实验四线性时不变离散时间系统的频域分析.

1、Name :Chen yifan 20112121006Section :Laboratory Exercise 4LINEAR, TIME-INVARIANT DISCRETE-TIME SY STEMS: FREQUENC Y-DOMAIN REPRESENTATIONS4.1 TRANSFER FUNCTION AND FREQUENCY RESPONSEProject 4.1Tran sfer Fun cti on An alysisAn swers:Q4.1The modified Program P3_1 to compute and plot the magnitude andp

2、hase spectra of a moving average filter of Eq. (2.13) for 0 _ 一 2“ is shown below :w=0:pi/511:2*pi;M=i nput(M二);num=o nes(1,M)/M;h=freqz( nu m,1,w);subplot(2,1,1);plot(w/pi,abs(h);grid;title(H(eAjomega)幅度谱); xlabel(omega八pi);ylabel(振幅);subplot(2,1,2);plot(w/pi,a ngle(h);grid;title(相位谱 H(eAjomega);xl

3、abel(omega/pi);ylabel(以弧度为单位的相位);This program was run for the following three different values ofM and the plots ofthe corresp onding freque ncy resp on ses are show n below:The types of symmetries exhibited by the magn itude and phase spectra are dueto -The type of filter represe nted by the movi n

4、g average filter isM=3H(e j )幅度谱0.20.40.60.811.21.41.61.82-相位谱H(e)50幅振o4 2 0 2 4- - 位相的位单为度弧以幅振H(ej )幅度谱0.20.40.60.811.21.41.61.82-200.40.60.811.21.41.61.82-/：相位谱H(ej )0.40.60.811.21.41.61.824 2 0-2位相的位单为度弧以20M=20;1.51IJ i / 1幅 振0.5H(ej )幅度谱000.20.40.60.811.21.41.61.8I：相位谱H(ej )4位相的位单为度弧以-4020-20.20

5、.40.60.811.21.41.61.827二The results of Question Q2.1 can now be explained as follows-By the graph, you can seethat it represents a low-pass filter.Q4.2 The plot of the frequency response of the causal LTI discrete-time system ofQuesti on Q4.2 obta ined using the modified program is give n below:w=0:

6、pi/511:pi;num=0.15 0 -0.15;den=1-0.5 0.7; h=freqz( nu m,de n, w);subplot(2,1,1);plot(w/pi,abs(h);grid;title(H(eAjomega)幅度谱);xlabel(omega/pi);ylabel(振幅);subplot(2,1,2);plot(w/pi,a ngle(h);grid;title(相位谱 H(e%omega); xlabel(omega八pi);ylabel(以弧度为单位的相位);位相的位单为度弧以H(ej )幅度谱0.5000.10.20.30.40.50.60.70.80.9相

7、位谱H(ej )2-1-200.10.20.30.40.50.60.70.80.9-/：The type of filter represented by this transfer function isobta ined by diagrams can be BPF-It says ban dpass filter isQ4.3 The plot of the frequency response of the causal LTI discrete-time system ofQuesti on Q4.3 obta ined using the modified program is g

8、ive n below:w=0:pi/511:pi;num=0.15 0 -0.15;den=0.7 -0.5 1h=freqz( nu m,de n, w);subplot(2,1,1);plot(w/pi,abs(h);grid;title(H(eAjomega)幅度谱);xlabel(omega/pi);ylabel(振幅);subplot(2,1,2);plot(w/pi,a ngle(h);grid;title(相位谱 H(e%omega); xlabel(omega八pi);H(ej )幅度谱!0.5ylabel(以弧度为单位的相位);000.10.20.30.40.50.60.7

9、0.80.9J：相位谱H(ej )4位的位单为度弧以-2相2-400.10.20.30.40.50.60.70.80.9I：The type of filter represe nted by this tran sfer fun cti on is- O causal li near time-i nvaria nt discretetime system freque ncy resp on se, determ ine the type filterThe differe nee betwee n the two filters of Questi ons 4.2 and 4.3 is-

10、 More on the topic offigure, the amplitude spectrum is the same, the phase spectrum, on the topic that is con ti nu ous, and kinds of phase jump lin e,.I shall choose the filter of Questi on Q4.3 for the followi ng reas on - the amplitudespectrum is the same, the phase spectrum, on the topic that is

11、 con ti nu ous, and kinds of phase jump line .Q4.6 The pole-zero plots of the two filters of Questions 4.2 and 4.3 developed using zpla ne are shown below : w=0:pi/511:pi;num=0.15 0 -0.15;den=1-0.5 0.7;h=zpla ne(nu m,de n);w=0:pi/511:pi;num=0.15 0 -0.15;den=0.7 -0.5 1;h=zpla ne(nu m,de n);-1-0.50.51

12、0Real Part10.80.60.40.20-0.2-0.4-0.6-0.8-11-1.5-1-0.500.511.5Real PartFrom these plots we make the follow ing observati ons: Different function, zero zero and pole polefigure with different relative position of the circle4.2 TYPES OF TRANSFER FUNCTIONSProject 4.2FiltersA copy of Program P4_1 is give

