1、实验4离散系统地分析报告实验四 离散系统分析一、 实验目的深刻理解离散时间系统的系统函数在分析离散系统的时域特性、 频域特性以及稳定性中的重要作用及意义,熟练掌握利用MATLA分析离散系统的时域响应、 频响特性和零极点的方法。掌握利用 DTFT和DFT确定系统特性的原理和方法。二、 实验原理可以在时域、复频域(Z域)及频域分析系统,在以上三种域表征系统固有特性 的量分别为:单位冲激响应h(n)(时域表征);系统函数H(z) ( Z域表征);频率响应H(ej )(频域表征)。MATLA吐要从以上三方面提供了许多可用于分析线性时不变系统的函数,包含 系统时域响应、系统函数、系统频域响应等分析函数。
2、本实验通过调用各种系统预置函数来求系统的以上几个表征量以及零极点图。三、 实验内容1.已知某LTI系统的差分方程为:yk 1.143yk 1 0.412yk 2 0.0675xk 0.1349xk 1 0.0675xk 2(1)初始状态y 1 1, y 2 2 ,输入xk uk,计算系统的全响应。程序段:N=40;b=0.0675,0.1349,0.0675;a=1,-1.143,0.412;x=on es(1,N);zi=filtic(b,a,1,2);y=filter(b,a,x,zi);stem(y)xlabel(k);title(yk);结果:川】7xdk cos(- k)uk;x2【
3、k cos k)uk; X3【k cos(- k)uk10 5 10程序N=30;k=O:N;b=0.0675,0.1394,0.0675;a=1,-1.143,0.412;x1=cos(pi*0.1.*k);x2=cos(pi*0.2*k);x3=cos(pi*0.7*k);y1=filter(b,a,x1);y2=filter(b,a,x2);y3=filter(b,a,x3);subplot(3,1,1);stem(y1)subplot(3,1,2);stem(y2)subplot(3,1,3); stem(y3): 结果:(3)该系统具有什么特性? 答:因果稳定。2.已知某因果LTI系
4、统的系统函数为:H(z)12 3 40.03571 0.1428z 1 0.2143z 2 0.1428z 3 0.03571z 41 1.035z 1 0.8264Z 2 0.2605z 3 0.04033z 4(1)计算系统的单位冲激响应 程序:N=50;k=1:N;b=0.0357,0.1428,0.2143,0.1428,0.0357; a=1,-1.035,0.8264,-0.2605,0.04033;y1=impz(b,a,N);stem(yl)结果为:(2)当信号通过系统时,c计算4系统的零状态2响应。 程序为:N=50;k=1:N;b=0.0357,0.1428,0.2143,
5、0.1428,0.0357; a=1,-1.035,0.8264,-0.2605,0.04033;x=o nes(1,N)+cos(pi*0.25*k)+cos(0.5*pi*k); y1=impz(b,a,N);y2=filter(b,a,x);subplot(2,1,1);stem(y1)subplot(2,1,2);stem(y2)结果为:CF0.2 -V :。百 r U 07乞七今七导 OiJO do O 牛? 导:卜y M t.- o cJO 4f soin3.已知LTI系统的输入输出序列分别为 xk uk cosqk)uk cos(? k)uk(a)xk (1)kuk, yk 1(
6、1)kuk (!)ku k2 4 2 4(b)xk (1)kuk, yk (4)kuk (4)k1uk 1(1)利用解析方法分别求解系统的单位取样响应。程序:N=20;a仁1,-0.75;b仁1.25,-19/16;h1=impz(b1,a1,N);subplot(2,1,1);stem(h1)b2=1,-1;a2=1;h2=impz(b2,a2,N);subplot(2,1,2);stem(h2)结果为:(2)利用系统辨识原理确定 并求出系统的单位脉冲响应hk。比较解析方法与系统辨识方法得到的系统单位冲激响应,分析误差原因。程序:N=50;k=O:N;x1=0.5.Ak;y1=0.25*0.
