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Control System Topics & Concepts Covered

Topics	Concepts
Feedback Control Systems	Closed-loop systems, Feedback loops
Control System Components	Controllers, Sensors, Actuators
Transfer Functions	System representation in the frequency domain
Time Response Analysis	Transient and steady-state response
Frequency Response Analysis	Bode plots, Nyquist plots
State-Space Analysis	State variables, State-space representation
Stability Analysis	Routh-Hurwitz criterion, Root locus analysis
Control System Design	Compensators, Pole placement
Digital Control Systems	Discrete-time systems, Z-transform

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Q1:Assessment Cover Sheet Assessment Title Designing a Speed Controller for a DC Motor Programme Title: Bachelor of Engineering Technology Course No .: EN7230T Course Title: Instrumentation and Automatic Control Student Name: Student ID: Section Number: Tutor Name: ➢ Assessor: Date of Marking: Grade/Mark: Assessment Weight: 60% of total mark Comments: Course Project Descriptor (EN7230T) Learning Outcomes The tasks included in this assignment fulfill the Learning Outcomes of the course as follows: 1) Demonstrate advanced knowledge of core principles of control and instrumentation. 2) Critically analyze and evaluate transient response characteristics of first and second order systems. 3) Apply and demonstrate instrumentation and control strategies for feedforward and feedback control systems. 4) Formulate, apply, and present tuning methods for finding suitable PID controller parameters to meet an industry standard requirement for a defined control design problem. The completion of this project carries a weightage of 60% towards your final grade. The project entails designing a speed controller for a DC motor using MATLAB/Simulink. The specific time frame for the project will be announced by your tutor. Deliverables By the project deadline, please submit: 1) The MATLAB/Simulink files used to design your proposed controller. 2) A MATLAB/Simulink simulation model of the complete control system. 3) A project report explaining how you determined the controller design and why you made those particular design choices. The report should include details of your calculations and reasoning behind the design. Simulation results and analysis can help show the effectiveness of the approach. 2 | Page Course Project Descriptor (EN7230T) Guidelines for the Project Report Prepare a comprehensive project report that encompasses the following tasks. Justify your selection of the controller and demonstrate how it meets the criteria for the transient response design specifications. Ensure that the report is clear and transparent by following these guidelines: . Present your results using appropriate MATLAB/Simulink plots and code (very, very strict rule). · Include all m-files (editable text, no pictures) and Simulink block diagrams in the appendix section of your report (not-so-strict rule). . Label the axes of each graph, including a figure legend when necessary. Specify the units in the label itself, such as "speed (m/s)" or "time in milliseconds [ms]," to provide clarity about the measured quantities (very, very strict rule). · To avoid sending multiple figures, consider using the subplot() function to group related figures (not-so-strict rule). . Add comments within each m-file to describe the purpose of the MATLAB code or command (very, very strict rule). . Give each graph a descriptive title and ensure that all axes are labeled and scaled accurately (very, very strict rule). . If a plot includes multiple lines, include a legend that explains each curve (very, very strict rule). . Do not include Simulink 'Scope' images in the report, as they lack proper labeling (very, very strict rule). · When creating Simulink block diagrams, minimize overlapping and crossing lines as much as possible. Rearrange icons to establish a clear left-to-right path (very, very strict rule). . Number your plots and graphs sequentially, such as Fig. 1, Fig. 2, and so on (very, very strict rule). DC Motor The DC motor is a widely used actuator in control systems. It is capable of producing rotary motion directly and, when combined with wheels, drums, or cables, can also generate translational motion. 3 | Page Course Project Descriptor (EN7230T) The electric equivalent circuit of the motor's armature and the free-body diagram of the rotor are depicted in the accompanying figure. Fixed field R L T + Armature circuit 0 v + e i - bė Rotor In this specific scenario, we will consider the voltage source (V) applied to the motor's armature as the input to the system, while the rotational speed of the shaft (0) will serve as the output. Both the rotor and the shaft are assumed to possess rigid characteristics. Additionally, we adopt a viscous friction model, meaning that the friction torque is proportionate to the angular velocity of the shaft. The physical parameters for our particular example are as follows: Parameter Description Value J moment of inertia of the rotor 0.05 kg·m2 b motor viscous friction constant 0.1 N.m.s Ke electromotive force constant 0.05 rad.s Kt motor torque constant V V 0.05 rad·s R electric resistance 0.5 02 L electric inductance 0.8 H 4 | Page Course Project Descriptor (EN7230T) System's Model Derivation Typically, the torque produced by a DC motor is directly related to both the armature current and the strength of the magnetic field. However, in this particular case, we will assume that the magnetic field remains constant, resulting in the motor torque being solely proportional