how to get the x value for a given y value?

I plot x-y graph (Gaussian function) and want to get the x-axis value (will be two points in this case) at a certain y-axis value (half of the maximum)

I tried this but it didn't work:

clc;
clear all;
Fs = 150; % Sampling frequency
t = -0.5:1/Fs:0.5; % Time vector of 1 second
x = 1/(sqrt(2*pi*0.01))*(exp(-t.^2/(2*0.01)));
figure;
plot(t,x);

xi = 0.5*max(x) ;
z=find(x==xi);
ti = x(z) ;
hold on
plot(ti,xi,'*r')

Answer 1

[Geoff Hayes]

ahmed - your code assumes that there is an x value that is identical to xi

z=find(x==xi);

This need not be true. And comparing doubles in this manner is not generally a good idea (due to precision, see Why is 0.3 - 0.2 - 0.1 (or similar) not equal to zero?). Usually a tolerance of some kind should be used (i.e. abs(x - y) < eps). In your case, a tolerance might not work as well because you will not know what that tolerance should be. You could try different values...the following seems to work for this dataset

z=find(abs(x-xi)< 0.10);
ti = t(z) ;
hold on
plot(ti,xi,'*r')

(Note how ti is obtained from the t array instead of x.)

[ahmed salah]

thank you for your help

Answer 2

[Jan]

You cannot expect that any of the points at t = -0.5:1/Fs:0.5 is exactly 0.5*max(x). Remember that you evaluate x at some time steps only and rounding errors have to be considered also.

There is an analytical solution also, but you can use fzero to find the searched points. But you have to find the maximum value of the curve at first. Setting the derivative to 0 will help you.

[Parmonangan Manalu]

clear;

a = 6.8;
b = 10;
weight = 767;

for i = 1:22
    if i>= a && i <= b
        gra = (1.4);
    else
        gra = 0;
    end
    GM(i) = gra.*weight;
    format long g
end
dis = 1:22;
plot(dis,GM)
grid on
ax = gca;
ax.XAxis.TickLabelFormat = '%,.1f';

Dear Jan,

can you please help with that coding, the y value is not exatcly as the x value when i plot the graph. I think the correct graph shall be the red line

thanks in advance

[Jan]

Please open a new thread for a new question. Posting new question is the section for comments to answers of another question is too confusing.

The code produces the diagram exactly as expected. Note that the diagram of x=[1,2,3], y = [0,1,0] is a triangle also. If you post this as new question, we will write some answers about how you can achieve, what you want.