comb(x) (or III(x), or Shah(x), whatever you want to call it) is DEFINED as the infinite sum of delta functions.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

III(x)=\sum_{n=-\infty}^{\infty}\delta(x-nT)

[/tex]

for some period T.

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

We also know that [itex]\delta[/itex] 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?

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# Comb(x) is confusing me.

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