数字信号处理第二章实验报告.docx

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1、实 验 报 告 课 程： 数字信号处理 专业班级： 学生姓名： 学 号： 年 月 日2.1对M=2，运行上述程序，生成输入xn=s1n+s2n的输出信号。输入xn的哪个分量被该离散时间系统抑制？% 程序 P2_1% 一个M点滑动平均滤波器的仿真% 产生输入信号n = 0:100;s1 = cos(2*pi*0.05*n); % 一个低频正弦s2 = cos(2*pi*0.47*n); % 一个高频正弦x = s1+s2;% M点滑动平均滤波器的实现M = input(滤波器所需的长度 = );num = ones(1,M);y = filter(num,1,x)/M;clf;subplot(2

2、,2,1);plot(n, s1);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(低频正弦);subplot(2,2,2);plot(n, s2);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(高频正弦);subplot(2,2,3);plot(n, x);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(输入信号);subplot(2,2,4);plot(n, y);axis(0, 100, -2, 2);xlabel(时间

3、序号n); ylabel(振幅);title(输出信号); axis;图形显示如下：答：输入部分的高频成分成分被抑制了。2.3对滤波器长度M和正弦信号s1n和s2n的频率取其他值，运行程序P2.1,算出结果。n = 0:100;s1=cos(2*pi*0.02*n);s2=cos(2*pi*0.46*n);x = s1+s2;% M点滑动平均滤波器的实现M = input(滤波器所需的长度 = );num = ones(1,M);y = filter(num,1,x)/M;clf;figure,subplot(2,2,1);plot(n, s1);axis(0, 100, -2, 2);xla

4、bel(时间序号n); ylabel(振幅);title(低频正弦);subplot(2,2,2);plot(n, s2);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(高频正弦);subplot(2,2,3);plot(n, x);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(输入信号);subplot(2,2,4);plot(n, y);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(输出信号); axis;num =1

5、,-ones(1,M-1);y = filter(num,1,x)/M;figure,subplot(2,2,1);plot(n, s1);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(低频正弦);subplot(2,2,2);plot(n, s2);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(高频正弦);subplot(2,2,3);plot(n, x);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(输入信号);sub

6、plot(2,2,4);plot(n, y);axis(0, 100, -2, 2);xlabel(时间序号n); ylabel(振幅);title(输出信号); axis;图形显示如下：答：运行结果如下图，可以看出输出信号保留了输入信号xn的高频分量，即保留了s2n分量，低频部分s1n被抑制了。2.5用不同频率的正弦信号作为输入信号，计算每个输入信号的输出信号。输出信号是如何受到输入信号频率的影响的？从数学上对你的结论加以证明。% 程序P2_2% 产生一个正弦输入信号clf;n = 0:200;f=input(Please input the value of f:)x = cos(2*pi

7、*f*n);% 计算输出信号x1 = x 0 0; % x1n = xn+1 x2 = 0 x 0; % x2n = xnx3 = 0 0 x; % x3n = xn-1y = x2.*x2-x1.*x3;y = y(2:202);% 画出输入和输出信号subplot(2,1,1)plot(n, x)xlabel(时间序列n);ylabel(振幅);title(输入信号)subplot(2,1,2)plot(n,y)xlabel(时间信号n);ylabel(振幅);title(输出信号);分别取F=0.05，F=0.47，F=0.5以及F=0，仿真结果如下所示： 证明：设输入信号为，则以及则答

8、：从图形中可以看出，输入频率越大，输出信号值越小。最后都逐渐趋于0。2.7运行程序P2.3，对由加权输入得到的yn在与相同权系数下输出y1n和y2n相加得到的ytn进行比较，这两个序列是否相等？该系统是线性系统么？clf;n = 0:40;a = 2;b = -3;x1 = cos(2*pi*0.1*n);x2 = cos(2*pi*0.4*n);x = a*x1 + b*x2;num=2.2403 2.4908 2.2403;den = 1 -0.4 0.75;ic = 0 0;y1 = filter(num,den,x1,ic);y2 = filter(num,den,x2,ic); y

