Analysis of Linear State Space Models

Transcription

1 Max f (lb) Max y (in) 1 ME313 Homework #7 Analysis of Linear State Space Models Last Updated November 6, 213 Repeat the car-crash problem from HW#6. Use the Matlab function lsim to perform the simulation. Before the simulation, compute the eigenvalues of [A], and the time constants and frequencies of oscillation. Set the time range for the simulation to be 5 times the longest time constant. Again consider the car-crash problem. Only now, write a Matlab script that will perform a simulation for a sequence of seat-belt stiffnesses k 2 ranging from k 2 =5 lb/ft to k 2 =5 klb/ft. After each simulation, extract the maximum force exerted by the seatbelt on the occupant, and the maximum relative displacement between the occupant and car. Plot both of these quantities versus the seat-belt stiffness k 2. In your script, use ODE45 to perform each simulation and pass the masses m 1, m 2 ; stiffnesses k 1, k 2 ; and damping coefficients c 1 and c 2 in a structure from the script to your state system using anonymous function syntax. Your plots should look like k2 (klb/ft) x k2 (klb/ft)

2 2 A small gantry crane is used to move parts. A x diagram of the crane is shown at the right. The part of weight 2 lb and mass m is suspended from the gantry by a cable of length l=2 ft. The horizontal displacement of the gantry is x, and the angular rotation of the mass m relative to vertical is as shown. Using the methodology used in class to model the inverted pendulum, the dynamic model for the angular movement of the mass m is I c mglsin a( t) mlcos, o T where I o =ml 2 is the moment of inertia of the mass about the pivot point on the gantry, c T =3 lb s models rotational friction of angular motion, and a( t) x is the horizontal acceleration of the gantry. If the angular motions of the mass m are small, it is relatively accurate to replace sin and cos 1 in the model above and obtain an approximation of the dynamic model as I c mgl a( t) ml. o T Note that the first model is nonlinear, while the approximate model is linear. In the following, you are to assume that the gantry is equipped with an apparatus that allows an engineer to control the acceleration of the gantry a(t). o An engineer wishes to move the part to the right. To achieve this motion, the engineer specifies the following acceleration of the gantry, a(t)=5 ft/s 2, t<1 sec, a(t)= ft/s 2, t>1 sec. Simulate the angular movement of the mass using initial conditions of zero angular displacement and velocity. Plot the angular displacement and angular velocity over the time range t 1 sec for this maneuver with the two models cited above. The plot should look like the one shown at the top of the following page.

3 Angular Velocity (deg/sec) Angular Displacement (deg) 3 1 Accel Time=1 sec, Accel Dist=2.5 ft, Vel After Accel=5 ft/sec wn=4.125 rad/sec, z=.1547 kp= kd= Mp= ts= sec Time (sec) o At t=1 sec and t=1 sec, how far has the gantry travelled, and what is the velocity of the gantry? o From the approximate model, compute the natural frequency n and the damping ratio. o An engineer wishes to reduce the angular oscillations of the mass m after the acceleration has been applied, i.e., the time range t>1 sec. Given that sensors are installed that are capable of measuring the angular displacement and angular velocity of the mass m, the engineer installs a feedback system that sets the acceleration of the cart a(t) to a( t) k p k d, after the acceleration has been applied, i.e., when t>1 sec. In the above control law, k p and k d are the overall proportional and derivative gains for the control apparatus between measurement of the angular displacement and velocity, and the acceleration a(t) of the gantry.

