Classical Control
Topics covered:
Modeling. ODEs. Linearization. Laplace transform. Transfer functions. Block diagrams. Mason’s Rule. Time resppponse specifications. Effects of zeros and poles. Stability via Routh-Hurwitz. Feedback: Disturbance rejection, Sensitivity, Steady -state tracking. PID controllers and Ziegler-Nichols tuning procedure. Actuator saturation and integrator wind-up.
Root locus. Frequency response--Bode and Nyquist diagrams. Stability Margins . Design of dynamic compensators.
Classical Control – Prof. Eugenio Schuster – Lehigh University 1 Classical Control
Text: Feedback Control of Dynamic Systems, 4th Edition, G .F . Franklin , J .D . Powel and A . Emami-Naeini Prentice Hall 2002.
Classical Control – Prof. Eugenio Schuster – Lehigh University 2 What is control? For any analysis we need a mathematical MODEL of the system
Model → Relation between gas pedal and speed: 10 mph change in speed per each degree rotation of gas pedal Disturbance → Slope of road: 5 mph change in speed per each degree change of slope
Block diagram for the cruise control plant:
w Slope (degrees)
0.5 y =10(u − 0.5w)
Control - Output speed (degrees) (mph) 10 u + y
Classical Control – Prof. Eugenio Schuster – Lehigh University 3 What is control? Open-loop cruise conttlrol:
w r u = PLANT 10 0.5
yol =10(u − 0.5w) - r 1/10 10 =10( − 0.5w) r u + yol 10 Reference = r − 5w (mph)
r = 65,w = 0 ⇒ eol = 0 eol = r − yol = 5w r = 65,w =1⇒ eol = 5mph,eol = 7.69% r − yol w eol[%] = = 500 OK when: r r 1- Plant is known exactly 2- There is no disturbance Classical Control – Prof. Eugenio Schuster – Lehigh University 4 What is control? Close d-loop cruise conttlrol:
w u = 20(r − ycl ) PLANT 0.5 ycl =10(u − 0.5w) 200 5 - + = r − w 10 201 201 1/10 u + r - ycl
1 5 1 ecl = r − ycl = r + w r = 65,w = 0 ⇒ e = % = 0.5% 201 201 cl 201 1 5 5 r − y 1 5 w r = 65,w =1⇒ e = + = 0.69% e [%] = cl = + cl 201 20165 cl r 201 201 r Classical Control – Prof. Eugenio Schuster – Lehigh University 5 What is control?
Feedback control can help:
• reference following (tracking) • disturbance rejection • changing dynamic behavior
LARGE gain is essential but there is a STABILITY limit
“TheThe issue of how to get the gain as large as possible to reduce the errors due to disturbances and uncertainties without making the system become unstable is what much of feedback control design is all about”
First step in this design process: DYNAMIC MODEL
Classical Control – Prof. Eugenio Schuster – Lehigh University 6 Dynamic Models
MECHANICAL SYSTEMS: F = ma Newton’s law
m&x& = u − bx& v = x& velocity a = v& = &x& acceleration
b u Vo 1 m TfTransfer FFtiunction v& + v = ⎯⎯⎯st ⎯⎯⎯st → = v=Voe ,u=Uoe d m m Uo s + b m → s dt
Classical Control – Prof. Eugenio Schuster – Lehigh University 7 Dynamic Models
MECHANICAL SYSTEMS: F = Iα Newton’s law θ 2 ml && = −lmg sinθ + Tc ωα = θ& angular velocity ω = & = θ&& angular acceleration 2 I = ml moment of inertia
g Tc g Tc θ&&+ sinθ = ⎯⎯θ⎯→θ&&+ θ = Linearization l ml 2 sin ≈θ l ml 2
Classical Control – Prof. Eugenio Schuster – Lehigh University 8 Dynamic Models θ g T &&+ θ = c l ml 2 Reduce to first order equations:
x1 = x2 x1 = θ & g Tc x2 = θ& x = − x + &2 l 1 ml 2
⎡ x ⎤ T ⎡ 0 1⎤ ⎡0⎤ x ≡ 1 ,u ≡ c ⇒ x = ⎢ g ⎥x + u State Variable ⎢x ⎥ ml 2 & − 0 ⎢1⎥ Representation ⎣ 2 ⎦ ⎣⎢ l ⎦⎥ ⎣ ⎦
General case: x& = Fx + Gu y = Hx + Ju Classical Control – Prof. Eugenio Schuster – Lehigh University 9 Dynamic Models
ELECTRICAL SYSTEMS: Kirchoff’s Current Law (KCL): The alge bra ic sum of currents enteri ng a nod e is zero at every instant Kirchoff’s Voltage Law (KVL) The algebraic sum of voltages around a loop is zero at every instant Resistors:
vR (t) = RiR (t) ⇔ iR (t) = GvR (t) i i R iC L Capacitors: + + + t v v dv (t) 1 τ R vC L i (t) = C C ⇔ v (t) = i ( )dτ + v (0) C C ∫ C C dt C 0 Inductors: di (t) 1 t v (t) = L L ⇔ i (t) = v (τ )dτ + i (0) L L ∫ L L dt L 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 10 Dynamic Models
ELECTRICAL SYSTEMS: OP AMP:
vO = A(vp − vn ), A → ∞ i + p + vp RO iO v vp = vn RI + O + in + i = i = 0 v A(vp-vn) p n n - -
To work in the linear mode we need FEEDBACK!!!
Classical Control – Prof. Eugenio Schuster – Lehigh University 11 Dynamic Models
ELECTRICAL SYSTEMS: KCL: R2
R 1 C dvO 1 1 v vO = vO − vI
1 dt R C R C - 2 1 +
1 t μ R = ∞ (OC) ⇒ v ()t = v 0 (− ) v ( )dμ 2 O O ∫0 I R1C
v v 1 1 K ∫ O K = − RC Inverting integrator
Classical Control – Prof. Eugenio Schuster – Lehigh University 12 Dynamic Models
ELECOCTRO-MECCSSSCCHANICAL SYSTEMS: DC Motor torque armature current
T = Ktia
e = Keθ&m
emf shaft velocity
Jmθ&&m = −bθ&m + T di − v + R i + L a + e = 0 a a a dt
Obtain the State Variable Representation
Classical Control – Prof. Eugenio Schuster – Lehigh University 13 Dynamic Models
HEAT-FLOW: Heat Flow Temperature Difference 1 q = (T −T ) R 1 2 1 T& = q C Thermal capacitance Thermal resistance
1 ⎛ 1 1 ⎞ T&I = ⎜ + ⎟(To −TI ) CI ⎝ R1 R2 ⎠
Classical Control – Prof. Eugenio Schuster – Lehigh University 14 Dynamic Models
FLUID-FLOW: Mass rate Mass Conservation law
m& = win − wout
Inlfllet mass flow Outlfllet mass flow
ρ 1 m = Ah& ⇒ h& = ()w − w & ρA in out
A: area of the tank ρ: density of fluid h: height of water
Classical Control – Prof. Eugenio Schuster – Lehigh University 15 Linearization
Dynamic System: x& = f (x,u)
0 = f (xo ,uo ) Equilibrium
Denote δx = x − xo ,δu = u − uo
δx& = f (xo + δx,uo + δu) Taylor Expansion ∂f ∂f δx ≈ f (x ,u )+ δx + δu & o o ∂x ∂u xo ,uo xo ,uo ∂f ∂f F ≡ ,G ≡ ⇒ δx ≈ Fδx + Gδu ∂x ∂u & xo ,uo xo ,uo Classical Control – Prof. Eugenio Schuster – Lehigh University 16 Linearization δx& ≈ Fδx + Gδu
⎡∂f1 ∂f1 ⎤ ⎡ ∂f1 ∂f1 ⎤ ⎢∂x L ∂x ⎥ ⎢∂u L ∂u ⎥ ∂f ⎢ 1 n ⎥ ∂f ⎢ 1 m ⎥ F ≡ = ⎢ M M ⎥ ,G ≡ = ⎢ M M ⎥ ∂x x ,u ∂u x ,u o o ⎢∂fn ∂fn ⎥ o o ⎢∂fn ∂fn ⎥ ⎢∂x L ∂x ⎥ ⎢∂u L ∂u ⎥ ⎣ 1 n ⎦ x ,u ⎣ 1 m ⎦ x ,u o o θ o o θ k g Example: Pendulum with friction &&+ & + sinθ = 0 m l ⎡ 0 1 ⎤ x = g k x & ⎢− cos x − ⎥ ⎢ 1 ⎥ ⎣ l m⎦ xo Classical Control – Prof. Eugenio Schuster – Lehigh University 17 Laplace Transform
• Function f(t) of time bt – Piecewise continuous and exponential order f (t) < Ke ∞ α + j∞ −st −1 1 st F(s) = ∫ f (t)e dt L []F()s = f (t) = ∫ F(s)e ds π 0- 2 j α − j∞
– 0- limit is used to capture transients and discontinuities at t=0 – s is a complex variable (σ+jω) • There is a need to worryyg about regions of conver gence of the inte gral – Units of s are sec-1=Hz • A frequency – If f(t) il()his volts (amps) then F(s) ilis volt-second(ds (amp-second)ds)
Classical Control – Prof. Eugenio Schuster – Lehigh University 18 Laplace transform examples • Step function – unit Heavyside Function ⎧0, for t < 0 • After Oliver Heavyside (1850-1925) u(t) = ⎨ ⎩1, for t ≥ 0 ∞ σ ∞ ∞ ∞ e−st e−(σ + jωω)t 1 F(s) = ∫u(t)e−stdt = ∫ e−stdt = − = − = if σ > 0 s + j s 0− 0− 0 0 • Exponential function α • After Oliver Exponential (1176 BC- 1066α BC) ∞ ∞ −(s+α )t ∞ − t −st −(s+α )t e 1 α σ F(s) = ∫∫e e dt = e dt = − = if > α s + s + 00 0 • Delta (impulse) function δ(t) ∞ st F(s) = ∫δ (t)e− dt =1 for all s 0−
Classical Control – Prof. Eugenio Schuster – Lehigh University 19 Laplace Transform Pair Tables Signal Waveform Transform impulse δ (t) 1 step u(t) 1 s 1 ramp tu(t) s2 1 exponential e−αtu(t) s+α t 1 damped ramp te−α u(t) (s+α )2 β sine sin(βt)u(t) s2+β 2 s cosine cos(βt)u(t) s2+β 2 damped sine −αt β e sin()βt u(t) α (s+ )2+β 2 damped cosine s+α e−αt cos(βt)u(t) α (s+ )2+β 2
Classical Control – Prof. Eugenio Schuster – Lehigh University 20 Laplace Transform Properties
Linearity: (absolutely critical property)
L{Aff1(t) + Bff2(t)}= AL{f1 (t)}+ BL{f2 (t)}= AF1(s) + BF2(s) ⎧t τ ⎫ F(s) Integration property: L⎨∫ f ( )dτ ⎬ = ⎩0 ⎭ s ⎧df (t)⎫ Differentiation property: L⎨ ⎬ = sF(s) − f (0−) ⎩ dt ⎭ ⎧ 2 ⎫ ⎪d f (t)⎪ 2 L = s F(s) − sf (0−) − f ′(0−) ⎨ 2 ⎬ ⎪⎩ dt ⎭⎪ ⎪⎧d m f (t)⎪⎫ = m F(s) − sm−1 f (0−) − sm−2 f ′(0−) − − f (m) (0−) L⎨ m ⎬ s L ⎪⎩ dt ⎭⎪
Classical Control – Prof. Eugenio Schuster – Lehigh University 21 Laplace Transform Properties
Translation properties:
s-domain translation: L{e−αt f (t)} = F(s +α)
−as t-domain translation: L{f (t − a)u(t − a)}= e F(s) for a > 0
Initial Value Property: lim f (t) = lim sF(s) t→0+ s→∞
Final Value Property: lim f (t) = lim sF(s) t→∞ s→0
If all poles of F(s) are in the LHP
Classical Control – Prof. Eugenio Schuster – Lehigh University 22 Laplace Transform Properties 1 s Time Scaling: L{ f (at)} = F( ) a a dF(s) Multiplication by time: L{tf (t)} = − ds τ t τ Convolution: L{ f ( )g(t − )dτ} = F(s)G(s) ∫0 σ 1 + jω Time product: L{ f (t)g(t)} = σ F(s)G(s − λ)dλ 2πj ∫ − jω
Classical Control – Prof. Eugenio Schuster – Lehigh University 23 Laplace Transform
Exercise: Find the Laplace transform of the following waveform 4(s + 2) f (t) = [2 + 2sin(2t) − 2cos(2t)]u(t) F(s) = s(s2 + 4)
Exercise: Find the Laplace transform of the following waveform
3 t s + 36s + 80 f (t) = e−4tu(t) + 5 sin()4x dx F(s) = ∫0 s()s + 4 ()s2 +16 −40t −40t d[5te ] 10s + 200 f (t) = 5e u(t) + u()t F(s) = dt ()s + 40 2 EiExercise: Find the Laplace transform of the following waveform 2 A(1− e−Ts ) f (t) = Au(t) − 2Au(t −T ) + Au(t − 2T ) F(s) = s
Classical Control – Prof. Eugenio Schuster – Lehigh University 24 Laplace transforms Linear system Complex frequency domain
ain) (s domain) mm Differential Laplace Algebraic do
t equation transform L equation in ( aa Classical Algebraic e dom techniques techniques mm Ti
Response Inverse Laplace Response signal transform L-1 transform
• The diagram commutes – Same answer whichever way you go Classical Control – Prof. Eugenio Schuster – Lehigh University 25 Solving LTI ODE’s via Laplace Transform
()n n−1 ( ) (m) (m−1) y + an−1 y +L+ a0 y = bmu + bm−1u +L+ b0u
(n−1) (m−1) Initial Conditions: y (0)(),K, y 0 ,u (0),K,u(0)
⎧d k f (t)⎫ k−1 Recall k (k−1− j) j L⎨ k ⎬ = s F(s) − ∑ f (0)s ⎩ dt ⎭ j=0
n−1 n−1 i−1 m i−1 n (n−1− j) j ⎡ i (i−1− j) j ⎤ ⎡ i (i−1− j) j ⎤ s Y (s) − ∑ y (0)s +∑ai ⎢s Y (s) − ∑ y (0)s ⎥ = ∑bi ⎢s U (s) − ∑u (0)s ⎥ j=0 i=0 ⎣ j=0 ⎦ i=0 ⎣ j=0 ⎦
n−1 i−1 m i−1 a y(i−1− j) (0)s j − b u(i−1− j) (0)s j m m−1 ∑ i ∑ ∑ i ∑ bms + bm−1s +L+ b1s + b0 i=0 j=0 i=0 j=0 Y (s) = n n−1 U (s) + n n−1 s + an−1s +L+ a1s + a0 s + an−1s +L+ a1s + a0
For a given rational U(s) we get Y(s)=Q(s)/P(s)
Classical Control – Prof. Eugenio Schuster – Lehigh University 26 Laplace Transform
Exercise: Find the Laplace transform V(s)
dv(t) 4 3 + 6v(t) = 4u(t) V (s) = − dt s()s + 6 s + 6 v(0−) = −3
Exercise: Find the Laplace transform V(s)
d 2v(t) dv(t) 5 2 + 4 + 3v(t) = 5e−2t V (s) = − dt 2 dt ()()()s +1 s + 2 s + 3 s +1 v(0−) = −2,v'(0−) = 2
What about v(()t)?
