Lecture Notes Introduction to Control Systems Instructor: Dr Huynh Thai Hoang Department of Automatic Control Faculty of Electrical & Electronics Engineering Ho Chi Minh City University of Technology Email: hthoang@hcmut.edu.vn huynhthaihoang@yahoo.com Homepage: www4.hcmut.edu.vn/~hthoang/ 24 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ Chapter SYSTEM STABILITY ANALYSIS 24 September 2011 © H T Hồng - ÐHBK TPHCM Content Stability concept Algebraic stability criteria Necessaryy condition Routh’s criterion Hurwitz’s criterion Root locus method Root locus definition Rules for drawing root loci Stability analysis using root locus Frequency response analysis Bode criterion Nyquist’ss stability criterion Nyquist 24 September 2011 © H T Hoàng - ÐHBK TPHCM Stability concept 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ BIBO stability A system is defined to be BIBO stable if every bounded input to the system results in a bounded output over the time interval [t0,+∞) for all initial times t0 u(t) y(t) System y(t) Stable system 24 September 2011 y(t) y(t) System at stability boundary © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Instable system Poles and zeros C id a system Consider t described d ib d by b the th transfer t f function f ti (TF): (TF) Y ( s ) b0 s m + b1s m −1 + K + bm −1s + bm G(s) = = U ( s ) a0 s n + a1s n −1 + K + an −1s + an Denote: A( s ) = a0 s n + a1s n−1 + K + an−1s + an ((TF’s denominator)) B ( s ) = b0 s m + b1s m−1 + K + bm−1s + bm (TF’ numerator) Zeros: are the roots of the numerator of the transfer function, i.e the roots of the equation B(s) = Since B(s) is of order m, the system has m zeros denoted as zi, i =1,2,…m =1 m Poles: are the roots of the denominator of the transfer function, i.e the roots of the equation A(s) = Since A(s) is of order n, the system has n poles denoted as pi , i =1,2,…n 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Pole – zero plot Pole – zero plot is a graph which represents the position of poles and zeros in the complex s-plane Pole Zero 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Stability analysis in the complex plane The stability of a system depends on the location of its poles If all the poles of the system lie in the left-half s-plane then the system is i stable bl If any of the poles of the system lie in the right-half s-plane then the system is unstable unstable If some of the poles of the system lie in the imaginary axis and the others lie in the left left-half half ss-plane plane then the system is at the stability boundary 24 September 2011 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ Characteristic equation Characteristic Ch t i ti equation: ti i the is th equation ti A(s) A( ) = Characteristic polynomial: is the denominator A(s) Note: Systems described d ib d by state equations Feedback systems Y(s) R(s) Yfb(s) A (t ) + Bu (t ) ⎧ x& (t ) = Ax ⎨ ⎩ y (t ) = Cx(t ) Characteristic equation Characteristic equation + G(s) H (s) = det (sI − A) = 24 September 2011 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ Algebraic stability criteria 24 September 2011 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 10 Graphical representation of frequency response (cont’) Bode diagram Nyquist plot Gain margin Gain margin Phase margin Phase margin 24 September 2011 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 58 Nyquist stability criterion Consider a unity feedback system shown below, suppose that we know the Nyquist plot of the open loop system G(s), the problem is to determine the stability of the closed closed-loop loop system Gcll(s) R(s) Y(s) Nyquist criterion: The closed-loop system Gcl(s) is stable if and only if the Nyquist plot of the open open-loop loop system G(s) encircles the critical point (−1, j0) l/2 times in the counterclockwise direction when ω changes from to +∞ (l is the number of poles of G(s) lying in the right-half right half s-plane) s plane) 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 59 Nyquist stability criterion – Example Consider an unity negative feedback system, system whose open-loop open loop system G(s) is