1
$\begingroup$

This question is in continuation of the one asked here.

Let's say that the measurement noise $w$ or any random variable is circularly Gaussian complex.

  • If the imaginary and real components each has a variance of 0.5, then what would be the total variance? Should I write: $w \sim CN(0,\sigma^2_w)$?

  • If on the other hand, if $w \sim CN(0,2\sigma^2_w)$ then does it mean that the real and imaginary components each have variance 1, so the total variance is 2. Is my understanding correct?

  • Is there a rule of thumb whether the variance should be 0.5 for each component or can it be anything?

$\endgroup$

1 Answer 1

3
$\begingroup$

If the imaginary and real components each has a variance of 0.5, then what would be the total variance?

The variance of the complex RV in that case would be $0.5 + 0.5 = 1$.

If on the other hand, if $w∼CN(0,2σ^2_w)$ then does it mean that the real and imaginary components each has variance 1, so the total variance is 2. Is my understanding correct?

Each has variance $2σ^2_w/2 = σ^2_w$.

In other words, when you add two RV, the variance of the result is the sum of the variances of each RV.

Is there a rule of thumb whether the variance should be 0.5 for each component?

When talking about Rayleigh fading coefficients, you want the variance per dimension to be 0.5 because that results in the average transmitted power being equal to the average received power (that is, the channel neither creates nor absorbs energy).

$\endgroup$
1
  • $\begingroup$ @SrishtiM When an RV is distributed as $CN(0,2\sigma_w^2)$, it means that its variance is $2\sigma_w^2$, and the variance per dimension is half that, or $\sigma_w^2$. If you want the variance per dimension to be 0.5, then you need to set $\sigma_w^2=0.5$. $\endgroup$
    – MBaz
    Commented Aug 17, 2017 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.