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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. Sistemas de control para ingenieria 3 edicion Norman Nise solucionario. Jordan Giraldo. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any from or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections or of the United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Rosewood Drive, Danvers, MA , , fax To order books please call 1 Using Table 2.

First find the mechanical constants. Substituting this value of vo into Eq. First derive the state equations for the transfer function without zeros. Since poles are at — The second-order approximation is valid, since the dominant poles have a real part of —2 and the higher-order pole is at —15, i. The second-order approximation is not valid, since the dominant poles have a real part of —1 and the higher-order pole is at —4, i.

Therefore, a second-order approximation is not valid. Therefore, a second-order approximation is valid. See Figure 4. Then, push to the left past the s pickoff point. Label nodes.

First find the eigenvalues. Make a Routh table. We encounter a row of zeros on the s 3 row. Replacing the row of zeros with the coefficients of the derivative yields the s 3 row. We also encounter a zero in the first column at the s 2 row. Thus the even polynomial 4th order has one right half-plane pole, one left half-plane pole, and 2 imaginary axis poles. From the top of the table down to the even polynomial yields one sign change.

The total for the system is two right half-plane poles, two left half-plane poles, and 2 imaginary poles. Thus, there are two rhp poles and one lhp pole. First check stability. Therefore, the system is stable. Stability also could be checked via Routh-Hurwitz using the denominator of T s. Instability could also be determined using the Routh-Hurwitz criteria on the denominator of T s.

Since the system is unstable, calculations about steady- state error cannot be made. System is stable for positive K. System is Type 0. First draw the vectors.

Since the angle is , the point is on the root locus. First, find the asymptotes. From which we evaluate the imaginary axis crossing at Searching for the minimum gain to the left of —2 on the real axis yields —7 at a gain of Thus the break-in point is at —7. Searching the real axis to the left of —6 for the point whose gain is Comparing this value to the real part of the dominant pole, The second-order approximation is not valid.

Find the closed-loop transfer function and put it the form that yields pi as the root locus variable. The following shows the root locus.

For the pole at — Compensator zero should be 20x further to the left than the compensator pole. Thus, for the compensated Thus the design point is 3 Thus, the compensator pole must contribute — A higher-order closed-loop pole is found to be at — This pole may not be close enough to the closed-loop zero at — Thus, we should simulate the system to be sure the design requirements have been met.

For a 10x improvement in steady- state error, Kv must be 8. Thus, the lag compensator zero should be 9. Summarizing the forward path with plant, compensator, and gain yields It would be advisable to simulate the system to see if there is indeed pole-zero cancellation. The configuration for the system is shown in the figure below. An active PID controller must be used. Using these values along with Eqs. Nyquist Diagrams 0. Thus, the resultant rotates — while its magnitude goes to zero.

The result is shown below. From Skill-Assessment Exercise Solution The phase angle is at a frequency of At this frequency the gain is — Therefore, the gain margin is The frequency is We calculate the phase margin to be — M and N circles, and on the b.

Nichols chart. Nichols Charts 0 dB 0. The open-loop frequency response plot goes through zero dB at a frequency of 9.

Hence, the phase margin is — Therefore, the system is Type 2. Taking the square root of the positive root, we find the 0 dB frequency to be 3. Therefore the phase margin is — Thus, the system is unstable. Drawing judicially selected slopes on the magnitude and phase plot as shown below yields a first estimate. We see first-order zero behavior on the magnitude and phase plots with a break frequency of about 5.

Thus, The magnitude at this frequency is 5. The Bode plot for this gain is shown below. Adding to compensate for the phase angle contribution of the lag, we use The frequency at which this phase occurs is At this frequency the magnitude plot must go through zero dB. Presently, the magnitude plot is Therefore draw the high frequency asymptote of the lag compensator at — Insert a break at 0.

The frequency of intersection will be the low frequency break or 0. The uncompensated Bode plot for this gain is shown below. We find that when the magnitude plot crosses 0 dB, the phase angle is We first find the desired characteristic equation.

We conclude that the system is controllable. First check controllability.


Engenharia de Sistemas de Controle, 6ª Edição.pdf

Guided missiles, automatic gain control in radio receivers, satellite tracking antenna. Yes - power gain, remote control, parame ter conversion; No - Expense, complexity. Motor, low pass filter, inertia supported between two bearings. Closed-loop systems compensate for disturbances by measuring the response, comparing it to.


Control Systems Engineering Sixth-edition-summits By Norman S. Nise


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