高斯白噪声频域统计表示推导及幅相分布依据问询

Great question! Let's walk through the reasoning step by step to understand why the phase (\theta(\omega)) is uniformly distributed, and clarify the distribution of the amplitude (a(\omega)) (since there's a common point of confusion here).

1. Background: Gaussian White Noise and Its Fourier Transform #

First, recall that Gaussian white noise (x(t)) is a zero-mean Gaussian random process where:

• Any two distinct time samples are uncorrelated (and thus independent, since it's a Gaussian process), with autocovariance (R(\tau) = \sigma^2\delta(\tau)).
• The Fourier transform (\hat{x}(\omega)) is a complex random variable for each frequency (\omega), which we write in magnitude-phase form:
  $$\hat{x}(\omega) = a(\omega)e^{i\theta(\omega)}$$
  where (a(\omega) = |\hat{x}(\omega)| \geq 0) is the real-valued amplitude, and (\theta(\omega) = \arg(\hat{x}(\omega))) is the phase.

We can decompose (\hat{x}(\omega)) into its real and imaginary parts:
$$\hat{x}(\omega) = X(\omega) + iY(\omega)$$
where:
$$X(\omega) = \int x(t)\cos(2\pi\omega t)dt, \quad Y(\omega) = \int x(t)\sin(2\pi\omega t)dt$$

2. Statistical Properties of (X(\omega)) and (Y(\omega)) #

Since (x(t)) is a zero-mean Gaussian process, linear operations (like these integrals) produce zero-mean Gaussian random variables (X(\omega)) and (Y(\omega)). Now let's compute their variances and covariance:

• Variance of (X(\omega)):
  $$\text{Var}(X) = \mathbb{E}\left[X(\omega)^2\right] = \int\int \mathbb{E}[x(t)x(s)]\cos(2\pi\omega t)\cos(2\pi\omega s)dsdt$$
  Substitute (R(\tau) = \sigma^2\delta(t-s)):
  $$\text{Var}(X) = \sigma^2 \int \cos^2(2\pi\omega t)dt = \frac{\sigma^2}{2}T$$
  (where (T) is the integration window length; for unit (T), this is (\sigma^2/2))
• Variance of (Y(\omega)):
  By identical reasoning, (\text{Var}(Y) = \text{Var}(X) = \sigma^2 T / 2)
• Covariance between (X(\omega)) and (Y(\omega)):
  $$\text{Cov}(X,Y) = \mathbb{E}[X(\omega)Y(\omega)] = \sigma^2 \int \cos(2\pi\omega t)\sin(2\pi\omega t)dt = 0$$
  The integral of (\sin(4\pi\omega t)) over any full period (or infinite window) is zero, so (X) and (Y) are uncorrelated. For Gaussian variables, uncorrelated implies independent.

So (X(\omega)) and (Y(\omega)) are independent, zero-mean Gaussian random variables with equal variance.

3. Phase (\theta(\omega)) is Uniformly Distributed #

To find the distribution of (\theta = \arctan2(Y,X)), we convert the joint probability density function (pdf) of (X) and (Y) to polar coordinates ((a, \theta)).

The joint pdf of (X) and (Y) is:
$$f_{X,Y}(x,y) = \frac{1}{2\pi\sigma_x^2}e{-(x^2 + y^{2)/(2\sigma_x}2)}$$
where (\sigma_x^2 = \sigma^2 T / 2).

Using the polar coordinate transformation (x = a\cos\theta), (y = a\sin\theta), the Jacobian determinant of the transformation is (a). The joint pdf in polar coordinates becomes:
$$f_{a,\theta}(a,\theta) = \frac{a}{2\pi\sigma_x^2}e{-a^2/(2\sigma_x2)}$$

To get the marginal pdf of (\theta), integrate over all non-negative values of (a):
$$f_\theta(\theta) = \int_0^\infty \frac{a}{2\pi\sigma_x^2}e{-a^2/(2\sigma_x2)}da$$
Let (u = a^2/(2\sigma_x2)), so (du = a/\sigma_x^2 da). The integral simplifies to:
$$f_\theta(\theta) = \frac{1}{2\pi} \int_0^\infty e^{-u}du = \frac{1}{2\pi}$$
This is constant for (\theta \in [-\pi, \pi]), so (\theta(\omega)) follows a uniform distribution (\mathcal{U}[-\pi, \pi]) — this matches the claim you mentioned.

4. Amplitude (a(\omega)): Rayleigh Distribution (Not Gaussian) #

Now for the amplitude (a = \sqrt{X^2 + Y^2}). To find its marginal pdf, integrate the joint polar pdf over all (\theta):
$$f_a(a) = \int_{-\pi}^\pi \frac{a}{2\pi\sigma_x^2}e{-a^2/(2\sigma_x2)}d\theta = \frac{a}{\sigma_x^2}e{-a^2/(2\sigma_x2)}$$
This is the Rayleigh distribution, not a Gaussian distribution (\mathcal{N}[0, \sigma]).

Why the Confusion with Gaussian? #

There are two common reasons this mix-up happens:

• Confusing amplitude with the complex Fourier transform itself: The complex variable (\hat{x}(\omega) = X + iY) is a zero-mean complex Gaussian variable (its real and imaginary parts are Gaussian), which is sometimes loosely referred to as having a "Gaussian amplitude" — but this is imprecise.
• Approximation for large amplitudes: For large (a), the Rayleigh distribution can be approximated by a Gaussian distribution (using the central limit theorem or a Taylor expansion), but this is only an approximation, not an exact result.

Additionally, note that the power spectral density condition (S(\omega) = \mathbb{E}[|a(\omega)|^2] = \sigma^2) holds for both cases:

• For the Rayleigh distribution, (\mathbb{E}[a^2] = 2\sigma_x^2 = 2 \cdot (\sigma^2 T / 2) = \sigma^2 T); normalizing by (T) (to get power per unit bandwidth) gives (S(\omega) = \sigma^2).
• If (a) were Gaussian (\mathcal{N}[0, \sigma]), (\mathbb{E}[a^2] = \text{Var}(a) = \sigma^2), which also matches — but this ignores the fact that amplitude is non-negative, making the Gaussian distribution physically unrealistic here.

Wrapping Up #

To summarize:

• The phase (\theta(\omega)) is exactly uniformly distributed (\mathcal{U}[-\pi, \pi]), derived directly from the independent Gaussian real/imaginary parts of the Fourier transform.
• The amplitude (a(\omega)) follows a Rayleigh distribution, not a Gaussian one — any claim of a Gaussian amplitude is either an approximation or a misstatement referring to the complex Fourier transform's real/imaginary components.

内容的提问来源于stack exchange，提问作者abalter