Control Systems Project Basil Hamed.

Control Systems Project Basil Hamed

Introduction: System Modeling
The first step in the control design process is to develop appropriate mathematical models of the system derived either from physical laws or experimental data. In this slide, we introduce the state-space and transfer function representations of dynamic systems. State-Space Representation For continuous (LTI) systems, state-space representation is given : Transfer Function Representation Basil Hamed

System Analysis Once appropriate mathematical models of a system have been obtained, either in state-space or transfer function form, we may then analyze these models to predict how the system will respond in both the time and frequency domains. To put this in context, control systems are often designed to improve stability, speed of response, steady-state error, or prevent oscillations The time response of a linear dynamic system consists of the sum of the transient response which depends on the initial conditions and the steady-state response which depends on the system input. These correspond to the free (homogeneous or zero input) and the forced (inhomogeneous or non-zero input). Basil Hamed

System Analysis Second Order Systems
Second order systems are commonly encountered in practice, and are the simplest type of dynamic system to exhibit oscillations. In fact many real higher order systems are modeled as second order to facilitate analysis. Examples include mass-spring-damper systems and RLC circuits. The second order transfer function has two poles at: Basil Hamed

PID The output of a PID controller, equal to the control input to the plant, in the time-domain is as follows: Basil Hamed

The Characteristics of P, I, and D Controllers
A proportional controller (Kp) will have the effect of reducing the rise time and will reduce but never eliminate the steady-state error. An integral control (Ki) will have the effect of eliminating the steady-state error for a constant or step input, but it may make the transient response slower. A derivative control (Kd) will have the effect of increasing the stability of the system, reducing the overshoot, and improving transient response. Basil Hamed

General Tips for Designing a PID Control
When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response. Obtain an open-loop response and determine what needs to be improved Add a proportional control to improve the rise time Add a derivative control to improve the overshoot Add an integral control to eliminate the steady-state error Adjust each of Kp, Ki, and Kd until you obtain a desired overall response. You can always refer to the table shown in this "PID Tutorial" page to find out which controller controls what characteristics. Lastly, please keep in mind that you do not need to implement all three controllers into a single system, if not necessary. For example, if a PI controller gives a good enough response, then you don't need to implement a derivative controller on the system. Keep the controller as simple as possible Basil Hamed

Lead/Lag Compensation
Lead/Lag compensation is very similar to PD/PI, or PID control. The lead compensator plays the same role as the PD controller, reshaping the root locus to improve the transient response. Lag and PI compensation are similar and have the same response: to improve the steady state accuracy of the closed-loop system. Both PID and lead/lag compensation can be used successfully, and can be combined.

Lead and Lag Compensator
Basil Hamed

State-Space Methods (Pole Placement) Design
let's build a controller for this system using pole placement. The schematic of a full-state feedback system is shown below. By full-state, we mean that all state variables are known to the controller at all times. For instance in this system, we would need a sensor measuring the ball position, another measuring velocity, and a third measuring current in the electro-magnet. Basil Hamed

Pole Placement for Plants in Phase-Variable Form
To apply pole-placement methodology to plants represented in phase-variable form, we take the following steps: Represent the plant in phase-variable form. Feed back each phase variable to the input of the plant through a gain, ki. Find the characteristic equation for the closed-loop system represented in Step 2. Decide upon all closed-loop pole locations and determine an equivalent characteristic equation. Equate like coefficients of the characteristic equations from Steps 3 and 4 and solve for ki. Basil Hamed

Since above matrix is in phase-variable form, the characteristic equation of the closedloop system can be written by inspection as Basil Hamed

Now assume that the desired characteristic equation for proper pole placement is where the di’s are the desired coefficients from which Now that we have found the denominator of the closed-loop transfer function, let us find the numerator. For systems represented in phase-variable form, we learned that the numerator polynomial is formed from the coefficients of the output coupling matrix, C. we conclude that the numerators of their transfer functions are the same Basil Hamed

