机电系统动力学仿真matlab课后答案刘白雁

上传人:鲁** 文档编号:469539092 上传时间:2023-11-16 格式:DOCX 页数:29 大小:946.93KB
返回 下载 相关 举报
机电系统动力学仿真matlab课后答案刘白雁_第1页
第1页 / 共29页
机电系统动力学仿真matlab课后答案刘白雁_第2页
第2页 / 共29页
机电系统动力学仿真matlab课后答案刘白雁_第3页
第3页 / 共29页
机电系统动力学仿真matlab课后答案刘白雁_第4页
第4页 / 共29页
机电系统动力学仿真matlab课后答案刘白雁_第5页
第5页 / 共29页
点击查看更多>>
资源描述

《机电系统动力学仿真matlab课后答案刘白雁》由会员分享,可在线阅读,更多相关《机电系统动力学仿真matlab课后答案刘白雁(29页珍藏版)》请在金锄头文库上搜索。

1、机电系统动力学仿真实验报告机工 1004班 201003130150宋康习题一1.3Abc wu_2004 两个是合法的。1.4(1)x Command Window? (12+2i(7-4)ans =Command WindowIdersort byShow厂SIhts.rnactionsthumlhowShow:! :$:$:function/zesDi recto ty -箭“ 审A =;回 Howto Add 回 Whats Nej/齢 A=1J:13;4J 5, 6 8, 9(3) clear; x=-8:0.5:8; y=x;X=ones(size(y)*x; Y=y*ones(s

2、ize(x);R=sqrt(X.A2+Y.A2)+eps;Z=sin(R)./R;mesh(X,Y,Z);colormap(hot)xlabel(x),ylabel(y),zlabel(z)5500-5-510.50-0.51010y-10 -10x1.71)9 WhatNewCommand Window-l=;2.*sin(0. 3*pi)/ (1+sqrt Jyi =si thumlx |8mmmnci wnwiigfiwI l=2*sin(0.*pi)/ (1+sqrt (5 l yi =0.5000 y2=,2Cos.(0. j*pi)/ (1+sqrt )1.115500-5-510

3、.50-0.51010y-10 -10x习题二2.12.2 x=linspace (Oj 2*pi, 50)x =Columns 1 through 160 0.12820.25650.384?0.51290. 64110.76940.8976Columns 17 through322. 05172. 17992.30812.43632.56462.69282. 82102. 9493Columns 33 through 484. 10334.23154.35984.48804.61624.74444.87275.0009Columns 49 through506.15506.28324r O

4、.1K/SOK/S1.02581. 15411.28231.41051. 53871.66701.79521.92343.07753.20573.33393.46223.59043 71863. 84683.97515.12915. 25745.38565.51385. 64205.77035. 89856.0267区域截图 x=0:2*pi/49:2*piColumns 1 through 1600. 12820.25650. 38470. 51290.64110.76940. 8976Columns17 through 322.05172.17992. 30812.43632. 56462

5、.69282.82102. 9493Columns33 through 484. 10334.23154.35984.48804.61624.74444.87275. 00091.02581.15411. 28231.41051.53871.66701. 79521. 92343.07753.20573.33393.46223. 59043.71863. 84683.9751区域截團5. 12915.25745.38565. 51385. 64205.77035.89856. 0267Columns 49 through 506.15506.28322.4甕“ ? x Command Wind

6、ow A=0J 3J4;lJ.i5J0; WD 5, 3; 1,5, 0,5;MB, A|B, AJA=BJ AB习题三3.1clf,clc,clear t=(0:18/100:18);xi=0.2:0.2:0.8;beita=sqrt(1-xi.A2);sita=atan(beita./xi);y=1-exp(-t*xi).*sin(t*beita+ones(101,1)*sita)./(ones(101,1)*beita); figure(1)plot(t,y(:,1),b:,t,y(:,2),r-.,t,y(:,3),g-,t,y(:,4),ko)legend(xi=0.2,xi=0.4

7、,xi=0.6,xi=0.8)3.2clf,clearx=-50:1:50;y=-40:1:60;z=1/(sqrt(1-x) 八2+y.A2)+sqrt(1+x) 八2+y2);figure(1)plot3(x,y,z),box onX,Y=meshgrid(x,y);Z=1/(sqrt(1-X) .A2+Y.A2)+sqrt(1+X) 八2+Y.T);figure(2)mesh(X,Y,Z)figure(3)surf(Z)5000-50-500.080.060.040.0201000.50.40.30.20.101000.40.30.101501001501005050003.3clc;

8、clear all; close all; t=0:pi/50:2*pi; x=sin(t); y=cos(t);z=t; plot3(x,y,z,g-)0-1 -1-0.51习题四4.1Command Window k=0;i=0;while iy c=tf (a b)Transfer function:4 + 35 53 + 291 s2 + 1093 s + 1700s5 + 289 事q + 254 s3 + 2541 s2 + 4684 s + 1700 z=-3; p= -1 -5 -15; k=15; sys=zpk (zj Pj k)Zero/pole/gain:15 (s+3

9、)(s+l) (s+5) (s+15) k=100; 2=0,-2,-2; p= -1 1; sys l=zpk (zj Pj k); a=l 3 2; b=l 2 5 2; sys2=tf (aj b); sys3=series(syslj sys2)Zero/pole/gain:100 s (s+2)(s+1)(s+1) (s+0.4668) (s-1) (s2+ 1.533s + 4.284)5.4 A=3 2 1 ;0 4 6;0 -3 3; B=l 2 3?: C=l 2 5; D=0; sys=ss (Aj Bj Cj D); SYS=tf(sys)Transfer functio

10、n:20 s2 - 123 s + 234- 10+ 51 s - 90 sys2=zpk(sys)Zero/pole/gain:20 (sA2 - 6. lbs + 11. 7)(s-3) (s2- 7s + 30)5.5 nun二1 15 50 500 den= 1 2 0; sys=tf(denj nun)Transf已匸 funct ion:s2 + 2 ss3 + 15 s2 + 50 s + 500 a=l 3 6 4; b= 4 0; sysl=tf (bj a)Transfer funct ion:4 s5.6k1k2k3k4si0100z2-2-0. 142900. 1429

11、k30001k400. 089290-0. 125ul k10k20k30k4 0. 1786k1 k2 k3 k4yl 0010d =ulyl 0Continuous-time model.Transfer function:0. 17862 + 0. 02551 s + 0. 3571sq + 0. 2679+ 2. 0052 + 0. 25 s5.7 ml=12; m2=38;k=1000;c=0. 1; sysl=tf(cj k3 ml c k)Transfer function:0.1 s + 100012 s2 + 0. 1 s + 1000 sys2=tf (12 0. 1 1000, 122, 0.1000*1-2), 0, 0)Transfer function:12 s2 + 0. 1 s + 1000456 54 + 5 s3 + 50

展开阅读全文
相关资源
相关搜索

当前位置:首页 > 建筑/环境 > 建筑资料

电脑版 |金锄头文库版权所有
经营许可证:蜀ICP备13022795号 | 川公网安备 51140202000112号