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 REP RESENTATIONS4.1 TRANSFER FUNCTION AND FREQUENCY RES PO NSEProject 4.1Tran sfer Fun cti on An alysisAn swers:P3_1 to compute and plot the magnitude and filter of Eq. (2.13) f

2、or 0 c 2兀 is shownQ4.1The modified Programphase spectra of a moving average below :w=0: pi/511:2* pi;M=i npu t(M=);num=o nes(1,M)/M;h=freqz( nu m,1,w);sub plot(2,1,1);pl ot(w/ pi,abs(h);grid; title(H(eAjomega)幅度谱); xlabel(omega八pi);ylabel(振幅); sub plot(2,1,2);p lot(w/pi,a ngle(h);grid; title(相位谱 H(e

3、Ajomega); xlabel(omega/pi);ylabel(以弧度为单位的相位);M and the plots ofThis p rogram was run for the followi ng three differe nt values of the corres ponding freque ncy res pon ses are show n below:The types of symmetries exhibited by the magn itude and p hase sp ectra are due to -The type of filter repr es

4、e nted by the moving average filter isM=3H(e jeO)幅度谱0幅振50嘔.振0.20.40.60.811.21.41.61.82相位谱H(e 00.20.40.60.811.21.41.61.82/兀O 4 2 0 0/ 位相的位单为度弧以度谱1/-i0.40.60.811.21.41.61.82相位谱H(e鬥 _ O4 2 0 0/位相的位单为度弧以0.40.60.811.21.41.61.82兀M=20;Hj幅度谱1.51、 T1/J!、厂、产 7-、八一厂、-厂、广,1i J/i1幅 振0.50.20.40.60.811.21.41.61.82

5、相位谱H(ej兮的位单为度弧以-41|1 1.1*1 * 1 1 kI | 1 1 1 VL11, 1 | 1 l20-200.20.40.60.811.21.41.61.82-By the graph, you can seeThe results of Questi on Q2.1 can now be explained as follows that it rep resents a low -p ass filter.Q4.2 The plot of the frequency response of the causal LTI discrete-time system ofQue

6、sti on Q4.2 obta ined using the modified pr ogram is give n below:w=0:p i/511: pi;num=0.15 0 -0.15;den=1 -0.5 0.7;h=freqz( nu m,de n, w);sub plot(2,1,1);pl ot(w/ pi,abs(h);grid;title(H(eAjomega)幅度谱);xlabel(omega八pi);ylabel(振幅);sub plot(2,1,2);plot(w/pi,a ngle(h);grid;titleC 相位谱 H(eAjomega); xlabel(o

7、mega八pi);ylabelC以弧度为单位的相位); sub plot(2,1,2); plot(w/pi,a ngle(h);grid;Hj幅度谱幅0.500.10.20.30.40.50.60.70.80.91相位谱H(ej兮位相的位单为度弧以-210-10.70.80.9100.10.20.30.40.50.6/HThe type of filter repr ese nted by this tran sfer fun cti on is obta ined by diagrams can be BPF-It says bandp ass filter isQ4.3 The plot

8、 of the frequency response of the causal LTI discrete-time system ofQuesti on Q4.3 obta ined using the modified pr ogram is give n below:w=O:p i/511: pi;num=0.15 0 -0.15;den=0.7 -0.5 1 h=freqz( nu m,de n, w);sub plot(2,1,1);pl ot(w/ pi,abs(h);grid; title(H(eAjomega)幅度谱); xlabel(omega八pi);ylabel(振幅);

9、titleC 相位谱 H(eAjomega); xlabel(omega八pi);ylabelC以弧度为单位的相位);H(ej勺幅度谱幅0.5振000.10.20.30.40.50.60.70.80.91相位谱H(e吗的位单为度弧以-2-400.10.20.30.40.50.60.70.80.91The type of filter rep rese nted by this tran sfer fun cti on is-O causal li near time-i nvaria nt discretetime system freque ncy respon se, determ ine

10、 the type filterThe differe nee betwee n the two filters of Questi ons 4.2 and 4.3 is-More on the topic offigure, the amp litude sp ectrum is the same, the p hase sp ectrum, on the topic that iscon ti nu ous, and kinds of p hase jump lin e,.I shall choose the filter of Questi on Q4.3 for the follow

11、ing reas on- the amp litudesp ectrum is the same, the p hase sp ectrum, on the topic that is continu ous, and kinds ofp hase jump line .Q4.6 The pole-zero plots of the two filters of Questions 4.2 and 4.3 developed using zplane are shown below :w=0: pi/511: pi;num=0.15 0 -0.15;den=1-0.5 0.7;h=z plan

