Question

PC Lab 7 Problem: See Sections: 13.3, 13.4, 13.5 & 13.6 for examples The block diagram of a sampled unity-feedback control system is shown below. R(s) – e7 2 TY() + 1 1) a) Calculate the z-transform of the closed-loop response, H(z) = Y(z)/R(z), of the above system as a function of sampling period, T. b) Calculate the value of the pole(s) and zero(s) of H(z) when T = 0.3 seconds. 2) Using the result of part (1), or otherwise, determine the range of sampling period, T, for which the closed-loop system shown above is stable. 3) Using the result of part (1), or otherwise, derive a difference equation for the closed loop system relating the sampled output, yk, to the sampled input, rk, when T = 0.3 seconds. a) Using the result of part (3), or otherwise, calculate the first 5 values of the output of the closed-loop system following a unit-step input, rk, with T = 0.3 seconds. b) Is the closed loop system with T = 0.3 seconds underdamped, overdamped, or critically damped? 5) Use MATLAB to check all answers

Answer 1

D a) Gp (2) - Gpo C1-es) a) GpCZ) - Z- pat dinit Gps) = 2 2. so sto where a= 1 4 Gp(o)= Limit 5- 0 :: 6p (7) 261- ēt) z - ě Y2 GACHI 201-e PUH teipzoti Z-e'+2-2e Le +2-32 b) H = 261-803) 7 +2- 3e -0.3 0.5184 -0.2225 T=0=3 one is located at 22 anakulis the From above z = -T -243e 3) For sampling time T> 1967,/21 > Then the pole gou out side of unit cirde. The system be comes Unstable for 1>1.099 " For T< 100g the system is stable. HZ) = 0.5184 2! - Y) 1-0.22757 RG =) Y(2) -0.2725 7'y (2) = 0.51843' AES Apply increase 2-transform, we haves ytn] -0.2225 yen -] = 0.5184 xth-1]. = ył n = 0.2225 y [n) 40.5184 x {n-1}.

a) y [o] = 0.2225 YED +0.5184 x(4) = 0 +0 -0. 0.2225 460) + 0-5184 *[0] y [1] = = 0.5184 y [2] = 0.2225 YTJ +0.5184 x[i] = 0.6337 y B] = 0.6594 y [4] = 0.6651 y[5) = 0.6664 4[6) = 0.6667 Y[7) = 06667 46 - b) any ouershoot. overdamped for The response does not experience Therefore the closed loop system is T=0.3.