How to calculate the value of (x) on xaxis in MATLAB ?
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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)?
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1604161/image.png)
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Réponses (2)
the cyclist
le 30 Jan 2024
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).
% 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)
6 commentaires
the cyclist
le 30 Jan 2024
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])
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1604241/image.png)
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
le 30 Jan 2024
@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')
Walter Roberson
le 30 Jan 2024
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)
Y1 = abs(Y)
This has two instanteneous peaks, one at pi/3 and the other at 659pi/1000
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