PI gains are fixed at K_c = 0.3 and T_i = 5 s. Simulation 0–30 s with dt = 0.05 s. When T_1 ≈ T_2 a tiny perturbation is added for numerical stability.
Red = response with RHP zero / gray dashed = plain first-order K(1−e^(−t/T1)) / yellow dots = reversal time t* and negative peak y_min.
Blue = closed-loop output y_cl(t) / gray dashed = setpoint r = 1. PI gains fixed at K_c=0.3, T_i=5 s. Note the transient dip opposite to the target.
Transfer function of a second-order process with an RHP zero:
$$G(s) = \frac{K\,(-T_z\,s + 1)}{(T_1\,s + 1)(T_2\,s + 1)},\quad T_z>0$$Analytic step response (for T_1 ≠ T_2):
$$y(t) = K + \frac{K}{T_2 - T_1}\left[(T_1 + T_z)\,e^{-t/T_1} - (T_2 + T_z)\,e^{-t/T_2}\right]$$Initial slope at t = 0⁺ is negative:
$$\left.\frac{dy}{dt}\right|_{t=0^+} = -\frac{K\,T_z}{T_1\,T_2} \lt 0$$Empirical upper bound on the closed-loop bandwidth set by the RHP zero:
$$\omega_{BW} \lesssim \frac{1}{2\,T_z}$$A larger $T_z$ makes the initial slope steeper and the reversal deeper; a smaller $T_z$ leaves only a shallow dip, and the response approaches a plain first-order curve.