to the armature current i. This relationship is expressed by the equation below, where the proportionality factor is denoted as Kt. This configuration is commonly referred to as an armature-controlled motor: T = Kti (1) The back electromotive force (emf), represented as e, is directly proportional to the angular velocity of the shaft. This relationship is governed by a constant factor Ke as follows: e = KęÒ (2) In the International System of Units (SI), the constants for motor torque and back electromotive force (emf) are equivalent, meaning that Kt is equal to Ke. For simplicity, we will denote this common constant as K to represent both the motor torque constant and the back emf constant. Using the information provided in the preceding figure, we can derive the following governing equations by applying Newton's second law and Kirchhoff's voltage law: JÄ + bÒ = Ki (3) L - + Ri = V - KÒ dt di (4) Transfer Function By employing the Laplace transform, we can represent the aforementioned modeling equations, given in (3)-(4), in terms of the Laplace variable s: s(Js + b)0(s) = KI(s) (5) (Ls+ R)I(s) =V(s) - Ks0(s) (6) By eliminating I(s) between the two equations mentioned in (5)-(6), we obtain the following open- loop transfer function: P(s)=Ó()- Ở(s) (Js+b)(Ls+R) + K2 K (7) In this transfer function, the rotational speed O(s) is regarded as the output, while the armature voltage V(s) is regarded as the input. 5 | Page Course Project Descriptor (EN7230T) Tasks Part 1: Open-Loop System In open-loop systems, there is no sensor to measure the output and compare it with the user-defined set-point via a feedback stream, and thus there is no compensator for the mismatch between input and output. This system can be illustrated as follows: V(s) ₿(s) P(s) 1.1 Substitute the numeric values of the system parameters in P(s) and re-express it again. 1.2 Using MATLAB, code the above transfer function and find the plot of its step response =. 1.3 From these plots, calculate the difference between the input signal and the output signal. Will the output reach the input, or will it have a steady-state error? For the step response mentioned in 1.2, each student should use the following amplitude: Student ID A Student ID A Student ID A 202306613 3.25 Part 2: Feedforward Control System One of the main objectives of embedding a control system is to ensure that the output signal, represented by O(s), closely resembles the input signal, represented by V(s). If we assume zero external disturbances and a perfect match between the model and the actual system, we can incorporate the following feedforward controller to achieve the objective. 6 | Page Course Project Descriptor (EN7230T) V(s) u(s) ₿(s) Kff P(s) 2.1 Find the value of the feedforward gain (Kff) that satisfies the condition O(s) = V(s). 2.2 Find the inverse of the final value of P(s); i.e., calculate P(s)-1 when s = 0. 2.3 Compare Kff with P(0)-1. Are they the same? Why? Part 3: Feedback Control System The main inherent weakness of the feedforward control scheme is its inability to mitigate uncertainty, such as unknown or unconsidered disturbances and/or imperfections in the model. Also, there is no control on the output performance (such as overshoot, undershoot, raise time, settling time, and offset or steady-state error). To achieve these criteria, a feedback control scheme is suggested. Assume the following block diagram represents the prospective feedback control scheme where a filter is placed in series with the sensor and the process is affected by an external disturbance: Step Disturbance Time Delay Rotational D(s) Speed Armature Voltage e -Tas d(s) Step Change in Set Point Compensator DC Motor V(s) e(s) + C(s) P(s) u(s) f(s) Ò(s) + + Filter m(s) H(s) The disturbance signal is modeled as a step function of magnitude 0.2 and delayed by a time constant of Ta = 0.1. Thus, the step disturbance in Laplace domain is: = 6021 - 0,2 0.2 (8) and the time delay can be expressed as a rational function using the 2nd order Pade approximation: e-0.15 ~ s2 + 60s + 1200 s2 - 60s + 1200 7 | Page (9) Course Project Descriptor (EN7230T) The Laplace transform of the measurement unit's filter is given as follows: H(s) = S+ 1 1 (10) Finally, the compensator is a PID controller whose general expression in Laplace transform is: C(s) = Kp + + + Kas (11) 3.1 Using Simulink, model this closed-loop system and connect the input and output signals to a two-channel scope. 3.2 Using MATLAB or by hand, find the overall transfer function between V(s) and Ò(s) and the overall transfer function between D(s) and ((s). 3.3 Apply the online Ziegler-Nichols (Z-N) tuning method to find the PID gains (Kp, Ki, and Ka). 3.4 Apply the online Tyreus-Luyben (T-L) tuning method to find the PID gains (Kp, Ki, and Ka). 