东南大学信号与系统MATLAB实践第二次作业.doc

练习二实验六一用MATLAB语言描述下列系统，并求出极零点、1. Ns=1;Ds=1,1;sys1=tf(Ns,Ds)实验结果：sys1 = 1 - s + 1 z,p,k=tf2zp(1,1,1)z = Empty matrix: 0-by-1p = -1k = 12. Ns=10Ds=1,-5,0sys2=tf(Ns,Ds)实验结果：Ns = 10Ds = 1 -5 0sys2 = 10 - s2 - 5 sz,p,k=tf2zp(10,1,-5,0)z = Empty matrix: 0-by-1p = 0 5k =10二已知系统的系统函数如下，用MATLAB描述下列系统。1 z=0;p=-1,-4;k=1;sys1=zpk(z,p,k)实验结果：sys1 = s - (s+1) (s+4) Continuous-time zero/pole/gain model.2. Ns=1,1Ds=1,0,-1sys2=tf(Ns,Ds)实验结果：Ns = 1 1Ds = 1 0 -1sys2 = s + 1 - s2 - 1 Continuous-time transfer function.3 Ns=1,6,6,0;Ds=1,6,8;sys3=tf(Ns,Ds)实验结果：Ns = 1 6 6 0Ds = 1 6 8sys3 = s3 + 6 s2 + 6 s - s2 + 6 s + 8 Continuous-time transfer function.六已知下列H(s)或H(z)，请分别画出其直角坐标系下的频率特性曲线。1. clear;for n = 1:400 w(n) = (n-1)*0.05; H(n) = (1j*w(n)/(1j*w(n)+1);endmag = abs(H);phase = angle(H);subplot(2,1,1)plot(w,mag);title(幅频特性)subplot(2,1,2)plot(w,phase);title(相频特性)实验结果：2. clear;for n = 1:400 w(n) = (n-1)*0.05; H(n) = (2*j*w(n)/(1j*w(n)2+sqrt(2)*j*w(n)+1);endmag = abs(H);phase = angle(H);subplot(2,1,1)plot(w,mag);title(幅频特性)subplot(2,1,2)plot(w,phase);title(相频特性)实验结果：3. clear;for n = 1:400 w(n) = (n-1)*0.05; H(n) = (1j*w(n)+1)2/(1j*w(n)2+0.61);endmag = abs(H);phase = angle(H);subplot(2,1,1)plot(w,mag);title(幅频特性)subplot(2,1,2)plot(w,phase);title(相频特性)实验结果：4. clear;for n = 1:400 w(n) = (n-1)*0.05; H(n) =3*(1j*w(n)-1)*(1j*w(n)-2)/(1j*w(n)+1)*(1j*w(n)+2);endmag = abs(H);phase = angle(H);subplot(2,1,1)plot(w,mag);title(幅频特性)subplot(2,1,2)plot(w,phase);title(相频特性)实验结果：实验七三已知下列传递函数H(s)或H(z)，求其极零点，并画出极零图。1. z=1,2;p=-1,-2;zplane(z,p)实验结果：2. z=1,2;p=-1,-2;zplane(z,p) num=1;den=1,0;z,p,k=tf2zp(num,den);zplane(z,p) num=1;den=1,0;z,p,k=tf2zp(num,den)zplane(z,p)实验结果：z = Empty matrix: 0-by-1p = 0k = 13. num=1,0,1;den=1,2,5;z,p,k=tf2zp(num,den)zplane(z,p)实验结果：z = 0 + 1.0000i 0 - 1.0000ip = -1.0000 + 2.0000i -1.0000 - 2.0000ik = 14. num=1.8,1.2,1.2,3;den=1,3,2,1;z,p,k=tf2zp(num,den)zplane(z,p)实验结果：z = -1.2284 0.2809 + 1.1304i 0.2809 - 1.1304ip = -2.3247 -0.3376 + 0.5623i -0.3376 - 0.5623ik =1.80005 clear;A=0,1,0; 0,0,1; -6,-11,-6;B=0;0;1;C=4,5,1;D=0;sys5=ss(A,B,C,D);pzmap(sys5)实验结果：五求出下列系统的极零点，判断系统的稳定性。1. clear;A=5,2,1,0; 0,4,6,0; 0,-3,-6,-1;1,-2,-1,3;B=1;2;3;4;C=1,2,5,2;D=0;sys=ss(A,B,C,D);z,p,k=ss2zp(A,B,C,D,1)pzmap(sys)实验结果：z = 4.0280 + 1.2231i 4.0280 - 1.2231i 0.2298 p = -3.4949 4.4438 + 0.1975i 4.4438 - 0.1975i 0.6074 k =28由求得的极点，该系统不稳定。4.z=-3P=-1,-5,-15所以该系统为稳定的。5. num=100*conv(1,0,conv(1,2,conv(1,2,conv(1,3,2,1,3,2);den=conv(1,1,conv(1,-1,conv(1,3,5,2,conv(1,0,2,0,4,1,0,2,0,4);z,p,k=tf2zp(num,den)实验结果：z = 0 -2.0005 + 0.0005i -2.0005 - 0.0005i -1.9995 + 0.0005i -1.9995 - 0.0005i -1.0000 + 0.0000i -1.0000 - 0.0000ip = 1.0000 0.7071 + 1.2247i 0.7071 - 1.2247i 0.7071 + 1.2247i 0.7071 - 1.2247i -1.2267 + 1.4677i -1.2267 - 1.4677i -0.7071 + 1.2247i -0.7071 - 1.2247i -0.7071 + 1.2247i -0.7071 - 1.2247i -1.0000 -0.5466 zplane(z,p)所以该系统不稳定。七已知反馈系统开环转移函数如下，试作其奈奎斯特图，并判断系统是否稳定。1. b=1;a=1,3,2;sys=tf(b,a);nyquist(sys);实验结果：由于奈奎斯特图并未围绕上-1点运动，同时其开环转移函数也是稳定的，由此，该线性负反馈系统也是稳定的。2 b=1;a=1,4,4,0;sys=tf(b,a);nyqui