4 Angular Velocity (deg/sec) Angular Displacement (deg) 4 Using the approximate linear dynamic model, choose the proportional and derivate gains so that the natural frequency and damping ratio of the controlled system will be n =1 rad/sec and =.7 respectively Perform a simulation of the system performance, including a(t)= 5 ft/s 2 for t 1 sec, and a( t) k p k d, for 1 t 1 sec. Ye response should look like the plot shown on the following page. 5 Accel Time=1 sec, Accel Dist=2.5 ft, Vel After Accel=5 ft/sec wn=1 rad/sec, z=.7 kp=167.8 kd= Mp= ts= sec Time (sec)

5 Simulation of Carcrash with lsim clear all;close all;clc; HW 7 Linear Simulation Problem ICS x1= Vehicle position x2= Vehicle velocity x3= Occupant position (inertial) x4= Occupant velocity (inertial) xo(1)=; xo(2)=-(528/36)*5; xo(3)=; xo(4)=-(528/36)*5; k1=1e3; c1=5; m1=3/32.2; k2=5e3; c2=5 ; m2=15/32.2; A=[ 1 ; -(k1+k2)/m1 -(c1+c2)/m1 k2/m1 c2/m1 ; 1 ; k2/m2 c2/m2 -k2/m2 -c2/m2]; B=[ ; ; ; ]; C=[1-1 ]; D=; [E,L]=eig(A) wn1=abs(l(1,1));z1=-real(l(1,1))/wn1;wd1=wn1*sqrt(1-z1^2);td1=(2*pi)/wd1; tau1=1/(z1*wn1); wn2=abs(l(3,3));z2=-real(l(3,3))/wn2;wd2=wn2*sqrt(1-z2^2);td2=(2*pi)/wd2; tau2=1/(z2*wn2); Set up time range tend=5*max([tau1 tau2]); N=5; t=linspace(,tend,n); dt=t(2)-t(1); Perform simulation u=zeros(1,length(t)); sys=ss(a,b,c,d); [y,t,x]=lsim(sys,u,t,xo); y=x(:,1)-x(:,3); f=(-k2*(x(:,3)-x(:,1))-c2*(x(:,4)-x(:,2))); subplot(3,1,1); plot(t,x(:,1)*12,t,x(:,3)*12) title('inertial Coordinates of Car and Occupant') axis([ tend -1*12 1*12]); legend('x1','x3'); ylabel('displacement (in)'); subplot(3,1,2); plot(t,y*12)

6 title('relative Displacement Between Car and Occupant') axis([ tend 1.1*min(y)*12 1.1*max(y)*12 ]); ylabel('displacement (in)'); subplot(3,1,3); plot(t,f) title('force of Seat-Belt on Occupant') axis([ tend 1.1*min(f) 1.1*max(f)]); ylabel('force (lb)'); xlabel('time'); Eigenvalues, frequencies of oscillation, damping ratios, time constants: L = i i i i >> wn1 wn1 = >> z1 z1 =.172 >> wd1 wd1 = >> Td1 Td1 =.1896 >> tau1 tau1 =.1729 >> wn2 wn2 = 1.94 >> z2 z2 =.2511 >> wd2

7 Force (lb) Displacement (in) Displacement (in) wd2 = >> Td2 Td2 =.6433 >> tau2 tau2 =.3946 >> 1 Inertial Coordinates of Car and Occupant x1 x Relative Displacement Between Car and Occupant Force of Seat-Belt on Occupant Time

8 Maximum Force and Displacement versus Seatbelt Stiffness clear all; close all; Compute maximum force exerted by seatbelt on occupant for a range of seatbelt stiffnesses N=5; tend=.5; t=linspace(,tend,n); dt=t(2)-t(1); k1=1e3; c1=5; m1=3/32.2; k2=5e3; c2=5 ; m2=15/32.2; P.k1=k1 ; P.c1=c1 ; P.m1=m1; P.k2=k2 ; P.c2=c2 ; P.m2=m2; Nk2=5; k2t=linspace(5,5e3,nk2); ICS x1= Vehicle position x2= Vehicle velocity x3= Occupant position (inertial) x4= Occupant velocity (inertial) xo(1)=; xo(2)=-(528/36)*5; xo(3)=; xo(4)=-(528/36)*5; for i1=1:nk2 k2=k2t(i1) ; P.k2=k2; [t,x]=ode45(@(t,x)carcrash1(t,x,p),t,xo); y=x(:,1)-x(:,3); ymax(i1)=max(abs(y)); f=(-k2*(x(:,3)-x(:,1))-c2*(x(:,4)-x(:,2))); fmax(i1)=max(abs(f)); end subplot(2,1,1); plot(k2t,ymax*12) title('maximum Occupant Displacement Versus Seatbelt Stiffness') ylabel('displacement (in)'); subplot(2,1,2); plot(k2t/1,fmax) ylabel('force (lb)'); xlabel('seatbelt Stiffness (klb/ft)')

