北京交通大学《信号与系统》 课后matlab作业.docx
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北京交通大学《信号与系统》课后matlab作业
Matlab课后作业
1.M2-1
(1)Matlab程序:
t=-5:
0.01:
5;
x=(t>0)-(t>2);
plot(t,x);
axis([-5,5,-2,2]);
仿真结果:
(8)Matlab程序:
t=-10:
0.01:
10;
pi=3.14;
x=sin(pi*t)./(pi*t).*cos(30*t);
plot(t,x);
仿真结果:
M2-2
Matlab程序:
t=-2:
0.001:
2;
x=(t>-1)-(t>0)+2*tripuls(t-0.5,1,0);
plot(t,x);
axis([-2,2,-2,2]);
仿真结果:
M3-3
(1)functionyt=f(t)
yt=t.*(t>0)-t.*(t>=2)+2*(t>=2)-3*(t>3)+(t>5);
(2)Matlab程序:
t=-10:
0.01:
11;
subplot(3,1,1);
plot(t,f(t));
title('x(t)');
axis([-1,6,-2,3]);
subplot(3,1,2);
plot(t,f(0.5*t));
axis([-1,11,-2,3]);
title('x(0.5t)');
subplot(3,1,3);
plot(t,f(2-0.5*t));
title('x(2-0.5t)');
axis([-9,5,-2,3]);
仿真结果:
M2-9
(1)Matlab程序:
k=-4:
7;
x=[-3,-2,3,1,-2,-3,-4,2,-1,4,1,-1];
stem(k,x);
仿真结果:
(2)Matlab程序:
k=-12:
21;
x=[-3,-2,3,1,-2,-3,-4,2,-1,4,1,-1];
N=length(x);
y=zeros(1,3*N-2);
y(1:
3:
end)=x;
stem(k,y);
仿真结果:
Matlab程序:
k=-1:
3;
x=[0,0,-3,-2,3,1,-2,-3,-4,2,-1,4,1,-1];
x1=x(1:
3:
end);
stem(k-1,x1);
仿真结果:
(3)Matlab程序:
k=-6:
5;
x=[-3,-2,3,1,-2,-3,-4,2,-1,4,1,-1];
stem(k,x);
仿真结果:
程序
>>k=-2:
9;
>>x=[-3,-2,3,1,-2,-3,-4,2,-1,4,1,-1];
>>stem(k,x);
结果
程序
>>k=-4:
7;
>>x=[-3,-2,3,1,-2,-3,-4,2,-1,4,1,-1];
>>xk=fliplr(x);
>>k1=-fliplr(k);
>>stem(k1,xk);
结果
M3-1
(1)程序
>>ts=0;te=5;dt=0.01;
>>sys=tf([21],[132]);
>>t=ts:
dt:
te;
>>x=exp(-3*t).*(t>=0);
>>y=lsim(sys,x,t);
>>plot(t,y);
>>xlabel('Time(sec)')
>>ylabel('y(t)')
结果
(2)程序
>>ts=0;te=5;dt=0.0001;
>>sys=tf([21],[132]);
>>t=sys:
dt:
te;
>>x=exp(-3*t).*(t>=0);
>>y=lsim(sys,x,t);
>>plot(t,y);
>>xlabel('Time(sec)')
>>ylabel('y(t)')
结果
M3-4
>>x=[0.85,0.53,0.21,0.67,0.84,0.12];
>>k1=-2:
3;
>>h=[0.68,0.37,0.83,0.52,0.71];
>>k2=-1:
3;
>>y=conv(x,h);
>>k=(k1
(1)+k2
(1)):
(k1(end)+k2(end));
>>stem(k,y)
结果
M6-1
(1)>>num=[1600];
>>den=[15.65698162262.7160000];
>>[r,p,k]=residue(num,den)
得r=
0.0992-1.5147i
0.0992+1.5147i
-0.0992+1.3137i
-0.0992-1.3137i
p=
-1.5145+21.4145i
-1.5145-21.4145i
-1.3140+18.5860i
-1.3140-18.5860i
k=
[]
所以可得
X(s)=
x(t)=3.0108e-1.5145tcos(21.4145t-1.5054)u(t)+2.635e-1.314tcos(18.586t+1.6462)u(t)
(2)X(s)=
解:
>>num=[1000];
den=conv([15],[1525]);
[r,p,k]=residue(num,den)
[angle,mag]=cart2pol(real(r),imag(r))
得r=
-5.0000+0.0000i
-2.5000-1.4434i
-2.5000+1.4434i
p=
-5.0000+0.0000i
-2.5000+4.3301i
-2.5000-4.3301i
k=
1
angle=
3.1416
-2.6180
2.6180
mag=
5.0000
2.8868
2.8868
所以X(s)=
x(t)=δ(t)+5e-5tu(t)+5.7736e-2.5tcos(4.3301t-2.618)u(t)
M6-2
程序
>>t=0:
0.1:
10;
>>y1=(2.5*exp(-t)-1.5*exp(-3*t)).*(t>=0);
>>y2=((1/3)+2*exp(-t)-(5/6)*exp(-3*t)).*(t>=0);
>>y=((1/3)+(9/2)*exp(-t)-(7/3)*exp(-3*t)).*(t>=0);
>>plot(t,y1,'r-',t,y2,'g--',t,y,'b-')
>>xlabel('Time');
>>legend('零输入响应','零状态响应','完全响应')
结果
M6-5
>>num=[12];
>>den=[1221];
>>sys=tf(num,den);
>>pzmap(sys)
>>num=[12];
den=[1221];
[r,p,k]=residue(num,den)
[angle,mag]=cart2pol(real(r),imag(r))
r=
1.0000+0.0000i
-0.5000-0.8660i
-0.5000+0.8660i
p=
-1.0000+0.0000i
-0.5000+0.8660i
-0.5000-0.8660i
k=
[]
angle=
0
-2.0944
2.0944
mag=
1.0000
1.0000
1.0000
所以H(s)=
系统冲激响应h(t)=e-tu(t)+2e-0.5tcos(0.866t-2.0944)u(t)
>>num=[12];
>>den=conv([10],[1221]);
>>[r,p,k]=residue(num,den)
r=
-1.0000+0.0000i
-0.5000+0.8660i
-0.5000-0.8660i
2.0000+0.0000i
p=
-1.0000+0.0000i
-0.5000+0.8660i
-0.5000-0.8660i
0.0000+0.0000i
k=
[]
[angle,mag]=cart2pol(real(r),imag(r))
angle=
3.1416
2.0944
-2.0944
0
mag=
1.0000
1.0000
1.0000
2.0000
所以Y(s)=
系统阶跃响应y(t)=e-tu(t)+2e-0.5tcos(0.866t+2.0944)u(t)
因为系统的冲激响应
h(t)=e-tu(t)-1.00001e-0.5tcos(0.866t)u(t)+1.73205e-0.5tsin(0.866t)u(t)
所以系统的频率响应
H(jω)=