Help on sign function argument

2015 3 25
1 个回答
回答已采纳
更新时间：2015 3 25
1 次查看（30 天）

0 个投票

Here's my function:
function xdot=gotp(t,x)
z=0.1; w=1;
A = [0 1; (-22*z*w*sign()-2*z*w) -w];
xdot=A*x;
Called by:
[t,x]=ode45('gotp',[0 10],[0 -0.025]);
I want the sign function to return 1 if x is above 0.010 and 0 if it is below 0.010. How can I do this? I've tried putting different integers into the sign function's parentheses, such as sign(10000) and sign(1), but it has no effect on the output. If I use the variable "x," I get the an error (Dimensions of matrices being concatenated are not consistent.)

请先登录，再回答此问题。

 采纳的回答

Andrew Newell
Andrew Newell 2015-3-25

0 个投票

Use
2*sign(x-0.01)-1

8 个评论
显示 6更早的评论 隐藏 6更早的评论

headlight
headlight 2015-3-25
Where am I putting this? Am I replacing the (-22*z*w*sign()-2*z*w) term with that?
Andrew Newell
Andrew Newell 2015-3-25
编辑：Andrew Newell 2015-3-25
It depends on what you wish to return 1 if x is above 0.01 etc. I was assuming that you'd replace sign() with this expression, i.e.,
s = 2*sign(x-0.01)-1;
A = [0 1; (-22*z*w*s-2*z*w) -w];
headlight
headlight 2015-3-25
My A matrix now reads this: A = [0 1; (-22*z*w*2*sign(x-0.01)-1 -2*z*w) -w]; I get the "Dimensions of matrices being concatenated are not consistent." error when I run it.
I want to return 1 if x is above 0.01 and 0 if it is below 0.01
Andrew Newell
Andrew Newell 2015-3-25
Ah, I hadn't noticed that your x is actually a vector of length 2. So if you put the whole vector in the sign expression, you're trying to fit 2 components in a space for 1 component. So which component(s) of x do you want to compare with 0.01?
headlight
headlight 2015-3-25
I may have made a mistake by making x a vector. In the command that calls the function, -0.025 is just the initial condition of x, or x0.
I changed the code to this:
function xdot=gotp(t,x) z=0.1; w=1; A = [0 1; (-22*z*w*2*sign(x-0.01)-1 -2*z*w) -w]; xdot=A*x;
called by [t,x]=ode45('gotp',[0 10],-0.025);
Now I'm getting an error that says "GOTP must return a column vector."
Am I on the right track?
Andrew Newell
Andrew Newell 2015-3-25
The problem is that, with x now a scalar, A*x is a 2x2 matrix. What are the differential equations you're trying to solve? I don't know what to suggest unless I know that.
headlight
headlight 2015-3-25
There are 2 cases. If x>0.01, x'' = -22zwx' - wx If x<0.01, x'' = -2zwx' - wx
Andrew Newell
Andrew Newell 2015-3-25
编辑：Andrew Newell 2015-3-25
O.k., the first case would translate into
x' = y
y' = -22zwy - wx
or, in matrix form,
[x'; y'] = [0 1; -w -22*z*w]*[x; y]
(you have the bottom two elements in the wrong order). Thus, in your vector X = [x; y] (capitalized to avoid confusion), the element you want to test is X(1), so in your code,
s = sign(x(1)-0.01);
A = [0 1; -w (-12-10*s)*z*w]
However, it might be more accurate to create a terminal event so the integration stops when x crosses 0.01. Then you can switch to the other function and continue. See ode45 for more information.

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Common Operations 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by