Adding Independent Noise

Question

Signal processing often involves managing and optimizing Signal to Noise Ratio (SNR). To achieve this, one may need to work with noise budgets and ensure that any additional additive noise sources are kept x dB below the current noise floor, with x typically in the range of 10 to 20 dB. These additive noise sources can include quantization noise resulting from an A/D converter or subsequent datapath truncation, aliasing caused by resampling operations, phase noise, front-end noise (which is usually factored into cascaded noise figure calculations), and any other additive noise sources- each reducing SNR.

In the process of having done this, I have committed to memory that adding additional noise 10 dB below the noise floor will result in increasing the noise floor by 0.4 dB, and 20 dB below will result in an increase of only 0.04 dB. We see this by converting the dB quantities to power terms, summing them and converting back to dB:

$$P_\Delta = 10\log_{10}(1+10^{-\text{dB}/10})$$

Where dB is the power in dB the added noise is below the existing noise floor.

However notice this interesting pattern from the formula given above:

$\text{ dB} \space\space\space\space\space\space\space\space\space\space\space P\Delta$
$-10 \text{ dB}: 0.414 \text{ dB}$
$-20 \text{ dB}: 0.0432 \text{ dB}$
$-30 \text{ dB}: 0.00434 \text{ dB}$
$-40 \text{ dB}: 0.000434 \text{ dB}$
$-50 \text{ dB}: 0.0000434 \text{ dB}$

Notice the result converges to $\approx 0.434\times10^{-n}$ as n increases (with -10 dB for n=0, -20 dB for n=1, etc). What is the exact answer for the result as n goes to infinity and what other common approximation can we relate this to? The answer I am looking for is not zero, but $x$ in $x\times10^{-n}$ as $n\rightarrow\infty$, together with an explanation as to how $x$ comes into this. The correct answer is exact and in a form that uses no numerical digits (no numerals 0 through 9).

This is a “DSP Puzzle”, please preface your answer with spoiler notation by typing ythe following two characters first ">!"

Answer 1

Let $\sigma_x^2$ and $\sigma_n^2$ be the signal and noise power, respectively. If (independent) noise with power $c\cdot\sigma_n^2$ is added to the existing noise, the resulting SNR becomes $$\textrm{SNR}=\frac{\sigma_x^2}{(1+c)\sigma_n^2}\tag{1}$$ In decibels we have $$\textrm{SNR}_{dB}=10\log\frac{\sigma_x^2}{\sigma_n^2}+10\log\frac{1}{1+c}\tag{2}$$ where the first term on the right-hand side of $(2)$ is the original SNR, and the second term is the change in SNR due to adding noise with a power that is a fraction of the original noise power.

With $c=10^{-n}$ and with the approximation $\ln(1+x)\approx x$ for small $x$, the change in $\textrm{SNR}_{dB}$ can be written as $$\begin{align}10\log\frac{1}{1+10^{-n}}&=-10\log(1+10^{-n})\\&=-\frac{10}{\ln (10)}\ln(1+10^{-n})\\&\approx-\frac{10}{\ln (10)}10^{-n},\qquad n\gg 1\tag{3}\end{align}$$

So for any additional noise with $10n$ dB less power than the original noise, the SNR is decreased by approximately $10/\ln (10)\cdot 10^{-n}\approx 4.342944819\cdot 10^{-n}$ dB. This approximation becomes better for increasing values of $n$.

Answer 2

The answer I was looking for was $\log(e)$ only because of it's perceived mathematical simplicity (and no use of the numerals 0 through 9 as I described) and it's direct relationship to the useful approximation $\ln(1+x) \approx x$ for small $x$. Obviously other than the factor of 10 likely due to confusion with my description, this is the same answer as Matt's without any real superiority. We can convert between the two answers readily using the following $\log$ to $\ln$ conversion: $$\ln(x) = \frac{\log(x)}{\log(e)}$$ Where $\log$ refers to a base 10 logarithm and $\ln$ refers to a base e logarithm.

This is how I came to the result, for large n: $$10\log(1+10^{-n})\approx 4.34 \times 10^{-n}$$ $$\log(1+10^{-n})\approx .434 \times 10^{-n}$$ As demonstrated in the post, the above appears to converge to something close to 0.434 and the question was what does it converge to for large $n$? So with that we have: $$\log(1+10^{-n})= x \times 10^{-n}$$ Convert to $\ln$ to use the simpler convergence of $\ln(1+x)$ for small $x$:$$\frac{\log(1+10^{-n})}{\log(e)}= \frac{x \times 10^{-n}}{\log(e)}$$ To get $$\ln(1+10^{-n})= \frac{x \times 10^{-n}}{\log(e)}$$ For large $n$, $10^{-n}$ is a small number so we can use the relation $$\lim_{n\rightarrow0}\ln(1+\alpha)=\alpha$$ Resulting in, as $n\rightarrow\infty$:$$\ln(1+10^{-n})= 10^{-n}=\frac{x \times 10^{-n}}{\log(e)}$$ Thus: $${\log(e)}\times10^{-n} = x \times 10^{-n}$$ and we have the result $$x = \log(e)$$

And of course we can convert this to or from Matt's answer to equivalently get $x=\frac{1}{\ln(10)}$