Exponential approximation for vector input

조회 수: 5 (최근 30일)
Eduardo
Eduardo 2023년 1월 31일
댓글: Eduardo 2023년 2월 1일
I was double checking the behaviour of a sigmoid function used in my Simulink model and I noticed that I was getting incorrect approximations when I made the computation for a vector of values
vect = [-5.0000 -5.0000 -5.0000 1.0000 0.9000 0.8000 0.7000 -5.0000 -5.0000];
y_vect = 1/(1+exp(-2*(vect'-1)));
% Value calculated using the vector
y_vect(4)
ans = 0
% Value calculated alone
y_val = 1/(1+exp(-2*(vect(4)-1)))
y_val = 0.5000
This approximation in my case causes great confussion due to the magnitude of the quantity expected.
Is there any way to solve this?

채택된 답변

Sulaymon Eshkabilov
Sulaymon Eshkabilov 2023년 1월 31일
You have overlooked one dot. Here is the corrected commands:
vect = [-5.0000 -5.0000 -5.0000 1.0000 0.9000 0.8000 0.7000 -5.0000 -5.0000];
y_vect = 1./(1+exp(-2*(vect-1)));
% Value calculated using the vector
y_vect(4)
ans = 0.5000
% Value calculated alone
y_val = 1/(1+exp(-2*(vect(4)-1)))
y_val = 0.5000
  댓글 수: 1
Eduardo
Eduardo 2023년 2월 1일
Oh nice to know!
I wrongly thought the broadcasting would be done automatically since we just had a scalar in the numerator

댓글을 달려면 로그인하십시오.

추가 답변 (1개)

Voss
Voss 2023년 1월 31일
vect = [-5.0000 -5.0000 -5.0000 1.0000 0.9000 0.8000 0.7000 -5.0000 -5.0000];
Using / (matrix right division), as you have it now:
y_vect = 1/(1+exp(-2*(vect'-1)));
disp(y_vect)
1.0e-05 * 0.6144 0 0 0 0 0 0 0 0
Using ./ (element-wise right division):
y_vect = 1./(1+exp(-2*(vect'-1)));
disp(y_vect)
0.0000 0.0000 0.0000 0.5000 0.4502 0.4013 0.3543 0.0000 0.0000

카테고리

Help CenterFile Exchange에서 Matrix Indexing에 대해 자세히 알아보기

제품


릴리스

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by