Answer Happy

Now, using the information above, explain how this solution
below was achieved, and if it was an accurate
solution.
Starting from the PD compensated system designed above, use root
locus techniques to design a PID compensator within K s( ) that
yields
• Zero steady-state error,
• No more than 20 second total settling time (for the PID
compensated system)
Problem Statement The figure below depicts the pitch angle control structure for an unmanned submersible. Reference Pitch Angle Controller Actuator Pitch Dynamics G() Pitch Angle Response 0,(s) - K(s) Gelev (s) + + 0(s) Gpr(s) Inner-Loop Sensor Dynamics 2.5 Gelev (s) = 5+2.5' 0.125(5 +0.435) G,(s)= (s +1.23)(s? +0.2265 +0.0169) Gp(s)=s pr Utilizing this control structure, please complete a controller design and analysis based on the following requirements, in the specified order.

0 (s) 0(s) + K(s) Gelev(s)G, (s) Gr(s) 0,(8) 0(5) K(s) Geley(s),(s) G 1+Golev(s),(s)G(s) 0,(s) 0(s) + K(s)Gelev(s)G(s) 1+Gelev(s),(s),(s) Figure 2: Unity Feedback Diagram

Gain of system without PD compensator: K(8 + 0.435) G(s) = (s +2.218) (s + 1.484) (s +0.1417)(8 +0.1091) Design Criteria: 4 ܠܕ OS < 10 T, < 15sec TS = Swn Solving for damping ratio using percent overshoot: - In (10/100) 142 + In? (10/100) = 0.591 Using damping ratio and transient setting time to solve for natural frequency: 4 15 = Wn = 0.4512 0.591 (wn) Solve for damped frequency = Wd Solve for the real value of the compensator poles: wnV1 – 52 = 0.364 od= 5 wn 0.267

Compensator Poles (complex value) s= -0.267 € j (0.364) Solving for the pole angles about the real axis: 0.364 Opı = tan 2.218 - 0.267 = 10.57° Opz = tan-1 = 0.364 1.484 – 0.267 16.65° 0.364 Opz = 180 – tan = 108.57° 0.267 – 0.1447 0.364 Opx = 180° – tan-1 = 113.45° 0.267 -0.1891 Using the known pole angles to solve for angle of the zero relative to the design point: : 0.364 -1 Oz1 = tan 65.22° 0.933 – 0.267 180 = 0 + 65.22 – 10.57 – 16.65 – 108.57 – 113.45 02 Az = 364.02° = 4.02° (0.93 = - = = ولا Use the damped frequency (0.364) and the angle of the zero (4.02) and real value of the compensator's poles (0.267) to find the location of the zero on the real axis: 0.364 Δσ = = 5.18 tan ez tan (4.02) Note that Ao = -5.18 because the zero is on the negative real axis. 02= Ao - Op= -5.18 – 0.267 = -5.45 The zero of the PD compensator is s= -5.45.

Solving for K constant of the PD compensator using the original transfer function poles and zeros as well as the new zero: 8 + 2.218||$+1.484||$+1.447||$ +0.1091| Kpd= 18 +5.458 +0.435 V1.9512 +0.36121.217 +0.3640.32 +0.3610.1546? +0.3642 5.1862 +0.3692 V0.1682 +0.3642 = 0.18208 Gain of system with PD Compensator: G(8) 0.18208 (s + 5.45) (8 +0.435) (s +2.218) (s +1.484) (8 +0.1417) (s +0.1091)

To achieve a PD compensator within K(s) that yields no more than 10% transient overshoot and no more than 15 sec transient settling time, the team utilized MATLAB to generate a root locus shown below: Root Locus Editor for Loop Transfer_C 2 S = -267.364 ZPD = -5.45 d Imag Axis -1 -2 3 -6 -1 -5 -4 -3 -2 Real Axis Figure 5: Root Locus for PD compensator Consequently, the PD Compensator's equation can be derived below: Gpp = .18208(s+5.45) The CLTF with the PD Compensator at the design point is shown in the equation below: 0.18208( s +5.446) (5 + 0.435) Тci T CL (s+ 2.435) ( s +0.992) (s2+0.530s + 0.200) =

The team was then able to generate a table of the poles and zeros of the system: Zeros Poles -5.446 -2.435 -0.435 -0.992 -0.265 +0.36li Table 1: Poles and Zeros of PD compensator system Using MATLAB, the team was then able to visually plot the step response of the completed PID compensator system to ensure that the requirements of no more than 10% transient overshoot and no more than 15 sec transient settling time were met. Step Response From: To: y 1.2 1 0.8 Amplitude 0.6 0.4 0.2 0 0 20 25 10 15 Time (seconds) Figure 6: Step response of completed PD compensator system

To achieve a PID compensator within K(s) that yields zero steady-state error and no more than 20 seconds of total settling time for the PID compensated system, the team utilized MATLAB to generate a root locus shown below: Root Locus Editor for Loop Transfer_C 3 3 2 So = -196 1.836j 1 Imag Axis -1 -2 -3 -6 -3 -2 Real Axis Figure 7: Root Locus for PID compensator By using MATLAB to visualize the PD compensated system the team could draw the line representing a settling time of 20 seconds and choose a point where the two lines intersect as the design point.

The CLTF with the PID Compensator at the design point is shown in the equation below: 2.0095( s + 5.446) ( s + 0.01901) (s+ 0.435) T CL (s+3.098) (s+ 0.455) (s+0.0108) ( s2+0.393s +3.41) = The team was then able to generate a table of the poles and zeros of the system. Zeros Poles -5.446 -3.098 -0.435 -0.455 -0.0109 -0.0108 -0.196 + 1.8361 Table 2: Poles and Zeros of PID compensator system Using MATLAB, the team was then able to visually plot the step response of the completed PID compensator system to ensure that the requirements of zero steady-state error and no more than 20 seconds of total settling time were met. Step Response From:r Toy 1.8 1.6 1.4 1.2 System: IoTransfer,2y VO:r to y Time (seconds): 19.8 Amplitude: 0.996 AN Amplitude 0.8 0.6 0.4 o 0.2 0 0 10 20 25 30 15 Time (seconds) Figure 8: Step response of completed PID compensator system