1、哈工大信号与系统第三次上机实验实验五 连续系统的频域分析及连续信号的采样与重构二、利用MATLAB 分析系统频响特性实验内容图5-4所示的电路为最平坦幅度型二阶低通滤波器。试用MATLAB 程序画出系统响应(1(2(j U j U j H =的幅度响应及相频响应,并与理论分析的结果进行比较。(j H 的截止频率 0=b=0 0 1;a=2 2.82 2;h,w=freqs(b,a,100;h1=abs(h;h2=angle(h;subplot(211;plot(w,h1;gridxlabel(角频率(W;ylabel(幅度;title(H(jw的幅频特性;subplot(212;plot(w,
2、h2*180/pi;gridxlabel(角频率(w;ylabel(相位(度;title(H(jw的相频特性; U1(tU2(t图5-4 二阶低通滤波器电路图01234567891000.20.40.6 0.8角频率(W幅度H(jw的幅频特性012345678910-200-150-100-50 角频率(w相位(度H(jw的相频特性三、连续信号的采样与重构实验内容 设t e t f 1000(-=,由于不是严格的带限信号,但其带宽m 可根据一定的精度要求做一近似。试根据以下三种情况用MATLAB 实现由(t f 采样信号(t f s 重构(t f 并求出两者误差,分析三种情况下的结果。(150
3、00=m ,m c =,m s T /=;运行程序:wm=5000*pi;wc=1*wm;Ts=pi/wm;ws=2*pi/Ts;n=-100:100;nTs=n*Ts;f=exp(-1000.*abs(nTs;Dt=0.005;t=-15:Dt:15;fa=f*Ts*wc/pi*sinc(wc/pi*(ones(length(nTs,1*t-nTs*ones(1,length(t; error=abs(fa-exp(-1000.*abs(t;t1=-15:0.5:15;f1=exp(-1000.*abs(t1;subplot(311;stem(t1,f1;ylabel(f(kTs;title
4、(exp(-1000.*abs(t的采样信号;subplot(312;plot(t,faylabel(fa(t;title(由exp(-1000.*abs(t的采样信号重构;grid;subplot(313;plot(t,error;ylabel(error(t;title(采样信号与原信号的误差error(t; f (k T s -15-10-5051015-101f a (t 由exp(-1000.*abs(t的采样信号重构 -15-10-5051015012-11e r r o r (t 过采样信号与原信号的误差error(t(210000=m ,m c 1.1=,m s T /=;运行
5、程序:wm=10000*pi;wc=1.1*wm;Ts=pi/wm;ws=2*pi/Ts;n=-100:100;nTs=n*Ts;f=exp(-1000.*abs(nTs;Dt=0.005;t=-15:Dt:15;fa=f*Ts*wc/pi*sinc(wc/pi*(ones(length(nTs,1*t-nTs*ones(1,length(t; error=abs(fa-exp(-1000.*abs(t;t1=-15:0.5:15;f1=exp(-1000.*abs(t1;subplot(311;stem(t1,f1;ylabel(f(kTs;title(exp(-1000.*abs(t的采样
6、信号;subplot(312;plot(t,faylabel(fa(t;title(由exp(-1000.*abs(t的过采样信号重构;grid;subplot(313;plot(t,error;ylabel(error(t;title(过采样信号与原信号的误差error(t;f (k T s exp(-1000.*abs(t的采样信号 -15-10-5051015-202f a (t 由exp(-1000.*abs(t的过采样信号重构 -15-10-505101500.0050.01e r r o r (t 过采样信号与原信号的误差error(t(32500=m ,m c 9.0=,m s
7、T /=;运行程序:wm=2500*pi;wc=0.9*wm;Ts=pi/wm;ws=2*pi/Ts;n=-100:100;nTs=n*Ts;f=exp(-1000.*abs(nTs;Dt=0.005;t=-15:Dt:15;fa=f*Ts*wc/pi*sinc(wc/pi*(ones(length(nTs,1*t-nTs*ones(1,length(t; error=abs(fa-exp(-1000.*abs(t;t1=-15:0.5:15;f1=exp(-1000.*abs(t1;subplot(311;stem(t1,f1;ylabel(f(kTs;title(exp(-1000.*ab
