Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 6: Generalized and Controller Design
Overview In this Lecture, you will learn: Generalized? What about changing OTHER parameters T D, T I, et c. mass, damping, et c. Compensation via Root-Locus Introduction Pole-Zero Compensation Lead-Lag M. Peet Lecture 6: Control Systems 2 / 22
Generalized We may want to know the effect of changing other parameters. Examples: Physics (e.g. Suspension System) Spring Constant Damper Wheel Mass m c m w x x 2 u Circuits (e.g. Toaster, Video Surveillance) Resisters Capacitors Inductors Maybe there is no control at all! M. Peet Lecture 6: Control Systems 3 / 22
as a General Tool What do parameters do? Suspension system: The full TF: K 2 (m c s 2 + cs + K ) m c m w s 4 + c(m w + m c )s 3 + (K m c + K m w + K 2 m c )s 2 + ck 2 s + K K 2 The Effect of Damping Constant: c No Feedback Only examine c Everything else is. m c x x 2 G(s) = Where are the Poles? s 2 + cs + s 4 + 2cs 3 + 3s 2 + cs + s 4 + 2cs 3 + 3s 2 + cs + = s 4 + 3s 2 + + c (2s 3 + s) = d(s) + c n(s) = m w Looks like the root locus! u M. Peet Lecture 6: Control Systems 4 / 22
as a General Tool What do parameters do? d(s) = s 4 + 3s 2 + n(s) = 2s 3 + s G(s) = s2 + cs + d(s) + c n(s) 2.5 The root locus is the set of roots of d(s) + kn(s).5.5 We plot the root locus for G c (s) = n(s) d(s) = 2s3 + s s 4 + 3s 2 +.5 2.5.5 Note that G c is totally fictional! G c must still be proper (n is lower degree than d). The effect of changing c is small. M. Peet Lecture 6: Control Systems 5 / 22
as a General Tool Suspension Example: Damping Ratio Root locus tells us: Changing c won t help with overshoot. We need Feedback! m c x x 2 Set c = and plot the root locus m w u G(s) = Examine the gain at s 2 + s + s 4 + 2s 3 + 3s 2 + s + s =.3536 +.922ı s2 =.7 +.83ı Find Crossover Points k = 3.58 k = 2.6 We ll want k = 3. 2.5.5.5.5 2.9.8.7.6.5.4.3.2. M. Peet Lecture 6: Control Systems 6 / 22
as a General Tool Suspension Example: Damping Ratio Closed Loop Transfer Function: kg(s) + kg(s) = k(s 2 + cs + ) k(s 2 + cs + ) + s 4 + 2cs 3 + 3s 2 + cs + Can damping ratio get us to 3% overshoot? With feedback Set k = 3 and plot root locus Closed Loop Transfer Function (k = 3): 3 2 G kc (s) = s 2 + cs + (s 4 + 6s 2 + 4) + c(2s 3 + 3s + s) Use rlocfind to pick off the best value of c. Choose the point s =.7 +.25ı. Yields c = d(s) n(s) =.44 3 2.5 2.5.5 2 M. Peet Lecture 6: Control Systems 7 / 22
as a General Tool Suspension Example: Damping Ratio Using c =.44 and k = 3, the closed-loop transfer function is kg c (s) + kg c (s) = 3s 2 + 4.24s + 3 s 4 + 2.8s 3 + 6s 2 + 5.7s + 4 Which has repeated poles at s,2,3,4 =.7 ±.2ı Corresponds to an overshoot of.4.2 Step Response M p = e πσ ω = e.7 π.2 =.8 Amplitude.8.6.4.2 The real overshoot is much bigger! 2 3 4 5 6 7 8 9 Time (sec) M. Peet Lecture 6: Control Systems 8 / 22
as a General Tool DIY Example G(s) = s 2 + b 2 s + b s 3 + (7 + b)s 2 + (2 + b)s + b Find the optimal value of b. Im(s) Re(s) M. Peet Lecture 6: Control Systems 9 / 22
Limitations of isn t perfect Can only study one parameter at a time. What to do if root locus doesn t go where we want? The Suspension Problem: 2.5.5.5 G(s) = s 2 + s + s 4 + 2s 3 + 3s 2 + s +.5 2 3 2.5 2.5.5.5 Adding a pole at the origin has a negative effect. 2.5 Question: Would adding a zero have a positive effect?.5.5.5 2 3 2.5 2.5.5.5 M. Peet Lecture 6: Control Systems / 22
Limitations of The Inverted Pendulum:.8 G(s) = s 2 +.5 We used PD feedback K(s) = k( + T D s) Puts a zero at s = T D Conclusion: Adding a zero at z = improves performance. Can we generalize this?.6.4.2.2.4.6.8.8.6.4.2.2.4.6.8.8.6.4.2.2.4.6.8 4 3.5 3 2.5 2.5.5.5 M. Peet Lecture 6: Control Systems / 22
