问题描述

根据傅里叶变换性质可得：

编程：
1、求h(t)在t=[-10,10]的近似图像
2、将得到的h(t)与x(t)卷积看看y(t)的样子

代码运行

T0 = 50;
intervalT = 0.001;
t = 0:intervalT:T0;

W0 = 20;
intervalW = 0.001;
w = 0:intervalW:W0;

functionW = zeros(1,length(w)); 
for i = 1:length(w)
    if sin(w(i))>1e-6 %零值判断
        functionW(i) = sin(0.5*w(i))^2/(w(i).*sin(w(i)));
    else 
        functionW(i) = 0.25; %趋于零的点取极限0.25
    end;
end;
%求h
h = zeros(1,length(t));
for i = 1:length(t)
    for j=1:length(functionW)
        h(i) = h(i)+functionW(j)*cos(w(j)*t(i))*intervalW;
    end;
end;
h=2*h/pi;

%由于h(t)是偶函数 因此将图像翻转到负半轴
t = [-t(end:-1:2),t];
h = [h(end:-1:2),h];

%画h(t)
subplot(2,1,1);
plot(t,h);

%这里验证求出的h(t)和x(t)卷积后的y(t)是什么样子
newT = -10:intervalT:10;
y = zeros(1,length(newT));
for i = 1:length(newT)
    time = newT(i);
    for tao = -20:0.001:20
        hindex = 1+round((tao-t(1))/intervalT);
        if time-tao>0 && time-tao<2 && hindex>=1 && hindex<=length(h)
            y(i) = y(i)+h(hindex)*0.001;
        end;
    end;
end;

%画y(t)
subplot(2,1,2);
plot(newT,y);

运行结果