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[编程技巧] 利用杜哈美积分求速度、加速度

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发表于 2006-5-8 14:04 | 显示全部楼层 |阅读模式

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请问,怎样利用杜哈美积分求速度、加速度,(求杜哈美积分的导函数)而不是利用v=(s2-s1)/t;a=(v2-v1)/t这样的差分法。

感谢happy上次提供的杜哈美积分的源程序,向具体问问A,B是什么?还有怎样用duhamel积分求速度和加速度。不用差分法。

[ 本帖最后由 ChaChing 于 2010-3-15 15:59 编辑 ]
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发表于 2006-5-13 16:04 | 显示全部楼层

回复:(assist)happy:请问杜哈美积分

正弦余弦成分的幅值<BR><BR>已知位移表达式速度加速度表达式你不知道吗?
 楼主| 发表于 2006-5-29 23:13 | 显示全部楼层

利用杜哈美积分求位移、速度和加速度

  1. %单自由度系统适用,初始条件任意,比happy提供的适应性强。对于多自由度系统,正
  2. %则化之后适用,加循环即可(有点修改,for t=0:dTaT end之间的不用改)。
  3. clc; clear
  4. n=1000; m=3; T=2; si=0.02; k=2700; w=30;
  5. dTao=T/n; wD=w*(1-si^2)^0.5;
  6. p2=[ones(1,333),zeros(1,668)]; % p2=[1000,zeros(1,1000)];
  7. %%%%%%%%%%%%%%%杜哈美积分开始%%%%%%%%%%%%%%%%
  8. x0=0; v0=0; j=1;
  9. for t=0:dTaT
  10. sysN=0; sysP=0;
  11. %%%%%%%%%%%%位移\速度\加速度公用部分%%%%%%%%%%%%%%
  12. for i=0:t/dTao
  13. sysN=sysN+p2(i+1)*exp(si*i*dTao*w)*cos(wD*i*dTao)*dTao;
  14. sysP=sysP+p2(i+1)*exp(si*i*dTao*w)*sin(wD*i*dTao)*dTao;
  15. end
  16. %%%%%%%%%%%%位移\速度\加速度公用部分%%%%%%%%%%%%%%
  17. %%%%%%%%%%%%%%%%%求位移%%%%%%%%%%%%%%%%%%%
  18. sysA=exp(-si*w*t)*(x0*cos(wD*t)+(si*w*x0+v0)/wD*sin(wD*t));
  19. sysB=exp(-si*w*t)/(m*wD)*(sin(wD*t)*sysN-cos(wD*t)*sysP);
  20. x(j)=sysA+sysB;
  21. %%%%%%%%%%%%%%%%%求位移%%%%%%%%%%%%%%%%%%%
  22. %%%%%%%%%%%%%%%%%%%%%%%%%求速度%%%%%%%%%%%%%%%%%%%%%%%%%%%
  23. %%%%%%%%%%%%%%%V=C+D+E+F%%%%%%%%%%%%%%%%%%
  24. sysC=exp(-si*w*t)*(-x0*wD*sin(wD*t)-(si*w)^2*x0*sin(wD*t)/wD);
  25. sysD=exp(-si*w*t)*(v0*cos(wD*t)-si*w*v0*sin(wD*t)/wD);
  26. %%%%%%%%%%%%%%%E=G+H%%%%%%%%%%%%%%%%%%%%%%
  27. sysG=sin(wD*t)*sysN;
  28. sysH=-cos(wD*t)*sysP;
  29. sysE=(sysN+sysP)*-si*w*exp(-si*w*t)/(m*wD);
  30. %%%%%%%%%%%%%%%F=I+J+K+L%%%%%%%%%%%%%%%%%%
  31. sysI=wD*cos(wD*t)*sysN;
  32. sysJ=sin(wD*t)*p2(j)*exp(si*w*t)*cos(wD*t);
  33. sysK=wD*sin(wD*t)*sysP;
  34. sysL=-cos(wD*t)*p2(j)*exp(si*w*t)*sin(wD*t);
  35. sysF=(sysI+sysJ+sysK+sysL)*exp(-si*w*t)/(m*wD);
  36. v(j)=sysC+sysD+sysE+sysF;
  37. %%%%%%%%%%%%%%%%%%%%%%%%%求速度%%%%%%%%%%%%%%%%%%%%%%%%%%%
  38. %%%%%%%%%%%%%%%%%%%%%%%%%求加速度%%%%%%%%%%%%%%%%%%%%%%%%%
  39. %%%%%%%%%%%%%%%%%%%sysdC%%%%%%%%%%%%%%%
  40. sysdC1=-x0*wD^2*cos(wD*t)-(si*w)^2*x0*cos(wD*t);
  41. sysdC=-si*w*sysC+exp(-si*w*t)*sysdC1;
  42. %%%%%%%%%%%%%%%%%%%sysdD%%%%%%%%%%%%%%%
  43. sysdD1=-v0*wD*sin(wD*t)-si*w*v0*cos(wD*t);
  44. sysdD=-si*w*sysD+exp(-si*w*t)*sysdD1;
  45. %%%%%%%%%%%%%%%%%%%sysdE%%%%%%%%%%%%%%%
  46. sysdG=wD*cos(wD*t)*sysN+sin(wD*t)*p2(j)*exp(si*w*t)*cos(wD*t);
  47. sysdH=wD*sin(wD*t)*sysP-cos(wD*t)*p2(j)*exp(si*w*t)*sin(wD*t);