9、= filter(num,den,x,ic);yt = a*y1 + b*y2; d = y - yt; subplot(3,1,1)stem(n,y);ylabel(振幅);title(加权输入: a cdot x_1n + b cdot x_2n的输出);subplot(3,1,2)stem(n,yt);ylabel(这幅);title(加权输出: a cdot y_1n + b cdot y_2n);subplot(3,1,3)stem(n,d);xlabel(时间序号 n);ylabel(振幅);title(差信号);图形显示如下：答：该仿真结果说明这两个序列相等，该系统是线性系统。2

10、.9当初始条件非零时重做习题Q2.7。令ic = 10 20；则仿真结论如下所示：答：该仿真结果说明这两个序列不相等，该系统不是线性系统。2.11假定另一个系统为yn=xnxn-1,修改程序P2.3，计算这个系统的输出序列y1n,y2n和yn。比较yn和ytn。这两个序列是否相等？该系统是线性系统吗？n=0:40;a=2;b=-3;x11=0 cos(2*pi*0.1*n) 0;x12=0 0 cos(2*pi*0.1*n);x21=0 cos(2*pi*0.4*n) 0; x22=0 0 cos(2*pi*0.4*n);y1=x11.*x12;y2=x21.*x22;yt=a*y1+b*y2

11、;y=(a*x11+b*x21).*(a*x12+b*x22);d=y-yt;subplot(3,1,1)stem(0 n 0,y);ylabel(振幅);title(加权输入: a cdot x_1n + b cdot x_2n的输出);subplot(3,1,2)stem(0 n 0,yt);ylabel(这幅);title(加权输出: a cdot y_1n + b cdot y_2n);subplot(3,1,3)stem(0 n 0,d);xlabel(时间序号 n);ylabel(振幅);title(差信号);图形显示如下：答：这两个序列不相等，该系统不是线性系统。2.13采用三个

12、不同的延时变量D的值重做习题Q2.12。D=2；D=6;D=12;显示图形如下：答：该系统是时不变系统，满足yn-D=ydn。2.15在非零的初始条件下重做习题Q2.12，该系统是时不变系统吗？clf;n = 0:40; D = 10;a = 3.0;b = -2;x = a*cos(2*pi*0.1*n) + b*cos(2*pi*0.4*n);xd = zeros(1,D) x;num = 2.2403 2.4908 2.2403;den = 1 -0.4 0.75;ic = 5 10; y = filter(num,den,x,ic);yd = filter(num,den,xd,ic)

13、;d = y - yd(1+D:41+D);subplot(3,1,1)stem(n,y);ylabel(); title( yn); grid;subplot(3,1,2)stem(n,yd(1:41);ylabel();title( xn-10); grid;subplot(3,1,3)stem(n,d);xlabel( n); ylabel();title();grid;图形显示如下：答：该仿真结果说明该系统是时变系统。2.17考虑另一个系统：yn=nxn+xn-1,修改程序P2.4，以仿真上面的系统并确定该系统是否为时不变系统。clf;n = 0:40; D = 10;a = 3.0;

14、b = -2;x = a*cos(2*pi*0.1*n) + b*cos(2*pi*0.4*n);xd = zeros(1,D) x;nd=0:length(xd)-1;y=(n.*x)+0 x(1:40);yd=(nd.*xd)+0 xd(1:length(xd)-1);d = y - yd(1+D:41+D);subplot(3,1,1)stem(n,y);ylabel(振幅); title(输出 yn); grid;subplot(3,1,2)stem(n,yd(1:41);ylabel(振幅);title(由于延时输入 xn-10的输出); grid;subplot(3,1,3)ste

