How do you calculate this transfer function?
5 ビュー (過去 30 日間)
古いコメントを表示
Dear friends I'm trying to make a filter for sound processing in matlab The transfer function is
H(e^(jω))=αjω.*exp(-βjω);
I have written this in matlab but seems not to work properly
w=0:pi/(2*Fs):pi; H=sqrt(-1)*w*a.*exp(-sqrt(-1)*w*b);
Do you think it's right?
0 件のコメント
回答 (3 件)
Honglei Chen
2014 年 9 月 12 日
You can use 1i for sqrt(-1) but mainly you need to set your w correctly, right now your step size is pi/(2*Fs). It could work but I don't know if that's what you want. Normally people decides the number of samples between 0 and pi as N and then the step size is pi/N, or pi/(N-1).
0 件のコメント
mohammad
2014 年 9 月 12 日
編集済み: mohammad
2014 年 9 月 12 日
first use approximation function instead of exponential: e^x = (1+(x/N))^N second instead of 'jω' use 's' and use 'tf' command. so you have (for N=1): H=αjω.*exp(-βjω)=αjω/(1+(βjω)) now you have: H=(α*s)/(1+β*s)
num=[α,0];
den=[β,1];
H=tf(num,den);
for plotting bode diagram use 'bodeplot'.
1 件のコメント
参考
カテゴリ
Help Center および File Exchange で Filter Analysis についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!