Course Outline. Closed Loop Stability. Stability. Amme 3500 : System Dynamics & Control. Nyquist Stability. Dr. Dunant Halim
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1 Amme 3 : System Dynamics & Control Nyquist Stability Dr. Dunant Halim Course Outline Week Date Content Assignment Notes 1 5 Mar Introduction 2 12 Mar Frequency Domain Modelling 3 19 Mar System Response 4 26 Mar Block Diagrams 5 2 Apr Feedback System Characteristics Assign 1 BREAK 6 16 Apr Root Locus 7 23 Apr Root Locus Apr Bode Plots Assign May Bode Plots May Nyquist Assign May State Space Design Techniques May Advanced Control Topics 13 4 June Review Assign 4 14 Spare Dr. Dunant Halim Amme 3 : Nyquist Slide 2 Stability In this lecture we will examine the stability of closed loop systems in more detail We will examine the relationship between the open-loop frequency response and closed-loop stability Methods for estimating closed loop stability will be presented in the form of the Nyquist diagram Dr. Dunant Halim Amme 3 : Nyquist Slide 3 Closed Loop Stability In general, a closed-loop transfer function will take the form Gs ( T( s 1 + GsHs ( ( Note: negative feedback is considered here. Be careful with the feedback sign! Dr. Dunant Halim Amme 3 : Nyquist Slide 4
2 Closed Loop Stability We will often have analytical expressions for the components of the system G(s and H(s The poles of the closed loop system, or roots of 1+G(sH(s, are generally more difficult to find Closed Loop Stability Consider a system with The open-loop transfer function is and NG ( s NH ( s G( s and H ( s D ( s D ( s NG( snh( s GsHs ( ( D ( s D ( s G G DG( s DH( s + NG( s NH( s 1 + GsHs ( ( D ( s D ( s H G H H Dr. Dunant Halim Amme 3 : Nyquist Slide 5 Dr. Dunant Halim Amme 3 : Nyquist Slide 6 Closed Loop Stability The closed-loop transfer function will be Gs ( NG( s DH( s T( s 1 + GsHs ( ( D( sd ( s + N ( sn ( s G H G H This leads us to conclude that The poles of 1+G(sH(s are the same as the poles of G(sH(s The zeros of 1+G(sH(s are the same as the poles of T(s Closed Loop Stability We would like to find a method for evaluating the closed-loop stability based on the location of open-loop poles and zeros The root locus provides us with one approach for estimating the stability of the system The Bode plots examined in the last lecture provide another means for evaluating stability For some system types, however, the frequency response may have multiple cross-over frequencies. This may make the rules for evaluating stability ambiguous Dr. Dunant Halim Amme 3 : Nyquist Slide 7 Dr. Dunant Halim Amme 3 : Nyquist Slide 8
3 Contour Mapping We will introduce a concept called contour mapping Given a series points on a contour A in the s- plane, the function F(s will map these points to another contour B Contour Mapping For example, if we have a function F(s of the form 2 Fs ( s + 2s+ 1 Then a point s 4+j3 will yield a mapping to the complex variable Fs j j 16 + j3 2 ( ( ( F(s s α.5 F(s Dr. Dunant Halim Amme 3 : Nyquist Slide 9 Dr. Dunant Halim Amme 3 : Nyquist Slide 1 For poles or zeros outside of the contour, the angle will not undergo a net change of 36 o This will result in a contour that does not encircle the origin Contour Mapping Dr. Dunant Halim Amme 3 : Nyquist Slide 11 Assuming clockwise contour A, for contours containing poles and/or zeros, the resulting contour will encircle the origin NP-Z times in the counter clockwise direction (when N> Contour Mapping P,Z1,N-1 P1,Z,N1 P1,Z1,N Dr. Dunant Halim Amme 3 : Nyquist Slide 12
