VIEWS: 15 PAGES: 12 CATEGORY: Technology POSTED ON: 3/11/2010
MEM 351 – Dynamic Systems Lab Representations: Transfer Functions Poles and Zeros Recall Damped Compound Pendulum Equations of Motion && + c θ + mL gd θ = 0 θ & J J d Linearized 2nd order differential θ equation assumes small angles c dθ dt L Bar length [m] L m g d Pivot to CG distance [m] L mL Mass of pendulum [kg] Moment of Inertia [kg ⋅ m ] 2 θ=0 J Nms c Viscous damping coefficient [ rad ] General 2nd order form & & θ& + 2ςω nθ + ω n θ = 0 2 Tedious Math: Time domain differential equation 2nd order damped system & & θ& + 2ζω nθ + ω n θ = 0 2 Yields complex roots s1, 2 = −ζω n ± j ⋅ ω n 1 − ζ 2 Time domain solution { ( θ c (t ) = e −ζ ωnt A1 cos ω n t 1 − ζ 2 )+ A sin (ω t 2 n 1−ζ 2 )} (1) Small real root will yield long settling times Can be shown: Time constant Tc = ω nζ 4 (2) 2% settling time t s = Tc Easier Math I: Laplace Domain • Voltage V (s ) applied to motor r • Propeller spins, creating lift force F (s) d F θ c dθ • Lift on lever arm r creates torque Τ(s) dt T = Fr • Pendulum rotates angle Θ(s ) motor-propeller mg Motorized Propeller 1/J T(s) Torque Θ (s) = θ V(s) K m 2 c m gd [Volts] [Nm] s + s+ L [ rad/sec ] [Nm/Volts] J J Compound Pendulum Calculating Constants Motorized Propeller 1/J T(s) Torque Θ (s) = θ V(s) K m 2 c m gd [Volts] [Nm] s + s+ L [ rad/sec ] [Nm/Volts] J J Compound Pendulum Km Theory: can calculate lift force if have propeller pitch and radius dimensions, air density and motor angular velocity. Experimentally, apply known voltage V and pendulum will eventually reach steady-state. Recall & & Jθ& + cθ + mL gd sin θ = T At steady-state angular acceleration and velocity are zero. The torque at this known voltage is calculated by: T T ss = m L gd sin θ ss And hence Km = ss V Open-Loop Transfer Function Motorized Propeller 1/J T(s) Torque Θ (s) = θ V(s) K m 2 c m gd [Volts] [Nm] s + s+ L [ rad/sec ] [Nm/Volts] J J Compound Pendulum Θ ( s) Km / J OLTF: = = Gol ( s) V ( s) c mL gd s + s+ 2 J J Given Km = 0.017 Nm/V d = 0.023 m Θ( s) 1.89 = 2 = Gol ( s) (3) J = 0.0090 kgm 2 V ( s) s + 0.039s + 10.77 m L = 0.43 kg Laplace domain OL Transfer function c = 0.00035 Nms/rad OLTF Simulations Simulink Simulation reveals long settling time. This is consistent with the low viscous damping coefficient. Poles of the characteristic equation reveal the large oscillations. Recall from (1) Θ( s) 1.89 = 2 = Gol ( s) V ( s) s + 0.039s + 10.77 Roots of the denominator (i.e. the poles) are: s1 = −0.0019 + j 3.28 s 2 = −0.0019 − j3.28 Small real root will yield long settling times In other words, the system is bordering on the margin of stability. Recall Equations (1) and (2). System Poles and Zeros – What are they? d θ dθ c dt L m g L θ=0 & & θ& + 2ςω nθ + ω n θ = 0 2 (1) Control Designer’s Goal: Create compensators that a ς 2π 1 X ln = = ln 1 (2A) yield desired damping and b 1 − ς 2 N X N +1 rise time. 2π In other words, place poles = ωn 1 − ς 2 (2B) where one wants them T Calculated the following: ωn = 2.50 rad/s & & θ& + 2ςω nθ + ω n2θ = 0 ζ = 0.0059 s + 0.0295 s + 6.25 = 0 2 s1,2 = −0.01475 ± 2.5 j poles s = a + jb 1 X ωn 1- ζ 2 ζ = sin θ ωn θ - ζ ωn X - ωn 1- ζ 2 s = a - jb 2 s = -0.01475 +j 2.50 1 X ωn 1- ζ 2 ζ = sin θ ωn θ - ζ ωn X - ωn 1- ζ 2 s = -0.01475 - j 2.50 θ = arctan 0.01475 = 0.0059 ζ = sin θ = sin 0.0059 = 0.0059 2.50 ω n = a 2 + b 2 = 0. 01475 2 + 2.50 2 = 2.50 Matches experimental data! Where are we going with this? It’s called the characteristic equation because it connotes system properties Poles are the roots of the characteristic equation. As such, the describe stability through ωn and ζ Question: can we alter the locations of the poles? If we can, then we change the characteristic of the system… Answer: This is exactly what the control engineer does. One popular method is called “pole placement” control