How to obtain max value of a plot

i was wondering how to obtain the data of the peak from a plot (max Y value and its X position) with: x = linspace(0,20); F = sin(x) P1 = plot (x,F);

ok so from this how can i write a command that shows me that the first peak is at F = 1 and X = 1.57

Answer 1

[Image Analyst]

Try this:

maxF = max(F);  % Find max value over all elements.
indexOfFirstMax = find(F == maxF, 1, 'first');  % Get first element that is the max.
% Get the x and y values at that index.
maxY = F(indexOfFirstMax);
maxX = x(indexOfFirstMax);

[ekko ekkosen]

Thanks but it still wont give me the first peak. I got a X value of 14,14 with this and that's the 3 peak, but i see now that i may need to write my full script here because i cant use max(F) command because i use syms z. this is the script. where i need to find the max value and position of plot 1 and 2

clear all
clear variables 
clc

H100 = 28.6083;  
Tmean = 15.6880;

%--------------------------------------------------------------------------
d = 100;                  % water depth
T = Tmean;                % wave period 
a = H100/2;    % or C100  % wave amplitude
g = 9.81;                 % acceleration of gravity             
                          % x = direction of propagation
                          % t = time
w = 2*pi/T;               % wave frequency

syms k
k1 = solve(g*k*tanh(k*d)-w^2 == 0, k);
k = sqrt(k1.^2);                             % wave number                   
L = simple(((g*T.^2)*tanh(k*d))/(2*pi));     % wave length

%--------------------------------------------------------------------------

t = linspace(0,30);
x = 0;

syms z
V1 = (((a*k*g*cosh(k*(z+d)))/(w*cosh(k*d)))*sin(w*t-k*x));         
A1 = ((a*k*g*cosh(k*(z+d)))/(cosh(k*d)))*cos(w*t-k*x);                    
%--------------------------------------------------------------------------

D = 2;
p = 1025;
Cm = 2;
Cd = 1.1;

Fm = (pi*p*Cm*A1*D.^2)/4;
Fd = ((p*Cd*D)/2)*V1.*(abs(V1));
F = int(Fm,z,-d,a)+int(Fd,z,-d,a);
P1 = plot(t,F);
grid on

% how to find max value and posision of P1

D = 20;
figure;
Fm = (pi*p*Cm*A1*D.^2)/4;
Fd = ((p*Cd*D)/2)*V1.*(abs(V1));
F = int(Fm,z,-d,a)+int(Fd,z,-d,a);
P2 = plot(t,F);
grid on

% how to find max value and posision of P2

[Image Analyst]

I don't have the symbolic toolbox so I can't help any further. I've added it to the Products list up top to alert other people. I know it works if you do it numerically:

close all;
x = sort(rand(1,50));
F = rand(1, 50);
plot(x, F, 'b-', 'LineWidth', 2);
grid on;
maxF = max(F);  % Find max value over all elements.
indexOfFirstMax = find(F == maxF, 1, 'first');  % Get first element that is the max.
% Get the x and y values at that index.
maxY = F(indexOfFirstMax)
maxX = x(indexOfFirstMax)
hold on;
plot(maxX, maxY, 'r*', ...
  'LineWidth', 2, 'MarkerSize', 20);

[ekko ekkosen]

i found out that i can use

max(double(F))

to find max value when i use a variable. so now i only need to find the x posision for that max value (x at max Y when i have Y)

[ekko ekkosen]

i found a solution from one of your earlier answers with some modifications. Thanks! i will be using:

[MaxValue, Index_At_F] = max(double(F))
t_Position_of_MaxValue = t(Index_At_F)

to find the x and y value of the peak.

[Image Analyst]

Ok, using max() is an alternate way - it gives you the same information as what I gave here except with a different function. And you changed the name of x to t, but otherwise the code is equivalent so I'm not sure why you said it didn't find the first peak. By the way, you don't need to cast F to double.