Lecture 10 Feedback Control Systems President UniversityErwin SitompulFCS 10/1 Dr.-Ing. Erwin Sitompul President University

President UniversityErwin SitompulFCS 10/2 Application to Control Design Root locus of G(s) = 1/[s(s+1) 2 ] Bode plot for KG(s) = 1/[s(s+1) 2 ], K = 1. Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/3 Application to Control Design I II III IV I II III IV The detour of C 1 excludes the pole on the imaginary axis. The detour of C 1 includes the pole on the imaginary axis. already done last time will be done now Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/4 Application to Control Design I II II I IV In this new detour of C 1, only the section IV-I is changed. As in previous evaluation, this path can be evaluated by replacing Performing it, Evaluation of G(s) forms a half circle with a very large radius. The half circle starts at 90° and ends at –90°, routing in counterclockwise direction. Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/5 Application to Control Design Combining all four sections, we will get the complete Nyquist plot of the system. III IV I II Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/6 Application to Control Design There is one encirclement of –1 in counterclockwise direction ( N = –1 ). There is one pole enclosed by C 1 since we choose the contour to enclose the pole at origin ( P = 1 ). The number of closed-loop pole in RHP Z = N + P = –1 + 1 = 0  No unstable closed-loop pole (the same result as when encircling the pole at origin to the right). Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/7 Application to Control Design If we choose K > 2, the plot will encircle –1 once in clockwise direction ( N = 1 ). Since P = 1, Z = N + P = 2  The system is unstable with 2 roots in RHP. Thus, the matter of choosing a path to encircle poles on the imaginary axis does not affect the result of analyzing Nyquist plot. Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/8 Stability Margin A large fraction of control systems behave in a pattern roughly similar to the system we just discussed: stable for small gain values and becoming unstable if the gain increases past a certain critical point. Two commonly used quantities that measure the stability margin for such systems are gain margin (GM) and phase margin (PM). They are related to the neutral stability criterion described by If |KG(jω)| < 1 at G(jω) = –180°, then the system is stable. “ ” Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/9 Stability Margin Gain margin (GM) is the factor by which the gain can be raised before instability results. A GM ≥ 1 is required for stability. Phase margin (PM) is the amount by which the phase of G(jω) exceeds –180° when |KG(jω)| = 1. A positive PM is required for stability. PM and GM determine how far the complex quantity G(jω) passes from –1. Within stable values of GM and PM, as can be implied from the figure, there will be no Nyquist encirclements. Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/10 Stability Margin Crossover frequency, ω c, is referred to as the frequency at which the gain is unity, or 0 db. Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/11 Stability Margin Chapter 6The Frequency-Response Design Method PM is more commonly used to specify control system performance because it is most closely related to the damping ratio of the system. The relation between PM, ζ, and furthermore M p for a second- order system can be summarized in the following two figures.

President UniversityErwin SitompulFCS 10/12 Stability Margin Conditionally stable system: a system in which an increase in the gain can make it stable. Several crossover frequencies exist, and the previous definition of gain margin is not valid anymore. Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/13 Example on Stability Margin As one example, determine the stability properties as a function of gain K for the system with the open-loop transfer function Chapter 6The Frequency-Response Design Method

President UniversityErwin SitompulFCS 10/14 Example on Stability Margin We again choose path IV-I as a half circle with a very small radius, routing from negative imaginary axis to positive imaginary axis in counterclockwise direction. This path can be evaluated by replacing Evaluation of G(s) forms an one-and- a-half circle with a very large radius. Inserting the value of θ, the half circle starts at 270° and ends at –270°, routing in clockwise direction. Chapter 6The Frequency-Response Design Method Performing it,

President UniversityErwin SitompulFCS 10/15 Example on Stability Margin Chapter 6The Frequency-Response Design Method The Nyquist plot was drawn for a stable value K = 7. There is one cc encirclement and one ccw encirclement of the –1 point Net encirclement equals zero  the system is stable.

President UniversityErwin SitompulFCS 10/16 Compensation Chapter 6The Frequency-Response Design Method As already discussed before, dynamic compensation is typically added to feedback controllers to improve the stability and error characteristics when a mere proportional feedback alone is not enough. In this section we discuss several kinds of compensation in terms of their frequency-response characteristics. To this point, the closed-loop system is considered to have the characteristic equation 1 + KG(s) = 0. With the introduction of compensation, the closed-loop characteristic equation becomes 1 + KD(s)G(s) = 0. All previous discussions pertaining to the frequency response of KG(s) applies now directly to the compensated case, where the response of KD(s)G(s) is of interest. We call this quantity L(s), the “loop gain” or open-loop transfer function, L(s) = KD(s)G(s).

