利用matlab进行信号与系统分析实验.pdf

该【利用matlab进行信号与系统分析实验 】是由【小屁孩】上传分享，文档一共【17】页，该文档可以免费在线阅读，需要了解更多关于【利用matlab进行信号与系统分析实验 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:..1【实验1】利用matlab进行信号的时域分析(1)指数信号f(t)?Ae?t>>A=1;>>a=-;1>>t=0::10;>>ft=A*exp(a*t);>>plot(t,ft);>>grid;>>axis([010-];00**********>>xlabel('t')t>>ylabel('ft')(2)正弦信号f(t)?Asin(?t??>>A=1;)>>w0=2*pi;>>phi=pi/6;1>>t=0::3;>>ft=A*sin(w0*t+phi);>>plot(t,ft);tftf0->>grid;-->>axis([03-]);--1>>xlabel('t')>>ylabel('ft')1:..2(3)抽样信号sin(t)f(t)?Sa(t)?t>>x=linspace(-20,20);1>>y=sinc(x/pi);>>plot(x,y);>>grid;>>axis([-2121-]);y>>xlabel('x')0>>ylabel('y')--20-15-10-505101520x(4)矩形脉冲信号f(t)?G(t)?A[?(t??)??(t??)]?01>>t=0::4;>>T=1;1>>ft=rectpuls(t-2*T,2*T);>>plot(t,ft);>>grid;>>axis([-15-]);>>xlabel('t')0-1012345>>ylabel('ft')t2:..3(5)阶跃信号?1(t?0)?(t)??0(t?0)?functionft=heaviside(t)ft=(t>0);>>t=-1::3;1>>ft=heaviside(t);>>plot(t,ft);>>grid;>>axis([-13-]);>>xlabel('t')0-1->>ylabel('ft')t(6)复指数信号的时域波形2f(t)?e?(t)3>>t=0::60;1>>f=exp(-*t).*sin(2/3*t);>>plot(t,f);>>grid;)t)t0(f(f->>axis([060-11]);->>xlabel('Time(sec)')-->>ylabel('f(t)')-10102030405060Time(sec)3:..4(7)加入随机噪声的正弦波>>t=0::50;1>>y=sin(2*pi*50*t);>>s=y+randn(size(t));0>>subplot(2,1,1);->>plot(t(1:100),y(1:100));->>grid;4>>subplot(2,1,2);2>>plot(t(1:100),s(1:100));0>>grid;-2-(8)>>A=1;1>>t=0::5;>>y=A*square(2*pi*t,20);0>>plot(t,y);>>grid;->>axis([05-]);-1-:..5(9)信号的基本运算>>symst;>>f=sym('(t/2+1)*(heaviside(t+2)-heaviside(t-2))');>>subplot(3,2,1),ezplot(f,[-3,3]);>>title('f(t)');f(t)f(t+2)>>grid;22>>y1=subs(f,t,t+2);11>>subplot(3,2,2),ezplot(y1,[-5,1]);00>>title('f(t+2)');-202-4-20tt>>grid;f(t-2)f(-t)>>y2=subs(f,t,t-2);22>>subplot(3,2,3),ezplot(y2,[-1,5]);11>>title('f(t-2)');00>>grid;024-202tt>>y3=subs(f,t,-t);f(2t)>>subplot(3,2,4),ezplot(y3,[-3,3]);2>>title('f(-t)');1>>grid;0-2-1012>>y4=subs(f,t,2*t);t>>subplot(3,2,5),ezplot(y4,[-2,2]);>>title('f(2t)');>>grid;5:..6【实验2】利用matlab进行系统的时域分析例1求系统y”(t)+2y’(t)+100y(t)=10f(t)的零状态响应,已知f(t)=(sin2πt)ε(t)。>>t=0::5;2>>sys=tf([10],[12100]);>>f=10*sin(2*pi*t);>>y=lsim(sys,f,t);0)t)t>>plot(t,y);(y(y->>grid;-1->>xlabel('Time(sec)')-2>>ylabel('y(t)')-(sec)例2求系统y”(t)+2y’(t)+100y(t)=10f(t)的零状态响应,已知f(t)=δ(t)。>>t=0::5;1>>sys=tf([10],[12100]);>>y=impulse(sys,t);>>plot(t,y);>>xlabel('Time(sec)'))t)t(h(h0>>ylabel('h(t)')-->>grid;--(sec)6:..7【实验3】利用matlab进行信号的频域分析[例1]实用matlab画出图示周期三角波信号的频谱f(t)1t-2-1012-1?4jn???sin()n?0C??n2?22n????0n?