Adding Poles and Zeroes PID control PID feedback: K(s) = k ( ) + T i s + T Ds = k T Ds 2 + s + T I s.8.6.4.2.2.4.6 Adds poles and zeros: Two zeros: z,2 = ± 4T D T I 2T D One pole: p = Question: Is there a systematic way to add poles and zeros?.8.8.6.4.2.2.4.6.8.5.5.5.5 2.5.5.5 M. Peet Lecture 6: Control Systems 2 / 22
Adding Poles and Zeroes How? How To Add a Pole/Zero? What does it mean? Error e K Input u=ke G Output y=gu Constraint: The plant is fixed. G(s) doesn t change. The pole/zero must come from the controller. e.g. What is a Controller? A system Maps e(t) to u(t) M. Peet Lecture 6: Control Systems 3 / 22
Adding Poles and Zeroes Zeros Goal: Add a Zero Like PD control. Controller: K(s) = k(s + z) Input to Controller: ê(s) Output from Controller: û(s) = k(s + z) = ksê(s) + kzê(s) Time-Domain: u(t) = k e (t) + kz e(t) Problem: Requires differentiation. e (t) = e(t) e(t τ) τ M. Peet Lecture 6: Control Systems 4 / 22
Adding a Pole Goal: Add a pole Controller: K(s) = k s + p 2.5 Input to Controller: ê(s) Output from Controller: û(s) = s+pê(s) k Internal Variable: x. Frequency Domain Time-Domain (s + p)x(s) = ke(s) u(s) = x(s) ẋ(t) = px(t) + ke(t) 2 3 2.5 2.5.5.5 u(t) = x(t) Adding a Pole: Requires us to construct a dynamical system whose output is the control. Easier than adding a zero, but less useful Zeros are better for attracting poles away from RHP. M. Peet Lecture 6: Control Systems 5 / 22.5.5.5
Pole-Zero Compensation The best way to modify the root locus is a using a pole-zero compensator. Adds a zero without differentiation K(s) = k s z s p Input to Controller: ê(s) Output from Controller: û(s) = k s pê(s) s z Doing long division: s z s p = + z p s p Im(s) Re(s) Hence û(s) = kê(s) + k z p s p ê(s) Internal Variable: ˆx(s) = k(z p) s p ê(s) M. Peet Lecture 6: Control Systems 6 / 22
Pole-Zero Compensation Internal Variable: ˆx(s) = k(z p) s p ê(s) Im(s) Frequency Domain: (s + p)x(s) = k(z p)e(s) u(s) = x(s) + kê(s) Re(s) Time-Domain: ẋ(t) = px(t) + k(z p)e(t) u(t) = x(t) + ke(t) Artificial Dynamics: Controller State: x(t) No differentiation of e(t)! A zero should always be combined with a pole. M. Peet Lecture 6: Control Systems 7 / 22
Lead Compensation Definition. A Lead Compensator is a pole-zero compensator where p < z. K(s) = s + z s + p Used when we really want a zero The pole has less effect than the zero. Im(s) Re(s) M. Peet Lecture 6: Control Systems 8 / 22
Lead Compensation Inverted Pendulum G(s) = s 2.5.8 4.6 3.4 2.2.2.4 2.6 3.8 4 3.5 3 2.5 2.5.5.5 Figure : K(s) = k(s + ) 4 4 3.5 3 2.5 2.5.5.5 Figure : K(s) = k s + s + 3 M. Peet Lecture 6: Control Systems 9 / 22
Lag Compensation Definition 2. A Lag Compensator is a pole-zero compensator where z < p. K(s) = s + z s + p Used when we really want a pole The zero has less effect than the pole. Doesn t increase the number of asymptotes. Im(s) Re(s) M. Peet Lecture 6: Control Systems 2 / 22
Lag Compensation Suspension Problem G(s) = s 2 + s + s 4 + 2s 3 + 3s 2 + s + 4 5 3 4 2 2 3 3 2 4 3.5 3 2.5 2.5.5.5.5 2 3 Figure : K(s) = k 5 6 5 4 3 2 2 s 4 Figure : K(s) = k s + 5 s +. M. Peet Lecture 6: Control Systems 2 / 22
Summary What have we learned today? Generalized? What about changing OTHER parameters T D, T I, et c. mass, damping, et c. Compensation via Root-Locus Introduction Pole-Zero Compensation Lead-Lag Next Lecture: Pole-Zero Compensation and Notch Filters M. Peet Lecture 6: Control Systems 22 / 22