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# Control system engineering solution manual pdf

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SOLUTION MANUAL Apago PDF Enhancer Solutions to Problems Student companion website Founded in , John Wiley & Sons, Inc. has been a. O N EIntroduction ANSWERS TO REVIEW QUESTIONS 1. Guided missiles, automatic gain control in radio receivers, satellite. Guided missiles, automatic gain control in radio receivers, satellite tracking of Solutions to Design Problems [s2+D(5)2s+ 1 4 (Control.

The G jwmax for the compensator is: The gain K is 2. Therefore, the compensator's zero must contribute FabricioJoseOliveira2 Obrigado pela ajuda!! The lead-compensated step response is shown below.

Revisiting Example 7. Case II: Based on this information, a rough sketch of the Nyquist plot can easily be made. Based on this information, a rough sketch of Nyquist plot can easily be made. With this information, Nyquist plot for the given system can easily be drawn. The Nyquist criterion is not applicable. With this information, a rough sketch of Nyquist plot can easily be made refer Fig. From this plot we see that the closed-loop system is table for all K.

From this plot we see that the closed-loop system is unstable for all K. The critical value of K is obtained by letting G j 2. As gain K is varied, we can visualize the Nyquist plot in this figure expanding increased gain or shrinking decreased gain like a balloon.

This is true if 1. Therefore, for stability 0. With this information, Nyquist plot can easily be constructed. From the Nyquist plot we find that the critical point is encircled twice in ccw if 0. For this range of K, the closed loop system is stable. The figure given below illustrates this result. A rough sketch of the Nyquist plot is shown in figure below. It is observed that the system is unstable for all values of K. At this point, the phase of the polar plot of e j D M J is found to be Therefore, http: The resulting Nyquist plot encircles the critical point once in clockwise direction and once in counter- clockwise direction.

Because the phase curve never reaches line, the gain margin is infinity.

## Solution Manual - Control Systems by Gopal

Therefore, the phase margin is The phase shift at this frequency is Therefore the phase margin is Phase never reaches line; the gain margin is infinity. The time-delay factor e 0. Compensated magnitude plot can easily be obtained by applying corrections. Hence the closed-loop system will be stable if the frequency response has a gain less than unity when the phase is This means that the gain can be increased by Thus the critical value of K for stability is 1.

## Control Systems Engineering Solution Manual | lesforgesdessalles.info

The relative stability of the system reduces due to the presence of dead-time. Therefore, 7 4 The system is therefore type-0; Ks is a factor of the transfer function with 20 log 0.

The transfer function http: From the phase curve and the asymptotic magnitude curve, we obtain. The magnitude of G jM H jM at this frequency is approximately 3.

The gain must be reduced by a factor of 3. It means that magnitude curve must be lowered by From the Bode plot, we find that lowering the magnitude plot by For this damping, required phase margin is This is achieved by reducing the gain by 5 dB. Therefore, the gain should be increased by 4. M and M g are satisfied if we increase the gain to the limit of zero GM with a phase margin Assuming a second-order approximation,. Therefore phase margin of the system without dead-time is 40; and with dead-time added, the phase margin becomes 3.

We find that with the dead-time added, the system becomes unstable. Therefore, the system gain must be reduced in order to provide a reasonable phase margin. We find from the Bode plot that in order to provide a phase margin of 30, the gain would have to be decreased by 5 dB, i.

We find that the dead-time necessitates the reduction in loop gain is order to obtain a stable response. The cost of stability is the resulting increase in the steady-state error of the system as the loop gain is reduced. The bandwidth of a system is defined as the frequency at which the magnitude of the closed-loop frequency response is 0. This frequency is found to be 3. The M r relationship has been derived for standard second-order systems with zero-frequency closed-loop gain equal to unity. The answer is based on the assumption that zero-frequency closed-loop gain for the system under consideration is unity. The following points are worth noting in the use of Nichols chart. One must do a little interpolation. We have made this assumption for the system under consideration. On a linear scale graph sheet, dB vs phase cure of the open-loop frequency response is plotted with data coming from Bode plot. On the same graph sheet, the 3dB contour using the given data is plotted.

Shifting the dB vs phase curve down by 5. Note that the 3dB bandwidth definition given by Eqn.

## Control Systems Engineering Solutions Manual

The given system does not satisfy this requirement. The answer, therefore, is an approximation of the bandwidth. For the gain margin to be 20 dB, the magnitude plot is to be brought down by 6 dB. This is achieved if magnitude plot is brought down by 7 dB. The corresponding value of M r read off from Bode plot is 0. The value of bandwidth M b is the frequency at which the dB-phase plot intersects the 3dB contour. Phase margin of the gain-compensated system is From the new Bode plot we find that the gain must be raised by 1.

