In this question, we are asked to calculate the minimum damping ratio required for overshoot using a specific formula. The data required for the calculation is missing, so we suggest verifying the formula with the provided notes. We provide the correct formula and calculate the minimum damping ratio. The result is 0.3183.
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Question 1.1: In the question it is asked to calculate the minimum damping ratio required for Mp≤ 10% from the formula mentioned in your notes. Data missing for calculation of overshootMp≤ 10%as the notes formula in unavailable data is missing. Though the following formula can be used for calculation of maximum overshoot, Mp. I would suggest you to verify the below formula used for the calculation with the mentioned notes formula. Mp=e −ζπ √¿¿¿¿ Mp≤10% 0.10=e−ζπ √¿¿¿¿ log0.10=−¿−ζπ √¿¿¿¿ −1=−ζπ √¿¿¿ 1−ζ2−ζ2π2=0 ζ2=1 1+π2 ζ=0.3183 The minimum damping ratio of the dominant poles for the overshoot, Mp is required to be 0.3183. Question 1.2Using the formula from your notes, calculate the minimum damped frequency,ωd=ωn p1 −ζ2, of the dominant poles for a peak timetp≤ 0.5sec. The mentioned formula for calculating the minimum damping frequency in terms of natural frequency is incorrect. Therefore I am writing the correct formula need to be used for tζhe calculation. I would suggest you to verify with the formula provided in your notes. As you can see there is no P present in the mention transfer function of the plant. wd=wn√(1−ζ2¿)¿ ζ=0.3183
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tp=π wn√(1−ζ2)≤0.05 wn=π tp√(1−ζ2) ¿66.27 wd=wn√(1−ζ2¿)=62.823¿ So, the minimum damping ratio can be calculated as 62.823 rad per seconds. Question 1.3Sketch the area in the s-plane where the dominant poles must lie in order to meet both the specifications. We know thatθ=cos−1ζ Nowθ=cos−10.3183=71.43° A line passing through the origin and making an angle of71.43°with negative real axis can be drawn on the root locus plot. The intersecting point on the root locus plot will give the value of poles present. As in the question it is asked to sketch the area so it is not possible to sketch it due to unavailability of resources, but the poles area will be on the left side of the s-plane which signifies that the system is stable. Though I have attached the matlab plot below. -18-16-14-12-10-8-6-4-20 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Pole-Zero Map Real Axis (seconds-1 ) Im a g in a r y A x is ( s e c o n d s- 1 ) Question 1.4Record the transfer function and comment on it. Record and comment on the plant poles.
0.956 Transfer Function, G=---------------------------- 0.006572 s^3 + 0.168 s^2 + s The above transfer function replies that the system is a third order continuous time system having three poles on its left plane. These poles are present on the negative side of the real axis. The pole locations are at 0, -16.1290, and- 9.4340, which says that the system is a stable system. -18-16-14-12-10-8-6-4-20 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Pole-Zero Map Real Axis (seconds-1) Imaginary Axis (seconds-1) Question 1.5What will happen if you try to obtain the plant open loop step response? Explain why.Obtain the open loop impulse response by impulse(G); The open loop step response of the plant is shown in the figure below. The steady state error possessed by a closed loop control system depends on the input and the open loop transfer function. The actual output of a plant control system may be in any of the physical form of position, velocity or acceleration. For evaluating steady state error the input function is specified as unit step for displacement, unit ramp for velocity and unit parabolic for acceleration. So applying unit step input will possess the result for the physical quantity displacement or voltage. 0510152025 0 5 10 15 20 25 Step Response Time (seconds) Amplitude 00.10.20.30.40.50.60.70.8 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Impulse Response Time (seconds) Amplitude
