INC 341 Feedback Control Systems: Lecture 3 Transfer Function of Dynamic Systems II Asst. Prof. Dr.-Ing. Sudchai Boonto Department of Control Systems and Instrumentation Engineering King Mongkut s University of Technology Thonburi
Transfer Functions for Systems with Gears Gears provide mechanical advantage to rotational system, e.g. a bicycle with gears. Gears are nonlinear. They exhibit backlash, which occurs from the loose fit between two meshed gears. In this course, we consider only the linearized version of gears. A gear system τ 1t) θ 1t) r 1 r 2 θ 2t) τ 2t) a small gear has radius r1 and N1 teeth is rotated through angle θ 1 t) due to a torque, τ 1 t). a big gear have radius r 2 and teeth responds by rotating through angle θ 2 t) and delivering a torque, τ 2 t). 2/19
Transfer Functions for Systems with Gears The gears turn, the distance traveled along each gear s circumference is the same. Thus r 1 θ 1 = r 2 θ 2 or θ 2 θ 1 = r 1 r 2 = If the gears are lossless, that is they do not absorb or store energy, the energy into Gear 1 equals the energy out of Gear 2. Since the translational energy of force times displacement becomes the rotational energy of torque time angular displacement. τ 1 θ 1 = τ 2 θ 2 or τ 2 τ 1 = θ 1 θ 2 = 3/19
Transfer Functions for Systems with Gears Transfer functions for lossless gears θ 1 θ 2 τ 1 τ 2 a) b) a) is a transfer functions for angular displacement in lossless gears b) is a transfer functions for torque in lossless gears. τ 1t) θ 1t) The first gear lossless) generates torque τ 1 ) to drive the second gear by τ 2, then Js 2 + bs + k ) Θ 2 s) = ˆτ 2 s) θ 2t) b τ 2 τ 1 = = θ 1 θ 2 J k Js 2 + bs + k ) Θ 1 s) = ˆτ 1 s) ) 2 ) 2 ) ) 2 N1 J s 2 N1 N1 + b s + k Θ 1 s) = ˆτ 1 s) 4/19
Transfer Functions for Systems with Gears Transfer functions for lossless gears τ 1t) θ 1t) J 1 b 1 θ 2t) b 2 J 2 k 2 Find the transfer function, Θ 2 s)/ˆτ 1 s). Assuming that τ et) is the torque generates at the first gear by the torque τ 1 t), the we have J1 s 2 + b 1 s ) Θ 1 s) + ˆτ es) = ˆτ 1 s) J1 s 2 + b 1 s ) Θ 2 s) + ˆτ es) = ˆτ 1 s) 5/19
Transfer Functions for Systems with Gears Transfer functions for lossless gears At the second gear, we have J2 s 2 + b 2 s + k 2 ) Θ2 s) = ˆτ 2 s) J2 s 2 + b 2 s + k 2 ) Θ2 s) = ˆτ es) J2 s 2 + b 2 s + k 2 ) Θ 2 s) = ˆτ es) Combining the equations of both gears, we have [ ) N2 J 1 s 2 N2 + b 1 [ J 1 N2 ) N1 s + J 2 ) 2 + J 2 ) s 2 + ) s 2 N1 + b 2 b 1 N2 ) N1 s + k 2 ) 2 + b 2 ) s + k 2 ] )] Θ 2 s) = ˆτ 1 s) Θ 2 s) = ˆτ 1 s) Θ 2 s) ˆτ 2 s) = / [ ) ) 2 ) ) ] 2 J N2 1 N + J2 s 2 + b N2 1 1 N + b2 s + k 2 1 6/19
Transfer Functions for Systems with Gears Transfer functions for lossless gears θ 1 θ 2 = N1 θ 1 N 3 θ 3 = N3 N 4 θ 2 = N1 N 3 N 4 θ 1 N 4 N 5 θ 4 = N5 N 6 θ 3 = N1 N 3 N 4 N 5 N 6 θ 1 N 6 A gear train can use instead of one large radii gear. This can be done by cascading smaller gear ratios. As shown in above figure, θ 4 = N 3 N 5 N 4 N 6 θ 1 7/19
Transfer Functions for Systems with Gears Transfer functions for lossless gears τ 1 θ 1 J 1, b 1 J 2, b 2 N 3 J 3 N 4 J 4 J 5 Starting from the left most of the gear, we can find the transfer function Θ 1 s)/ˆτ 1 s) as follow: J1 s 2 + b 1 s ) Θ 1 s) + ˆτ e1 s) = ˆτ 1 s) [ J2 + J 3 ) s 2 + b 2 s ] Θ 2 s) + ˆτ e2 s) = ˆτ 2 s) [ J4 + J 5 ) s 2] Θ 4 s) = ˆτ 4 s) Transform all torques and angle to be in terms of ˆτ 1 s) and Θ 1 s) respectively. 8/19
Transfer Functions for Systems with Gears Transfer functions for lossless gears [ J4 + J 5 ) s 2] N 3 N 4 Θ 2 s) = N 4 N 3 ˆτ e2 s) Substituting ˆτ e2 s) to one above equation, we have [ ) 2 J 2 + J 3 ) s 2 N3 + J 4 + J 5 ) s 2 + b 2 s] Θ 2 s) = ˆτ 2 s) N 4 [ ) 2 J 2 + J 3 ) s 2 N3 + J 4 + J 5 ) s 2 + b 2 s] Θ 1 s) = ˆτ e1 s) N 4 Substituting ˆτ e1 s) to one above equation, we have [ ) 2 ) ) N1 N1 ) ) ] 2 3 J 1 + J 2 + J 3 ) + J 4 + J 5 ) s 2 N1 + b 1 + b 2 s Θ 1 s) = ˆτ 1 s) N 4 Θ 1 s) ˆτ 1 s) = 1 [ ) 2 J 1 + J 2 + J 3 ) N1 N + J4 + J 5 ) 2 N1 N 3 N 4 ) 2 ) s 2 + ) ) ] 2 b 1 + b N1 2 s 9/19
Electromechanical System Transfer Functions An electromechanical systems is a hybrid system of electrical and mechanical variables. This system has a lot of application for examples an antenna azimuth position control system robot and robot arm controls sun and star trackers disk-drive position controls Industrial robot arm 10/19
