BENJAMIN KUO OTOMATIK KONTROL SISTEMLERI PDF

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To browse Academia. Skip to main content. By using our site, you agree to our collection of information through the use of cookies. To learn more, view our Privacy Policy. Log In Sign Up. Automatic Control Systems by Benjamin C. Kuo Solution. HwiYoung Lee. Hence, 0. Then increase resolution if desired. Then plot the time response by selecting the parameter values. Or use toolbox Stability of a linear system is independent of its initial conditions. Find the time response and apply the initial conditions.

Converting the Cartesian components with polar inertial components and replace x, y, z by T, D, and L. M-m is the mass of the block alone. Also, the current is changing with the rate of changes in displacement. Substituting equation 3 , 4 , 5 and 6 into equation 7 and 8 gives the model of the system. The state model of the system is given by substituting equations 2 , 3 , and 6 into these equations give.

The state equations can be rewritten by substituting P2, Pv, Ps and Q2-v from other equations. 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. 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: a. Considering an amplifier gain of 2 and K b 0. 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. 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: a. 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: 9 G s , 0. 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.

Additional load inertia effect: a. 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. The nonlinearities such as friction and saturation cause these differences. For example, the chattering phenomenon and flatness of the response at the beginning can be considered as some results of nonlinear elements in Virtual Lab software.

Comparing this plot with the previous one without integral gain, results in less steady state error for the case of controller with integral part. As seen in the figure, for higher proportional gains the effect of saturations appears by reducing the frequency and damping property of the system.

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.

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: a The steady state speed: This is the time constant of the motor. The current d 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.

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. 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.

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