# Combining compressed measurements from the same source

Suppose I want to measure a signal $$x \in \mathbb{R}^n$$ subject to i.i.d. noise $$\epsilon$$. In traditional Nyquist Sampling, I can increase my signal-to-noise ratio by measuring $$x + \epsilon$$ for $$k$$ times and averaging over the measurements.

$$\overline{x} = \frac{1}{K}\sum_{k=1}^{K}{(x +\epsilon_k)}$$

Instead of $$x$$ i now have $$k$$ compressed measurements $$y \in \mathbb{R}^m$$ and their corresponding k differing measurement matrices $$\Psi \in \mathbb{R}^{m*n}$$ with $$m, that contain e.g. gaussian random entries so that $$y = \Psi*(x+\epsilon_k)$$. Assuming $$x$$ is sufficiently sparse so that reconstruction via $$l_1$$-Norm minimization is possible.

Can I also improve my SNR in the second case?

1. What are the noise implications if I reconstruct each of the $$k$$ measurement pairs $$(y,\Psi)$$ and take the average of the reconstructed signals?

2. What will be the noise implications of taking the average of all measurements $$\overline{y}$$ and $$\overline{\Psi}$$ and recover the signal from the single pair?

• In your first, classic sampling equation, do you perhaps mean that all $K$ samples $x_k$ are in fact identical $x_k = x, \forall k$ and it is the noise realisations $\epsilon$ that actually differ over $k$? I mean: $$\bar x = \frac1K \sum_{k=1}^K (x + \epsilon_k)$$ – Thomas Arildsen Aug 22 '19 at 6:40
• yes that's true, I corrected it – Mr Vinagi Aug 22 '19 at 9:04
• Do you know how sparse $x$ should be? – Royi Aug 23 '19 at 11:28
• Yes, in my case I can assume, that the sparsity is known. – Mr Vinagi Aug 26 '19 at 8:40

If you have $$k$$ individual measured vectors $$\mathbf y$$, each taken with an individual measurement matrix $$\mathbf \Psi$$, you actually have an interesting, third option: $$\mathbf y = \begin{bmatrix} \mathbf y_1 \newline \vdots \newline \mathbf y_k \newline \vdots \newline \mathbf{y}_K \end{bmatrix} = \begin{bmatrix} \mathbf \Psi_1 \newline \vdots \newline \mathbf \Psi_k \newline \vdots \newline \mathbf \Psi_K \end{bmatrix} \mathbf x$$ So now you effectively have a (much) taller measurement matrix, so you are effectively "under-sampling less" and that will improve your estimate $$\mathbf x$$ based on the above equation. How much, I cannot recall an expression for off the top of my head.
Note also that you may have a tall measurement matrix now due to the stacking of the $$\mathbf \Psi_k$$'s. That is, you do not necessarily have an under-determined system anymore. This gives you the option of for example using least-squares estimation instead of $$\ell_1$$-norm optimisation. If you know that $$\mathbf x$$ is sparse, then it still makes sense to use sparse estimation ($$\ell_1$$-norm optimisation etc.), but if $$\mathbf x$$ is only approximately sparse, compressible then I would try least squares estimation as well.
• Ah very interesting. So given $m*k \geq n$ least squares is the smartest way to go – Mr Vinagi Aug 22 '19 at 10:05