【正文】 ndamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 711 Frequency response for Example . Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 716 “Hello” passed through b filter. Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 720 Magnitude response for Example . Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 729 Filter shape for Example . Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 731 Filter shapes for Example . Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 722 Magnitude response for fs = 10 kHz for Example . Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. 模擬頻率 f 與數(shù)字頻率 Ω 數(shù)字濾波器的形狀 |H(Ω)|設(shè)計可不依賴采樣頻率(sampling frequency)，但所選的采樣頻率將影響濾波器輸入頻率的范圍。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 713 Frequency response for Example . Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 79 Frequency response for Example . Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。因為相位差為 86186。 （無單位，或 dB 20log|H(Ω)| ) Θ(Ω) 是數(shù)字濾波器在數(shù)字頻率 Ω 處的相位差 (phase difference)。 ∞ N=∞ 令 m=n – n0 , n=m+n0 ∑ x[m] ej(m+n0)Ω = ejn0Ω ∑ x[m] ejmΩ = ejn0Ω X(Ω) ∞ N=∞ ∞ N=∞ 時域中延遲 n0 在頻率中引入一個復(fù)指數(shù) ejn0Ω 周期性： X(Ω+2π )= ∑ x[n] ejn(Ω+2π) = ∑ x[n] ejnΩ? ejn2π ∞ N=∞ ∞ N=∞ 歐拉公式： ejn2πn =cos(2πn) – jsin(2πn) = 1 ∴ X(Ω+2π ) = ∑ x[n] ejnΩ = X(Ω) ∴ DTFT 是周期的，周期為 2π ，也就是 DTFT對于所有的 Ω，每 2π 重復(fù)一次。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 71 Signal resonance for the discrete time Fourier transform. Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 71 Signal resonance for the discrete time Fourier transform. Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。 例 求濾波器 的頻率響應(yīng)，它的傳輸函數(shù) (transfer function)是： H(z)= 1 – 2 1+ + 解： 頻率響應(yīng)為： H(Ω)= 1 – j2Ω 1+ + 頻率響應(yīng)和脈沖響應(yīng) (impulse response) 圖 描述濾波器方法 濾波器的傳輸函數(shù) H(z) 是脈沖響應(yīng) h[n] 的 z變換 x[n] δ[n] X(Ω)= ∑ δ[n] ejnΩ =1 y[n] h[n] Y(Ω)= ∑ h[n] ejnΩ H(Ω)= = Y(Ω) ∴ 頻率響應(yīng) H(Ω) 與脈沖響應(yīng) h[n] 的 DTFT 一樣。 Y(Ω)=|Y(Ω)|ejΘy(Ω) =|X(Ω)|ejΘx(Ω) |H(Ω)|ejΘ(Ω) =|X(Ω)||H(Ω)|ej(Θx(Ω)+Θ(Ω)) 幅值計算不能用分貝，都要轉(zhuǎn)成線形計算。 例：一系統(tǒng)的頻率響應(yīng)為 求該系統(tǒng)的幅度響應(yīng)和相位響應(yīng)，并畫出圖。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458