Can I add two variances directly? I have my noise variance $\sigma_{noise}=-10 \ \mathrm{dB}$ and signal variance, $\sigma_{signal}=40 \ \mathrm{dB}$. So, can I say the total variance is $\sigma_{total}=30 \ \mathrm{dB}$?

I will use this total variance to generate random variables using a normal distribution with $\mu =0$ and $\sigma = \sigma_{total}$.


If the signal and noise are uncorrelated then the variance of the sum of the two equals the sum of their variances. So, yes, you can add variances (of uncorrelated signals). However, you cannot add dB-values! First of all, a variance in dB does not make sense because dB quantifies the ratio between two values. What you probably mean is dBm or dBW. So before adding variances you have to compute their actual (linear) values from the given values in dB(m/W).

  • $\begingroup$ thank you for your valuable answer @Matt L. I will convert the dBm to m/W. But, in one of the paper the authors have mentioned the variance_signal to be in terms of dB. $\endgroup$ – roshni Feb 25 '17 at 16:52

Caveat: it can be misleading to write $\sigma$ for the noise variance instead of $\sigma^2$.

The formula for a linear combination of "signals" (noise included), considered as a sum of random variables, is given in SE.Stats: Variance of Uncorrelated Variables: for observation $Y$ composed of signal $X$ and noise $N$:


one has:


If the signal and the noise are correlated, you do not a lot of options. if not, based on power formulas like:

$$\sigma^\mathrm{dB} = 10 \log \left(\frac{\sigma}{\sigma^\mathrm{ref}}\right)$$

you can get:

$$\sigma_Y^\mathrm{dB} = 10 \log \left(\frac{10 ^ \left(\frac{\sigma_X^\mathrm{dB}}{10}\right)+10 ^ \left(\frac{\sigma_N^\mathrm{dB}}{10}\right)}{\sigma^\mathrm{ref}}\right)$$

In your case, if I am not making a huge mistake, this gives about 40 dB, since the noise is very low with respect to the signal.

What is not clear to me is your purpose. If you create a noise with this total "variance", I am afraid you won't reach the expected result.


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