Classical Control – Prof. Eugenio Schuster – Lehigh University 27 Transfer Functions
()n (n−1) (m−1) y + an−1 y +L+ a0 y = bm−1u +L+ b0u
Assume all Initial Conditions Zero:
n n−1 m−1 (s + an−1s +L+ a1s + a0 )Y (s) = (bm−1s +L+ b1s + b0 )U (s) Output Input m−1 bm−1s +L+ b1s + b0 B(s) Y (s) = n n−1 U (s) = U (s) s + an−1s +L+ a1s + a0 A(s) m−1 Y (s) bm−1s +L+ b1s + b0 H (s) = = n n−1 U (s) s + an−1s +L+ a1s + a0 (s − z )(s − z ) (s − z ) = K 1 2 L m (s − p1)(s − p2 )L(s − pn )
Classical Control – Prof. Eugenio Schuster – Lehigh University 28 Rational Functions
• We shall mostly be dealing with LTs which are rational functions – ratios of polynomials in s m m−1 bms + bm−1s +L+ b1s + b0 F(s) = n n−1 ans + an−1s +L+ a1s + a0 (s − z )(s − z ) (s − z ) = K 1 2 L m (s − p1)(s − p2 )L(s − pn )
• pi are the poles and zi are the zeros of the function • K is the scale factor or (sometimes) gain • A proper rational function has n≥m •A strictly proper rational function has n>m • An improper rational function has n Classical Control – Prof. Eugenio Schuster – Lehigh University 29 Residues at simple poles FiFunctions of a compl ex variiblable wiihth iildsolated, finite order poles have residues at the poles (s − z1)(s − z2 )L(s − zm ) k1 k2 kn F(s) = K = + +L+ (s − p1)(s − p2 )L(s − pn ) (s − p1) (s − p2 ) (s − pn ) k1(s − pi ) k2 (s − pi ) kn (s − pi ) (s − pi )F(s) = + +L+ ki +L+ (s − p1) (s − p2 ) (s − pn ) Residue at a simple pole: ki = lim(s − pi )F(s) s→ pi Classical Control – Prof. Eugenio Schuster – Lehigh University 30 Residues at multiple poles Compute residues at poles of order r: (s − z )(s − z ) (s − z ) k k k F(s) = K 1 2 L m = 1 + 2 + + r (s − p )r (s − p ) (s − p )2 L (s − p )r 1 r − j 1 1 1 1 d k = lim ⎡(s − p )r F(s)⎤, j =1 r j (r − j)! r − j ⎣⎢ i ⎦⎥ L s → pi ds 2s2 + 5s 2 1 3 Example: = + − (s +1)3 s +1 (s +1)2 (s +1)3 3 2 ⎡ 3 2 ⎤ 2 ⎡ 3 2 ⎤ (s+1) (2s +5s) d (s+1) (2s +5s) 1 d (s+1) (2s +5s) = lim =−3 lim ⎢ ⎥ =1 lim 2 ⎢ 3 ⎥ 2 s→−3 (s+1)3 s→−1ds⎣⎢ (s+1)3 ⎦⎥ 2!s→−1ds ⎣⎢ (s+1) ⎦⎥ ⎛ 2s2 + 5s ⎞ ⎛ 2 1 3 ⎞ L−1⎜ ⎟ L−1⎜ ⎟ e−t (2 t 3t 2 )u(t) ⎜ 3 ⎟ = ⎜ + 2 − 3 ⎟ = + − ⎝ ()s +1 ⎠ ⎝ s +1 ()s +1 ()s +1 ⎠ Classical Control – Prof. Eugenio Schuster – Lehigh University 31 Residues at complex poles • Compute residues at the poles lim (s − a)F(s) s→a • Bundle complex conjugate pole pairs into second-order terms if α you want β α β • but you will need to be careful α α (s − − j )(s − + j ) = [s2 − 2 s + ( 2 + β 2 )] • Inverse Laplace Transform is a sum of complex exponentials • The answer will be real Classical Control – Prof. Eugenio Schuster – Lehigh University 32 Inverting Laplace Transforms in Practice • We have a table of inverse LTs • Write F(s) as a partial fraction expansion b sm + b sm−1 + + b s + b F(s) = m m−1 L 1 0 n n−1 ans + an−1s +L+ a1s + a0 (s − z )(s − z ) (s − z ) = K 1 2 L m (s − p1)(s − p2)L(s − pn ) α α α α α α = 1 + 2 + 31 + 32 + 33 + ...+ q 2 3 (s − p1) (s − p2) (s − p3) (s − p ) (s − p ) (s − pq) 3 3 • Now appeal to linearity to invert via the table – Surprise! – Nastiness: computing the partial fraction expansion is best dbllihiddone by calculating the residues Classical Control – Prof. Eugenio Schuster – Lehigh University 33 Example 9-12 20(s + 3) • Find the inverse LT of F(s) = (s +1)(s2 + 2s + 5) k k k* F(s) = 1 + 2 + 2 s +1 s +1− j2 s +1+ j2 20(s + 3) k = lim (s +1)F(s) = =10 1 2 s → −1 s + 2s + 5 s = −1 5 20(s + 3) j π k = lim (s +1− 2 j)F(s) = = −5 − 5 j = 5 2e 4 2 s → −1+ 2 j (s +1)(s +1+ 2 j) s = −1+ 2 j 5 5 ⎡ (−1+ j2)t+ j π (−1− j2)t− j π ⎤ f (t) = ⎢10e−t + 5 2e 4 + 5 2e 4 ⎥u(t) ⎢ ⎥ ⎣ ⎦ ⎡ 5π ⎤ = 10e−t +10 2e−t cos(2t + ) u(t) ⎣⎢ 4 ⎦⎥ Classical Control – Prof. Eugenio Schuster – Lehigh University 34 Not Strictly Proper Laplace Transforms s3 + 6s2 +12s + 8 • Find the inverse LT of F(s) = s2 + 4s + 3 – Convert to polynomial plus strictly proper rational function • Use polynomial division s + 2 F(s) = s + 2 + s2 + 4s + 3 0.5 0.5 = s + 2 + + s +1 s + 3 • Invert as normal dδ (t) f (t) = ⎡ + 2δ (t) + 0.5e−t + 0.5e−3t ⎤u(t) ⎢⎣ dt ⎦⎥ Classical Control – Prof. Eugenio Schuster – Lehigh University 35 Block Diagrams Series: G1 G2 G = G1G2 G1 Parallel: + + G2 G = G1 + G2 Classical Control – Prof. Eugenio Schuster – Lehigh University 36 Block Diagrams Negative Feedback: R Reference input R(s) + E(s) C(s) G E = R − B Error signal - C = GE Output B(s) H B = HC Feedback signal C G C = GR − GHC ⇒ (1+ GH )C = GR ⇒ = R (1+ GH ) E 1 E = R − HGE ⇒ (1+ GH )E = R ⇒ = R (1+ GH ) Rule: Transfer Function=Forward Gain/(1+Loop Gain) Classical Control – Prof. Eugenio Schuster – Lehigh University 37 Block Diagrams Positive Feedback: R Reference input R(s) + E(s) C(s) G E = R + B Error signal + C = GE Output B(s) H B = HC Feedback signal C G C = GR + GHC ⇒ (1− GH )C = GR ⇒ = R (1− GH ) E 1 E = R + HGE ⇒ (1− GH )E = R ⇒ = R (1− GH ) Rule: Transfer Function=Forward Gain/(1-Loop Gain) Classical Control – Prof. Eugenio Schuster – Lehigh University 38 Block Diagrams Moving through a branching point: R(s) C(s) R(s) C(s) G G ≡ B(s) B(s) 1/G Moving through a summing point: R(s) C(s) R(s) C(s) + G ≡ G + + + B(s) G B(s) Classical Control – Prof. Eugenio Schuster – Lehigh University 39 Block Diagrams Example: H1 - R(s) + + C(s) G1 G2 G3 - H 2 R(s) G1G2G3 C(s) 1+ H1G2G3 + H 2G1G2 Classical Control – Prof. Eugenio Schuster – Lehigh University 40 Mason’s Rule H 4 H6 + + U (s) + + + + Y (s) H1 H 2 H3 + + H5 H7 Signal Flow Graph H nodes 4 bbaranc hes H6 1 H 2 H 3 H 4 U (s) 1 2 3 Y (s) H5 H7 Classical Control – Prof. Eugenio Schuster – Lehigh University 41 Mason’s Rule PthPath: a sequence of connect ed branc hes in the directi on of the signal flow without repetition Loop: a closed path that returns to its starting node Forward path: connects input and output Y (s) 1 G(s) = = ∑GiΔi U (s) Δ i Gi = gain of the ith forward path Δ = the system determinant =1-∑(all loop gains) + ∑(gain products of all possible two loops that do not touch) − ∑(gain products of all possible three loops that do not touch) +L Δi = value of Δ for the part of the graph that does not touch the ith forward path Classical Control – Prof. Eugenio Schuster – Lehigh University 42 Mason’s Rule H 4 H6 1 H 2 H 3 H 4 U (s) 1 2 3 Y (s) H5 H7 Y (s) H H H + H − H H H = 1 2 3 4 4 2 6 U (s) 1− H1H5 − H 2H6 − H3H7 − H 4H7H6H5 + H1H5H3H7 Classical Control – Prof. Eugenio Schuster – Lehigh University 43 Impulse Response τ ∞ δ Dirac’s delta: u( ) (t −τ )dτ = u(t) ∫0 Integration is a limit of a sum ⇓ u(t) is represented as a sum of impulses By superposition principle, we only need unit impulse response h()t −τ Response at t to an impulse applied at τ System Response: u(t) h y(t) ∞ y(t) = u(τ )h(t −τ )dτ ∫0 Classical Control – Prof. Eugenio Schuster – Lehigh University 44 Impulse Response t-domain: u(t) h y(t) τ Impulse response ∞ τ y(t) = u( )h(t − )dτ u(t) = δ (t) ⇒ y(t) = h(t) ∫0 The system response is obtained by convolving the input with the impulse response of the system. ∞ Convolution: L{ u(τ )h(t −τ )dτ} = H (s)U (s) ∫0 s-domain: U (s) H Y(s) Impulse response Y (s) = H (s)U (s) u(t) = δ (t) ⇒U (s) =1⇒ Y (s) = H (s) The system response is obtained by multiplying the transfer function and the Laplace transform of the input. Classical Control – Prof. Eugenio Schuster – Lehigh University 45 Time Response vs. Poles 1 Real pole: H (s) = ⇒ h(t) = e−σt Impulse s +σ Response σ > 0 Stable σ < 0 Unstable 1 τ = Time Constant σ Classical Control – Prof. Eugenio Schuster – Lehigh University 46 Time Response vs. Poles Real pole: σ H (s) = ⇒ h(t) = σe−σt Impulse s +σ Response 1 τ = Time Constant σ σ 1 Y (s) = ⇒ y(t) =1− e−σt Step s +σ s Response Classical Control – Prof. Eugenio Schuster – Lehigh University 47 Time Response vs. Poles 2 ζωωn IlImpulse Complex poles: H (s) = 2 2 s + 2 ns + ωn Response ζω 2 ω nω = 2 2 2 ()s + n + n ()1−ζ ωn : Undamped natural frequency ζ : Dampingσ ratio ω ω 2 H (s) = n σ ()()s + + j d s + − jωd ω 2 σ = n ζω σ 2 2 ()s + ω + ωd ω 2 = n , d = n 1−ζ Classical Control – Prof. Eugenio Schuster – Lehigh University 48 Time Response vs. Poles Complex poles: ω 2 ω H (s) = n → h(t) = n e−σt sin()ω t 2 2 2 2 d (s + ζωn ) + ωn (1−ζ ) 1−ζ Impulse Response σ > 0 Stable σ < 0 Unstable Classical Control – Prof. Eugenio Schuster – Lehigh University 49 Time Response vs. Poles Complex poles: ω 2 ω H (s) = n → h(t) = n e−σt sin()ω t 2 2 2 2 d (s + ζωζ n ) + ωn (1−ζ ) 1−ζ Impulse Response Classical Control – Prof. Eugenio Schuster – Lehigh University 50 ω ζω ω ζ Time Response vs. Poles Complex poles: ω σ 2 1 ⎡ ⎤ Y(s) n y(t) 1 