# Variance calculation from PSD

I have a receiver which is left to open for sensing. I'm using the periodogram method for estimating the PSD. I need to calculate the SNR of the incoming for which I need to calculate the variance. Can anyone suggest me ways to calculate the variance from the PSD?

• So your main goal is to calculate the SNR of the received signal?
– Deve
Feb 14 '13 at 10:39
• Yes, calculating SNR is my interest, but variance is the heart of my problem. So in any case I need the variance to be calculated @ Deve. Feb 14 '13 at 10:52

For a wide-sense-stationary random process, all the random variables comprising the process have the same mean $\mu$ and variance $\sigma^2$, and the variance is the integral of the power spectral density $S(f)$ less the square of the mean:
$$\sigma^2 = \int_{-\infty}^{\infty} S(f)\,\,\mathrm df - \mu^2.$$
Generally, the mean is $0$ and $\sigma^2$ is just the integral of $S(f)$, but when $\mu \neq 0$, then $S(f)$ includes an impulse of $\mu^2\delta(f)$ at $f=0$ which contributes $\mu^2$ to the integral, and this gets subtracted off by the $-\mu^2$ term in the above formula. For discrete-time signals, the integrals should be replaced by the appropriate sums. But when there is a (deterministic) signal component and a (random, zero-mean) noise component, calculation of the signal-to-noise ratio SNR (which has many definitions including one which says it is the ratio of the signal energy to the noise variance times the signal duration), requires separating out the two numbers needed from knowledge of their sum (which is what the periodogram will give you). This is a lot trickier and the answer will typically depends on details of the signal that you have not shared with us.