1
$\begingroup$

The autocorrelation function of AWGN is an impulse response at t=0, and its fourier transform or its spectral power density has a flat infinite BW

Now any receiver will have bandpass or low pass filters that will limit the noise BW

Does that mean that in reality we are actually dealing with "colored" noise and considering noise to be "white" within the filter BW is an assumption made for convenience purpose?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Yes.

However, if you are dealing with a discrete-time signal, if your A/D is proceeded by an ideal low-pass filter with cutoff frequency $F_s/2$, with unity passband gain, the output power spectral density before sampling is: $$ S_y(f) = |H(f)|^2 S_x(f) = \frac{N_0}{2} \operatorname{rect}(f/F_s) $$ This means that the autocorrelation of the filtered noise will be: $$ R_y(\tau) = \mathscr F^{-1}\{S_y(f)\} = \frac{N_0}{2T_s}\operatorname{sinc}(\tau/T_s) $$ where $T_s = 1/F_s$. If the noise is zero-mean, Gaussian, and sampled at time instants $t_n = n T_s$, then each sample will be uncorrelated, and since they are Gaussian, will also be independent. In this case, in the digital domain, the power spectrum will be flat for all frequencies.

This is not the only way to generate noise that is white in the discrete-time domain, but is one way, albeit a physically unrealizable one.

Improve this answer
$\endgroup$
4
  • $\begingroup$ i hope it was okay for me to tweek your answer a little Carlos. (it's a very good answer.) i'll just add that since (theoretically) white noise has non-zero power spectral density (of height $\frac{N_0}{2}$), then it must have infinite power (which is the area underneath). so, even in an analog setting, true white noise is an impossibility. but for the output of an random number generator (samples of finite variance that are totally uncorrelated to each other) the bandlimited model above is correct. the pre-sampled bandlimited power spectrum has to be that. $\endgroup$
    – robert bristow-johnson
    Nov 24, 2018 at 2:12
  • $\begingroup$ so, assuming no DC (that is that the mean of the random numbers coming out of the RNG is zero) and if the variance is $\sigma^2$ which, if this is a uniform RNG then it's $\sigma^2 = \frac{\Delta^2}{12}$, then that variance, which is mean power, must be the same as $\frac{N_0}{2} \times 2 \times \frac{F_s}{2}$. i believe i got this right. $\endgroup$
    – robert bristow-johnson
    Nov 24, 2018 at 2:17
  • $\begingroup$ So why talk about a whitening filter when it comes to an MMSE equalizer? $\endgroup$
    – Hatem Tawfik
    Nov 24, 2018 at 2:31
  • $\begingroup$ whiten it over the bandwidth of interest. but white noise of unlimited bandwidth has infinite power. we can only really deal with white noise over a finite bandwidth. $\endgroup$
    – robert bristow-johnson
    Nov 24, 2018 at 3:27

Your Answer

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

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