P13.4. A closed loop system has a zero-order hold and process. 2 (a) Determine G(Z) with T=1s and πΊπ π = π +2 (b) Let r(t) be a unit step input and calculate the response of the system by synthetic division Answer: (a) πΊ π = π΅ 1−π −π π 2 π π +2 = 1 − π −1 π΅ 1 π 1 − π +2 = 1 − π −1 (b) The Z-transform of unit step input is π π = π π = 0.865 π π π−1 π 0.865 − π−π −2π = π−0.135 π π−1 πΊ(π) 0.865 = 1 + πΊ(π) π + 0.37 0.865π So π π = π π π π = π+0.37 × π−1 = π 2 −0.27π−0.73 π π = 0.865π −1 + 0.234π −2 +βββ= ∞ π=0 π¦ ππ π −π ⇒ π 0 = 0, π π = 0.865, π 2π = 0.234 P13.10 A unit feedback loop system has a zero-order hold and a process πΊπ π = 1 , π (π +10) the sampling period is T=0.1s (a) (b) (c) (d) (e) (f) Let D(z)=K and determine the transfer function π· π πΊ(π) Determine the characteristic equation of the closed loop system Calculate the maximum value of K for a stable system Determine K such that the overshoot is less than 30% Based on the K got from (d), determine the closed loop transfer function Determine the location of the closed loop roots and the overshoot if K is one-half of the value determined in (c) Answer: 1 − π −π π 1 π΄ π΅ πΆ × = 1 − π −1 π΅ 2 + + π π π + 10 π π π + 10 1 1 1 1 1 1 = 1 − π −1 π΅ × − + × 10 π 2 100 π 100 π + 10 1 10ππ π π = 1 − π −1 − + 2 100 (π − 1) π − 1 π − π −10π πΊ π = πΊπ πΊπ π = π΅ (a) With T=0.1s 1 π π π 1 − π −1 − + 2 100 (π − 1) π − 1 π − π −1 −1 −2 1 π π − 2π + 1 0.0037π + 0.0026 =πΎ× × 2 = πΎ × 100 π − 1 − π −1 π + π −1 π 2 − 1.368π + 0.368 π· π πΊ π =πΎ× (b) The characteristic equation of the closed loop system is 1+π· π πΊ π =0 (c) Based on the characteristic equation, we have π π = π 2 − 1.368π + 0.368 + πΎ0.0037π + 0.0026πΎ = 0 Because the q(Z) is quadratic and has real coefficients, the necessary and sufficient conditions for q(Z) to have all roots within the unit circle are π(0) < 1, π(1) > 0, πππ π(−1) > 0 ⇒ πΎ < 240 (d) πΎ π (π +10) = πΎ/10 π (0.1π +1) ⇒ π = 0.1, π π = 1, πππ ππ ππ π‘βπ πΉπππ’ππ 13.19 We have K=75 (e) The closed loop transfer function is π π = 0.2759π+0.1982 π 2 −1.092π+0.5661 (f) K=119.5, the roots are π§ = 0.4641 ± 0.6843π. Based on the Figure 13.19, we have overshoot is 0.55 10 AP13.4 A closed loop system has a zero-order hold and a process πΊπ π = π +1. Determine the range of sampling period T for which the system is stable. Answer: πΊ = πΊπ πΊπ π = 10(1 − π −π π ) π (π + 1) 10 1 − π −π π 10 10 πΊ π =π΅ = 1 − π −1 π΅ − = 10 1 − π −1 π π +1 π π +1 10(1 − π −π ) = π − π −π The closed loop transfer function is π π − π − 1 π − π −π π π = For stability, πΊ(π) 10 − 10π −π 10 − 10π −π = = 1 + πΊ(π) π − π −π + 10 − 10π −π π − (11π −π − 10) 11π −π − 10 < 1 ⇒ 0 < π < 0.2 πΎ E13.14 A unity feedback system has a process πΊπ π = π (π +3) , with T=0.5 (a) Determine whether the system is stable when K=5 (b) Determine the maximum value of K for stable Answer: πΎ 1 − π −π π 1/3 1/9 1/9 = 1 − π −1 π΅ 2 − + 2 π π +3 π π π +3 1 3ππ π π =πΎ× ×[ − + ] 9 (π − 1)2 π − 1 π − π −3π πΊ π = πΊπ πΊπ π = π΅ The characteristic equation of the closed loop transfer function is π π =1+πΊ π =0 πΎ π π = π − 1 π − π −3π + 9 [3π π − π −3π − π − 1 π − π −3π + (π − 1)2 ] Because π(0) < 1, π(1) > 0, πππ π(−1) > 0, ⇒ 0 < πΎ < 6.32 So, when K =5, the system is stable.