12、e(nu m,de n);w=0: pi/511: pi;num=0.15 0 -0.15;den=0.7 -0.5 1;h=z plane(nu m,de n);10.80.60.40.20-0.2-0.4-0.6-0.8-1-0.500.51Real Part-110.50-0.5-1-1.5-1-0.50.511.50Real PartDifferent function, zero zero and pole poleFrom these p lots we make the followi ng observati ons figure with different relative

13、 p osition of the circle 4.2 TYPES OF TRANSFER FUNCTIONSP roject 4.2FiltersA copy of P rogram P 4_1 is give 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); xIabelC时间序号 n); ylabel(振幅);gridAn swers:N=13Q4.7 The plot of the imp ulse response

14、of the appr oximati on to the ideal low pass filter obta ined using P rogram P 4_1 is show n below:clf;fc=0.25;n=-6.5:1:6.5;y=2*fc*si nc(2*fc* n);k=n+6.5; stem(k,y);title(N=13);axis(0 13 -0.2 0.6); xIabelC时间序号 n);ylabelC振幅);gridn=-6.5:1:6.5;-Fc p arameters con trol the cutoffThe len gth of the FIR l

15、ow pass filter is -14The stateme nt in P rogram P 4_1 deter mining the filter len gth isThe p arameter con trolli ng the cutoff freque ncy isfreque ncyQ4.8 The required modifications to response of the FIR low pass filter freque ncy of 汪=0.45Pr ogram P4_1 of Project 4.2 are as in dicated below :to c

16、ompute and plot the impulsewith a length of 20 and a cutoffclf;fc=0.45/(2* pi);n=-9.5:1:9.5;y=2*fc*s in c(2*fc* n);k=n+9.5;stem(k,y);title(N=20);axis(0 20 -0.2 0.6); xIabelC时间序号 n);ylabel(振幅);gridThe plot gen erated by running the modified pr ogram is give n belowN=20Q4.9 The required modifications

17、to Program P4_1 response of the FIR low pass filter of P roject 4.2 freque ncy of % = 0.65 are as in dicated below :to compute andwith a length ofplot the impulse15 and a cutoffclf;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);

18、ylabel(振幅);gridThe plot gen erated by running the modified pr ogram is give n below0.6!o.2-0.20JJrN=150.50.40.30.10-0.12410126 8时间序号n14Q4.11using P rogram P 4_2 is show n belowA plot of the gai n response of a len gth-2 movi ng average filter obta inedunction g,w=gain(num,den) -gain 函数w=0: pi/255: p

19、i; 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);p lot(w/pi,g);grid;axis(0 1 -50 0.5)xlabel(omega八pi);ylabel(单位为 db 的增益); title(M= ,nu m2str(M)0 5 0 52 2 3 3 - - - - 益增Soo为位单-40-45-5000.10.20.30.40.50.60.7C/H0.80.9From the plot it can be see n that

20、 the 3-dB cutoff freque ncy is at-3dB。Q4.12The required modificationsto Program P4gain res ponse of a cascade ofto compute and plot theK length-2 moving average filters are given below:The plot of the gain response for a cascade of 3 sections obtained using the modified pr ogram is show n below:-5-1

21、0-15-20-25-30-35-40-45Q4.19-5000.10.20.30.40.50.60.70.80.9From the plot it can be see n that the 3-dB cutoff freque ncy of the cascade is atwc=0.3* piA copy of Pr ogram P 4_3 is give n belowclf;b=1 -8.5 30.5 -63;num1=b 81 fli plr(b);num2=b 81 81 fli plr(b);num3=b 0 -fli plr(b);num4=b 81 -81 -fli pl

22、r(b);n1=0:le ngth( num1) -1;n2=0:le ngth( num2) -1;sub plot(2,2,1); stem( n1,n um1);振幅);grid;xlabel(时间序号n); ylabel(ti tle(1型有限冲激响应滤波器);sub plot(2,2,2); stem( n2,n um2);xlabel(时间序号n); ylabel(振幅);grid;ti tle(2型有限冲激响应滤波器);sub plot(2,2,3); stem( n1,n um3);xlabelC 时间序号 n); ylabel(振幅);grid;title(3型有限冲激响应滤

