32 MOTOR SERVO WITH BACKLASH 114 32 Motor Servo with Backlash Consider the following: a motor is attached to a load through a gearbox, and the gearbox has a backlash behavior. Physically, backlash is a decoupling of gear teeth (due to imprecise meshing), so that the motor-side and the load-side teeth are not always in contact. Contacts are characterized by the condition |θm − θl| >h, where the θ’s are the motor and load angles respectively, and h is the half-width of the backlash. During contact, a small deflection material occurs according to a linear spring model. In the forward direction, that is, when θm leads θl by a positive amount, τt = k(θm − θl − h)when θm − θl >h, where τt is the torque transmitted from the motor side to the load. A similar condition holds in the reverse direction. Conversely, whenever θm is within h of the angle θl, the teeth are out of contact and no torque is passed between the motor and the load. h Tm Tl 2h (motor side gear) (load side gear) The rotary mass on the motor side is Jm; on the load side, we have rotary mass Jl as well as a rotary damping b. It is desired to position the load via a servo, so a simple feedback controller is implemented of the form τm = −gθl,where τm is the motor torque. We assume that we can control the torque in the motor directly through a current amplifier. All the physical properties in a particular system built by Company XYZ are well understood, except for the load-side damping b. This varies slowly over time because the bearings break in and then age, and because the load side is connected to an unpredictable environment. For the purposes of this problem, we assume that b is a random variable, described by a uniform probability density function, with minimum value zero and maximum value 0.02 Nm/(rad/s). The other parameters in the model are Jm = 0.003 ; % motor-side inertia, kg-m^2 Jl = 0.01 ; % load-side effective inertia, kg-m^2� � 32 MOTOR SERVO WITH BACKLASH 115 k = 10000 ; % stiffness of tooth contact point, Nm/rad h = 0.02 ; % backlash half-width, rad g = 6 ; % position feedback gain, Nm/rad The standard transient response that we are interested in has the initial condition of zero speeds, zero motor position, and a load position of h. We take as characteristic outputs of any particular model run the two-norms of the load-side position error, and the commanded torque. The two-norm of a signal over a time interval [t1,t2] is defined as: t2 ||z(t)||2 = z2(t)dt t1 These norms are to be computed over the time range of 2.5-5.0 seconds after the feedback system turns on. Question: What are the means and standard deviations of the two characteristic outputs defined above? The attached code models the dynamic behavior, and runs a number of cases along the uniformly-divided range of damping b. I used a trapezoid rule for the integration of the mean, and just computed the straight standard deviation of the samples. It would also have been possible to make these calculations with a Gauss quadrature or a Monte Carlo technique. A typical response is shown below versus time, as is the norm ||z(t)||2 versus b. The mean and standard deviation of the load-side position are 0.0101 and 0.00527 rad, respectively. The mean and standard deviation of the motor torque are 0.0609 and 0.0316 Nm, respectively. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Backlash sensitivity problem % FSH MIT Mechanical Engineering April 2008 clear all; global Jm Jl k h g b ; Jm = 0.003 ; % motor-side inertia, kg-m^2 Jl = 0.01 ; % load-side inertia, kg-m^2 k = 10000 ; % stiffness of tooth contact point, Nm/rad h = 0.02 ; % backlash half-width, rad g = 6 ; % position feedback gain, Nm/rad bvec = 0:.0005:.02 ; % vector of dampings to apply, % on load side, Nm/(rad/s) = kg*m^2/s32 MOTOR SERVO WITH BACKLASH 116 rad Time response with b = 0.01 Nm/(rad/s). Angle:angle response with b = 0.01 Nm/(rad/s). 0.03 motor angle load angle 0.04 0.02 0.03 0.01 0.02 load angle, rad0.01 0 0 −0.01 −0.02 −0.01 −0.02 −0.03 −0.03 −0.04 −0.05 −0.04 0 1 2 3 4 5 −0.03 −0.02 −0.01 0 0.01 0.02 time, s motor angle, rad Control action vs. time with b = 0.01 Nm/(rad/s). 0.2 0.16 Nm Norm of load position, rad Norm of commanded torque, Nm0.15 0.1 0.05 0 −0.05 −0.1 −0.15 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 0 1 2 3 4 50 0.005 0.01 0.015 time, s b, Nm/(rad/s) tfinal = 5 ; % final simulation time tcut = 2.5; % time after which we will computing the signal norm plotFlag = 0 ; % set to one to get all plots of trajectories ct = 0 ; % counter for b = bvec, % loop through all the b’s disp(sprintf(’Approx. closed-loop damping ratio: %g’, ... b/2/sqrt((Jm+Jl)*g) )); % run the simulation - note initial error in load position % equal to the backlash half-width (states are listed in deriv.) [t,y] = ode45(’backlashderiv’,tfinal,[0 0 0 h]); tauMotor = -g*y(:,4); % recreate the motor torque command if plotFlag | b == bvec(end)/2, % show some intermediate results 0.0232 MOTOR SERVO WITH BACKLASH 117 figure(1);clf;hold off; subplot(’Position’,[.2 .2 .5 .5]); plot(t,y(:,[2,4]),’LineWidth’,1); xlabel(’time, s’); ylabel(’rad’); legend(’motor angle’, ’load angle’,4); title(sprintf(’Time response with b = %g Nm/(rad/s).’, b)); figure(2);clf;hold off; subplot(’Position’,[.2 .2 .5 .5]); plot(y(:,2),y(:,4),’r’,’LineWidth’,1); hold on; plot(y(:,2),y(:,2),’k--’,y(:,2),y(:,2)+h,’k--’,... y(:,2),y(:,2)-h,’k--’); axis(’tight’); xlabel(’motor angle, rad’); ylabel(’load angle, rad’); title(sprintf(’Angle:angle response with b = %g Nm/(rad/s).’, b)); figure(3);clf;hold off; subplot(’Position’,[.2 .2 .5 .5]); plot(t,tauMotor); xlabel(’time, s’); ylabel(’Nm’); title(sprintf(’Control action withb=%g Nm/(rad/s).’, b)); end; ct = ct+1 ; % calculate the norm of the load motion and the torque, after tcut [dum,i] = sort(abs(t-tcut)) ; energyNorm(ct) = sqrt(trapz(t(i(1):end),y(i(1):end,4).^2)); tauMotorNorm(ct) = sqrt(trapz(t(i(1):end),tauMotor(i(1):end).^2));
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