3.5 Check these tuned PID parameters in your Simulink model and compare between the two tuning methods. Which one has a better result? Why? 3.6 Can you reach a better result? How (state some references)? Show it numerically. Note: there are 3 parts, each task of part 1 weighs 1 mark, each task of part 2 weighs 1.5 mark, and each task of part 3 (except the last task) weighs 2 marks, and the last task of part 3 weighs 2.5 marks. Plagiarism & Collusion WARNING ON PLAGIARISM & COLLUSION Plagiarism - is defined as the misrepresentation of another person's ideas, thoughts or words as though they were your own. All types of deliverables are covered by this definition, including, written work, diagrams, designs, engineering drawings and pictures. At Bahrain Polytechnic, we encourage independent thinking, and you are entitled to criticize other people's work and include it in your report. However, you MUST acknowledge your sources. This can be achieved by quoting and citing the published or unpublished work of others, which include sources from the internet, books, journal articles, conference proceedings, and any other credible or uncredible sources. A full reference to the source must be provided in an appropriate format and should be included in your References/Bibliography section of your report. Collusion - is defined as an unauthorized conscious collaboration between two or more students in producing work that is deemed to be identical or largely similar in content and representing the work as though they were their own. If a tutor suspects a plagiarism and/or collusion offence in a report, then this may result in an allegation of cheating. Such cases will be dealt with under the Bahrain Polytechnic's procedure and may result in a severe penalty being taken against any student found guilty and it may result in the student being expelled from the course indefinitely. Departments can give advice about the appropriate use and correct acknowledgements of other sources in your own work. 8 | PageSee Answer
Q2:Department of Engineering Technology EET 441/GET 431 Controls Lab Spring, 2024 Lab #3: 1st Order Systems: Open-loop/Closed-loop harmonic response of an RC circuit Due Date: Laboratory #3 compares the open-loop and closed-loop response of a first order system to harmonic input. A range of frequencies are applied and the results used to construct Bode plots. The goal of this laboratory is to verify the measured responses with theory, observe the differences between the open-loop and closed-loop responses, and show how the time constant can be estimated from the amplitude and phase data. Harmonic Response Procedure. 1. Measure the resistance and capacitance of the plant. 2. Monitor the input on oscilloscope channel 1 and output on channel 2. Set up the measurement functions on the oscilloscope to measure channel 1 and channel 2 peak-to- peak voltage and the phase shift between channels. The oscilloscope screen should have these measurements displayed across the bottom of the screen as shown in figure 1. RIGOL TD 160ml TTTTTTTTIT VPF(2)= 2.984 UPP(1)= 5.12V CH1= 1.00V Pha_A= 14º Time 1.000ms 0+0.0000s 4-365 500ml Figure 1. Oscilloscope screen showing channel measurements and phase. 3. Adjust the feedback circuit to obtain an open-loop gain of 0.5. Measure and record the open-loop input amplitude, output amplitude and phase at each frequency of the attached datasheet. 4. Reattach the feedback to the circuit to provide a closed-loop control system. Repeat the measurements over all frequencies as listed on the datasheet. 5. Repeat steps 3 and 4, adjusting the open-loop gain each time to values of 1.0, 2.0, and 4.0. Report 1. Prepare a written report in memorandum format (short report format) that describes the experiment goal, equipment, procedure, and results. Present the underlying mathematical models and describe how well the model fits the experimental data. Describe any errors observed and deviations of experimental data from the expectations predicted in the model. Prepare your figures carefully. Make sure the axes are labeled correctly with units and a legend to identify the curves. Each figure must be numbered, have a description, and be referenced in the body text. 2. Include a Bode plot with amplitude and phase figures for the open-loop and closed-loop test results. Determine the break frequency (time constant) from each curve in each figure. Explain the process for these calculations and compare these results with the measured value of the plant time constant (RC) and the response time constant, [T/(1 + Kp)]. Discuss your results.See Answer
Q3:5. Consider a feedback control system shown below. Let's say the state space form of the plant's differential equation (from u to y) is x = Ax + Bu , and y = Cx + Du, where x is the plant state vector. And let's say the controller is a proportional controller so that u = k(r - y). If you were to write the closed loop differential equation (from the input r to output y) in the state space form, it will have the form: 2 = Acu2 + Bcur , and y = CCLZ + DcIT, where ž is the closed loop state vector. Write ACL, BCL, CCL, DeL and z in terms of A, B, C, D, x and k. R(s) + E(s) U(s) G(s) Y(s KSee Answer
Q4:4. As mentioned in the lectures, there are multiple A, B, C and D matrices possible for a given differential equation from u to y. What is the reason for this? How can you check if A1, B2, C1, D1 and A2, B2, C2, D2 correspond to the same differential equation?See Answer
Q5:3. If the state space equation for a plant from the input u to the output y, is x = Ax + Bu , and y = Cx + Du. Find the plant transfer function from I to Y in terms of matrices A, B, C, D and s.See Answer
Q6:2. Convert the following state space form to a transfer function from u to y. A = [1 2; 3 4] B = [1 ; 3] C = [1 4] D = 0 What are the poles of the TF, can you find the poles without having to find the TF first?See Answer
Q7:Problem 3 [(a)-4pts, (b)-6pts, (c)-2pts, (d)-6pts] In the closed-loop control system in the figure below, the fixed plant or process is given to be: 1 G(s)= (s+2)(s+10) To examine the effect of feedback performance in the system, a unit step input is applied to the system with gain K. (a) Determine the sensitivity: ST(s) in its simplest form, where T(s) = (b) For a gain K = 12, find the system response c(t), and sketch its graph, clearly labeled. (c) For a gain K = 12, what the settling time Ts of the feedback system? (d) For a gain K = 12, determine the value of the steady state error e (c) of the system response. R(s) K G(s) -> C(s) - Solution:See Answer