9 Force (lb) Displacement (in) 6 Maximum Occupant Displacement Versus Seatbelt Stiffness x Seatbelt Stiffness (klb/ft)

10

11 clear all; close all; clc; Gantry Control Problem N=5; tend=1; t=linspace(,tend,n); dt=t(2)-t(1); ICS (first set are for linear problem, second set are for nonlinear problem xo(1)=; xo(2)=; xo(3)=xo(1); xo(4)=xo(2); System Parameters- Uncompensated system P.m=2/32.2; P.d=2; P.g=32.2; P.cT=3; P.Io=P.m*P.d^2;P.T=1;P.a=5; P.kp=; P.kd=; wn=sqrt((p.m*p.g*p.d)/p.io) ; z=p.ct/(2*sqrt(p.io*p.m*p.g*p.d)); Mp=1*exp(-pi*z*sqrt(1-z^2)); ts=1/(z*wn); Control gains for compensated system wn=1 ; z=.7; P.kp=(P.Io*wn^2)/(P.m*P.d)-P.g; P.kd=(P.Io/(P.m*P.d))*(2*z*wn-(P.cT/P.Io)); Mp=1*exp(-pi*z*sqrt(1-z^2)); ts=1/(z*wn); [t,x]=ode45(@(t,x) gantry(t,x,p),t,xo); subplot(2,1,1) ; plot(t,x(:,1)*(18/pi),t,x(:,3)*(18/pi)) xg=(1/2)*p.a*p.t^2; vg=p.a*p.t; title([' Accel Time=',num2str(P.T),' sec,',... ' Accel Dist=',num2str(xg),' ft,',... ' Vel After Accel=',num2str(vg),' ft/sec']) axis([ tend -1*12 1*12]); legend('x1','x3'); ylabel('angular Displacement (deg)'); grid on; xg=(1/2)*p.a*p.t^2; subplot(2,1,2);plot(t,x(:,2)*(18/pi),t,x(:,4)*(18/pi)) title([' wn=',num2str(wn),' rad/sec,',... ' z=', num2str(z),... ' kp=', num2str(p.kp),... ' kd=', num2str(p.kd),... ' Mp=', num2str(mp),... ' ts=',num2str(ts),' sec']) xlabel('time (sec)'); ylabel('angular Velocity (deg/sec)'); grid on;

12 function xdot=gantry(t,x,p) x1= Vehicle position x2= Vehicle velocity x3= Occupant position (nonlinear) x4= Occupant velocity (nonlinear) Define the parameters m=p.m; d=p.d; g=p.g; ct=p.ct; Io=P.Io;T=P.T;kp=P.kp;kd=P.kd; Define the input if (t < T) a=p.a; u=; else a=; u=kp*x(1)+kd*x(2); end Define the state equations xdot(1)=x(2); xdot(2)=(1/io)*(-m*g*d*sin(x(1))-ct*x(2)-p.m*p.d*(a+u)); xdot(3)=x(4); xdot(4)=(1/io)*(-m*g*d*x(3)-ct*x(4)-p.m*p.d*(a+u)); xdot = xdot' ;

13 Angular Velocity (deg/sec) Angular Displacement (deg) Angular Velocity (deg/sec) Angular Displacement (deg) 1 Accel Time=1 sec, Accel Dist=2.5 ft, Vel After Accel=5 ft/sec wn=4.125 rad/sec, z=.1547 kp= kd= Mp= ts= sec Time (sec) 5 Accel Time=1 sec, Accel Dist=2.5 ft, Vel After Accel=5 ft/sec wn=1 rad/sec, z=.7 kp=167.8 kd= Mp= ts= sec Time (sec)