MATLAB實(shí)習(xí)教學(xué)提綱

1、Good is good, but better carries it.精益求精，善益求善。MATLAB實(shí)習(xí)-1矩陣運(yùn)算矩陣建立，轉(zhuǎn)置、加、減、乘、求逆等運(yùn)算。123456101 -120113211計(jì)算A+B，AB、|A|、A-1、AB.A+Bans=043569342det(A)ans=12A.-1ans=1.00000.50000.33330.25000.20000.16671.00000.33331.0000ABans=1.1667-1.83331.50000.66670.66670-1.16670.8333-0.5000A*Bans=77913192146102.編寫M文件的編寫與運(yùn)

2、行輸出閏年i=1;for(j=1000:3000)if(rem(j,100)=0)if(rem(j,400)=0)a(i)=ji=i+1;endelseif(rem(j,4)=0)a(i)=ji=i+1;endendk=i-1end部分?jǐn)?shù)據(jù)：Columns463through4692908291229162920292429282932Columns470through4762936294029442948295229562960Columns477through4832964296829722976298029842988Columns484through485299229963數(shù)據(jù)文件的建

3、立與讀取例：建立班級一門課成績數(shù)據(jù)庫，編程讀取并排序、另存。學(xué)號123456789成績859685875665968575S.sc=859685875665968575,S.num=123456789S=sc:859685875665968575S=sc:859685875665968575num:123456789fori=1:9forj=i:9ifS.sc(i)1)=0;%在滿足x2+y21的網(wǎng)格點(diǎn)上的Z值變成0mesh(X,Y,abs(Z)%畫圖title(球面x.2+y.2+z.2=1)%加標(biāo)題axis(-12-1101)%設(shè)置X,Y,Z軸坐標(biāo)范圍5多項(xiàng)式運(yùn)算例：多項(xiàng)式運(yùn)算求值、求根。

4、x3+x2+x-3=0a=111-3a=111-3r=poly(a)r=10-68-3q1=poly2str(a,x)q1=x3+x2+x3f=inline(x3+x2+x-3,x)f=Inlinefunction:f(x)=x3+x2+x-3f(3)ans=366曲線擬合例：數(shù)據(jù)的多項(xiàng)式擬合X0.51.01.52.02.53.0y1.752.453.814.807.008.60使用了MATLAB中curvefitting，進(jìn)行了該多形式的擬合。由圖可知這些點(diǎn)是五階多項(xiàng)式linearmodelPoly5:f(x)=p1*x5+p2*x4+p3*x3+p4*x2+p5*x+p6Coefficie

5、nts:p1=-1.6p2=13.74p3=-44.07p4=65.67p5=-42.63p6=11.357.線性規(guī)劃例：線性規(guī)劃求解（如教材中）。Min-5x1-4x2-6x3s.tx1-x2+x3=20;3x1+2x2+4x3=42;3x1+2x2=0A=1-11;324;320A=1-11324320b=20;42;30b=204230 x,fval,exitflag,output,lambda=linprog(f,A,b,lb)Optimizationterminated.x=0.000015.00003.0000fval=-78.0000exitflag=1output=iterat

6、ions:6algorithm:large-scale:interiorpointcgiterations:0message:Optimizationterminated.lambda=ineqlin:3x1doubleeqlin:0 x1doubleupper:3x1doublelower:3x1doublelambda.ineqlinans=0.00001.50000.50008微分方程數(shù)值解例：常微分方程數(shù)值解。求解該微分方程y=-y+x+1y0=1 ,x0,1X精確解RK解誤差01101/5,1.01873075,1.07001364,.512828904e-12/5,1.070320

7、05,1.15254655,.822265010e-13/5,1.14881164,1.24639295,.975813184e-14/5,1.24932896,1.35046324,.1011342741,1.36787944,1.46377272,.958932744e-16/5,1.50119421,1.58543154,.842373280e-17/5,1.64659696,1.71463557,.680386042e-18/5,1.80189652,1.85065811,.487615877e-19/5,1.96529889,1.99284245,.275435594e-12,2.1

8、3533528,2.14059514,.525985426e-211/5,2.31080316,2.29337988,.174232823e-113/5,2.67427358,2.61215359,.621199906e-114/5,2.86081006,2.77730880,.835012599e-13,3.04978707,2.94581999,.1039670769插值問題例：樣條插值。X1.02.03.04.05.0y3.54.65.53.22.6xx=1:1:5yx=3.22xxi=1:0.5:5f0=interp1(xx,yx,xxi)f1=interp1(xx,yx,xxi,li

