 Solution

Write $x=X/S$, $y=Y/S$, $z=Z/S$, $h=K/S$ and $s=\alpha t$. The repressilator in the stated logic approximation isAssume $0<h<1$. Initially low Z allows X production; high Y blocks Z; and Y is initially at its unrepressed steady level. X first rises through $h$ at $s_1=a=-\ln (1-h)$, turning Y production off. Y then decays through $h$ at $s_2=a+b$, where $b=-\ln h$, allowing Z to rise. Z reaches $h$ at $s_3=2a+b$ and switches X production off. At this point $x_3=1-h(1-h{)}^2$, so X falls through $h$ at $s_4=s_3+\ln (x_3/h)$ and releases Y production.

Y's rise then switches Z off, Z's fall releases X again, and the sequence repeats. Every segment is exponential relaxation towards one if its production gate is on, or towards zero if off. For a rising segment starting at $v_0<h$, the next threshold time is $\ln [(1-v_0)/(1-h)]$; for a falling segment starting at $v_0>h$, it is $\ln (v_0/h)$. These rules determine the whole trace without pretending the concentrations are square waves. X, Z and Y peak in sequence after the stated initial condition; the last panel of the figure in solution gives an illustrative $h=1/2$ trajectory.

The delayed sequence of three inhibitory links supplies negative feedback with finite response times. Equal rates make the cycling symmetric after the initial transient, not simultaneous. If $h\ge 1$, repression is never reliably engaged and the proteins approach their unrepressed levels instead; oscillation is not implied for arbitrary thresholds or for every smooth regulatory law replacing the ideal relay.