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楼主 |
发表于 2016-6-8 09:12
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来自: 中国黑龙江哈尔滨
这是我在论坛里找到的,得到阶次和幅值关系,没有频率的关系。但是程序不太对,kn维数没法相乘。也得不出来。
N=201;
n=0:N-1;
t=0:0.1/2/pi:2*pi;
xn=ss(:,1);
k=0:N/2;
WN=exp(j*2*pi/N);
kn=k*n;
WNnk=WN^kn;
Xk=WNnk*xn*2/N;
stem(k,abs(Xk))
abs(Xk);
yy1=abs(Xk(2))*sin(t+pi/2-angle(Xk(2)));
yy2=abs(Xk(3))*sin(2*t+pi/2-angle(Xk(3)));
yy3=abs(Xk(4))*sin(3*t+pi/2-angle(Xk(4)));
yy4=abs(Xk(5))*sin(4*t+pi/2-angle(Xk(5)));
yy5=abs(Xk(6))*sin(5*t+pi/2-angle(Xk(6)));
yy6=abs(Xk(7))*sin(6*t+pi/2-angle(Xk(7)));
yy7=abs(Xk(8))*sin(7*t+pi/2-angle(Xk(8)));
yy8=abs(Xk(9))*sin(8*t+pi/2-angle(Xk(9)));
yy9=abs(Xk(10))*sin(9*t+pi/2-angle(Xk(10)));
yy10=abs(Xk(11))*sin(10*t+pi/2-angle(Xk(11)));
yy11=abs(Xk(12))*sin(11*t+pi/2-angle(Xk(12)));
yy12=abs(Xk(13))*sin(12*t+pi/2-angle(Xk(13)));
yy13=abs(Xk(14))*sin(13*t+pi/2-angle(Xk(14)));
yy14=abs(Xk(15))*sin(14*t+pi/2-angle(Xk(15)));
yy15=abs(Xk(16))*sin(15*t+pi/2-angle(Xk(16)));
yy16=abs(Xk(17))*sin(16*t+pi/2-angle(Xk(17)));
yy17=abs(Xk(18))*sin(17*t+pi/2-angle(Xk(18)));
yy18=abs(Xk(19))*sin(18*t+pi/2-angle(Xk(19)));
yy19=abs(Xk(20))*sin(19*t+pi/2-angle(Xk(20)));
a1=abs(Xk(2))
a2=abs(Xk(3))
a3=abs(Xk(4))
a4=abs(Xk(5))
a5=abs(Xk(6))
a6=abs(Xk(7))
a7=abs(Xk(8))
a8=abs(Xk(9))
a9=abs(Xk(10))
a10=abs(Xk(11))
a11=abs(Xk(12))
a12=abs(Xk(13))
a13=abs(Xk(14))
a14=abs(Xk(15))
a15=abs(Xk(16))
a16=abs(Xk(17))
a17=abs(Xk(18))
a18=abs(Xk(19))
a19=abs(Xk(20))
pi/2-angle(Xk(2))
figure(1)
plot(n*2*pi/201,xn,t,yy1,t,yy3,t,yy11,t,yy13,t,yy17,t,yy19); |
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