23、波器);sub plot(2,2,4); stem( n2,n um4);xlabel(时间序号n); ylabel(振幅);grid;title(4型有限冲激响应滤波器);p ausesub plot(2,2,1); zplane(n um1,1);ti tle(1型有限冲激响应滤波器);sub plot(2,2,2); zplane(n um2,1);ti tle(2型有限冲激响应滤波器);sub plot(2,2,3); zplane(n um3,1);ti tle(3型有限冲激响应滤波器);sub plot(2,2,4); zplane(n um4,1);ti tle(4型有限冲激响应

24、滤波器);dis p(1型有限冲激响应滤波器的零点是);dis p(r oots (n um1);dis p(2型有限冲激响应滤波器的零点是);dis p(r oots (n um2);dis p(3型有限冲激响应滤波器的零点是);dis p(r oots (n um3);dis p（4型有限冲激响应滤波器的零点是）;dis p( roots (n um4);The pl ots of the imp ulse respon ses of the four FIR filters gen erated by P rogram P 4_3 are give n below :running101

25、0050-50-100 0时间序号n3型有限冲激响应滤波器nT246时间序号n5时间序号n10From the pl ots we make the follow ing observati onsFilter #1 is of le ngth 8Typelinear-p hase FIR filterwith a80i mp ulse res ponse and is therefore aFilter #2 is of le ngth 10_a Typelin ear -p hase FIR filterwith a80imp ulse res ponse and is therefore

26、Filter #3 is of le ngth 8Typelinear-p hase FIR filterwith a60i mp ulse res ponse and is therefore aFilter #4 is of le ngth 10a Typelin ear -p hase FIR filterwith a80imp ulse res ponse and is thereforeFrom the zeros of these filters gen erated by P rogram P 4_3 we observe thatFilter #1 has zeros at z

27、 =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 - 0.3530i,0.2289 +4.7627 ,1.6279 + 3.0565i,1.6279 - 3.0565i,-0.0922i,0.2289

28、 - 0.0922iFilter #3 has zeros at z1.0000 ,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,0.3159 - 0.3020i,0.2929P lots of the p hase response of each of these filters obta

29、ined using MATLAB are shown below :1型有限冲激响应滤波器2型有限冲激响应滤波器xurap ntrap ya卩昌卩1 18 ,110-1-130 1 2Real Part3型有限冲激响应滤波器a P y a m；9 - 丿X1 !210-1-2-240 2Real Part4型有限冲激响应滤波器-8-20-2-26y a n g mJ9 1B10-1-13024Real Part0 1 2Real PartFrom these pl ots we con clude that each of these filters have phase .lin earQ

30、4.20The pl ots of the imp ulse res pon ses of the four FIR filters gen erated byrunning P rogram P 4_3 are give n below :1型有限冲激响应滤波器2型有限冲激响应滤波器-1-123型有限冲激响应滤波器时间序号n4型有限冲激响应滤波器时间序号nFrom the pl ots we make the follow ing observati onsFilter #1 is of le ngth 8Type linear-phase FIR filterwith a80i mp ul

31、se res ponse and is therefore aFilter #2 is of le ngth 10_a Typelin ear -p hase FIR filterwith a80imp ulse res ponse and is thereforeFilter #3 is of le ngth 8Typelinear-p hase FIR filterwith aimp ulse response and is therefore aFilter #4 is of le ngth 10_a Typelin ear -p hase FIR filterwith a80imp u

32、lse res ponse and is thereforeFrom the zeros of these filters gen erated by P rogram P 4_3 we observe thatFilter #1 has zeros at z = 2.3273 + 2.0140i,2.3273 - 2.0140i,-1.2659 +2.0135i,-1.2659 - 2.0135i, -0.2238 + 0.3559i,-0.2238 - 0.3559i,0.2457 + 0.2126i,0.2457 - 0.2126iFilter #2 has zeros at z =2.

33、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 + 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

34、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 - 0.2399iP lots of the p hase response of each of these filters obta ined using MATLAB are shown below :1型有限冲激响应滤波器2型有限冲激响应滤波器r a P y a n g m/ 0 0

35、X.1 ,1 /10210-1-2-2r a P y a n g mX-1/宀1 9二卩 丿a丨C71/、一-广210-1-2-20 2Real Part3型有限冲激响应滤波器0 2Real Part4型有限冲激响应滤波器AZ(IA,0、 8 II.、4-11j1广X二 / .11 JU J10.50-0.51aPya n gmC-I9、 r；/ 、20 丁(-210-1-2-20Real Part0Real PartFrom these pl ots we con clude that each of these filters have phase .linearAn swers:4.3STABILITY TESTA copy of P rogram P4_4 is give n below:clf;den=input(分母系数=;ki=po ly2rc(de n);disp（稳定