 Solution

For a specified negative-feedback loop with return ratio $L(s)$, let $P$ be its number of open-right-half-plane poles. Let $N_{\mathrm{cw}}$ be the signed clockwise encirclement count of $-1$ by the standard Nyquist stability criterion contour. Provided the contour does not hit a singularity or the critical point, the number of right-half-plane closed-loop poles is $Z=P+N_{\mathrm{cw}}$. Thus closed-loop stability requires $N_{\mathrm{cw}}=-P$ and no closed-loop pole on the imaginary axis, with the usual well-posedness and absence of hidden unstable cancellations. For an already stable return ratio, there must be no encirclement of $-1$ and no passage through it.

The open cavity itself, for $\gamma >0$, has state eigenvalue and transfer-function pole $-\gamma /2$. Its initial transient decays and its impulse response is a direct-feedthrough term plus a decaying exponential. Hence the open-loop cavity is asymptotically stable and its transfer function is BIBO stable. The right-half-plane zero at $+\gamma /2$ does not make the open cavity unstable. On the frequency axis,Its unit-circle plot passes through $-1$ at $\omega =0$. A Nyquist stability criterion is a feedback criterion; this passage is not by itself a test of open-plant stability.

If the intended interpretation is a unit negative-feedback loop with $L=G$, thenThat loop has a pole at zero, so unit negative feedback is marginal rather than asymptotically stable, and the critical-point passage correctly prevents a strict Nyquist stability conclusion. Without a specified feedback interconnection, the unambiguous conclusion is the stable open cavity, not that particular closed loop.