e−σt cos()t ω sin()t = 2 2 2 → = − ⎢ d + ωd ⎥ ()s + n + n (1− )s ⎣ d ⎦ Step Response Classical Control – Prof. Eugenio Schuster – Lehigh University 51 Time Response vs. Poles 2 ζωωn Complex poles: H (s) = 2 2 s + 2 ns + ωn ζω 2 ω nω = 2 2 2 ()s + n + n ()1−ζ CASES: ζ ζ 2 2 ζ = 0: s +ζωω Undamped n ω ζ 2 2 2 <1:()s + n ζ+ n (1−ζ ) Underdamped 2 ω Critically damped ζ =1:(s + ωn ) ζ ζ 2 2 Overdamped >1:[]s + ( + −1) n [s + ( − −1)ωn ] Classical Control – Prof. Eugenio Schuster – Lehigh University 52 Time Response vs. Poles Classical Control – Prof. Eugenio Schuster – Lehigh University 53 Time Domain Specifications Classical Control – Prof. Eugenio Schuster – Lehigh University 54 Time Domain Specifications 1- The rise time tr is the time it takes the system to reach the vicinity of its new set point 2- The settling time ts is the time it takes the system transients to decay 3- The overshoot Mp is the maximum amount the system oversh oot its fina l va lue div ide d by its final value 4- The peak time tp is the time it takes the system to reach the maximum overshoot point ω π 1.8 t = ζ t ≅ p 2 r n 1− ωn π ζ − 1−ζ 2 4.6 M p = e ts = ζωn Classical Control – Prof. Eugenio Schuster – Lehigh University 55 Time Domain Specifications Design specification are given in terms of tr ,t p ,M p ,ts These sppgppecifications give the position of the poles ωn ,ζ ⇒ σ ,ωd Example: Find the pole positions that guarantee tr ≤ 0.6sec,M p <10%,ts ≤ 3sec Classical Control – Prof. Eugenio Schuster – Lehigh University 56 Effect of Zeros and Additional poles Additional poles: 1- can be neglected if they are sufficiently to the left of the dominant ones. 2- can increase the rise time if the extra pole is within a factor of 4 of the real part of the complex poles. Zeros: 1- a zero near a pole reduces the effect of that pole in the time response. 2- a zero in the LHP will increase the overshoot if the zero is within a factor of 4 of the real part of the complex poles (due to differentiation). 3- a zero in the RHP (nonminimum phase zero) will depress the overshoot and may cause the step response to start out in the wrong direction. Classical Control – Prof. Eugenio Schuster – Lehigh University 57 Stability m m−1 Y (s) bms + bm−1s +L+ b1s + b0 = n n−1 R(s) s + an−1s +L+ a1s + a0 Y (s) (s − z )(s − z ) (s − z ) = K 1 2 L m R(s) (s − p1)(s − p2 )L(s − pn ) Y (s) k1 k2 kn = + +L+ R(s) (s − p1) (s − p2 ) (s − pn ) Impulse response: k1 k2 kn R(s) =1⇒ Y (s) = + +L+ (s − p1) (s − p2 ) (s − pn ) p1t p2t pnt y(t) = k1e + k2e +L+ kne Classical Control – Prof. Eugenio Schuster – Lehigh University 58 Stability p1t p2t pnt y(t) = k1e + k2e +L+ kne We want: pit e ⎯t⎯→⎯∞→0 ∀i =1Kn Definition: A system is asymptotically stable (a.s.) if Re{pi} < 0 ∀i n n−1 Characteristic polynomial: a(s) = s + an−1s +L+ a1s + a0 Characteristic equation: a(s) = 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 59 Stability Necessary condition for asymptotical stability (a.s.): ai > 0 ∀i Use this as the first test! If any ai<0, the the system is UNSTABLE! Example: s2 + s − 2 = 0 (s + 2)(s −1) = 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 60 Routh’s Criterion Necessary and sufficient condition Do not have to find the roots pi! Routh’s Array: n s 1 a2 a4 L Depends on whether an n−1 n is even or odd s a1 a3 a5 L n−2 s b1 b2 b3 a a − a a a − a a a − a n−3 b = 1 2 3 , b = 1 4 5 , b = 1 6 7 s c1 c2 c3 1 2 3 L a1 a1 a1 n−4 b a − a b b a − a b s d1 d2 1 3 1 2 1 5 1 3 c1 = , c2 = , L b1 b1 M c b − b c c b − b c d = 1 2 1 2 , d = 1 3 1 3 , 0 1 c 2 c L s an 1 1 M M Classical Control – Prof. Eugenio Schuster – Lehigh University 61 Routh’s Criterion How to remember this? Routh’s Array: n , s m11 m12 m13 L m1, j = a2 j−2 n−1 s m21 m22 m23 L m2, j = a2 j−1, sn−2 m m m 31 32 33 L mi−2,1 mi−2, j+1 sn−3 m m m 41 42 43 L mi−1,1 mi−1, j+1 mi, j = − ,∀i ≥ 3 M M M M mi−1,1 Classical Control – Prof. Eugenio Schuster – Lehigh University 62 Routh’s Criterion The criterion: • The system is asymptotically stable if and only if all the elements in the first column of the Routh’s array are positive • The number of roots with positive real parts is equal to the number of sign changes in the first column of the Routh array Classical Control – Prof. Eugenio Schuster – Lehigh University 63 Routh’s Criterion - Examples Example 1: 2 s + a1s + a2 = 0 3 2 Example 2: s + a1s + a2s + a3 = 0 Example 3: s6 + 4s5 + 3s4 + 2s3 + s2 + 4s + 4 = 0 Example 4: s3 + 5s2 + (k − 6)s + k = 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 64 Routh’s Criterion - Examples Example: Determine the range of K over which the system is stable s 1 R(s) + K + Y (s) - s(s −1)(s + 6) Classical Control – Prof. Eugenio Schuster – Lehigh University 65 Routh’s Criterion Special Case I: Zero in the first column We replace the zero with a small positive constant ε>0 and proceed as before. We then apply the stability criterion by taking the limit as ε→0 Example: s4 + 2s3 + 4s2 + 8s +10 = 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 66 Routh’s Criterion Special Case II: Entire row is zero This indicates that there are complex conjugate pairs. If the ith row is zero, we form an auxiliary equation from the previous nonzero row: i+1 i−1 i−3 a1(s) = β1s + β2s + β3s +L Where βi are the coefficients of the (i+1)th row in the array. We then replace the ith row by the coefficients of the derivative of the auxiliary polynomial. Example: s5 + 2s4 + 4s3 + 8s2 +10s + 20 = 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 67 Properties of feedback Disturbance Rejection: Open loop w + r + y Ko A y = Ko Ar + w Closed loop w + r + + y Kc A - K A 1 y = c r + w 1+ Kc A 1+ Kc A Classical Control – Prof. Eugenio Schuster – Lehigh University 68 Properties of feedback Disturbance Rejection: Choose control s.t. for w=0,y≈r 1 Open loop: K = ⇒ y = r + w o A 1 Closed loop: K >> ⇒ y ≈ r + 0w = r c A Feedback allows attenuation of disturbance without h(h)having access to it (without measuring it)!!! IMPORTANT: High gain is dangerous for dynamic response!!! Classical Control – Prof. Eugenio Schuster – Lehigh University 69 Properties of feedback Sensitivity to Gain Plant Changes Open loop w + r + y Ko A ⎛ y ⎞ To = ⎜ ⎟ = AKo ⎝ r ⎠o Closed loop w + r + + y Kc A - ⎛ y ⎞ AKc Tc = ⎜ ⎟ = ⎝ r ⎠c 1+ AKc Classical Control – Prof. Eugenio Schuster – Lehigh University 70 Properties of feedback Sensitivity to Gain Plant Changes Let the plant gain be A + δA δT δA Open loop: o = To A δT δA 1 δA δT Closed loop: c = << = o Tc A 1+ AKc A To Feedback reduces sensitivity to plant variations!!! dT /T A dT Sensitivity: ST = = A dA/ A T dA 1 Example: Tc To SA = ,SA =1 1+ AKc Classical Control – Prof. Eugenio Schuster – Lehigh University 71 PID Controller PID: Proportional – Integral – Derivative P Controller: Y (s) C(s)G(s) R(s) + E(s) U (s) Y (s) = , C(s) = K p G(s) R(s) 1+ C(s)G(s) - E(s) 1 = . R(s) 1+ C(s)G(s) u(t) = K pe(t), U (s) = K p E(s) Step Reference: 1 1 1 1 R(s) = ⇒ ess = limsE(s) = lims = s→0 s→0 s 1+ K pG(s) s 1+ K pG(0) •Proportional gain is high e = 0 ⇔ K G(0) → ∞ True when: ss p •Plant has a pole at the origin High gain proportional feedback (needed for good tracking) results in underdamped (or even unstable) transients. Classical Control – Prof. Eugenio Schuster – Lehigh University 72 PID Controller P Controller: Example (lecture06_a.m) R(s) + E(s) U (s) A Y (s) K p 2 - s + s +1 Y (s) K pG(s) K p A = = 2 R(s) 1+ K pG(s) s + s + (1+ K p A) 2 ωn =1+ K p A 1 1 ⇒ ζ = = ⎯⎯⎯→0 K p →∞ 2ζζωn =1 2ωn 2 1+ K p A 9 Underdamped transient for large proportional gain 9 Steady state error for small proportional gain Classical Control – Prof. Eugenio Schuster – Lehigh University 73 PID Controller PI Controller: Y (s) C(s)G(s) R(s) + E(s) K I U (s) Y (s) = , C(s) = K p + G(s) R(s) 1+ C(s)G(s) - s E(s) 1 = . R(s) 1+ C(s)G(s) t ⎛ KI ⎞ u(t) = K pe(t) + KI e(τ )dτ , U (s) = ⎜ K p + ⎟E(s) ∫ ⎝ s ⎠ Step Reference: 0 1 1 1 1 R(s) = ⇒ ess = limsEE(s) = lims = lim = 0 s s→0 s→0 ⎛ K ⎞ s s→0 ⎛ K ⎞ 1+ ⎜ K + I ⎟G(s) 1+ ⎜ K + I ⎟G(s) ⎝ p s ⎠ ⎝ p s ⎠ • It does not matter the value of the proportional gain • Plant does not need to have a pole at the origin. The controller has it! Integral control achieves perfect steady state reference tracking!!! Note that this is valid even for Kp=0 as long as Ki≠0 Classical Control – Prof. Eugenio Schuster – Lehigh University 74 PID Controller PI Controller: Example (lecture06_b.m) R(s) + E(s) K I U (s) A Y (s) K p + 2 - s s + s +1 ⎛ KI ⎞ ⎜ K p + ⎟G(s) Y (s) s (K ps + KI )A = ⎝ ⎠ = R(s) ⎛ K ⎞ s3 + s2 + (1+ K A)s + K A 1+ ⎜ K + I ⎟G(s) p I ⎝ p s ⎠ DANGER: for large Ki the characteristic equation has roots in the RHP 3 2 s + s + (1+ K p A)s + KI A = 0 Analysis by Routh’s Criterion Classical Control – Prof. Eugenio Schuster – Lehigh University 75 PID Controller PI Controller: Example (lecture06_b.m) 3 2 s + s + (1+ K p A)s + KI A = 0 Necessary Conditions: 1+ K p A > 0, KI A > 0 This is satisfied because A > 0, K p > 0, KI > 0 Routh’s Conditions: 3 s 1 1+ K p A 1+ K p A − KI A > 0 2 s 1 KI A ⇓ s1 1+ K A − K A 1 p I K < K + 0 I P A s KI A Classical Control – Prof. Eugenio Schuster – Lehigh University 76 PID Controller PD Controller: Y (s) C(s)G(s) R(s) + E(s) U (s) Y (s) = , C(s) = K p + KDs G(s) R(s) 1+ C(s)G(s) - E(s) 1 = . R(s) 1+ C(s)G(s) de(t) u(t) = K e(t) + K , U (s) = (K + K s)E(s) p D dt p D Step Reference: 1 1 1 1 R(s) = ⇒ ess = limsE(s) = lims = s→0 s→0 s 1+ ()K p + KDs G(s) s 1+ K pG(0) •PtilProportional gain is hihhigh e = 0 ⇔ K G(0) → ∞ True when: ss p •Plant has a pole at the origin PD controller fixes problems with stability and damping by adding “anticipative” action Classical Control – Prof. Eugenio Schuster – Lehigh University 77 PID Controller PD Controller: Example (lecture06_c.m) R(s) A Y (s) + E(s) C(s) = K + K s U (s) p D 2 - s + s +1 Y (s) (K p + KDs)G(s) A(K p + KDs) = = 2 R(s) 1+ (K p + KDs)G(s) s + (1+ KD A)s + (1+ K p A) 2 ωn =1+ K p A 1+ K A 1+ K A ⇒ ζ = D = D 2ζζωn =1+ KD A 2ωn 2 1+ K p A 9 The damping can be increased now independently of Kp 9 The steady state error can be minimized by a large Kp Classical Control – Prof. Eugenio Schuster – Lehigh University 78 PID Controller PD Controller: Y (s) C(s)G(s) R(s) + E(s) U (s) Y (s) = , C(s) = K p + KDs G(s) R(s) 1+ C(s)G(s) - E(s) 1 = . R(s) 1+ C(s)G(s) de(t) u(t) = K e(t) + K , U (s) = (K + K s)E(s) p D dt p D NOTE: cannot appl y p ure differentiation. In practice, KDs is implemented as KDs τ Ds +1 Classical Control – Prof. Eugenio Schuster – Lehigh University 79 PID Controller PID: Proportional – Integral – Derivative R(s) + E(s) ⎛ 1 ⎞ U (s) Y (s) K p ⎜1+ + TDs⎟ G(s) - ⎝ TI s ⎠ ⎡ 1 d de(t)⎤ K u(t) = K e(t) + e(τ )dτ + T K = p , K = K T p ⎢ ∫ D ⎥ I D p D ⎣ TI 0 dt ⎦ TI U (s) ⎛ 1 ⎞ = K p ⎜1+ + TDs⎟ E(s) ⎝ TI s ⎠ PID Controller: Example (lecture06_d.m) Classical Control – Prof. Eugenio Schuster – Lehigh University 80 PID Controller: Ziegler-Nichols Tuning • Empirical method (no proof that it works well but it works well for simple systems) • Only for stable plants • You do not need a model to apply the method Class of plants: Y (s) Ke−td s = U (s) τs +1 Classical Control – Prof. Eugenio Schuster – Lehigh University 81 PID Controller: Ziegler-Nichols Tuning METHOD 1: Based on step response, tuning to decay ratio of 0.25. Tuning Table: τ P : K p = td τ td PD : K p = 0.9 ,TI = td 0.3 τ PID : K p =1.2 ,TI = 2td ,TD = 0.5td td Classical Control – Prof. Eugenio Schuster – Lehigh University 82 PID Controller: Ziegler-Nichols Tuning METHOD 2: Based on limit of stability, ultimate sensitivity method. R(s) + E(s) U (s) Y (s) Ku G(s) - • Increase the constant gain Ku until the response becomes ppyurely oscillatory y( (no decay –marggyinally stable – pure imaginary poles) • Measure the period of oscillation Pu Classical Control – Prof. Eugenio Schuster – Lehigh University 83 PID Controller: Ziegler-Nichols Tuning METHOD 2: Based on limit of stability, ultimate sensitivity method. Tuning Table: P : K p = 0.5Ku P PD : K = 0.45K ,T = u p u I 1.2 P P PID : K = 0.6K ,T = u ,T = u p u I 2 D 8 The Tuning Tables are the same if you make: τ Ku = 2 , Pu = 4td td Classical Control – Prof. Eugenio Schuster – Lehigh University 84 PID Controller: Integrator Windup Actuator Saturates: - valve (fully open) - aircraft rudder (fully deflected) u (Input of the plant) uc (Output of the controller) Classical Control – Prof. Eugenio Schuster – Lehigh University 85 PID Controller: Integrator Windup R(s) + E(s) K I Uc (s) U (s) Y (s) K p + G(s) - s What happens? - large step input in r - large e -large uc → u saturates - eventually e becomes small - uc still large because the integrator is “charged” - u still at maximum - y overshthoots for a long time Classical Control – Prof. Eugenio Schuster – Lehigh University 86 PID Controller: Anti-Windup Plaaount without Anti-Windup: Plant with Anti-Windup: Classical Control – Prof. Eugenio Schuster – Lehigh University 87 PID Controller: Anti-Windup In saturati on, the plant bbhehaves as: For large Ka, this is a system with very low gain and very fast decay rate, i.e., the integration is turned off. Classical Control – Prof. Eugenio Schuster – Lehigh University 88 Steady State Tracking The Unity Feedback Case R(s) E(s) U (s) Y (s) + C(s) G(s) - E(s) 1 = R(s) 1+ C(s)G(s) t k Test Inputs: r(t) = 1(t) k=0: step (position) k! k=1: ramp (velocity) 1 R(s) = sk+1 k2k=2: parabblola ((laccelera ti)tion) Classical Control – Prof. Eugenio Schuster – Lehigh University 89 Steady State Tracking Type n The Unity Feedback Case System R(s) E(s) GU((ss)) Y (s) + C(s) o G(s) n - s Go (s) 1 1 C(s)G(s) = n , E(s) = R(s), R(s) = k +1 s Go (s) s 1+ n Steady State Error: s Final Val ue Theorem 1 1 sn 1 sn−k ess = lime(t) = limsE(s) = lims = lim = lim t→∞ s→0 s→0 G (s) sk+1 s→0 sn + G (s) sk s→0 sn + G (0) 1+ o o o sn Classical Control – Prof. Eugenio Schuster – Lehigh University 90 Steady State Tracking Typype n The Unity Feedback Case System R(s) + E(s) Go (s) Y (s) n n−k - s s ess = lim s→0 n s + Go (0) Steady State Error: Input (k) Type (n) Step (k=0) Ramp (k=1) Parabola (k=2) 1 1 1 Type 0 = = ∞ ∞ 1+ Go (0) 1+ limC(s)G(s) 1+ K p s→0 1 1 1 Type 1 0 = = ∞ Go (0) limsC(s)G(s) Kv s→0 1 1 1 Type 2 0 0 = 2 = Go (0) lims C(s)G(s) Ka s→0 Classical Control – Prof. Eugenio Schuster – Lehigh University 91 Steady State Tracking Position Constant K p = limC(s)G(s) n = 0 s→0 Velocity Constant Kv = limsC(s)G(s) n =1 s→0 2 Acceleration Constant Ka = lims C(s)G(s) n = 2 s→0 n: Degree of the poles of CG(s) at the origin (the number of integrators in the loop with unity gain feedback) • Applying integral control to a plant with no zeros at the origin makes the system type ≥ I • All this is true ONLY for unity feedback systems • Since in Type I systems ess=0 for any CG(s), we say that the system type is a robust property. Classical Control – Prof. Eugenio Schuster – Lehigh University 92 Steady State Tracking t k w(t) = 1(t) The Unity Feedback Case k! 1 W (s) W (s) = sk+1 R(s) E(s) U (s) + Y (s) + G(s) C(s) + - Set rr0=0. Want Y(s)/W(s)=0. Y (s) G(s) = = T (s) = snT (s) W (s) 1+ C(s)G(s) o Steady State Error: e=r-y=-y Final Value Theorem 1 sn − ess = yss = lim y(t) = limsY (s) = lim sT (s) = limTo (s) t→∞ s→0 s→0 sk +1 s→0 sk Classical Control – Prof. Eugenio Schuster – Lehigh University 93 Steady State Tracking The Unity Feedback Case W (s) R(s) E(s) U (s) + Y (s) + G(s) C(s) + - Steady State Output: Disturbance (k) Type (n) Step (k=0) Ramp (k=1) Parabola (k=2) Type 0 * ∞ ∞ Type 1 0 * ∞ Type 2 0 0 * 0 < * < ∞ Classical Control – Prof. Eugenio Schuster – Lehigh University 94 Steady State Tracking Example: W (s) + A R(s) + E(s) K I U (s) Y (s) K p + + - s s()τs +1 KI ≠ 0 ⇒ type 1 to w KP ≠ 0, KI = 0 ⇒ type 0 to w Classical Control – Prof. Eugenio Schuster – Lehigh University 95 Root Locus Controller Plant R(s) E(s) U (s) Y (s) + C(s) G(s) - H (s) Sensor Y (s) C(s)G(s) C(s)G(s) C(s) = KD(s) ⇒ = = R(s) 1+ C(s)G(s)H (s) 1+ KL(s) Writing the loop gain as KL(s) we are interested in tracking the closed-loop poles as “gain” K varies Classical Control – Prof. Eugenio Schuster – Lehigh University 96 Root Locus Characteristic Equation: 1+ KL(s) = 0 The roots (zeros) of the characteristic equation are the closed-loop poles of the feedback system!!! The closed-loop poles are a function of the “gain” K Writing the loop gain as m m−1 b(s) s + b1s +L+ bm−1s + bm L(s) = = n n−1 a(s) s + a1s +L+ an−1s + an The closed loop poles are given indistinctly by the solution of: b(s) 1 1+ KL(s) = 0, 1+ K = 0, a(s) + Kb(s) = 0, L(s) = − a(s) K Classical Control – Prof. Eugenio Schuster – Lehigh University 97 Root Locus RL = zeros{1+ KL(s)}= roots{den(L) + Knum(L)} when K varies from 0 to ∞ (positive Root Locus) or from 0 to -∞ (negative Root Locus) ⎧ 1 1 ⎪ L(s) = Magnitude condition K > 0: L(s) = − ⇔ ⎨ K K ⎩⎪∠L(s) =180o Phase condition ⎧ 1 1 ⎪ L(s) = − Magnitude condition K < 0: L(s) = − ⇔ ⎨ K K ⎩⎪∠L(s) = 0o Phase condition Classical Control – Prof. Eugenio Schuster – Lehigh University 98 Root Locus by Characteristic Equation