CONTROL SYSTEMS ENGINEERING BY NAGRATH AND GOPAL EBOOK
Control Systems Engineering By I.J. Nagrath, M. Gopal provides an integrated treatment of continuous-time and discrete-time systems. It emphasises the. this is a recommendation for you >> CONTROL SYSTEMS: ENGINEERING, 5th Edition. I. J. Nagrath (Author), Madan Gopal (Author). out of 5 This item:Control Systems Engineering (Old Edition) by I. J. Nagrath Paperback Rs. In stock.
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[PDF] Control Systems Engineering By I.J. Nagrath, M. Gopal Book Free little serious about your studies, you should never consider eBooks/Books in PDF. Title, Textbook Of Control Systems Engineering (Vtu). Authors, I. J. Nagrath, Madan Gopal. Publisher, New Age International, ISBN, CONTROL SYSTEMS: ENGINEERING, 5th Edition I. J. Nagrath, M. Gopal. This book frequency domain techniques of the analysis and design of control systems have been discussed at length. Consisting of . Gopal ebook PDF download.
The final value theorem is applicable: The final value theorem is not applicable; the function sY s has a pole on the jM-axis. Newtons For this value of force, we get from Fig. R ; this gives Q m 17 It is easy to calculate with long division that 1 2 1 2. G 1 and Q 1 are obtained from the following energy-balance equations at steady state:
By magnitude criterion, the value of K at this point is found to be 1. It turns out that real part of complex pair away from ju-axis is approximately four times as large as that of the pair near ju-axis. Therefore, the transient response term due to the pair away from the ju-axis will decay much more rapidly than the transient response term due to the pair near ju-axis.
By magnitude criterion, the value of K at this point is found to be The greatest value of K that can be used before continuous oscillations occur is the value of K at the point of intersection with the ju-axis.
Draw a line from the origin tangential to the circular part of the locus. This line corresponds to damping ratio for maximum oscillatory response.
Complex-root branches form a circle. By determining a sufficient number of points that satisfy the angle criterion, it can be found that the root locus is a circle with centre at 1. For minimum steady- state error we should select larger value of K. Complex-conjugate roots can then be found by long division. The value of K at this point, obtained by magnitude criterion, is 1.
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As per the result of Problem 7. Hence the desired value of velocity feedback gain K t is 0. This is not the case. In fact the closed-loop system does not have a zero at the origin: The value of K at this point, obtained by magnitude criterion, is 4.
The value of K at this point is We should therefore never attempt to cancel poles in the right-half plane, since any inexact cancellation will result in an unstable closed-loop system.
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The value of K at s d is D s improves transient response considerably, and there is no effect on steady-state performance. The gain of the compensated system, obtained by magnitude criterion, is The K v of the compensated system is less than the desired value of Therefore, we must repeat the design procedure for a second choice of the desired root. The value of K, found by magnitude criterion, is The value of is 0. The velocity error constant is 0.
The desired dominant closed-loop poles are 0. The lag compensator http: The angle contribution of this compensator at 0. Since the angle contribution is not small, there will be a small change in the new root locus near the desired dominant closed-loop poles.
The undamped natural frequency reduces to 0. This implies that the transient response of the compensated system is slower than the original.
The gain at 0. The value of K is found as The angle contribution of this compensator at 3. Clearly the roots of the system when K v requirement is satisfied are a long way from satisfying the requirement. It will be difficult to bring the dominant http: Therefore, we will attempt to satisfy the K v and requirements by a lag compensator. For this choice, we find that the angle contribution of the compensator at 2. The gain K is 2. 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. 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.
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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. 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.
Solution Manual - Control Systems by Gopal
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. 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.
The design approach may be that of achieving a portion of the desired phase margin improvement by each compensator stage. We first add a single-stage lead compensator that will provide a phase lead of about Notice that the single-stage lead compensator not only improves the phase margin, but also reduces the slope of the phase curve of the gain crossover frequency. Is our service is satisfied, Anything want to say? Cancel reply.
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