Terocamo,

Perhaps your confusion stems from the fact that there are two different ways to represent the ideal sampling of a continuous time function $f(t)$. One way is to represent the sampled function as another function of continuous time, $f_s(t)$ , for which we will want do to standard continuous time Fourier transforms and other operations such as integration and/or differentiation. For this representation we sample with the Shah function:
<br /> f_s(t)=\sum_{n=-\infty}^{\infty}f(t)\delta(t-nT)=\sum_{n=-\infty}^{\infty}f(nT) \delta(t-nT)<br />
This is useful for doing things like proving the sampling rate required to avoid aliasing, etc. For this case, you shouldn't think about the "value" of $f_s(t)$ at any particular time - thinking about "infinities" isn't helpful here.

On the other hand, we can take our continuous time function and represent the sampled function as a discrete time function, $f_s[m]$. This is just a sequence of numbers: one number for each integer $m$. The ideal sampling in this case is simply $f_s[m] = f(mT)$. We cannot take the continuous time Fourier transform of this, or integrate it, etc. But we can do discrete-time Fourier transforms, discrete Fourier transforms (DFT), Z-transforms, etc. This discrete-time representation is usually what we think of when dealing with actual sampled and quantized data on a computer.

jason

 

[jasonRF]

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?

The Dirac delta is indeed well defined in mathematics - it is included in the theory of distributions (also called generalized functions). Chapter 4 of the lecture notes by Osgood give an nice outline of distribution theory that is accessible to engineers. If you are really interested you should read the notes.

jason

 

[Terocamo]

The Dirac delta is indeed well defined in mathematics - it is included in the theory of distributions (also called generalized functions). Chapter 4 of the lecture notes by Osgood give an nice outline of distribution theory that is accessible to engineers. If you are really interested you should read the notes.

jason

I came across a page which brings me a whole new of perspective towards Dirac Delta. I shouldn't have treated Dirac Delta simply like a function but, as you said, a distribution. It is then clear to me that the "value" of Dirac Delta itself doesn't seems to be important and that the integral is kind of like a heuristic property given to overcome physical problems. Thanks man

 

[BvU]

Good conclusion, I would say. The dirac thing is a very useful tool and you can deal with it almost as if it were a function. Now that you know what you can and can't do with it it is a good extension of your 'toolbox'!