Cruise Control: System Modeling
Physical setup Automatic cruise control is an excellent example of a feedback control system found in many modern vehicles. The purpose of the cruise control system is to maintain a constant vehicle speed despite external disturbances, such as changes in wind or road grade. This is accomplished by measuring the vehicle speed, comparing it to the desired or reference speed, and automatically adjusting the throttle according to a control law. Basil Hamed

System equations With these assumptions we are left with a first-order mass-damper system. Summing forces in the x-direction and applying Newton's 2nd law, we arrive at the following system equation: Since we are interested in controlling the speed of the vehicle, the output equation is chosen as follows y= v Basil Hamed

System parameters For this example, let's assume that the parameters of the system are: (m) vehicle mass 1000 kg (b) damping coefficient 50 N.s/m (u) nominal control force 500 N State-space model Transfer function model Basil Hamed

Cruise Control: System Analysis
Performance specifications The next step is to come up with some design criteria that the compensated system should achieve. When the engine gives a 500 Newton force, the car will reach a maximum velocity of 10 m/s (22 mph), see open-loop step response section below. An automobile should be able to accelerate up to that speed in less than 5 seconds. In this application, a 10% overshoot and 2% steady-state error on the velocity are sufficient Keeping the above in mind, we have proposed the following design criteria for this problem: Rise time < 5 s Overshoot < 10% Steady-state error < 2% Basil Hamed

Open-loop step response The open-loop response of the system, without any feedback control, to a step input force of 500 Newtons is simulated in MATLAB as follows: Basil Hamed

We see that the open-loop system exhibits no overshoot or oscillations, and does reach the desired steady-state speed of 10 m/s; however, the rise time is much too slow, ~60 s. Therefore we need to design a feedback controller which speeds up the response significantly without negatively affecting the other dynamic performance metrics Basil Hamed

Cruise Control: PID Controller Design
(r) reference speed m/s We can define a PID controller in MATLAB using the transfer function directly: Kp = 1; Ki = 1; Kd = 1; s = tf('s'); C = Kp + Ki/s + Kd*s C = s^2 + s + 1 s Basil Hamed

Cruise Control: PID Alternatively, we may use MATLAB's pid controller object to generate an equivalent continuous time controller as follows: C = pid(Kp,Ki,Kd) Basil Hamed

Cruise Control: Proportional Control
The first thing to do in this problem is to find a closed-loop transfer function with a proportional control (C = Kp) added. By reducing the unity feedback block diagram, the closed-loop transfer function with a proportional controller becomes: Recall from the Introduction: PID Controller Design page, a proportional controller, Kp, decreases the rise time, which is desirable in this case. For now, use Kp equal 100 and a reference speed of 10 m/s. Create a new m-file and enter the following commands. Basil Hamed

100 T= 1000 s + 150 Basil Hamed

Basil Hamed

You can increase the proportional gain, Kp, to reduce the rise time and the steady-state error. Change the existing m-file so that Kp equals 5000 and rerun it in the MATLAB command window. You should see the following plot. Basil Hamed

The steady-state error is now essentially zero, and the rise time has been reduced substantially. However, this response is unrealistic because a real cruise control system generally can not change the speed of the vehicle from 0 to 10 m/s in less than 0.5 seconds due to power limitations of the engine and drivetrain. Actuator limitations are very frequently encountered in practice in control systems engineering, and consequently, the required control action must always be considered when proposing a new controller. We will discuss this issue much more in subsequent tutorials. The solution to this problem in this case is to choose a lower proportional gain, Kp, that will give a reasonable rise time, and add an integral controller to eliminate the steady-state error Basil Hamed

Cruise Control: PI control
The closed-loop transfer function of this cruise control system with a PI controller (C = Kp + Ki/s) is: Basil Hamed

Cruise Control: PI control
Now adjust both the proportional gain, Kp, and the integral gain, Ki, to obtain the desired response. When you adjust the integral gain, Ki, we suggest you to start with a small value since a large Ki can destabilize the response. When Kp equals 800 and Ki equals 40, the step response will look like the following: Basil Hamed

