【正文】 All rights reserved. FIGURE 710 Frequency responses for mon filters. 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 715 Pulse passed through b filter. Joyce Van de Vegte Fundamentals of Digital Signal Processing Copyright 169。根據(jù)上式，可將 0~ π 弧度的數(shù)字頻率用 0 ~fs/2 Hz的模擬頻率代替。2022 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. 由上可擴(kuò)展到具有多個(gè)極零點(diǎn)的濾波器 對(duì)于 弧度的頻率 , 離濾波器極點(diǎn)越近 ,離零點(diǎn)越遠(yuǎn) ,則幅度就越大 .同樣 ,靠近單位圓的極點(diǎn) ,將導(dǎo)致濾波器形狀在某一頻率上有非常大的幅值 ,而靠近單位圓的零點(diǎn)將導(dǎo)致濾波器形狀在某一頻率上有非常小的幅值 .這個(gè)幅值大小的劇烈變化可增強(qiáng)濾波器的選擇性 . 001()()( ) ( )K z zHzz p z p????0 00101(()(((()(jjjK e z KzHe p e pKH???? ?? ?? ??????到 的 距 離 ）到 p 的 距 離 ） 到 p 的 距 離 ）到 零 點(diǎn) 距 離 的 乘 積 ）到 極 點(diǎn) 距 離 的 乘 積 ）0~?je?例 :推斷濾波器的形狀 ,濾波器的傳輸函數(shù)為 1()0 . 4 5Hz z? ?FIGURE 726 Polezero plot 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. [ ] [ 1 ] [ 2 ] [ ]y n y n y n x n??? ? ? ? ?CHAPTER SUMMARY ? The discrete time Fourier transform (DTFT) of a signal x[n] is given by ? x(Ω)= ∑ x[n] ejnΩ. It reports the frequencies present in a signal. ? The DTFT of a signal x[n] gives the signal’s spectrum X(Ω ). ? The DTFT of a system h[n] gives the system’s frequency response H(Ω). ? The DTFT is periodic with period 2 π. ? A difference equation can be expressed as a frequency response. ? A transfer function can be expressed as a frequency response. ? The frequency response H(Ω) is the DTFT of the impulse response h[n]. ? A frequency response H(Ω) is a plex number and may be expressed in polar form in terms of a gain |H(Ω)| and a phase difference ?(?) as H(Ω)= |H(Ω)| e j ?(?) ? The frequency response can be used to find a filter’s output for a sinusoidal input. The output is a sinusoid with the same frequency as the input,but with an amplitude multiplied by the gain of the filter and a phase shifted by the phase difference of the filter. For the input x[n]=Acos(n Ω0+ ? x),with a digital frequency Ω0 ,the output is y[n]=H Acos(n Ω0+ ? +? x),where the gain of the filter is H(Ω)= |H(Ω0)| and the phase difference is ?? ?(Ω0). ? The gains applied by the filter at each frequency form the magnitude response,plotted as |H(Ω)| versus Ω, or 20log |H(Ω)| versus Ω. The phase differences applied by the filter at each frequency form the phase response, plotted as ?(?) , in degrees or radians, versus ?. ? It is sufficient to plot a frequency response for a system for the digital frequencies between 0 and π radians, as ?? π radians corresponds to the Nyquist frequency or fs/2. ? The magnitude response shows the shape of a filter:low pass,high pass, band pass or band stop. Some filters, such as b filters, have more plicated shapes. ? The magnitude and phase responses, |H(Ω)| and ?(?) ,are normally plotted against digital frequency Ω in radians. They can be plotted instead against analog frequency f in Hz using the equation ? f= Ωfs/2 π ? With this substitution , the magnitude response may be plotted as |H(f)| versus f, and the phase response as ?(f) versus f. ? The shape of a filter can be deduced from its polezero plot. As Ω increases, the plex number e j ? moves around the unit circle. Proximity to a zero tends to reduce the magnitude of the filter, while proximity to a pole tends to increase it. The closer the poles and zeros are to the unit circle, the more selective the filter is.