4 Contour Mapping We can apply this result to find the number of poles of the closed loop system in the RHP This will allow us to determine the stability of the closed-loop system P1,Z2, so N1-2-1 We extend the contour to enclose the entire RHP Applying the contour mapping techniques we can determine if any closed loop poles exist in the RHP by examining the encirclements of the origin for 1+KG(sH(s Im(s Contour at infinity Re(s Dr. Dunant Halim Amme 3 : Nyquist Slide 13 Dr. Dunant Halim Amme 3 : Nyquist Slide 14 P, Z so N No encirclement around -1 P, Z2 so N clockwise encirclements around -1 Dr. Dunant Halim Amme 3 : Nyquist Slide 15 This is equivalent to finding the zeros of 1+KG(sH(s (Remember: they are also closed loop poles! To simplify this procedure, we can map through KG(sH(s since the poles and zeros of this quantity are often known The resulting contour is the same as mapping through 1+KG(sH(s but translated one unit to the left Any clockwise encirclement of the point -1 indicates a closed loop pole in RHP, i.e. unstable closed loop! Dr. Dunant Halim Amme 3 : Nyquist Slide 16
5 Effectively we are observing the change in magnitude and phase as a function of increasing frequency by varying s along the jω axis The resulting polar plot represents this change in a slightly different fashion to that of the Bode plot Sketching the Nyquist plot consists of: Plot KG(sH(s for -j s j starting with KG(jωH(jω for ω The magnitude (and hence radius from the origin will be small at high frequency for any physical system Reflect the plot about the real axis as the plot will be symmetric Dr. Dunant Halim Amme 3 : Nyquist Slide 17 Dr. Dunant Halim Amme 3 : Nyquist Slide 18 Once the plot is complete Evaluate the number of counter clockwise encirclements, N, of -1 Determine the number, P, of unstable poles of G(s or G(sH(s, i.e. the open loop system The number of unstable closed-loop roots Z will be given by Z P - N Example 1 Dr. Dunant Halim Amme 3 : Nyquist Slide 19 Dr. Dunant Halim Amme 3 : Nyquist Slide 2
6 Example 1 For this system we find Gs ( ( s+ 1( s+ 3( s+ 1 The closed-loop transfer function is T( s ( s+ 1( s+ 3( s+ 1 + It is not immediately clear if this is stable Dr. Dunant Halim Amme 3 : Nyquist Slide 21 Example 1 We can consider the evaluation of the contour as complex arithmetic using the vectors of G(s drawn to the points along the contour We consider sections AC and CD. Note: Analysis for section AD is not required since the Nyquist plot is symmetrical about the real axis Dr. Dunant Halim Amme 3 : Nyquist Slide 22 Example 1: Mapping of AC For s along the imaginary axis G( jω ( s+ 1( s+ 3( s+ 1 s jω ω + + j ω ω 2 3 ( 14 3 (43 Multiplying by the complex conjugate of the denominator yields 2 3 ( 14ω + 3 j(43 ω ω G( jω ( 14 ω (43 ω ω Example 1: Mapping of AC We can evaluate this as a function of ω 2 3 ( 14ω + 3 j(43 ω ω G( jω ( ( ω + + ω ω For zero frequency G(jω /3 5/3 (Point A As ω increases, the real part remains + ve and the imaginary part remains - ve At ωsqrt(3/14 the real part becomes ve, the imaginary part remains - ve At ωsqrt(43 the plot crosses the ve real axis, the imaginary part becomes + ve. Substitute this ω value to give G(jω-.874 At infinite frequency G(jω is close to j/ω 3 (approximately zero (Point C Dr. Dunant Halim Amme 3 : Nyquist Slide 23 Dr. Dunant Halim Amme 3 : Nyquist Slide 24
7 Example 1: Mapping of CD We can also obtain G(s via the contribution of each complex number in polar form: G( s ( R e 1 jθ 1 jθ 3 ( R e ( R ( θ + θ + θ e jθ 1 ( θ 1 + θ 3 + θ 1 Fortunately, Matlab allows us to generate these plots systf(,[1,1]*tf(1,[1,3]* tf(1,[1,1] nyquist(sys Example 1 jθ i where R i e ( s + i consisting of the magnitude and angle of the complex number At point C, all angles are 9 o o, so point C is 27 At point D, all angles are -9 o o, so point D is 27 Dr. Dunant Halim Amme 3 : Nyquist Slide 25 Dr. Dunant Halim Amme 3 : Nyquist Slide 26 Example 1 Since all open loop poles of -1, -3, -1 are stable (i.e. not in RHP, we can conclude that this system is stable as there is no encirclement of -1 That is: P, ZP-N-. No unstable closed loop poles. Example 2: What if there are open loop poles along the imaginary axis? To avoid undetermined solution in the mapping, we have to make an infinitesimally small detour around the poles Dr. Dunant Halim Amme 3 : Nyquist Slide 27 Dr. Dunant Halim Amme 3 : Nyquist Slide 28
