数字信号处理实验7

Laboratory Exercise 7DIGITAL FILTER DESIGN7.1 DESIGN OF IIR FILTERSProject 7.1 Estimation of IIR Filter OrderAnswers:Q7.1The normalized passband edge angular frequency Wp is -0.2The normalized stopband edge angular frequency Ws is -0.4The desired passband ripple Rp is -0.5dBThe desired stopband ripple Rs is -40dB(1) Using these values and buttord we get the lowest order for a Butterworth lowpass filterto be - 8The corresponding normalized passband edge frequency Wn is - 0.2469 or 0.2469pi(2) Using these values and cheb1ord we get the lowest order for a Type 1 Chebyshevlowpass filter to be -5The corresponding normalized passband edge frequency Wn is - 0.2000(3) Using these values and cheb2ord we get the lowest order for a Type 2 Chebyshevlowpass filter to be -5[N, Wn] = cheb2ord(0.2,0.4,0.5,40).The corresponding normalized passband edge frequency Wn is - 0.4000(4) Using these values and ellipord we get the lowest order for an elliptic lowpass filter tobe – 4[N, Wn] = ellipord(0.2,0.4,0.5,40).From the above results we observe that the Elliptic filter has the lowest order meeting thespecifications.Q7.2The normalized passband edge angular frequency Wp is - 0.6000The normalized stopband edge angular frequency Ws is - 0.3429The desired passband ripple Rp is -1dBThe desired stopband ripple Rs is -50dB(1) Using these values and buttord we get the lowest order for a Butterworth highpass filterto be -8[N, Wn] = buttord(Wp,Ws,Rp,Rs).The corresponding normalized passband edge frequency Wn is - 0.5647(2) Using these values and cheb1ord we get the lowest order for a Type 1 Chebyshevhighpass filter to be – 5[N,Wn] = cheb1ord(Wp,Ws,Rp,Rs).The corresponding normalized passband edge frequency Wn is - 0.6000(3) Using these values and cheb2ord we get the lowest order for a Type 2 Chebyshevhighpass filter to be -5[N,Wn] = cheb2ord(Wp,Ws,Rp,Rs).The corresponding normalized passband edge frequency Wn is – 0.3429(4) Using these values and ellipord we get the lowest order for an elliptic highpass filter tobe -4[N,Wn] = ellipord(Wp,Ws,Rp,Rs).The corresponding normalized passband edge frequency Wn is –Wn = 0.6000,From the above results we observe that the Elliptic filter has the lowest order meeting thespecifications.Project 7.2 IIR Filter DesignA copy of Program P7_1 is given below:%程序p7.1%巴特沃斯带阻滤波器的设计ws=[0.4 0.6];wp=[0.3 0.7];rp=0.4;rs=50;%估计滤波器的阶数[N1,wn1]=buttord(wp,ws,rp,rs);%设计滤波器[num,den]=butter(N1,wn1,'stop');%显示传输函数disp('分子系数是');disp(num);disp('分母系数是');disp(den);%计算增益响应[g,w]=gain(num,den);%绘制增益响应plot(w/pi,g);gridaxis([0 1 -60 5]);xlabel('\omega/\pi');ylabel('增益，dB');title('巴特沃斯带阻滤波器的设计');Answers:Q7.5The coefficients of the Butterworth bandstop transfer function generated by running Program P7_1 are as follows:分子系数是Columns 1 through 60.0330 0.0000 0.2972 0.0000 1.1889 0.0000Columns 7 through 122.7741 0.0000 4.1611 0.0001 4.1611 0.0000Columns 13 through 182.7741 0.0000 1.1889 0.0000 0.2972 0.0000Column 190.0330分母系数是Columns 1 through 61.0000 0.00002.6621 0.0000 4.1451 0.0001 Columns 7 through 124.1273 0.0001 2.8977 0.0000 1.4381 0.0000 