Cruise Control: PID control
For this particular example, no implementation of a derivative controller was needed to obtain the required output. However, you might want to see how to work with a PID control for the future reference. The closed-loop transfer function for this cruise control system with a PID controller (C = Kp + Ki/s + Kd*s) is: Basil Hamed

Cruise Control: PID control
Let Kp equal 1, Ki equal 1, and Kd equal 1 and enter the following commands into an new m-file. Plot the step response and adjust all of Kp, Kd, and Ki until you obtain satisfactory results. We will leave this as an exercise for you to work on. Suggestion: Usually choosing appropriate gains requires a trial and error process. The best way to attack this tedious process is to adjust one variable (Kp, Ki, or Kd) at a time and observe how changing one variable influences the system output Basil Hamed

Cruise Control: Lead/Lag controller
System parameters For this example, let's assume that the parameters of the system are (m) vehicle mass kg (b) damping coefficient N.s/m (r) reference speed m/s and the block diagram of a typical unity feedback system is shown below. Performance specifications Rise time < 5 sec Overshoot < 10% Steady-state error < 2% Basil Hamed

To reduce the steady-state error, a lag controller will be added to the system. The transfer function of the lag controller is The closed-loop transfer function (not including Kp) now becomes Finally, including the loop gain Kp, the closed-loop transfer function becomes: Basil Hamed

Lead/Lag controller Lead and Lag Compensator Design, the pole and the zero of a lag controller need to be placed close together. Also, it states that the steady-state error will be reduced by a factor of zo/po. For these reasons, let zo equal 0.3 and po equal Create a new m-file, and enter the following commands. Basil Hamed

Using the rlocfind command again, we can choose a new loop gain Kp. Enter the code [Kp,poles]=rlocfind(C_lag*P_cruise) into the command window and click on the real axis around -0.4 as shown in the following figure. Basil Hamed

After doing this, you should see the following output in the MATLAB. Select a point in the graphics window selected_point = i Kp = e+03 poles = , We can then generate the new closed-loop step response as follows As you can see, the steady-state error has been reduced to near zero. The overshoot is a result of the zero added in the lag controller. For now all of the design criteria have been met and no further iterations are needed; however, you should experiment with different zo and po values to see what their effect is on the closed-loop system response Basil Hamed

Cruise Control: Pole Placement
The equations of motion in state-space form are as follows: where (m) vehicle mass kg (b) damping coefficient N.s/m (u) nominal control force N (v) vehicle velocity where y=v is the system output Design requirements Rise time < 5 s Overshoot < 10% Steady-state error < 2% Basil Hamed

Control design using pole placement Basil Hamed

we can use a "pole placement" technique to obtain the desired output. Poles of a closed-loop system can be found from the characteristic equation: the determinant of the [sI-(A-B*K)] matrix. If the poles the system can be placed in the desired location by designing an appropriate control matrix (K), then the desired output can be obtained Basil Hamed

K= 1450 As you can see, the rise time is satisfactory, but the steady-state error is too large. Basil Hamed

Reference input Once again from the State-Space Tutorial page, a scaling factor called Nbar (the schematic is shown below) can be used to eliminate the steady-state error. We can use the rscale function to compute the scaling factor. Download it here, rscale.m. The input is already multiplied by 500, and we want the steady-state speed to be 10 m/sec, so we need to account for these factors as well Basil Hamed

As you can see, the steady-state error has been eliminated. The rise time is less than 5 seconds and the overshoot is, in fact, zero. All the design requirements are satisfied. Basil Hamed

DC Motor Position: System Modeling
Physical setup A common actuator in control systems is the DC motor. It directly provides rotary motion and, coupled with wheels or drums and cables, can provide translational motion. The electric equivalent circuit of the armature and the free-body diagram of the rotor are shown in the following figure. Basil Hamed