stable and has the Nyquist plots below (three cases) Analyze the stability of the closed-loop system 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 60 Nyquist stability criterion – Example Solution The number of poles of G(s) lying in the right-half s-plane is because G(s) ( ) is stable Then according g to the Nyquist yq criterion, the closed-loop system is stable if the Nyquist plot G(jω) does not encircle the critical point (−1, j0) Case Case Case : G(jω) does not encircle (−1, j0) ⇒ the close close-loop loop system is stable stable : G(jω) pass (−1, j0) ⇒ the close-loop p system y is at the stability y boundary; y : G(jω) encircles (−1, j0) ⇒ the close-loop system is unstable 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 61 Nyquist stability criterion – Example A l Analyze th stability the t bilit off a unity it negative ti feedback f db k system t whose h open loop transfer function is: K G ( s) = s (T1s + 1)(T2 s + 1)(T3 s + 1) Solution: Nyquist plot: Depending on the values of T1, T2, T4 andd K, K the h N Nyquist i plot l off G(s) could be one of the three curves 1,, or 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 62 Nyquist stability criterion – Example (cont’) The h number b off poles l off G(s) G( ) lying l i in i the h right-half i h h lf s-plane l i is because G(s) is stable Then according to the Nyquist criterion, the closed-loopp system y is stable if the Nyquist yq plot G(j p (jω) does not encircle the critical point (−1, j0) Case Case Case : G(jω) does not encircle (−1, j0) ⇒ the close-loop system is stable : G(jω) pass (−1, ( j0) ⇒ the close-loop system is at the stability boundary; : G(jω) encircles ((−1, 1, j0) ⇒ the close-loop system is unstable 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 63 Nyquist stability criterion – Example Given an unstable open-loop systems which have the Nyquist plot as below In which cases the closed-loop system is stable? Unstable Stable 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 64 Nyquist stability criterion – Example (cont’) Given an unstable open-loop systems which have the Nyquist plot as below In which cases the closed-loop system is stable? Unstable 24 September 2011 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 65 Nyquist stability criterion – Example (cont’) Given an unstable open-loop systems which have the Nyquist plot as below In which cases the closed-loop system is stable? S bl Stable 24 September 2011 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ U Unstable bl 66 Nyquist stability criterion – Example Gi Given a open-loop l system which hi h has h the h transfer f function: f i K G( s) = (Ts T + 1) n (K>0, T>0, n>2) Find the condition of K and T for the unity negative feedback closedloop system to be stable stable Solution: K Frequency response of the open-loop system: G ( jω ) = (Tjω + 1) n K Magnitude: M (ω ) = n 2 T ω +1 ( Phase: 24 September 2011 ) ϕ (ω ) = −ntg −1 (Tω ) © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 67 Nyquist stability criterion – Example (cont’) Nyquist plot: Stability condition: the Nyquist plot of G(jω) does not encircle the critical point (−1,j0) According to the Nyquist plot, this requires: M (ω−π ) < 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 68 Nyquist stability criterion – Example (cont’) We have: ϕ (ω−π ) = −ntg −1 (Tω−π ) = −π ⇒ ⇒ Then: tg −1 (Tω−π ) = ω−π π ⇒ n ⎛π ⎞ = tg ⎜ ⎟ T ⎝n⎠ M (ω−π ) < ⇔ ⎛ ⎞ π ⎛ ⎞ ⇔ K < ⎜ tg ⎟ g + ⎜ ⎟ ⎜ ⎟ n ⎝ ⎠ ⎝ ⎠ 24 September 2011 ⎛π ⎞ (Tω−π ) = tg ⎜ ⎟ ⎝n⎠ n K ⎛ ⎞ π ⎡ ⎤ ⎛ ⎞ ⎜ T tg ⎟ t + ⎢ T ⎜ n ⎟⎥ ⎟ ⎜ ⎝ ⎠⎦ ⎣ ⎝ ⎠ © H T Hồng - www4.hcmut.edu.vn/~hthoang/ n ⇔ The closed - loop system is stable ⎨ ⎩Φ M > 24 September 2011 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 70 Bode criterion – Example Consider a unity y negative g feedback system y whose open-loop p p system y has the Bode diagram as below Determine the gain margin, phase margin of the open-loop system Is the closed-loop system stable or not? Bode diagram: ωc = ω −π = L(ω−π ) GM L(ω −π ) = 35dB ϕ (ω c ) = −2700 GM = −35dB ΦM = 180 + (−270 ) = −900 −180 ϕ(ωC) 24 September 2011 ΦM ω−π ωC Because GM