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Application to Control Design

Chapter 6. The Frequency-Response Design Method. 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 6. The Frequency-Response Design Method. Application to Control Design. II. II. I. I. IV. IV.

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Application to Control Design

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  1. Chapter 6 The Frequency-Response Design Method 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.

  2. Chapter 6 The Frequency-Response Design Method Application to Control Design II II I I IV IV III III • The detour of C1includes the pole on the imaginary axis. • The detour of C1excludes the pole on the imaginary axis. will be done now already done last time

  3. Chapter 6 The Frequency-Response Design Method Application to Control Design • In this new detour of C1, only the section IV-I is changed. • As in previous evaluation, this path can be evaluated by replacing II I IV • Performing it, III • 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.

  4. Chapter 6 The Frequency-Response Design Method Application to Control Design • Combining all four sections, we will get the complete Nyquist plot of the system. IV II III I

  5. Chapter 6 The Frequency-Response Design Method Application to Control Design • There is one encirclement of –1 in counterclockwise direction(N=–1). • There is one pole enclosed by C1 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).

  6. Chapter 6 The Frequency-Response Design Method 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.

  7. Chapter 6 The Frequency-Response Design Method 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. “ ”

  8. Chapter 6 The Frequency-Response Design Method 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.

  9. Chapter 6 The Frequency-Response Design Method Stability Margin • Crossover frequency, ωc, is referred to as the frequency at which the gain is unity, or 0 db.

  10. Chapter 6 The Frequency-Response Design Method Stability Margin • 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 Mp for a second-order system can be summarized in the following two figures.

  11. Chapter 6 The Frequency-Response Design Method 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.

  12. Chapter 6 The Frequency-Response Design Method 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

  13. Chapter 6 The Frequency-Response Design Method 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 • Performing it, • 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.

  14. Chapter 6 The Frequency-Response Design Method Example on Stability Margin • 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.

  15. Chapter 6 The Frequency-Response Design Method Compensation • 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).

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

  17. Chapter 6 The Frequency-Response Design Method PD Compensation • 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

  18. Chapter 6 The Frequency-Response Design Method Lead Compensation • 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

  19. Chapter 6 The Frequency-Response Design Method Lead Compensation • 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

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

  21. Chapter 6 The Frequency-Response Design Method Lead Compensation • 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? • How?

  22. Chapter 6 The Frequency-Response Design Method First Design Using Lead Compensation 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.

  23. Chapter 6 The Frequency-Response Design Method First Design Using Lead Compensation 45° • 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

  24. Chapter 6 The Frequency-Response Design Method First Design Using Lead Compensation • However, adding a compensator zero would shift the crossover frequency to the right also  requires extra additionalPM 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

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

  26. Chapter 6 The Frequency-Response Design Method Homework 10 • No.1, FPE (6thEd.), 6.25. • No.2 • Design a compensation for the system which will yield an overall phase margin of 45° and the same gain crossover frequency ω1 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. • Due: Wednesday, 11.12.2013.

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