Columns 13 through 180.5027 0.0000 0.1178 0.0000 0.0167 0.0000 Column 190.0011The exact expression for the transfer function is–The gain response of the filter as designed is given below:ω/π增益，d B巴特沃斯带阻滤波器的设计From the plot we conclude that the design 符合the specifications.The plot of the unwrapped phase response and the group delay response of this filter is given below :Here is the program to find and plot the unwrapped phase response and group delay: % 程序 Q7_5B% 巴特沃斯帯阻滤波器的设计% Plot the unwrapped phase and the group delay. Ws = [0.4 0.6]; Wp = [0.2 0.8]; Rp = 0.4; Rs = 50; % 估计滤波器节数[N1, Wn1] = buttord(Wp, Ws, Rp, Rs); % 设计滤波器[num,den] = butter(N1,Wn1,'stop'); %显示传输函数，绘制不卷绕相位 wp = 0:pi/1023:pi;wg = 0:pi/511:pi; Hz = freqz(num,den,wp); Phase = unwrap(angle(Hz)); figure(1); plot(wp/pi,Phase); grid;% axis([0 1 a b]);xlabel('\omega /\pi'); ylabel('Unwrapped Phase (rad)');title('Unwrapped Phase Response of a Butterworth Bandstop Filter'); % 全延时GR = grpdelay(num,den,wg); figure(2); plot(wg/pi,GR); grid;%axis([0 1 a b]);xlabel('\omega /\pi'); ylabel('Group Delay (sec)'); title('Group Delay of a Butterworth Bandstop Filter');ω /πG r o u p D e l a y (s e c )Group Delay of a Butterworth Bandstop Filterω /πG r o u p D e l a y (s e c )Group Delay of a Butterworth Bandstop FilterProject 7.4 Estimation of FIR Filter OrderAnswers: Q7.13The estimated order of a linear-phase lowpass FIR filter with the following specifications: ωp = 2 kHz, ωs = 2.5 kHz, δp = 0.005, δs = 0.005, and F T = 10 kHz obtained using kaiord is -The purpose of the command ceil is - To round the estimated order up to thenext largest integer; the order has to be integer, so if the formula returns a fraction it needs to be rounded up to the next whole number.The purpose of the command nargin is -To detect if kaiord has been called withfourarguments or with five. If five, it’s assumed that all the frequ encies are analogand thatthe last argument is the sampling frequency. If four, then the sampling frequency defaults to 2, implying that the other frequency arguments are in units of cycles per sample.Project 7.5 FIR Filter DesignAnswers:Q7.20The MATLAB program to design and plot the gain and phase responses of a linear-phase FIR filter using fir1 is shown below. The filter order is estimated using kaiserord. Theoutput data are the filter coefficients.% 程序Q7_20% 线性相位FIR低通滤波器的设计% 满足Q7.13的参数% 写出分子系数% 传输函数% - 增益响应% - 相位响应% - 绘制未卷绕相位响应%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear;Fp = 2*10^3;Fs = 2.5*10^3;FT = 10*10^3;Rp = 0.005;Rs = 0.005;%估计阶数N = kaiord(Fp,Fs,Rp,Rs,FT)Wp = 2*Fp/FT; % 规划角频率Ws = 2*Fs/FT;Wn = Wp + (Ws - Wp)/2;h = fir1(N,Wn);disp('Numerator Coefficients are ');disp(h);[g, w] = gain(h,[1]);figure(1);plot(w/pi,g);grid;%axis([0 1 -60 5]);xlabel('\omega /\pi'); ylabel('Gain in dB');title('Gain Response');% 频率响应w2 = 0:pi/511:pi;Hz = freqz(h,[1],w2);MagH = abs(Hz);T1 = 1.005*ones(1,length(w2));T2 = 0.995*ones(1,length(w2));T3 = 0.005*ones(1,length(w2));figure(4);plot(w2/pi,MagH,w2/pi,T1,w2/pi,T2,w2/pi,T3);grid;% 相位figure(2);Phase = angle(Hz);plot(w2/pi,Phase);grid;xlabel('\omega /\pi'); ylabel('Phase (rad)');title('Phase Response');figure(3);UPhase = unwrap(Phase);plot(w2/pi,UPhase);grid;xlabel('\omega /\pi'); ylabel('Unwrapped Phase (rad)');title('Unwrapped Phase Response');The coefficients of the lowpass filter corresponding to the specifications given in Question 7.20 are as shown below–0.0010 -0.0004 -0.0015 0.0000 0.0024 0.0010 -0.0038 -0.0032 0.00490.0071 -0.0050 -0.0128 0.0026 0.0202 0.0038 -0.0284 -0.0166 0.0366 0.0404 -0.0436 -0.0909 0.0483 0.3129 0.4498 0.3129 0.0483 -0.0909 -0.0436 0.0404 0.0366 -0.0166 -0.0284 0.0038 0.0202 0.0026 -0.0128 -0.0050 0.0071 0.0049 -0.0032 -0.0038 0.0010 0.0024 0.0000 -0.0015 -0.0004 0.0010The generated gain and phase responses are given below:From the gain plot we observe that the filter as designed ___DOES NOT__ meet thespecifications.AS SHOWN in the two detail plots above, with N=46, neither the passbandspec at wp= 0.4 (normalized frequency) nor the stopband spec at ws = 0.5(normalized frequency) are met. So this design DOES NOT meet the spec.The filter order that meets the specifications is – N=66Q7.23The MATLAB program to design and plot the gain and phase responses of a linear-phase FIR filter using fir1 and kaiser is shown below. The filter order N is estimated using Eq.(7.37) and the parameter is computed using Eq. (7.36). The output data are the filtercoefficients.% 程序 Q7_23% 使用凯泽共识设计FIR低通滤波器% - 增益响应%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% clear;% 参数满足Q7.23.Wp = 0.31;Ws = 0.41;Wn = Wp + (Ws-Wp)/2;As = 50;Ds = 10^(-As/20);Dp = Ds;if As > 21N = ceil((As-7.95)*2/(14.36*(abs(Wp-Ws)))+1)elseN = ceil(0.9222*2/abs(Wp-Ws)+1)end% (7.36) 计算Bif As > 50BTA = 0.1102*(As-8.7);elseif As >= 21BTA = 0.5842*(As-21)^0.4+0.07886*(As-21); elseBTA = 0;endWin = kaiser(N+1,BTA);h = fir1(N,Wn,Win);% 分子系数disp('Numerator Coefficients are ');disp(h); %增益响应[g, w] = gain(h,[1]);figure(1);plot(w/pi,g);grid;axis([0 1 -80 5]);xlabel('\omega /\pi'); ylabel('Gain in dB');title('Gain Response');%频率响应w2 = 0:pi/511:pi;Hz = freqz(h,[1],w2);figure(2);Phase = angle(Hz);plot(w2/pi,Phase);grid;xlabel('\omega /\pi'); ylabel('Phase (rad)');title('Phase Response');figure(3);UPhase = unwrap(Phase);plot(w2/pi,UPhase);grid;xlabel('\omega /\pi'); ylabel('Unwrapped Phase (rad)');title('Unwrapped Phase Response');The coefficients of the lowpass filter corresponding to the specifications given in Question 7.23 are as shown below–0.0003 0.0008 0.0003 -0.0011 -0.0017 0.0000 0.0026 0.0027 -0.0010 -0.0049-0.0035 0.0033 0.0080 0.0034 -0.0074 -0.0119 -0.0018 0.0140 0.0161 -0.0027-0.0241 -0.0201 0.0127 0.0406 0.0236 -0.0354 -0.0754 -0.0258 0.1214 0.28710.3597 0.2871 0.1214 -0.0258 -0.0754 -0.0354 0.0236 0.0406 0.0127 -0.0201-0.0241 -0.0027 0.0161 0.0140 -0.0018 -0.0119 -0.0074 0.0034 0.0080 0.0033-0.0035 -0.0049 -0.0010 0.0027 0.0026 0.0000 -0.0017 -0.0011 0.0003 0.00080.0003From the gain plot we observe that the filter as designed ___DOES__ meet the specifications.The filter order that meets the specifications is - N=60.The generated gain and phase responses are given below:。