Hybrid Sampling Scheme Idea?

Question

I am facing a problem which I could not find any robust solution for . Assume we have a signal, $x$, composed of sum of a few sinusoidal. e.g. $$x=A_1\sin(\omega_1t+\phi_1)+A_2\sin(\omega_2t+\phi_2)+A_3\sin(\omega_3t+\phi_3)+\ldots +A_n\sin(\omega_nt+\phi_n)$$ We have sampled the signal, ($30\%$) of samples are high resolution 10-Bits and the other $70\%$ are low resolution 3-4 Bits samples. Also,assume a sub-set of frequencies are known(for example $\omega_1, \omega_5$ and $\omega_{10}$ out of $\omega_1,\ldots , \omega_5$ are known).

1. How can we reconstruct the original signal?
2. Is there any method which can reconstruct the signal from these samples?

Moreover, we have these assumptions:

• The sinusoidals have unknown amplitudes ($A_n$)
• of course, we know when the high resolution samples occur
• We know maximum number of possible independent frequencies in the summation
• and spacing of sampling of high resolution sampling is reconfiguration and can be programmed to be regular or random (whichever is suitable?)

For examples, the below is a signal sampled uniformly at Nyquist rate($50\textrm{ Hz}$), with two components of $25\textrm{ Hz}$ and $12\textrm{ Hz}$. Green samples are high resolution samples, red samples are samples with considerable error (low resolution) and the blue curve is the original signal.

Following code provides an example of mentioned data:

Fs=1000; %sampling frequency 
t=(0:800).*(1/Fs);
x=0.5*sin(2*pi*300*t)+0.6*cos(2*pi*460*t)+0.2*cos(2*pi*60*t)+0.8*cos(2*pi*40*t)+1.6*cos(2*pi*12*t); %Original Signal
NoisySample=x+0.1*randn(size(x));           %Uniform Noisy Samples
ClearSamples=x(1:10:end); %Clear Samples

I would appreciate any hint and direction.

Answer 1

i will reframe the question a little:

we have

$$\begin{align} x(t) &= \sum\limits_{n=1}^N r_n \cos(\omega_n t + \phi_n) \ + \ \epsilon(t) \\ \\ &= \sum\limits_{n=1}^N a_n \cos(\omega_n t) + b_n \sin(\omega_n t) \ + \ \epsilon(t) \\ \end{align}$$

where

$$\begin{align} a_n &= r_n \cos(\phi_n) \\ b_n &= -r_n \sin(\phi_n) \\ \\ r_n &= \sqrt{a_n^2 + b_n^2} \\ \\ \phi_n &= \arg\{a_n - j b_n \} \\ &= -\operatorname{atan2}(b_n,a_n) \\ \end{align}$$

and $\epsilon(t)$ is the unknown error signal.

first you have to assume that $\epsilon(t)$ contains no components having frequencies of any of the $\omega_n$ in it, otherwise that component will just team up with the component having amplitude $r_n$.

if you know all of the $\omega_n$ in advance, you can calculate all of the $a_n, b_n$ by using an inner product:

$$\begin{align} a_n &= \frac{\omega_n}{M\pi} \int\limits_{t_0}^{t_0+\tfrac{2M\pi}{\omega_n}} x(t) \cos(\omega_n t) \ dt \\ \\ b_n &= \frac{\omega_n}{M\pi} \int\limits_{t_0}^{t_0+\tfrac{2M\pi}{\omega_n}} x(t) \sin(\omega_n t) \ dt \\ \end{align}$$

$t_0$ can be the most convenient time for you and $M$ should be a large positive integer (and need not be the same $M$ for each frequency component $\omega_n$), but the wider your integration is, the better error rejection you will have.

from the $a_n$ and $b_n$ coefficients, you can calculate $r_n$ and $\phi_n$.