Q8:Problem 2 [(a)-10pts, (b)-10pts] (a) Use block diagram transformation to rearrange the feedback system block diagram in Fig. A to the form in Fig. B and identify G(s). (b) Draw the signal flow graph (SFG) of the feedback block diagram in Fig. A and use Mason's Gain Rule to find the transmittance- C from your SFG. R + + + R G G -c H1 H2 Fig. A R(s)- + G(s) - C(s) Fig. B Solution:See Answer
Q9:USD ELEC 460 Dr MLD Lumori Mid Term Exam 1 Problem 1(a) [10 points] The student-professor learning process is in fact a feedback process intended to reduce the system error to a minimum. The desired output is the knowledge being studied and the student may be considered to be the process. Construct a feedback control model of the learning process and identify each block of the system and all the inputs and outputs in each part of the system. Solution 1(a):See Answer
Q10:Problem 3 |(0)-8pts, (ii)-4pts, (Iii)-8pts) The Figure below is a block diagram of a system used in maintain a high level of consistency in a paper-making process. Given that for this wystere; HL)-1. G.(s) =- K 1 (1) Derive that: 7(s) -- 32+ +128+1+K R() YU) (II) Determine the sensitivity: SE in its simplest for. (ill) Find the value of K required for an allowable steady-state error: ca(10) < 1% [NOTE: Make sure K accommodates this range (ce(o) = 1%) in your answer to obtain full pointa !!!!. Solution 3:See Answer
Q11:Problem 2(a) [10 polars) Liuc block diagram trans nuations to roarrange the foodbank tysum block diagram in 1'ig. 2(a) to the form in Fig. 2(b) below and identify Go) 1 Fig. 2(a) ution 2(a): Fig. 2(b)See Answer
Q12:Problem 5 Consider the open-loop system: G(s) : s(s+2) 10 Design a Proportional Derivative (PD) controller with the specifications: a rise time < 0.3 sec, overshoot < 10% and steady state error < 2%. The controller D(s) will be implemented in the feed-forward path as shown below. Note that the transfer function of the PD controller is of the following form D(s) = KDS + Kp ELECTRICAL ENGINEERING DEPARTMENT EE 380 Control Engineering I Second Semester 2023/2024 E U G(s) > R D(s)See Answer
Q13:Problem 3. For the unit step response shown below, find the transfer function of the system. 1.4 1.2 1.0 Response 0.8 0.6 0.4 0.2 0 5 10 15 20 25 Time (seconds)See Answer
Q14:Problem 2. For the system shown below, do the following: (a) Find the transfer function G(s) = X(s)/F(s). (b) Find 5, @n, % OS, Ts, Tp and Tr. 28 N/m -x(1) 5 kg - f(1) 5 N-s/mSee Answer
Q15:Supervisor Dr Hayley Wyatt Laboratory H Location S0.14 (CUSP Laboratory) HI Control Laboratory --- Liquid Level Control 1. INTRODUCTION The liquid level control laboratory aims to introduce second year students to control theory and more importantly the use of control in industrial applications. Specific aims however, are to give students an understanding of: · The importance of sensor characteristics in control applications . The benefit of using closed loop control when compared with open loop (feed forward) control systems · The effects of proportional and integral feedback on control systems 2. EXPERIMENTAL EQUIPMENT Prior to starting the experimental programme it is advised you familiarise yourself with the experimental equipment used in the experiment additional handouts will be provided by lab support. 2.1 The CE120 Controller The controller is an instrument comprising a wide selection of amplifiers, signal conditioning circuits and power supplies which can be connected together in a variety of ways to provide control for any system. 2.2 The Coupled Tanks Apparatus This apparatus relates specifically to fluid transport and liquid level problems as they would occur in process control industries. It comprises two separate vertical tanks which are connected by a flow channel. In this laboratory only the left hand tank (tank A) is considered. Figure 1 shows the front panel of the apparatus, which provides a schematic functional detail of the unit, and easy access, via 2 mm terminals, to the flow-rate transducers, liquid level transducers and pump control circuits. 3. CONTROL THEORY To complete this laboratory you will need to know some basic control theory. This will be taught to you during the control and instrumentation module however as this section is not lectured until later in the term the basic theory is outlined below. 3.1 System Modelling - 1st Order System In order to control a system we must understand the dynamics and express them mathematically. It can be shown that the transfer function of the system is given by: yı(s)_G Vi(s) TS+1 1 TANK A TANK B H Water Level Calibration Output from Output from Scales Right-Hand Level Left-Hand Level Sensor Sensor Output from Output from Left-Hand Flow Right-Hand Flow Sensor (Optional) Sensor Vale A 0 to +10V 0 to +10V Variable D.C. 1.1 Variable D.C. Motor Control Motor Control Voltage Voltage (Optional 1 1 Common Earth) Valve B Valve C Terminal Prismatic Level Indicators Figure 1 Detail of the Coupled Tanks Apparatus (where y (s) = level sensor output, vi (s) = set point voltage or input) and that G = K1KsKb and T = AKb (where K1 = pump input gain, Ks = level sensor gain, Kb = output characteristics for a particular operating level and A = cross sectional area of tank.) The step response method of determining G and t is simple and accurate. The gain G is determined by applying a step input of U volts, the steady state change in output is then U x G. The time constant t is defined as the time required for the system to reach 0.632 of its final value. Y(1) UXG 0.632 × UXG T Time Figure 2 Step response for first order system 2 Common Earth) Terminal 3. 