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There are so many definitions of the minimum phase transfer function, and these are two of them.

  1. The transfer function of the system which has no zeros or poles at right half plane.
  2. The transfer function which has the minimum phase angle range among the systems which has the same magnitude characteristic.

And these two sentences are describing the same thing. So I want to prove that these statements are equivalent. How to prove it?

PS : Suppose that the system is continuous.

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  • $\begingroup$ two things. Matt's answer suffices. a quick way to think of it is that any system with poles in the right half-plane is unstable, so let's leave that out. and any system with poles in the left half-plane and some zeros in the right half-plane, that system can be thought of as a system with all the zeros in the left half-plane in series with a system that is an all-pass filter. the poles of the APF cancel some zeros in the left and reflect those zeros in the right. the APF does not change the magnitude, but makes the phase shift greater. no APF means the means the minimum phase. $\endgroup$ – robert bristow-johnson Nov 26 '18 at 10:00
  • $\begingroup$ the second thing is, there is a property of minimum phase systems that is sometimes used as another definition, that is that the phase, measured in radians, is the negative Hilbert transform of the log magnitude, expressed in nepers. that is that the real part and imaginary part of the natural logarithm of the transfer function are a Hilbert pair. i asked a rigorous question about why that is the case and MattL answered it. $\endgroup$ – robert bristow-johnson Nov 26 '18 at 10:07
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To your second definition it should be added that you only consider causal transfer functions, because it is not difficult to find a smaller phase lag with a non-causal system:

A minimum-phase system is a causal and stable system with a phase lag that is smaller than the phase lag of any other causal and stable system with the same magnitude response.

Note that a real-valued zero in the left half-plane contributes a phase change of $\pi/2$ to the total phase as we move along the frequency axis from $\omega=0$ to $\omega\rightarrow\infty$:

$$0\le\arg\{j\omega+a\}<\frac{\pi}{2},\qquad a>0,\quad\omega\in[0,\infty)\tag{1}$$

A complex conjugate pair of zeros contribute a phase change of $\pi$.

On the other hand, a real-valued zero in the right half-plane contributes a phase change of $-\pi/2$ as $\omega$ moves from zero to infinity:

$$-\frac{\pi}{2}<\arg\{a-j\omega\}\le 0,\qquad a>0,\quad\omega\in[0,\infty)\tag{2}$$

Note that I've chosen the sign of the term $s\pm a$ such that in both cases the phase is zero for $\omega=0$, which is necessary for a fair comparison between the two cases.

Consequently, exchanging a zero in the left half-plane for a zero in the right half-plane (without changing the magnitude of the frequency response) will always result in an additional phase lag of $\pi$ for $\omega\rightarrow\infty$.

As an example, consider two first-order transfer functions:

$$H_1(s)=\frac{s+2}{s+1}\quad\text{and}\quad H_2(s)=\frac{2-s}{s+1}\tag{3}$$

The signs of $H_1(s)$ and $H_2(s)$ were chosen such that their phases are both zero for $s=0$ (and not $\pm\pi$). The figure below shows the phase plots ($\arg\{H_1(j\omega)\}$ in blue, and $\arg\{H_2(j\omega)\}$ in green):

enter image description here

The pole contributes a phase change of $-\pi/2$ as $\omega$ moves from zero to infinity. The left half-plane zero of $H_1(s)$ contributes a phase change of $\pi/2$, resulting in a net phase change of zero, whereas the right half-plane zero of $H_2(s)$ contributes a phase change of $-\pi/2$, resulting in a total phase change of $-\pi$.

Another way to see the same thing is to note that any causal and stable transfer function can be written as the product of the minimum-phase transfer function with the same magnitude and a causal and stable allpass:

$$H(s)=H_m(s)H_a(s)\tag{4}$$

It can be shown that the phase of a causal and stable allpass is always non-positive for $\omega\in[0,\infty)$, and, consequently, the phase lag of the minimum-phase system is always less than or equal to the phase lag of any other causal and stable system with the same magnitude response.

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  • $\begingroup$ I understood about the additional phase lag. I was wondering why $$\sup_{\omega}H_m(j\omega)-\inf_{\omega}H_m(j\omega)>\sup_{\omega}H(j\omega)-\inf_{\omega}H(j\omega) $$, because the phase angle value can be larger than the value when $\omega=0$ and be smaller than the value when $\omega=\infty$. Thanks. $\endgroup$ – Kim Jaewoo Nov 26 '18 at 1:36
  • $\begingroup$ I'm not sure I understand your comment, but your inequality seems wrong to me. It's just the minimum-phase system that has the minimum range of phase. But maybe I don't see what you mean. What is $H_m(j\omega)$? I assume you mean the phase response. $\endgroup$ – Matt L. Nov 26 '18 at 8:05
  • $\begingroup$ I was wrong about my inequality. The inequality I want to know is why $$\sup_{\omega} {\angle H_m(j\omega)}-\inf_{\omega} {\angle H_m(j\omega)}<\sup_{\omega} {\angle H(j\omega)}-\inf_{\omega} {\angle H(j\omega)}$$ $\endgroup$ – Kim Jaewoo Nov 26 '18 at 9:25
  • $\begingroup$ @KimJaewoo: But this is equivalent to saying that the phase lag is minimum for a minimum-phase system, because the maximum phase lag is just the maximum range of the phase assuming that the phase at $\omega=0$ equals zero, which can always be achieved by choosing the sign of $H(\omega)$ accordingly. $\endgroup$ – Matt L. Nov 26 '18 at 9:34

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