8 * N.S. Nise (24 Control Systems Engineering Wiley & Sons Example 2: System: ( s + 2 G( s 2 s We consider sections AB, BCD and EFA, being mapped into A B, B C D and E F A An infinitesimally small detour is made around 2 poles at the origin Dr. Dunant Halim Amme 3 : Nyquist Slide 29 Ex 2: Mapping of AB & BCD Section AB is along the + ve imaginary axis jω with ω from to infinity. Thus, substitute s jω into G(s: ( jω G( jω 2 ω 9 For section BCD: G( s Very low frequency, point A Very high frequency, point B 9 9 ( 9 ( 9 R 2 θ 2 ( R θ ( R θ ( ( 9 9 ( 9 ( 9 Dr. Dunant Halim Amme 3 : Nyquist Slide 3 Point B Point C Point D Ex 2: Mapping of EFA Mapping of section DE is trivial because it is a mirror image of the section AB. For section EFA (Note: ε is an infinitesimal positive number: 2 18 Point E ( ε 9 ( ε 9 R 2 θ 2 2 G( s Point F ( R θ ( R θ ( ε ( ε 2 18 Point A ( ε 9 ( ε 9 Note that point A has been calculated before using a different approach. Point E does not need to be calculated again since it is a mirror image of point A with respect to real axis. Range of gain for Stability What if we had a control parameter for the previous system? Can the Nyquist plot tell us anything about the range of gains for which the system is stable? K Dr. Dunant Halim Amme 3 : Nyquist Slide 31 Dr. Dunant Halim Amme 3 : Nyquist Slide 32
9 Stability Range Stability Range Changes in gain will result in a corresponding change in the Nyquist plot K1 K2 ZP-N- for small gain Stable closed loop ZP-N-(-22 for large gain Unstable closed loop K.5 ZP-N2-2 for small gain Unstable closed loop ZP-N2-2 for small gain Stable closed loop Dr. Dunant Halim Amme 3 : Nyquist Slide 33 Dr. Dunant Halim Amme 3 : Nyquist Slide 34 Gain and Phase Margins As we saw, the Nyquist plot will change as a function of the gain parameter In the lecture on Bode plots we also saw that the Bode magnitude will rise or lower based on the system gain K We defined the gain margin as the factor by which the gain can be raised before instability results The phase margin is the change in phase shift at unity gain that will result in instability Gain and Phase Margin The gain and phase margins can also be read from the Nyquist plot * N.S. Nise (24 Control Systems Engineering Wiley & Sons Dr. Dunant Halim Amme 3 : Nyquist Slide 35 Dr. Dunant Halim Amme 3 : Nyquist Slide 36
10 Nyquist and Bode It is apparent from the previous development that there is a strong relationship between the Nyquist and Bode plots The Nyquist plot is effectively a polar plot of the magnitude and phase with frequency encoded along the path ω6.5 ω1.5 Nyquist and Bode ω.6 ω Dr. Dunant Halim Amme 3 : Nyquist Slide 37 Dr. Dunant Halim Amme 3 : Nyquist Slide 38 Nyquist and Bode For simple, LTI systems there is little difference between the Nyquist and Bode plots for estimating the stability of the system For more complex systems, such as multiinput/multi-output systems we can t use the Bode plot but may be able to use Nyquist to determine stability These techniques are important in more advanced control studies Conclusions We have looked at another method of determining the stability of a closed loop system based on its open loop characteristics We have presented rules for sketching the Nyquist plot given the open loop transfer function Dr. Dunant Halim Amme 3 : Nyquist Slide 39 Dr. Dunant Halim Amme 3 : Nyquist Slide 4
11 Nise Sections Franklin & Powell Section Further Reading Dr. Dunant Halim Amme 3 : Nyquist Slide 41
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