MATLAB仿真及電子信息應(yīng)用答案

1、MATLAB仿真及電子信息應(yīng)用第二單元1. >> a=zeros(2)a = 0 0 0 0>> a(1,2)=2a = 0 2 0 02. >> a=1 0 -2 3;b=0 1 2 1;>> a+bans = 1 1 0 4>> a.*bans = 0 0 -4 3 /-4+3 =-1>> dot(a,b)ans =-13. >> a=randn(4)a = 0.8156 1.1908 -1.6041 -0.8051 0.7119 -1.2025 0.2573 0.5287 1.2902 -0.0198 -

2、1.0565 0.21930.6686 -0.1567 1.4151 -0.9219>> sum(diag(a)ans = -2.3652>> reshape(a,2,8)ans = 0.8156 1.2902 1.1908 -0.0198 -1.6041 -1.0565 -0.8051 0.21930.7119 0.6686 -1.2025 -0.1567 0.2573 1.4151 0.5287 -0.92194. >> a=rand(3,5)a = 0.9501 0.4860 0.4565 0.4447 0.9218 0.2311 0.8913 0.0

3、185 0.6154 0.73820.6068 0.7621 0.8214 0.7919 0.1763>> a(:,4)=a(:,4)+0.2a = 0.9501 0.4860 0.4565 0.6447 0.9218 0.2311 0.8913 0.0185 0.8154 0.73820.6068 0.7621 0.8214 0.9919 0.1763>> lt(a,0.5)ans = 0 1 1 0 0 1 0 1 0 0 0 0 0 0 15. >> a=randn(3,2);b=randn(3,2);>> cat(1,a,b) /按列方向

4、ans = 0.7310 0.6771 0.5779 0.5689 0.0403 -0.2556 -0.3775 -0.2340 -0.2959 0.1184 -1.4751 0.3148>> cat(2,a,b) /按行方向ans = 0.7310 0.6771 -0.3775 -0.2340 0.5779 0.5689 -0.2959 0.11840.0403 -0.2556 -1.4751 0.31486. >> 'Hello everyone'ans =Hello everyone7. >> str1='MATLAB '

5、;str2 ='is strong'str3=str1,str2str3 =MATLAB is strong8.(1)直接法>> a='Function' 'Variable''y=f(x)' 'x''6' 1:5a = 'Function' 'Variable' 'y=f(x)' 'x' '6' 1x5 double(2)函數(shù)法>> a=cell(3,2);a1,1='Function&

6、#39;a1,2='Variable'>> a2,1='y=f(x)'a2,2='x'>> a3,1='6'a3,2=1:5;>> aa = 'Function' 'Variable' 'y=f(x)' 'x' '6' 1x5 double9. >>student1.Name='Mary'student1.Age='Eighteen'>>student1.Sco

7、re='502'student1.Class='1'student1student1 = Name: 'Mary' Age: 'Eighteen' Score: '502'Class: '1'>> student2.Name='Mike'student2.Age='Nighteen'>> student2.Score='498'student2.Class='2' student2student2 = Name:

8、 'Mike' Age: 'Nighteen' Score: '498'Class: '2'>> student3.Name='Joe'student3.Age='Eighteen'>> student3.Score='520'student3.Class='3'student3student3 = Name: 'Joe' Age: 'Eighteen' Score: '520' Class: &

9、#39;3'第三單元3-1clear allclcp=5 4 1 2 1;roots(p) %求根y=polyval(p,5) %求在5處函數(shù)的函數(shù)值3-2clear allclcp=2 0 -2 5;x=1 2 4;-1 0 3;5 2 1;f=polyder(p) %多項(xiàng)式求導(dǎo)polyvalm(p,x)3-3clear allclcf=factor(sym('2*x4-5*x3+4*x2-5*x+2')3-4clear allclcf1=sym('(x2+1)(x-2)');f2=sym('2*x2+1');f=expand(f1*f

10、2)3-5clear allclc>>syms x y;>>f=log(1+x);>>g=2*sin(y);>>compose(f,g)>>compose(g,f)3-6clear allclc>>syms x ;>>finverse(cos(1+x)3-7（1）（2）clear allclc>>syms x y k;>>f1=(1/2x)+(-1)x/x2);>>f2=1/(1+k2) 1/(2*k+1)>>s1=symsum(f1,1,inf)>>

11、s2=symsum(f2,1,inf)（3）（4）（5）（6）>> clear allclcsyms x ;f1=(1+2x)(1/x);f2=(x2)*exp(1/x2);f3=log(tan(x)/sin(x);f4=(x/(2*x+1)(x-1);limit(f1,inf)limit(f2,0)limit(f3,x,0,'right')limit(f4,inf)（7）（8）clear allclcsyms x y z a b;f1=1/(1+x2);f2=(x2+y2+z2);int(f1,x,a,b)int(int(int(f2,z,sqrt(x*y),x*

12、y),y,sqrt(x),x),x,0,1)3-8clear allclcsyms x;f=x*exp(x) log(sin(x);1/(1+x2) x(3/2);diff(f,2)3-9（1）clear allclcf=sym('x3-2*x+5');solve(f)（2）clear allclcsyms x1 x2 x3;f1=2*x1-5*x2+3;f2=5*x1-2*x2+18;x1 x2=solve(f1,f2)（3）方法一：clear allclcsyms x1 x2 x3;f1=3*x1+11*x2-2*x3-8;f2=1*x1+1*x2-2*x3+4;f3=x1-x2+x3-3;x1 x2 x3=solve(f1,f2,f3)方法二：clear allclcA=3 11 -2;1 1 -2;1 -1 1;B=8;-4;3;x=AB（4）clear allclcsyms x1 x2 x3;f1=2*x1-x2-exp(-x1);f