Suppose we have an added discrete noise signal defined as:

$$y[n]=x[n]+w[n], $$ where $w$ is zero mean white Gaussian noise.


  1. When we say white noise, is it sufficient to say that it is always be uncorrelated with signal $x$, no matter what the distribution of signal is? (I think yes)

  2. Can we conclude that both $x$ and $w$ are independent IF noise is white Gaussian BUT the signal have unknown distribution ? (I am Confused about it)

  3. Lastly, What the information we glean-out of the model (intuitively) if we are informed about noise that besides being zero mean, white and Gaussian, it is circular symmetric complex as well? (I do know white noise has impulse auto-correlation but confused with circular-symmetric term).

  • $\begingroup$ 1. No. 2. No/Yes 3. Depends on whether $x$ and $w$ are complex or real or _complex-baseband representations of narrowband bandpass signals. You might want to read this answer and the link contained therein) and then edit your question. $\endgroup$ Dec 7, 2014 at 4:23

1 Answer 1


To expand on Dilip's comment:

  1. The noise being white implies only a certain property of the noise itself, not its correlation with any other signal.

  2. In digital communications, the independence and/or lack of correlation between the signal and the noise is assumed, if not explicitly stated. The reason is that this is what is observed in practice. Even if you can mathematically conclude that a given message signal and a given noise are independent and uncorrelated, there is not much point in doing so.

  3. What is meant by circularly symmetric Gaussian noise is that the noise looks the same in all directions. Imagine a certain Gaussian complex noise that (for some physical reason) produces larger noise in specific directions; for example, along a line at an angle of $\pi$/4 on the complex plane. You could then design a QAM constellation (and detector) that takes advantage of this behavior to get better than expected performance given the noise power in the channel. When a problem states that the noise is circularly symmetric, it's saying that you can't play this kind of game.

  • $\begingroup$ But dsp.stackexchange.com/questions/16714/… in the comments Jason says something different. $\endgroup$
    – kaka
    Dec 11, 2014 at 0:07
  • $\begingroup$ @kaka, I don't see how that is relevant. Jason's comment is about the joint correlation/independence of white noise samples. In your questions 1 and 2 above, you asked about the correlation/independence between noise samples and signal samples. $\endgroup$
    – MBaz
    Dec 11, 2014 at 2:13
  • $\begingroup$ @ So in the first point if noise is said to be 'white', then it means that noise samples are uncorrelated from each other plus noise possess spectrum broader than signal-spectrum. $\endgroup$
    – kaka
    Dec 19, 2014 at 1:40
  • $\begingroup$ @kaka Just adding a bit of precision: whiteness means that the power density spectrum is flat, just like white light's. This means that noise samples are independent and, in most signal processing scenarios, this also imples they are uncorrelated. The bandwidth is at least equal to the signal's. $\endgroup$
    – MBaz
    Dec 19, 2014 at 2:04

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.