>>N=8;>>n1=-N:-1;>>c1=-4*j*sin(n1*pi/2)/pi^2./n1.^2;0-8-6-4-202468>>c0=0;2>>n2=1:N;1位>>c2=-4*j*sin(n2*pi/2)/pi^2./n2.^2;相0的n>>cn=[c1c0c2];C-1-2>>n=-N:N;-8-6-4-202468?/?0>>subplot(2,1,1);>>stem(n,));>>ylabel('Cn的幅度');>>subplot(2,1,2);>>stem(n,));>>ylabel('Cn的相位');>>xlabel('\omega/\omega0');7:..8[例2]求周期矩形脉冲的Fourier级数表达式。并用matlab求出由前N项Fourier级数系数得出的信号近似波形。fT(t)A??t?T?T22?A??A?2?An??C?Sa(n??/2)?jn?t?0?n0f(t)?Ce0??Sa()cosntTn0TT2n???n?1取A=1,T=2,τ=1,ω0=πN?f(t)??Sa(n?/2)cos(n?t)Nn?1t=-2::2;N=5N==input('Numberofharmonics=');=;=c0*ones(1,length(t));=1:2:N---2--1--2--1-=xN+cos(pi*n*t)*sinc(n/2);N=50N=(t,xN);(['N='num2str(N)]);00---2--1--2--1-:..9[例3]试用数值方法计算三角波信号的频率f(t)11t?101解:图示三角波可表示为f(t)?(1?t),t?11三角波信号频谱的理论值为F(jw)=Sa2(w/2)functiony=sf1(t,w);y=(t>=-1&t<=1).*(1-abs(t)).*exp(-j*w*t);>>w=linspace(-6*pi,6*pi,512);>>N=length(w);>>F=zeros(1,N);>>fork=1:NF(k)=quad('sf1',-1,1,[],[],w(k));end>>figure(1);1>>plot(w,real(F));>>title('')>>xlabel('\omega');)?(j(F>>ylabel('F(j\omega)');--20-15-10-505101520?9:..10【实验四】利用matlab进行系统的频域分析例1三阶归一化的Butterworth低通滤波器的系统函数为1H(?)?试画出|H(w)|和?(w)。(j?)3?2(j?)2?2(j?)?1>>w=linspace(0,5,200);>>b=[1];>>a=[1221];>>h=freqs(b,a,w);>>subplot(2,1,1);42>>plot(w,abs(h));0-2>>subplot(2,1,2);->>plot(w,angle(h));[例2]周期方波通过RC系统的响应。f(t)1Ry(t)f(t)Ct?5?3?101351/j?C11H(?)??cos(n?t??)R?1/j?C1?j?RC1?jnRC?00A??n???c?Sa?0?????nnT2????ejn?t???0y(t)????nRe??1?jnRC???n?1010:..11>>T=4;>>w0=2*pi/T;>>RC=;>>t=-6::6;--5-3-10135>>N=51;timeRC=>>c0=;>>xN=c0*ones(1,length(t));>>forn=1:2:-5-3-10135H=abs(1/(1+j*RC*w0*n));timeRC=1phi=angle(1/(1+j*RC*w0*n));xN=xN+H*cos(w0*n*t+phi)*sinc(n*);end>>plot(t,xN);>>xlabel(['timeRC=',num2str(RC)]),grid;>>set(gca,'xtick',[-5-3-10135]);例3>11:..12>b=2;a=[156];100>>w=logspace(-1,1);edudutiti10-1ngngaMaM>>figure(1);freqs(b,a,w);10-210-1100101Frequency(rad/s)>>H=freqs(b,a,w);0)s)s-50eeeergrg>>mag=abs(H);eded-100(e(e(esasa-150hPhP-200>>phase=angle(H)*180/pi;10-1100101Frequency(rad/s)>>figure(2);100>>subplot(2,1,1),loglog(w,mag)ededutut10-1iningagaM>>xlabel('Frequency(rad/s)');10-210-1100101Frequency(rad/s)>>ylabel('Magnitude');grid;0)s)s-50eeeergrg>>subplot(2,1,2),semilogx(w,phase)eded-100(e(esasah-150P>>xlabel('Frequency(rad/s)');-20010-1100101Frequency(rad/s)>>ylabel('Phase(degrees)');