Solution R(s) E(s) U (s) 1 Y (s) Example: + K - ()()s +10 s +1 Y (s) K = R(s) s2 +11s + (10 + K) 2 Closed-loop poles: 1+ L(s) = 0 ⇔ s +11s + (10 + K) = 0 s = −1,−10 K=0 81− 4K 81− 4K s = −5.5 ± 81−4K>0 s = −5.5 ± 2 2 s = −5.5 81−4K=0 4K − 81 s = −5.5 ± i 81−4K<0 Classical Control – Prof. Eugenio Schuster – Lehigh University 2 99 Root Locus by Characteristic Equation Solution We need a systematic approach to plot the closed-loop poles as function of the gain K → ROOT LOCUS Classical Control – Prof. Eugenio Schuster – Lehigh University 100 Root Locus by Phase Condition Example: R(s) + E(s) U (s) s +1 Y (s) K 2 - s()s + 5 (s + 4s + 8) s +1 L(s) = s(s + 5)(s2 + 4s + 8) s +1 = s(s + 5)(s + 2 + 2i)(s + 2 − 2i) so = −1+ 3i belongs to the locus? Classical Control – Prof. Eugenio Schuster – Lehigh University 101 Root Locus by Phase Condition 45o o 90 108.43o 36.87o 78.70o 90o − [108.43o + 36.87o + 45o + 78.70o ]≈ −180o ⇒ so = −1+ 3i belongs to the locus! Note: Check code lecture09_a.m Classical Control – Prof. Eugenio Schuster – Lehigh University 102 Root Locus by Phase Condition so = −1+ 3i We need a systematic approach to plot the closed-loop poles as function of the gain K → ROOT LOCUS Classical Control – Prof. Eugenio Schuster – Lehigh University 103 Root Locus RL = zeros{1+ KL(s)}= roots{den(L) + Knum(L)} when K varies from 0 to ∞ (positive Root Locus) or from 0 to -∞ (negative Root Locus) 1 1+ KL(s) = 0 ⇔ L(s) = − ⇔ a(s) + Kb(s) = 0 K Basic Properties: • Number of branches = number of open-loop poles • RL begins at open-loop poles K = 0 ⇒ a(s) = 0 • RL ends at open-loop zeros or asasmptotesymptotes ⎧ b(s) = 0 K = ∞ ⇒ L(s) = 0 ⇔ ⎨ ⎩s → ∞ (n − m > 0) • RL symmetrical about Re-axis Classical Control – Prof. Eugenio Schuster – Lehigh University 104 Root Locus Rule 1: The n branches of the locus start at the poles of L(s) and m of these branches end on the zeros of L(s). n: order of the denominator of L(s) m: order of the numerator of L(s) Rule 2: The locus is on the real axis to the left of and odd number of poles and zeros. In other words, an interval on the real axis belongs to the root locus if the total number of poles and zeros to the right is odd. This rule comes from the phase condition!!! Classical Control – Prof. Eugenio Schuster – Lehigh University 105 Root Locus Rule 3: As K→∞, m of the closed-loop poles approach the open-loop zeros, and n-m of them approach n-m asymptotes with angles π φ = (2l +1) , l = 0,1, ,n − m −1 l n − m K and centered at b − a poles − zeros α = 1 1 = ∑ ∑ , l = 0,1, ,n − m −1 n − m n − m K Classical Control – Prof. Eugenio Schuster – Lehigh University 106 Root Locus Rule 4: The locus crosses the jω axis (looses stability) where the Routh criterion shows a transition from roots in the left half-plane to roots in the right-half plane. Example: s + 5 G(s) = s(s2 + 4s + 5) K = 20,s = ± j5 Classical Control – Prof. Eugenio Schuster – Lehigh University 107 Root Locus s +1 Example: G(s) = s4 + 3s3 + 7s2 + 6s + 4 Classical Control – Prof. Eugenio Schuster – Lehigh University 108 Root Locus Design dangers revealed by the Root Locus: • High relative degree: For n-m≥3 we have closed loop instability due to asymptotes. s +1 G(s) = s4 + 3s3 + 7s2 + 6s + 4 • Nonminimum phase zeros: They attract closed loop poles into the RHP s −1 G(s) = s2 + s +1 Note: Check code lecture09_b.m Classical Control – Prof. Eugenio Schuster – Lehigh University 109 Root Locus Viete’s formula: When the relative degree n-m≥2, the sum of the closed loop poles is constant a1 = −∑closed loop poles m m−1 b(s) s + b1s +L+ bm−1s + bm L(s) = = n n−1 a(s) s + a1s +L+ an−1s + an Classical Control – Prof. Eugenio Schuster – Lehigh University 110 Phase and Magnitude of a Transfer Function m m−1 bms + bm−1s +L+ b1s + b0 G(s) = n n−1 s + an−1s +L+ a1s + a0 (s − z )(s − z ) (s − z ) G(s) = K 1 2 L m (s − p1)(s − p2 )L(s − pn ) The fact ors K, (s-zj) and (s-pk) are complex numbers: z z iφ j (s − z j ) = rj e , j =1Km p p iφk (s − pk ) = rk e , k =1L p K K = K eiφ φ φ φ z i z z i z z iφ z K 1 φ2 m iφ r1 e r2 e Lrme G(s) = K e p p p p i 1 p i 2 p iφn r1 e r2 e Lrn e Classical Control – Prof. Eugenio Schuster – Lehigh University 111 Phase and Magnitude of a Transfer Function φ z iφ z z iφ z z iφ z K r e 1 r e φ2 r e m φiφ 1 2 L m G(s) = K e p p p p i 1 p i 2 p iφn r e r e φ r e 1 2 Lφ n z z φ z z z z i(1 + 2 +L+ m ) K r r r e φ i 1 2 L mφ φ = K e φ p p p p p p i()1 +φ 2 +L+φn r1 r2 Lrn e φ φ z z z φ r r r K z z z p p p 1 2 L m i[]+()1 + 2 +L+ m −()1 + 2 +L+φn = K p p p e r1 r2 Lrn Now it is easyφ to give the phase and magnitude φ ooeaseuof the transfer function:φ r zr z r z G(s) = K 1 2 L m , φ p p p φ r1 r2 Lrn φ K z z z p p p ∠G(s) = + ( 1 + 2 +L+ m )− ( 1 + 2 +L+φn ) Classical Control – Prof. Eugenio Schuster – Lehigh University 112 Phase and Magnitude of a Transfer Function (s + 6.735) Example: G(s) = ()()s +1 s + 5 (s + 20) s = s = −7 + 5i o r z G(s) = 1 φ p p p r1 r2 rφ3 p z p p r3 r1 r2 r1 z p φ p p ∠G(s) = 1 − ()1 + 2 +φ3 p z p p φ3 φ1 φ2 φ1 Classical Control – Prof. Eugenio Schuster – Lehigh University 113 Root Locus- Magnitude and Phase Conditions RL = zeros{1+ KL(s)}= roots{den(L) + Knum(L)} when K varies from 0 to ∞ (positive Rootφ Locus) or φ from 0 to -∞ (negative Root Locus) φ φ φ z z z φ (s − z )(s − z ) (s − z ) r r r K p z z z p p p 1 2 L m 1 2 L m i[ +()1 + 2 +L+ m −()1 + 2 +L+φn ] L(s) = K p = K p p p p e (s − p1)(s − p2 )L(s −φpn ) r1 r2 Lrn φ φ r zr z r z 1 L(s) = K 1 2 L m = φ 1 p p p p φ K > 0 : L(s) = − ⇔ r1 r2 Lrn K K φ K p z z z p p p o ∠L(s) = + ()()1 + 2 +L+ m − 1 + 2 +L+φn =180 r zr z r z 1 L(s) = K 1 2 L m = − 1 p p p p K < 0 : L(s) = − ⇔ r1 r2 Lrn K K K p z z z p p p o ∠L(s) = φ + (φ1 +φ2 +L+φm )− (φ1 +φ2 +L+φn )= 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 114 Root Locus Selecting K for desired closed loop poles on Root Locus: If so belongs to the root locus, it must satisfies the characteristic equation for some value of K 1 L(s ) = − o K Then we can obtain K as 1 K = − L(so ) 1 K = L(so ) Classical Control – Prof. Eugenio Schuster – Lehigh University 115 Root Locus 1 Example: L(s) = G(s) = ()()s +1 s + 5 1 so = −3 + i4 ⇒ K = = so +1 so + 5 = − 3 + i4 +1 − 3 + i4 + 5 L(so ) = ()− 2 2 + 42 ()2 2 + 42 = 20 Using MATLAB: sys=tf(1,poly([-1 -5])) so=-3+4i [K,POLES]=rlocfind(sys,so) Classical Control – Prof. Eugenio Schuster – Lehigh University 116 Root Locus 1 Example: L(s) = G(s) = s = −3 + i4 (s +1)(s + 5) o so = −7 + i5 ⇓ 1 K = = 42.06 L(so ) so = −7 + i5 When we use the absolute value formula we are assuming that the point belongs to the Root Locus! Classical Control – Prof. Eugenio Schuster – Lehigh University 117 Root Locus - Compensators 1 Example: L(s) = G(s) = ()()s +1 s + 5 Can we place the closed loop pole at so=-7+i5 only varying K? NO. We need a COMPENSATOR. 1 1 L(s) = G(s) = L(s) = D(s)G(s) = (s +10) (s +1)(s + 5) (s +1)(s + 5) The zero attracts the locus!!! Classical Control – Prof. Eugenio Schuster – Lehigh University 118 Root Locus – Phase lead compensator Pure derivative control is not normally practical because of the amplification of the noise due to the differentiation and must be approximated: s + z D(s) = , p > z Phase lead s + p COMPENSATOR When we study frequency response we will understand why we call “Phase Lead” to this compensator. s + z 1 L(s) = D(s)G(s) = , p > z s + p ()()s +1 s + 5 How do we choose z and p to place the closed loop pole at so=-7+i5? Classical Control – Prof. Eugenio Schuster – Lehigh University 119 Root Locus – Phase lead compensator s + z 1 Example: L(s) = D(s)G(s) = , p < z s + p (s +1)(s + 5) Phase lead COMPENSATOR o o 21.04o 111.80 140.19 z φ1 ? K p z z z p p p o ∠L(s) = φ + (φ1 +φ2 +L+φm )− (φ1 +φ2 +L+φn )=180 z o o o o o o φ1 =180 +140.19 +111.80 + 21.04 = 453.03 = 93.03 ⇒ z = −6.735 Classical Control – Prof. Eugenio Schuster – Lehigh University 120 Root Locus – Phase lead compensator s + 6.735 1 Example: L(s) = D(s)G(s) = s + 20 ()()s +1 s + 5 Phase lead COMPENSATOR so = −7 + i5 K =117 Classical Control – Prof. Eugenio Schuster – Lehigh University 121 Root Locus – Phase lead compensator Selecting z and p is a trial an error procedure. In general: • The zero is ppglaced in the neighborhood of the closed- loop natural frequency, as determined by rise-time or settling time requirements. • The poles is placed at a distance 5 to 20 times the value of the zero location. The pole is fast enough to avoid modifying the dominant pole behavior. The exact position of the pole p is a compromise