作业帮下面是CIC滤波器频率响应图的M代码.特别是那几个函数的运用搞清楚D=5; r=D;fs=1e5;\x05S1_cic=ones(1,D);[h1,f1]=freqz(ones(1,D),1,1000,fs);hold on;plot(f1/(fs/2),20*log10(abs(h1))-max(20*log10(abs(h1)))
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/28 23:57:40
![下面是CIC滤波器频率响应图的M代码.特别是那几个函数的运用搞清楚D=5; r=D;fs=1e5;\x05S1_cic=ones(1,D);[h1,f1]=freqz(ones(1,D),1,1000,fs);hold on;plot(f1/(fs/2),20*log10(abs(h1))-max(20*log10(abs(h1)))](/uploads/image/z/12274632-0-2.jpg?t=%E4%B8%8B%E9%9D%A2%E6%98%AFCIC%E6%BB%A4%E6%B3%A2%E5%99%A8%E9%A2%91%E7%8E%87%E5%93%8D%E5%BA%94%E5%9B%BE%E7%9A%84M%E4%BB%A3%E7%A0%81.%E7%89%B9%E5%88%AB%E6%98%AF%E9%82%A3%E5%87%A0%E4%B8%AA%E5%87%BD%E6%95%B0%E7%9A%84%E8%BF%90%E7%94%A8%E6%90%9E%E6%B8%85%E6%A5%9AD%3D5%3B+r%3DD%3Bfs%3D1e5%3B%5Cx05S1_cic%3Dones%281%2CD%29%3B%5Bh1%2Cf1%5D%3Dfreqz%28ones%281%2CD%29%2C1%2C1000%2Cfs%29%3Bhold+on%3Bplot%28f1%2F%28fs%2F2%29%2C20%2Alog10%28abs%28h1%29%29-max%2820%2Alog10%28abs%28h1%29%29%29)
下面是CIC滤波器频率响应图的M代码.特别是那几个函数的运用搞清楚D=5; r=D;fs=1e5;\x05S1_cic=ones(1,D);[h1,f1]=freqz(ones(1,D),1,1000,fs);hold on;plot(f1/(fs/2),20*log10(abs(h1))-max(20*log10(abs(h1)))
下面是CIC滤波器频率响应图的M代码.特别是那几个函数的运用搞清楚
D=5;
r=D;
fs=1e5;\x05
S1_cic=ones(1,D);
[h1,f1]=freqz(ones(1,D),1,1000,fs);hold on;
plot(f1/(fs/2),20*log10(abs(h1))-max(20*log10(abs(h1)))
下面是CIC滤波器频率响应图的M代码.特别是那几个函数的运用搞清楚D=5; r=D;fs=1e5;\x05S1_cic=ones(1,D);[h1,f1]=freqz(ones(1,D),1,1000,fs);hold on;plot(f1/(fs/2),20*log10(abs(h1))-max(20*log10(abs(h1)))
这段代码是级联积分梳状滤波器中梳状滤波器的部分.
[h1,f1]=freqz(ones(1,D),1,1000,fs)在matlab help中查freqz有详细解释及举例.该句用于分析长度为5的FIR滤波器的频率响应,返回值h1为频率响应,f1为频率轴,已经是以Hz为单位了(0~fs/2)
因此下句也就用错了
plot(f1/(fs/2),20*log10(abs(h1))-max(20*log10(abs(h1)))应改为
plot(f1,20*log10(abs(h1))-max(20*log10(abs(h1)))
此时画出的是幅频图,幅度进行了归一化,幅度单位是dB.