AUTOMATIC CONTROL SYSTEM BY BC KUO PDF
BENJAMIN C. KUO. Automatic. Control Systems. THIRD EDITION. 2 Control Systems. Automatic EHER bo-. CO-O. EDITION. THIRD. PRENTICE. HALL. So lu t io ns M an ua l Automatic Control Systems, 9th Edition A Chapter 2 Solution ns Golnarraghi, Kuo C Chapter 2 2 2 1 (a) 10; Poless: s = 0, 0, 1, (b) Poles: s. Automatic Control Systems by Benjamin C. Kuo - Ebook download as PDF File . pdf), Text File .txt) or read book online.
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Automatic Control. Systems. FARID GOLNARAGHI. Simon Fraser University. BENJAMIN C. KUO. University of Illinois at Urbana-Champaign. WILEY. Download Automatic Control Systems By Benjamin C. Kuo, Farid Golnaraghi – Automatic Control Systems provides engineers with a fresh new controls book. PDF | On Jan 1, , B C Kuo and others published Automatic Control System.
The heat flow-out can be defined as: The displacement of actuator is changing proportionally with the temperature differences: The heat flow through the sides is: Substituting equation 3 , 4 , 5 and 6 into equation 7 and 8 gives the model of the system. The changes if the temperature of heat sink is supposed to be zero, then: The state model of the system is given by substituting equations 2 , 3 , and 6 into these equations give. The behaviour of the valve in this system can be written as: The state equations can be rewritten by substituting P2, Pv, Ps and Q2-v from other equations.
About Automatic Control Systems By Benjamin C. Kuo, Farid Golnaraghi
The dynamic for the well can be written as two pipes separating by mass m: For the walking beam: According to the equation of angular motion: Thus, N 9. See Chapter 6. Thus e ss f. Since the system is linear, then the effect of X s is the summation of effect of each individual input. Thus Z n Thus, Z n Thus Kt 0. Z n The coefficients are ordered in descending powers.
Other parts are the same. Therefore, the equation of motion is rewritten as: For description refer to Chapter 9. Therefore the natural frequency range in the region shown is around 2. Overshoot increases with K. No need to adjust parameters. For the sake simplicity, this problem we assume the control force f t is applied in parallel to the spring K and damper B. We will not concern the details of what actuator or sensors are used. Lets look at Figure and equations and This is now an underdamped system.
The process is the same for parts b, c and d. Use Example as a guide. Use Acsys to do that as demonstrated in this chapter problems. Also Chapter 2 has many examples. You may look at the root locus of the forward path transfer function to get a better perspective. For a better design, and to meet rise time criterion, use Example 5- Open loop speed response using SIMLab: The form of response is like the one that we expected; a second order system response with overshoot and oscillation.
Considering an amplifier gain of 2 and K b 0. To find the above response the systems parameters are extracted from: Study of the effect of viscous friction: The above figure is plotted for three different friction coefficients 0, 0. As seen in figure, two important effects are observed as the viscous coefficient is increased. First, the final steady state velocity is decreased and second the response has less oscillation.
Both of these effects could be predicted from Eq. Additional load inertia effect: As the overall inertia of the system is increased by 0. The above results are plotted for 5 V armature input. Study of the effect of disturbance: As seen, the effect of disturbance on the speed of open loop system is like the effect of higher viscous friction and caused to decrease the steady state value of speed.
Using speed response to estimate motor and load inertia: Using first order model we are able to identify system parameters based on unit step response of the system.
The final value of the speed can be read from the curve and it is 8. Considering Eq. Based on this time and energy conservation principle and knowing the rest of parameters we are able to calculate B.
However, this method of identification gives us limited information about the system parameters and we need to measure some parameters directly from motor such as Ra , K m , K b and so on. So far, no current or voltage saturation limit is considered for all simulations using SIMLab software.
Open loop speed response using Virtual Lab: Then the system time constant is obviously different and it can be identified from open loop response. Identifying the system based on open loop response: Open loop response of the motor to a unit step input voltage is plotted in above figure. Using the definition of time constant and final value of the system, a first order model can be found as: In both experiments 9 and 10, no saturation considered for voltage and current in SIMLab software.
