1、2.*给出一种计算关于质量矩阵M和刚度矩阵K的按模最小的非零广义特征值的方法(误差小于3%);3.试估计前五个特征值所对应的特征向量,(此时可利用已给的测量特征值数据),并给出估计结果与测量数据之间的相对误差;4*.一般说来,测量数据应满足KX=MXD,但题目中给出的数据并不严格满足这一条件,试给出一种方法降低误差,要求不能改变X,D,可以改变K,M.问题解答1 K=redgongwithstiff; M=redgongwithmass; max(eig(K)ans = 9.8915e+009 max(eig(M)2.04142 AA,BB=qz(K,M);% Q*A*Z = AA, Q*B*
2、Z = BB aa=diag(AA);%将值集中到对角线上 bb=diag(BB); cc=aa./bb;Warning: Divide by zero. min(cc) 3.2587e+0033M文件:function t,x = chengmi(A,v)w0=v;y0=w0/norm(w0,inf);w1=A*y0;y1=w1/norm(w1,inf);while(abs(norm(w1,inf)-norm(w0,inf)=0.001) y0=y1; w0=w1; w1=A*y0; y1=w1/norm(w1,inf);endt=norm(w1,inf);x=y1;%-乘幂法求解矩阵A的按
3、模最小特征值和特征向量在COMMAND输入以下程序就可得答案:M=redgongwithmass;X=redgongwithmode;D=gongwitheig; v=rand(240,1); t1,x1=chengmi(K,v); TeZhengXiangLiang=x1;x2x3x4x5;%- K2=K-t1.*(x1*x1)/(x1*x1); t2,x2=chengmi(K2,v); K3=K2-t2.*(x2*x2)/(x2*x2); t3,x3=chengmi(K3,v); K4=K3-t3.*(x3*x3)/(x3*x3); t4,x4=chengmi(K4,v); K5=K4-t
4、4.*(x4*x4)/(x4*x4); t5,x5=chengmi(K5,v); TeZhengZhi=t1,t2,t3,t4,t5;t1 = 9.8915e+009t2 = 9.8469e+009t3 = 9.7729e+009t4 = 9.6699e+009t5 = 9.5385e+009%- eig(K) 1.0e+009 * 9.5385 9.6699 9.7729 9.8469 9.8915注1:这里是通过MATLAB直接求解特征值所得的最大的五个特征值。注2:1) x1、x2、x3、x4、x5的解请看附录2)此处的特征值和特征向量都已保存在文件夹中附录X1= -0.0775 0 0
5、.1545 -0.2307 0.3054 -0.3783 0.4489 -0.5168 0.5816 -0.6429 0.7003 -0.7536 0.8023 -0.8461 0.8849 -0.9184 0.9463 -0.9686 0.9850 -0.9955 1.0000 -0.9985 0.9910 -0.9775 0.9582 -0.9331 0.9023 -0.8662 0.8248 -0.7785 0.7275 -0.6721 0.6127 -0.5496 0.4832 -0.4139 0.3421 -0.2682 0.1927 -0.1161 0.0388x2 = -0.3054 -0.5816 -0.8023 -0.9463 -1.0000 -0.9582 -0.8248 -0.6127 -0.3421 -0.0388 0.2682 0.5496 0.7785 0.9331 0.9985 0.9686 0.8461 0.6429 0.3783 0.0775%-x3 = 0.2310 -0.4496 0.6439 -0.8035 0.9198 -0.9865 -0.9596 0.8675 -0.7286 0.5504 -0.3426 0.1163
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