# Proof of the Kramers-Kronig relation for amplitude and phase

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.

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.

• I've read that paper, and unfortunately, that doesn't work. The integration at infinity won't vanish when we take the log(). That is why in the proof of the forward direction we need to use second order in the denominator. Commented Jan 16, 2023 at 8:46
• Why wouldn't it vanish? Commented Jan 16, 2023 at 13:45
• A strictly proper H will go to zero at infinity. When taking log(H), it goes to infinity at infinity. Commented Jan 18, 2023 at 1:45
• What matters here is the function $\log(H(\omega) )/(\omega-\omega ')$. This one goes to $0$ at infinity, so you can apply Cauchy theorem Commented Jan 18, 2023 at 14:02
• That function is the integrand, and will be multiplied with the $d\omega$. Since the integrand goes to zero at the rate of $1/\omega$, you cannot justify the convergence of the integral at infinity. If the integrand has second order denominator, then that would be fine. Commented Jan 19, 2023 at 3:18