13、 n belowclf;fc=0.25;n=-6.5:1:6.5;y=2*fc*s in c(2*fc* n); k=n+6.5;stem(k,y); title(N=13); axis(0 13 -0.2 0.6);xlabel(时间序号 n); ylabel(振幅);gridN=13An swers:Q4.7 The plot of the impulse response of the approximation to the ideal lowpass filter obta ined using Program P4_1 is show n below:clf;fc=0.25;n=-

14、6.5:165;y=2*fc*s in c(2*fc* n);k=n+6.5;ste m( k,y);title(N=13);axis(0 13 -0.2 0.6);xlabel(时间序号 n);ylabel(振幅);gridThe len gth of the FIR lowpass filter is -14The stateme nt in Program P4_1 determ ining the filter len gth is- n=-6.5:165;The parameter con trolli ng the cutoff freque ncy is-Fc parameter

15、s con trol the cutofffreque ncyQ4.8 The required modifications to Program P4_1 to compute and plot the impulse response of the FIR lowpass filter of Project 4.2 with a length of 20 and a cutoff frequency of c = 0.45 are as indicated below :clf;fc=0.45/(2*pi);n=-9.5:1:9.5;y=2*fc*si nc(2*fc* n);k=n+9.

16、5;stem(k,y);title(N=20);axis(0 20 -0.2 0.6); xlabel(时间序号 n);ylabel(振幅);gridThe plot gen erated by running the modified program is give n below:0.60.50.40.30.10-0.1-0.2142010 12时间序号n幅 0 2振16 18QI1 1N=20to compute andwith a length ofplot the impulse15 and a cutoffQ4.9 The required modifications to Pro

17、gram P4_1 response of the FIR lowpass filter of Project 4.2freque ncy of，c = 0.65are as in dicated belowclf;fc=0.65/(2*pi);n=-7.5:1:6.5;y=2*fc*s in c(2*fc* n);k=n+7.5;stem(k,y);title(N=15);axis(0 14 -0.2 0.6);xlabel(时间序号 n);ylabel(振幅);gridThe plot gen erated by running the modified program is give n

18、 belowN=15Q4.11A plot of the gain resp onse of a len gth-2 movi ng average filter obta inedusing Program P4_2 is show n below : unction g,w=gain(num,den) -gain 函数w=0:pi/255:pi;h=freqz( nu m,de n, w); g=20*log10(abs(h);M=2;-滑动平均低通滤波器的增益响应程序num=o nes(1,M)/M;g,w=gai n(n um,1); plot(w/pi,g);grid;axis(0

19、1 -50 0.5)xlabel(omega/pi);ylabel(单位为 db 的增益);title(M= ,nu m2str(M)M= 20-40-45 11 -50 :00.10.20.30.40.50.60.70.80.91From the plot it can be seen that the 3-dB cutoff frequency is at- 3dB。Q4.12The required modifications to Program P4_2 to compute and plot thegain response of a cascade of K length-2 m

20、oving average filters are given below:fun cti on g,w=ga in li nk(nu m,de n) w=0:pi/255:pi;h=freqz( nu m,de n, w);H=h.*h.*h;g=20*log10(abs(H);wc=2*acos(2A(-1/(2*3)wc =0.9430wc =0.3*piThe plot of the gain response for a cascade of 3 sections obtained using the modified program is show n below :0-5-10-

21、15-20-25-30-35-40-45-5000.10.20.30.40.50.60.70.80.9From the plot it can be see n that the 3-dB cutoff freque ncy of the cascade is atwc=0.3*piQ4.19A copy of Program P4_3 is give n belowCopy from m-file(s) and paste. In sert program code here.clf;b=1 -8.5 30.5 -63;num1=b 81 fliplr(b);num2=b 81 81 fli

22、plr(b);num3=b 0 -fliplr(b);num4=b 81 -81 -fliplr(b);n1=0:length(num1) -1;n2=0:le ngth( num2) -1;subplot(2,2,1); stem(n1,numl);xlabel(时间序号n); ylabel(振幅);grid;ti tle(1型有限冲激响应滤波器)；subplot(2,2,2); stem( n2, num2);xlabel(时间序号n); ylabel(振幅);grid;ti tle(2型有限冲激响应滤波器)；subplot(2,2,3); stem(n1,num3);xlabel(时间序

23、号n); ylabel(振幅);grid;ti tle(3型有限冲激响应滤波器)；subplot(2,2,4); stem(n2,num4);xlabel(时间序号n); ylabel(振幅);grid;ti tle(4型有限冲激响应滤波器)；pausesubplot(2,2,1); zpla ne( nu m1,1);ti tle(1型有限冲激响应滤波器)；subplot(2,2,2); zpla ne(n um2,1);ti tle(2型有限冲激响应滤波器)；subplot(2,2,3); zpla ne(n um3,1);ti tle(3型有限冲激响应滤波器)；subplot(2,2,4

