How to get an x coordinate from a given y?

Question

Hieu Nguyen 2017-10-24

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/363050-how-to-get-an-x-coordinate-from-a-given-y

回答： Hieu Nguyen 2017-10-25

采纳的回答： Guillaume

HW7_R2 copy.jpg

在 MATLAB Online 中打开

Hello, I am working a problem that asks me to graph the square of the Amplitude(A) of an oscillation that is a function of omega. I successfully found a way to extract the maximum value of A^2, but have not found a way to find the position omega where A^2 is half of its maximum value. Also, I do not know how to let value x, which is in this case, take all value in real number instead of just the range [0,1000].

x = 0:1000;
y = ((159)./sqrt((1000.^2 - x.^2)+(2*50*x).^2)).^2;
plot(x,y)
xlabel('Omega in rad/second');
ylabel('Amplitude A^2 in meter square(m^2)');
title('Rolling 2 HW7');
max(y);
indexmax = find(max(y) == y);
xmax = x(indexmax);
ymax = y(indexmax);
strmax = ['Max = ',num2str(ymax)];
text(xmax,ymax,strmax,'HorizontalAlignment','right');
[maxY, indexOfMaxY] = max(y);
value = find(y == 0.5*ymax);
display(value);

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Guillaume 2017-10-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/363050-how-to-get-an-x-coordinate-from-a-given-y#answer_287553

编辑：Guillaume 2017-10-24

在 MATLAB Online 中打开

One usually uses

indexmax = find(max(y) == y);

when there can be several y values that are equal to the max, as indexmax will be a vector of indices of all these max values. e.g.

y = [3 2 1 1 2 3 2 1 1 3 2 1]
indexmax = find(max(y) == y);

returns

indexmax == [1 6 10]

However, later on, your code (the strmax = ...) assumes that indexmax is scalar. In that case, one would use the simpler syntax

[~, indexmax] = max(y);

This will always return just one index. In case of duplicate max, it's the index of the first max value.

As for your question, there's no guarantee that the exact value of half maximum will be present in your array so you can't search for that exact value. Instead you can search for the nearest. The position of that nearest value is the index of the absolute minimum of the difference between y and the half maximum, so:

[~, indexhalfmax] = min(abs(y - max(y)/2));

edit: fixed that last line of code.

7 个评论
显示 5更早的评论隐藏 5更早的评论

Guillaume 2017-10-24

在 MATLAB Online 中打开

Again, isn't

x(indexhalfmax)

the value you're looking for. (i.e. the x that correspond to the y value nearest to the actual half max value)

Hieu Nguyen 2017-10-25

I don't know why it is giving that but it does not make sense. I plug the number x = 10 into my formula: y = (159^2)/[(1000^2-x^2)^2+4*50^2*x^2. The result is not 0.01264.

请先登录，再进行评论。