15、m(n,d);xlabel(时间序号 n); ylabel(振幅);title(差值信号);grid;图形显示如下：答：从仿真结果看，该系统是时变系统。2.19运行程序P2.5，生成式（2.15）所给的离散系统的冲激响应。clf;N=40;num=2.2403 2.4908 2.2403;den=1 -0.4 0.75;y=impz(num,den,N);%画出冲激相应stem(y);xlabel(时间序号n);ylabel(振幅); title(冲激响应);grid;图形显示如下：2.21利用filter命令编写一个MATLB程序，生成式（2.17）给出的因果线性时不变系统的冲激响应，计算并

16、画出前40个的样本。把你的结果和习题Q2.20中得到的结果相比较。clf;N = 40;num = 0.9 -0.45 0.35 0.002;den = 1.0 0.71 -0.46 -0.62;% input: unit pulsex = 1 zeros(1,N-1);% outputy = filter(num,den,x);% Plot the impulse response% NOTE: the time axis will be WRONG; h0 will% be plotted at n=1; but this will agree with% the INCORRECT pl

17、otting that was also done% by program P2_5.stem(y);xlabel(Time index n); ylabel(Amplitude);title(Impulse Response); grid;图形显示如下：答：两题结果一样。2.23运行程序P2.6，计算输出序列yn和y2n以及差值信号dn。yn和y2n相等吗？% Program P2_6% Cascade Realizationclf;x = 1 zeros(1,40); % Generate the inputn = 0:40;% Coefficients of 4th order syst

18、emden = 1 1.6 2.28 1.325 0.68;num = 0.06 -0.19 0.27 -0.26 0.12;% Compute the output of 4th order systemy = filter(num,den,x);% Coefficients of the two 2nd order systemsnum1 = 0.3 -0.2 0.4;den1 = 1 0.9 0.8;num2 = 0.2 -0.5 0.3;den2 = 1 0.7 0.85;% Output y1n of the first stage in the cascadey1 = filter

19、(num1,den1,x);% Output y2n of the second stage in the cascadey2 = filter(num2,den2,y1);% Difference between yn and y2nd = y - y2;% Plot output and difference signalssubplot(3,1,1);stem(n,y);ylabel(Amplitude);title(Output of 4th order Realization); grid;subplot(3,1,2);stem(n,y2)ylabel(Amplitude);titl

20、e(Output of Cascade Realization); grid;subplot(3,1,3);stem(n,d)xlabel(Time index n);ylabel(Amplitude);title(Difference Signal); grid;图形显示如下：答：y(n)和y2(n)相等。2.25用任意的非零初始向量ic，ic1和ic2来重做习题Q2.23。clf;x=sin(2*pi*0.2*n); % Generate the inputn = 0:40;% Coefficients of 4th order systemden = 1 1.6 2.28 1.325 0

21、.68;num = 0.06 -0.19 0.27 -0.26 0.12;xi=1 2 3 4;% Compute the output of 4th order systemy = filter(num,den,x,xi);% Coefficients of the two 2nd order systemsnum1 = 0.3 -0.2 0.4;den1 = 1 0.9 0.8;num2 = 0.2 -0.5 0.3;den2 = 1 0.7 0.85;xi1=1 2;% Output y1n of the first stage in the cascadey1 = filter(num

22、1,den1,x,xi1);xi2=3 4;% Output y2n of the second stage in the cascadey2 = filter(num2,den2,y1,xi2);% Difference between yn and y2nd = y - y2;% Plot output and difference signalssubplot(3,1,1);stem(n,y);ylabel(Amplitude);title(Output of 4th order Realization); grid;subplot(3,1,2);stem(n,y2)ylabel(Amp

23、litude);title(Output of Cascade Realization); grid;subplot(3,1,3);stem(n,d)xlabel(Time index n);ylabel(Amplitude);title(Difference Signal); grid;图形显示如下：答：y(n)和y2(n)不相等。2.27用任意非零初始向量ic，ic1和ic2来重做习题Q2.26。% Program P2.27% Cascade Realizationclf;x = 1 zeros(1,40); % Generate the inputn = 0:40;% Coeffici

