Question

Question #4 (25 points): Consider the open loop system that has the following transfer function 1 G(S) = 10s+ 35 Using Matlab: a) Plot the step response of the open loop system and note the settling time and steady state 15 pts error. b) Add proportional control K 300 and simulate the step response of the closed loop 15 pts system. Note the settling time, %OS and steady state error. c) Add proportional derivate control Kp 300, Ko 10 and simulate the step response of the 15 pts closed loop system. Note the settling time, %OS and steady state error. d) Add PID control K loop system. Note the settling time, %OS and steady state error. 350, Ko-50, Ki-300 and simulate the step response of the closed [10 pts]

Answer 1

Part-A Simple Open loop system

1 2+10s+35 Step Input Output Open loop Transfer Fon

From Output wave as given below, the Settling time is 2.4 second and steady state error is 0.003

0.03 0.025. 002 0,015 0.01 tin 0.005

Part-B- Close loop system with Kp=300

P(s) 2+10s+35 Step Input Output P Controller Open loop Transfer Fon Kp 300 Gain

From Output wave given as below, The Settling time is 2.0 second and Steady state error is 0.06

1.4 1.2 1 0.8 0.6 0.4 0.2 1 2 3 7 10 4

Part-C , Close loop system with PD Controller

1 PD(s) $2+10s+35 Step Input Output PD Controller Open loop Transfer Fcn Kp 300 & Kd-10 Gain

From Output wave given as below, The Settling time is 1.5 second and Steady state error is 0.14

1.4 1.2 0.8 0.6 0.4 0.2 2 7 9 10

Part-D , Close System with PID Controller

PID(s) $2+10s+35 Step Input Output PID Controller Open loop Transfer Fon Kp 350, Kd- 50 & Ki- 300 Gain

From Output wave given as below, The Settling time is 5.1 second and Steady state error is 0.00001

1.4 1.2 0.8 0.6 0.4 0.2 3 5 10