between: • Noise suppression (we want a small value for p) • Compensation effectiveness (we want large value for p) Classical Control – Prof. Eugenio Schuster – Lehigh University 122 Root Locus – Phase lag compensator s + 6.735 1 Example: L(s) = D(s)G(s) = s + 20 ()()s +1 s + 5 s + 6.735 1 −2 K p = lim L(s) = lim D(s)G(s) = lim = 6.735×10 s→0 s→0 s→0 s + 20 ()()s +1 s + 5 What can we do to increase Kp? Suppose we want Kp=10. s + z s + 6.735 1 L(s) = D(s)G(s) = , p < z s + p s + 20 ()()s +1 s + 5 Phase lag COMPENSATOR z 1 We choose: = ×103 =148.48 p 6.735 Classical Control – Prof. Eugenio Schuster – Lehigh University 123 Root Locus – Phase lag compensator s + 0.14848 s + 6.735 1 Example: L(s) = D(s)G(s) = s + 0.001 s + 20 ()()s +1 s + 5 so = −6.94 + i5.03 K =18.31 Classical Control – Prof. Eugenio Schuster – Lehigh University 124 Root Locus – Phase lag compensator Selecting z and p is a trial an error procedure. In general: • The ratio zero/pole is chosen based on the error constant specification. • We pick z and p small to avoid affecting the dominant dynamic of the system (to avoid modifying the part of the locus representing the dominant dynamics) • Slow transient due to the small p is almost cancelled by an small z. The ratio zero/pole cannot be very big. The exact position of z and p is a compromise between: • Steady state error (we want a large value for z/p) • The transient response (we want the pole p placed far from the origin) Classical Control – Prof. Eugenio Schuster – Lehigh University 125 Root Locus - Compensators s + z Phase lead compensator: D(s) = , z < p s + p s + z Phase lag compensator: D(s) = , z > p s + p We will see why we call “phase lead” and “phase lag” to these compensators when we study frequency response Classical Control – Prof. Eugenio Schuster – Lehigh University 126 Frequency Response • We now know how to analyze and design systems via s-domain methods which yield dynamical information • The responses are described by the ex ponential modes – The modes are determined by the poles of the response Laplace Transform • We next will look at describing cct performance via frequency response methods – This guides us in specifying the system pole and zero positions Classical Control – Prof. Eugenio Schuster – Lehigh University 127 Sinusoidal Steady-State Response CidConsider a stbltable transf er ftifunction with a sinusoidal input: Aω u(t) = Acos(ωt) ⇔ U (s) = s2 + ω 2 The Laplace Transform of the response has poles • Where the natural modes lie –These are in the open left half plane Re(s)<0 • At the input modes s=+jω and s=-jω (s − z1)(s − z2 )L(s − zm ) Aω Y (s) = G(s)U (s) = K 2 2 (s − p1)(s − p2 )L(s − pn ) (s + ω ) Only the response due to the poles on the imaginary axis remains after a sufficiently long time This is the sinusoidal steady-state response Classical Control – Prof. Eugenio Schuster – Lehigh University 128 Sinusoidal Steady-State Response • Input u(t) = Acos(ωt +φ) = Acosωt sinφ − Asinωt cosφ s ω • Transform U (s) = −Acosφ + Asinφ s2 + ω 2 s2 + ω 2 • RTfResponse Transform * k k k1 k2 kN Y (s) = G(s)U (s) = + + + +L+ s − jω s + jω s − p1 s − p2 s − pN • Response Signal forced response natural response jωt * − jωt p1t p2t pNt y(t) = ke + k e + k1e + k2e +L+ kN e 1442443 1444442444443 forced response natural response • Sinusoidal Steady State Response t → ∞ jωt * − jωt ySS (t) = ke + k e 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 129 Sinusoidal Steady-State Response • Calculating the SSS response to u(t) = Acos(ω+t φ) • Residue calculation k = liω m[(s − jω)Y(s)]= lim[(s − jω)G(s)U(s)] ω s→ j ω s→ j ω φ ω φ ω ⎡ φ scos − sinφ ⎤ ⎡ j cos − sinφ ⎤ = lim ⎢G(s)(s − j )A ω ⎥ = G( j )A⎢ ⎥ s→ jω⎣ ω (s − jω)(s + jω)⎦ ⎣ 2 jω ⎦ 1 1 φ = AG( j ) e j = AG( jω)e j( +∠G( jω)) 2 2 •Signal calculation * −1⎧ k ω k ⎫ yss (t) = L ⎨ ω + ⎬ ⎩s − j s + jω⎭ = k e j∠Ke j t + k e− j∠Ke− jωt = 2k cos()ωt + ∠K yss (t) = AG( jω) cos(ωt +φ + ∠G( jω)) Classical Control – Prof. Eugenio Schuster – Lehigh University 130 Sinusoidal Steady-Stateω Response ω • Response to u(t) = Acos(ωt +φ) φ •is yss = G( j ) Acos( t + + ∠G( jω)) – Output frequency = input frequency – Outppput amplitude = inp pput amplitude × |G(jω)| – Output phase = input phase +∠ G(jω) • The F requency R esponse of th e transf er f uncti on G(s ) iiis given by its evaluation as a function of a complex variable at s=jω • We speak of the amplitude response and of the phase response – They cannot independently be varied » Bode’s relations of analytic function theory Classical Control – Prof. Eugenio Schuster – Lehigh University 131 Frequency Response • Find the steady state output for v1(t)=Acos(ωt+φ) + sL + R V (s) V1(s) _ 2 - • CtthCompute the s-ditfftiT()domain transfer function T(s) ω R – Voltage divider T (s) = sL + R • Compute the frequency response R ω −1⎛ωL ⎞ T ( j ) = , ∠T ( jω) = − tan ⎜ ⎟ R2 + ( L)2 ⎝ R ⎠ ω • Compute the steady state output φ AR −1 v2SS (t) = cos[t + − tan ()ωL / R ] R2 + (ωL)2 Classical Control – Prof. Eugenio Schuster – Lehigh University 132 Bode Diagrams • Log-log plot of mag(T), log-linear plot of arg(T) versus ω Bode Diagram 0 )) -5 -10 -15 itude (dB agnitude (dB)agnitude nn MM -20 Mag -25 0 -45 e (deg) (deg) ss ee Pha Phas -90 4 5 6 10 10 10 Frequency (rad/sec) Classical Control – Prof. Eugenio Schuster – Lehigh University 133 Frequency Response u(t) = Acos(ωt +φ) G(s) yss = G( jω) Acos(ωt +φ + ∠G( jω)) Stable Transfer Function • After a transient, the output settles to a sinusoid with an amplitude magnified by G ( j ω ) and phase shifted by ∠ G ( j ω ) . • Since all signals can be represented by sinusoids (Fourier series and transform), the quantities G ( j ω ) and ∠ G ( j ω ) are extremely important. • Bode developed methods for quickly finding G ( j ω ) and ∠ G ( j ω ) for a given G ( s ) and for using them in control design. Classical Control – Prof. Eugenio Schuster – Lehigh University 134 Frequency Response • Find the steady state output for v1(t)=Acos(ωt+φ) + sL + R V (s) V1(s) _ 2 - • CtthCompute the s-ditfftiT()domain transfer function T(s) ω R – Voltage divider T (s) = sL + R • Compute the frequency response R ω −1⎛ωL ⎞ T ( j ) = , ∠T ( jω) = − tan ⎜ ⎟ R2 + ( L)2 ⎝ R ⎠ ω • Compute the steady state output φ AR −1 v2SS (t) = cos[t + − tan ()ωL / R ] R2 + (ωL)2 Classical Control – Prof. Eugenio Schuster – Lehigh University 135 Frequency Response - Bode Diagrams • Log-log plot of mag(T), log-linear plot of arg(T) versus ω Bode Diagram 0 )) -5 -10 -15 itude (dB agnitude (dB)agnitude nn MM -20 Mag -25 0 -45 e (deg) (deg) ss ee Pha Phas -90 4 5 6 10 10 10 Frequency (rad/sec) []ω = rad /sec, ω = 2πf , [f ]= Hz Classical Control – Prof. Eugenio Schuster – Lehigh University 136 Bode Diagrams (s − z )(s − z ) (s − z ) G(s) = K 1 2 L m (s − p1)(s − pφ2 )L(s − pn ) φ φ φ φ z z z φ r r r K z z z p p p 1 2 L m i[]+()1 + 2 +L+ m −()1 + 2 +L+φn G(s) = K p p p e r1 r2 Lrn The magnitudeω and phase of G(s) when s=jω is given by: φ φ φ Nonlinear in the magnitudes r zr z r z G( jω) = K 1 2 L m , φ p p p φ r1 r2 Lrn φ K z z z p p p ∠G( j ) = + ( 1 + 2 +L+ m )− ( 1 + 2 +L+φn ) Linear in the phases Classical Control – Prof. Eugenio Schuster – Lehigh University 137 Bode Diagrams Why do we express G ( jω ) in decibels? ω G( jω) dB = 20log G( jω) z z z r1 r2 Lrm G( j ) = K p p p ⇒ G( jω) dB = ? r1 r2 Lrn By properties of the logarithm we can write: z z z p p p 20logG(s) = 20log K + (20logr1 + 20logrm +L+ 20logrm )− (20logr1 + 20logr2 +L+ 20logrn ) The magnitude and phase of G(s) when s=jω is given by: Linear in the magnitudes (dB) G(s) = K + (r z + r z + + r z )− (r p + r p + + r p ) dB dB 1 dB 2 dB L m dB 1 dB 2 dB L n dB K z z z p p p ∠G(s) = φ + (φ1 +φ2 +L+φm )− (φ1 +φ2 +L+φn ) Linear in the phases Classical Control – Prof. Eugenio Schuster – Lehigh University 138 Bode Diagrams Why do we use a logarithmic scale? Let’s have a look at our example: R R 1 ωT (s) = ⇒ T ( jω) = = 2 2 2 sLL + R R + (ωL) ⎛ωL ⎞ 1+ ⎜ ⎟ ⎝ R ⎠ Expressing the magnitude in dB: ω 2 2 ⎛ L ⎞ ⎡ ⎛ωL ⎞ ⎤ T ( j ) dB = 20log1− 20log 1+ ⎜ ⎟ = −10log⎢1+ ⎜ ⎟ ⎥ ⎝ R ⎠ ⎣ ⎝ R ⎠ ⎦ Asymptotic behavior: ω ωω → 0 : T( jω) dB → 0 ⎛ ω ⎞ R → ∞ : T ( jω) dB → −20log⎜ ⎟ = 20log()R / L − 20log = − 20logω ⎝ R / L ⎠ L dB LINEAR FUNCTION in logω!!! We plot as a function of logω. G( jω) dB Classical Control – Prof. Eugenio Schuster – Lehigh University 139 Bode Diagrams Decade: Any frequency range whose end points have a 10:1 ratio A cutoff frequency occurs when the gain is reduced from its maximum passb an d val ue