- #1
skynelson
- 57
- 4
- TL;DR
- Repeated convolution tends, under certain conditions, to Gaussian distribution. Why doesn't this apply to sinc(x)?
Hello,
I've read that repeated convolution tends, under certain conditions, to Gaussian distribution. I found this description helpful, and Wikipedia's version of this says:
The central limit theorem states that if x is in L1 and L2 with mean zero and variance ##σ^2##, then
$$P\left(\frac{x^{*n}}{\sigma \sqrt{n}}\right)\rightarrow \Phi(\beta)$$
where ##\Phi(\beta)## is a standard normal distribution on the real line.
But if I apply a low pass filter, ##rect(t/\Delta t)## to an arbitrary distribution, ##\Psi##, twice in a row, I obtain,
$$rect(t/\Delta t)rect(t/\Delta t)\Psi = rect(t/\Delta t)\Psi$$ because the rect function acting twice doesn't change anything.
But through the convolution theorem this is equivalent to
$$\mathcal{F}^{-1}(sinc(\omega \Delta t) \ast sinc(\omega \Delta t) \ast \tilde{\Psi})$$
where ##\tilde{\Psi}## is the Fourier transform of the distribution.
So, because the rect function acting twice does nothing, the sinc function convolving twice doesn't change either. So ##n## convolutions of ##sinc()## will always spit back the same (non-Gaussian) result.
Is my understanding of the convergence to Gaussian incorrect or incomplete?
I've read that repeated convolution tends, under certain conditions, to Gaussian distribution. I found this description helpful, and Wikipedia's version of this says:
The central limit theorem states that if x is in L1 and L2 with mean zero and variance ##σ^2##, then
$$P\left(\frac{x^{*n}}{\sigma \sqrt{n}}\right)\rightarrow \Phi(\beta)$$
where ##\Phi(\beta)## is a standard normal distribution on the real line.
But if I apply a low pass filter, ##rect(t/\Delta t)## to an arbitrary distribution, ##\Psi##, twice in a row, I obtain,
$$rect(t/\Delta t)rect(t/\Delta t)\Psi = rect(t/\Delta t)\Psi$$ because the rect function acting twice doesn't change anything.
But through the convolution theorem this is equivalent to
$$\mathcal{F}^{-1}(sinc(\omega \Delta t) \ast sinc(\omega \Delta t) \ast \tilde{\Psi})$$
where ##\tilde{\Psi}## is the Fourier transform of the distribution.
So, because the rect function acting twice does nothing, the sinc function convolving twice doesn't change either. So ##n## convolutions of ##sinc()## will always spit back the same (non-Gaussian) result.
Is my understanding of the convergence to Gaussian incorrect or incomplete?