How to know dimensions to give in trapz function

4 ビュー (過去 30 日間)
J K
J K 2019 年 7 月 11 日
コメント済み: Rena Berman 2019 年 9 月 19 日
Hi,
Let's say I have a matrix (see below). How do I know what number to put in the dimension variable? Matlab just gives some odd explanation without telling how to determine the dimension. Could someone please help to understand how to get the value for dim?
[Kx, W, Ky] = meshgrid(kx, w, ky);
for j = 1:9;
functione = exp(-((W-w_o).^2)/deltaW.^2).*exp(-(Kx.^2+Ky.^2)/(deltaK.^2)).*exp(1i.*sqrt((W/c).^2-(Kx.^2+Ky.^2)).*z(j));
normw = trapz(functione.^2,dim)*dw
normkx = trapz(functione.^2,dim)*dkx
normky = trapz(functione.^2,dim)*dky
  2 件のコメント
TADA
TADA 2019 年 7 月 11 日
It is up to you along which dimension you want it to integrate
Rena Berman
Rena Berman 2019 年 9 月 19 日
(Answers Dev) Restored edit

サインインしてコメントする。

回答 (1 件)

TADA
TADA 2019 年 7 月 11 日
let's say you have a column vector x and you calculate y according to y = ax^2
now if a is a vector, you will get a matrix y:
x = linspace(0,10,5)';
a = [1 10 100];
y = a.*(x.^2)
y =
1.0e+04 *
0 0 0
0.0006 0.0063 0.0625
0.0025 0.0250 0.2500
0.0056 0.0563 0.5625
0.0100 0.1000 1.0000
This y matrix represents 3 possible column vectors ax^2 with different parameter value
Now to integrate correctoly, you should integrate along the first dimension (same dimension as the x column vector)
dim = 1;
A = trapz(x,y,dim)
A =
1.0e+04 *
0.0344 0.3438 3.4375
  7 件のコメント
J K
J K 2019 年 7 月 12 日
I understand it all now. Thank you so much for your extensive answers!
TADA
TADA 2019 年 7 月 14 日
Cheers

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeNumerical Integration and Differentiation 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