显示 更早的评论
how that can be happen when i increase filter order the message shape change to worst
code:
function [op]=amm(m,fc,fm)
%am modulation
for i=1:3
t=0:.0001:1;
c(i,:)=2*cos(2*pi*fc(1,i).*t);
tx(i,:)=m(i,:).*c(i,:);
end
%tansmitte signal
rx=tx.*c;
fs=10000;
fnyq=fs/2;
for i=1:3
wn(1,i)=fm(1,i)/fnyq;
[b a]=butter(1,wn(1,i));
op(i,:)=filter(b,a,rx(i,:));
end
figure;
subplot 311;plot(t,op(1,:));
subplot 312;plot(t,op(2,:));
subplot 313;plot(t,op(3,:));
return
3 个评论
Walter Roberson
2012-1-7
Please keep in mind that there are signals for which any _finite_ order fitting will have error that grows without bounds.
For example, no polynomial of finite order can ever fit any (non-zero) periodic signal such as sine.
Jan
2012-1-7
@mohamed al-asklany: If you give an example for the inputs m, fc and fm, we could reproduce, what you are talking about.
Wayne King
2012-1-8
@mohamed al-asklany, I agree with Jan's comment. Also, you have your order fixed at 1 in your call to butter(). How can you obtain a useable filter with that order? You are not changing your filter order based on any input with this code.
回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Fourier Analysis and Filtering 的更多信息
2012-1-7
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
