matlab在信号与系统系统中

Matlab在信号与系统中的应用1、连续时间信号的波形内容：1、绘制以下信号的波形：a.3e4tplot(t,y)606050403020109x109x105876543210b.5cos(2t)3 4t=-pi:0.01:pi;plot(t,y)543210-1-2-3-4-5-4 -3 -2 -1 0 1 2 3 4Sa(t)y=sin(t)./t;plot(t,y)10.80.60.40.20-0.2-0.4-10 -8 -6 -4 -2 0 2 4 6 8 10t=-5:0.1:5;x=(t>=0);plot(t,x)10.90.80.70.60.50.40.30.20.10-5 -4 -3 -2 -1 0 1 2 3 4 5y=(t==0);plot(t,y)10.90.80.70.60.50.40.30.20.10-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 12、连续时间信号的时域分析内容：1、绘制以下信号的波形：ysincost 3 3t=-5:0.01:5;y=sin(5*pi*t)+cos(pi/3*t);plot(t,y)21.510.50-0.5-1-1.5-2-5 -4 -3 -2 -1 0 1 2 3 4 54ye2tsin2t4 t=-5:0.01:5;x104x10421.510.50-0.5-1-5 -4 -3 -2 -1 0 1 2 3 4 52)

u(t)x(t)y(t)的图形，t-35，0.01。t=-3:0.01:5;z=(t>0);x=exp(-0.5*t).*z;y=x.*(1.5*t+3);plot(t,x,t,y)32.521.510.50-3 -2 -1 0 1 2 3 4 53

f(t)eatu(t)(a=1f2t)sintut)f(t)f1(tf2(t)1的时间范围取为0-10，步长值取为0.1。绘制三个信号的波形。1function[w,tw]=conv1(u,tu,v,tv)T=tu(2)-tu(1);w=T*conv(u,v);tw=tu(1)+tv(1)+T*[0:length(u)+length(v)-2]';t=[0.1:0.1:10]';a=1;u=heaviside(t);f1=exp((-a).*t).*u;f2=sin(t).*u;[f,t1]=conv1(f1,t,f2,t);plot(t,f1,'r',t,f2,'g',t1,f)10.80.60.40.20-0.2-0.4-0.6-0.8-10 2 4 6 8 10 12 14 16 18 203、连续时间系统的时域分析内容：1、 已知系统的微分方程为r''(t)4r'(t)3r(t)e(t)当系统的输入信号为e(t)4cos(2t)，求系统的零状态响应，并画出图形。a=[1,4,3];b=[1];sys=tf(b,a);lsim(sys,e,t)LinearSimulationResultsInput:In(1)LinearSimulationResultsInput:In(1)Time(sec):0.755Amplitude:0.111321p0pmA-1-2-3-40 1 2 3 4 5Time(sec)

6 7 8 9 10r''(t4r'(t3r(te'(t2e(t，求系统的冲激响应，同时求出理论解，绘图比较。a=[1,4,3];b=[1,2];sys=tf(b,a);t=[0:0.1:5]';[h1,t1]=impulse(sys,t);plot(t1,h1)10.90.80.70.60.50.40.30.20.100 0.5 1 1.5 2 2.5 3 3.5 4 4.5 53.求下列微分方程描述的系统的阶跃响应，同时求出理论解，绘图比较。r''(t)'(t)2r(t)e'(t)3e(t)a=[1,3,2];b=[1,3];sys=tf(b,a);t=[0:0.1:5]';step(sys)StepResponse1.5StepResponse1ppmA0.500 1 2 3Time

4 5 64、连续时间信号的频谱分析内容：1、求以下信号的傅里叶变换及绘制频谱图a.f(t)cost 2 2symstF=cos((pi*t)/2);f=fourier(F);w=[0:0.1:5];f=pi*(dirac(-pi/2-w)+dirac(pi/2-w));x=abs(f);y=angle(f);plot(w,x,'*',w,y,'p')10.80.60.40.20-0.2-0.4-0.6-0.8-10 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5b.f(t)2e0.4tu(t)symstu=heaviside(t);F=2*exp(-2/5*t).*u;f=fourier(F);w=[0:0.1:3];f=2./(i*w+2/5);x=abs(f);y=angle(f);plot(w,x,w,y)543210-1-200.511.522.53c.f(t)u(t2)u(t2)symstf=fourier(heaviside(t+2)-heaviside(t-2));f=(i./w-pi.*dirac(-w))./exp(2*i*w)-exp(2*i.*w).*(i./w-pi*dirac(-w));abs(f);w=0:0.1:5;angle(f);plot(w,abs(f),w,angle(f))43210-1-2-3-40 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5d.f(t)etsymstF=exp(-abs(t));f=fourier(F);f=2./(w.^2+1);abs(f);w=0:0.1:5;angle(f);plot(w,abs(f),'*',w,angle(f),'r')21.81.61.41.210.80.60.40.200 0.5 1 1.5 2 2.5 3 3.54 4.555、连续时间系统的s域分析内容：1．a. 3s(s4)(s2)symssS=(3*s)/((s+4)*(s+2));f=ilaplace(S)f=6/exp(4*t)-3/exp(2*t)b.symss

4s5s25s6S=(4*s+5)/(s^2+5*s+6);f=ilaplace(S)f=7/exp(3*t)-3/exp(2*t)2.LTIa.r'(t3r(t)3e(t)a=[1,3];b=[3];freqs(b,a)uM

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101010100-20-40-60-8010

100Frequency(rad/s)100Frequency(rad/s)

101101b.3r''(t)4r'(t)r(t)e''(t)5e(t)a=[3,4,1];b=[1,0,5];freqs(b,a)uM

102100101010-2 10-1 100 101Frequency(rad/s)e(eaP

500-50-100-150

10

Frequency

100

101分别画出它们的系统频率响应的幅值和相位特性曲线。3、有一模拟滤波器，其传递函数为：H(s)

0.2s20.3s1s20.4s1用freqs函数画出它的幅频特性和相位特性曲线。b=[0.2,0.3,1];a=[1,0.4,1];freqs(b,a)uM10-1 Frequency(rad/s)e(hP

0-20-40-60-80-100-12010-1

Frequency(rad/s)4.已知系统函数H（s）为：s1H(s)

s27s10求该系统的零极点，并画出零极点分布图。b=[1,1];a=[1,7,10];pzmap(sys)MapMap0.80.6iAmI

0.40.20-0.2-0.4-0.6-0.8-1-5 -4.5 -4 -3.5 -3 -2.5RealAxis

-2 -1.5 -1 -0.5 06、离散时间信号的产生与运算内容：n=[-10:10]范围内产生离散信号：2nf(n)0

3n3其余n=[0:30]范围内画出下列信号：naf(n)cos2( )2 43、画出如下信号的波形 a.ysinn cos 2 3 4b.y(2)nsinn4 4、两个有限长序列分别如下图所示，求两个信号的卷积和。x(x(n)13221121 2nx(n)243211101 23n