2
$\begingroup$

Problem statement:

I am designing a NP detector for the following detection problem:

$\mathcal H_{0}: x[n] = A_0\cos(2\pi f_0n) + w[n]$
$\mathcal H_{1}: x[n] = A_1\cos(2\pi f_0n) + w[n]$

where:

$f_{0}: \text{known parameter}$
$w[n] \sim \mathcal N\left(0, \sigma^2\right), \text {WGN}$
$A_{0}\ \ \sim \mathcal N\left(a_0, \sigma_0^2\right)$
$A_{1}\ \ \sim \mathcal N\left(a_{1},\sigma_1^2\right)$

What didn't work for me:

o I have tried to directly apply Neyman-Pearson theorem in the way it's done in "Fundamentals of Statistical Signal Processing, Volume II, Detection Theory" by Steven M. Kay. But I have failed to derive test statistics for optimal detector.
o I have tried to use the amplitude of the spectral component for $f_{0}$ as a test statistics and estimate threshold of making decision from given $P_{FA}$ (probability of false alarm). It worked fine for me, but I'm still concerned that this detector is not optimal.

Question:

Is there some way to design optimal NP detector for such problem?
Any ideas, advices, links will be much appreciated.

$\endgroup$
  • $\begingroup$ Are both your null hypothesis $\mathcal H_0$ and the alternative hypothesis $\mathcal H_1$ signal + noise ? $\endgroup$ – Gilles Feb 29 '16 at 22:27
  • $\begingroup$ @Gilles, Yes, you are correct. Both hypothesis are signal (sine signal with amplitude described by random variable) + noise. Difference between hypothesis is in random variables that describe amplitudes. $\endgroup$ – Nikolai Popov Mar 1 '16 at 8:26
  • $\begingroup$ any updates since ? I'm curious. $\endgroup$ – Gilles May 18 '16 at 11:01
2
$\begingroup$

Since both the null hypothesis $\mathcal H_0$ and the alternative hypothesis $\mathcal H_1$ are signal + noise, this is not only the detection of a random signal in WGN, but also a discrimination problem (i.e. "classification") of one signal from the other both embedded in WGN. The Neyman-Pearson detector decides $\mathcal H_1$ if:

$$\frac{p(\mathbf x; \mathcal H_1)}{p(\mathbf x; \mathcal H_0)}> \gamma$$

Which is:

$$ \frac{\displaystyle\frac{1}{\left[2\pi\left(\sigma_1^2 + \sigma^2\right)\right]^\frac N2}\exp\left[-\frac{1}{2\left(\sigma_1^2 + \sigma^2\right)}\sum_{n=0}^{N-1}\left(x[n]-a_1\cos\left(2\pi f_0 n\right)\right)^2\right]}{\displaystyle\frac{1}{\left[2\pi\left(\sigma_0^2 + \sigma^2\right)\right]^\frac N2}\exp\left[-\frac{1}{2\left(\sigma_0^2 + \sigma^2\right)}\sum_{n=0}^{N-1}\left(x[n]-a_0\cos\left(2\pi f_0 n\right)\right)^2\right]}>\gamma $$

Taking the logarithm on the LHS you get:

\begin{align} T(\mathbf x)&=\frac N2 \ln\left(\frac{\sigma_0^2 + \sigma^2}{\sigma_1^2 + \sigma^2}\right) +\frac{\sum_{n=0}^{N-1}\left(x[n]-a_0\cos\left(2\pi f_0 n\right)\right)^2}{2\left(\sigma_0^2 + \sigma^2\right)}-\frac{\sum_{n=0}^{N-1}\left(x[n]-a_1\cos\left(2\pi f_0 n\right)\right)^2}{2\left(\sigma_1^2 + \sigma^2\right)}\\ &=\frac N2 \ln\left(\frac{\sigma_0^2 + \sigma^2}{\sigma_1^2 + \sigma^2}\right) +\left[\frac{\left(\sigma_1^2-\sigma_0^2\right)}{2\left(\sigma_0^2+\sigma^2\right)\left(\sigma_1^2+\sigma^2\right)}\right]\cdot\sum_{n=0}^{N-1}x^2[n] \\&\quad+ \left[\frac{\left(a_1-a_0\right)\sigma^2-a_0\sigma_1^2+a_1\sigma_0^2}{\left(\sigma_0^2+\sigma^2\right)\left(\sigma_1^2+\sigma^2\right)}\right]\cdot\sum_{n=0}^{N-1}x[n]\cos\left(2\pi f_0 n\right)\\&\quad+\left[\frac{a_0^2}{2\left(\sigma_0^2 + \sigma^2\right)}-\frac{a_1^2}{2\left(\sigma_1^2 +\sigma^2\right)}\right]\cdot\sum_{n=0}^{N-1} \cos^2\left(2\pi f_0 n\right) \end{align}

