# How to make colored noise white?

I am studying linear minimum mean squared error (MMSE) receivers. The technique used in it suggests to make colored noise white by multiplying it with the invertible covariance matrix $K_z^{-\frac{1}{2}}$, with $K$ given by

$$K_{zk} = N_0 I_nr +\sum_{i≠k}^{n_t}P_ih_ih_i^*$$

where $h_i$ and $h_i^*$ are the channel gain and channel gain conjugate transpose, respectively, and $n_t$ is the number of transmit antennas. The colored noise is because of interference from other users (I guess so!).

I am confused as how the multiplication of $y = hx + z$ with $K_z^{-\frac{1}{2}}$ can transform colored noise to white though I can guess that may be multiplying with inverse cancel out the colored part and the noise is white now. Am I correct?

• It's more likely to be the case that you're observing colored noise because of some of the characteristics of your physical device or channel – often, some of the noise power goes through a frequency-selective system and gets "filtered" to be colored. – Marcus Müller Oct 17 '16 at 9:08

Given the linear model $$y = hx + z$$ where $z \sim \mathcal{N}(0,K_z)$. We say that noise is colored when $K_z \neq \Sigma$, a diagonal matrix. To uncolor the noise, you can pre-multiply $y$ by $K_z^{-\frac{1}{2}}$, so $$K_z^{-\frac{1}{2}}y = K_z^{-\frac{1}{2}}hx +K_z^{-\frac{1}{2}}z = K_z^{-\frac{1}{2}}hx +\tilde{z}$$ Notice that the noise is now $\tilde{z} = K_z^{-\frac{1}{2}}z$, which is uncolored since $$E(\tilde{z} \tilde{z} ^T) = K_z^{-\frac{1}{2}}E({z} {z} ^T)K_z^{-\frac{1}{2}} = K_z^{-\frac{1}{2}}K_zK_z^{-\frac{1}{2}} = K_z^{-\frac{1}{2}}K_z^{+\frac{1}{2}}K_z^{+\frac{1}{2}}K_z^{-\frac{1}{2}} = I$$ So $\tilde{z} \sim \mathcal{N}(0,I)$

• Can you please explain how colored noise is non diagonal‚ I mean what I know is filtered white noise is colored but how is that noise has correlation between samples???? Further for wireless channel does this mean that noise of one stream is linked with another stream? – Userhanu Oct 18 '16 at 10:47
• This is the definition of colored noise, to the best of my knowledge. And yes filtered white noise is colored. Say you have white noise $n$ with covariance $I$, and you filter by $F^{\frac{1}{2}}$, the new noise is $F^{\frac{1}{2}}n$, with covariance $F$. As for wireless channels, you mean $y$ is a vector of streams ? – Ahmad Bazzi Oct 18 '16 at 11:38
• Yes y is a vector for streams we are receiving where x is the vector we are transmitting, I want to know why noise is getting colored here why not AWGN ??? If you can help! – Userhanu Oct 18 '16 at 11:45
• I do not really know what you're talking about, unless you can refer me to something. Probably, it's because of some filtering going on. – Ahmad Bazzi Oct 18 '16 at 12:12
• I mean to say that is that interference from other users that is making this noise colored from white, $$K_{zk} = N_0 I_nr +\sum_{i≠k}^{n_t}P_ih_ih_i^*$$ The Pi part in this expression is making this Identity matrix of noise colored. Am I right???? – Userhanu Oct 18 '16 at 13:02

The other answer is very good and mathy-- I will atempt a less-mathematical explanation in case that is helpful.

The idea behind colored noise (in the time-domain) is that there is correlation between time-samples. In the case of a random vector, it means that the covariance matrix is no longer diagonal (white noise has a diagonal covariance matrix). In the case of a random process, colored noise means that the autocovariance function is not a unit impulse, implying that samples are correlated with each other.

To "whiten" the noise, premultiplying the sample vector by the appropriate inverse scaling "undoes" the correlation and yields a diagonal covariance matrix.

• Can you please explain how colored noise is non diagonal‚ I mean what I know is filtered white noise is colored but how is that noise has correlation between samples???? Further for wireless channel does this mean that noise of one stream is linked with another stream? – Userhanu Oct 18 '16 at 11:29
• @Ankit The elements of the correlation matrix are the correlation between different time indices. The main diagonal elements represent the autocorrelation values i.e. the correlation with zero delay. If a signal is white then the non-diagonal elements will be zero - therefore of a signal is correlated (non-white) then the off diagonal elements will become non-zero. – David Oct 18 '16 at 12:14