Lecture Outline Systeem- en Regeltechniek II Previous lecture: PID controller design, lead and lag compen- Lecture 10 – Nyquist plot and stability criterium sators. Robert Babuˇ ska Today: Delft Center for Systems and Control • Nyquist plot. Faculty of Mechanical Engineering Delft University of Technology • Nyquist stability criterion. The Netherlands e-mail: r.babuska@dcsc.tudelft.nl www.dcsc.tudelft.nl/ ˜ babuska tel: 015-27 85117 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 1 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 2 Frequency Domain Methods Frequency Domain Methods The key of frequency domain design: Bode Diagrams 0 -10 Phase (deg); Magnitude (dB) provide sufficient phase at the crossover frequency -20 Bode plot -30 transfer -40 -5 0 -1 0 0 function -1 5 0 10 -1 1 0 0 1 0 1 Frequency (rad/sec) ( = get the closed-loop far enough from the point of becoming un- frequency response stable) Nyquist plot data (experiment) ⇒ Bode plots are well suited as a design and analysis tool. . . . , so, do we need yet another kind plot? In fact, we do, let’s have a look why . . . Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 3 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 4
Motivating Example Deficiency of Bode Plots 10 For systems with poles in right half-plane, the Bode plot alone G ( s ) = sketch the bode plot, indicate PM s − 1 does not provide any good indication of stability / instability. 20 magnitude [dB] → In the above example, the phase will never cross − 180 ◦ , and 0 yet, for K < 0 . 1 , the closed loop becomes unstable (check the -10 root locus!). 1 frequency [rad/s] Is there a method for frequency domain design, considering stabil- -90 phase [deg] ity for all kinds of systems? PM Yes, the Nyquist plot and stability criterion. -180 1 frequency [rad/s] Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 5 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 6 Complex Numbers as Vectors Euler Representation s Im s Im s - p s - p p p Re s − p = | s − p | e jφ Re Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 7 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 8
Transfer Function Nyquist (Polar) Plot Let s = jω for ω ∈ [0 , ∞ ) and plot Im[ G ( jω )] against Re[ G ( jω )] . s − z 1 | s − z 1 | | s − p 1 | · | s − p 2 | e j ( ψ 1 − φ 1 − φ 2 ) G ( s ) = ( s − p 1 )( s − p 2 ) = Nyquist Diagram 5 4 3 s Im 2 1 Imaginary Axis x p 1 0 −1 −2 z 1 Re −3 −4 −5 x p 2 −6 −12 −10 −8 −6 −4 −2 0 2 Real Axis Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 9 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 10 Relation Bode Plot – Nyquist Plot Argument Principle (I) 1 s-plane s Im G ( s ) = ( s + 1)( s + 5) x 0 Re magnitude [dB] x G s -plane ( ) Im Im 0 G s ( ) phase [deg] Re Re -180 No zero/pole encirclement ⇒ no origin encirclement = Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 11 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 12
Argument Principle in General Argument Principle (II) s-plane For a clockwise contour in the s -plane, denote: s Im P number of poles encircled in the s -plane x Z number of zeros encircled in the s -plane Re x N number of clockwise encirclements of the origin by G ( s ) G s -plane ( ) N = Z - P Im G s ( ) Recall: Re � � � � ∠ G ( s ) = ∠ ( s − z i ) − ∠ ( s − p j ) = ψ i − φ j Encirclement of one zero or pole i j i j ⇒ one encirclement of the origin = Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 13 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 14 Argument Principle for Stability Analysis Argument Principle for Stability Analysis G cl ( s ) = Y ( s ) KG ( s ) Given the Nyquist plot of KG ( s ) , we want to determine whether R ( s ) = 1 + KG ( s ) the closed loop: G cl ( s ) = Y ( s ) KG ( s ) Poles of G cl ( s ) are the solutions of 1 + KG ( s ) = 0 , i.e.: R ( s ) = 1 + KG ( s ) is stable. poles of G cl ( s ) are the zeros of (1 + KG ( s )) in addition, as: • Closed-loop stability ⇐ ⇒ G cl ( s ) has no poles in RHP. 1 + K b ( s ) a ( s ) + Kb ( s ) 1 + KG ( s ) = 0 → a ( s ) = 0 → = 0 a ( s ) • Poles are given by 1 + KG ( s ) , so let us study 1 + KG ( s ) . poles of G ( s ) are the poles of (1 + KG ( s )) Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 15 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 16
Argument Principle for Stability Analysis Argument Principle for Stability Analysis So, we want to find out, if 1 + KG ( s ) has no RHP zeros. Im Im Im s-plane 1 + KG s ( ) KG s ( ) 0 -1 Re Re Re encircle the entire RHP Draw the Nyquist diagram of the loop TF L ( s ) = KG ( s ) , = draw the Nyquist diagram for frequencies ω ∈ ( −∞ , ∞ ) count clockwise encirclements of − 1 : N = Z − P Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 17 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 18 Argument Principle for Stability Analysis Nyquist Stability Criterion N = Z - P Z = N + P In words: Z = number of RHP zeros of (1 + KG ( s )) P = number of RHP poles of (1 + KG ( s )) number of RHP closed-loop poles = clockwise encirclements + number of RHP open-loop poles Given that: a ( s ) + Kb ( s ) 1 + KG ( s ) = 0 → = 0 a ( s ) we have: Z = number of RHP poles of G cl ( s ) . . . CL poles P = number of RHP poles of G ( s ) . . . OL poles Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 19 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 20
Stability Margins in Nyquist Plot Why Don’t We Ride These Bikes? Im 1/GM -1 0 Re PM Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 21 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 22 Simple Bicycle Models Front Steering: Nyquist Plot Nyquist Diagram Front steering: 1.5 δ ( s ) = K s + v G ( s ) = φ ( s ) a s 2 − g 5G(s) h 1 Rear steering: 0.5 δ ( s ) = K − s + v G ( s ) = φ ( s ) Imaginary Axis G(s) a s 2 − g 0 h −0.5 a – distance of COM to fixed wheel center −1 h – height of COM above ground −1.5 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0 Real Axis Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 23 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 24
Rear Steering: Nyquist Plot Nyquist: Homework Assignments Nyquist Diagram • Read Section 6.3 (Nyquist stability criterion). 1.5 5G(s) 1 • Work out examples in this section and verify the results by using Matlab. 0.5 Imaginary Axis G(s) • Work out problems 6.18 – 6.22 and verify your results by using 0 Matlab. −0.5 −1 −1.5 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0 Real Axis Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 25 Robert Babuˇ ska Delft Center for Systems and Control, TU Delft 26
Recommend
More recommend
