How to plot exact user defined function?

5 ビュー (過去 30 日間)
Agniswar
Agniswar 2014 年 3 月 25 日
編集済み: Serge Boisse 2018 年 10 月 27 日
I have defined a function like this
function z=agni(x)
if x > 0
z=x;
else
z = x.^2;
end
But in the time of plotting plot(x,agni(x)) with x =[-2:.01:2] generate only the curve of x^2. WHY?

採用された回答

Walter Roberson
Walter Roberson 2014 年 3 月 25 日
if x > 0
means
if all(x > 0)
so if you pass in a vector of x then if any x in the vector is <= 0 then the "if" will not be true.
You should research logical indexing.
  1 件のコメント
Agniswar
Agniswar 2014 年 3 月 25 日
Then what can I do to plot a composite function?

サインインしてコメントする。

その他の回答 (1 件)

Serge Boisse
Serge Boisse 2018 年 10 月 27 日
編集済み: Serge Boisse 2018 年 10 月 27 日
Hello, You should understand that your function accepts a vector of values, not a scalar. then it must return a vector of the same length (that will be plotted) So you may rewrite your function as follows :
function z=agni(x)
z = x; % this is a vector !
ind = (x<0); % logical array : 1 if corresponding element of x is <0
z(ind) = z(ind).^2;
end
Hope this helps. If you prefer to code functions "the classical way", have a look at symbolic functions.

カテゴリ

Help Center および File Exchange2-D and 3-D Plots についてさらに検索

Community Treasure Hunt

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

Start Hunting!

Translated by