本人小白，现在有九个矩阵t1~t9（矩阵里含有k，k=3.14*f/340，其余参数已给）想把九个矩阵相乘得到t ，f是自变量，怎么实现f从50到520，步长为5，然后得到很多个t，最后把数组t再代入公式y=****，最后得到f与y的函数图像？
l1=1;
l3=0.7;
l4=0.8;
l5=1.2;
l6=0.4;
l7=1.1;
l9=1;
m=1.53;
p=0.53;
c0=520;
s1=0.2;s5=0.2;s9=0.2;
s3=1.3;s7=1.3;
se=1.1;o=6.5;
c=340;
aa=0;
for f=50:5:520
aa=aa+1;
k(aa)=2*3.1415926*f/c;
t1(aa)=[cos(k(aa)*l1/(1-m^2)),1i*(p*c0/s1)*sin(k(aa)*l1/(1-m^2));1i*(s1/p/c0)*sin(k(aa)*l1/(1-m^2)),cos(k(aa)*l1/(1-m^2))];
t2(aa)=[1,(-2*m/o/s1)*p*c0*(1-1/o);0,1];
t3(aa)=[cos(k(aa)*l3/(1-m^2)),1i*(p*c0/s3)*sin(k(aa)*l3/(1-m^2));1i*(s3/p/c0)*sin(k(aa)*l3/(1-m^2)),cos(k(aa)*l3/(1-m^2))];
t4(aa)=[1-(se/s5)*o^2*m*tan(k(aa)*l4),(1-o^2)*m*p*c0/s5;1i*(s5/p/c0)*tan(k(aa)*l4),1+1i*(se/s5)*m*tan(k(aa)*l4)];
t5(aa)=[cos(k(aa)*l5/(1-m^2)),1i*(p*c0/s5)*sin(k(aa)*l5/(1-m^2));1i*(s5/p/c0)*sin(k(aa)*l5/(1-m^2)),cos(k(aa)*l5/(1-m^2))];
t6(aa)=[1-(se/s5/o)*m*tan(k(aa)*l6),(-2*m/o*s5)*p*c0*(1-1/o);j(s5/p/c0)*tan(k(aa)*l6)*(1+1i*(se*(o-2)*m)/s5/o)*tan(k(aa)*l6),1+1i*(se*(o-2)*m/s5/o)*tan(k(aa)*l6)];
t7(aa)=[cos(k(aa)*l7/(1-m^2)),1i*(p*c0/s7)*sin(k(aa)*l7/(1-m^2));1i*(s7/p/c0)*sin(k(aa)*l7/(1-m^2)),cos(k(aa)*l7/(1-m^2))];
t8(aa)=[1,(m/s9)*p*c0*(1-o^2);0,1];
t9(aa)=[cos(k(aa)*l9/(1-m^2)),1i*(p*c0/s9)*sin(k(aa)*l9/(1-m^2));1i*(s9/p/c0)*sin(k(aa)*l9/(1-m^2)),cos(k(aa)*l9/(1-m^2))];
t(aa)=t1(aa)*t2(aa)*t3(aa)*t4(aa)*t5(aa)*t6(aa)*t7(aa)*t8(aa)*t9(aa);
end