2 Open Loop Control Consider the system shown in Figure 3; a pump fills the tank at a constant rate. An outlet, controlled by a valve at a set position, allows water to leave the tank such that eventually the liquid will achieve a constant level. At this point, PUMP FLOWRATE = TANK DISCHARGE RATE or Water in = Water out and the system is in equilibrium or steady state. This is an example of OPEN LOOP CONTROL, i.e. no information concerning the liquid level is fed back to the pump circuit to compensate for changes in liquid level. If this system were used, it would require an operator to vary the pump voltage to account for any changes in the system. Water Level Tank Pump Water Water Out In Discharge Valve Pump Motor Amplifier Figure 3 Single tank system A more acceptable control system would use a transducer to produce a signal proportional to liquid level. Electronic circuits would generate an error signal equal to the difference between the measured signal and the reference signal (or set point) which is chosen to achieve the desired liquid level or OUTPUT. Figure 4 shows a typical arrangement for a closed loop system in which the use of a signal measured at output is used to adjust the input. Error Signal Differencing Reference Signal. (Set Level) Amplifier Liquid Level Level Transducer Signal Liquid Level -Tank Pump Water Water .. In Discharge Out Valve Pump Motor Feedback Amplifler Controller Figure 4 Diagram of a closed loop control system The schematic diagram shown in Figure 5 represents this system. H 3 Error Actuation Signal Signal Differencing Amplifier Output Reference + Controller System (Process Signal Varlable (Set Point) Feedback Loop Figure 5 Block diagram of control system If the system is in equilibrium and the output valve is opened, the liquid level will fall and therefore the error signal will increase. This will increase the supply to the pump motor and hence the flow rate. As the liquid returns to the reference or set level, the error signal reduces causing the energy supplied to the motor to be reduced. In theory, the motor supply would become so small that the pump would stop. In practice the pump rate reduces until a new equilibrium is produced where pump rate = discharge rate and the error achieves a new value. The difference between the Actual level and the Set level is the Steady State Error of the system (Figure 6). If the gain of the controller is increased (termed proportional control) the steady state error can be reduced, i.e. the system level will get closer to the required level. The time constant t would also be reduced. If too much gain were introduced the system may become unstable (i.e. the level would oscillate and not achieve equilibrium). Proportional control is so called since the output from the controller is proportional to the input, so in order to maintain a non-zero input to the pump drive there must be a non-zero error signal at the controller. In order to operate with zero error we must introduce integral control. The integrator is designed such that its output is proportional to the integral of its input. In practice proportional and integral control are used together (termed PI control) to achieve zero error as shown below. With a low level of integral control, the system will be slow to respond to change and is termed over damped. With a high level of integral control, the system will be under damped, will overshoot when responding to change and may even become unstable Controlled Varlable Set-Point Steady State Error - P Only PI P Only Time Figure 6 PI Control H 4 4 EXPERIMENTATION In this laboratory the given units are sometimes not expressed as standard SI units - this is for convenience. 4.1 Circuit Calibration. For an effective control system all aspects within the system need to be calibrated. Calibration allows systems to be 'controlled' in terms of voltages. It is evident that a control system can not utilise a level of water in mm or a flow rate in cm3/min but require voltages, in proportion to the level or flow rate. Tables 1 and 2 present the calibration data for the pump and the water level respectively. The pump calibration constant K1 has been found to be approximately 500 cm3.min-1. V-1 whilst the level sensor constant Ks has been found to be approximately 0.04V.mm-1. Plot the calibration curves using the data provided (flow rate against supply voltage and tank height against voltage output) Table 1: Calibration data for the pump system Flow Rate [cm3/min] Voltage [V] 0 1.95 0 0 1000 3.15 50 1.93 2000 4.95 100 3.94 3000 6.95 150 5.88 4000 8.92 3000 6.87 200 7.87 2000 4.90 1000 3.07 Table 2: Calibration data for the water level Tank Level Voltage [mm] [V] Using the plotted curves to: · Demonstrate that the constants described in section 4.1 are correct. · Write an expression for Q (flowrate) in terms of the Gain constant of the pump K1, the pump supply voltage Vs, and Vmin. (Hint y=mx + c). . Can you identify why the flow rate vs. supply voltage plot does not pass through zero? 4.2 Open and closed loop systems H 5 4.2.1 Open-loop control The single tank is an example of a first order system, whose transfer function can be written in the general form G 1+TS G is the gain constant of the system and t is the time constant of the system. G and t are to be determined by examining the system's open loop response to a step input. Connect the equipment as shown in Figure 7 (supplied by lab support). The connections to the supply potentiometer allow us to supply the system with