8、s(t的采样信号;subplot(312;plot(t,faylabel(fa(t;title(由exp(-1000.*abs(t的采样信号重构;grid;subplot(313;plot(t,error;ylabel(error(t;title(采样信号与原信号的误差error(t;f (k T s exp(-1000.*abs(t的采样信号 -15-10-5051015-101f a (t 由exp(-1000.*abs(t的采样信号重构 -15-10-505101500.010.02e r r o r (t 采样信号与原信号的误差error(t实验六 拉普拉斯变换及其逆变换实验内容1、求
9、解下述信号的拉普拉斯变换,并利用MATLAB 绘制拉普拉斯变换的曲面图:(1(2cos(t u t t f =a=-0.5:0.08:0.5; b=-1.99:0.08:1.99;a,b=meshgrid(a,b;d=ones(size(a; c=a+i*b; e=c.*c; f=e+d; c=c./f; c=abs(c; mesh(a,b,c; surf(a,b,c;axis(-0.5,0.5,-2,2,0,15;title(单边余弦信号拉氏变换曲面图; colormap(hsv; -0.50.5单边余弦信号拉氏变换曲面图(2(sin(2t u t e t f t -= a=-10:0.1:
10、5; b=-5:0.1:5;a,b=meshgrid(a,b;c=a+i*b;c=(c+2.*(c+2; c=1./(c+1; c=abs(c; mesh(a,b,c; surf(a,b,c;axis(-10,5,-5,5,0,6;title(正弦信号拉氏变换曲面图; colormap(hsv; -105正弦信号拉氏变换曲面图2、已知连续时间信号(3t u e t f t -=,试求出该信号的拉普拉斯变换(s F 和傅立叶变换(j F ,用MATLAB 绘出拉普拉斯变换曲面图(s F 及幅频曲线(j F ,观察曲面图在虚轴上的剖面图,并将其与幅频曲线相比较,分析频域与复频域的对应关系。 x1=
11、-10:0.5:10; y1=-10:0.5:10;x,y=meshgrid(x1,y1; s=x+i*y;fs=abs(1./(s+3; mesh(x,y,fs; surf(x,y,fs;title(单位阶跃信号拉氏变换曲面图; colormap(hsv;axis(-10,10,-10,10,0,2; rotate3d; -1010单位阶跃信号拉氏变换曲面图b=0 0 1; a=1 6 9;h,w=freqs(b,a,100; h1=abs(h; subplot(1,1,1; plot(w,h1; gridxlabel(角频率(W; ylabel(幅度;title(H(jw的幅频特性;012
12、3456789100.020.040.060.080.1 0.12角频率(W幅度H(jw的幅频特性w=-20:0.1:20;Fw=1./(i*w+3.*(i*w+3; plot(w,abs(Fwtitle(傅里叶变换(振幅频谱曲线 xlabel(频率w-20-15-10-50510152000.020.040.060.080.1 0.12傅里叶变换(振幅频谱曲线频率w3、已知信号的拉普拉斯变换如下所示,试用MATLAB 绘制曲面图,观察拉普拉斯变换零极点分布对曲面图的影响。 (13(2(4(1(+=s s s s s s Fclf;a=-6:0.48:6; b=-6:0.48:6;a,b=me
13、shgrid(a,b; c=a+i*b;d=(c+1.*(c+4; e=c.*(c+2.*(c+3; c=d./e; c=abs(c; mesh(a,b,c; surf(a,b,c;axis(-6,6,-6,6,0,4.5;title(拉普拉斯变换曲面图; colormap(hsv; view(-25,30 -666拉普拉斯变换曲面图(244(22+-=s s s F clf;a=-6:0.48:6; b=-6:0.48:6;a,b=meshgrid(a,b; c=a+i*b; d=c.*c-4; d=c.*c+4; c=d./e; c=abs(c; mesh(a,b,c; surf(a,b,
14、c; axis(-6,6,-6,6,0,16; title(拉普拉斯变换曲面图; colormap(hsv; view(-25,30 拉普拉斯变换曲面图 15 10 5 0 6 4 2 0 -2 -4 -6 -6 -4 -2 0 2 4 6 4、试用 MATLAB 求下列信号的拉普拉斯逆变换 （1） F ( s = s + 5s + 4 2 s + 5s + 6s 3 2 运行程序： a=1 5 6 0; b=1 5 4; r,p,k=residue(b,a 运行结果： r= -0.6667 1.0000 0.6667 p= -3.0000 -2.0000 0 k= （2） F ( s = 1 s + 2s + 2s + 1 3 2 运行程序： a=1 2 2 1; b=1; r,p,k=residue(b,a 运行结果： r= 1.0000 -0.5000 - 0.2887i -0.5000 + 0.2887i p= -1.0000 -0.5000 + 0.8660i -0.5000 - 0.8660i k=
