离散系统分析.doc

第二篇系统分析与设计 实验二离散系统分析 1. 已知一个LTI系统的差分方程为： y[n]-1.143*y[n-1]+0.4128*y[n-2]=0.0675*x[n]+0.1349*x[n-1]+0.0675*x[n-2] 1、(1)初始条件y(-1)=1,y(-2)=2，输入x(n)=u(n)，计算系统的零输入响应 N=11; n=0:N-1; x=[n=0]; b=[0.0675,0.1349,0.0675];a=[1,-1.143,0.4128]; zi=filtic(b,a,[1,2]); y=filter(b,a,zi); stem(y) (2)当下面三个信号分别通过系统，分别计算系统的响应： 1.输入信号x1[n]=cos((pi/10)*n)*u[n] N=11; n=0:N-1 x1=cos((pi/10)*n) b=[0.0675,0.1349,0.0675];a=[1,-1.143,0.4128]; zi=filtic(b,a,[1,2]); y=filter(b,a,x1) stem(n,y) 2. 输入信号:x2[n]=cos((pi/5)*n)*u[n] N=11; n=0:N-1 x2=cos((pi/5)*n) b=[0.0675,0.1349,0.0675];a=[1,-1.143,0.4128]; zi=filtic(b,a,[1,2]); y=filter(b,a,x2,zi) stem(n,y) 3.输入信号:x3[n]=cos((7*pi/10)*n)*u[n] N=11; n=0:N-1 x3=cos((7*pi/10)*n) b=[0.0675,0.1349,0.0675];a=[1,-1.143,0.4128]; zi=filtic(b,a,[1,2]); y=filter(b,a,x3,zi) stem(n,y) （3）系统特性分析 b=[0.0675,0.1349,0.0675];a=[1,-1.143,0.4128]; z=roots(b) p=roots(a) zplane(b,a) 此系统为因果稳定系统 2.(1)计算系统的单位脉冲响应 N=11; b=[0.03571,0.1428,0.2143,0.1428,0.03571];a=[1,-1.035,0.8264,-0.2605,0.04033]; x=[1,zeros(1,N-1)]; y=filter(b,a,x) stem(n,y) (2)当信号通过系统：x[n]=1+2cos(pi/4*n)+cos(pi/2*n)*u[n]，计算系统的响应。 N=11; n=0:N-1 b=[0.03571,0.1428,0.2143,0.1428,0.03571];a=[1,-1.035,0.8264,-0.2605,0.04033]; x=[1,zeros(1,N-1)]; h=filter(b,a,x) %plot(h) x=1+cos(pi/4*n)+cos(pi/2*n) y=filter(b,a,x) stem(n,y) 3. 根据零极点分布对系统频率特性的影响试设计一个单极点单零点、因果实系数滤波器，满足下列指标，并画出其幅频特性曲线： 单极点单零点低通滤波器: %低通滤波器 c=1.1119 y=[-1]; x=[c]; k=(c-1)/2; b=k*poly(y); a=poly(x); w=linspace(0,pi,1000); h=freqz(b,a,w); plot(w,abs(h)) (2)%高通滤波器 c=1.1119 y=[1]; x=[c]; k=(c-1)/2; b=k*poly(y); a=poly(x); w=linspace(0,pi,512); h=freqz(b,a,w); plot(w,abs(h)); (3) %带阻滤波器 b=1.031*[1,0,1]; a=[1,0,1.0621]; zplane(b,a); [H,w]=freqz(b,a,250); plot(w/pi,abs(H));