by a fact or 1 / 2 : ⎛ 1 ⎞ 20log⎜ T ⎟ = 20logT − 20log 2 ≈ 20logT − 3dB ⎝ 2 MAX ⎠ MAX MAX Bandwith: frequency range spanned by the gain passband Let’s have a look at our example:ω 1 ⎧ = 0 T ( jω) =1 T ( jω) = ⇒ 2 ⎨ ⎛ωL ⎞ ⎩ω = R / L T ( jω) =1/ 2 1+ ⎜ ⎟ ⎝ R ⎠ This is a low-pass filter !!! Passb and gai n=1, Cuto ff frequency=R/L The Bandwith is R/L! Classical Control – Prof. Eugenio Schuster – Lehigh University 140 Generalω Transfer Function (Bode Form) 2 q ω ⎡⎛ jω ⎞ jω ⎤ G()j = K j ()(m jωτ +1 n ⎢⎜ ω )⎟ + 2ζ +1⎥ o ⎜ ⎟ ω ⎣⎢⎝ n ⎠ n ⎦⎥ The magnitude (dB) (phase) is the sum of the magnitudes (dB) (phases) of each one of the terms. We learn how to plot each term, we learn how to plot the whole magnitude and phase Bode Plot. Classes of terms: 1- G(jω)= Ko ω 2- G()j = (jω)m 3- ω G()j = (jωτ +1)n 2 q 4- ⎡⎛ jω ⎞ jω ⎤ G(jω)= ⎢⎜ ⎟ + 2ζ +1⎥ ⎜ ω ⎟ ω ⎣⎢⎝ n ⎠ n ⎦⎥ Classical Control – Prof. Eugenio Schuster – Lehigh University 141 General Transfer Function: DC gain G(jω)= Ko Magnitude and Phase: G(jω)dB = 20log Ko ⎧ 0 if Ko > 0 ∠G()jω = ⎨ G(s)= −10 ⎩± π if Ko < 0 40 200 180 35 160 30 140 25 120 20 e (deg) 100 tude (dB) ss Pha 80 15 Magni 60 10 40 5 20 0 0 -1 0 1 2 -1 0 1 2 10 10 10 10 10 10 10 10 Frequency (rad/sec) Frequency (rad/sec) Classical Control – Prof. Eugenio Schuster – Lehigh University 142 General Transfer Function: Poles/zeros at origin G(jω)= (jω)m Magnitude and Phase: G(jω)dB = m⋅ 20logω ω 1 π m = −1,G(s)= ∠G()j = m s 2 20 0 10 -20 G(1)dB = 0 0 -40 -10 -60 itude (dB) se (deg) nn aa Ph Mag -20 dB -80 m⋅ 20 -30 dec -100 -40 -120 -1 0 1 2 -1 0 1 2 10 10 10 10 10 10 10 10 Frequency (rad/sec) Frequency (rad/sec) Classical Control – Prof. Eugenio Schuster – Lehigh University 143 General Transfer Function: Real poles/zeros ω G(j )= (jωτ +1)n Magnitude and Phase: ω 2 2 G()jω ωdB = n ⋅10log( τ +1) ∠G(j )= n tan−1(ωτ ) Asymptoticω behavior: ω ω o τ ∠G( jω) ⎯⎯τ⎯→0 G( jω) dB ⎯⎯<<τ1⎯/ →0 <<1/ ω ∠G( jω) ⎯ω⎯→⎯ n ⋅90o G( j ) dB ⎯⎯→>>1⎯/τ n ⋅ dB + n ⋅ 20logω >>1/τ Classical Control – Prof. Eugenio Schuster – Lehigh University 144 General Transfer Function: Real poles/zeros 10 n = −1,τ =1/10 5 0 1 dB G()s = -5 n ⋅ 20 s B) n ⋅3dB dec +1 dd -10 10 -15 -20 agnitude ( M -25 G( j0) dB = 0dB -30 G( j1/τ ) dB = n ⋅3dB -35 G(∞) = sgn(n)∞dB -40 dB -1 0 1 2 3 10 10 10 10 10 Freqqy()uency (rad/sec) Classical Control – Prof. Eugenio Schuster – Lehigh University 145 General Transfer Function: Real poles/zeros 10 n = −1,τ =1/10 0 -10 1 G()s = -20 s +1 )) -30 10 -40 -50 hase (deg o PP -60 n ⋅ 45 -70 ∠G( j0) = 0o -80 o -90 ∠G( j1/τ ) = n ⋅ 45 -100 o -1 0 1 2 3 ∠G( j∞) = n ⋅90 10 10 10 10 10 Frequency (rad/sec) Classical Control – Prof. Eugenio Schuster – Lehigh University 146 General Transfer Function: Complex poles/zeros 2 q ⎡⎛ jω ⎞ jω ⎤ G()jω = ⎢⎜ ω ⎟ + 2ζ +1⎥ ⎜ ⎟ ω ω ⎣⎢⎝ n ⎠ n ⎦⎥ Magnitude and Phase: 2 2 ⎡⎛ ω 2 ⎞ ⎛ ω ⎞ ⎤ G()j = q ⋅10log⎢⎜1− ⎟ + ⎜2 ζ ⎟ ⎥ dB ⎜ ω 2 ⎟ ⎜ ω ⎟ ⎣⎢⎝ n ⎠ ⎝ n ⎠ ⎦⎥ ζω ⎛ 2 /ω ⎞ ( ) −1⎜ n ⎟ ∠G jω = q ⋅ tan ⎜ 2 2 ⎟ ⎝1−ω /ωn ⎠ Asymptotic behavior: G( jω) ⎯ω⎯⎯→0 ∠G( jω) ⎯ω⎯→⎯ 0o dB <<ωn <<ωn o G( jω) ⎯ω⎯>>⎯ω →−2q ⋅ωn + q ⋅ 40logω ∠G( jω) ⎯⎯⎯→q ⋅180 dB n dB ω>>ωn Classical Control – Prof. Eugenio Schuster – Lehigh University 147 General Transfer Function: Complex poles/zeros 20 q = −1,ωn =1,ζ = 0.05 q ⋅ 2ζ 0 dB 1 -20 dB G(s)= 2 q ⋅ 40 s + 0.1s +1 dB) dec (( -40 -60 Magnitude Magnitude ω G( j0) = 0dB -80 ω dB G( j n ) dB = q ⋅()3dB + ζ dB -100 ω G( j∞) dB = sgn(q)∞dB -120 -1 0 1 ζ 2 3 10 10 10 10 10 Frequency (rad/sec) ζ MAX 2 ω ωn G( j ) = G( j r ) = q ⋅(2 1− ) ⇔ = ωr = dB dB dB 1−ζ 2 Classical Control – Prof. Eugenio Schuster – Lehigh University 148 General Transfer Function: Complex poles/zeros q = −1,ω =1,ζ = 0.05 20 n 0 -20 1 G(s)= 2 -40 s + 0.1s +1 )) -60 -80 o -100 ∠G( j0) = 0 hase (deg q ⋅90o PP -120 ∠G( j1/τ ) = q ⋅90o -140 ∠G( j∞) = q ⋅180o -160 -180 -200 -2 -1 0 1 2 10 10 10 10 10 Frequency (rad/sec) Classical Control – Prof. Eugenio Schuster – Lehigh University 149 Frequency Response 2000(s + 0.5) Example: G()s = s(s +10)(s + 50) Classical Control – Prof. Eugenio Schuster – Lehigh University 150 Frequency Response: Poles/Zeros in the RHP • Same G ( jω ) . • The effect on ∠ G ( j ω ) is opposite than the stable case. An unstable pole behaves like a stable zero An “unstable” zero behaves like a “stable” pole 1 Example: G(s)= s − 2 This frequency response cannot be found experimentally but can be computed and used for control design. Classical Control – Prof. Eugenio Schuster – Lehigh University 151 First order LOW PASS K T (s) = s +α ⇓ K T ( jω) = jω +α Gain and Phase: ω K T ( j ) = ω 2 +α 2 θ (ω) = ∠K − tan−1(ω /α ) ⎧ 0 K > 0 ∠K = ⎨ ⎩−180o K < 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 152 First order LOW PASS ω K T ( j ) = ω 2 +α 2 θ (ω) = ∠K − tan−1(ω /α ) K T (0) = ,T (∞) = 0 α K K α T (0) ω T ( jα) = = = ⇒ c = α Cutoff frequency α 2 +α 2 2 2 K T ( jω) ⎯ω⎯→⎯ <<α α K K = ⇔ ω = ω = α Cutoff frequency K α ω c T ( jω) ⎯ω⎯→⎯ >>α ω B = ωc Bandwith Classical Control – Prof. Eugenio Schuster – Lehigh University 153 First order LOW PASS ω K T ( j ) = ω 2 +α 2 θ (ω) = ∠K − tan−1(ω /α ) θ (0) = ∠K θ (α) = ∠K − tan−1(1) = ∠K − 45o θ (ω) ⎯ω⎯<<⎯α →∠K θ ω o (ω) ⎯⎯→>>⎯α ∠K − 90 Classical Control – Prof. Eugenio Schuster – Lehigh University 154 First Order HIGH PASS Ks T (s) = s +α ⇓ ω Kjω T ( j ) = jω +α Gain and Phase: K ω T ( jω) = ω 2 +α 2 θ (ω) = ∠K + 90o − tan−1(ω /α ) ⎧ 0 K > 0 ∠K = ⎨ ⎩−180o K < 0 Classical Control – Prof. Eugenio Schuster – Lehigh University 155 First order HIGH PASS K ω T ( jω) = ω 2 +α 2 θ (ω) = ∠K + 90o − tan−1(ω /α ) T (0) = 0, T (∞) = K K α K T (∞) ω T ( jα) = = = ⇒ c = α Cutoff frequency α 2 +α 2 2 2 T ( jω) ⎯ω⎯⎯→ K ω /α <<α K ω = K ⇔ ω = ωc = α Cutoff frequency ω T ( jω) ⎯⎯>>⎯α → K α B = ∞ Bandwith Classical Control – Prof. Eugenio Schuster – Lehigh University 156 First order HIGH PASS K ω T ( jω) = ω 2 +α 2 θ (ω) = ∠K + 90o − tan−1(ω /α ) θ (0) = ∠K + 90o θ (α) = ∠K + 90o − tan−1(1) = ∠K + 45o θ ω o (ω) ⎯⎯→<<⎯α ∠K + 90 −1 ω o θ (ω) ⎯⎯>>⎯α →∠K + 90 − tan (∞) = ∠K Classical Control – Prof. Eugenio Schuster – Lehigh University 157 First Order BANDPASS ⎛ K1αs ⎞⎛ K2 ⎞ T (s) = T1(s)×T2 (s) = ⎜ ⎟⎜ ⎟ ⎝ s + 1 ⎠⎝ s +α2 ⎠ ⇓ ω ⎛ K1 jω ⎞⎛ K2 ⎞ T ( j ) = ⎜ ⎟⎜ ⎟ ⎝ jω +α1 ⎠⎝ jω +α2 ⎠ ω ⎛ K ω ⎞⎛ K ⎞ T ( j ) = ⎜ 1 ⎟⎜ 2 ⎟ ⎜ 2 2 ⎟⎜ 2 2 ⎟ ⎝ ω +α1 ⎠⎝ ω +α2 ⎠ ω K1 K2 ω T ( jω) ⎯⎯→α⎯⎯⎯ ω << 1<<α2 α K K ω K K 1α2 1 2 1 2 H = ⇒ ω = ωc = α1 α1α2 α2 α K1 K2 T ( j ) ⎯⎯→⎯ω ⎯⎯ ω 1<< <<α2 K K K K ω α2 1 2 1 2 L ω = ⇒ = c = α2 α ω α K K 2 ω α 1 2 T ( jω) ⎯⎯→⎯⎯⎯ L H α 1<< 2 <<ω ω B = c − c = 2 −α2 Passband Classical Control – Prof. Eugenio Schuster – Lehigh University 158 ω First Order BANDSTOPω ω ⎛ α ⎞ ⎛ ⎞ K1 K2 T ( j ) = ⎜ ⎟ + ⎜ ω ⎟ ω⎜ 2 2 ⎟ ⎜ 2 2 ⎟ ⎝ + 1 ⎠ ⎝ +α2 ⎠ ω K2 T ( j ) ⎯⎯→α⎯⎯⎯ << 2 <<α1 α2 α α T ( jω) ⎯⎯⎯⎯⎯→ K1 2 << 1<<ω K2 = K1 α2 K2 α T ( jω) ⎯⎯⎯<ω→ 2 ω K2 K1 ω α1 K2 α ⇒ = α ⇒ ω = = 1α2 K ω ω K T ( jω) ⎯⎯⎯→ 1 1 1 ω<α1 α1 B = α1 −α2 Bandstop Classical Control – Prof. Eugenio Schuster – Lehigh University 159 Second Order BANDPASS ζω ω Ks ω Kζωjω T (s) = 2 2 ⇒ T ( jω) = 2 ω 2 s + 2 os + o − + 2 o j + ωo ωo : NtNatura l Frequency ζ : Damping Ratio Classical Control – Prof. Eugenio Schuster – Lehigh University 160 Second Order BANDPASS ζω ω Ks ω Kjζωω T (s) = 2 2 ⇒ T ( jω) = 2 ω 2 s + 2 os + o − + 2 o j + ωo K ω T ( jω) ⎯⎯⎯→ = ω ω<<ωo 2 ωo K ω Kω ω 2 = ⇒ = ωo K o T ( jω) ⎯⎯→⎯ = ω>>ωo ω K /ωo K /ωo T ( jω) = ⎯ω⎯⎯→T ( jω) = ζ =ωo MAX ⎛ ω ωo ⎞ 2ζ 2 + j⎜ − ⎟ ⎝ωo ω ⎠ Classical Control – Prof. Eugenio Schuster – Lehigh University 161 Second Order BANDPASS ω ω K /ωo K /ωo T ( jω) = ⎯ω⎯→⎯ T ( jω) = =ωo MAX ⎛ o ⎞ 2ζ 2ζ + j⎜ − ⎟ ⎝ωo ω ⎠ K /ωo ω ζ ω K /ωo ζ 2ζ T ( jω) MAX T ( jω) ω = ⇒ T ( jω) = = − ωo =±2ζ 2 + j2 2 2 o ω ω ω ωo ω The roots of − = ±2ζ are the cutoff frequencies!!! ω o ζ ω 2 2 ω C1 = o (− + 1+ ζ ) = ω Center Frequency ⇒ o C1 C 2 2 Bandwith ωC 2 = ωo (+ ζ + 1+ ζ ) B = ωC 2 −ωC1 = 2ζ Classical Control – Prof. Eugenio Schuster – Lehigh University 162 Second Order LOWPASS ζω ω ω ω K ζωK T (s) = 2 2 ⇒ T ( j ) = 2 ω 2 s + 2 os + o − + 2 o j + ωo ωo : NtNatura l Frequency ζ : Damping Ratio Classical Control – Prof. Eugenio Schuster – Lehigh University 163 ζω ω Second Orderω LOWPASS ω K ζωK T (s) = 2 2 ⇒ T ( j ) = 2 ω 2 s + 2 os + o − + 2 o j + ωo K T ( jω) ⎯ω⎯→⎯ = T (0) <<ωo 2 ωo K K ω 2 = 2 ⇒ ω = ωo K o ω ω T ( jω) ⎯⎯→⎯ = 2 >>ωo ω K /ω T (0) T ( jω ) = o = o 2ζ 2ζ T (0) 2 T ( jω) = ⇔ ω = ωMAX = ωo 1−ζ MAX 2ζ 1−ζ 2 Classical Control – Prof. Eugenio Schuster – Lehigh University 164 Second Order HIGHPASS ζω ω Ks2 ω −ζωKω 2 T (s) = 2 2 ⇒ T ( jω) = 2 ω 2 s + 