Understanding the Confusion Behind Comb(x) and III(x) Definitions in Mathematics

[scholzie]

comb(x) (or III(x), or Shah(x), whatever you want to call it) is DEFINED as the infinite sum of delta functions.
<br /> III(x)=\sum_{n=-\infty}^{\infty}\delta(x-nT)<br />
for some period T.

We know that \delta(x-a) is the shifted delta function where a is some constant.

We also know that \delta is infinite in magnitude, concentrated at a single point. Therefore III(x) should be defined as an infinite train of infinite spikes, with period T.

However, III(x) is also used as a sampling function where multiplying by f(x) simply gives you the value of f(x) at each spike. This implies that III(x) has a height of unity, and NOT infinity.

This really troubles me, because by definition III(x) is an infinite train of delta functions, but it's used as a train of unit-impulse functions. I understand that when you integrate f(x)delta(x) over infinity you sample f(0), so it should make sense that the INTEGRAL of f(x)III(x) over infinity gives you sample points of f(x), but that's not how it's used in practice.

Is this simply a case of people using two definitions for the same thing?

 

[scholzie]

I just thought of something. If the FT of delta is 1, then the FT of comb(x) would be a train of 1s, not a train of infinities, wouldn't it? I know that the FT of comb is another comb, but is it fair to say that one is a train of infinite points while the other is a train of 1s?

That doesn't sit well with me, but at least it makes some sort of sense.

 

[yyat]

However, III(x) is also used as a sampling function where multiplying by f(x) simply gives you the value of f(x) at each spike.

No, you have to "multiply" with III(x) and then integrate over R (as with the delta function). You should be aware though, that this is not mathematically rigourous unless you use the theory of distributions.

 

[scholzie]

No, you have to "multiply" with III(x) and then integrate over R (as with the delta function). You should be aware though, that this is not mathematically rigourous unless you use the theory of distributions.

That makes more sense, but it seems people leave the integration step out when explaining or using III(x). I've looked it up in numerous places and find the same conclusion.

Brad Osgood of Stanford even states this in the course reader for his course "The Fourier Transform and its Applications"

To summarize:
• Convolving a function with III (with IIIp) produces a periodic function with period 1 (with period p).
• Multiplying a function by III (by IIIp) samples the function at the integer points (at the points pk).

That seems pretty cut and dry to me, and he doesn't mention integration in his summation. See page 219 of the http://www.stanford.edu/class/ee261/book/all.pdf . He states
<br /> f(x)III(x)=\sum_{k=-\infty}^{\infty}f(x)\delta(x-k)=\sum_{k=-\infty}^{\infty}f(k)\delta(x-k)<br />
Looks similar to an integral to me, but dealing with integers rather than all values in R.

He never explicitly mentions integrating f(x)III(x) over the reals.In fact, this is the entirety of my confusion. I understand that one should multiply f(x) with III(x) and then integrate from -infinity to +infinity in order to get the summation of f(x) at the period of III(x), but I *never* see it written that way.

Furthermore, I see too many people defining the Dirac Comb as an infinite train of UNIT impulses, not INFINITE impulses. I have no idea what to believe. Is this simply an artifact of removing the integral from the process?