We know that the lag window estimator of the spectral density is defined as: $$S^{(lw)}(f)=\int_{-f_{(N)}}^{f_{(N)}}W_m(f-\phi)\hat{S}^{p}(\phi)d\phi=\sum\limits_{\tau=-(N-1)}^{\tau=N-1}w_{\tau,m}s^{(p)}_{\tau}e^{-2\pi f\tau}$$ Where $W_m(f)$ (the spectral window) is the Fourier transform of $w_{\tau,m}$( the lag window), and $\hat{S}^{p}$ (The periodogram ) is the Fourier transform of $s^{(p)}_{\tau}$ (the covariance function).

The covariance between $S^{(lw)}(f)$ and $S^{(lw)}(f')$ is approximatly $0$ if $|f-f'|$ is larger than the width (or bandwidth) of the lag window $W_m$. But if we write the following: $$cov\{S^{(lw)}(f),S^{(lw)}(f')\}=E\{S^{(lw)}(f)S^{(lw)}(f')\}- S(f)S(f')$$ $S(f)$ being the true spectral density at frequency $f$. We can write then: $$E\{S^{(lw)}(f)S^{(lw)}(f')\} =\frac{1}{N}\sum\limits_ {f,f'}W_m(f)W_m(f')E\{\hat{S}^{(p)}(f)\hat{S}^{(p)}(f')\}$$ $$=\frac{1}{N}\sum\limits_ {f,f'}W_m(f)W_m(f')E\{\hat{S}^{(p)}(f)\}E\{\hat{S}^{(p)}(f')\}=S(f)S(f')$$ because we know that the periodogram ordinates are uncorrelated for $|f-f'|> \frac{1}{N}$. From what I have just written the ordinates of the lag window estimator are uncorrelated onlu if the peridodogram ordinates are uncorrelated ($|f-f'|> \frac{1}{N}$), which is inconsistent with the result that we already know. Where am I mistaken?

  • $\begingroup$ Hi Tony: I'm not clear on how the last line is obtained but I think you are assuming that the lag window estimator is an unbiased estimator of the true spectral density which might not be an ok assumption. Similarly, the equation two lines up from the last also assumes the same thing, namely that the expected value of the lag window estimator is equal to the true spectral density. I haven't entered the spectral world much yet ( I'm self studying DSP ) but I know that time domain estimates of the spectral density are prone to problems so the unbiasedness assumption may not be valid. $\endgroup$ – mark leeds Jan 18 '18 at 9:09

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