9、near)f2=interp1(xx,yx,xxi,cubic)f3=interp1(xx,yx,xxi,spline)f4=interp1(xx,yx,xxi,neareat)f0,f1,f2,f3,f4yx=3.50004.60005.50003.20002.0000 xxi=1.00001.50002.00002.50003.00003.50004.00004.50005.0000f0=3.50004.05004.60005.05005.50004.35003.20002.60002.0000f1=3.50004.05004.60005.05005.50004.35003.20002.6

10、0002.0000f2=3.50004.07624.60005.17385.50004.54713.20002.48412.0000f3=3.50003.77344.60005.37665.50004.59533.20002.07972.0000f4=3.50004.60004.60005.50005.50003.20003.20002.00002.0000f0=3.50004.05004.60005.05005.50004.35003.20002.60002.0000f1=3.50004.05004.60005.05005.50004.35003.20002.60002.0000f2=3.5

11、0004.07624.60005.17385.50004.54713.20002.48412.0000f3=3.50003.77344.60005.37665.50004.59533.20002.07972.0000f4=3.50004.60004.60005.50005.50003.20003.20002.00002.0000plot(xxi,f1,xxi,f2,xxi,f3,xxi,f4)legend(線性插值,三次插值,樣條插值,最近區(qū)域差值)10綜合練習(xí)：計(jì)算機(jī)模擬例1追逐問題:如圖,正方形ABCD的四個(gè)頂點(diǎn)各有一人.在某一時(shí)刻,四人同時(shí)出發(fā)以勻速v=1米/秒按順時(shí)針方向追逐下一人,如

12、果他們始終保持對準(zhǔn)目標(biāo),則最終按螺旋狀曲線于中心點(diǎn)O.試求出這種情況下每個(gè)人的行進(jìn)軌跡.）模擬步驟（1）建立HYPERLINK/z/Search.e?sp=S%E5%B9%B3%E9%9D%A2%E7%9B%B4%E8%A7%92%E5%9D%90%E6%A0%87%E7%B3%BB&ch=w.search.yjjlink&cid=w.search.yjjlinkt_blank平面直角坐標(biāo)系：A(x1,y1)，B(x2,y2)，C(x3,y3)，D(x4,y4)。（2）取時(shí)間間隔為t，在每一時(shí)刻t計(jì)算每人在下一時(shí)刻t+t的坐標(biāo)。例如，甲的追逐目標(biāo)是乙，在t時(shí)刻，甲的坐標(biāo)為(x1,y1)，乙的坐

13、標(biāo)為(x2,y2)，則甲在t+t時(shí)刻的坐標(biāo)為(x1+vtcosa,y+vtsina)，其中cosa=(x2-x1)/d，sina=(y2-y1)/d，d=(x2-x1)2 + (y2-y1)2（3）選取足夠小的，模擬到距離d0.1)x(5)=x(1);y(5)=y(1);fori=1:4d=sqrt(x(i+1)-x(i)2+(y(i+1)-y(i)2);x(i)=x(i)+v*dt*(x(i+1)-x(i)/d;y(i)=y(i)+v*dt*(y(i+1)-y(i)/d;plot(x(i),y(i),.)holdonendendend軌跡如圖例2導(dǎo)彈追蹤問題：設(shè)位于坐標(biāo)原點(diǎn)的甲艦向位于x軸上

14、點(diǎn)A(1,0)處的乙艦發(fā)射導(dǎo)彈，導(dǎo)彈頭始終對準(zhǔn)乙艦.如果乙艦以最大的速度(是常數(shù))沿平行于y軸的直線行駛，導(dǎo)彈的速度是5，模擬導(dǎo)彈運(yùn)行的軌跡.又乙艦行駛多遠(yuǎn)時(shí)，導(dǎo)彈將它擊中？查資料知道：艦艇的速度在每小時(shí)40節(jié)左右，即80kM/h,而導(dǎo)彈在2400km/h.他們速度之比為30:1.于此令乙減速度為a=0.167.導(dǎo)彈的速度v=5.在某一時(shí)刻導(dǎo)彈的水平速度為51+t2a2.（1）豎直速度5at1+t2a2（2）速度的積分路程，通過（1）（2）對時(shí)間t積分得到某一時(shí)刻的位置X=5*log(a2*t/(a2)(1/2)+(1+a2*t2)(1/2)/(a2)(1/2)Y=5/a*(1+a2*t2)(1/2)-29.9401dt=