24、); zpla ne(n um4,1);ti tle(4型有限冲激响应滤波器)；disp(1型有限冲激响应滤波器的零点是);disp(roots (n um1);disp(2型有限冲激响应滤波器的零点是);disp(roots (n um2);disp(3型有限冲激响应滤波器的零点是);disp(roots (n um3);disp（4型有限冲激响应滤波器的零点是 ）;disp(roots (n um4);runningThe plots of the impulse resp on ses of the four FIR filters gen erated by Program P4_3

25、are give n below : I nsert MATLAB figure(s) here.Copy from figure wi ndow(s) and paste.1型有限冲激响应滤波器2型有限冲激响应滤波器3型有限冲激响应滤波器4型有限冲激响应滤波器From the plots we make the followi ng observati ons:Filter #1 is of length 8with a 80i mpulse response and is therefore aType linear-phase FIR filter .Filter #2 is of le

26、ngth 10with a 80impulse response and is thereforea Typelin ear-phase FIR filter .Filter #3 is of length 8with a 60i mpulse response and is therefore aType linear-phase FIR filter.Filter #4 is of length 10with a 80impulse response and is thereforea Typelin ear-phase FIR filter.From the zeros of these

27、 filters gen erated by Program P4_3 we observe that:Filter #1 has zeros at z =2.9744,2.0888 ,0.9790 + 1.4110i,0.9790 -1.4110i, 0.3319 + 0.4784i, 0.3319 - 0.4784i,0.4787,0.3362Filter #2 has zeros at z =3.7585 + 1.5147i,3.7585 - 1.5147i,0.6733 +2.6623i,0.6733 - 2.6623i,-1.0000 ,0.0893 + 0.3530i,0.0893

28、 - 0.3530i,0.2289 + 0.0922i,0.2289 - 0.0922iFilter #3 has zeros at z =4.7627 ,1.6279 + 3.0565i,1.6279 - 3.0565i,- 1.0000 ,1.00000.1357 + 0.2549i,0.1357 - 0.2549i,0.2100Filter #4 has zeros at z =3.4139 ,1.6541 + 1.5813i,1.6541 - 1.5813i,-0.0733 + 0.9973i,-0.0733 - 0.9973i,1.0000 ,0.3159 + 0.3020i,0.3

29、159 - 0.3020i,0.2929Plots of the phase resp onse of each of these filters obta ined using MATLAB areshown below :trdp 卩tr3p WFaoa 卩Q4.20The plots of the impulse resp on ses of the four FIR filters gen erated byrunning Program P4 3 are given below :1型有限冲激响应滤波器2型有限冲激响应滤波器3型有限冲激响应滤波器4型有限冲激响应滤波器From the

30、 plots we make the follow ing observati onsFilter #1 is of le ngth8with a80i mpulse resp on se a nd is therefore aTypelin ear-phase FIR filter .Filter #2 is of le ngth10with a80impulse resp onse and is thereforea Typelin ear-phase FIR filter .Filter #3 is of le ngth8with a5impulse resp on se a nd is

31、 therefore aTypelin ear-phase FIR filter .Filter #4 is of le ngth10with a80impulse resp onse and is thereforea Type linear-phase FIR filter .From the zeros of these filters gen erated by Program P4_3 we observe that:Filter #1 has zeros at z = 2.3273 + 2.0140i,2.3273 - 2.0140i,-1.2659 +2.0135i,-1.265

32、9 - 2.0135i, -0.2238 + 0.3559i,-0.2238 - 0.3559i,0.2457 + 0.2126i,0.2457 - 0.2126iFilter #2 has zeros at z =2.5270 + 2.0392i, 2.5270 - 2.0392i,-1.0101 +2.1930i.-1.0101 - 2.1930i,-1.0000 ,-0.1733 + 0.3762i,-0.1733 - 0.3762i,0.2397 + 0.1934i,0.2397 - 0.1934iFilter #3 has zeros at z =-1.0000 ,0.2602 +

33、1.2263i,0.2602 - 1.2263i,1.0000 ,0.6576 + 0.7534i,0.6576 - 0.7534i,0.1655 + 0.7803i,0.1655 - 0.7803iFilter #4 has zeros at z = 2.0841 + 2.0565i,2.0841 - 2.0565i,-1.5032 +1.9960i,-1.5032 - 1.9960i,1.0000 ,-0.2408 + 0.3197i,-0.2408 - 0.3197i,0.2431 + 0.2399i,0.2431 - 0.2399iPlots of the phase resp ons

34、e of each of these filters obta ined using MATLAB are shown below -Trap VTanrga 卩E，!E/ 0 I8-i1型有限冲激响应滤波器21o1-2-Trap VTanrga 卩iuii91QT-I2型有限冲激响应滤波器210-1-2-2 0 2Real Part3型有限冲激响应滤波器-2 0 2Real Part4型有限冲激响应滤波器811-Si/i-i 11 5 0 5 1 o o- .Trap VFanrgaXLrap VFanrga 卩f o-9U1F210-1-2-20 2Real Part-1 0 1Real PartFrom these plots we con elude that each of these filters havelinearphase .An swers:4.3 STABILITY TESTA copy of Program P4_4 is give n below:clf;den=input(分母系数=;ki=poly2rc(de n);