I have generated colored noise from the white gaussian noise using auto-regressive model (by solving Yule-walker equation). The covariance matrix that I used to generate the colored noise is [1 0.2;0.2 1].

It shows that variance(power) of white noise and colored noise are same. So if I don't whiten the colored noise, the BER performance of the white noise and colored noise should be same because power of both noise are same (1 in this case).

But the simulation shows that the BER performance of colored noise is nearly 1.4 dB degraded as compared to the white noise. It seems the covariance term (0.2) is causing this BER degradation.

Can anyone please explain why this happened? I used the matched filter as the original pulse shape in both cases. Does this affect the performance?

I am considering my threshold as: if the matched filter output at signal interval >0 then bit 1 is transmitted else bit 0 is transmitted.

  • $\begingroup$ wow! complex problem. However, this really reads like a giant runaway sentence. Could you use empty lines to mark logical paragraphs, so that it's clearer which deductions base on what? Thank you! $\endgroup$ Jun 12, 2018 at 19:24
  • $\begingroup$ Thanks for the response. I have edited the question. Hope the question becomes clear now. $\endgroup$
    – San
    Jun 12, 2018 at 19:38

1 Answer 1


Even though the variance of the noise is the same, the power spectral density of the noise has changed. Thus the fraction of the noise power that makes it through the matched filter has changed. If the variance remains the same but more of the noise power is concentrated outside the bandwidth of the matched filter, then the performance may improve because the noise power within the signal bandwidth has decreased. If the variance remains the same but more of the noise power is concentrated inside the bandwidth of the matched filter, then the BER performance may degrade because the noise power within the signal bandwidth has increased.

When the noise is not white, the filter matched to the pulse shape does not maximize SNR anymore. The Wikipedia article on matched filter addresses this situation and shows the new matched filter to be proportional to $R_v^{-1}s$, where $R_v$ is the noise covariance matrix and $s$ is the signal vector. Using the same notation as the Wikipedia article, if you kept the filter matched to the original pulse shape $s$, your signal component would be $\alpha s^{H}s$, your noise component would be $\alpha s^{H} v$, and the SNR would be

$$\frac{(s^{H}s)^2}{s^{H}R_vs} = \frac{s^Hs}{\bar{s}^HR_v\bar{s}}=\frac{E_s}{\bar{s}^HR_v\bar{s}}$$

Here $E_s$ is the energy per symbol and $\bar{s}=\frac{s}{\sqrt{s^Hs}}$ is the signal normalized to unit energy. This just expresses as an equation what we already concluded: that the SNR will depend on the noise correlation (or power spectral density) with respect to the signal.

If you applied a noise whitening filter, then $R_v$ returns to a scaled identity matrix but you have now introduced intersymbol interference.

  • $\begingroup$ Thanks a lot for a wonderful explanation. It seems that if we denoise our received signal using the prewhitening filter, the filter converts the colored noise into white but also distorts the desired signal. Thus we now correlate with the distorted version of the signal. Thus the conclusion is: even if the noise has been whitened, the BER performance of the resulting matched filter will still be poorer than the original matched filter for white noise process. $\endgroup$
    – San
    Jun 16, 2018 at 17:15
  • $\begingroup$ To be exact, applying a whitening filter does not denoise the signal; rather, it decorelates the noise at the output of the white filter. But yes, whitening the noise introduces distortion in the form of intersymbol interference. If you apply the whitening filter prior to downsampling to 1 sample/symbol, the whitening filter causes the pulse shape to longer be orthogonal: i.e., no longer be a Nyquist pulse. If you apply the whitening filter at 1 sample/symbol, it causes each symbol to contain a linear combination of neighboring symbols. An equalizer can be used in these cases. $\endgroup$ Jun 18, 2018 at 19:28

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