1、(2) 乘:f(t)=f1(t)(3) 延时或平移:f(t)f(t-t0) t00时右移;t00。2;y=sin(2*pi*(t);y1=sin(2*pi*(t-0.2);plot(t, y, -,t, y1, -tf(t)信号的移位信号及其移位结果如下图所示。4)翻转信号的翻转就是将信号的波形以纵轴为对称轴翻转180。将信号f(t)中的自变量t替换成-t即可得到其翻转信号。0.02:1;t1=-1:0;g1=3*t;g2=3*(-t1);grid on;plot(t, g1, ,t1, g2);g(t)信号的反折信号及其反折结果如下图所示。2.程序设计实验(1)已知信号f1(t)=(-t+4
2、)U(t)-U(t-4),f2(t)=sin(2t),用MATLAB绘出下列信号的时域波形。要求写出全部程序,并绘制出信号的时域波形。(a)f3(t)f1(-t)+f1(t)(b)f4(t)-f1(-t)f1(t)(c)f5(t)f2(t)(d)f6(t)f1(t)f2(t)clearall;t=-5:u=stepfun(t,-4);u1=stepfun(t,0);u2=stepfun(t,4);f1=(-t+4).*(u1-u2);f2=sin(2*pi*t);g=(t+4).*(u-u1);f3=g+f1;plot(t,f3);f3(t)f3(t)=f1(-t)+f1(t)gridon;f
3、4=-f3;plot(t,f4);f4(t)f4(t)=-f1(-t)+f1(t)f5=f2.*f3;plot(t,f5);f5(t)f5(t)=f2(t)f6=f1.*f2;plot(t,f6);f6(t)f6(t)=f1(t)f2(t)信号时域波形如下图所示。(3)若f1(t)=(t),f2(t)=U(t),f3(t)=U(t)-U(t-4) 试证明卷积满足如下结论:f1(t)*f2(t)=f2(t)*f1(t)f1(t)*f2(t)+f3(t)=f1(t)*f2(t)+f1(t)*f3(t)第一问MATLAB程序如下:a=1000;t1=-5:1/a:f1=stepfun(t1,-1/a
4、)-stepfun(t1,1/a);f2=stepfun(t1,0);y1=conv(f1,f2);r=2*length(t1)-1;t=-10:10;subplot(121);plot(t,y1);axis(-5,5,0,2.4);f1(t)*f2(t)y1(t)y2=conv(f2,f1);subplot(122);plot(t,y2);f2(t)*f1(t)y2(t)运行结果如下图所示。从图中可以清晰的看出结论一成立。第二问MATLAB程序如下:f3=stepfun(t1,0)-stepfun(t1)-4,0);f4=f2+f3;y1=conv(f1,f4);f1(t)*f2(t)+f3(t)y2=conv(f1,f2)+conv(f1,f3);f1(t)*f2(t)+f1(t)*f3(t)从图中可以清晰的看出结论二成立。