matlab及在電子信息類課程中的應(yīng)用習(xí)題答案電子科大第二版-資料下載頁
2025-06-19 14:18本頁面
【正文】 freqz(b,a)dimpulse(b,a,100)wp=100。ws=200。rp=2。rs=15。Fs=500。wp=wp/(Fs/2)。ws=ws/(Fs/2)。[N,wc]=buttord(wp,ws,rp,rs)。[b,a]=butter(N,wc,39。z39。)freqz(b,a)1b=fir1(48,[ ])。a=1。freqz(b,a)1b=fir1(37,)。a=1。freqz(b,a)1低通：F=[0:1/56:1]。A=[ones(1,50),zeros(1,5649)]。b=fir2(56,F,A)。a=1。freqz(b,a)帶通：F=[0:1/56:1]。A=[ones(1,25),zeros(1,25),ones(1,7)]。b=fir2(56,F,A)。a=1。freqz(b,a)高通F=[0:1/56:1]。A=[zeros(1,50),ones(1,5649)]。b=fir2(56,F,A)。a=1。freqz(b,a)%濾波器設(shè)計（橢圓濾波器）wp1=300。wp2=700。ws1=301。ws2=699。rp=。rs=50。Fs=2000。wp=[wp1]/(Fs/2)。ws=[ws1]/(Fs/2)。[N,wc]=ellipord(wp,ws,rp,rs,39。z39。)[b,a]=ellip(N,rp,rs,wc,39。high39。) %濾波器特性分析 freqz(b,a)%信號采樣及時頻變換Fs=2000。t=0:1/Fs:10。yt=sin(2*pi*200*t)+sin(2*pi*500*t)+sin(2*pi*800*t)。figure(2)。subplot(3,1,1)。plot(t,yt)figure(2)。subplot(3,1,2)。plot(t,yt)。axis([0,5,5])yf=abs(fft(yt))。f=(1:length(yf))/length(yf)*Fs。figure(2)。subplot(3,1,3)。plot(f,yf)%信號濾波及時頻分析yt1=filter(b,a,yt)。figure(3)。subplot(3,1,1)。plot(t,yt1)figure(3)。subplot(3,1,2)。plot(t,yt1)。axis([0,5,5])yf1=abs(fft(yt1))。f=(1:length(yf1))/length(yf1)*Fs。figure(3)。subplot(3,1,3)。plot(f,yf1)
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