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If a system has a zero somewhere along the imaginary axis of the s-plane (i.e. it's not on the right half of the plane), does this still make the system nonminimum phase ? Or does the zero have to be strictly on the right half side ?

Also does having a zero on the imaginary axis (especially at the origin) has a significant physical meaning ?

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    [Did you prove that](https://en.wikipedia.org/wiki/BIBO_stability#Continuous-time_signals) if a Laplace transform $H(s) = \int_0^\infty h(t) e^{-st}dt$ of a function $h(t)$ converges for $Re(s)$ large enough and is a rational function, that is $H(s) = C\frac{\prod_{m=1}^M (s-b_m)}{\prod_{m=1}^N (s-a_m)}$ then (stability) $\int_0^\infty |h(t)|dt < \infty$ if and only if $Re(a_m) < 0$ ? – 2017-01-30
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    The purely imaginary zeros represent a marginal case and hence cannot be identified as (non-)minimum phase. Regarding your second question: each zero decreases the system's output degree. However, this holds true for all kinds of zeros. I cannot figure out any other physical meaning, but would be interested if somebody can come up with an idea. – 2017-02-01

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The definition as stated on the Wikipedia page of Minimum phase reads as follows:

In control theory and signal processing, a linear, time-invariant system is said to be minimum-phase if the system and its inverse are causal and stable.

A causal system means that the system doesn't rely on future inputs for the current state of the system (more zeros than poles to put it really easily for now). And (BIBO) stable meaning that the states are bounded or $\forall t:|x(t)|<\infty$. Note that this doesn't require that the states to converge to zero (which should be the case if you're requiring a-symptotic stability).

Having a system with for example the transfer function $H(s)=\frac{s^2+10}{s^2+1}$ satisfies both these criteria, as well as that the inverse satisfies these criteria, meaning a minimum phase system, but it's easy to see that in both cases the poles of $H(s)$ and $H_{inv}(s)$ have poles on the imaginary axis.

(in case of $H(s)$, $p_{1,2}=+/- i$, and in case of $H_{inv}(s)$, $p_{1,2}=+/- i\sqrt{10}$.)

The physical meaning of this is that the system oscilates around a specific point with a certain frequency.

Always try to think of solutions of systems in exponential powers (since systems are written as differential equations, and the solutions to differential equations normally are in the form of exponential powers, in state-space description $\dot{x}(t)=Ax$ has as a solution $x(t) = e^{At}$). This is also why you want your poles to be in the left half plane, such that you can have solutions of the form $x(t) = e^{p t}$, where if $p<0$ the exponent converges to zero, and if $p$ is complex/purely imaginary you can use Euler's formula to rewrite the solution as sines and cosines.

Which is the case if you have purely imaginary poles you get solutions of the form $e^{ict} = c (\cos(t)+i \sin(t)$.

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    Not that the definitions of causality and (BIBO) stability are by no means formal definitions, but quick definitions to get the point across ;) – 2017-02-26
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    If you're familiar with mass-spring-damper systems you can see the meaning of an imaginary pole as the absence of the damper, so the mass will always keep oscilating due to the spring. – 2017-02-26
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    Your example is not BIBO stable. The output is bounded for unit step input but not for $cos(t)$ for example. See http://www.wolframalpha.com/input/?i=transfer+function+(s%5E2%2B10)%2F(s%5E2%2B1)+response+to+cos(t) – 2017-03-01