a constant voltage and then increase this instantaneously, i.e. a step input. By monitoring the fluid level sensor output, we are monitoring the system response as illustrated in Figure 8. INPUT OUTPUT Vo Vi Vi G Vo 1 + TS t t Figure 8 Response of first order system to step input Valve settings: Valve A - 0, Valve B - 3, Valve C - 5. Control settings: Upper potentiometer 5V, lower potentiometer 1V. Apply a constant voltage from the upper potentiometer only, and wait for system to reach equilibrium (steady state) (15 minutes). Attach the output from appropriate points on the apparatus to the input channels of the data logger as follows (as shown in Figure 7), i) The pump input voltage ii) The level sensor output iii) The flow-meter output (n.b. gain = 2.16x10-3 V. cm-3.min-1) Start recording and after a few seconds apply a step increase of 1 V by making the link as shown as a dotted line in Figure 7. Keep the test running until the steady state is again reached, and print the responses of above measured signals. Name graph as the control scheme used and mark each line in the graph. Using the plotted graph: · Determine G (gain of system) and t (time constant). · Can you identify what factors affect G and T ? . Can you think of another way to calculate G and t ? H 6 4.2.2 Closed-loop proportional control This section of the laboratory will demonstrate the clear benefits of closed loop proportional control. Connect the equipment as shown in Figure 9 as supplied by the laboratory support. Valve settings: Valve A - 0, Valve B - 3, Valve C - 5. Controller settings: Proportional gain Kp = 1. (Note: Integral and differential switches off) Build a circuit for calculating the control error, the difference of the set point and the measured liquid level, and connect the error signal to the inlet of the PID controller. Connecting the outputs from appropriate points on the apparatus to the input channels of the data logger as follows, i) The set point (desired output of the liquid level sensor) ii) The level sensor output iii) The flow-meter output iv) The PID controller output Set the set point to 5 volts and wait until the liquid level is in steady state. Start recording and after a few seconds apply a step increase of 1 V. Keep the test running until the steady state is reached, and print the responses of above measured signals. Name graph as the control scheme used and mark each line in the graph. Repeat for values of Kp of 5 and 10. Using the plotted graphs can you: · Comment on the effectiveness of proportional control? · Discuss how increasing the proportional gain Kp influences system parameters such as the time constant t and the steady-state error? · Determine what would happen if too large proportional gain was used? . Can you identify an industry that could use closed loop control to monitor volumes of liquid in tanks? · Can you think why it would not be possible to have zero steady state error using only proportional gain? · Verify that the graphs plotted are in accordance with your expectations and briefly describe the key features of graphs, and how they change as the proportional gain in increased. From the completed work can you: · Draw a block diagram to represent the system, inserting the values calculated from the graph recorded at Kp = 5. 4.2.3 Closed-loop proportional and integral control H 7 Experiment 4.2.2 demonstrated steady state error under proportional control. Proportional control will not remove the error in a system, no matter how high the gain Kp. In order to obtain zero error we must also include an integrator in the feedback loop, which will respond to changes in the error signal such that it will attempt to reduce the error to zero. Connect the equipment as for the previous, and switch on the integral controller. Conduct test on Kp = 5 & Ki = 1; and repeat with Kp & 5, Ki = 5 Using the plotted graphs can you on: · Discuss the effect of integral action on steady state error. · Comment on How increasing the integral gain Ki affects the system? · What would happen if too much integral gain was used? · Verify that the graphs plotted by the data logger are in accordance with your expectations and briefly describe the key features of the graphs. 5. Control Laboratory Short Report Hints Make good laboratory notes, but do not repeat anything already detailed in this handout, you will not get a chance to immediately write a long report, and hence you may need to refer back to your notes at a later date. Ensure all figures are labelled correctly (titles, units, headings etc) remembering titles for figures go below the plot/image but above for tables. Don't forget to leave space at the start of your report for an abstract. An abstract should include information on the work you have completed and your main conclusions. A good abstract would allow a reader to know if the report is applicable to their work and worth further reading. Your discussion can be in the form of bullet points in your lab notebook but should be detailed enough to allow a fuller discussion if you are requested to submit a long report on this laboratory. Please add a conclusion section in your lab notebook, again in bullet points if desired. If you need help ask! 8See Answer
Q16:Problem 1 [5-pts] In a given system, the sink Y(s) is related to the source X(s) as follows: (s)= R & 1-1 ()-(s)- 1 (s) (i) Draw and clearly label the SFG for this system (3 points) (ii) Use Mason's Rule to determine the transmittance, T(s) from your SFG. (Note: Show all the steps (variables, etc) involving Mason's Rule) (2 points) Solution to Problem 1: (1)See Answer
