Unstable 1st order system (bioreactor): proportional control

Antonio Sala, UPV

Difficulty: ** ,       Relevance: ,      Duration: 18:59

*Enlace a Spanish version

Summary:

The videos [bio1modEN] and [bio1mod2EN] described the basic features of an unstable first-order bioreactor model: open-loop poles and equilibrium points. It was addressed both in ‘internal representation’ (differential equation) $\frac{dx}{dt}=0.4x+u$ and in ‘transfer function’ $x(s)=\frac{1}{s-0.4}u(s)$.

Since the equilibrium is unstable, in order to be used for maintaining a constant biomass, the system must be stabilized. To do this, in the present video, the behavior of a proportional control $u(x)=K_c\cdot (r-x)$ will be analyzed. It is verified that the closed loop poles are $0.4-K_c$, so the closed loop is stable with $K_c>0.4$, and that the proportional control has a position error since the final output value is $x_{eq}=\frac{K_c}{K_c-0.4}\cdot r$, which for $K_c>0.4$ is greater than $r$ (steady-state overshoot).

Detailed derivations are made in both the time domain (differential equations) and the Laplace transform domain (transfer functions). Simulations in the final part of the video confirm the correctness of the theoretical calculations.

The correction of the steady-state error and the response to disturbances will be analyzed in forthcoming materials.

*Link to my [ whole collection] of videos in English. Link to larger [ Colección completa] in Spanish.