Info

La pregunta está cerrada. Vuélvala a abrir para editarla o responderla.

New to MATLAB so need help with following

1 visualización (últimos 30 días)
Muhammad Islam
Muhammad Islam el 28 de Jun. de 2019
Cerrada: MATLAB Answer Bot el 20 de Ag. de 2021
Given the following equation,
Capture.PNG
Given, d0=0.1 & d1=0.01 and model depends on value of b.
If x0=10 I need to investigate behavior of xn for large n for b = 0.1, 0.2, 2.3, 2.6, 3
Can anyone help me with the syntax I need to do this? There are many more situations I need to consider but if I can get this one syntax, I will be able to understand the rest.
  2 comentarios
darova
darova el 28 de Jun. de 2019
I CAN help you. What difficulties do you have?

Respuestas (2)

Basil C.
Basil C. el 29 de Jun. de 2019
Editada: Basil C. el 29 de Jun. de 2019
You could first declare an array of x element knowing that you need to find value upto X(6)
clc
clear all
close all
N=6; %if you want to find the value of x(6)
x=zeros(1,N); %declare an array
Declare the given data
b=0.1; % change the value of b accordingly
d0=0.1;
d1=0.01;
x(1)=10; %this stores the value of x0
Then use a for loop to find the value
for n=1:N-1
x(n+1)=x(n)+b*x(n)-d0*x(n)-d1*x(n)^2;
end
The value upto x6 will be stored in the array 'x'

Image Analyst
Image Analyst el 29 de Jun. de 2019
You need a loop over b outside the computation of the x vector:
d0=0.1;
d1=0.01;
x(1)=10; % First x
N = 5; % or however long you want the final x to be.
b = [0.1, 0.2, 2.3, 2.6, 3]
for k = 1 : length(b)
for n = 1 : length(x) - 1
x(n+1) = x(n) + b(k) * x(n) - d0*x(n) - d1 * x(n)^2;
end
% Now do something with x, like plot it or whatever.
end

La pregunta está cerrada.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by