7、5.Ak+0.25.Ak;X1=fft(x1);丫仁fft(y1);H仁 Y1/X1;h1=ifft(H1)n=1:50;x2=0.25.a n;y2=0.25.A n-4*0.254 n;X2=fft(x2);Y2=fft(y2);H2=Y2/X2;h2=ifft(H2)结果:hi 二 H(ej )1. 10F1 十 O.OOOOi h2 二-3.0000 - O.OOOOil误差原因:取样点数不够多引起的误差,而且不易与 k对应4.已知某离散系统的输入输出序列。输入序列:2, 0.8333,0.3611 ,0.162,0.0748,0.0354,0.017,0.0083,0.0041, 0
8、.002,0.001,0.0005,0.0002,0.0001,0.0001,后面的数值均趋于 0;输出序列:0.0056,-0.0259,0.073,-0.1593,0.297,-0.4974 ,0.7711 ,-1.1267 ,1.5702 ,-2.1037 ,2.724 ,-3.4207 ,4.174 ,-4.9528, 5.7117, -6.3889 , 6.9034 ,-7.1528 , 7.012 ,-6.3322 , 4.9416 ,-2.648 , -0.7564 , 5.4872 , -11.7557 , 19.7533 , -29.6298 , 41.4666 , -55.
9、2433 , 70.7979, -87.7810(1)绘出输入输出信号的波形。程序:x=2,0.8333,0.3611,0.162,0.0748,0.0354,0.017,0.0083,0.0041,0.002, 0.001,0.0005,0.0002,0.0001,0.0001,zeros(1,16);y=0.0056,-0.0259,0.073,-0.1593,0.297,-0.4974,0.7711,-1.1267,1.5702,-2.1037,2.724,-3.4207,4.174,-4.9528,5.7117,-6.3889,6.9034,-7.1528,7.012,-6.3322,
10、4.9416,-2.648,-0.7564,5.4872,-11.7557,19.7533, -29.6298,41.4666,-55.2433,70.7979,-87.7810; subplot(2,1,1);stem(x)subplot(2,1,2); stem(y) 结果:100(2)计算该系统的频率响应,并绘出其幅频特性。 程序:x=2,0.8333,0.3611,0.162,0.0748,0.0354,0.017,0.0083,0.0041,0.002, 0.001,0.0005,0.0002,0.0001,0.0001,zeros(1,16);y=0.0056,-0.0259,0.
11、073,-0.1593,0.297,-0.4974,0.7711,-1.1267,1.5702,-2.1037,2.724,-3.4207,4.174,-4.9528,5.7117,-6.3889,6.9034,-7.1528,7.012,-6.3322,4.9416,-2.648,-0.7564,5.4872,-11.7557,19.7533,-29.6298,41.4666,-55.2433,70.7979,-87.7810;X=fft(x,128);Y=fft(y,128);H=Y./X;H0=abs(H);plot(H0)结果为:3)计算该系统的单位冲激响应,并绘出其波形程序:x=2,
12、0.8333,0.3611,0.162,0.0748,0.0354,0.017,0.0083,0.0041,0.002, 0.001,0.0005,0.0002,0.0001,0.0001,zeros(1,16);y=0.0056,-0.0259,0.073,-0.1593,0.297,-0.4974,0.7711,-1.1267,1.5702,-2.1037,2.724,-3.4207,4.174,-4.9528,5.7117,-6.3889,6.9034, -7.1528,7.012,-6.3322,4.9416,-2.648,-0.7564,5.4872,-11.7557,19.7533
13、, -29.6298,41.4666,-55.2433,70.7979,-87.7810;X=fft(x,64);Y=fft(y,64);H=Y./X;H0=abs(H);h0=ifft(H0)stem(h0)结果为:hO =Colujims 1 through 566.52S1 -35.5614 20.0669 -12.2746 6.7&I6CoIujips 昌 through 10-3. 29760, 8766Q.691S-1.6B712.2493CqIujiuis 11through15-2. 50372*552945372. 23S-2.0228CoIujuis 16throneh20
