Root mean square (RMS) of a sinc function

[redtree]

Given that the root mean square (RMS) of a sine function is as follows:

RMS of (a*sin($$\omega$$*r) = a / $$\sqrt{}2$$

Let a = 1/$$\omega$$

Thus

RMS of ((1/$$\omega$$)*sin($$\omega$$*r)) = 1 / ($$\omega$$*$$\sqrt{}2$$)

But for sinc($$\omega$$*x), what is formula for the RMS?

 

[adriank]

The rms value of sinc is zero.

I'll define rms as
$$f_{rms} = \sqrt{ \lim_{a \to \infty} \frac{1}{2a} \int_{-a}^{a} f(x)^2 \,dx }.$$
Let's find the rms for sinc(x). The integral here is
$$\int_{-a}^{a} \frac{\sin^2 x}{x^2} \,dx$$
$$= \int_{-a}^{-1} \frac{\sin^2 x}{x^2} \,dx + \int_{-1}^{1} \frac{\sin^2 x}{x^2} \,dx + \int_{1}^{a} \frac{\sin^2 x}{x^2} \,dx$$
$$\le \int_{-a}^{-1} \frac{1}{x^2} \,dx + \int_{-1}^{1} 1 \,dx + \int_{1}^{a} \frac{1}{x^2} \,dx$$
$$= 4 - \frac2a < 4.$$

Thus, the rms value of sinc is
$$\sqrt{ \lim_{a \to \infty} \frac{1}{2a} \int_{-a}^{a} \frac{\sin^2 x}{x^2} \,dx } \le \sqrt{ \lim_{a \to \infty} \frac{1}{2a} 4 } = 0.$$

 

[redtree]

Thanks

 

[redtree]

I understand the case when a approaches infinity, but what about the case where "a" is a finite number?

 

[uart]

I understand the case when a approaches infinity, but what about the case where "a" is a finite number?

The sinc function is not periodic, in other words it's period is infinity. That's why adriank found it's RMS value to be zero. Think of it this way, it's got a finite total energy that is spread over an infinite time interval, the result is an RMS value of zero.

In many ways it doesn't really make a lot of sense to even talk about the RMS value of an aperiodic function like that. What exactly is your application?

 

[redtree]

I want to find a good approximation for the following function:
a = k * sinc( pi k r) = sin (pi k r) / (pi r)

One approximation would be the following
a = 1 / (pi r)

But I think a better approximation is based on RMS. My feeling is that the sinc funciton above is a periodic function, though with decreasing amplitude as a function of r and therefore can be approximated using RMS.

 

[adriank]

Well, the rms value of sin is $1/\sqrt{2}$, so you might guess
$$a = \frac{1}{\sqrt{2} \lvert \pi r \rvert}.$$
It doesn't seem very useful though. It's not square-integrable, for instance.