  48. sysdE=(si*w)^2*exp(-si*w*t)*(sysG+sysH)/(m*wD)-si*w*exp(-si*w*t)*(sysdG+sysdH)/(m*wD);
  49. %%%%%%%%%%%%%%%%%%%sysdF%%%%%%%%%%%%%%%
  50. sysdI=-wD^2*sin(wD*t)*sysN+wD*cos(wD*t)*p2(j)*exp(si*w*t)*cos(wD*t);
  51. %%%%%%%%%%%%怎样考虑dP2,暂时取0%%%sysdJ%%%%%%%%%%
  52. sysdJ1=wD*cos(wD*t)^2*p2(j)*exp(si*w*t);
  53. sysdJ2=sin(wD*t)*0*exp(si*w*t)*cos(wD*t);
  54. sysdJ3=sin(wD*t)*p2(j)*si*w*exp(si*w*t)*cos(wD*t);
  55. sysdJ4=-wD*sin(wD*t)^2*p2(j)*exp(si*w*t);
  56. sysdJ=sysdJ1+sysdJ2+sysdJ3+sysdJ4;
  57. %%%%%%%%%%%%%%%%%%%sysdK%%%%%%%%%%%%%%%
  58. sysdK=wD^2*cos(wD*t)*sysP+wD*sin(wD*t)*p2(j)*exp(si*w*t)*sin(wD*t);
  59. %%%%%%%%%%%%怎样考虑dP2,暂时取0%%%sysdL%%%%%%%%%%
  60. sysdL1=wD*sin(wD*t)^2*p2(j)*exp(si*w*t);
  61. sysdL2=cos(wD*t)*0*exp(si*w*t)*sin(wD*t);
  62. sysdL3=-cos(wD*t)*p2(j)*si*w*exp(si*w*t)*sin(wD*t);
  63. sysdL4=-wD*cos(wD*t)^2*p2(j)*exp(si*w*t);
  64. sysdL=sysdL1+sysdL2+sysdL3+sysdL4;
  65. %%%%%%%%%%%%%%%%%sysdF%%%%%%%%%%%%%%
  66. sysdF=-si*w*exp(-si*w*t)*(sysI+sysJ+sysK+sysL)/(wD*m)+exp(-si*w*t)*(sysdI+sysdJ+sysdK+sysdL)/(wD*m);
  67. %%%%%%%%%%%%%%%%%acc%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  68. acc(j)=sysdC+sysdD+sysdE+sysdF;
  69. %%%%%%%%%%%%%%%%%%%%%%%%%求加速度%%%%%%%%%%%%%%%%%%%%%%%%%
  70. j=j+1;
  71. end
  72. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%杜哈美积分完毕%%%%%%%%%%%%%%%%%%%%%%%%%
  73. figure;
  74. t=0:dTaT;
  75. subplot(3,1,1);plot(t,x);grid on;
  76. subplot(3,1,2);plot(t,v);grid on;
  77. subplot(3,1,3);plot(t,acc);grid on;
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[ 本帖最后由 ChaChing 于 2009-4-18 21:28 编辑 ]
发表于 2006-12-7 09:48 | 显示全部楼层
好东西!我要研究研究!谢了!
发表于 2007-6-11 17:26 | 显示全部楼层
好东西,收藏
发表于 2007-10-27 10:28 | 显示全部楼层
好东西!我要研究研究!谢了!
发表于 2007-12-20 18:37 | 显示全部楼层
杜哈梅积分,求任意激励的响应,振动力学里的,我也编写过,你这个好难看懂!
这是我写的duhamel积分,完全是按照西安交大版《振动力学》上的公式写的,调试无误。
% duhamel        duhamel integral
% y=duhamel(x,h,fs)
% inputs:     x:analysed signal(vector)
%             h:impulse response
%             fs:frequency of sampling
% outputs:    y:duhamel integral of analysed signal x
function y=duhamel(x,h,fs)
    n=length(x);
    for i=1:n
        xx=x(1:i);
        hh=fliplr(h(1:i));
        temp=xx.*hh;
        y(i)=trapz(temp)*(1/fs);
    end

[ 本帖最后由 ChaChing 于 2009-4-18 21:26 编辑 ]
 楼主| 发表于 2007-12-21 09:30 | 显示全部楼层

我这个没有使用任何matlab系统函数,完全的一步一步下来的,如果你对duhamel积分公式研究一下是能看懂的。
发表于 2007-12-21 11:29 | 显示全部楼层
支持非matlab算法程序!
发表于 2009-1-27 00:08 | 显示全部楼层
好好学习一下
发表于 2009-1-27 10:21 | 显示全部楼层
恩,好东西,楼主新年给大家送好礼啊:lol
发表于 2009-4-18 14:47 | 显示全部楼层
:@) :@) :@) :@) 经典,就是不出结果啊!
发表于 2009-5-24 17:40 | 显示全部楼层
for t=0:dTaT
dTaT是什么意思?提示没定义呀?
发表于 2009-6-1 16:46 | 显示全部楼层
支持!!!!
发表于 2009-6-8 10:00 | 显示全部楼层
for t=0:dTaT
dTaT是什么意思?提示没定义呀?
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