用MATLAB的向量表示法描述f(t)=cos(πt/2)[u(t)-u(t-4)] t=-2:0.01:8;f1=[zeros(1,2),ones(1,5),zeros(1,4)];f=cos(pi*t/2).*f1;plot(t,f)axis([-2,8,-2,2])?Error using ==> timesMatrix dimensions must agree.怎么改能得到以下图形?一定要
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用MATLAB的向量表示法描述f(t)=cos(πt/2)[u(t)-u(t-4)] t=-2:0.01:8;f1=[zeros(1,2),ones(1,5),zeros(1,4)];f=cos(pi*t/2).*f1;plot(t,f)axis([-2,8,-2,2])?Error using ==> timesMatrix dimensions must agree.怎么改能得到以下图形?一定要
用MATLAB的向量表示法描述f(t)=cos(πt/2)[u(t)-u(t-4)]
t=-2:0.01:8;
f1=[zeros(1,2),ones(1,5),zeros(1,4)];
f=cos(pi*t/2).*f1;
plot(t,f)
axis([-2,8,-2,2])
?Error using ==> times
Matrix dimensions must agree.
怎么改能得到以下图形?一定要用向量表示法.
用MATLAB的向量表示法描述f(t)=cos(πt/2)[u(t)-u(t-4)] t=-2:0.01:8;f1=[zeros(1,2),ones(1,5),zeros(1,4)];f=cos(pi*t/2).*f1;plot(t,f)axis([-2,8,-2,2])?Error using ==> timesMatrix dimensions must agree.怎么改能得到以下图形?一定要
t=-2:0.01:8;
f1=[zeros(1,length(-2:0.01:0)),ones(1,length(0.01:0.01:4)),zeros(1,length(4.01:0.01:8))];
f=cos(pi*t/2).*f1;
plot(t,f)
axis([-2,8,-2,2])
你那u（t）是啥表达式？啥叫向量表达法？f1=[zeros(1,2),ones(1,5),zeros(1,4)];这句跟逻辑数组作用是相同的，搞这么麻烦干啥？fi和t的大小不同，导致f=cos(pi*t/2).*f1;这句的乘法没法运行，你这到底要干啥呀？
clear all
t=-2:0.01:8;
f=cos(pi*t/2).*(t>=0&t<=4);
plot(t,f),axis([-2,8,-1.2,1.2]);