# Validity of an argument that two transfer functions are minimum-phase based on their ratio being minimum-phase

Update I think the essence of my question below is this: If the ratio of two transfer functions may be represented exactly as a minimum-phase filter (MPF) plus a pure delay (in the title, I simply refer to this representation as “minimum-phase” for brevity, even though it’s not accurate), does that imply that each transfer function in the ratio must also be represented exactly as an MPF plus a pure delay?

I was reading a paper where an argument is made in the Appendix regarding the validity of representing a transfer function as a minimum-phase filter (MPF) plus a pure delay. In particular, the goal is to show that two transfer functions $$H_L$$ and $$H_R$$ can each be represented as a minimum-phase filter (MPF) plus a pure delay. That is, if we have $$H_L = |H_L| \text{e}^{\text{i}\phi_L}$$ and $$H_R = |H_R| \text{e}^{\text{i}\phi_R}$$, where $$\phi_L$$ and $$\phi_R$$ are their respective phase responses, then given that these are rational transfer functions that can be decomposed into a cascade of minimum-phase and all-pass transfer functions, we must show that $$\phi_L = \phi_L^{\text{mp}} + \omega\,t_L$$ and $$\phi_R = \phi_R^{\text{mp}} + \omega\,t_R$$, for some constants $$t_L$$ and $$t_R$$ (i.e., the all-pass phase response is purely linear), and where the superscript "mp" stands for minimum-phase.

As I understand it, the argument by the authors of the paper goes as follows:

Give $$H_L$$ and $$H_R$$ as above, we have $$\ln \left(\frac{H_R}{H_L}\right) = \underbrace{\left(n_R-n_L\right)}_{\Delta n} + \text{i}\underbrace{\left(\phi_R-\phi_L\right)}_{\Delta \phi},$$ where $$n_R = \ln |H_R|$$ and $$n_L = \ln |H_L|$$. Since, in general, we have $$\phi_L = \phi_L^{\text{mp}} + \phi_L^{\text{ap}}$$ and $$\phi_R = \phi_R^{\text{mp}} + \phi_R^{\text{ap}}$$, where the superscripts "mp" and "ap" stand for "minimum-phase" and "all-pass", respectively, this means $$\Delta \phi = \Delta \phi^{\text{mp}} + \Delta \phi^{\text{ap}} = \mathcal{H}\left(\Delta n\right) + \Delta \phi^{\text{ap}}$$, where $$\mathcal{H}$$ denotes the Hilbert transform operator. The authors show that this may be expressed as $$\Delta \phi = \mathcal{H}\left(\Delta n\right) + \omega t_0$$ for a constant, $$t_0$$. Let us assume that this equation (which corresponds to Eq. (A8) in their paper) is valid. The authors then go on to suggest (incorrectly in my opinion) that since $$\Delta \phi^{\text{ap}} = \omega t_0$$, $$H_L$$ and $$H_R$$ must each always be representable exactly as an MPF plus pure delay (when this is, in fact, not known).

I think this argument is flawed because it assumes that $$\Delta \phi^{\text{ap}} = \phi_R^{\text{ap}}-\phi_L^{\text{ap}} = \omega t_0$$ implies that $$\phi_R^{\text{ap}}$$ and $$\phi_L^{\text{ap}}$$ are also each linear functions of $$\omega$$ (since only then do $$H_L$$ and $$H_R$$ each have MPF plus delay representations).

However, this is not the only possibility correct? I think a function like $$\omega t_0$$, which is linear in $$\omega$$ can, in general, be obtained as the difference of two non-linear functions of $$\omega$$, each corresponding to either $$\phi_R^{\text{ap}}$$ or $$\phi_L^{\text{ap}}$$, which would mean that an MPF plus delay representation for $$H_L$$ and $$H_R$$ as described earlier is not always necessary, right? For example, if $$\phi_R^{\text{ap}} = \alpha \omega^2$$ and $$\phi_L^{\text{ap}} = \alpha \omega^2 + t_0 \omega$$, for some scalar $$\alpha$$, then $$\Delta \phi^{\text{ap}} = \omega t_0$$. Am I going wrong somewhere?

Thanks in advance for any inputs!

$$H(z) = \frac{p-1/z}{1-p/z} = H_{MP}(z) \cdot H_{AP}(z)$$ which has a pole at $$p$$ and a zero at $$1/p$$. The "minimum phase" part, $$H_{MP}(z)$$ is simply unity, i.e. $$H_{MP}(z)=1$$ and the all pass part is just the original all pass. The all pass phase response is NOT that a of a pure delay, we clearly have $$\Phi_{MP}(\omega) \neq \omega \cdot t_0$$