remove noise from measured data
6 次查看(过去 30 天)
显示 更早的评论
Hi All,
I am reading this page from Matlab, and I have the following question from it. Can you please explain? It says: "The filter is designed to allow all frequency components with a period less than 48 hours and block all components higher than that." So my question is: based on what it says, how did it come up with the following values?
% Filter Order
N = 3;
% Passband frequency
Fpass = 1/(24*60*60);
% Stopband frequency
Fstop = 1/(6*60*60);
% Ripple Factor and Attenuation in stop band
Rp = 0.5;
Astop = 50;
Thanks
3 个评论
John BG
2017-6-29
1/(24*60*60) is the slowest fastest signal that the filter is going to let in, or should let in, without significant attenuation. 1/(6*60*60) is a 6 hours cycle signal that the filter will block. 6 hours cycle is the frequency (f=1/T) where the 1st notch of the filter is going to be placed to attenuate faster signals.
The question reads 'let through frequencies lower than 1/2days' the 48 hours, but the pass / stop frequencies chosen are not that stringent.
回答(0 个)
社区
更多回答在 ThingSpeak Community
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Filter Analysis 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!