The inequality becomes $T(\mathbf x) > \ln\gamma$. Keeping data-dependent terms on the LHS and defining $\gamma'$ as:

$$ \gamma'=\ln\gamma-\frac N2 \ln\left(\frac{\sigma_0^2 + \sigma^2}{\sigma_1^2 + \sigma^2}\right)-\left[\frac{a_0^2}{2\left(\sigma_0^2 + \sigma^2\right)}-\frac{a_1^2}{2\left(\sigma_1^2 +\sigma^2\right)}\right]\cdot\sum_{n=0}^{N-1}{\cos^2\left(2\pi f_0 n\right)} $$

We get:

$$ \underbrace{\overbrace{\frac{\left(\sigma_1^2-\sigma_0^2\right)}{2\left(\sigma_0^2+\sigma^2\right)\left(\sigma_1^2+\sigma^2\right)}\sum_{n=0}^{N-1}x^2[n]}^{\rm Quadratic} + \overbrace{\frac{\left(a_1-a_0\right)\sigma^2-a_0\sigma_1^2+a_1\sigma_0^2}{\left(\sigma_0^2+\sigma^2\right)\left(\sigma_1^2+\sigma^2\right)}\sum_{n=0}^{N-1}{x[n]\cos\left(2\pi f_0 n\right)}}^{\rm Linear}}_{T'(\mathbf x)}> \gamma' $$

The test statistic $T'(\mathbf x)$ contains both a linear term and a quadratic term, the detection should then be based on both the variance characterized by the quadratic term and the "mean" characterized by the linear term. More like a composite energy detector. That's my two cents for now.

By the way, this problem looks like a special case the general Gaussian detection problem where the signal has a deterministic part and a random part. And this under $\mathcal H_1$ is modeled as the Bayesian general linear model as follows:

$$ \mathbf{ x = H\theta + w}\tag{1} $$

Where $\mathbf H$ is a $n\times p$ deterministic matrix , $\mathbf \theta$ is a $p\times 1$ random vector with PDF $\mathcal N\left(\mu_\theta, \sigma_\theta^2\right)$ and independent of $\mathbf w$.

If the problem was to detect the presence of a signal (any) versus noise, then one could transform the initial definition to for instance $\mathcal H_0$ noise only versus $\mathcal H_1$ signal 1 + signal 2. And this could be brought to the case in $(1)$.

UPDATE:

Note two special cases:

  • When the random amplitudes have equal variances but different means, that is if $\sigma_0^2 = \sigma_1^2$ but $a_0 \neq a_1$, the detection problem reduces to the one of simply tracking the "mean" as the quadratic term (i.e energy part) cancels out and the test statistic becomes

$$ T'(\mathbf x)=\left(\frac{a_1-a_0}{\sigma_0^2+\sigma^2}\right)\sum_{n=0}^{N-1}{x[n]\cos\left(2\pi f_0 n\right)} $$

  • And for the case where $a_0 = a_1$ but $\sigma_0^2 \neq \sigma_1^2$ you get both the quadratic and linear term with the same coefficient by a factor of $-2a_0$.

$$ T'(\mathbf x)=\frac{\left(\sigma_1^2-\sigma_0^2\right)}{2\left(\sigma_0^2+\sigma^2\right)\left(\sigma_1^2+\sigma^2\right)}\sum_{n=0}^{N-1}x^2[n] + \frac{\left(\sigma_0^2-\sigma_1^2\right)a_0}{\left(\sigma_0^2+\sigma^2\right)\left(\sigma_1^2+\sigma^2\right)}\sum_{n=0}^{N-1}{x[n]\cos\left(2\pi f_0 n\right)} $$

So, the random amplitudes $A_0$ and $A_1$ play part in the detection not only through their respective means $a_0$ and $a_1$ but also through their variances $\sigma_0^2$ and $\sigma_1^2$.

$\endgroup$
  • $\begingroup$ Thank you for interest to the question. You are technically correct, so I'm accepting you answer. About updates: I have derived same test statistics a while ago but implemented spectral component version of the detector in the system for simplicity reasons. Experiments showed that it works pretty well. $\endgroup$ – Nikolai Popov May 20 '16 at 8:22
  • $\begingroup$ You're welcome. Good to hear that it worked. Please see my update. Did you take into account both the means and variances ? $\endgroup$ – Gilles May 20 '16 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.