24、ents of 4th order systemden = 1 1.6 2.28 1.325 0.68;num = 0.06 -0.19 0.27 -0.26 0.12;ic=4 10 2 12% Compute the output of 4th order systemy = filter(num,den,x,ic);% Coefficients of the two 2nd order systemsnum1 = 0.3 -0.2 0.4;den1 = 1 0.9 0.8;num2 = 0.2 -0.5 0.3;den2 = 1 0.7 0.85;% Output y1n of the

25、first stage in the cascadey1 = filter(num2,den2,x,ic(1:2);% Output y2n of the second stage in the cascadey2 = filter(num1,den1,y1,ic(3:4);% Difference between yn and y2nd = y - y2;% Plot output and difference signalssubplot(3,1,1);stem(n,y);ylabel(Amplitude);title(Output of 4th order Realization); g

26、rid;subplot(3,1,2);stem(n,y2)ylabel(Amplitude);title(Output of Cascade Realization); grid;subplot(3,1,3);stem(n,d)xlabel(Time index n);ylabel(Amplitude);title(Difference Signal); grid;图形显示如下：答：由结果表明，两个输出结果有没差别。有差别证明级联的顺序不可以互换，这里是一个验证程序。2.29修改程序P2.7，计算长度为15的序列hn和长度为10的序列xn的卷积，重做问题Q2.28。hn和xn的样本值你自己给定

27、。% Program P2.29clf;h = 3 2 1 -2 1 0 -4 0 3 1 5 4 0 3 5; % impulse responsex = 1 -2 3 -4 3 2 1 5 6 1; % input sequencey = conv(h,x);n = 0:23;subplot(2,1,1);stem(n,y);xlabel(Time index n); ylabel(Amplitude);title(Output Obtained by Convolution); grid;x1 = x zeros(1,14);y1 = filter(h,1,x1);subplot(2,1

28、,2);stem(n,y1);xlabel(Time index n); ylabel(Amplitude);title(Output Generated by Filtering); grid;图形显示如下：答：xn后面补零数应为x(n)和hn序列的长度之和减一，为14.2.31使用命令break的目的是什么？答：使用命令break是使当在k未到最后一个数值是此时的值已经小于时，跳出for循环。2.33考虑用差分方程描述的离散时间系统。修改程序P2.8，计算并画出上述系统的冲激响应。该系统稳定吗？% values of the impulse response samplesclf;num

29、= 1 -4 3; den = 1 -1.7 1;N = 200;h = impz(num,den,N+1);parsum = 0;for k = 1:N+1; parsum = parsum + abs(h(k); if abs(h(k) 10(-6), break, endend% Plot the impulse responsen = 0:N;stem(n,h)xlabel(Time index n); ylabel(Amplitude);% Print the value of abs(h(k) disp(Value =);disp(abs(h(k);图形显示如下：答：该系统不稳定。

30、2.35修改程序P2.9，将输入序列改变成扫频正弦序列（长度为301、最低频率为0、最高频率为0.5）。那个滤波器能更好的抑制输入信号xn的高频分量？% Program P2_9% Generate the input sequenceclf;n = 0:300;a=pi/600;arg=a*n.*n;x=cos(arg);%x1 = cos(2*pi*10*n/256);%x2 = cos(2*pi*100*n/256);%x = x1+x2;% Compute the output sequencesnum1 = 0.5 0.27 0.77;y1 = filter(num1,1,x); %

31、 Output of System #1den2 = 1 -0.53 0.46;num2 = 0.45 0.5 0.45;y2 = filter(num2,den2,x); % Output of System #2% Plot the output sequencessubplot(2,1,1);plot(n,y1);axis(0 300 -2 2);ylabel(Amplitude);title(Output of System #1); grid;subplot(2,1,2);plot(n,y2);axis(0 300 -2 2);xlabel(Time index n); ylabel(Amplitude);title(Output of System #2); grid;图形显示如下：答：由输出结果图形可以看出：y2滤波器能更好地抑制输入信号xn的高频分量。