How can I solve these recursive equations ?

8 次查看（过去 30 天）
Fox
Fox 2016-1-13
评论： Fox 2016-1-15
Hello I want to take the solution of a equation and use this as a new variable and so on like in the following demonstrated.
x1=a+bx0
x2=a+bx1
x3=a+bx2 ......
How can I solve this by a loop or so because I have to do this until 743 and I need every of the x values, so in the end I want to have a x matrix with 743x1 dimension.
  1 个评论
Stephen23
Stephen23 2016-1-14
编辑：Stephen23 2016-1-14
Whatever you do, do not create the variable names dynamically, just save the values in a vector instead. Here is an explanation of why it is very poor programming practice to create variable names dynamically:
http://www.mathworks.com/matlabcentral/answers/196952-how-to-create-new-variables-in-batches-with-strcat

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Torsten
Torsten 2016-1-13
xn = a*(1-b^n)/(1-b) + b^n*x0.
Now insert n=743.
Best wishes
Torsten.
  6 个评论
显示 4更早的评论
Torsten
Torsten 2016-1-14
The formula is only valid for constant a. For a depending on n you will have to refer to the loop solution from above:
x(1) = some value;
for m=1:742
x(m+1)=a(m+1)+b*x(m);
end
Best wishes
Torsten.

请先登录，再进行评论。

更多回答（1 个）

Guillaume
Guillaume 2016-1-14
The filter function allows you to compute all your elements in one go for both use cases (a constant or variable):
  • constant a and b:
numelemrequired = 743;
x = filter(1, [1 -b], [x0 repmat(a, 1, numelemrequired)])
  • variable a and b:
x = filter(1, [1 -b], [x0 a]) %where a is a vector of length 743
Note that your a, b, and x are not the same a, b, and x used by the documentation of filter.

类别

在 Help Center 和 File Exchange 中查找有关 Digital Filter Analysis 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by