2 os + o − + 2 o j + ωo ωo : NtNatura l Frequency ζ : Damping Ratio Classical Control – Prof. Eugenio Schuster – Lehigh University 165 ζω ω Second Order ωHIGHPASS ω ω Ks2 −ζωK 2 T (s) = 2 2 ⇒ T ( j ) = 2 ω 2 s + 2 os + o − + 2 o j + ωo K ω 2 T ( jω) ⎯⎯⎯→− 2 ω<<ωo 2 K ω ωo = K ⇒ ω = ω ω 2 o T ( jω) ⎯ω⎯→⎯ = K = T (∞) o >>ωo K T (∞) T ( jω ) = = o 2ζ 2ζ T (∞) ωo T ( jω) = ⇔ ω = ωMAX = MAX 2ζ 1−ζ 2 1−ζ 2 Classical Control – Prof. Eugenio Schuster – Lehigh University 166 Frequency Response u(t) = Acos(ωt +φ) G(s) yss = G( jω) Acos(ωt +φ + ∠G( jω)) Stable Transfer Function ω j∠G( jω) G( j ) = G( jω) e BODE plots G( jω) = Re{}G( jω) + j Im{G( jω)} NYQUIST plots Classical Control – Prof. Eugenio Schuster – Lehigh University 167 ω ω Frequency Response ω G( j ) = Re{}G( j ) + j Im{G( j )}= G( jω) e j∠G( jω) How are the Bode and Nyquist plots related? They are two way to represent the same information j Im{}G( jω) G( jω) ∠G( jω) Re{G( jω)} Classical Control – Prof. Eugenio Schuster – Lehigh University 168 Frequency Response • Find the steady state output for v1(t)=Acos(ωt+φ) + sL + R V (s) V1(s) _ 2 - • CtthCompute the s-ditfftiT()domain transfer function T(s) ω R – Voltage dividerT (s) = sL + R • Compute the frequency response R ω −1⎛ωL ⎞ T ( j ) = , ∠T ( jω) = − tan ⎜ ⎟ R2 + ( L)2 ⎝ R ⎠ ω • Compute the steady state output φ AR −1 v2SS (t) = cos[t + − tan ()ωL / R ] R2 + (ωL)2 Classical Control – Prof. Eugenio Schuster – Lehigh University 169 Frequency Response - Bode Plots • Log-log plot of mag(T), log-linear plot of arg(T) versus ω Bode Diagram 0 )) -5 10 G(s) = -10 s +10 -15 itude (dB agnitude (dB)agnitude nn MM R / L =10 -20 Mag -25 0 -45 e (deg) (deg) ss ee Pha Phas -90 4 5 6 10 10 10 Frequency (rad/sec) [ω]= rad /sec, ω = 2πf , [f ]= Hz Classical Control – Prof. Eugenio Schuster – Lehigh University 170 Frequencyω Response – Nyquist Plots Rω R R − jωL R2 − jωRL T ( j ) = = = R + j L R + jωL R − jωL R2 + ω 2L2 R −1⎛ωL ⎞ T ( jω) = , ∠T ( jω) = − tan ⎜ ⎟ R2 + (ωL)2 ⎝ R ⎠ ω R2 ωRL Re{}T ( jω) = , Im{}T ( j ) = − R2 + ω 2L2 R2 + ω 2L2 ω ω → 0 : T ( jω) →1, ∠T ( jω) → 0 T ( jω) =1 1- ω ω R 2- → ∞ : T( j ) → 0, ∠T ( jω) → −90o T ( jω) ⎯⎯→⎯ − j ⎯⎯→⎯ 0 →∞ ωL ω→∞ 3- Re{}T ( jω) = 0 ⇔ ω = ∞ 4- Im{T( jω)}= 0 ⇔ ω = 0,ω = ∞ Classical Control – Prof. Eugenio Schuster – Lehigh University 171 Frequency Response - Nyquist Plots Im{G( jω)} vs. Re{G( jω)} 10 G(s) = s +10 R / L =10 Classical Control – Prof. Eugenio Schuster – Lehigh University 172 Nyquist Diagrams General procedure for sketching Nyquist Diagrams: • Find G(j0) • Find G(j∞) • Fin d ω* such that Re{ G(jω*)}0}=0; II{m{G(jω*)} is th e intersection with the imaginary axis. • Find ω* such that Im{G(jω*)}0;}=0; Re{G(jω*)} is the intersection with the real axis. • Connect the points Classical Control – Prof. Eugenio Schuster – Lehigh University 173 Frequency Response - Nyquist Plots 1 Example: G(s) = s()s +1 2 ω ω 1 1 (− j )(1− jω 2 ) − 2 + j(ω 2 −1) G( jω) = 2 = 2 2 = 2 jω(jω +1) jω(jω +1) (− jω)(1− jω) ω(ω 2 +1) 1- ω → 0:G( jω) = −2 − j∞ ω ω 1 2- → ∞ :G( jω) ⎯⎯→⎯ j ⎯⎯→⎯ 0 →∞ ω3 ω→∞ 3- Re{}G( jω) = 0 ⇔ ω = ∞ 1 Im{G( jω)}= 0 ⇔ ω =1,ω = ∞ Re{G( j1)}= − 4- 2 Classical Control – Prof. Eugenio Schuster – Lehigh University 174 Frequency Response - Nyquist Plots 1 Example: G(s) = s()s +1 2 Classical Control – Prof. Eugenio Schuster – Lehigh University 175 Nyquist Plots based on Bode Plots 1 G(s) = s()s +1 2 dB − 20 dec dB − 60 dec − 90o −180o − 270o = 90o Classical Control – Prof. Eugenio Schuster – Lehigh University 176 Nyquist Stability Criterion U (s) + G(s) Y (s) - When is this transfer function Stable? NYQUIST: The closed loop is asymptotically stable if the number of counterclockwise encirclements of the point (-1+j0) that the Nyquist curve of G(jω) is equal to the number of poles of G(s) with positive real parts (unstable poles) Corollary: If the open-loop system G(s) is stable, then the closed-loop system is also stable provided G(s) makes no encirclement of the point (-1+j0). Classical Control – Prof. Eugenio Schuster – Lehigh University 177 Nyquist Stability Criterion 1 1 G(s) = G(s) = s4 + 2s3 + 3s2 + 3s +1 s4 + 5s3 + 3s2 + 3s +1 Classical Control – Prof. Eugenio Schuster – Lehigh University 178 Nyquist Stability Criterion U (s) + G(s) Y (s) - K When is this transfer function Stable? NYQUIST: The closed loop is asymptotically stable if the number of counterclockwise encirclements of the point (-1/K+j0) that the Nyquist curve of G(jω) is equal to the number of poles of G(s) with positive real parts (unstable poles) Classical Control – Prof. Eugenio Schuster – Lehigh University 179 Neutral Stability U (s) + KG(s) Y (s) - 1 G(s) = s(s +1)2 Root locus condition: KG(s) =1, ∠G(s) = −180o At points of neutral stability RL condition hold for s=jω KG( jω) =1, ∠G( jω) = −180o Stability: At ∠G( jω) = −180o KG( jω) <1 If ↑K leads to instability KG( jω) >1 If ↓K leads to instability Classical Control – Prof. Eugenio Schuster – Lehigh University 180 Stability Margins The GAIN MARGIN (GM) is the factor by which the gain can be raised before instability results. UNSTABLE SYSTEM GM <1(GM dB < 0)⇒ GM is equal to at the frequency 1/ KG( jω) (− KG( jω) dB ) where ∠ G ( jω ) = − 180 o . The PHASE MARGIN (PM) is the value by which the phase can be raised before instability results. PM < 0 ⇒ UNSTABLE SYSTEM PM is the amount by which the phase of G ( j ω ) exceeds -180° when KG( jω) =1 (KG( jω) dB = 0) Classical Control – Prof. Eugenio Schuster – Lehigh University 181 Stability Margins 1 G(s) = s()s +1 2 1/GM PM Classical Control – Prof. Eugenio Schuster – Lehigh University 182 Stability Margins 1 G(s) = s()s +1 2 GM PM Classical Control – Prof. Eugenio Schuster – Lehigh University 183 Specifications in the Frequency Domain 1. The crossover frequency ωc, which determines bandwith ωBW, rise time tr and settling time ts. 2. The phase margin PM, which determines the damping coefficient ζ and the overshoot Mp. 3. The low-frequency gain, which determines the steady-state error characteristics. Classical Control – Prof. Eugenio Schuster – Lehigh University 184 Specifications in the Frequency Domain The phase and the magnitude are NOT independent! Bode’s Gain-Phase relationship: 1 +∞dM ∠G()jω = W ()u du o π ∫−∞ du M = ln G(jω) u = ln(ω /ωo ) W (u) = ln()coth u / 2 Classical Control – Prof. Eugenio Schuster – Lehigh University 185 Specifications in the Frequency Domain The crossover frequency: ωc ≤ ωBW ≤ 2ωc Classical Control – Prof. Eugenio Schuster – Lehigh University 186 Specifications in the Frequency Domain The Phase Margin: PM vs. Mp Classical Control – Prof. Eugenio Schuster – Lehigh University 187 Specifications in the Frequency Domain PM The Phase Margin: PM vs. ζ ζ ≅ 100 Classical Control – Prof. Eugenio Schuster – Lehigh University 188 Frequency Response – Phase Lead Compensators Ts +1 D(s) = , α <1 αTs +1 1− sinφ α =ω MAX 1+ sinφMAX 1 ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞⎤ log MAX = ⎢log⎜ ⎟ + log⎜ ⎟⎥ 2 ⎣ ⎝T ⎠ ⎝αT ⎠⎦ Classical Control – Prof. Eugenio Schuster – Lehigh University 189 Frequency Response – Phase Lead Compensators 1. Determine the open-loop gain K to satisfy error or bandwidth requirements: - To meet error requirement, pick K to satisfy error constttants (Kp, Kv, Ka) so ththtat ess specifica tion is met . - To meet bandwidth requirement, pick K so that the open-loop crossover frequency is a factor of two below the desired closed-loop bandwidth. 2. Determine the needed phase lead →αbased on the PM specification. 3. Pick ωMAX to be at the crossover frequency. 4. Determine the zero and pole of the compensator. 5. Draw the compensated frequency response and check PM. 6. Iterate on the design Classical Control – Prof. Eugenio Schuster – Lehigh University 190 Frequency Response – Phase Lag Compensators Ts +1 D(s) = α , α >1 αTs +1 Classical Control – Prof. Eugenio Schuster – Lehigh University 191 Frequency Response – Phase Lag Compensators 1. Determine the open-loop gain K that will meet the PM requirement without compensation. 2. Draw the BBdode plltot of the uncompensat tded system with crossover frequency from step 1 and evaluate the low- frequency gain. 3. Determine α to meet the low frequency gain error requirement. 4. Choose the corner frequency ω=1/T (the zero of the compensator) to be one decade below the new crossover freqqyuency ωc. 5. The other corner frequency (the pole of the compensator) is then ω=1/ α T. 6. Iterate on the design Classical Control – Prof. Eugenio Schuster – Lehigh University 192