Question 1.6Explain this response characteristics. The curve shows unit impulse response of a third order open loop transfer function. The system represents an overdamped response with having a larger value of steady state point. It has aSettling time of 0.506 sec with no overshoot and peak time. Proportional controller design by Bode plot: Question 2.1Explain the Bode plot characteristics. For Kp = 10, we obtain the following bode plot. This response shows the system is closed loop stable system with GM as 8.15 dB at frequency 12.3 rad per sec. And PM is 28.3 degree. 10-1100101102103 -270 -225 -180 -135 -90 Phase (deg) Bode Diagram Frequency (rad/s) -150 -100 -50 0 50 System: G Gain Margin (dB): 8.15 At frequency (rad/s): 12.3 Closed loop stable? Yes Magnitude (dB) Question 2.2Find the limiting value ofkpfor system stability. Show the workings on the Bode plot. The limiting value of kp for this system stability is 0 < kp < 26.75
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Bode plot taking kp= 29, beyond the limiting value. The system is unstable. 10 -110 010 110 210 3 -270 -225 -180 -135 -90 P h a s e( d e g ) Bode Diagram Frequency (rad/s) -150 -100 -50 0 50 100System: G Gain Margin (dB): -1.1 At frequency (rad/s): 12.3 Closed loop stable? No Ma g n it u d e( d B ) Question 2.3From the Bode plot, determine the value ofkprequired forMp= 10%assuming PM≈ 100ζforPM <70deg wherePMis the phase margin. Show the workings on the Bode plot. Theζvalue for a 10% overshoot can be calculated as 0.3183. Therefore, the value of kp should be 2.2 to get an overshoot of 10%. The bode plot for overshoot 10% is shown below. Bode Diagram Frequency (rad/s) 10-1100101102103 -270 -225 -180 -135 -90 System: G Phase Margin (deg): 69.8 Delay Margin (sec): 0.572 At frequency (rad/s): 2.13 Closed loop stable? Yes Phase (deg) -150 -100 -50 0 50System: G Gain Margin (dB): 21.3 At frequency (rad/s): 12.3 Closed loop stable? Yes Magnitude (dB)
Question 2.4Record and comment on the closed loop transfer function. Comment on the response. RecordMp andtp. CompareMpto the value predicted from the Bode plot. Does your design meet the specifications? The closed loop transfer function is4.84 ----------------------------------- 0.006572 s^3 + 0.168 s^2 + s + 4.84 The value of Mp and Tp got from the step response are 17.2% and 0.64 Sec respectively. The design meet the specification for the GM< 31.83 and PM<70 degree. But it could not satisfy the MP value as 10 %. Though the value of tp is within the acceptable range. Step Response Time (seconds) Amplitude 00.20.40.60.811.21.41.61.82 0 0.2 0.4 0.6 0.8 1 1.2 1.4 System: T Peak amplitude: 1.17 Overshoot (%): 17.2 At time (seconds): 0.64 Question 2.5Record the gain margin, phase margin, and associated frequencies. Comment on the Bode plot of the open loop compensated system. Show the gain and phase margins on the plot. GM = 5.2816 PM = 50.8700 Wcg = 12.3353 Wcp = 4.2639 GM= 14.4553 decibels