Q17:GILDART HAASE SCHOOL OF COMPUTER SCIENCES AND ENGINEERING University College: Arts – Sciences – Professional Studies Fairleigh Dickinson University EENG 4355 Analog & Digital Control Homework # 1 Date:__________________ Student’s Name:_________________ For the feedback control system represented by the block diagram given below, find the transfer function of the system i.e., the relationship between the input and the output: C(S) R(S) Convert the block diagram into a signal flow graph and find the transfer function using Mason’s rule.See Answer
Q18:FeedBack Control System Signal Flow Graph Homework For the flow graph shown below find the transfer function: G s = C(s) R(s) using Mason’s gain formula. Good luckSee Answer
Q19:Following figure showing the scatter plot of provided data temperature vesus time plot 900 800 700 temperature (oF) 600 500 400 300 -@ 1-A AOF > 1-P AF 200 100 0 7:54:37 PM 7:54:55 PM 7:55:12 PM 7:55:29 PM 7:55:47 PM 7:56:04 PM 7:56:21 PM time This figure needs work. 1. There is no figure number or caption. 2. The time axis should be measured in seconds and the origin positioned near the start of the cooling process. 3. The legend doesn't explain what the curves represent. 4. The time axis has no units. 0 0 10 20 30 40 50 60 70 80 time figure 1 response curve for predicted temperature with time. responce temperatureT(t) vesus time(t) plot 1000 temperature (oF) 800 600 400 -- experimental 200 - analytical curve 0 0 10 20 30 40 50 60 time,t(s) 4. figure 2 response curve for the experimental data and analytical data with time. (Figure 1 & 2 should be combined and presented as a single figure with two curves: the experimental and analytical response. "response" is spelled wrong in figure 2. Figure 2 is too small; the vertical scale should be longer such as this: responce temperatureT(t) vesus time(t) plot 900 800 measuring the value of R ment data analysis for 2 > 0 0 10 20 30 40 50 60 time,t(s) So mathematical modelling of the first order circuit can defined as ... ????? (Did you leave something out here? The input output differential equation of the RC circuit can be defined as dVo W Fo dt + -Vo = 1 Vi .... .eqn.1 Here T = RC time constant In LaPlace-Laplace domain write this equation as below I 1 sVo + -Vo = - Vi T IT 1 (Should this equation be numbered ?? ) eqn.2 Vo 1 T Vi Here Vi= Test input and Vo= output response for input. There are following case for test input. Cose 1 Impulse Input Case I filipuise Input Materials and Procedure · Input impulse function as Vi(t)-kô(t) is taken where k= is constant and ô(t) is unit impulse So, the time domain solution of the output is (Should this equation be numbered ?? ) Vo - Ket € -t · It is required to configure function generator to provide a pulse at frequency 10 Hz and duty cycle of · 1% and output need to turn on. · The trigger and channel gains on the oscilloscope was adjusted and input and output were recorded. · Required data was recorded as per given time span. Impulse Input 1.20E+01 1.00E+01 8.00E+00 6.00E+00 4.00E+00 2.00E+00 D. FO Wi Fo In Voltage (V) 0.00E+00 0.00E+00 5.00E-01 1.00E+00 1.50E+00 2.00E+00 2.50E+00 3.00E+00 3.50E+00 4.00E+00 4.50E+00 5.00E+00 Time (ms) Fig.5 (Output impulse response waveform from given data) This figure needs work: 1) Do not use scientific notation for the numerical values, 2) The vertical scale is too long. It should span 0 to 0.3, not 0 to 0.7. This leaves too much wasted space at the top of the figure, 3) The legend doesn't describe anything, 4) The time scale is too long. It could be shorted to 0 to 4 ms. Analysis for time=1-time constant output can fall can to 36 % its initial value (Awkward !! ) so the valid time is less than 0.001 s so required time constant for above case should be 0.0015s from above analysis. So, the required value of gain K= 10 and t = 0.00115 s So, for the predicted analysis output value can obtained from (Should this equation be numbered?) 1 _t Vo = = = = k T Predictive Response 5 4.5 4 0.00E+00 0.00E+00 5.00E-01 1.00E+00 1.50E+00 2.00E+00 2.50E+00 3.00E+00 3.50E+00 4.00E+00 -2.00E+00 Time (ms) Fig.8 (Output step response waveform from given data) I Analysis and Result For time = 1-time constant output response curve reached at 63 % of its final value which is 5. It took approximately 0.00099 s to reach it. So, time constant from above analysis is 0.0011 s (I don't understand this; why isn't the time constant 0.00099 s?) So, the required value of gain K= 10 ?? and z = 0.0011 s So, for the predicted analysis output value can obtained from (Should this equation be numbered ?? ) t =V0 = k (1 - e =) Predictive output curve Chart Area 1.00E+01 8.00E+00 Accessibility: Investigate xt Predictions: On 1 Q Search T W PRE Time (ms) Fig.11 (Output ramp response -waveform from given data) Result and analysis So, the response time constant is again obtained as 0.00079 by measuring the response error after many times for given input. (Need a more complete explanation of how to estimate the time constant here.) The gain K= 5 (The value of k represents the slope for the ramp function. It is not 5 in this case.) adjusted as per input data analysis. So the required value of gain K= 10 ?? and t = 0.0015 s ? ? So, for the predicted analysis output value can obtained from 2 ??? Is something missing here ??? Ramp predictive curve Predictive Response 1 0.9 0.8 0.7 E 0.6 5% Accessibility: Investigate On Predictive Response 1 0.9 0.8 0.7 Amplitude (V) 0.6 0.5 0.4 0.3 0.2 0.1 0 300 350 400 450 500 550 600 Time (ms) Fig.11 (Output ramp response waveform from predictive data analysis) Something is really messed up here. The total range of voltage