If we use the calculation of phase and magnitude in both SIMLab and Virtual Lab we will find that as input frequency increases the magnitude of the output decreases and phase lag increases. Because of existing saturations this phenomenon is more sever in the Virtual Lab experiment In this experiments we observe that M 0.
Apply step inputs SIMLab In this section no saturation is considered either for current or for voltage. The same values selected for closed loop speed control but as seen in the figure the final value of speeds stayed the same for both cases. As seen, the effect of disturbance on the speed of closed loop system is not substantial like the one on the open loop system in part 5, and again it is shown the robustness of closed loop system against disturbance.
Also, to study the effects of conversion factor see below figure, which is plotted for two different C. Apply step inputs Virtual Lab a. Comments on Eq. In experiments 19 through 21 we observe an under damp response of a second order system. According to the equation, as the proportional gain increases, the damped frequency must be increased and this fact is verified in experiments 19 through Experiments16 through 18 exhibits an over damped second order system responses.
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In following, we repeat parts 16 and 18 using Virtual Lab: Study the effect of integral gain of 5: In order to find the current of the motor, the motor constant has to be separated from the electrical component of the motor. The response of the motor when 5V of step input is applied is: This is the time constant of the motor. The current When Jm is increased by a factor of 2, it takes 0.
This means that the time constant has been doubled. The motor achieves this speed 0. It does not change Part 2: It does not change. This is the same as problem It does not change d As TL increases in magnitude, the steady state velocity decreases and steady state current increases; however, the time constant does not change in all three cases.
If there is saturation, the rise time does not decrease as much as it without saturation. Also, if there is saturation and Kp value is too high, chattering phenomenon may appear. The torque generated by the motor is 0. It takes 0. In order to get the same result as Problem , the Kp value has to increase by a factor of 5.
Automatic Control Systems by Farid Golnaraghi, benjamin C.kuo
It has less steady state error and a faster rise time than Problem , but has larger overshoot. As the proportional gain gets higher, the motor has a faster response time and lower steady state error, but if it the gain is too high, the motor overshoot increases. If the system allows for overshoot, the best proportional gain is dependant on how much overshoot the system can have. As the derivative gain increases, overshoot decreases, but rise time increases. Also, the amplitude of the output starts to decrease when the frequency increases above 0.
Repeat the process for other frequency values and use the calculated gain and phase values to plot the frequency response of the system. Enter the Step Input and Controller Gain values by double clicking on their respective blocks.
Observe the steady state value change with K. Simulate the response and show the desire variables. There are no asymptotes, since the number of zeros of P s and Q s are equal. The angles of asymptotes: Departure angles from: Breakaway Points: Root Locus 15 10 Imaginary Axis 5 0 -5 Root Locus 1. Asymptotes angle: Therefore, Departure angle from: Asymptotes angles: The break away points: These intersection points are shown for part a where the corresponding gain is Root Locus 15 10 System: Angle of asymptotes: Then d and Poles: Angles of asymptotes: Root Locus 60 40 System: Root Locus 0.
Root Locus 2. Root locus diagram, part a: Imaginary Axis locus diagram.
Root Locus 5 4 3 Imaginary Axis 2 1 0 -1 -2 -3 -4 -5 -4 Breakaway points: Root Locus Imaginary Axis 0 Real Axis Root locus diagram, part b: Root Locus 10 8 6 Imaginary Axis 4 2 0 -2 -4 -6 -8 Answers to True and False Review Questions: Bode Diagram 20 Magnitude dB 0 System: To change the crossover frequency requires adding gain as: As , if GH is rearranged as: Bode diagram: L j Nyquist Diagram 1.
Nyquist Diagram 0. Nyquist diagram is added to the results of sisotool.
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Higher values of K resulted in unstable Nyquist diagram. Following is the Nyquist diagram at margin of stability. Part a , Nyquist at margin of stability: Nyquist Diagram 1 0. Using Routh criterion, the coefficient table is as follows: Location of poles in root locus diagram of the second figure will also verify that. Nyquist Diagram 20 15 10 Imaginary Axis 5 0 -5 -1 Inf Freq: Inf Stable loop -1 -1 The result and approach is similar to part a , a sample of Nyquist diagram is presented for his case as follows: Nyquist Diagram 15 10 Imaginary Axis 5 0 -5 -1 Using MATLAB sisotool, the transfer function gain can be iteratively changed in order to obtain different phase margins.
