How to calculate the value of (x) on xaxis in MATLAB ?

I have this code in Matlab.

% Clear workspace.
clear
% Clear command window.
clc
% Close all windows.
close all

% Define the parameters.

pt1 = 2 ;

pt2 = 3.954 ;


h = 0.4 ; % wave length
dx = 1 ;
z = 30 ;
N = 10 ; % No.of samples

% Imaginary part.
j = sqrt( -1 ) ;

% Initialize the vector to store the values.
c = zeros(1, 20);

% Make for loop.
for L = 1: N
    
    a1 = exp( j*( 2*pi/h ) * ( (L-1)*dx ) *( pt1/z ) ) ;
    a2 = exp( j*( 2*pi /h ) *( (L-1)*dx ) *( pt2/z ) ) ;
    
    c(L) = a1 + a2  ;
    
end

% Compute FFT.
Y = fft(c,256);

% Compute Magnitude FFT
Y1 = abs(Y);

We can see in the attached picture the value of (x=44) and so on......

My question is: What is the relationship or mathematical formula that I can use to calculate the value of (x) ?

I don't care, I just need to know the mathematical formula or equation used to calculate the value of (x)?

Answer 1

[the cyclist]

How, specifically, do you want to define that point? It is not the highest peak, which is at index=85 (as shown in the last line of code I added, at the end of yours).

You could perhaps use the findpeaks function.

% Define the parameters.
pt1 = 2 ;
pt2 = 3.954 ;
h = 0.4 ; % wave length
dx = 1 ;
z = 30 ;
N = 10 ; % No.of samples
% Imaginary part.
j = sqrt( -1 ) ;
% Initialize the vector to store the values.
X = zeros(1, 20);
% Make for loop.
for L = 1: N
    
    x1 = exp( j*( 2*pi/h ) * ( (L-1)*dx ) *( pt1/z ) ) ;
    x2 = exp( j*( 2*pi /h ) *( (L-1)*dx ) *( pt2/z ) ) ;
    
    X(L) = x1 + x2  ;
    
end
% Compute FFT.
Y = fft(X,256);
% Compute Magnitude FFT
Y1 = abs(Y);

% Plot it
plot(Y1)

% Max value of Y1, and its index
[maxY1,indexOfMaxY1] = max(Y1)

maxY1 = 10.3835

indexOfMaxY1 = 85

[Muhammad Salem]

@the cyclist

No, no, please read my question carefully so that you understand what is meant.

My question is: What is the relationship or mathematical formula that I can use to calculate the value of (x) ?

I mean, how was Matlab able to calculate the value of X=44? What is the relationship or mathematical formula that was used?

[Dyuman Joshi]

That value is the index of the corresponding y-value shown.

[the cyclist]

@Muhammad Salem, the way you plotted your data is

• x-axis values are 1, 2, 3, 4, 5, ... 256
• y-axis values are the values in the vector Y1

When you hover over the data point with the black square (in your figure), MATLAB is showing that the value has (x,y) = (44,10.38)

That value of x has nothing to do with the value of X in your program. It is just the index of Y1 vector.

[Muhammad Salem]

@the cyclist

I mean the value of x in the picture

I need the mathematical formula that Matlab uses to give me the value of x=44?

[the cyclist]

It is not a mathematical formula. MATLAB is telling you that are hovering over the 44th value of Y1.

Look at this code, and my screenshot of the figure when I run it locally:

plot([5 5 8 5 5 5])

When I hover over that top data point, MATLAB reports (3,8).

There is no mathematical formula. Rather, when I use the shorthand

plot([5 5 8 5 5 5])

MATLAB is implicitly plotting

plot([1 2 3 4 5 6],[5 5 8 5 5 5 ])

So, when I hover over the point that has y=8, it tells me (x,y) = (3,8). It is using that implicit value of x. There is no mathematical formula.

[Steven Lord]

@the cyclist is correct. From the part of the Description section on the plot documentation page where it explains the plot(Y) syntax: "If Y is a vector, the x-coordinates range from 1 to length(Y)."

If you have X data associated with the Y data, call plot with two input arguments. If you do, the scale of the X axis (and the data tips) will reflect the X data you specified in your plot call. Compare:

x = [0 1.5 2 6 9];
y = x.^2;
figure
plot(y, 'o-')
% Horizontal lines at the values in x WILL NOT line up with the points
xline(x, 'r') 
title('Y data alone -- X coordinates are 1:length(Y)')

figure
plot(x, y, 'o-')
% Horizontal lines at the values in x WILL line up with the points
xline(x, 'r') 
title('X and Y data -- X coordinates are from the variable x')

Answer 2

[Walter Roberson]

Q = @(v) sym(v);
pt1 = Q(2) ;
pt2 = Q(3.954) ;
h = Q(0.4) ; % wave length
dx = Q(1) ;
z = Q(30) ;

syms L
x1 = exp( j*( 2*pi/h ) * ( (L-1)*dx ) *( pt1/z ) ) ;
x2 = exp( j*( 2*pi /h ) *( (L-1)*dx ) *( pt2/z ) ) ;
x = x1 + x2;

Y = fourier(x)

Y = 

Y1 = abs(Y)

Y1 = 

This has two instanteneous peaks, one at pi/3 and the other at 659pi/1000

[Muhammad Salem]

@Walter Roberson

I mean the value of x in the picture

I need the mathematical formula that Matlab uses to give me the value of x=44?