Question

Control Assignment 2017/18 In order to complete this assignment you will have to consider a system which is unique to you. Th parameters of the system are derived from your student mumber and so if you use the wrong mumber you will get the wrong answers and will lose marks. Use the last two digits of your student mumber where Ki-last but_one_digit+I and K-last digit+1. Thus the mumber 201656789 would give K9 and K 10. When you submit your report, make sure it is clear what your student mumber is and what KI and K2 are. 1. A system has the transfer fiunction ga(5) -,2KSHK H er function gi (s) = d+2K sakur2. Substitute the values for K, and Substitute the values for Ki and 100 K: from your student number. This would give gi(s) 18 your number is 201656789. Write this transfer function in polynomial form and pole-zero form. Sketch the positions of the poles of the system on the complex plane and sketch the impulse response corresponding to each pole. Derive an expression for the impulse response of the system with zero initial conditions, sketch this response and compare it to the response predicted using MATLAB. s2+18s+181 mark] 3 marks] [4 marks] 4. A second system has the transfer function g2 (5)tKhstKUse MATLAB to plot 2. 3. the step response of this system and use the Zeigler-Nichols Reaction Curve method to determine the approximate parameters for a PID controller. Using MATLAB plot the step response of the system with this PID controller and determine the values of the percentage overshoot, 5% settling time and steady state error. By adjusting the PID parameters, attempt to improve the step response of the system. [4 marks] 5. A third systemgs()(+(CK, +Ky)s+K,k2 is formed into a closed loop system as shown below: R(s) + Sketch the root locus for this system and compare your sketch to the plot generated using MATLAB. 4 marks] Determine the value of gain, k which would lead to marginal stability. Using MATLAB, plot the impulse response of the closed loop system with this value of gain. Comment on this response. 6. 4 marks You should submit a brief report giving your solutions to the above problems. Your report should exceed 6 pages, including figures. Submission deadline 2pm Thursday 7e December 2017

Answer 1

Given data f(t) = 2 we know L(k) = k/s Constant L[f(t)] = L(2) = 2/s b) f(t) = e^-K2t sin (k, t) we know L[e^-at f(b)] f bar (s0 = L[f(k)] Here a = K_2 d[f(t)] = sin(k, t) dL (sin (K, T)] = k_1/s^2 + k^2_1 Applying property (1) therefore L[e^k^2t sin (k, b)] = k_1/(s_1 + k_2)^2 + k_1^2 \r\n f(t) = {0 e^-k, t t < k_2 greaterthanorequalto k_2 By definition of L_0 place transform L[f(t)] = integral^infinity_0 e^-st f(t) dt = integral^k_2_0 e^-3t (0) dt - integral^infinity_K_2 e^3t e^-k_1 t dt {integrated by parts} = 0 + integral^0_K_2 e^-(K_1 + s)^t dt = e[-K_1 + s)t/-k_1 + s]^infinity_k_2 = e^-infinity/-(k + s) - e^-(k_1 + s) k_2 -(k_1 + 1) {e^-infinity = 0 L[f(t) = e^-k_2 (k_1 + s/(k_1 + s) \r\n f(t) = e^k_1 t cos(k_t) l_4(t)