Following two figures illustrate the sisotool and Nyquist results at margin of stability for part a.
Similar methodology applied as in part a. Following diagrams correspond to margin of stability: In this case, we need 0 CCW encirclements. See alternative solution to Root Locus 30 20 Imaginary Axis 10 0 0 10 20 Real Axis Part c , Gain and frequency that instability occurs: The Part e: Nyquist Diagram 8 6 4 Imaginary Axis 2 0 -2 -4 -6 -8 -1 Both of these values are consistent with the results of part a from sisotool. CL Gain: Use G to obtain the gain-phase plots and Gm and Pm. Use the Bode plot to graphically obtain Mr.
Sisotool Result shows that by changing K between 0 and inf. Therefore, the system is stable for all positive K. Inf 0 Stable loop 0 0 0 P. PD controller design: The open-loop transfer function of a system is: The open loop transfer function of a system is: The transfer functions are generated and imported in sisotool as in According to the requirements the gain must be greater than must be less than or 0.
The Lead compensator T. To obtain a slightly higher PM, lead compensator zero was re-tuned, where the zero is pulled closer to imaginary axis from -5 to To include some integral action, Ki is set to 1.
The open loop bode shows as PM of By try and error, 2 compensators a double lead compensator each with phase lead of 55 deg was found suitable. Considering the change in cross over frequency after applying the lead filters, overall, a PM of 52 deg was obtained as seen in the bode diagram of compensated loop: Double Lead filter design: This was due to the shape of phase diagram affected by integral action i.
Bode diagram of compensated loop transfer function can be observed in the following figure, showing a PM pf The bode of the loop transfer function shows a PM of deg at 3. Considering the change in cross over frequency after applying the lead filters, overall, a PM of Bode diagram of compensated loop can be observed in the following figure, showing a PM pf In order to add some integral action, First, the bode plot of the Loop transfer function is obtained demonstrating a PM of By try and error, a double lead compensator, each with phase lead of 48 deg was found suitable.
The effect is similar to adding a Zero at K K. At K results: The root locus diagram can be seen as: The zero and the gain of the PI controller needs to be designed. The place of the zero and the overall gain is iteratively changed in the MATLAB sisotool to achieve the crossover frequency of Now required to achieve PM 45 10 10,then PM 1 As a result, 91o phase lead is The crossover frequency is Therefore, This pole is usually placed at least 1 decade lower frequency wise than the slowest existing poles of the system.
In this case, since KI K 0. Use the transfer function for the open-loop system, and a series PID compensator in a unity feedback system. Assume a small electric time constant or small inductance and simplify to Equation The resulting zero in the right hand plane is troubling. Looking at the TF poles, it seems prudent to design the controller by placing its zero farther to LHS of the s-plane.
Now, we can choose pole p far enough from pole dominant of second order. Mathematical Model: Draw free body diagrams Assume both xc and xw are positive and are measured from equilibrium.
Refer to Chapter 4 problems for derivation details. For a better design, and to meet rise time criterion, use Example and Chapter 9 PD design examples. You get the next window. Enter the A,B,C, and D values. Note C must be entered here and must have the same number of columns as A. We us [1,1] arbitrarily as it will not affect the eigenvalues.
Characteristic Polynomial: Transfer function: The A matrix is: If The solution s for x is , then or 0 , theerefore: The 2nd order desired characteristic equation of the 2 2 0 1 On the other hand: For overshoot of 4. First convert the transfer function to a unity feedback system to make compatible to the format used in the Control toolbox. You can adjust K values to obtain alternative results by repeating this process.
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Jump Linear Systems in Automatic Control.This is the the matrix condition of complete state controllability. A are the eigenvalues of the 2.
There are many situations where several variables are to be controlled simulmultivariabk taneously by a number of inputs. Ai 1n-l A2 A3 ln Matrix Determinant An with n rows array of numbers or elements columns. They are of the same order. Both of these effects could be predicted from Eq. Designed to excel as both a text for students and a self-teaching reference for the professional engineer, Automatic Control Systems will be the one resource you keep at your side for years to come as you meet the challenges and rewards of real-world control systems design.
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