format long x=load('time.dat');y=load('30um.dat');x=x;xx=[2.510384615000000:0.009:8.336272727000001];sst=interp1(x,y,xx,'spline');variance = std(sst)^2;sst = (sst - mean(sst))/sqrt(variance) ;n = length(sst)pausedt =0.009 ;time = [0:length(sst)-1]*dt

来源：学生作业帮助网 编辑：作业帮 时间：2024/05/07 09:23:22
format long x=load('time.dat');y=load('30um.dat');x=x;xx=[2.510384615000000:0.009:8.336272727000001];sst=interp1(x,y,xx,'spline');variance = std(sst)^2;sst = (sst - mean(sst))/sqrt(variance) ;n = length(sst)pausedt =0.009 ;time = [0:length(sst)-1]*dt

format long x=load('time.dat');y=load('30um.dat');x=x;xx=[2.510384615000000:0.009:8.336272727000001];sst=interp1(x,y,xx,'spline');variance = std(sst)^2;sst = (sst - mean(sst))/sqrt(variance) ;n = length(sst)pausedt =0.009 ;time = [0:length(sst)-1]*dt
format long
x=load('time.dat');
y=load('30um.dat');
x=x;
xx=[2.510384615000000:0.009:8.336272727000001];
sst=interp1(x,y,xx,'spline');
variance = std(sst)^2;
sst = (sst - mean(sst))/sqrt(variance) ;
n = length(sst)
pause
dt =0.009 ;
time = [0:length(sst)-1]*dt + 2.510384615000000 % construct time array
pause
xlim = [2.500,8.500]; % plotting range
pad = 1;
dj = 1/8;
s0 = 40*dt; j1 = 7/dj;
lag1 = 0.72;
mother = 'Morlet';
[wave,period,scale,coi] = wavelet(sst,dt,pad,dj,s0,j1,mother);
power = (abs(wave)).^2 ;
levels = [5,10,20,40,80,160,320,640,1280] ;
Yticks = 2.^(fix(log2(min(period))):fix(log2(max(period))));
contourf(time,log2(period),log2(power),log2(levels));
在这里levels是什么意思,起到什么作用,Yticks这样操作是什么意思?

format long x=load('time.dat');y=load('30um.dat');x=x;xx=[2.510384615000000:0.009:8.336272727000001];sst=interp1(x,y,xx,'spline');variance = std(sst)^2;sst = (sst - mean(sst))/sqrt(variance) ;n = length(sst)pausedt =0.009 ;time = [0:length(sst)-1]*dt
证明：在OP,OQ上A,B使得OA=OB,过A做AD⊥L于D,连接BD
由已知得：△AOD≌△BOB
BD⊥L
设AD=BD=X
OA=OB=根号2X
AB=X
COS∠POQ=(2X^2+2X^2-X^2)/4X^2=3/4
∠POQ的余弦值为3/4

level从在contourf函数里的参数位置来看，代表所画坐标图的标尺。
就是坐标系中，坐标轴上标示出数值的点

level从在contourf函数里的参数位置来看，代表所画坐标图的标尺。
就是坐标系中，坐标轴上标示出数值的点。
Yticks你后面还没用上呢，不过看意图，应该是限制纵轴范围。
后面可以用ylim(Yticks)限制y轴范围。

format long x=load('time.dat');y=load('30um.dat');x=x;xx=[2.510384615000000:0.009:8.336272727000001];sst=interp1(x,y,xx,'spline');variance = std(sst)^2;sst = (sst - mean(sst))/sqrt(variance) ;n = length(sst)pausedt =0.009 ;time = [0:length(sst)-1]*dt Matlab中format long g表示什么? matlab显示格式>> format short>> pians =3.1416>> format long>> pians =3.14159265358979>> format short e>> pians =3.1416e+000>> format long e>> pians =3.141592653589793e+000书上说short是4位小数,long是15为数,short e是5位科学计数法,l 英语翻译bb=[………………]';load=[………………]';flow=[………………]';X = [ones(size(load)) flow load.*flow load.*load.*load ];b = regress(bb,X);scatter3(load,flow,bb,'filled');hold onloadfit = min(load):500:max(load);flowfit = m 一个有关Fortran程序的问题,部分程序1001 FORMAT(4I5)1002 FORMAT(6X,'TOTAL NUMBER OF FILMENTS' 15/& 24X,'NOODES=',I5/25X,'LOADS-',I5/22X,'FIXITIES'I5)1003 FORMAT(/5X,'ELEMENT DEFINITIONS'/& 5X,'ELEMENT',2X,'NOOD1',2X,'NOOD2',2N/)1004 FORMAT load load LOAD FORTRAN语言中,FORMAT(1X,E8. FORTRAN 101 format(1X,F7. 牛顿法求x.^3-x-1=0在1.5附近的解,MATLAB程序出错求解function y=func1_1(x)y=x.^3-x-1; function y=func1_1_1(x)y=3*x.^2-1; x=1.5;format long;x1=x-func1_1(x)/func1_1_1(x);if(abs(x1)1e-6||abs(func1_1(x))>1e-6)x=x1; x1=x-func1_1(x)/func1_1_1 请高手帮我找下这个Lingo程序的错误!下面是Lingo程序model:sets:hotel/1..10/;format/1..3/;link(hotel,format):x,y,c;endsetsmin=@sum(link:x*y*c);@for(format(j):@sum(hotel(i):(y(i)*r(i,j)))=num(j)/2);@for(format(j):@sum(hotel(i):(y(i)*r1( matlab出错 “Inner matrix dimensions must agree.”function result = XYZtoBL(X,Y,Z)format long ga=6378137.0000000000; %a为地球椭圆长半径e2=0.00669438002290; %e为地球第一偏心率 %进行迭代初值赋值if(X>0)L=atand(Y/X);elseL=atan matlab 关于inline的问题我不懂matlab,想用它求个积分,但遇到个很久结的问题,我想用matlab求函数y=x2（x的平方）的定积分,积分区间从1到2,下面是我的程序：format long syms x;w=x.^2;f=inline('w');a=quad(f,1, 我知道牛顿迭代法解方程的Matlab程序如何编写了,但不知道输入什么命令才能运行!求详解求12-3x+2cosx=0在【2,4】内的解,Matlab程序如下：clear; %清除所有内存数据；f=inline('12-3*x+2*cos(x)');format long matlab看不懂.function y=f(x) y=1./x+2*sin((2*x).^0.5);end >> format long>> x=3.3;h=0.1;g=(f(x+h)-2*f(x)+f(x-h))/(h^2) g = -0.009399348448680 >> x=3.3;h=0.001;g=(f(x+h)-2*f(x)+f(x-h))/(h^2) g = -0.009377437715230 >> x=3.3;h=0.001;g=(f(x+h)-2* FORTRAN初学者,有个问题不懂, PROGRAM MAIN IMPLICIT NONE INTEGER I,J,K I=570 J=3410 K=5069 PRINT 10,I,J,K PRINT 20,I,J,K PRINT 30,I,J,K 10 FORMAT(1X,I4) 20 FORMAT(1X,2I5) 30 FORMAT(1X,2(I5/2X)) END问一下倒数第 matlab int积分太慢,积不出来,怎么办?clear clc syms x format long H=[0:0.01:1.2]; v=pi*4.1/180;a=0.4;b=2.05;n=1.2;m=1.78; for i=1:15 h=H(i); S2=(pi/2+(h+(a-x)*tan(v)-n/2)/n*2.*sqrt(1-((h+(a-x)*tan(v)-n/2)/n*2).^2)+asin((h+(a-x)*tan(v)-n/2)/n