White noise going through a Hilbert filter.

[MathematicalPhysicist]

Now, assume I have a white noise, $$n(t)\tilde \ N(0,1)$$, i.e gaussian with zero mean and variance 1, and it goes through a Hilbert filter, i.e we get:

$$ \hat{n}(t) = \int_{-\infty}^{\infty} \frac{1}{t-\tau} n(\tau) d\tau $$

I read that $$\hat{n}$$ should be also a gaussian, because this is an LTI system.

But when I want to calculate its variance I get zero, so I guess I did something wrong here.

$$ E[\hat{n}^2(t) ] =\int \int_{\mathbb{R}^2} \frac{1}{(t-\tau_1)(t-\tau_2)} E[n(\tau_1)n(\tau_2)]d\tau_1 d\tau_2 $$

Now because the noise has zero mean value and variance 1 we should have: $$E[n(\tau_1)n(\tau_2)]=\delta(\tau_1 - \tau_2)$$

But when I plug I get zero. I read that we should take the principal value of the integral but I don't think it changes this.

Where did I get it wrong?

Thanks in advance.

 

[mmwave]

I can understand your confusion and frustration with this problem. It can be difficult to wrap your head around mathematical concepts, especially when it comes to complex systems like the one described in this forum post. Let's break down the problem step by step to see where the mistake may have occurred.

First, let's review the properties of a Gaussian distribution. A Gaussian distribution, also known as a normal distribution, is a probability distribution that is symmetric and bell-shaped. It is characterized by two parameters, the mean (μ) and the standard deviation (σ). In this case, the mean is 0 and the standard deviation is 1.

Next, let's look at the Hilbert filter. This is a type of filter that is used to create a complex signal from a real signal. It is a linear, time-invariant (LTI) system, meaning that its output is only dependent on its input and not on the time at which the input is applied. This is an important characteristic to keep in mind.

Now, let's examine the calculation of the variance of the filtered signal, \hat{n}(t). The variance is a measure of how spread out a distribution is. In this case, we are trying to calculate the variance of \hat{n}(t) which is the output of the Hilbert filter. This means that we need to consider the distribution of the input to the filter, n(t).

Based on the properties of the Gaussian distribution, we know that the variance of n(t) is 1. This means that the input to the filter has a spread of 1. However, when we apply the Hilbert filter, we are essentially taking the integral of n(t) with respect to time. This will cause the spread of the distribution to increase, resulting in a larger variance for the output.

So, where did the mistake occur in your calculation? It seems that you have assumed that the variance of \hat{n}(t) is 0 because the input to the filter has a variance of 1. However, as we have seen, the output of the filter will have a larger variance due to the integral operation.

In summary, the filtered signal, \hat{n}(t), should still be a Gaussian distribution, but with a larger variance than the input signal, n(t). I hope this helps clarify the problem for you. Keep exploring and learning, and don't be afraid to ask for help when needed. Good luck!