The increase in M g implies that lead compensation increases speed of response. In this particularly simple example, specifications could be met by either compensation. In more realistic situations, there are additional performance specifications such as bandwidth and there are constraints on loop gain. Had there been additional specifications and constraints, it would have influenced the choice of compensator lead or lag. The settling time and peak overshoot requirements on performance may be translated to the following equivalent specifications: The phase margin of the uncompensated system is 0 because the double integration results in a constant phase lag.

Therefore, we must add a 45 phase lead at the gain crossover frequency of the compensated magnitude curve. The phase margin of the compensated system is found to be From the Nichols chart analysis of the compensated and uncompensated systems, we find that the lead compensator has increased the bandwidth from 9. To realize a phase margin of 45, the gain crossover frequency should be moved to M g where the phase angle of the uncompensated system is: The attenuation necessary to cause M g to be the new gain crossover frequency is 20 dB.

As a final check, we numerically evaluate the phase margin and the bandwidth of the compensated system. It is found that that. The reader should, in fact, try a single-stage lead compensator. For the present system, in which the desired K v is , a phase lead of more than 85 is required. For phase leads greater than 60, it is advisable to use two or more cascaded stages of lead compensation refer Fig. Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available.

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## Control Systems Engineering, 4th Edition - Solutions Manual

Bookmark it to easily review again before an exam. The best part? As a Chegg Study subscriber, you can view available interactive solutions manuals for each of your classes for one low monthly price. Why buy extra books when you can get all the homework help you need in one place? You bet! Just post a question you need help with, and one of our experts will provide a custom solution. AlonsoRuiz20 por fin el puto solucionario, muchas gracias hermano! Show More. Otani Atsushi. Seunghyun Lee. Sadman Alam. No Downloads. Views Total views.

Actions Shares. Embeds 0 No embeds. No notes for slide. Guided missiles, automatic gain control in radio receivers, satellite tracking antenna 2. Yes - power gain, remote control, parameter conversion; No - Expense, complexity 3. Motor, low pass filter, inertia supported between two bearings 4.

Closed-loop systems compensate for disturbances by measuring the response, comparing it to the input response the desired output , and then correcting the output response. Under the condition that the feedback element is other than unity 6. Actuating signal 7.

Multiple subsystems can time share the controller. Any adjustments to the controller can be implemented with simply software changes. Stability, transient response, and steady-state error 9.

Steady-state, transient It follows a growing transient response until the steady-state response is no longer visible. The system will either destroy itself, reach an equilibrium state because of saturation in driving amplifiers, or hit limit stops. Transient response True Transfer function, state-space, differential equations Five turns yields 50 v. Introduction 2. Solutions to Problems 3 6. Introduction 8. Solutions to Problems 5 9. Introduction b.

The solution is 7. The final solution is b. Solutions to Problems 9 Solving for the arbitrary constants, x.

Also, the derivative of the solution is dx dt Solving for the arbitrary constants, x. The final solution is c. The final solution is Solutions to Problems 11 Transfer Functions Finding each transfer function: But, the resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since the voltage across the inductor is zero at dc.

Therefore, multiplying the transfer function by 6. Transfer function 2. Linear time-invariant 3. Laplace 4. Initial conditions are zero 6.

Equations of motion 7. Free body diagram 8. There are direct analogies between the electrical variables and components and the mechanical variables and components. Mechanical advantage for rotating systems Armature inertia, armature damping, load inertia, load damping Multiply the transfer function by the gear ratio relating armature position to load position.

Solving for X s , The transfer function is C s R s The block diagram is then, Solutions to Problems 19 Polynomial Transfer function: Transfer function: Modeling in the Frequency Domain Thevenizing, Writing the nodal equations yields, Solutions to Problems 29 b. Let X1 s be the displacement of the left member of the spring and X3 s be the displacement of the mass. Writing the equations of motion Reflecting impedances and applied torque to respective sides of the spring yields the following equivalent circuit: Reflecting impedances and applied torque to respective sides of the viscous damper yields the following equivalent circuit: Reflect all impedances on the right to the viscous damper and reflect all impedances and torques on the left to the spring and obtain the following equivalent circuit: Reflect impedances to the left of J5 to J5 and obtain the following equivalent circuit: Draw the freebody diagrams, Draw a freebody diagram of the translational system and the rotating member connected to the translational system.

The parameters are: The following torque-speed curve can be drawn from the data given: Thus, From Eqs. Thus, the equivalent inertia at the load is 9, and the equivalent damping at the load is 9.

Reflecting these back to the armature, yields an equivalent inertia of 9 4 and an equivalent damping of 9 4.