实验四IIR数字滤波器设计及软件实现实验报告.docx

信号不相等呢？谁能解释一下？谢谢 程序如下：

clearelf

% (1)构造原始信号Fs二10000; T二1/Fs; %先设定采样频率

t=0:T:0. 1;n=length(t);s=cos (2*pi*250*t). *cos(2*pi*25*t)+cos(2*pi*500*t).

*cos(2*pi*50*t) +... cos (2*pi*1000*t)・ *cos(2*pi*100*t);

subplot (2, 1, 1), plot (t, s), axis ([0 0. 08

xlabel (' t/s'), ylabel (' s (t)')% (2)画出s(t)的频谱

[N, WC]= ellipord(Wp , Ws , Ap , As ,'s');[bLPs , aLPs]=ellip(N , Ap, As, Wc,' s');

[H, w]=freqs (bLPs, aLPs);db =20*logl0(abs(H)):

figure, subplot (2, 1, 1), plot (w/2/pi, db);axis ([0 1600 -80 5]),grid

title ('模拟低通滤波器的幅度特性');xxxxxx

yyyyyy (

[bLPz, aLPz] = bilinear (bLPs , aLPs , Fs);w 二 linspace (0, pi, 1000);

h = freqz(bLPz, aLPz , w);subplot (2,1, 2), plot (w*Fs/2/pi, 20*logl0(abs(h)));

axis ([0 1600 -80 5]),grid titlef数字低通滤波器的幅度特性');xxxxxx ('f (Hz)');

yyyyyyCdB')；

xxxxxx (' f (Hz)');yyyyyy (' dB')

%%3. 2b将模拟带通滤波器转换为数字带通滤波器[bBPz,aBPz] = bilinear (bBPs ,

aBPs , Fs);

w = linspace (0, pi, 1000);h = freqz (bBPz, aBPz , w); subplot (2, 1, 2), plot(w*Fs/2/pi,

20*logl0(abs(h)));

axis([0 1600 -80 5]),grid titleC数字低通滤波器的幅度特性')； xxxxxx ('f

(Hz)');yyyyyy ('dB')； %%3. 3a设计模拟高通滤波器

fp=800;fs=700;Ap=0. 1;As=60; wp二

2*pi*fp/Fs;ws=2*pi*fs/Fs;Wp=(2*Fs)*tan(wp/2);Ws=(2*Fs)*tan(ws/2);

[N, Wc]= ellipord (Wp , Ws , Ap , As ,'s');[bHPs , aHPs]=ellip (N, Ap, As, Wc ,'

%%3. 3b将模拟高通滤波器转换为数字高通滤波器[bHPz, aHPz] = bilinear(bHPs ,

aHPs , Fs);

w = linspace(0, pi, 1000);h = freqz (bHPz, aHPz , w);

subplot (2, 1, 2), plot(w*Fs/2/pi, 20*logl0(abs(h))); axis([0 1600 -80 5]),grid title

('数字高通滤波器的幅度特性')；xxxxxx ('f (Hz)'); yyyyyy('dB')；

% (4)用得到的滤波器进行滤波，分离出三路信号，观察时域波形和幅频特 性％%4. 1

滤波si = filter(bLPz, aLPz, s); s2 二 filter(bBPz, aBPz, s); s3 = filter

(bHPz, aHPz, s);

%%4. 2时域波形figure, subplot (3, 1, 1), plot (t, si) xlabel (' t/s' ), ylabel (' si

(t)') subplot (3, 1, 2), plot (t, s2) xlabel (' t/s'), ylabel (' s2 (t)') subplot (3,

abs(ft3),' Marker

始信号ss=sl+s2+s3;%合成信号记作ss isequal (s, ss)%判断合成信号是否和原信号相等

，结果不相等，why?

[本帖最后由cwjy于2010-5-318:19编辑]