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Suppose we have a time dependent noise signal $\theta(t)$. We sample $\theta(t)$ with a device which has 1 bit of resolution and is itself stochastic in the sense that the readout value $u$ of the device follows a probability distribution

$$P(u) = \sin(\theta(t)/2)^2 \delta(u - 1) + \cos(\theta(t)/2)^2 \delta(u + 1) \, .$$

In other words, we get $u=1$ with probability $\sin(\theta(t)/2)^2$ and $u=-1$ with probability $\cos(\theta(t)/2)^2$.

We sample at discrete times $n\delta t$ where $\delta t$ is the sampling interval. This yields a discrete sequence of random variables with distributions

$$P_n(u) = \sin(\theta_n/2)^2 \delta(u - 1) + \cos(\theta_n/2)^2 \delta(u + 1) \, .$$

where $\theta_n \equiv \theta(n \delta t)$. We record a series of discrete samples $u_n$ from the measurement device, and from that can compute the discrete Fourier transform of the series $u_n$.

How is the spectral density of $u_n$ related to the spectral density of $\theta(t)$?

This problem is difficult because the response of the measurement device is not linear, which means that the usual transfer function methods don't work. We can assume, if it helps, that

$$\theta(t) = \frac{\pi}{2} + \delta \theta(t)$$

where $\delta \theta$ is understood to be small. With this assumption, we have $$\sin(\theta/2)^2 \approx \frac{1}{2}(1 + \delta \theta) \quad \text{and} \quad \cos(\theta/2)^2 \approx \frac{1}{2}(1 - \delta \theta) \, .$$


My attempt

I tried solving this by writing out the expression for the discrete Fourier transform

$$U_k \equiv \sum_{n=0}^{N-1} u_n e^{-i 2 \pi k n / N} \, .$$

Taking the real part and using some basic relations of probability distributions we can write the probability distribution of the real part of $U_k$ as $$P_{\text{Re} U_k}(x) = \sum_{n=0}^{N-1} \frac{1}{\cos(2 \pi k n / N)} P_n \left(\frac{x}{\cos(2\pi k n / N)} \right) \, .$$ We can next use the fact that the probability distribution of a sum of random variables is equal to the convolution of the distributions of the summands, and then additionally use the fact that convolution turns into products in the frequency domain. Doing this yields

\begin{align} \mathcal{FT}[P_{\text{Re}U_k}](k) &= \prod_{n=0}^{N-1} \frac{1}{\cos(2 \pi k n / N)} \\ &\left( \sin(\theta_n/2)^2 e^{-i k \cos(2\pi k n / N)} + \cos(\theta_n/2)^2 e^{i k \cos(2\pi k n / N)} \right) \, . \end{align} The notation is considerably simpler if we define $c_{n,k} \equiv \cos(2\pi k n / N)$, giving $$ \mathcal{FT}[P_{\text{Re}U_k}](k) = \prod_{n=0}^{N-1} \frac{1}{c_{n,k}} \left( \sin(\theta_n/2)^2 e^{-i k c_{n,k}} + \cos(\theta_n/2)^2 e^{i k c_{n,k}} \right) \, . $$ If we use the linearization mentioned above this simplifies to \begin{align} \mathcal{FT}[P_{\text{Re}U_k}](k) &= \prod_{n=0}^{N-1} \frac{1}{2 c_{n,k}} \left( (1 + \delta \theta_n) e^{-i k c_{n,k}} + (1 - \delta \theta_n) e^{i k c_{n,k}} \right) \\ &= \prod_{n=0}^{N-1} \frac{1}{c_{n,k}} \left( \cos(k c_{n,k}) - i \delta \theta_n \sin(k c_{n,k}) \right) \, . \end{align} This is where I'm stuck. I realize there may be a considerably easier way to solve this. I presented this strictly mathematical approach as a point of reference and to demonstrate a possible way of thinking about the problem.

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    $\begingroup$ Without having worked out the math myself, here's a thought: you might instead try to calculate the autocorrelation function of the random process $\theta_n$ first. Then, by the Wiener-Khinchin Theorem, the power spectral density is just the Fourier transform of the autocorrelation function. This may or may not be easier to compute. Note that this does assume that the process is wide-sense stationary. I didn't see anything in your question that would indicate this isn't true, but you'll need to think about it. $\endgroup$ – Jason R Aug 4 '15 at 11:31
  • $\begingroup$ @JasonR Thanks for the suggestion. I think that is probably a good approach and I've used it before so I know exactly what you mean. $\endgroup$ – DanielSank Aug 4 '15 at 15:09
  • $\begingroup$ @JasonR I misread your comment the first time. The spectral density of $\theta$ is presumed known. The hard part is connecting that to the spectral density of $u$. $\endgroup$ – DanielSank Aug 4 '15 at 16:05
  • $\begingroup$ Indeed, but you should still be able to apply the principles suggested above. Find the autocorrelation function of the random process $u_n$. This will be a function of the statistics of the underlying random process $\theta(t)$. $\endgroup$ – Jason R Aug 4 '15 at 16:34
  • $\begingroup$ @JasonR yes, totally agree. I'm pretty sure in the linearized case this is really easy. In the fully nonlinear case the best I can probably hope for is a series expansion. If nobody else posts an answer I'll a self-answer. $\endgroup$ – DanielSank Aug 4 '15 at 16:37

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