ENGR202 2018 Q A2Past paperOld spec ENGR2022:225 marks30 min

ENGR202 Summer 2018 Q A2[VALID] official answers

Answer ALL parts (a) - (d).

Figure A2: PI control system, controller (k1 s + k2)/s and plant 2/(s^2 + 5s + 3) with unity negative feedback.
Figure A2: PI control system, controller (k1 s + k2)/s and plant 2/(s^2 + 5s + 3) with unity negative feedback.
Formulas you may need
  • Laplace with zero initial conditions: dx/dt→sXdx/dt \to sX (learn this)
  • Stable if all poles have negative real parts; a single pole at the origin is marginally stable; repeated poles on the imaginary axis are unstable (learn this)
  • Necessary conditions: all coefficients exist and have the same sign (learn this)
  • Steady state gain lim⁡s→0G(s)\lim_{s \to 0} G(s) (learn this)
  • Unity feedback: CG1+CG\dfrac{CG}{1 + CG} (learn this)
  1. (a(i))
    Consider one of the joints of a robotic manipulator, for which the joint angle and an applied voltage represent the output and input respectively. Define the terms open-loop and closed-loop control, using the robotic manipulator as an example.
    [3]
  2. (a(ii))
    Briefly explain the term stability. What are the likely practical consequences of using an unstable robotic manipulator control system?
    [3]
  3. (a(iii))
    Using the robotic manipulator as an example, explain the term dead-zone nonlinearity.
    [2]
  4. (b)
    A simplified model for the robotic manipulator (without control) is as follows: dxdt=Ku(t)\frac{dx}{dt} = Ku(t) where xx is the joint angle and u(t)u(t) is the input. Determine the Transfer Function form of this model. Find the pole, plot its position on the complex plane and state the stability condition. Write down an expression for how to determine the steady state gain of this model, and comment on the physical interpretation of your answer.
    [6]
  5. (c)
    State the stability condition of the following characteristic equations, giving the reason for your answer in each case. s3+s+1=0s^3 + s + 1 = 0 s4+2s3−s2+2s+0.01=0s^4 + 2s^3 - s^2 + 2s + 0.01 = 0 s3+2s2+3s=0s^3 + 2s^2 + 3s = 0 (s+2)(s+3)=0(s + 2)(s + 3) = 0 s2=0s^2 = 0 (s−0.5)(s+0.7)(s+1)=0(s - 0.5)(s + 0.7)(s + 1) = 0
    [3]
  6. (d(i))
    Figure A2 shows a Proportional-Integral (PI) control system. Is the proportional gain k1k_1 or k2k_2?
    [1]
  7. (d(ii))
    Using algebra or the rules of block diagram manipulation, show that the Closed Loop Transfer Function is X(s)=2(k1s+k2)s3+5s2+(2k1+3)s+2k2 V(s)X(s) = \frac{2(k_1 s + k_2)}{s^3 + 5s^2 + (2k_1 + 3)s + 2k_2}\,V(s)
    [4]
  8. (d(iii))
    Find the steady state gain of the Closed Loop Transfer Function and comment on the significance of your answer.
    [3]