For this example, we will assume the following values for the physical parameters. These values were derived by experiment from an actual motor in Carnegie Mellon's undergraduate controls lab. (J) moment of inertia of the rotor E-6 kg.m^2 (b) motor viscous friction constant E-6 N.m.s (Kb) electromotive force constant V/rad/sec (Kt) motor torque constant N.m/Amp (R) electric resistance Ohm (L) electric inductance E-6H In this example, we assume that the input of the system is the voltage source (V) applied to the motor's armature, while the output is the position of the shaft (theta). The rotor and shaft are assumed to be rigid. We further assume a viscous friction model, that is, the friction torque is proportional to shaft angular velocity. Basil Hamed

Transfer Function State Space Model Basil Hamed

DC Motor Position: Design Requirements
We will want to be able to position the motor very precisely, thus the steady-state error of the motor position should be zero when given a commanded position. We will also want the steady-state error due to a constant disturbance to be zero as well. The other performance requirement is that the motor reaches its final position very quickly without excessive overshoot. In this case, we want the system to have a settling time of 40 ms and an overshoot smaller than 16%. If we simulate the reference input by a unit step input, then the motor position output should have: Settling time less than 40 milliseconds Overshoot less than 16% No steady-state error, even in presence of a step disturbance input Basil Hamed

DC Motor Position: System Analysis
Open-loop response From the above plot, we can see that when 1 volt is applied to the system the motor position grows unbounded. This is obviously at odds with the given requirements, in particular, that there be no steady-state error. The open-loop response of the system is not even stable Basil Hamed

Closed-loop response Let's now consider the closed-loop response of the system where the system schematic has the following structure. The closed-loop transfer function for the above with the controller C(s) simply set equal to 1 can be generated using the MATLAB command feedback as shown below. sys_cl = feedback(P_motor,1) Basil Hamed

sys_cl = e-12 s^ e-05 s^ s The corresponding unit step response can be generated by adding the above and following command to your m-file. The annotations for the peak response, settling time, and final value can be added to the plot from the right-click menu under Characteristics. step(sys_cl,t) Examining the above closed-loop step response, the addition of feedback has stabilized the system. In fact, the steady-state error appears to be driven to zero and the overshoot is less than 16%, though the settle time requirement is not met Basil Hamed

DC Motor Position: PID From the main problem, the open-loop transfer function of the DC Motor is given as follows. The structure of the control system has the form shown in the figure below. Basil Hamed

DC Motor Position: Proportional control
Let's first try using a proportional controller with gain ranging from 1 to 21. An array of LTI models, each with a different proportional gain, can be built using a for loop. The closed-loop transfer functions can be generated using the feedback command. Add the following code to the end of your m-file and run it in the MATLAB command window: Now let's see what the step responses look like. Add the following code to the end of your m-file and again run it in the command window. You should generate the plot shown in the figure below. Basil Hamed

DC Motor Position: P Control
Let's also consider the system's response to a step disturbance. In this case, we will assume a reference of zero and look at the how the system responds to the disturbance by itself Basil Hamed

DC Motor Position: P Control
The above plots show that the system has no steady-state error in response to the step reference by itself, no matter the choice of proportional gain Kp. This is due to the fact that the plant has an integrator, that is, the system is type 1. However, the system has significant steady-state error when the disturbance is added. Specifically, the response due to the reference and disturbance applied simultaneously is equal to the sum of the two graphs shown above. Basil Hamed

DC Motor Position: PI Control
Let's first try a PI controller to get rid of the steady-state error due to the disturbance. We will set Kp = 21 and test integral gains Ki ranging from 100 to 500. Change your m-file to the following and run in the command window. You should generate a figure like the one shown below. Basil Hamed

Basil Hamed

Now let's see what happened to the step disturbance response. Change the following commands in your m-file and re-run in the command window. You should generate a plot like the one shown in the figure below. Basil Hamed

