フィルターのクリア

Function function/ passing function/ varargin help?

1 回表示 (過去 30 日間)
Gabrielle Bartolome
Gabrielle Bartolome 2020 年 10 月 24 日
コメント済み: Stephen23 2020 年 10 月 26 日
Develop a function function M-file that returns the difference between the passed function's maximum and minimum value given a range of the independent variable. In addition, have the function generate a plot of the function for the range. Test it for the following cases:
f(t) = a*exp^(-b*t)*sin(t-c) from t = 0 to 6*pi
1. a = 8, b = 0.25, c = 2 below is what I have so far. I keep getting an error. I don't know what I am doing wrong. .I would appreciate any help that you could provide. Thank you.
%input:
% a = lower bound of range
% b = upper bound of range
% n = number of intervals
%output:
% fdiff = difference between max and minimum of a function
function fdiff = funcdiff(f, a, b, n, varargin)
x = linspace(a,b,n);
y = f(x, varargin{:});
fdiff = max(y) - min(y);
fplot(f, [a b], varargin{:})
end
% clear clc
% diff = @(t) a*exp^(-b*t)*sin(t-c);
% funcdiff(diff, 0, 6*pi, 180, 8, 0.25, 2)

回答 (1 件)

KSSV
KSSV 2020 年 10 月 24 日
編集済み: KSSV 2020 年 10 月 24 日
a = 8 ;
b = 0.25 ;
c = 2 ;
n = 180 ;
f = @(t) a*exp(-b*t).*sin(t-c);
t = linspace(a,b,n);
y = f(t);
fdiff = max(y) - min(y);
plot(t,y)
  2 件のコメント
Gabrielle Bartolome
Gabrielle Bartolome 2020 年 10 月 24 日
I am not sure I really understand the code you placed. It seems to work by itself but not as a funciton function and it doesn't use varargin.
Cole Fontana
Cole Fontana 2020 年 10 月 26 日
編集済み: Cole Fontana 2020 年 10 月 26 日
Don't worry I'm just as stuck as you are lol
Professor Huang said that we can't use fplot when using varagin.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeApp Building 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