信号与系统上机实验Q10<=n<=31x1(n)=sin(pi*n/4)*cos(pi*n/4);x2(n)=cos(pi*n/4)*cos(pi*n/4)x3(n)=sin(pi*n/4)*cos(pi*n/8)分别画出图形,求出其周期。（1）x1(n)=sin(pi*n/4)*cos(pi*n/4);程序如下：n=0:31x1=sin(pi*n/4).*cos(pi*n/4)stem(n,x1)MATLAB画出图形如下：0.60.40.20-0.2-0.4-0.6-0.835302550101520由上图图可知周期T=4（2）x2(n)=cos(pi*n/4)*cos(pi*n/4)程序如下：n=0:31x2=cos(pi*n/4).*cos(pi*n/4)stem(n,x2)MATLAB画出图形如下：10.0.0.0.0.50.40.30.20.1035253020051015T=4由上图可知周期x3(n)=sin(pi*n/4)*cos(pi*n/8)3（）程序如下：n=0:31x3=sin(pi*n/4).*cos(pi*n/8)stem(n,x3)MATLAB画出图形如下：0.80.0.0.-0.-0.4-0.6-0.835253050101520由上图可知周期T=16Q2当0<=n<=5时，h(n)=n;其他h(n)=0;x(n)=h(n);求y(n)=x(n)*h(n);用stem函数画出y(n).程序如下：n=0:5;y=n.^2;stem(y)画出图形如下：MATLAB．252115065.54.552.533.54211.5Q3:LTI系统，描述差分方程y(n)-0.8y(n-1)=2x(n)-x(n-2)表征的因果a1（a）.定义用向量和b1个等份频率上频率响4pi之间是在freqz定义H10和）中的系数向量，利用(b).用在（aomega1应的值，是这些频率值。个等份频率上频率42*pi之间和定义用在（(c).a）中的系数向量，利用freqzH2是在0omega2是这些频率值。响应的值，程序如下：n=4a1=[5,0,-4]b1=[10,0,-5][H1,W1]=freqz(b1,a1,n)[H2,W2]=freqz(b1,a1,n,'whole')输出结果如下：n=4a1=50-4b1=100-5H1=5.0000+0.0000i1.7073-0.3659i1.6667+0.0000i1.7073+0.3659iW1=00.78541.57082.3562H2=5.00001.66675.00001.6667W2=01.57083.14164.7124Q4X1(n)=u(n)-u(n-8);其周期N1=8,X2(n)=u(n)-u(n-8);其周期N2=16,X3(n)=u(n)-u(n-8);其周期N3=32,(1)画出这些周期信号在0<=n<=63的图形程序如下：N=64;n=0:63;x1=zeros(1,64);x2=zeros(1,64);x3=zeros(1,64);flag1=0;flag2=0;i=1:64forx1(i)=1;endi=1:64for((i>=1i<=8)||(i>=17i<=24)||(i>=33i<=40)||(i>=49i<=56))ifx2(i)=1;elsex2(i)=0;endendi=1:64for((i>=1i<=8)||(i>=33i<=40))ifx3(i)=1;elsex3(i)=0;endendstem(n,x1);figure;stem(n,x2);figure;stem(n,x3);画出图形分别如下：MATLAB用．10.0.0.0.0.0.0.0.0.1070506010203040010.90.80.70.60.50.40.30.20.1001020304050607010.0.0.0.0.0.0.0.0.10706030405010020并画图。分别为a1,a2,a3)(2)求其对应的付氏级数，(程序如下：N=64;n=0:63;x1=zeros(1,64);x2=zeros(1,64);x3=zeros(1,64);flag1=0;flag2=0;i=1:64forx1(i)=1;endi=1:64for((i>=1i<=8)||(i>=17i<=24)||(i>=33i<=40)||(i>=49i<=56))ifx2(i)=1;elsex2(i)=0;endendi=1:64for((i>=1i<=8)||(i>=33i<=40))ifx3(i)=1;elsex3(i)=0;endenda1=fft(x1)figure;stem(a1)figure;a2=fft(x2)stem(a2)figure;a3=fft(x3)stem(a3)画出波形如下：MATLAB．7060504030201007050601020304002520151050-5-10-15-20-25706050200103040151050-5-10-15010203040506070(3)综合x3a.x31(n)=a3(1)*exp(j*2*pi*n/32)+a3(2)*exp(j*2*2*pi*n/32)+a3(30)*exp(j*2*30*pi*n/32)+a3(31)*exp(j*2*31*pi*n/32)+a3(32)*exp(j*2*32*pi*n/32)b.x32(n)=a3(1)*exp(j*2*pi*n/32)+……+a3(8)*exp(j*2*8*pi*n/32)+a3(24)*exp(j*2*24*pi*n/32)+……+a3(32)*exp(j*2*32*pi*n/32)c..x33(n)=a3(1)*exp(j*2*pi*n/32)+……+a3(12)*exp(j*2*12*pi*n/32)+a3(20)*exp(j*2*20*pi*n/32)+……+a3(32)*exp(j*2*32*pi*n/32)c..x34(n)=a3(1)*exp(j*2*pi*n/32)+……+a3(32)*exp(j*2*32*pi*n/32)yi=abs(x3i),用stem对yi作图，试比较其于x3的区别。程序如下：x31=a3(1)*exp(1i*2*pi*n/32)+a3(2)*exp(1i*2*2*pi*n/32)+a3(30)*exp(1i*2*30*pi*n/32)+a3(31)*exp(1i*2*31*pi*n/32)+a3(32)*exp(1i*2*32*pi*n/32)stem(abs(x31))figure;x32=sum(a3(1:8))*exp(1i*2*8*pi*n/32)+sum(a3(24:32))*exp(1i*2*32*pi*n/32)．stem(abs(x32))figure;x33=sum(a3(1:12))*exp(1i*2*12*pi*n/32)+sum(a3(20:32))*exp(1i*2*32*pi*n/32)stem(abs(x33))x34=a3(1)*exp(1i*2*pi*n/32)+a3(32)*exp(1i*2*32*pi*n/32)figure;stem(abs(x3...