Transcription of 1.伝達関数とインパルス応答 - トップページ
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(t) h(t) f(t) g(t)L 1/8 (t) h(t) f(t) g(t)L f(t) g(t) Lf(t)=g(t)( ) f1(t),f2(t) Laf1(t)+bf2(t)=ag1(t)+bg2(t)( ) Lf(t t0)=g(t t0)( ) t0 (t) h(t) L (t)=h(t)( ) h(t) f(x) Lf(x) (t x)=f(x)h(t x)( ) Lf(x) (t x)dx =f(x)h(t x)dx ( ) Lf(t) ( ) g(t) g(t)=f(x)h(t x)dx =f(t) h(t)( ) 2/8 h(t) H( ) G( )=F( )H( )( ) H( ) (Transfer Function) H( ) f(t) g(t) g(t)=kf(t td)( ) G( )=kF( )e j td( ) H( ) H( )=k H( )= td( ) 2.
スペクトルと信号処理 6/8 = 1 2 lim T 1 T F - F ej t d となり,ここで時間たたみ込みとフーリエ変換の性質 Fft gt = F G f(-t) F(- ) を用いると F-1 S = 1 2 lim
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