3
$\begingroup$

I have a signal and its derivative simultaneously measured, both including additive noise. The measurement is completed before the analysis, so it can be looked ahead. Now I want to reconstruct a less noisy version of the signal. I'm looking for pointers to algorithms I should look into.

Kalman filter seems to be on the right track, but the implementations I see so far are trying to estimate based on previous measurements only, while I probably should use both previous and coming measurements for optimal results at each point.

Ideas?

$\endgroup$
2
$\begingroup$

This is a really nice problem.

Problem Formulation

I will formulate it as following:

Let $ x \in \mathbb{R}^{n} $ be a signal. Given $ y \in \mathbb{R}^{n} $ which is a noisy measurement of $ x $ such that $ y = x + v $ and $ z $ be a noisy measurement of the derivative of $ x $ such that $ z = F x + w $ where $ F $ is the finite differences operator. It is known that both $ v $ and $ w $ are White Noise which is independent form each other. Find a reasonable estimator of $ x $.

OK, actually, I did the hard part by formulating the problem the way I did (Formulating connection between $ x $ and its derivative).
I find this formulation to be reasonable. It doesn't assume too much but gives us enough to work with.

So my first of choice would be something like:

$$ \arg \min_{x} \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \frac{\lambda}{2} {\left\| F x - z \right\|}_{2}^{2} $$

This is really easy to solve.
The tricky part is $ \lambda $. Well, I put it there to allow playing with the ratio between the noise level of the measurements. If the $ \sigma $ of both noises is the same, set $ \lambda = 1 $. Else, set it as $ \lambda = \frac{ {\sigma}_{v}^{2} }{ {\sigma}_{w}^{2} } $. The intuition is straight forward, if the noise level of $ w $ is smaller then $ v $ we want $ \lambda $ to be bigger than $ 1 $ in order to maximize the knowledge coming from $ z $.

Remark
The model above is actually the Maximum Likelihood Estimator of the model above if you assume the noises are AWGN.
I don't want to get into the Math of deriving it, but Least Squares is almost (Always) the ML for the AWGN model.

Solution to the Model

As mentioned above:

$$ \hat{x} = \arg \min_{x} \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \frac{\lambda}{2} {\left\| F x - z \right\|}_{2}^{2} $$

This is a strictly convex problem hence the solution is given in the stationary point.

$$\begin{align*} \frac{\mathrm{d}}{\mathrm{d} \hat{x}} \frac{1}{2} {\left\| \hat{x} - y \right\|}_{2}^{2} + \frac{\lambda}{2} {\left\| F \hat{x} - z \right\|}_{2}^{2} & = 0 && \text{Definition of Stationary Point} \\ & = \hat{x} - y + \lambda {F}^{T} \left( F x - z \right) && \text{} \\ & \Leftrightarrow \left( I + \lambda {F}^{T} F \right) = y + \lambda {F}^{T} z && \text{} \\ & \Leftrightarrow \hat{x} = {\left( I + \lambda {F}^{T} F \right)}^{-1} \left( y + \lambda {F}^{T} z \right) \end{align*}$$

MATLAB Simulation

The model I chose is using an harmonic signal - sine.
We use Finite Differences as the Derivative Model.

Here are the signals and the estimation using $ \lambda = \frac{ {\sigma}_{v}^{2} }{ {\sigma}_{w}^{2} } $:

enter image description here

It can be seen that the estimation is much better then any of the given signals.

Let's verify the optimality of $ \lambda $:

enter image description here

It is hard to see 2 circles at the bottom but they are close to verify the optimality of $ \lambda $.

The code is available on my StackExchange Signal Processing Q52150 GitHub Repository.

$\endgroup$
0
$\begingroup$

One alternative track could be that, when sampling at the same time a band-limited analog signal and its derivative, it can be sampled at half the standard Nyquist sampling rate. I have learn quite recently that it was used in geophysics when recording at the same location with a vibration and and pressure sensor.

Hence there is room to improve your signal's quality (which I am still looking for). Meanwhile, a few references to start from:

$\endgroup$

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.