Passive cavity input-output transfer function ($G(s)=(s-\gamma /2)/(s+\gamma /2)$)

For $\dot{a}=-\gamma a/2-\sqrt{\gamma }{b}_0$ and $b_1=\sqrt{\gamma }a+b_0$, zero-initial-data Laplace transforms give the displayed transfer function. For $\gamma >0$ its pole is in the open left half-plane, while $|G(i\omega )|=1$ and its zero lies in the right half-plane. Passing through $-1$ in this plant's frequency plot is not evidence of an unstable open cavity: a Nyquist stability criterion needs an explicitly specified feedback return ratio.

With $b_0=\beta b_{\mathrm{in}}-\alpha b_1$ and $b_2=\alpha b_{\mathrm{in}}+\beta b_1$, real ${\alpha }^2+{\beta }^2=1$ and $|\alpha |<1$, eliminate the internal fields to obtain $\dot{a}=-{\gamma }_{\mathrm{eff}}a/2-c{b}_{\mathrm{in}}$, where $c=\beta \sqrt{\gamma }/(1+\alpha )$ and $c^2={\gamma }_{\mathrm{eff}}$. The internal output transfer is $b_1/b_{\mathrm{in}}=\beta G/(1+\alpha G)$; the external output transfer is $b_2/b_{\mathrm{in}}=(\alpha +G)/(1+\alpha G)=(s-{\gamma }_{\mathrm{eff}}/2)/(s+{\gamma }_{\mathrm{eff}}/2)$. The loop is asymptotically stable in this nondegenerate range. At $\alpha =1$ the cavity mode is undamped and decoupled; at $\alpha =-1$ the instantaneous algebraic feedback connection is singular.