长江大学信号与系统matlab实验答案

以下部分非实验相关内容

RC频率特性：

f0=0;f1=10000;f=f0:20:f1; RC=2;

fc1=round(1./(RC*2*pi)); H=1./(1+j*2*pi*f*RC); Hw1=abs(H); P1=angle(H); RC=0.1;

fc2=round(1./(RC*2*pi)); H=1./(1+j*2*pi*f*RC); Hw2=abs(H); P2=angle(H); RC=0.001;

fc3=round(1./(RC*2*pi)); H=1./(1+j*2*pi*f*RC); Hw3=abs(H); P3=angle(H); RC=0.0001;

fc4=round(1./(RC*2*pi)); H=1./(1+j*2*pi*f*RC); Hw4=abs(H); P4=angle(H); RC=0.00002;

fc5=round(1./(RC*2*pi)); H=1./(1+j*2*pi*f*RC); Hw5=abs(H); P5=angle(H);

subplot(1,2,1),plot(f,Hw1,'r--',f,Hw2,'k:',f,Hw3,'g-.',f,Hw4,'m-.',f,Hw5,'linewidth',2);

xlabel('f(Hz)');title('幅频特性'),grid

set(gca,'Xtick',[fc1,fc2,fc3,fc4,fc5,f1],'Ytick',[0,0.707,1]) legend('RC=2','RC=0.1','RC=1e-3','RC=1e-4','RC=2e-5',1)

subplot(1,2,2),plot(f,P1*180/pi,'r--',f,P2*180/pi,'k:',f,P3*180/pi,'g-.',f,P4*180/pi,'m-.',f,P5*180/pi,'linewidth',2);

xlabel('f(Hz)');title('相频特性'),grid

set(gca,'Xtick',[fc1,fc2,fc3,fc4,fc5,f1],'Ytick',[-90,-45,0]) legend('RC=2','RC=0.1','RC=1e-3','RC=1e-4','RC=2e-5',1)

结果：

幅频特性1 RC=2RC=0.1RC=1e-3RC=1e-40.707RC=2e-50159159220 795810000f(Hz)相频特性0 RC=2RC=0.1RC=1e-3RC=1e-4RC=2e-5-45-90159159220 795810000f(Hz)