grid;>>figure(3);>>subplot(2,1,1),plot(w,mag)00**********Frequency(rad/s)0>>xlabel('Frequency(rad/s)');)s)s-50eeeergrgeded-100(e(esasahPhP-150>>ylabel('Magnitude'),grid-2000**********Frequency(rad/s)>>subplot(2,1,2),plot(w,phase)>>xlabel('Frequency(rad/s)'),ylabel('Phase(degrees)'),grid12:..13练****某一LTI连续时间系统由y(3)(t)?8y''(t)?6y'(t)?4y(t)?10x'(t)?5x(t)表征,求该系统的频率响应。>>b=[105];>>a=[18-64];101>>w=logspace(-1,1);e100dutiti>>H=freqs(b,a,w);ng-1a10M>>mag=abs(H);10-210-1100101Frequency(rad/s)>>phase=angle(H)*180/pi;200)s)s100>>subplot(2,1,1),loglog(w,mag)eergrge0d(e(es>>xlabel('Frequency(rad/s)');a-100hP-20010-1100101>>ylabel('Magnitude'),gridFrequency(rad/s)>>subplot(2,1,2),semilogx(w,phase)>>xlabel('Frequency(rad/s)');>>ylabel('Phase(degrees)'),grid13:..14【实验五】利用matlab进行连续系统的复频域分析例1进行Laplace变换>>symstsazw>>f1=t^4;>>L1=laplace(f1)L1=24/s^5>>f2=1/sqrt(s);>>L2=laplace(f2)L2=pi^(1/2)/t^(1/2)>>f3=exp(-a*t);>>L3=laplace(f3,z)L3=1/(z+a)>>f4=abs(t)*exp(-a*abs(w));>>L4=laplace(f4,w,z)L4=abs(t)/(z+a)14:..15例22s/(s2+1)2>>symss;>>F=sym('2*s/(s^2+1)^2');->>f=ilaplace(F);-4-3-2-101234stsin(t)2>>subplot(2,1,1);ezplot(F);0-2>>subplot(2,1,2);ezplot(f);-4-6-4-20246ts?2F(s)?[例1]用部分分式展开法求F(s)的反变换s3?4s2?3s>>formatrat>>num=[12];>>den=[1430];>>[r,p]=residue(num,den)r=-1/6-1/22/3p=-3-102/3??1/6F(s)???故F(s)可展开为ss?1s?3211f(t)?u(t)?e?tu(t)?e?3tu(t)326[例2]用部分分式展开法求F(s)的反变换2s3?3s2?5F(s)?(s?1)(s2?s?2)>>num=[2305];15:..16>>den=conv([11],[112]);>>[r,p,k]=residue(num,den)>>magr=abs(r)>>angr=angle(r)r=-2+2024/1785i-2-2024/1785i3p=-1/2+1012/765i-1/2-1012/765i-1k=2magr=7895/34347895/34343angr=1972/751-1972/75101[例3]试画出系统H(s)?的零极点分布图,求其单位冲s3?2s2?2s?1激响应h(t)和频率响应H(jw)。ImpulseRespone>>num=[1];den=[1221];>>sys=tf(num,den);->>poles=roots(den)-4-3-2-101234sPole-ZeroMappoles=--1/2+1170/1351ininigagam--1-1----------1/2-1170/1351iRealAxis>>figure(1);pzmap(sys);>>t=0::10;>>h=impulse(num,den,t);00>>figure(2);plot(t,h)>>title('ImpulseRespone')1600?:..17>>[H,w]=freqs(num,den);>>figure(3);plot(w,abs(H))>>xlabel('\omega')>>title('MagnitudeRespone')练****gt;>num=[1-45];>>den=[32-46];sixAsixAsixAsixAyrayrayrayra0nignignigamam>>sys=tf(num,den);I>>poles=roots(den)0poles=010--+--210-101101010Frequency(rad/s)>>figure(1),pzmap(sys);50>>w=logspace(-1,1);)see0rgrge>>figure(2),freqs(num,den,w);d((e-50sahP-10010-1100101Frequency(rad/s)17