The power spectral density (PSD) of an AR(p) model excited by a white Gaussian noise input $u(t)$ of variance $\sigma_u^2$ is $P_{xx}(f:\theta) = \frac{\sigma_u^2}{|A(f)|^2}$

where $\theta = [a[1]a[2]...a[p]\sigma_u^2]^T$ and $A(f) = 1+ \sum_{m=1}^p a[m]\exp{(-j2\pi fm)}$. Can somebody please show the steps on how to get the partial derivative of $\frac{\partial \ln P_{xx}(f;\theta)}{\partial a[k]}$

  • $\begingroup$ You're missing the $a[m]$ terms in front of the exponential terms. At first, I thought it was your edit. Sorry. $\endgroup$ – Gilles Nov 13 '14 at 20:20
  • $\begingroup$ @Gilles: Sorry for the typo, I have included $a[m]$ terms. $\endgroup$ – SKM Nov 13 '14 at 20:33

\begin{align} \displaystyle \frac{\partial \ln P_{xx}(f;\theta)}{\partial a[k]} = &\frac{\partial \left[ \ln \left(\displaystyle\frac{\sigma_u^2}{\left|A(f)\right|^2}\right)\right]}{\partial a[k]}\\ =&\displaystyle\frac{\partial\left[\ln\left(\sigma_u^2\right) - \ln\left|A(f)\right|^2\right]}{\partial a\left[k\right]}\\ =&-\displaystyle \frac{\partial\ln\left|A(f)\right|^2}{\partial a\left[k\right]}\\ =&-\displaystyle \frac{1}{\left|A(f)\right|^2}\cdot\frac{\partial \left|A(f)\right|^2}{\partial a\left[k\right]}\\ =&-\displaystyle \frac{1}{\left|A(f)\right|^2}\cdot\frac{\partial\left[A(f)A^*(f)\right]}{\partial a\left[k\right]}\\ =&-\displaystyle \frac{A^*(f)\exp(-j2\pi fk) + A(f)\exp(j2\pi fk)}{\left|A(f)\right|^2}\\ \end{align}


Some clarity before the "final" step:

$\displaystyle \frac{\partial\left[A(f)A^*(f)\right]}{\partial a\left[k\right]} = \frac{\partial A(f)}{\partial a\left[k\right]}\cdot A^*(f) + \frac{\partial A^*(f)}{\partial a\left[k\right]}\cdot A(f)$

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  • $\begingroup$ Thank you for your reply. I am weak in calculus and unable to understand how the derivative of $A(f) = A(f)\exp(-j2\pi fk)$. Can you please shed some light or resource as this was the part that I could not solve. $\endgroup$ – SKM Nov 13 '14 at 19:54
  • 1
    $\begingroup$ No. It's the partial derivative of the product $A(f)A^*(f)$ and not of $A(f)$ that gives the numerator above. Remembering the product rule for derivatives: $[f(x)g(x)]^{'} = f^{'}(x)g(x) + f(x)g^{'}(x)$, keeping that in mind and noting that all partial derivatives w.r.t $a[k]$ are zero except the $k^{\textrm{th}}$ terms. So you only get the exponential $k^{\textrm{th}}$ terms that you use in the product rule. $\endgroup$ – Gilles Nov 13 '14 at 20:10
  • $\begingroup$ Right, so the partial derivative of a[k] becomes A(f)? And does the $*$ mean conjugate? $\endgroup$ – SKM Nov 13 '14 at 20:25
  • $\begingroup$ Is $*$ the Conjugate transpose? en.wikipedia.org/wiki/Conjugate_transpose $\endgroup$ – SKM Nov 13 '14 at 20:35
  • $\begingroup$ Yes, it's for the conjugate. No, with the partial derivatives you only remain with the exponential terms which you then multiply with the remaining terms in the product(see my edit). I have edited the answer, I hope this helps. $\endgroup$ – Gilles Nov 13 '14 at 20:42

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