Proof of the Kramers-Kronig relation for amplitude and phase

Question

I learned that the the amplitude and phase of a transfer function of a minimum phase system are related by the Hilbert transform, specifically,

$$\arg(H(\Omega))=-\frac{1}{\pi} \mathrm{PV}\int_{-\infty}^{\infty}\frac{\ln|H(\omega)|}{\Omega-\omega}d\omega $$

where $\mathrm{PV}$ denotes principal value. A proof I found will be provided at the end of this post. My question concerns the other direction: We can also obtain the amplitude from the phase by Hilbert transform, up to a constant multiplicative factor $c$, i.e.,

$$ \mathcal{H} \{\arg(H(\omega))\}=\ln{|cH(\omega)|} $$

However, I am unable to find a rigorous proof. Does anyone know how to prove this? Thanks.

Appendix: Proof of the forward direction (amp-to-phase). Referring to the contour below. By minimum phase, the transfer function $H$ has no zero at the lower half $\omega$-plane, and thus $\ln{H(\omega)}$ is analytic there. Then by Cauchy integral theorem, we have $$\int_{C_P+C_R+C_0}\frac{\ln{H(\omega)}}{\omega^2-\Omega^2}d\omega=0$$

Integration along $C_0$ vanishes as the radius goes to infinity, because the denominator is second order. (It is not the case if the denominator is only first order, as in the standard proof of the real-imag form of KK relation, because the log function goes to $\infty$ in the lower half $\omega$-plane.) Hence we have

$$\mathrm{PV}\int_{-\infty}^{\infty}\frac{\ln{H(\omega)}}{\omega^2-\Omega^2}d\omega=-\int_{C_R}\frac{\ln{H(\omega)}}{\omega^2-\Omega^2}d\omega\\=-i\pi\{\mathrm{res}_{\Omega}[\frac{\ln{H(\omega)}}{\omega^2-\Omega^2}]+\mathrm{res}_{-\Omega}[\frac{\ln{H(\omega)}}{\omega^2-\Omega^2}]\}\\=-i\pi\{\frac{\ln{H(\Omega)}}{2\Omega}+\frac{\ln{H(-\Omega)}}{-2\Omega}\}=\frac{\pi}{\Omega}\arg(H(\Omega))$$

Since the RHS is real, we only have to consider the real part of the LHS. By writing $\frac{1}{\omega^2-\Omega^2}=\frac{1}{2\Omega}[\frac{1}{\omega-\Omega}-\frac{1}{\omega+\Omega}]$, and noting that $|H(\omega)|$ is an even function of $\omega$, we have

$$\frac{\pi}{\Omega}\arg(H(\Omega))=\frac{1}{2\Omega}\mathrm{PV}\int_{-\infty}^{\infty}\frac{\ln{|H(\omega)|}}{\omega-\Omega}-\frac{\ln{|H(\omega)|}}{\omega+\Omega}d\omega\\ =-\frac{1}{\Omega}\mathrm{PV}\int_{-\infty}^{\infty}\frac{\ln{|H(\omega)|}}{\Omega-\omega}d\omega$$

as was to be shown.

Answer 1

Take a look at this nice paper John Bechhoefer , "Kramers–Kronig, Bode, and the meaning of zero", American Journal of Physics 79, 1053-1059 (2011) https://doi.org/10.1119/1.3614039

https://arxiv.org/pdf/1107.0071.pdf

You start by taking the logarithm of the transfer function

$\log \left(H \left( \omega\right)\right) = \log \left| H \left(\omega\right)\right| + i \arg H \left(\omega\right)$

The transfer function is analytic (no poles) in the upper half plane because of causality. If you furthermore suppose it has no zeroes, $\log H(\omega)$ will be analytic in the upper-half plane as well. This last hypothesis, as explained in the paper, is not always verified. If it holds, you can apply K-K relations between real and imaginary part of $\log\left(\right)$ and you obtain Bode relations, that is the equations you were looking for.