Bode Diagram Frequency(rad/s) -150 -100 -50 0 50System: untitled1 Gain Margin (dB): 14.5 At frequency (rad/s): 12.3 Closed loop stable? Yes M a g n itu d e ( d B ) 10 -110 010 110 210 3 -270 -225 -180 -135 -90 System: untitled1 Phase Margin (deg): 50.9 Delay Margin (sec): 0.208 At frequency (rad/s): 4.26 Closed loop stable? Yes P h a s e ( d e g ) Question 3.1What will be the system type if a PID controller is used on this plant? For zero steady state error (to unity step references) for this plant, is an integrator required in the controller? If a PID controller is used in this plant it will be an Industrial automation system. Yes to get a zero steady state error integrator is required. Because by using only proportional controller the system cannot reach the steady state value. It remains some amount of offset. Question 3.2What are the advantages and disadvantages of not using integral action (i.e., using P+D control)? Disadvantage: ï‚·Steady state error cannot be reduced to zero. ï‚·Some offset always remain. ï‚·Cannot apply for fast moving process control like flow, pressure. Advantage: ï‚·System stability increases. ï‚·Maximum overshoot, Mp decreases. ï‚·Settling time, ts decreases. Question 3.3From the plot, measureMpandtp. Does your design meet the specifications? Record the final values ofkpandkd. Record the transfer functionK(s), the system open loop transfer functionL(s)and the closed loop transfer functionT(s). From the plot we got the value of Mp as 10.3% and tp as 0.155 sec. Yes, this design meets the specification. The final value of kp =2.2 and kd = 3.8. 3. 4 84s + 2.2
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transfer functionK(s) =------------- 0.02 s + 1 3.675 s + 2.103 L(S) =-------------------------------------------- 0.0001314 s^4 + 0.009932 s^3 + 0.188 s^2 + s 3.675 s + 2.103 T(S) =---------------------------------------------------------- 0.0001314 s^4 + 0.009932 s^3 + 0.188 s^2 + 4.675 s + 2.103 Step Response Time (seconds) Amplitude 012345678 0 0.2 0.4 0.6 0.8 1 1.2 1.4System: T Peak amplitude: 1.1 Overshoot (%): 10.3 At time (seconds): 0.155 Question 3.4the Bode plot of controller transfer function characteristics.
10-210-1100101102103 0 30 60 90 Phase (deg) Bode Diagram Frequency (rad/s) 0 10 20 30 40 50 Magnitude (dB) Question 3.5Record the gain margin, phase margin, and associated frequencies. Show the gain margin and phase margin on the Bode plot. Bode Diagram Frequency (rad/s) -150 -100 -50 0 50System: L Gain Margin (dB): 10.8 At frequency (rad/s): 37.3 Closed loop stable? Yes Magnitude (dB) 10-210-1100101102103104 -270 -180 -90 0 System: L Phase Margin (deg): 44.7 Delay Margin (sec): 0.0414 At frequency (rad/s): 18.9 Closed loop stable? Yes Phase (deg)
GM = 3.4851 PM = 44.6978 Wcg = 37.2854 Wcp =18.8625 GM =10.8444 decibels Question 3.6Record and comment on the closed loop poles and zeros. Pole-Zero Map Real Axis (seconds-1) Imaginary Axis (seconds-1) -70-60-50-40-30-20-100 -25 -20 -15 -10 -5 0 5 10 15 20 25 System: T Pole : -6.73 + 22.8i Damping: 0.283 Overshoot (%): 39.6 Frequency (rad/s): 23.8 System: T Pole : -6.73 - 22.8i Damping: 0.283 Overshoot (%): 39.6 Frequency (rad/s): 23.8 System: T Zero : -0.572 Damping: 1 Overshoot (%): 0 Frequency (rad/s): 0.572 System: T Pole : -61.7 Damping: 1 Overshoot (%): 0 Frequency (rad/s): 61.7 Pole locations on s-plane are -61.6522 + 0.0000i, -6.7263 +22.8314i, -6.7263 -22.8314i, - 0.4581 + 0.0000i and zero location is -0.5723. All the poles are present on the left side of s- plane. The system is stable. Reference: Antsaklis, P.,Gao, Z. (2004) The Electronics Engineers' Handbook: Control System [online].Availablefrom:https://www3.nd.edu/~pantsakl/Publications/348A- EEHandbook05.pdf[Accessed 17 April 2019] Control Tutorials for MATLAB and Simulink (2019) Control Tutorials for MATLAB and Simulink[online].Availablefrom:http://ctms.engin.umich.edu/CTMS/index.php? aux=Home [Accessed 20 April 2019]. Control Tutorials for MATLAB and Simulink (2019) Suspension PID controller Design [online].Availablefrom:http://ctms.engin.umich.edu/CTMS/index.php? example=Suspension§ion=ControlPID [Accessed 21 April 2019].
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