试图在 MATLAB 中创建子图，这段代码有意义吗？答案

【问题标题】：trying to create a subplot in MATLAB, does this code make sense?试图在 MATLAB 中创建子图，这段代码有意义吗？
【发布时间】：2021-09-05 23:14:59
【问题描述】：

我正在尝试在 matlab 中创建一个子图，但是此代码似乎无法使用。你对如何让它工作有什么建议吗？我是一个编码新手，所以如果它不应该在这里，任何关于在哪里搜索这样的东西的帮助或建议将不胜感激。

 Uinput=inputdlg({'width:',...
                 'thickness:',...
                 'length:',...
                 'density:',...
                 'natural frequency 1:',...
                 'natural frequency 2:',...
                 'natural frequency 3:',...
                 'natural frequency 4:',...
                 'damping ratio:'},...
                 'input',1,{'','','','','','','','',''});
w=eval(Uinput{1});
t=eval(Uinput{2});
L=eval(Uinput{3});
rho=eval(Uinput{4});
f1=eval(Uinput{5});
f2=eval(Uinput{6});
f3=eval(Uinput{7});
f4=eval(Uinput{8});
zn=eval(Uinput{9});
mnum=eval(Uinput{6});
%%
%COMMENT FROM HERE TO END
fn=[2.3481 14.7152 41.2033 80.7406]; % natural frequencies taken from task 1
kn=((2*pi.*fn).^2)*(L*w*t*rho); % equation to calculate the 
zn1=0.01;
zn=[zn1 zn1 zn1 zn1];
phif=[0.011 0.063 0.159 0.294]; %force potential amplitudes and the applied position %mode shapes for the forcing location   %nodes and antinode locations  %YOU NEED TO CHANGE THESE. mode shapes for the forcing location for each mode
%50mm, 302mm, 600mm 
phir=[0.3464 0.7112 -0.0083 -0.7455]; % response potential %modeshapes for the response locations     %YOU NEED TO CHANGE THESE. mode shapes for the response location for each mode
f=linspace(0,100,5000);
for n = 1:length(fn),
    r = f/fn(n);
    hi(:,n) = phif(n)*phir(n)/kn(n)./(1-r.^2+j*2*zn(n)*r);
    phin=hi(:,n)/(max(abs(hi(:,n)));
    subplot(2,2,n);
    plot(x,phin, '-o')
    grid on
    xlabel('position')
    ylabel('normalised shape')
    title(['Mode ' num2str(mnum) ', ' num2str(fn) ' Hz'])
end
h=abs(sum(hi,2)); 
loglog(f,h)
ylabel('H')
xlabel('frequency (Hz)')

【问题讨论】：

标签： matlab subplot

【解决方案1】：

第 34 行缺少括号。第 36 行，“x”没有在任何地方定义。

在解决了这 2 个错误之后，这段代码可以正常运行并绘图。

【讨论】：

【解决方案2】：

#请试试这个修改后的代码！ #Line 34 - 缺少括号 #第 36 行 - 代替 'x' 变量，请使用 'f' 变量然后绘图 #plot(f,phin, '-o')

 Uinput=inputdlg({'width:',...
                 'thickness:',...
                 'length:',...
                 'density:',...
                 'natural frequency 1:',...
                 'natural frequency 2:',...
                 'natural frequency 3:',...
                 'natural frequency 4:',...
                 'damping ratio:'},...
                 'input',1,{'','','','','','','','',''});
w=eval(Uinput{1});
t=eval(Uinput{2});
L=eval(Uinput{3});
rho=eval(Uinput{4});
f1=eval(Uinput{5});
f2=eval(Uinput{6});
f3=eval(Uinput{7});
f4=eval(Uinput{8});
zn=eval(Uinput{9});
mnum=eval(Uinput{6});
%%
%COMMENT FROM HERE TO END
fn=[2.3481 14.7152 41.2033 80.7406]; % natural frequencies taken from task 1
kn=((2*pi.*fn).^2)*(L*w*t*rho); % equation to calculate the 
zn1=0.01;
zn=[zn1 zn1 zn1 zn1];
phif=[0.011 0.063 0.159 0.294]; %force potential amplitudes and the applied position %mode shapes for the forcing location   %nodes and antinode locations  %YOU NEED TO CHANGE THESE. mode shapes for the forcing location for each mode
%50mm, 302mm, 600mm 
phir=[0.3464 0.7112 -0.0083 -0.7455]; % response potential %modeshapes for the response locations     %YOU NEED TO CHANGE THESE. mode shapes for the response location for each mode
f=linspace(0,100,5000);
for n = 1:length(fn)
    r = f/fn(n);
    hi(:,n) = phif(n)*phir(n)/kn(n)./(1-r.^2+ j*2*zn(n)*r);
    phin=hi(:,n)/(max(abs(hi(:,n))))
    subplot(2,2,n);
    plot(f,phin, '-o')
    grid on
    xlabel('position')
    ylabel('normalised shape')
    title(['Mode ' num2str(mnum) ', ' num2str(fn) ' Hz'])
end
h=abs(sum(hi,2)); 
loglog(f,h)
ylabel('H')
xlabel('frequency (Hz)')

【讨论】：