数字信号处理第二章上机作业.pdf

第二章上机作业

1、ljdt(A,B)函数定义

functionljdt(A,B)

p=roots(A);

q=roots(B);

p=p;

q=q;

x=max(abs([pq1]));

x=x+0.1;

y=x;

clf

holdon

axis([-xx-yy])

w=0:pi/300:2*pi;

t=exp(i*w);

plot(t)

axis(square)

plot([-xx],[00])

plot([00],[-yy])

text(0.1,x,jIm[z])

text(y,1/10,Re[z])

plot(real(p),imag(p),x)

plot(ral(q),imag(q),o)

title(pole-zerodiagramfordiscretesystem)

holdoff

例2.26

a=[3-10001];

b=[11];

ljdt(a,b)

p=roots(a)

q=roots(b)

pa=abs(p)

程序运行结果如下：

P=

0.7255+0.4633i

0.7255+0.4633i

-0.1861+0.7541i

-0.1861-0.7541i

-0.7455

q=

-1

pa=

0.8608

0.8608

0.7768

0.7768

0.7455

例2.27

b=[0121];a=[1-0.5-0.0050.3];

subplot311

zplane(b,a);xlabel(实部);ylabel(虚部);

num=[0121];den=[1-0.5-0.0050.3];

h=impz(num,den);

subplot312

stem(h);xlabel(k);title(单位脉冲响应);

[H,w]=freqz(num,den);

subplot313

plot(w/pi,abs(H));

xlabel(频率频率

title(频率响应)

例2.28

a=[1,-1];

b=[1];

subplot321

impz(b,a);

a1=[1,-0.8];

b1=[1];

subplot322

impz(b1,a1,10);

a2=[1,-2];

b2=[1];

subplot323

impz(b2,a2,10);

a3=[1,-2*0.8*cos(pi/4),0.8^2];

b3=[1];

subplot324

impz(b3,a3,20);

a4=[1,-2*0.8*cos(pi/8),1];

b4=[1];

subplot325

impz(b4,a4,20);

a5=[1,-2*1.2*cos(pi/4),1.2^2];

b5=[1];

subplot326

impz(b5,a5,20);

例2.29

b=[1,0,-1];a=[1,0,-0.81];

figure(1)

subplot(2,1,1);dimpulse(b,a,50);ylabel(h(n));

subplot(2,1,2);dstep(b,a,50);ylabel(g(n));

figure(2)

w=[0:1:500]*pi/500;

freqz(b,a,w)

例2.30

b=[1,0,0,0,0,0,0,0,-1];

a0=1;

a1=[1,0,0,0,0,0,0,0,-(0.8)^8];

a2=[1,0,0,0,0,0,0,0,-(0.9)^8];

a3=[1,0,0,