The integral control has reduced the steady-state error to zero, even when a step disturbance is present; that was the goal for adding the integral term. For the response to the step reference, all of the reponses look similar with the amount of oscillation increasing slightly as Ki is made larger. However, the response due to the disturbance changes significantly as the integral gain Ki is changed. Specifically, the larger the value of Ki employed, the faster the error decays to zero. We will choose Ki = 500 Basil Hamed

DC Motor Position: PID Control
Adding a derivative term to the controller means that we now have all three terms of the PID controller. We will investigate derivative gains Kd ranging from 0.05 to Go back to the m-file and make the following changes. Running the altered m-file will generate a graph like the one shown below. Basil Hamed

Basil Hamed

Let's see what happened to the step disturbance response, change the following commands in your m-file and re-run at the command line. Basil Hamed

It looks like when Kd = 0.15, we can meet our design requirements. To determine the precise characteristics of the step response you can use the right-click menu of the step response plot, or you can use the MATLAB command stepinfo as shown below. stepinfo(sys_cl(:,:,2)) ans = RiseTime: Settling Time: Overshoot: From the above, we see that the response to a step reference has a settling time of roughly 34ms (< 40 ms), overshoot of 12% (< 16%), and no steady-state error. Additionally, the step disturbance response also has no steady-state error. So now we know that if we use a PID controller with Kp = 21, Ki = 500, and Kd = 0.15, Basil Hamed

DC Motor Position: Pole Placement Design
From the main problem, the dynamic equations in state-space form are given below. Since all of the state variables in our problem are very easy to measure (simply add an ammeter for current, a tachometer for speed, and a potentiometer for position), we can design a full-state feedback controller for the system without worrying about having to add an observer. The control law for a full-state feedback system has the form u = r - Kc x. The associated block diagram is given below. Basil Hamed

Recall that the characteristic polynomial for this closed-loop system is the determinant of sI-(A-B*Kc) where s is the Laplace variable. Since the matrices A and B*Kc are both 3x3 matrices, there should be 3 poles for the system. This fact can be verified with the MATLAB command order. If the given system is controllable, then by designing a full state-feedback controller we can move these three poles anywhere we'd like. Whether the given system is controllable or not can be determined by checking the rank of the controllability matrix [ B AB A^2B ...]. Basil Hamed

The MATLAB command ctrb constructs the controllability matrix given A and B. Additionally, the command rank determines the rank of a given matrix, though it can be numerically unreliable sys_order = 3 determinant = e+24 From the above, we know that our system is controllable since the determinant of the controllability matrix is not zero and hence we can place the system's closed-loop poles anywhere in the s-plane. We will first place the poles at -200, i and i Basil Hamed

Kc = Basil Hamed

Disturbance response In order to observe the system's disturbance response, we must provide the proper input to the system. In this case, a disturbance is physically a load torque that acts on the inertia of the motor. This load torque acts as an additive term in the second state equation (which gets divided by J, as do all the other terms in this equation). We can simulate this simply by modifying our closed-loop input matrix, B, to have a 1 / J in the second row assuming that our current input is only the disturbance. Add the following lines to your m-file and re-run. Basil Hamed

Notice that the error due to the step disturbance is non-zero. Therefore, this will also need to be compensated for. Basil Hamed

Adding integral action From prior examples, we know that if we put an extra integrator in series with the plant it can remove the steady-state error due to a step reference. If the integrator comes before the injection of the disturbance, it will also cancel a step disturbance input in steady state. This changes our control structure so that it now resembles the block diagram shown in the following figure Basil Hamed

These equations represent the dynamics of the system before the loop is closed. We will refer to the system matrices in this equation that are augmented with the additional integrator state as Aa, Ba, Ca, and Da. The vector multiplying the reference input r will be referred to as Br. We will refer to the state vector of the augmented system as xa. Note that the reference, r, does not affect the states (except the integrator state) or the output of the plant Basil Hamed