should be about 10 v. The time scale should span about 5-7 ms. Result & Discussion So, from above experimental data analysis it makes it easy to measure and calculate the value of required time constant for different test input (Impulse, Step and Ramp) to the RC first order circuit. As measured value of R=10000 and C=100*10-9 F. The required measured time constant for the RC first order circuit should be 0.001 s for given value of R & C. So, from above experimental data analysis it was measured the time constant of RC circuit for the different test input impulse, step gation X document Q and ramp input be 0.00115s, 0.0011 s and 0.0015 s respectively. (The estimated time constants should be much closer to 1 ms than what you have reported.) So, it proves that for different test input time constants are approximately similar value and very closed to the measured time constant value 0.001 s. So, in conclusion that- the time constant is the-dependent on system characteristics parameters and does not change by supply- the input to the system, but output response has been affected but time constant remains same. It cam- can also obtained predicted the output response for different test inputs using the time constant and gain values obtained by the experiment, and observed that nature of the The observed transient response of the output-is very similar to the experimental data response curve . So, the predictive curve is similar in nature shows the correct data to the experimental analysis- response of the first order RC circuit. You have most of the requirements in here; you just need to correct some formulations, clean up the structure, and edit for clarity. Format/structure: 2.7/3 - You have some headings in the report but they are not very consistent and tend to confuse the reader instead of guide them through the report. Some sections seem to be devoid of discussion and rely simply of formulas. The equations should be explained in the text. Equations by themselves is bad writing. Figures: 1.5/3 - There should be 4 figures in the report. Any more is just confusing. 1. The thermal system figure has two curves: the measured response and the analytical response. 2. The Impulse test results figure should two curves: the measured response and the analytical response. 3. The step test results figure should have three curves: the input step, the measured response and the analytical response. 4. The ramp test response figure should have three curves: the input ramp, the measured response and the analytical response. Do not use scientific notation for values on the axes. Include units on the axes. Each figure needs a figure number and a short descriptive statement. Each figure should be discussed in the body of the report. Report body: 2/3 - Number your equations correctly. Explain how the time constant is estimated from each of the four tests. Report the value of the estimated time constant from each test along with the other coefficients in equations used to create analytical curve fits. There are several places where the test makes no sense. Edit carefully. Content: 0.9/1 - Most of the requirements for content are there! Total: 7.1/10 words English (United States) Text Predictions: On Accessibility: Investigate Js Pages Results = x duction erials and Method ults erials and Procedure erials and Procedure erials and Procedure D' FO 1ºC Q Search 1 T W You have most of the requirements in here; you just need to correct some formulations, clean up the structure, and edit for clarity. Format/structure: 2.7/3 - You have some headings in the report but they are not very consistent and tend to confuse the reader instead of quide them through the report. Some sections seem to be devoid of discussion and rely simply of formulas. The equations should be explained in the text. Equations by themselves is bad writing. Figures: 1.5/3 - There should be 4 figures in the report. Any more is just confusing. 1. The thermal system figure has two curves: the measured response and the analytical response. 2. The Impulse test results figure should two curves: the measured response and the analytical response. 3. The step test results figure should have three curves: the input step, the measured response and the analytical response. 4. The ramp test response figure should have three curves: the input ramp, the measured response and the analytical response. Do not use scientific notation for values on the axes. Include units on the axes. Each figure needs a figure number and a short descriptive statement. Each figure should be discussed in the body of the report. Report body: 2/3 - Number your equations correctly. Explain how the time constant is estimated from each of the four tests. Report the value of the estimated time constant from each test along with the other coefficients in equations used to create analytical curve fits. There are several places where the test makes no sense. Edit carefully. Content: 0.9/1 - Most of the requirements for content are there! Total: 7.1/10 d States) Text Predictions: On Accessibility: Investigate [D] Fo Q Search 1 T W PRESee Answer
Q20:Problem 4.a Determine a state-space representation for the following transfer function using MATLAB: G(s) = s3 + 5s2 + 7s + 8 s+ + 5s3 + 2s2 + 7s + 6 ELECTRICAL ENGINEERING DEPARTMENT EE 380 Control Engineering I Second Semester 2023/2024 February 13, 2024 Due Date: February 20, 2024 Problem 4.b Determine a transfer function representation for the following state-space model using MATLAB: [5 -5 2] 17 *= 1 -1 2 x+ 8r 16 -3 -4] 15 y =[-2 -9 7]xSee Answer

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