14、1. 7587-1, 4926k 2395-1.0850-B048Coltunns 21through25-0. 63060,4E63-0. 37030.230-0.2136Colmns ZB through 300*3604-0.13530. H5o-0. 1071CL 1033Colrnns 31through35-fl.1027C.1030-0. 10340- 1030-0. 1027Columns 36-through400-1033*0.10710. LI56-0. 13530. 1664匚 oliimns 41through45-0-21360.2804-0, 37030. 486
15、3-0. 6306Colwns 46throvieh500. 8048-1.C0B51. 2395-i 19261+ 7537Colunins 1 through &5-2. 02282.2639-2. 4637Z.5529K5087CdIujiltls 56t hrough602.2493-1.68710.691S0.3766-3. 2976ColUUITLS 61through646+ 7916 -12. 274S20. 0669-38.56U图形为:5.利用load mtlb命令读入一段语音信号得到序列 ,然后在该段语音信号中加入500Hz的正弦型干扰信号得到信号,利用 FFT分析其频谱
16、。(1)下列数字滤波器能够滤除信号中 500Hz正弦型干扰信号。利用zplane命令做出其零极点分布图,利用freqz命令分析该滤波器的幅频特 性和相频特性,比较零极点分布与滤波器频率特性的关系。程序:a=1,-3.594,5.17,-3.494,0.945;b=0.6877,-2.509,3.664,-2.509,0.6877;z=roots(b);p=roots(a);subplot(2,2,1);zpla ne(b,a)title( 零极点分布)H,w=freqz(b,a); subplot(2,2,2); plot(w,abs(H) xlabel(omega(rad); ylabel(
17、HO);title( 幅频特性); subplot(2,2,3); plot(w,a ngle(H) xlabel(omega(rad); ylabel(phi );title( 相频特性); 结果:关系:系统零点越接近 1贝肿畐频响应的波谷越低,即滤波器的滤波衰减最低点 越低;系统极点越接近 1贝肿畐频响应的波峰越高,即滤波器的滤波通带最高点 越高。极点主要影响频率响应的峰值,极点愈靠近单位圆,峰值愈尖锐;零点主 要影响频率特性的谷值,零点愈靠近单位圆,谷值愈深,当零点在单位圆上时, 频率特性为零,一个传递函数有几个极点幅度响应就有几个峰值, 对应出现一些 谷值。(2)利用该数字滤波器滤除信
18、号中的噪声,利用 FFT观察其频谱,利 用sou nd函数播放处理前后的信号,比较处理前后的效果。程序;load mtlbN=150;x=mtlb(1:N);k=1:N;subplot(1,2,1);plot(k,x); title( 处理前); X=fft(x,150); h=ifft(H); H=fft(H,150); Y=X.*H; y=ifft(Y); subplot(1,2,2); plot(1:N),y); title(处理后);结果:实验思考题1.系统函数的零极点对系统频率特性有何影响?ANS系统零点越接近1,贝肿畐频响应的波谷越低,即滤波器的滤波衰减最低 点越低;系统极点越接近
19、1,贝肿畐频响应的波峰越高,即滤波器的滤波通带最高 点越高。极点主要影响频率响应的峰值,极点愈靠近单位圆,峰值愈尖锐;零点 主要影响频率特性的谷值,零点愈靠近单位圆,谷值愈深,当零点在单位圆上时, 频率特性为零,一个传递函数有几个极点幅度响应就有几个峰值, 对应出现一些 谷值。2.对于因果稳定实系数的低通、高通、带通、带阻数字滤波器,零极点分布有何 特点?ANS:因为是因果稳定系统,所以极点都在单位圆内。若为最小相位系统,其 零点也在单位圆内。3.离散系统的系统函数的零极点对系统脉冲响应有何影响?ANS系统函数的极点位置决定序列包络的变化趋势和变化频率, 极点的半径 决定了序列包络的变化趋势,而极点的幅角决定序列包络的变化频率, 而零点位 置只影响冲激响应的幅度大小和相位。4.若某因果系统不稳定,有哪些主要措施可使之稳定?ANS改变参数,让极点在单位圆内。5.从频域利用DFT确定离散LTI系统的特性,一般会产生哪些误差,如何改善? 大全ANS:频谱混叠,对于带限连续信号,只要提高抽样频率使之满足时域抽样 定理;对于非带限信号,更具实际情况对其进行低通滤波,使之成为带限信号; 频谱泄露,时域加窗使之成为有限长序列;栅栏现象,在序列后补零,构成新序 列后再求频谱6.若使用DFT对连续LTI系统进行辨识,需要解决哪些问题?ANS循环卷积长度的确定。
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