In order to find the closed-loop equations, we have to look at how the input, u, affects the plant. In this case, it affects the system in exactly the same manner as in the unaugmented equations except now u = -Kc x - Ki w. We can also rewrite this in terms of our augmented state as u = -Ka xa where Ka = [ Kc Ki ]. Substituting this u into the equations above provides the following closed-loop equations In the above, the integral of the error will be fed back, and will result in the steady-state error being reduced to zero. Now we must redesign our controller to account for the augmented state vector. Since we need to place each pole of the system, we will place the pole associated with the additional integrator state at -300, which will be faster than the other poles Basil Hamed

Add the following lines to your m-file which reflect the closed-loop equations presented above. Note that since the closed-loop transition matrix Aa-Ba*Ka depends on Ba, it will be used in the place command rather than Br. Running your m-file will then produce the plot shown below. Basil Hamed

To observe the disturbance response, we use a similar approach to that used without the integral action. We can see that all of the design specifications are close to being met by this controller. The settle time may be a little large, but by placing the closed-loop poles a little farther to the left in the complex s-plane, this requirement can also be met. Basil Hamed

Ball & Beam: Lead Controller Design
, Ball & Beam: Lead Controller Design A ball is placed on a beam, see figure below, where it is allowed to roll with 1 degree of freedom along the length of the beam. A lever arm is attached to the beam at one end and a servo gear at the other. As the servo gear turns by an angle $\theta$, the lever changes the angle of the beam by $\alpha$. When the angle is changed from the horizontal position, gravity causes the ball to roll along the beam. A controller will be designed for this system so that the ball's position can be manipulated. Basil Hamed

System parameters For this problem, we will assume that the ball rolls without slipping and friction between the beam and ball is negligible. The constants and variables for this example are defined as follows: (m) mass of the ball 0.11 kg (R) radius of the ball (d) lever arm offset 0.03 m (g) gravitational acceleration 9.8 m/s^2 (L) length of the beam 1.0 m (J) ball's moment of inertia 9.99e-6 kg.m^2 (r) ball position coordinate (alpha) beam angle coordinate (theta) servo gear angle Design criteria Settling time < 3 seconds Overshoot < 5%

Transfer Function: Open a new m-file and add the following code to create a transfer function model in MATLAB. Basil Hamed

Open-loop step response Now, we would like to observe the ball's response to a step input on the motor servo gear angle theta (1-radian step). To do this you will need to add the following line to your m-file. step(P_ball) From this plot it is clear that the system is unstable in open-loop causing the ball to roll right off the end of the beam. Therefore, some method of controlling the ball's position in this system is required Basil Hamed

A first order lead compensator tends to shift the root locus into the left-hand plane. For a more detailed description of lead compensators refer to the Lead & Lag Compensator Design. A lead compensator has the form given below: where, the magnitude of z0 is less than the magnitude of p0. Now, let us add the controller to the plant and view the root locus. We will position the zero near the origin to cancel out one of the poles. The pole of our compensator will be placed to the left of the origin to pull the root locus further into the left-hand plane. Add the following lines of MATLAB code to your m-file. Run your m-file in the MATLAB command window and you should see the following: Basil Hamed

Selecting the gain Now that we have moved the root locus into the left-hand plane, we may select a gain that will satisfy our design requirements. We can use the rlocfind command to help us do this. Add the code [k,poles]=rlocfind(C*P_ball) onto the end of your m-file. Basil Hamed

Then go to the plot and select a point near those indicated by the cross marks on the plot below. After doing this, you should see the following output in the MATLAB command window. Basil Hamed

selected_point = i k = poles = i i Plotting the closed-loop response This value of Kc can be put into the system and the closed-loop response to a step input of 0.25 m can be obtained. Add the following lines to your m-file to perform this analysis. Run your m-file and select a point on the root locus similar to the selected point above. The step response should look like the following. Basil Hamed

From this plot we see that when a 0.25m step input is given to the system both the settling time and percent overshoot design criteria are met. Note: A design problem does not necessarily have a unique answer. Using this method (or any other) may result in many different compensators. Basil Hamed

Control Systems Project Basil Hamed.

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