Signal-to-Noise Ratio (SNR) needed to discern two superimposed signals

Question

Say I am trying to measure a exponentially decaying signal ($S_y$), that constitutes a portion $\gamma$ (about 5% so $\gamma=\frac{1}{20}$) of total signal measured($S=S_y+S_b$ and $S_b=S_0 \beta e^{\frac{-t}{C_b}}$ and $S_y=S_0 \gamma e^{\frac{-t}{C_y}}$). Where $\gamma+\beta=1$ and $C_Y$ is about $\frac{1}{4}$ of $C_b$. I want to determine the least Signal-to-Noise Ratio (SNR) needed to be able to discern that signal. The noise in our signal is additive white noise. Should the SNR be $>\gamma=\frac{1}{20}$ or is there a better more formal way to decide the SNR limit? $$S(t)=S_0(\gamma e^{\frac{-t}{C_y}}+\beta e^{\frac{-t}{C_b}}) + n(t)$$

So my question is how do I determine the minimal SNR when I measure signal $S$ over time $t$, but the thing I am interested in is fluctuation (max 10%) in $S_y$. All constants are known to me, however i only know the mean of $C_y$, as it fluctuates and so $S_y$ does too. Thanks for any help. If I need to clarify any details please tell me and I will do so.

Answer 1

This is a work-in-progress until more information is added to the question.

First, I'm assuming that you know the magnitudes of your decaying exponentials, but you don't know their time constants (either $C_x$ or $C_y$) or your signal level $S_0$.

Second, you say that there is Gaussian noise, but not whether it's additive. I'm going to assume it's additive. That means your signal is really:

$$ S(t)=S_0(\frac{1}{20}e^{\frac{-t}{C_y}}+\frac{19}{20}e^{\frac{-t}{C_x}}) + n(t) $$

where $n(t)$ is additive, white, Gaussian noise with zero mean and variance $\sigma^2$.

Then, to set up the problem, we need to postulate the estimated values of $\hat{C}_x$, $\hat{C}_y$, $\hat{S}_0$, and $\sigma^2$ in $\hat{S}(t)$:

$$ \hat{S}(t; \hat{C}_x, \hat{C}_y, \hat{S}_0)=\hat{S}_0(\frac{1}{20}e^{\frac{-t}{\hat{C}_y}}+\frac{19}{20}e^{\frac{-t}{\hat{C}_x}}) $$

Then a least squares approach to solving it would define the error, $E$, as:

$$ E(\hat{C}_x, \hat{C}_y, \hat{S}_0) = \int_{I} \left| S(t) - \hat{S(t)}\right|^2 dt $$

where $I$ is your time interval of interest, and minimize this with respect to $\hat{C}_x$, $\hat{C}_y$, and $\hat{S}_0$.

I interpret your question as asking: what value $\frac{S_0^2}{\sigma^2}$ does this work over?

The answer is: it depends on what you mean by "work". You can get an estimate for any noise level. The question is, what error can you tolerate in that estimate. Also, it will depend somewhat on your integration length ($I$).

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EDIT

OK, I see you've changed notation so it's $C_b$ and $C_y$ now.

I've written a short scilab script to try to see how things change with noise level. I've assumed EVERYTHING is precisely known except the $C_y$ parameter.

The error in estimating the parameter versus the value of $\sigma$ is shown in the plot below. Code below generates it.

---

 S0 = 1;
 Cb = 201;
 Cy = 100;
 sigma = 0.1;
 mb = 1/20;
 my = 19/20;

 T=1000;
 t=[0:T-1];
 Sb = mb*exp(-t/Cb);
 Sy = my*exp(-t/Cy);

 clf
 subplot(311)
 plot(Sb)
 plot(Sy,'g')
 plot(S,'r')

 ERRORS = [];
 sigmaRange = [0.0 0.1 0.2 0.5 1 2 5 10];
 for sigma=sigmaRange,
 CyhatRange = 90:.01:110;
 estimates=[]
 NRuns = 100;
 for n=1:NRuns
     S = S0*(Sb + Sy) + sigma*rand(1,T,'normal');

     ERR= [];
     for Cyhat = CyhatRange,
         S_hat = S0*(Sb + my*exp(-t/Cyhat))

         ERR = [ERR; sum((S-S_hat).^2)];
     end

     [mx,ix] = min(ERR);

     CyhatEst = CyhatRange(ix);
     //disp(CyhatEst);
     estimates = [estimates; CyhatEst];

     subplot(312);
     plot(CyhatRange,ERR)
 end

 disp(mean((estimates-Cy).^2))

 ERRORS = [ERRORS; mean((estimates-Cy).^2)];
 end

 subplot(313)
 plot(sigmaRange,ERRORS);