Question

3. Consider the following mass-spring-damper system. Let m= 1 kg, b = 10 Ns/m, and k = 20 N/m. b m F k a) Derive the open-loop transfer function X(S) F(s) Plot the step response using matlab. b) Derive the closed-loop transfer function with P-controller with Kp = 300. Plot the step response using matlab. c) Derive the closed-loop transfer function with PD-controller with Ky and Ka = 10. Plot the step response using matlab. d) Derive the closed-loop transfer function with PID-controller with Kp = 350, Kp = 300 and Kd = 50. Plot the step response using matlab. e) Discuss the step responses of a), b), c), and d) by comparing performance indices: overshoot, rise time, settling time, and steady-state error.

Answer 1

a)

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$\tiny m\ddot x = F - kx - b\dot x$

$\tiny m\ddot x +kx+ b\dot x= F$

Take the Laplace transform.

$\tiny (ms^2 +bs+k)X(s)= F (s)$

1 X(s) F(s) (ms2 + bs + k) 2

Substitute the given values.

$\tiny \frac{X(s)}{F (s)} = \frac{1}{(s^2 +10s+20)}$

>> G = tf(1, [1 10 20]) G = 1 s^2 + 10 S + 20 Continuous-time transfer function. >> step (G);

Step Response 0.05 0.045 0.04 0.035 0.03 Amplitude 0.025 0.02 0.015 0.01 0.005 0 0 0.5 1 2 2.5 3 1.5 Time (seconds)

b)

>> G = tf(1, [1 10 20]); >> C = pid (300,0,0); >> T = feedback (G*C,1) T = 300 S^2 + 10 s + 320 Continuous-time transfer function. >> step (T)

Step Response 1.4 1.2 1 0.8 Amplitude 0.6 0.4 0.2 0 0.2 0.4 1 1.2 1.4 0.6 0.8 Time (seconds)

c)

>> G = tf(1, [1 10 20]); >> C = pid (300,0,10) C = Kp + Kd U with Kp 300, Kd = 10 Continuous-time PD controller in parallel form. >> T = feedback (G*C,1) T = 10 S + 300 s^2 + 20 s + 320 Continuous-time transfer function. >> step (T)

Step Response 1.2 1 0.8 Amplitude 0.6 0.4 0.2 0 0.1 0.2 0.4 0.5 0.6 0.3 Time (seconds)

d)

>> G = tf(1, [1 10 20]); >> C = pid (350, 300,50) C = 1 Kp + Ki * + Kd * S 5 with Kp = 350, Ki 300, Kd = 50 Continuous-time PID controller in parallel form. >> T = feedback (G*C,1) T = 50 s^2 + 350 S 300 S^3 + 60 s^2 + 370 S + 300 Continuous-time transfer function. >> step (T)

Step Response 1 0.9 0.8 0.7 0.6 Amplitude 0.5 0.4 0.3 0.2 0.1 0 0 0.5 1 2 2.5 1.5 Time (seconds)

e)

Performance characteristics of a:

>> G = tf(1, [1 10 20]) G = 1 s^2 + 10 s + 20 Continuous-time transfer function. >> stepinfo (G) ans struct with fields: RiseTime: 0.8843 SettlingTime: 1.5894 SettlingMin: 0.0452 SettlingMax: 0.0500 Overshoot: 0 Undershoot: 0 Peak: 0.0500 PeakTime: 3.2966

Performance characteristics of b:

>> G = tf(1, [1 10 20]); с pid (300,0,0); T = feedback (G*C, 1) T = 300 S^2 + 10 s + 320 Continuous-time transfer function. >> stepinfo (T) ans = struct with fields: RiseTime: 0.0727 SettlingTime: 0.7724 SettlingMin: 0.7871 SettlingMax: 1.3131 Overshoot: 40.0588 Undershoot: 0 Peak: 1.3131 PeakTime: 0.1842

Performance characteristics of c:

>> G = tf(1, [1 10 20]); C = pid (300,0,10); T = feedback (G*C,1) T = 10 s + 300 s^2 + 20 S + 320 Continuous-time transfer function. >> stepinfo (T) ans struct with fields: = RiseTime: 0.0777 SettlingTime: 0.2897 SettlingMin: 0.8490 SettlingMax: 1.0813 Overshoot: 15.3418 Undershoot: 0 Peak: 1.0813 PeakTime: 0.1704

Performance characteristics of d:

>> G = tf(1, [1 10 20]); C = pid (350, 300,50); feedback (G*C,1) T = T = 50 s^2 + 350 S + 300 S^3 + 60 s^2 + 370 S + 300 Continuous-time transfer function. >> stepinfo (T) ans = struct with fields: RiseTime: 0.0548 SettlingTime: 0.8308 SettlingMin: 0.9006 SettlingMax: 0.9959 Overshoot: 0 Undershoot: 0 Peak: 0.9959 PeakTime: 2.4775

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The original system(a) is overdamped. It does not have any overshoot.

Adding a proportional controller (b) gives an overshoot to the system and it becomes underdamped.

In (c), the derivative part reduces the overshoot and oscillations in the system by increasing the damping ratio.

Finally, the integral part in (d) removes the overshoot in the system makes it rise up quickly. But this increases the settling time of the system.