Nyquist Criterion For Stability of Closed Loop System

Transcription

1 Nyquist Criterion For Stability of Closed Loop System Prof. N. Puri ECE Department, Rutgers University Nyquist Theorem Given a closed loop system: r(t) + KG(s) = K N(s) c(t) H(s) = KG(s) +KG(s) = KN(s) +KN(s) n: The order of denominator polynomial ; KG(s) = Open Loop Transfer Function; D(S) = Open Loop Characteristic polynomial with O unstable roots (unstable open loop poles) and n O stable roots; P (s) = + KN(s) = Closed Loop Characteristic polynomial with C unstable roots (unstable closed loop poles); KG(jω) = Open Loop frequency response; + KG(jω) = Shifted Plot. The plot of ( + KG(jω)) as ω varies from ω = to ω = is referred to as Nyquist Locus in ( + KG(jω)) plane. Plot of Nyquist Locus in ( + KG(jω)) plane determines the stability of the Closed Loop System. Let N = number of counter clockwise encirclements of the origin (0 + 0j) by the Nyquist Locus as ω varies from to, in ( + KG(jω)) plane. Nyquist Theorem state: So this implies: N = O C ()

2 if C = 0, the conclusion is: Feedback control system is stable, if the number of counterclockwise encirclements of the origin in ( + KG(jω) plane is equal to the number of open loop unstable poles O. if O = 0 and C = 0, the conclusion is: Feedback control system is stable if the Nyquist Locus in ( + KG(jω) does not encircle the origin. Origin in ( + KG(jω)) transform to + j0 point in KG(jω) plane. ( + KG(jω)) Plane KG(jω) Plane (0,0) (-,0) O O origin in ( + KG(jω)) plane O, (0, 0) in ( + KG(jω)) plane is Mapped into O, (, 0) in KG(jω) plane, the point O is referred to as the critical point. 2 Graphical Interpretation of G(s) Concept of G(s) as a Mapping form s plane to G(s) plane. Im s s = σ + jω s Re s Im G(s) G(s) Re G(s) A point s in s plane is Mapped into a point G(s) in G(s) plane. Example: Consider function G(s) = + s 2, when s = + j, then G(s) = + s 2 = + ( + j) 2 = + + ( ) + 2j = + 2j Point A in s plane has been mapped in B in G(s) plane. Let s be moving on a curve c as c, c 2,, c K, then corresponding points move in G(s) plane along d, d 2,, d K. Thus the curve c in s plane has been mapped to curve d in G(s) plane via the function G(s). 2

3 A 2 B c K c c 2 c d K d d 2 d G(s) s=jω = G(jω) implies that s is allowed to take values only in the imaginary axis of the s plane. Thus as ω is varied from ω = to ω = +, the whole of the jω axis in the s plane is mapped into a locus in the G(jω) plane. +j j jω e d c b a e ω = a G(jω) d c b G(jω) can be considered as a phaser. G(jω) Plane X(ω) G(jω) A(ω) ϕ(ω) R(ω) G(jω) = R(ω) + jx(ω) G(jω) = R(ω) + jx(ω) = [ R 2 (ω) + X 2 (ω) ] 2 3 X(ω) tan R(ω) = A(ω) ϕ(ω)

4 And it has the following properties. If given G (jω), G 2 (jω) as then G (jω) = A (ω) ϕ (ω) = R (ω) + jx (ω) G 2 (jω) = A 2 (ω) ϕ 2 (ω) = R 2 (ω) + jx 2 (ω) G (jω) + G 2 (jω) = [R (ω) + R 2 (ω)] + j [X (ω) + X 2 (ω)] G (jω)g 2 (jω) = A (ω)a 2 (ω) [ϕ ω + ϕ 2 (ω)] G (jω) = A (ω) G 2 (jω) A 2 (ω) [ϕ ω ϕ 2 (ω)] 3 Proof of Nyquist Criterion Restate the theorem here: Given: KG(s) = Open Loop Transfer Function with O unstable poles; H(s) = KG(s) +KG(s) = Closed Loop Transfer Function with C unstable poles; N = number of counterclockwise encirclements of ( +j0) point by the locus of G(jω), < ω < + ; n = degree of. Nyquist Theorem states that: C = N + O, and C = 0 implies stability of the closed loop system. This implies that For a system to be closed loop stable, the number of encirclements of ( + j0) point by the locus of G(jω), < ω < + in the counterclockwise direction is equal to the number of unstable open loop poles. clockwise counterclockwise Proof: F (s) = + KG(s) = + KN(s) = F (s) = + KN(s) ni= (s p ic ) nk= (s p ko ), F (jω) = ni= (jω p ic ) nk= (jω p ko ) = P (s) 4

5 then n n F (jω) = F (jω) ϕ F (ω), ϕ F (ω) = ϕ ic (ω) ϕ ko (ω) i= k ϕ F (ω) + ω= = n i= ϕ ic (ω) + ω= n k= ϕ ko (ω) + ω= p jω G(jω) L.H.S. R.H.S. L.H.S. R.H.S. jω p jω p stable half unstable half stable half unstable half p in L.H.S. (jω p) = jω p ϕ p (ω) + ω= = +π ϕ p (ω) p in R.H.S. (jω p) = jω p ϕ p (ω) ϕ p (ω) + ω= = π Then for F (s) = +KN(s), it has n zeros and n poles. And it has zeros which are same as Closed Loop poles and has poles which are same as open loop poles. Now { (n C) zeros in L.H.S. and C zeros in R.H.S.. F (s) has (n O) poles in L.H.S. and O poles in R.H.S.. Thus where ϕ F (ω) + ω= = n i= ϕ ic (ω) + ω= n k= ϕ ko (ω) + ω= = [π(n C) πc] [π(n O) πo] = 2π(C O) = 2πN (2) N = counterclockwise encirclements of origin in ( + G(jω)) plane = counterclockwise encirclements of ( + j0) in G(jω) plane 5