Hello!
Could you, please, suggest an answer:
How to define function f (x), which for x from -1 to 1, f (x) = 1; and outside these limits f (x) = 0.
Thank in advance!

3 件のコメント

Rik
Rik 2020 年 8 月 11 日
This is your homework. What have you tried so far? And do you want a normal (or anonymous) function, or a symbolic function?
Viktoriia Buliuk
Viktoriia Buliuk 2020 年 8 月 11 日
I try to define Window function. I have tried with if.
KSSV
KSSV 2020 年 8 月 11 日
Show us your code.

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 採用された回答

hosein Javan
hosein Javan 2020 年 8 月 11 日

0 投票

function y = f(x)
y = zeros(size(x));
y(abs(x) < 1) = 1;
end

5 件のコメント

hosein Javan
hosein Javan 2020 年 8 月 11 日
the advantage of such definition is that you can pass inputs to the function as array rather than scalars. it should really help in dealing with fourier transform which requires element-wise product.
hosein Javan
hosein Javan 2020 年 8 月 11 日
however this method could lead to unnecessary computer calculations becasue when you multiply the window function to your signal, the elements that are multiplied eaither by 1 or 0 which is unnecessary. multiplication is for a non-rectangular window function like gaussian,etc,.. . but in the case of rectangular. you only have to select the values of your function that meet the condition by array indexing. if you are planning to do heavy simulation with rectangular window, indexing is more efficient but if it's just a simple calculation function multiplication should be fine.
Rik
Rik 2020 年 8 月 11 日
編集済み: Rik 2020 年 8 月 11 日
In this case you can simply convert the logical to double:
f=@(x) double(abs(x)<1);
In most other cases you will need multiplication:
%g={x^2 for -1<x<1
% {8 for x>2
% {0 otherwise
g=@(x) ...
x.^2.*(-1<x & x<1)+...
8.*(x>2)+...
0;
hosein Javan
hosein Javan 2020 年 8 月 11 日
ofcourse unless you want to use function file, you need this multiplication only to define your function in anonymous form. correct.
Viktoriia Buliuk
Viktoriia Buliuk 2020 年 9 月 4 日
Thank you very much!

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