Control Pid Ejercicios Resueltos Jun 2026

u(k)=u(k−1)+Kp⋅[e(k)−e(k−1)]+Ki⋅Ts⋅e(k)+KdTs⋅[e(k)−2e(k−1)+e(k−2)]u open paren k close paren equals u open paren k minus 1 close paren plus cap K sub p center dot open bracket e open paren k close paren minus e open paren k minus 1 close paren close bracket plus cap K sub i center dot cap T sub s center dot e open paren k close paren plus the fraction with numerator cap K sub d and denominator cap T sub s end-fraction center dot open bracket e open paren k close paren minus 2 e open paren k minus 1 close paren plus e open paren k minus 2 close paren close bracket 0;1ce1;Donde 0;1a80; es el error actual ( 0;1cdb;) y 0;ee;0;41a; es la señal de control (válvula). Error = Setpoint - NivelActual; 0;58e; AccionP = Kp * Error; Integral = Integral + (Ki * Error * Ts); Derivativo = Kd * (Error - ErrorAnterior) / Ts; 0;979;

[ \lim_s \to 0 G_c(s)G(s) = \lim_s \to 0 \left(4 + \frac2s\right) \cdot \frac1s+2 = \lim_s \to 0 \frac4(s+2) + 2s(s+2) = \lim_s \to 0 \frac4s+8+2s(s+2) = \lim_s \to 0 \frac4s+10s(s+2) ] Este límite tiende a infinito debido al polo en (s=0) del integrador. control pid ejercicios resueltos

Kd=2⋅(0.707)⋅4=5.656cap K sub d equals 2 center dot open paren 0.707 close paren center dot 4 equals 5.656 es el error actual ( 0

u(0) = 2 * 10 cm + 0,2 * ∫10 cmdt + 1 * d(10 cm)/dt = 20 cm + 0 (la integral y la derivada son cero en t=0) AccionP = Kp * Error

Y(s)=T(s)⋅R(s)=10s+20s(s2+13s+20)cap Y open paren s close paren equals cap T open paren s close paren center dot cap R open paren s close paren equals the fraction with numerator 10 s plus 20 and denominator s open paren s squared plus 13 s plus 20 close paren end-fraction