存在哪些数学工具来理解调制噪声？

假设我们有一个包含高斯白噪声的信号 $n$ 。如果我们通过乘以调节这个信号 $\sin 2\omega t$ ，产生的信号仍然有一个白色的功率谱，但很明显的噪声现在时间“束身”。这是一个循环平稳过程的例子。

x (t) = n (t) \sin 2 ω t

$x(t) = n(t) \sin2\omega t$

假设我们现在通过与正弦和余弦本机振荡器混合以形成I和Q信号来解调频率为 $\omega$ 的信号：

I = x (t) \times \sin ω t

$I = x(t) \times \sin\omega t$

Q = x (t) \times \cos ω t

$Q = x(t) \times \cos\omega t$

天真地观察到的功率谱（在大于的时间间隔内获取）是白色的，我们可以预期和都包含相同幅度的白色高斯噪声。但是，真正发生的是，正交选择性地对具有高方差的时间序列的部分进行采样，而相位相差90度的对较低方差的部分进行采样： $x(t)$ $1/f$ $I$ $Q$ $I$ $x(t)$ $Q$

调制噪声描述

结果是我的噪声频谱密度是是倍。 $\sqrt{3}$ $Q$

显然，在功率谱之外必须有一些对描述调制噪声有用的东西。我所在领域的文献中有许多描述上述过程的可访问论文，但我想学习信号处理/ EE社区如何更普遍地对待它。

有哪些有用的数学工具可用于理解和控制循环平稳噪声？ 任何参考文献也将不胜感激。

参考文献：

Niebauer等人，“非平稳散粒噪声及其对干涉仪灵敏度的影响”。物理修订版A 43，5022-5029。

noise modulation cyclostationary-random-process

— 尼伯特
source

为了让你展现的结果，你的解调器必须由相同的载波频率，下变频

，而不仅仅是

。

2 ω

$2 \omega$

ω

$\omega\$

— 杰森R

@Jason R，啊，我看我弄错与原始

调制。这是由于从泊松噪声转换为高斯噪声时的错误。

2 ω

$2\omega$

— nibot

我不确定您在这里找什么。通常通过功率谱密度或等效的自相关函数来描述噪声。随机过程的自相关函数及其PSD是傅立叶变换对。例如，白噪声具有脉冲自相关。这将在傅立叶域中转换为平坦的功率谱。

您的示例（而有些不切实际）是类似于在的载波频率观察载波调制的白噪声的通信接收机 $2 \omega$ 。示例接收机非常幸运，因为它的振荡器与发射机的振荡器相干。在调制器和解调器产生的正弦波之间没有相位偏移，从而允许“完美”下变频至基带。单靠这并不是不切实际的。相干通信接收机的结构很多。但是，通常将噪声建模为通信信道的附加元素，该元素与接收器试图恢复的调制信号无关。发射机很少实际发送噪声作为其调制输出信号的一部分。

不过，通过这种方式，看看示例背后的数学可以解释您的观察。为了获得您描述的结果（至少在原始问题中如此），调制器和解调器均具有在相同参考频率和相位下工作的振荡器。调制器输出以下内容：

\begin{aligned} n (t) & \sim N (0, σ^{2}) \\ x (t) & = n (t) \sin (2 ω t) \end{aligned}

$\begin{align} n(t) &\sim \mathcal{N}(0, \sigma^2) \\ x(t) & = n(t) \sin(2\omega t) \end{align}$

接收器生成下变频的I和Q信号，如下所示：

\begin{aligned} I (t) & = x (t) \sin (2 ω t) = n (t) \sin^{2} (2 ω t) \\ Q (t) & = x (t) \cos (2 ω t) = n (t) \sin (2 ω t) \cos (2 ω t) \end{aligned}

$\begin{align} I(t) &= x(t) \sin(2 \omega t) = n(t) \sin^2(2 \omega t)\\ Q(t) &= x(t) \cos(2 \omega t) = n(t) \sin(2 \omega t) \cos(2 \omega t) \end{align}$