Electromechanical System Transfer Functions R a L a Fixed field + τ mt) θ mt) e at) + iat) v b t) = K e θ M Rotor Armature circuit dθ mt) v b t) = K b, where v b t) is the back electromotive force back emf); K b is a dt constant of proportionality called the back emf constant. The relationship between the armature current, i at), the applied armature voltage, e at), and the back emf, v b t) is R ai at) + L a di at) dt + v b t) = e at) 11/19
Electromechanical System Transfer Functions The torque developed by the motor is proportional to the armature current; thus τ mt) = K ti at), where τ mt) is the torque developed by the motor, and K t is a constant for proportionality, called the motor torque constant. Taking the Laplace transform of both relationship and substituting I as) into the mesh equation, we have R a + L as)ˆτ ms) K t + K b sθ ms) = E as) τ mt) θ mt) J m b m The figure shows a typical equivalent mechanical loading on a motor. We have ˆτ ms) = J ms 2 + b ms ) Θ ms) 12/19
Electromechanical System Transfer Functions Substituting ˆτ ms) into the armature equation yields R a + L as) J ms 2 + b ) ms Θ ms) + K b sθ ms) = E as) K t Assuming that the armature inductance, L a is small compared to the armature resistance, R a, the equation become [ ] Ra J ms + b m) + K b sθ ms) = E as) K t After simplification, the desired transfer function is Θ ms) E = K t/r aj m) [ )] as) s s + 1 b J m + K tk b m R a 13/19
Electromechanical System Transfer Functions DC motor driving a rotational mechanical load θ mt) τ 1 t), Motor θ L t) J a, b a b L J L, τ 2 t) A motor with inertia J a and damping b a at the armature driving a load consisting of inertia J L and damping b L. Assuming that J a, J L, b a, and b L are known. Then, we have Jas 2 + b as ) Θ ms) + ˆτ 1 s) = 0. At the load side, we obtain JL s 2 + b L s ) Θ L s) = ˆτ 2 s) JL s 2 + b L s ) Θ ms) = ˆτ 1 s) 14/19
Electromechanical System Transfer Functions DC motor driving a rotational mechanical load Substituting the ˆτ 1 s) back to the motor side equation, the equivalent equation is [ ) ) 2 ) ) ] 2 N1 J a + J L s 2 N1 + b a + b L s Θ ms) = 0 Or the equivalent inertial, J m, and the equivalent damping, b m, at the armature are ) 2 ) 2 N1 N1 J m = J a + J L ; b m = b a + b L Next step, we going to find the electrical constants by using a dynamometer test of motor. This can be done by measuring the torque and speed of a motor under the condition of a constant applied voltage. Substituting V b s) = K b sθ ms) and ˆτ ms) = K ti as) in to the Laplace transformed armature circuit, with L a = 0, yields R a K t ˆτ ms) + K b sθ ms) = E as) 15/19
Electromechanical System Transfer Functions DC motor driving a rotational mechanical load Taking the inverse Laplace transform, we get R a K t τ mt) + K b ω mt) = e at) If e at) is a DC voltage, at the steady state, the motor should turn a a constant speed, ω m, with a constant torque, τ m. With this, we have R a τ m + K b ω m = e a τ m = KtK b ω m + Kt e a K t R a R a τ stall τ m Torque-peed curve e a1 ω m = 0, the value of torque is called the stall torque, τ stall. Thus τ stall = Kt R a e a Kt R a = τ stall e a τ m = 0, the angular velocity becomes no-load speed, ω no-load. Thus e a2 ω no-load ω m ω no-load = ea K b K b = ea ω no-load 16/19
Transfer Function DC Motor and Load R a Fixed field e at) + i at) + M θ mt) τ 1t), = 100 θ L t) b L = 800 N-m s/rad 500 N-m τ m e a = 100 V J a = 5kg-m 2 b a = 2 N-m s/rad = 1000, τ 2t) J L J L = 700 kg-m 2 50 rad/s ω m Find Θ L s)/e as) from the given system and torque-speed curve. The total inertia and the total damping at the armature of the motor are ) 2 1 = 5 + 700 10 N1 J m = J a + J L ) 2 N1 b m = b a + b L = 2 + 800 1 10 ) 2 = 12 ) 2 = 10 17/19
Transfer Function DC Motor and Load Next, we find the electrical constants K t/r a and K b from the torque-speed curve. Hence, and We have K t R a = τ stall e a = 500 100 = 5 K b = ea = 100 ω no-load 50 = 2 Θ ms) E = K t/r aj m) 0.417 [ as) s s + 1 b m + KtK )] = b ss + 1.667). J m R a Using the gear ratio, / = 0.1 Θ L s) E = 0.0417 as) ss + 1.667) 18/19
Reference 1. Norman S. Nise, Control Systems Engineering,6 th edition, Wiley, 2011 2. Gene F. Franklin, J. David Powell, and Abbas Emami-Naeini, Feedback Control of Dyanmic Systems,4 th edition, Prentice Hall, 2002 19/19