President UniversityErwin SitompulFCS 10/17 PD Compensation Chapter 6The Frequency-Response Design Method The PD compensation transfer function is given by Frequency-Response of PD Compensation The stabilizing influence is apparent by the increase in phase and the corresponding 20 dB/dec- slope at frequencies above the break point 1/T D. We use this compensation by locating 1/T D so that the increased phase occurs in the vicinity of crossover (that is, where |KD(s)G(s)| = 1  Effect: PM is increased.

President UniversityErwin SitompulFCS 10/18 PD Compensation Chapter 6The Frequency-Response Design Method Note: The magnitude of the compensation continues to grow with increasing frequency. This feature is undesirable because it amplifies the high-frequency noise. Frequency-Response of PD Compensation

President UniversityErwin SitompulFCS 10/19 Lead Compensation Chapter 6The Frequency-Response Design Method In order to alleviate the high- frequency amplification of the PD compensation, a first order pole is added in the denominator at frequencies substantially higher than the break point of the PD compensator. Effect: The phase lead still occurs, but the amplification at high frequency is limited. This resulting lead compensation has a transfer function of where 1/ α is the ratio between the pole/zero break-point frequencies. Frequency-Response of Lead Compensation

President UniversityErwin SitompulFCS 10/20 Lead Compensation Chapter 6The Frequency-Response Design Method Note: A significant amount of phase lead is still provided, but with much less amplification at high frequencies. A lead compensator is generally used whenever a substantial improvement in damping of the system is required. The phase contributed by the lead compensation is given by It can be shown that the frequency at which the phase is maximum is given by The maximum phase contribution, i.e. the peak of D(s) curve, corresponds to

President UniversityErwin SitompulFCS 10/21 Lead Compensation Chapter 6The Frequency-Response Design Method For example, a lead compensator with a zero at s = –2 (T = 0.5) and a pole at s = –10 ( α T = 0.1, α = 0.2) would yield the maximum phase lead at

President UniversityErwin SitompulFCS 10/22 Lead Compensation Chapter 6The Frequency-Response Design Method The amount of phase lead at the midpoint depends only on α and is plotted in the next figure. We could increase the phase lead up to 90° using higher values of the lead ratio, 1/ α. However, increasing values of 1/ α also produces higher amplifications at higher frequencies. A good compromise between an acceptable PM and an acceptable noise sensitivity at high frequencies must be met. How?

President UniversityErwin SitompulFCS 10/23 First Design Using Lead Compensation Chapter 6The Frequency-Response Design Method Find a compensation for G(s) = 1/[s(s+1)] that will provide a steady-state error of less than 0.1 for a unit-ramp input. Furthermore, an overshoot less than 25% is desired. KD(0) must be greater than 10, so we pick K = 10.

President UniversityErwin SitompulFCS 10/24 First Design Using Lead Compensation Chapter 6The Frequency-Response Design Method Sufficient PM to accommodate overshoot requirement (25%) is taken to be 45°. From the read of Bode Plot, the existing PM is 20°  additional phase of 25° at crossover frequency of ω = 3 rad/s 45°

President UniversityErwin SitompulFCS 10/25 First Design Using Lead Compensation Chapter 6The Frequency-Response Design Method However, adding a compensator zero would shift the crossover frequency to the right also  requires extra additional PM or gain attenuation. To be save, we will design the lead compensator that supplies a maximum phase lead of 40°. 5 As can be seen, 1/ α = 5 will accomplish the goal. The maximum phase lead from the compensator must occur at the crossover frequency. This requires trial-and-error in placing the compensator’s zero and pole. The best result will be obtained when placing zero at ω = 2 rad/sec and the pole at ω = 10 rad/s. K = 10

President UniversityErwin SitompulFCS 10/26 Design Procedure of Lead Compensation Chapter 6The Frequency-Response Design Method 1.Determine the open-loop gain K so that the steady-state errors are within specification. 2.Evaluate the PM of the uncompensated system using the value of K obtained above. 3.Allow for extra margin (about 10°) and determine the needed phase lead φ max. 4.Determine α from the graph. 5.Pick ω max to be at the wished crossover frequency, thus the zero is at 1/T and the pole is at 1/ α T. 6.Draw the frequency response of the compensated system and check the new PM. 7.Iterate on the design if the requirement is not met. This is done by changing the position of the pole and zero.

President UniversityErwin SitompulFCS 10/27 Homework 10 No.1, FPE (6 th Ed.), No.2 Design a compensation for the system Due: Wednesday, Chapter 6The Frequency-Response Design Method which will yield an overall phase margin of 45° and the same gain crossover frequency ω 1 as the uncompensated system. Hint: Use MATLAB to draw the Bode plots before and after the compensation. Check the PMs. Submit also the two Bode plots.