一些三角恒等式可以使和 $I(t)$ $Q(t)$ ：

\begin{aligned} \sin^{2} (2 ω t) & = \frac{1 - \cos (4 ω t)}{2} \\ \sin (2 ω t) \cos (2 ω t) & = \frac{\sin (4 ω t) + \sin (0)}{2} = \frac{1}{2} \sin (4 ω t) \end{aligned}

$\begin{align} \sin^2(2 \omega t) &= \frac{1 - \cos(4 \omega t)}{2}\\ \sin(2 \omega t) \cos(2 \omega t) &= \frac{\sin(4 \omega t) + \sin(0)}{2} = \frac{1}{2} \sin(4 \omega t) \end{align}$

Now we can rewrite the downconverted signal pair as:

\begin{aligned} I (t) & = n (t) \frac{1 - \cos (4 ω t)}{2} \\ Q (t) & = \frac{1}{2} n (t) \sin (4 ω t) \end{aligned}

$\begin{align} I(t) &= n(t) \frac{1 - \cos(4 \omega t)}{2}\\ Q(t) &= \frac{1}{2} n(t) \sin(4 \omega t) \end{align}$

The input noise is zero-mean, so $I(t)$ and $Q(t)$ are also zero-mean. This means that their variances are:

\begin{aligned} σ_{I (t)}^{2} & = E (I^{2} (t)) = E (n^{2} (t) {[\frac{1 - \cos (4 ω t)}{2}]}^{2}) = E (n^{2} (t)) E ({[\frac{1 - \cos (4 ω t)}{2}]}^{2}) \\ σ_{Q (t)}^{2} & = E (Q^{2} (t)) = E (n^{2} (t) \sin^{2} (4 ω t)) = E (n^{2} (t)) E (\sin^{2} (4 ω t)) \end{aligned}

$\begin{align} \sigma^{2}_{I(t)} &= \mathbb{E}(I^2(t)) = \mathbb{E}\left(n^2(t) \left[\frac{1 - \cos(4 \omega t)}{2}\right]^2\right) = \mathbb{E}\left(n^2(t)\right) \mathbb{E}\left(\left[\frac{1 - \cos(4 \omega t)}{2}\right]^2\right) \\ \sigma^{2}_{Q(t)} &= \mathbb{E}(Q^2(t)) = \mathbb{E}\left(n^2(t) \sin^2(4 \omega t)\right) = \mathbb{E}\left(n^2(t)\right) \mathbb{E}\left(\sin^2(4 \omega t)\right) \end{align}$

You noted the ratio between the variances of $I(t)$ and $Q(t)$ in your question. It can be simplified to:

\frac{σ_{I (t)}^{2}}{σ_{Q (t)}^{2}} = \frac{E ({[\frac{1 - \cos (4 ω t)}{2}]}^{2})}{E (\sin^{2} (4 ω t))}

$\frac{\sigma^{2}_{I(t)}}{\sigma^{2}_{Q(t)}} = \frac{\mathbb{E}\left(\left[\frac{1 - \cos(4 \omega t)}{2}\right]^2\right)}{\mathbb{E}\left(\sin^2(4 \omega t)\right)}$

The expectations are taken over the random process $n(t)$ 's time variable $t$ . Since the functions are deterministic and periodic, this is really just equivalent to the mean-squared value of each sinusoidal function over one period; for the values shown here, you get a ratio of $\sqrt 3$ , as you noted. The fact that you get more noise power in the I channel is an artifact of noise being modulated coherently (i.e. in phase) with the demodulator's own sinusoidal reference. Based on the underlying mathematics, this result is to be expected. As I stated before, however, this type of situation is not typical.

