实验三 信号与系统&Matlab 连续时间信号的频域分析
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连续时间信号的频域分析
题目:

第一题:
clear all
clc
m = input('m=');
t = -2*pi:0.01:2*pi;
n = round(length(t)/8);
f = [ones(n,1);-1*ones(n,1);-1*ones(n,1);ones(n,1);ones(n,1);-1*ones(n,1);-1*ones(n+1,1);ones(n,1)];
y = zeros(m+1,max(size(t)));
y(m+1,:)=f';
figure(1);
plot(t/pi,y(m+1,:));
grid;
axis([-2 2 -1.5 1.5]);
title('周期方波');
xlabel('单位pi','Fontsize',8);
x = zeros(size(t));
kk = '1';
for k=1:2:2*m-1
pause;
x = x+4/pi*cos(k*t)*sin(k*pi/2)/k;
y((k+1)/2,:) = x;
plot(t/pi,y(m+1,:));
hold on;
plot(t/pi,y((k+1)/2,:));
hold off;
grid;
axis([-2 2 -1.5 1.5]);
title(strcat('第',kk,'次谐波叠加'));
xlabel('单位pi','Fontsize',8);
kk = strcat(kk,'、',num2str(k+2));
end
pause;
plot(t/pi,y(1:m+1,:));grid;
axis([-2 2 -1.5 1.5]);
title('各次谐波叠加波形');
xlabel('单位pi','Fontsize',8);
结果:

第二题:
(1):
clear all;
clc
syms t w;
f=e
xp(-2*t)*heaviside(t);
t1=-2:0.01:2;
f1=subs(f,t,t1)
subplot(1,2,1)
plot(t1,f1)
title('f(t)')
F=fourier(f,t,w)
w1=-2:0.01:2;
F1=subs(F,w,w1)
subplot(1,2,2)
plot(w1,abs(w1))
title('|F(w)|')
结果:

(2):
clear all;
clc
syms t w;
f=exp(-2*abs(t));
t1=-2:0.01:2;
f1=subs(f,t,t1)
subplot(1,2,1)
plot(t1,f1)
title('f(t)')
F=fourier(f,t,w)
w1=-2:0.01:2;
F1=subs(F,w,w1)
subplot(1,2,2)
plot(w1,abs(w1))
title('|F(w)|')
结果:

(3):
clear all;
clc
syms t w;
f=t.*(t>0.5)-t.*(t>(-1*0.5));
t1=-6:0.01:6;
f1=subs(f,t,t1)
subplot(1,2,1)
plot(t1,f1)
title('f(t)')
F=fourier(f,t,w)
w1=-2:0.01:2;
F1=subs(F,w,w1)
subplot(1,2,2)
plot(w1,abs(w1))
title('|F(w)|')
结果:

(4):
clear all;
clc
syms t w;
f = (1-abs(t)).*(abs(t)<=1)
t1=-2:0.01:2;
f1=subs(f,t,t1)
subplot(1,2,1)
plot(t1,f1)
title('f(t)')
F=fourier(f,t,w)
w1=-2:0.01:2;
F1=subs(F,w,w1)
subplot(1,2,2)
plot(w1,abs(w1))
title('|F(w)|')
结果:

(5):
clear all;
clc
syms t w;
f = sin(t)/t;
t1=-10*pi:0.001:10*pi;
f1=subs(f,t,t1)
subplot(1,2,1)
plot(t1,f1)
title('f(t)')
F=fourier(f,t,w)
w1=-10*pi:0.001:10*pi;
F1=subs(F,w,w1)
subplot(1,2,2)
plot(w1,abs(w1))
title('|F(w)|')
结果:

(6):
syms t w;
F=2/(1+w^2) ;
f=ifourier(F,t)

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