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3

Like you mentionned, you cannot cancel a right-half-plane zero (or a zero outside the unit circle) by placing a pole on it. A unstable pole in your compensator will make the command of your controller unbounded (i.e. it will reach infinity). There are no ways to cancel a right-half-plane zero. It's sometimes possible to "remove" a right-half-plane zero by ...

1

Marginally stable means that an otherwise stable system has one or more simple poles on the unit circle (in discrete time), or on the imaginary axis (in continuous time). The consequence of that is that transients don't decay, but they also don't grow without bounds. Marginally stable systems are unstable in the bounded-input bounded-output (BIBO) sense.

0

A real continuous time system has more restrictions than a discrete time system. The reason is that a first order continuous system of the form $\dot{x} = f(x)$ cannot pass to the 'other side' of an equilibrium. A discrete time system has no such limitations. It has nothing to do with the poles of the system, which need not be linear. The discrete time ...

5

For continuous-time systems, a pole at location $s_0=\sigma_0+j\omega_0$ will create a time-domain contribution of the form $$e^{s_0t}=e^{\sigma_0t}e^{j\omega_0t}\tag{1}$$ which is a damped oscillation if the pole is in the left half-plane (i.e., $\sigma_0<0$), and if the pole is not on the real axis (i.e., $\omega_0\neq 0)$. For $\omega_0=0$ there is ...

1

In the first case you have a pole at the location (-0.7). In the second case, your pole is at 0.7. Having a pole at -0.7 means that the natural frequency of your system is fs/2, that's why you have an oscillation at fs/2. Since the pole is stable, i.e. inside the unit circle, the oscillation eventually dies down. Edit : You happen to have a pole at fs/2, ...

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Note that the imaginary axis is not inside the ROC because $|H(j\omega)|\to\infty$ for $\omega\to\infty$. So the system is definitely unstable, regardless of causality, which confirms your intuition. Concerning causality, the ROC is neither a right half-plane nor a left half-plane (both because of the pole at $s\to\infty$), which is consistent with a (...

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