Although you didn't directly ask about it, I wanted to note that this type of operation (modulation by a sinusoidal carrier followed by demodulation of an identical or nearly-identical reproduction of the carrier) is a fundamental building block in communication systems. A real communication receiver, however, would include an additional step after the carrier demodulation: a lowpass filter to remove the I and Q signal components at frequency $4 \omega$ . If we eliminate the double-carrier-frequency components, the ratio of I energy to Q energy looks like:

\frac{σ_{I (t)}^{2}}{σ_{Q (t)}^{2}} = \frac{E ((\frac{1}{2})^{2})}{E (0)} = \infty

$\frac{\sigma^{2}_{I(t)}}{\sigma^{2}_{Q(t)}} = \frac{\mathbb{E}\left((\frac{1}{2})^2\right)}{\mathbb{E}(0)} = \infty$

This is the goal of a coherent quadrature modulation receiver: signal that is placed in the in-phase (I) channel is carried into the receiver's I signal with no leakage into the quadrature (Q) signal.

Edit: I wanted to address your comments below. For a quadrature receiver, the carrier frequency would in most cases be at the center of the transmitted signal bandwidth, so instead of being bandlimited to the carrier frequency $\omega\$ , a typical communications signal would be bandpass over the interval $[\omega - \frac{B}{2}, \omega + \frac{B}{2}]$ , where $B$ is its modulated bandwidth. A quadrature receiver aims to downconvert the signal to baseband as an initial step; this can be done by treating the I and Q channels as the real and imaginary components of a complex-valued signal for subsequent analysis steps.

With regard to your comment on the second-order statistics of the cyclostationary $x(t)$ , you have an error. The cyclostationary nature of the signal is captured in its autocorrelation function. Let the function be $R(t, \tau)$ :

R (t, τ) = E (x (t) x (t - τ))

$R(t, \tau) = \mathbb{E}(x(t)x(t - \tau))$

R (t, τ) = E (n (t) n (t - τ) \sin (2 ω t) \sin (2 ω (t - τ)))

$R(t, \tau) = \mathbb{E}(n(t)n(t - \tau) \sin(2 \omega t) \sin(2 \omega(t - \tau)))$

R (t, τ) = E (n (t) n (t - τ)) \sin (2 ω t) \sin (2 ω (t - τ))

$R(t, \tau) = \mathbb{E}(n(t)n(t - \tau)) \sin(2 \omega t) \sin(2 \omega(t - \tau))$

Because of the whiteness of the original noise process $n(t)$ , the expectation (and therefore the entire right-hand side of the equation) is zero for all nonzero values of $\tau$ .

R (t, τ) = σ^{2} δ (τ) \sin^{2} (2 ω t)

$R(t, \tau) = \sigma^2 \delta(\tau) \sin^2(2 \omega t)$

The autocorrelation is no longer just a simple impulse at zero lag; instead, it is time-variant and periodic because of the sinusoidal scaling factor. This causes the phenomenon that you originally observed, in that there are periods of "high variance" in $x(t)$ and other periods where the variance is lower. The "high variance" periods are selected by demodulating by a sinusoid that is coherent with the one used to modulate it, which stands to reason.

— Jason R
source

Re: "This is the goal of a coherent quadrature modulation receiver..." -- this is true only if the original signal is band-limited to frequencies less than the carrier frequency, right?

— nibot

Re: "Noise is typically described via its power spectral density, or equivalently its autocorrelation function". This cyclostationary noise (

n (t) \cdot \sin ω t

$n(t)\cdot\sin\omega t$ ) is spectrally white and has a

δ (t)

$\delta(t)$ autocorrelation function, just like regular (stationary) Gaussian noise. I am looking for a description that encapsulates its cyclostationary nature.

— nibot

I edited the answer to talk about your two comments.

— Jason R

@Jason, good post. I am trying to understand however, the part where you talk about the cyclostationarity process. I am having a hard time understanding why 't' here is a function of R... - after the expectation operator, there is no 't' (time) variable anymore... only a function of tau.

— Spacey

@Jason nevermind, I just realized 't' must be there since the statistics change with time, (albeit cyclically), and so therefore the autocorr function will also be a function of time and delay... but what I do not understand in this case is how you got the delta*sin^2 ... does this warrant an actual question for me to post?

— Spacey